o
10
FIGURE - 0
(continued)
INTRODUCTION TO MACHINERY PRINCIPLES 23
/
~
'
'
'
20 30 40 50 ]00 200
Magnetizing intensity H (A ' turnslm)
I"
"-
300
'" i'-
500
between flux and magnetomotive force. The slope of the curve of flux density ver-
sus magnetizing intensity at any value of H in Figure I- I Db is by definition the per-
meability ofthe core at that magnetizing intensity. The curve shows that the penne-
ability is large and relatively constant in the unsaturated region and then gradually
drops to a very low value as the core becomes heavily saturated.
1000
Figure I- IDe is a magnetization curve for a typical piece of steel shown in
more detail and with the magnetizing intensity on a logarithmic scale. Only with
the magnetizing intensity shown logarithmically can the huge saturation region of
the curve fit onto the graph.
The advantage of using a ferromagnetic material for cores in electric ma-
chines and transfonners is that one gets many times more flux for a given magne-
tomotive force with iron than with air. However, if the resulting flux has to be pro-
portional, or nearly so, to the applied magnetomotive force, then the core must be
operated in the unsaturated region of the magnetization curve.
Since real generators and motors depend on magnetic flux to produce volt-
age and torque, they are designed to produce as much flux as possible. As a result,
most real machines operate near the knee of the magnetization curve, and the flux
in their cores is not linearly related to the magnetomotive force producing it. This](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-47-320.jpg)
![46 ELECIRIC MACHINERY FUNDAMENTALS
(d) This task is ideally suited for MATLAB. We can take advantage ofMATLAB's
vectoriled calculations to detennine the velocity of the bar for each value of
force. The MATLAB code to perform this calculation is just a version of the
steps that were performed by hand in part c. The program shown below calcu-
lates the current, induced voltage, and velocity in that order, and then plots the
velocity versus the force on the bar.
% M
- f il e, exl _ 10 .m
% M
- f il e to ca l cu l ate and p l ot the vel oci t y of
f unc t i on of l oad .
% a linear motor as a
VB = 120;
r = 0 . 3;
% Battery vol tage (V)
% Res i s tance (ohms )
1 = 1 ;
B = 0 . 6;
% Sel ect the
F = 0, 10,50;
•Ca l cu l ate
1 0
, ./ (1
•Ca l cu l ate
e i nd = VB -
•Ca l cu l ate
v_ba r 0 e i nd
% Bar l ength (m)
% Flux density (T)
for ces to appl y t o
•
the bar
Force (N)
'he current s f l owi ng 1 0 'he mot or.
• B) ;
•CUrrent (A )
'ho i nduced vol tages 0 0 'ho bar.
i • c, •I nduced vol tage
'ho vel oci t i es of the bar.
. / (1 • B) ; % Vel oci t y (m/ s)
% Pl ot the vel oci t y of the bar ver s u s f or ce .
p l ot (F,v_ba r ) ;
t i t l e ('Pl ot of Vel oci t y ve r s u s Appli ed For ce');
x l abel (' Force (N) ' ) ;
y l abel ('Vel oci t y (m/ s)');
axi s ( [0 500 200 ] ) ;
(V)
The resulting plot is shown in Figure 1- 28. Note that the bar slows down more
and more as load increases.
(e) If the bar is initially unloaded, then eind = VB. If the bar suddenly hits a region
of weaker magnetic field, a transient will occur. Once the transient is over,
though, eind will again equal VB.
This fact can be used to determine the final speed of the bar. The initial speed was
120 rnls. The final speed is
VB = eind = vllBl
V,
v.. = Bl
120 V
= (0.08 TXIO m) = 150 mls
Thus, when the flux in the linear motor weakens, the bar speeds up. The same behavior oc-
curs in real dc motors: When the field flux of a dc motor weakens, it turns faster. Here,
again, the linear machine behaves in much the same way as a real dc motor.](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-70-320.jpg)
![TRANSFORMERS 71
where ()s is the angle between the secondary voltage and the secondary current.
Since voltage and current angles are unaffected by an ideal transfonner, ()p - ()s = ().
The primary and secondary windings of an ideal transfonner have the same power
factor.
How does the power going into the primary circuit of the ideal transformer
compare to the power coming out of the other side? lt is possible to find out
through a simple application ofthe voltage and current equations [Equations (2-4)
and (2- 5)]. 1lle power out of a transformer is
Poot = Vsls cos () (2-8)
Applying the turns-ratio equations gives Vs = Vp /a and Is = alp, so
-""
P out - a (alp) cos ()
I Poot - VpIp cos () - P;n (2- 9)
Thus, the output power ofan ideal transfonner is equal to its input power.
The same relationship applies to reactive power Q and apparent power S:
IQ
;" VpIp sin () VsIs sin () Qoo, I (2- 10)
and Is;, ~ Vplp = VsIs = Soot I (2- 11 )
Impedance Transformation through a Transformer
The impedance of a device or an element is defined as the ratio of the phasor volt-
age across it to the phasor current flowing through it:
V,
ZL = 1; (2- 12)
One of the interesting properties of a transfonner is that, since it changes voltage
and current levels, it changes the ratio between voltage and current and hence the
apparent impedance of an element. To understand this idea, refer to Figure 2- 5. If
the secondary current is calJed Is and the secondary voltage Vs, then the imped-
ance of the load is given by
V,
ZL = 1;
The apparent impedance of the primary circuit of the transfonner is
V
Z' = --.f..
, Ip
Since the primary voltage can be expressed as
Vp = aVs
(2- 13)
(2- 14)](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-95-320.jpg)
![TRANSFORMERS 95
It is customary to select two base quantities to define a given per-unit sys-
tem. The ones usually selected are voltage and power (or apparent power). Once
these base quantities have been selected, all the other base values are related to
them by the usual electrical laws. In a single-phase system, these relationships are
(2- 54)
(2- 55)
(2- 56)
and (2- 57)
Once the base values of S (or P) and V have been selected, all other base values
can be computed easily from Equations (2- 54) to (2- 57).
In a power system, a base apparent power and voltage are selected at a spe-
cific point in the system. A transfonner has no effect on the base apparent power
of the system, since the apparent power into a transfonner equals the apparent
power out of the transfonner [Equation (2-11 )] . On the other hand, voltage
changes when it goes through a transformer, so the value of VI>a.. changes at every
transformer in the system according to its turns ratio. Because the base quantities
change in passing through a transfonner, the process of referring quantities to a
common voltage level is automatically taken care of during per-unit conversion.
EXllmple 2-3. A simple power system is shown in Figure 2- 22. This system con-
tains a 480-V generator connected to an ideal I: 10 step-up transfonner, a transmission line,
an ideal 20: I step-down transformer, and a load. The impedance of the transmission line is
20 +j60 n, and the impedance of the load is IOL30on. The base values for this system are
chosen to be 480 V and 10 kVA at the generator.
(a) Find the base voltage, current, impedance, and apparent power at every point in
the power system.
(b) Convert this system to its per-unit equivalent circuit.
(c) Find the power supplied to the load in this system.
(d) Find the power lost in the transmission line.
YG 480LOo y
'-''-'
Region I RegIOn 2
FIGURE 2-22
The power system of Example 2- 3.
RegIOn 3](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-119-320.jpg)
![TRANSFORMERS 99
Ip,po R., jXoq I" po
- -
+
j
+
,
0.012 fJ·06
I~H1 I'm
V R, JXm V"po
,,~
49.7 jl2
FIGURE 2-15
The per-unit equivalent circuit ofExample 2-4.
Sba.se t
(P, Q, S)poon base 2 = (P, Q, S)poon base ]-
S--
"=,
v.: v.: Vbase ]
P" on base 2 = po on base ]
Vbase2
(R, X, Z)P" 00 base 2 =
(Vbase t?(Sbase 2)
(R, X, Z)pu on base t(l< )'(S )
base 2 base ]
(2- 58)
(2- 59)
(2-60)
Example 2-4. Sketch the approximate per-unit equivalent circuit for the trans-
fonner in Example 2- 2. Use the transformer's ratings as the system base.
Solutioll
The transfonner in Example 2- 2 is rated at 20 kVA, 8()(x)/240 V. The approximate equiva-
lent circuit (Figure 2- 21) developed in the example was referred to the high-voltage side of
the transfonner, so to convert it to per-unit, the primary circuit base impedance must be
fOlUld. On the primary,
Therefore,
V!>Me I = 80CXl V
Sbo>e I = 20,(XXl VA
(Vb... t)l (8()(x) V)2
z....•• t= S = 20 00Cl VA =3200 0
~, '
_ 38.4 + jl92 0 _ .
ZsE,po - 3200 0 - 0.012 + jO.06 pu
159 ill
Rc.pu = 3200 0 = 49.7 pu
38.4 kO
ZM.pu = 3200 0 = 12 pu
The per-unit approximate equivalent circuit, expressed to the transfonner's own base, is
shown in Figure 2- 25.](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-123-320.jpg)
![11 2 ELECTRIC MACHINERY RJNDAMENTALS
NSE + Nc
- N IH
C
"'
I, NSE + Nc
IH
-
Nc
The Apparent Power Rating Advantage
of Autotransformers
(2- 78)
(2- 79)
It is interesting to note that not all the power traveling from the primary to the sec-
ondary in the autotransformer goes through the windings. As a result, if a con-
ventional transformer is reconnected as an autotransfonner, it can handle much
more power than it was originally rated for.
To understand this idea, refer again to Figure 2- 32b. Notice that the input
apparent power to the autotransformer is given by
Sin = VLIL
and the output apparent power is given by
(2-80)
(2-81)
It is easy to show, by using the voltage and current equations [Equations (2- 76) and
(2- 79)], that the input apparent power is again equal to the output apparent power:
(2-82)
where SIO is defined to be the input and output apparent powers ofthe transformer.
However, the apparent power in the transfonner windings is
(2-83)
1lle relationship between the power going into the primary (and out the sec-
ondary) of the transformer and the power in the transfonner's actual windings can
be found as follows:
Sw = Vcl c
Using Equation (2- 79), we get
= VL(JL - IH)
= VLIL - ~IH
-s
- 10NsE + Nc
(2-84)
(2-85)](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-136-320.jpg)
![TRANSFORMERS 131
A major disadvantage of this connection is that a very large return current
must flow in the neutral of the primary circuit.
The Scott-T Connection
The Scott-T connection is a way to derive two phases 90° apart from a three-phase
power supply. In the early history of ac power transmission, two-phase and three-
phase power systems were quite common. In those days, it was routinely neces-
sary to interconnect two- and three-phase power systems, and the Scott-T trans-
fonner connection was developed for that purpose.
Today, two-phase power is primarily limited to certain control applications,
but the Scott T is still used to produce the power needed to operate them.
The Scott T consists of two single-phase transformers with identical ratings.
One has a tap on its primary winding at 86.6 percent of full-load voltage. 1lley are
connected as shown in Figure 2-43a. TIle 86.6 percent tap of transfonner T2 is
connected to the center tap of transformer Tl . The voltages applied to the primary
winding are shown in Figure 2-43b, and the resulting voltages applied to the pri-
maries of the two transformers are shown in Figure 2- 43c. Since these voltages
are 90° apart, they result in a two-phase output.
It is also possible to convert two-phase power into three-phase power with
this connection, but since there are very few two-phase generators in use, this is
rarely done.
The Three-Phase T Connection
The Scott-T connection uses two transformers to convert three-phase power to
two-phase power at a different voltage level. By a simple modification of that
connection, the same two transfonners can also convert three-phase power to
three-phase power at a different voltage level. Such a connection is shown in Fig-
ure 2- 44. Here both the primary and the secondary windings of transformer Tl are
tapped at the 86.6 percent point, and the taps are connected to the center taps of
the corresponding windings on transformer Tl . In this connection T] is called the
main transfonner and Tl is called the teaser transfonner.
As in the Scott T, the three-phase input voltage produces two voltages 90°
apart on the primary windings of the transformers. TIlese primary voltages pro-
duce secondary voltages which are also 90° apart. Unlike the Scott T, though, the
secondary voltages are recombined into a three-phase output.
One major advantage of the three-phase T connection over the other three-
phase two-transfonner connections (the open-delta and open-wye-open-delta) is
that a neutral can be connected to both the primary side and the secondary side of
the transfonner bank. This connection is sometimes used in self-contained three-
phase distribution transformers, since its construction costs are lower than those
of a full three-phase transformer bank.
Since the bottom parts of the teaser transfonner windings are not used on ei-
ther the primary or the secondary sides, they could be left off with no change in
perfonnance. 1llis is, in fact, typically done in distribution transformers.](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-155-320.jpg)
![TRANSFORMERS 137
% Initia liz e va lues
VM 325;
NP = 850;
% Max imum vo lt age (V)
% Prima r y turn s
% Ca l c ula t e a ngula r ve l oc ity f or 60 Hz
fr eq = 60; % Freq (Hz )
w = 2 * p i .. fr eq;
% Ca l c ula t e flu x ver s u s time
time 0, 1 /300 0, 1 /3 0; % 0 t o 1 /3 0 sec
flu x = - VM /(w*NP ) * cos(w .* time ) ;
% Ca l c ula t e the mmf correspo nd ing t o a g i ven flu x
% u s ing the flu x ' s int e r po l a tion func tion.
mmf = int e r p l ( flu x_dat a, mmf_dat a, flu x ) ;
% Ca l c ula t e the magne tiz a tio n c urre nt
im = mmf 1 NP ;
% Ca l c ula t e the rms va lue o f the c urre nt
irms = sqrt (sum ( im. A
2 )/ l e ngth ( im )) ;
d i sp( ['The rms c urre nt a t 60 Hz i s " num2str (inns) ] ) ;
% Plot the magne tiz a tion c urre nt.
fi gure ( l )
s ubp l ot (2, 1 , 1 ) ;
p l ot (time, im ) ;
title ('b fMagne tiz a tion Curre nt a t 60 Hz' ) ;
x l abe l ('b fTime (s) ' ) ;
y l abe l ('b f itI_( m) rm (A) ' ) ;
ax i s( [ O 0.04 - 2 2 ] ) ;
g rid on ;
% Ca l c ula t e a ngula r ve l oc ity f or 50 Hz
fr eq = 50; % Freq (Hz )
w = 2 * p i .. fr eq;
% Ca l c ula t e flu x ver s u s time
time 0, 1 / 2500, 1 / 25; % 0 t o 1 / 25 sec
flu x = - VM /(w*NP ) * cos(w . * time ) ;
% Ca l c ula t e the mmf correspo nd ing t o a g i ven flu x
% u s ing the flu x ' s int e r po l a tion func tion.
mmf = int e r p l ( flu x_dat a, mmf_dat a, flu x ) ;
% Ca l c ula t e the magne tiz a tio n c urre nt
im = mmf 1 NP ;
% Ca l c ula t e the rms va lue o f the c urre nt
irms = sqrt (sum ( im. A
2 )/ l e ngth ( im )) ;
d i sp( ['The rms c urre nt a t 50 Hz i s " num2str (inns) ] ) ;](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-161-320.jpg)
![138 ELECTRIC MACHINERY RJNDAMENTALS
~
'"
-'
1.414
0.707
0
-0.7(17
- 1.414
0 0.005 om om5 om 0.025 0.Q3 0.Q35 0.04
Time (s)
,,'
1.414, - , - -"'C---,--,--,--"'__---,----,
0.707 50Hz
o
-0.7(17
-1.414:'---o~~~~cc!~--c'~--o+.co-~o-cc!=~
o 0.005 om om5 om 0.025 0.Q3 0.Q35 0.04
Time (s)
,b,
""GURE 2-47
(a) Magnetization current for the transformer operating at 60 Hz. (b) Magnetization current for the
transformer operating at 50 Hz.
% Pl ot the magnet i zat i on curre nt.
s ubpl ot (2, 1 ,2);
p l ot (t i me , i m) ;
tit l e (' b f Magnet i zat i on Curren t at 50 Hz' ) ;
x l abe l (' b fTime (s) ' ) ;
y l abe l (' b f it I _{m) nn (A) ' ) ;
axi s( [O 0 . 04 - 2 2 ] ) ;
gri d on;
When this program executes, the results are
,.. mag_ current
The nns c urrent at 60 Hz i s 0 .4 894
The nns c urrent at 50 Hz i s 0 . 79252
The resulting magnetization currents are shown in Figure 2-47. Note that the nns magne-
tization current increases by more than 60 percent when the frequency changes from 60 Hz
to 50 Hz.
The Apparent Power Rating of a Transformer
TIle principal purpose of the apparent power rating of a transfonner is that, to-
gether with the voltage rating, it sets the current flow through the transformer
windings. TIle current now is important because it controls the jlR losses in trans-
fonner, which in turn control the heating of the transformer coils. It is the heating](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-162-320.jpg)
![150 ELECTRIC MACHINERY RJNDAMENTALS
transfonner, the following values are measured on the primary (480-V) side of the
transformer:
Open,circuit tcst
Voe = 480 V
loe = 0.41 A
Poe = 38W
Short,circuit test
Vsc = 10.0 V
Isc = 1O.6A
Psc = 26W
(a) Find the per-unit equivalent circuit of this transfonner when it is connected in the
conventional ma/Uler. What is the efficiency of the transfonner at rated condi-
tions and lUlity power factor? What is the voltage regulation at those conditions?
(b) Sketch the transfonner connections when it is used as a 600/480-V step-down
autotransfonner.
(c) What is the kilovoltampere rating of this transformer when it is used in the au-
totransformer connection?
(d) Answer the questions in a for the autotransformer connection.
2-23. Figure P2-4 shows a power system consisting of a three-phase 480-V 60-Hz gener-
ator supplying two loads through a transmission line with a pair of transfonners at
either end.
Generator
41lO V
T, ~~~~~_ _ _ T,
480114.400 V
](XXl kVA
R "'O.OlOpu
X ",0.040pu
""GURE "2-4
Line
ZL",1.5+jIOO 14.4001480 V
500 kVA
R", 0.020 pu
X =0.085 pu
Lood I
ZLood t-
0.45L36.87°n
Y- connected
Lood2
ZLood 2 '"
-jO.8ll
Y- connected
A one-tine diagram of the power system of Problem 2- 23. Note that some impedance values are
given in the per-unit system. while others are given in ohms.
(a) Sketch the per-phase equivalent circuit of this power system.
(b) With the switch opened, find the real power P, reactive power Q, and apparent
power S supplied by the generator. What is the power factor of the generator?
(c) With the switch closed, find the real power P, reactive power Q, and apparent
power S supplied by the generator. What is the power factor of the generator?
(d) What are the transmission losses (transformer plus transmission line losses) in
this system with the switch open? With the switch closed? What is the effect of
adding load 2 to the system?](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-174-320.jpg)
![INTRODUCTION TO POWER ELECTRONICS 153
I. 1lle diode
2. 1lle two-wire thyristor (or PNPN diode)
3. 1lle three-wire thyristor [or silicon controlled rectifier (SCR)]
4. 1lle gate turnoff (GTO) thyristor
5. 1lle DlAC
6. 1lle TRIAC
7. 1lle power transistor (PTR)
8. 1lle insulated-gate bipolar transistor (IGBT)
Circuits containing these eight devices are studied in this chapter. Before the cir-
cuits are examined, though, it is necessary to understand what each device does.
The Diode
A diode is a semiconductor device designed to conduct current in one direction
only. The symbol for this device is shown in Figure 3-1. A diode is designed to
conduct current from its anode to its cathode, but not in the opposite direction.
The voltage-current characteristic of a diode is shown in Figure 3- 2. When
a voltage is applied to the diode in the forward direction, a large current flow re-
sults. When a voltage is applied to the diode in the reverse direction, the current
flow is limited to a very small value (on the order of microamperes or less). If a
large enough reverse voltage is applied to the diode, eventually the diode will
break down and allow current to flow in the reverse direction. 1llese three regions
of diode operation are shown on the characteristic in Figure 3- 2.
Diodes are rated by the amount of power they can safely dissipate and by the
maximum reverse voltage that they can take before breaking down. The power
v,
Anode
PIV 'D
Cathode
FIGURE 3-1 FIGURE 3-2
The symbol of a diode. Voltage-current characteristic of a diode.](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-177-320.jpg)
![166 ELECTRIC MACHINERY RJNDAMENTALS
f or ii = l:nv a l s
t emp = t e mp + wavef o rm (ii )A 2;
end
rms = sqrt (t emp / nva l s) ;
~ Ca l c ula t e rippl e f act or
r = sqrt (( rms / aver age )A2 - 1 ) * 100 ;
FlUlction ripp l e can be tested by writing an m-file to create a half-wave rectified wave-
fonn and supply that wavefonn to the function. The appropriate M-file is shownbelow:
~ M-file: t est _ ha lfwave .m
~ M-file t o ca l c ula t e the rippl e on the output o f a h a lf-wave
~ wave r ectifie r.
~ Firs t , gene r a t e the out put o f a ha lf-wave r ectifi e r
wave f orm = ze r os( 1 , 128 ) ;
f or ii = 1 : 128
wave f orm (ii ) = ha lfwave (ii*pi / 64 ) ;
end
~ Now ca l c ula t e the rippl e f a c t o r
r = rippl e (wave f orm ) ;
~ Print out the r esult
s tring = ['The ripp l e i s ' num2str (r ) '%.'] ;
d i sp(string ) ;
The output of the half-wave rectifier is simulated by flUlction h a l f wave .
func tion vo lt s = ha lfwave(wt )
~ Func tion t o s imula t e the output o f a ha lf-wave r ectifi e r.
~ wt = Phase in r adi a n s (=omega x time)
~ Convert input t o the r a nge 0 <= wt < 2*p i
while wt >= 2*p i
"
0
" 2*pi ;
end
while
" < 0
"
0
" • 2*pi ;
end
~ Simula t e the out put o f the h a lf-wave r ectifi e r
if wt >= 0 & wt <= p i
vo lt s = s in (wt ) ;
e l se
vo lt s = 0 ;
end
When t es t _ h a l f wave is executed, the results are:
,. te s t ha1fwave
The rippl e i s 121.1772% .
This answer agrees with the analytic solutioncalculated above.](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-190-320.jpg)
![176 ELECTRIC MACHINERY RJNDAMENTALS
where A and B are constants depending on the initial conditions in the circuit.
Since vC<O) = 0 V and vC<D<» = Yoc, it is possible to solve for A and B:
A = vc(D<» = Yoc
A + B = vdO) = 0 => B = - Yoc
Therefore,
(3--4)
The time at which the capacitor will reach the breakover voltage is found by
solving for time t in Equation (3--4) :
(3- 5)
In this case,
120V-75V
tl = - (1ookOXI/.!F)ln 120V
= 98ms
Similarly, the equation for the voltage on the capacitor as a flUlction of time dur-
ing the discharge portion of the cycle turns out to be
vc(t) = VBQe- ·,R,C
so the current flow through the PNPN diode becomes
V,a
i(t) = -- e- tlR,c
R,
(3-6)
(3-7)
If we ignore the continued trickle of current through R], the time at which i(t)
reaches IH and the PNPN diode turns off is
Therefore, the total period of the relaxation oscillator is
T = tl + tl = 98 ms + 2 ms = lOOms
and the frequency of the relaxation oscillator is
/= 1 = 10Hz
T
(b) If R] is increased to 150 kO, the capacitor charging time becomes
Yoc - ¥ BO
t] = - RIC in V
oc
= -(150 kO)( 1 F) I 120 V - 75 V
/.! n 120V
= 147 ms
(3-8)](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-200-320.jpg)
![INTRODUCTION TO POWER ELECTRONICS 177
The capacitor discharging time remains unchanged at
1,/1,
t2 = - RlC ln -V = 2 ms
'0
Therefore, the total period of the relaxation oscillator is
T = t] + t2 = 147ms + 2ms = 149ms
and the frequency of the relaxation oscillator is
1
f= 0.149 s = 6.71 Hz
Pulse Synchronization
In ac applications, it is important that the triggering pulse be applied to the con-
trolling SCRs at the same point in each ac cycle. TIle way this is nonnally done is
to synchronize the pulse circuit to the ac power line supplying power to the SCRs.
This can easily be accomplished by making the power supply to the triggering cir-
cuit the same as the power supply to the SCRs.
If the triggering circuit is supplied from a half-cycle of the ac power line, the
RC circuit will always begin to charge at exactly the beginning of the cycle, so the
pulse wi ll always occur at a fIXed time with respect to the beginning of the cycle.
Pulse synchronization in three-phase circuits and inverters is much more
complex and is beyond the scope of this book.
3.4 VOLTAGE VARIATION BY
AC PHASE CONTROL
The level of voltage applied to a motor is one of the most common variables in
motor-control applications. The SCR and the TRIAC provide a convenient tech-
nique for controlling the average voltage applied to a load by changing the phase
angle at which the source voltage is applied to it.
AC Phase Control for a DC Load Driven from an
AC Source
Figure 3- 28 illustrates the concept of phase angle power control. The figure
shows a voltage-phase-control circuit with a resistive dc load supplied by an ac
source. TIle SCR in the circuit has a breakover voltage for iG = 0 A that is greater
than the highest voltage in the circuit, while the PNPN diode has a very low
breakover voltage, perhaps 10 V or so. The full-wave bridge circuit ensures that
the voltage applied to the SCR and the load will always be dc.
If the switch SI in the picture is open, then the voltage VI at the tenninals of
the rectifier will just be a full-wave rectified version of the input voltage (see Fig-
ure 3- 29).
If switch SI is shut but switch S2 is left open, then the SCR will always be
off. TIlis is true because the voltage out of the rectifier will never exceed VBO for](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-201-320.jpg)
![INTRODUCTION TO POWER ELECTRONICS 191
;" [
n n n ,
;" [
n n n ,
i i!l
SCR I
vc(l)
i!2
L +
co ~ vc(l)
- SCR,
~
-
+
Inductive
D
load
I
~~-~
-
~
.=
-!--~~--~- ,
Ready
'"
-
fire
(a)
(bj
FIGURE 3-44
(a) A series-capacitor forced-commutation chopper cin:uit with improved capacitor recovery time.
(b) The resulting capacitor and load voltage waveforms. Note that the capacitor discharges much
more rapidly, so SCR[ could be refired sooner than before.
quickly than the resistor would. TIle inductor in series with SCR1 protects SCR1
from instantaneous current surges that exceed its ratings. Once the capacitor dis-
charges, SCR1 turns off and SCR] is ready to fi re again.
Parallel-Capacitor Commutation Circuits
The other common way to achieve forced commutation is via the parallel-
capacitor commutation scheme. A simple example of the parallel-capacitor
scheme is shown in Figure 3-45. In this scheme, SCR] is the main SCR, supply-
ing power to the load, and SCR2 controls the operation of the commutating capac-
itor. To apply power to the load, SCRI is fired. When this occurs, a current flows
through the SCR to the load, supplying power to it. Also, capacitor C charges up
through resistor R to a voltage equal to the supply voltage Voc.
When the time comes to turn off the power to the load, SCR2 is fired. When
SCR1 is fired, the voltage across it drops to zero. Since the voltage across a](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-215-320.jpg)
![192 ELECTRIC MACHINERY RJNDAMENTALS
•
R
D ,~
vOC
L
"
-
- C.
SCR]
~ ~
""GURE 3-45
A parallel-<:apacitor forced-commuta.tion chopper circuit.
•
R
L
D ,~
L
oc
I o---J
v
_ Vc +
"
SCR]
~ ~
""GURE 3-46
A parallel-<:apacitor forced-commuta.tion chopper circuit with improved capacitor charging time.
SCRJ permits the load power to be turned off more quickly thaD it could be with the basic parallel-
capacitor circu it.
capacitor cannot change instantaneously, the voltage on the left side of the capaci-
tor must instantly drop to -Voc volts. This turns off SCRb and the capacitor
charges through the load and SCR2 to a voltage of Voc volts positive on its left side.
Once capacitor C is charged, SCRl turns off, and the cycle is ready to begin again.
Again, resistor R] must be large in order for the current through it to be less
than the holding current of SCR2. But a large resistor Rt means that the capacitor
will charge only slowly after SCR] fi res. This limits how soon SCR] can be turned
off after it fires, setting a lower limit on the on time of the chopped waveform.
A circuit with a reduced capacitor charging time is shown in Figure 3-46. In
this circuit SCR1 is triggered at the same time as SCR] is, and the capacitor can](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-216-320.jpg)
![196 ELECTRIC MACHINERY RJNDAMENTALS
Current source invener Voltage source invener
L, I,
L,
-
I I r + I
Main circuit
~ ~ ~
y,( f--o
configuration ~ ~ ~ c f--o
~ ~ ~
f--o
I I I
-
I
Rectifier Inverter Rectifier Inverter
Type of source Current source - Is almost constant Voltage source - Vs almost constam
Output impedance High Low
Line
/"""-... , ', ' Vi r-
Line
voltage / voltage
lO'J"ilf a If 21f
(1800
conduction)
Output waveform
=t1
,
CUrrent OJ
/tv Currem
(1200
conduction)
I. Easy to control overcurrent I. Difficult to limit current
Characteristics
conditions with this design because of capacitor
2. Output voltage varies widely 2. Output voltage variations
with changes in load small because of capacitor
""GURE 3-49
Comparison of current source inveners and voltage source inverters.
instantaneously, this forces the voltage at the top of the capacitor to instantly be-
come 3Voc, turning off SCRt . At this point, the voltage on the bottom half of the
transfonner is built up positive at the bottom to negative at the top of the winding,
and its magnitude is Voc. The voltage in the bottom half induces a voltage Voc in
the upper half of the transformer, charging the capacitor C up to a voltage of 2Voc,
oriented positive at the bottom with respect to the top of the capacitor. The condi-
tion of the circuit at this time is shown in Figure 3- 5Oc.
When SCR] is fired again, the capacitor voltage cuts off SCR2, and this
process repeats indefinitely. The resulting voltage and current wavefonns are
shown in Figure 3- 51.](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-220-320.jpg)
![+
(a)
+
(b'
+
«,
oc
L
' ",-
~-
FIGURE 3-50
INTRODUCTION TO POWER ELECTRONICS 197
+ Ji...(t)
C
",
- SCR
2
_4----'
SCRl
-
c-':= ;, Voc
+
+
;,
if.
;~
Lo,'
•
-
• ;~
(a) A simple single-phase inverter circuit. (b) The voltages and currems in the circuil when SCR[ is
triggered. (c) The voltages and currents in the circuit when SCR1is lriggered.
A Three-Phase Current Source Inverter
Figure 3- 52 shows a three-phase current source inverter. In this circ uit, the six
SCRs fire in the order SCR], SC~, SCR2, SC~, SCR1, SCR5. Capacitors Cl
through C6 provide the commutation required by the SCRs.](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-221-320.jpg)
![198 ELECTRIC MACHINERY RJNDA MENTALS
3Voc
SCR2 turned SCR] turned
~ off
~]
I~
"
2Voc
"
I - - - -
I I
I I
I ;
I I
o
- ,----"
vif),
;,.(1)
2Voc
Voc
o
I
-
H GURE 3-51
I
/
~
,
SCR2 turned SCR] tu
~ off off
I~
"
I I
I -
I I
I I
I ;
I I
,
----"
~
I,
I,
I No<
t
~d
- - - "'SCRI cathode
- - - "'SCR2 cathode
_ _ v)/)
- - - jefl)
Plots of the voltages and current in the inverter circuit: VI is the voltage al the cathode of seRlo and
V1 is the voltage at the cathode of SeRlo Since the voltage at their anodes is Voc. any time VI or V1
exceeds Voc. that SCR is turned off. il.ood is the current supplied to the invener's load.](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-222-320.jpg)
![N
Q
Q
Three-
p,,",
input
-
I, I
L,
Rectifier
FlG URE J-..5 3
---.
SCR]
J
D,
L-
D,
SCR4
J
SCR2 SCR3
C, J
+ ,;-
C, J
"
+ -
D, C] D3
" -
b
Motor
I - ./
c
"-
D, D.
C, C,
"
,,~
+ "- -" +
" C.
"
SCR,
~
SCRo
-'
,,'
T _~
p"'~
input
I, I
L,
Rectifier
SCR ]
J
D,
D,
SCR.,
J
SCR2 SCR1
C, J
+ ,,'-
C, ~
"
+ -
D,
"
C1 D3
D, D.
C,
~,'
+ - v~ _11 +
~,'
"
SCR, SCRo
-' ~
,b,
"
b
Motor
c ./
'.WhenSC~fires.
"'6 ---+ O. Therefore
the anode voltage of
SCR~ ('~ ) becomes
negative. and
SC R~ turns off.
1be operation of the three-phase CSI. (3) Initially. SCRI and SCRI are conducting. Note how the commutating capacitocs have charged up. (b) 1be situation when SC~ fires. 1be
voltage at the aoode of SCR6 falls almost instantaneously to zero. Since the voltage across capacitor Cl cannot change instantaneously. the voltage at the anode ofSeRl will become
negative. arxI SCRI will tum off.](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-224-320.jpg)
![Three-
p,,",
input
I, I
L,
Rectifier
---.
SCR]
J
D,
I
D,
SCR4
J
N FlGURE 3-S3 (concluded)
-
SCR2 SCR3
C, J C, -'
+ ,,-
"
+ -
D, ~3 D3
D, D.
C, C,
" + "
" C.
+ "
SCI<,
~ J
SCR,
..i
,"
" -
b
MOlOr
,
I -
V
Gale pulses
SCR
1
1 6 2 4 3 S 6
conducting I SCR t SCR2 SCRJ SCR t
nlerva/s
SCI<, SCR, SCR, SCI<, SCR,
,
- ,
'.(1'1 I I ,
I, I
I I
- I,
]1----,--------'---'----,-------,----
"(:~II------,-----,-------'----'--------,-----
- I,
,d,
8: (c) Now SCR[ and SC~ are conducting, and the commutating capacitors charge up as shown, (d) The gating pulses, SCR conducting intervals, and the output current from this
invener,](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-225-320.jpg)
![210 ELECTRIC MACHINERY RJNDAMENTALS
D,
D,
VA(I) '" VMsin WI V
VB(I) '" VMsin (ro/- 120") V
Vc(l) '" VMsin (&1- 240°) V
fo'IGURE 3-59
D, D,
Lo.d
D, D,
+
}~(')
-
A three-phase full-wave diode bridge ci['(;uit connected to a resistive load.
diode with the highest voltage applied to it at any given time will conduct, and it
will reverse-bias the other two diodes in the section. In the negative half, the diode
with the lowest voltage applied to it at any given time will conduct, and it will
reverse-bias the other two diodes in the section. TIle resulting output voltage is
shown in Figure 3--60.
Now suppose that the six diodes in the bridge circuit are replaced by six
SCRs as shown in Figure 3--61. Assume that initially SCR] is conducting as shown
in Figure 3--61b. nlis SCR will continue to conduct until the current through it falls
below IH. Ifno other SCR in the positive halfis triggered, then SCR[ will be turned
ofT when voltage VA goes to 7..ero and reverses polarity at point 2. However, if SCRl
is triggered at any time after point I, then SCR[ will be instantly reverse-biased and
turned off. TIle process in which SCR2 forces SCR[ to turn off is calJed/orced com-
mutation; it can be seen that forced commutation is possible only for the phase an-
gles between points I and 2. 1lle SCRs in the negative half behave in a similar
manner, as shown in Figure 3--61c. Note that ifeach of the SCRs is fired as soon as
commutation is possible, then the output of this bridge circuit will be the same as
the output of the full-wave diode bridge rectifier shown in Figure 3- 59.
Now suppose that it is desired to produce a linearly decreasing output volt-
age with this circuit, as shown in Figure 3--62. To produce such an output, the con-
ducting SCR in the positive half of the bridge circuit must be turned off whenever
its voltage falls too far below the desired value. 1llis is done by triggering another
SCR voltage above the desired value. Similarly, the conducting SCR in the nega-
tive half of the bridge circuit must be turned ofT whenever its voltage rises too far
above the desired value. By triggering the SCRs in the positive and negative
halves at the right time, it is possible to produce an output voltage which de-
creases in a manner roughly corresponding to the desired wavefonn. It is obvious
from examining Figure 3--62 that many harmonic components are present in the
resulting output voltage.](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-234-320.jpg)
![(,)
(q)
(.)
'Pro1 ;KJ] 0] P"!1ddil "iil'HOA 1C10] ;KU (:l) 's<PpOfP
J111Q-JAnl1ii"U ;KJ] WOJJ "iiC1IOA mdlnO "IlL (q) 's"POfP JI11Ij-JAmsod JI/I UJO.IJ Jiil1l[OA ]ndlnO JIlL (e)
09-£ :nm::
>1.'1
, , " , ' ,,- - ~.... ,,'--..... ,--......... ,,-- ~ ....
/' " /" ...., /" "v" ...., / ',// ...., /"
~ )- /' X /, <
, /~ / ' / ' / /~ I ,
~ / , / ' / ' I~ I , /~
' / , 1 ' ,/~/,I,
' 1 , 1 , 1 1 1
1 ,I ,I 1 '/ 1
, I ,I ' I .1 'I vI ,
Y A v .~ f ,
/ " 1/ / / / /
1 I , 1 ' 1 1 I
1 , 1 1 1 I
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1 ' 1 1 , I
1 ' I , ' I , I " 1
I , 1 ' / ' I 'I , I ,I
I '( ) I~ I )1 ~
, I 1 I 1 1
1 " / I I / I I
// '// ,/ 'x'/ ',,/ ';1// 1
o , ,
I ' / ' 1 I ' I , 1
1 , I , I ' 1 I '
f I , I~/ 1 , 1 '
,
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,
III'I/~/
I /1 ./ ',/ / ~./ ,
,
,
iiu!I:lnpuOJ ta iiunJnpUOJ 9
a l!unJnpUO:l 'a
/.... " /'.
/ ' / , /,
/ ' " ~ / ,
'" / ~ I ,
I / '. / • / , ~
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/ .... 1 ' / ~/
/ , / ' / , /
/ , I / ' /
/ , " , / ,"
, .( A /,
.... / , / ' / '
, / , / , / ,
, / " .... / ....
',,-----' ',,-----''' ',------''' '
(/j-J~ (lfI~ (I)V~
..... ..... --
'-, /"....... ",,- -......... ,,- --- ~ .....
, / / . . . "
'" / .... / ....".
, / ' / , , /
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/ 1 t ( . 1 , ~
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/ , / , /1 ~
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;
;
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l!unJnpUO:l 'a
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~
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l!unJnpUO:l Za l!ufpnpUOJ la
II Z S:JINOlli.:J3"l3 M3MOd 0.1 NOIl.JOamUNI
,
, ;
, ;
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"
(!)V~
" A-
" A
IVA £L'I
(!)JlOO'I~
" A
(11"",
" A
(!)rod~](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-235-320.jpg)
![224 ELECTRIC M ACHINERY RJNDA MENTALS
+ 21
D,
• • I +
)'~(')
")
Y. Lood
S, /-
(-~I SCR
0,
v,...(l) '" 339 sin 3771 V C2 C]
FlGURE P3-1
The cin:uit of Problems 3- 3 through 3--6.
3-4. What would the nns voltage on the load in the circuit in Figure P3-1 be if the firing
angle of the SCR were (a) 0°, (b) 30°, (c) 90°?
'3-5. For the circuit in Figure P3-1, assume that VBO for the DlAC is 30 V, Ct is I p.F, R
is adjustable in the range I to 20 ill, and switch SI is open. What is the firing angle
of the circuit when R is 10 kO? What is the rillS voltage on the load lUlder these con-
ditions? (Caution: This problem is hard to solve analytically because the voltage
charging the capacitor varies as a function of time.)
3-6. One problem with the circuit shown in Figure P3-1 is that it is very sensitive to vari-
ations in the input voltage v.At). For example, suppose the peak. value of the input
voltage were to decrease. Then the time that it takes capacitor C] to charge up to the
breakover voltage of the DIAC will increase, and the SCR will be triggered later in
each half-cycle. Therefore, the rillS voltage supplied to the load will be reduced both
by the lower peak voltage and by the later firing. This same effect happens in the
opposite direction if voc(t) increases. How could this circuit be modi fied to reduce its
sensitivity to variations in input voltage?
3-7. Explain the operation of the circuit shown in Figure P3- 2, and sketch the output
voltage from the circuit.
3-8. Figure P3- 3 shows a relaxation oscillator with the following parameters:
R] = variable
C= IJ.tF
VBO = 30V
R2 = 1500 0
Voc = 100 V
lH = 0.5 mA
(a) Sketch the voltages vc(t), vo(t), and rrJt) for this circuit.
(b) If R, is currently set to 500 kO, calculate the period of this relaxation oscillator.
3-9. In the circuit in Figure P3-4, T] is an autotransformer with the tap exactly in the
center of its winding. Explain the operation of this circuit. Assruning that the load is
inductive, sketch the voltage and current applied to the load. What is the purpose of
SCR2? What is the purpose of D2? (This chopper circuit arrangement is known as a
Jones circuit.)
*The asterisk in front of a problem number indicates that it is a more difficult problem.](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-248-320.jpg)
![INTRODUCTION TO POWER ELECTRONICS 225
• "~---T
~
•
•
c, Lood
--
.,'----.L T,
T,
FIGURE 1'3-2
The inverter circuit of Problem 3- 7.
+~-----;
Voc'" lOOV
" J
+
+
Vc (I) (
'>'0 (/)
R2", 1500 n
C'" 1.0~
FIGURE 1'3-3
The relaxation oscillator circuit of Problem 3--8.
+
J
SCR] ,icC
~CR2
D,
•
T] (Autotnlnsformer)
D, Lo,d
FlGURE P3-4
The chopper ci["(;uit of Problem 3- 9.](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-249-320.jpg)
![226 ELECTRIC MACHINERY RJNDAMENTALS
3-10. A series-capacitor forced commutation chopper circuit supplying a purely resistive
load is shown in Figure P3- 5.
Voc = 120 V
lH = 8 rnA
VBO = 200 V
Rl =20 kll
Rlood = 250 0
C = 150 /lF
(a) When SCRl is turned on. how long will it remain on? What causes it to tum off?
(b) When SCRl turns off. how long will it be until the SCR can be turned on again?
(Assume that 3 time constants must pass before the capacitor is discharged.)
(c) What problem or problems do these calculations reveal about this simple series-
capacitor forced-commutation chopper circuit ?
(d) How can the problem(s) described in part c be eliminated?
+
SCR
0---/ +
R C
)"
-
voc
/
D ,~
~ Rwm Lo,'
""GURE 1'3-5
The simple series-capacitor forced-commulation cin:uit of Problem 3-10.
3-11. A parallel-capacitor forced-conunutation chopper circuit supplying a purely resis-
tive load is shown in Figure P3-6.
Voc = 120 V
lH = 5 rnA
VBO = 250V
R] =20kfi
Rlood = 250 0
C= 15 /lF
(a) When SCR] is turned on, how long will it remain on? What causes it to IlUll off?
(b) What is the earliest time that SCR] can be turned off after it is turned on?
(Assume that 3 time constants must pass before the capacitor is charged.)
(c) When SCR] turns off, how long will it be until the SCR can be tlUlled on again?
(d) What problem or problems do these calculations reveal about this simple parallel-
capacitor forced-commutation chopper circuit?
(e) How can the problem(s) described in part d be eliminated?
3-12. Figure P3- 7 shows a single-phase rectifier-inverter circuit. Explain how this circuit
ftmctions. What are the purposes of C] and C2? What controls the output frequency
of the inverter?](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-250-320.jpg)
![ACMACHINERYFUNDAMENTALS 239
°
,
°b'
FIGURE 4- 8
11",, -(/)
11",-(/)
" ",,(0
0"
"I
o a'
(a) A simple three-phase stator. Currents in this stator are assumed positive if they flow into the
unprimed end and out the primed end of the coils. The magnetizing intensities produced by each coil
are also shown. (b) The magnetizing intensity vector H.... (/) produced by a current flowing in coil 00'.
A · turns/ m (4-22b)
A·turns/ m (4- 22c)
The flux densities resulting from these magnetic field intensities are give n
by Equation ( 1-2 1):
They are
Baa' (1) = BM sin wl L 0° T
Bbb' (1) = BM sin (wl- 120°) L 120°
BC<", (1) = BM sin (wl- 240°) L 240°
T
T
( 1-21)
(4- 23a)
(4-23b)
(4- 23c)
where BM = J1HM
. 1lle currents and their corresponding flux densities can be ex-
amined at specific times to detennine the resulting net magnetic field in the stator.
For example, at time wt = 0°, the magnetic field from coil ad will be
B"",,= 0
The magnetic field from coil bb' will be
BbI>' = BM sin (_1 20°) L 1200
and the magnetic field from coil ee' wi] I be
BC<", = BM sin (_240°) L 240°
(4- 24a)
(4-24b)
(4- 24c)](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-263-320.jpg)
![ACMACHINERYFUNDAMENTALS 241
The resulting magnetic field is shown in Figure 4- 9b. Notice that although the di-
rection of the magnetic field has changed, the magnitude is constant. TIle mag-
netic field is maintaining a constant magnitude while rotating in a counterclock-
wise direction.
Proof of the Rotating Magnetic Field Concept
At any time t, the magnetic field wi ll have the same magnitude I.5BM, and it wi ll
continue to rotate at angular velocity w. A proof of this statement for all time t is
now given.
Refer again to the stator shown in Figure 4-8. In the coordinate system
shown in the fi gure, the x direction is to the right and the y direction is upward.
TIle vector:l1 is the unit vector in the horizontal direction, and the vector S' is the
unit vector in the vertical direction. To find the total magnetic flux density in the
stator, simply add vectorially the three component magnetic fields and detennine
their sum.
The net magnetic nux density in the stator is given by
B••(I) ~ B_, (I) + B~, (I) + B~, (I)
= BM sin wt LO° + BMsin (wt - 120°) L 120° +BM sin (wi_ 240°) L 2400T
Each of the three component magnetic fields can now be broken down into its x
and y components.
Bnet(t) = BMsin wt x
- [O.5BMsin (wt - 1200)]x + ['] BMsin (wt - 1200)]y
- [O.5BMsin (wt - 2400)]x - ['] BM sin (wt - 2400)]y
Combining x and y components yields
Dnet(t) = [BMsin wi - O.SBMsin (wt - 120°) - O.5BMsin (wt - 240°)] x
+ ['7BMsin(wt - 120°) - '7BMsin (wt - 2400)]y
By the angle-addition trigonometric ide ntities,
BnetCt) = [BMsin wi + iBM sin wt + 1BM cos wi + i BMsin wt -1BMcos wt]x
+ [-1BMsinwt - ~BMCOSWt + ~BM sinwt - ~BMCOSWt]S'
IBneln = (1.5BMsin wt):I1 - (1.5BMcos wI)y I (4- 25)
Equation (4- 25) is the final expression for the net magnetic flux density. Notice
that the magnitude of the field is a constant I.SBMand that the angle changes con-
tinually in a counterclockwise direction at angular velocity w. Notice also that at](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-265-320.jpg)
![244 ELECTRIC MACHINERY RJNDAMENTALS
fe = 2fm four poles
W e = 2wm four poles
(4- 29)
(4- 30)
Ingeneral, if the number of magnetic poles on an ac machine stator is P, then
there are PI2 repetitions of the winding sequence a-c '-b-a '-e-b' around its inner
surface, and the electrical and mechanical quantities on the stator are related by
le, ~ iem l (4- 3 1)
(4- 32)
(4- 33)
Also, noting that fm = n,,/60, it is possible to relate the electrical frequency in
hertz to the resulting mechanical speed of the magnetic fields in revolutions per
minute. nlis relationship is
Reversing the Direction of Magnetic
Field Rotation
(4- 34)
Another interesting fact can be observed about the resulting magnetic field.lfthe
current in any two of the three coils is swapped, the direction of the mngnetie
field's rotation will be reversed. This means that it is possible to reverse the direc-
tion of rotation of an ac motor just by switching the connections on any two of the
three coils. lllis result is verified below.
To prove that the direction of rotation is reversed, phases bb' and ee' in Fig-
ure 4-8 are switched and the resulting flux density Bn
.. is calculated.
llle net magnetic flux density in the stator is given by
B...,(t) = B"".(t) + Bw(t) + BeAt)
= BM sin wi L 0° + BM sin (wi- 240°) L 120° + BM sin (wI- 120°) L 240° T
Each of the three component magnetic fie lds can now be broken down into its x
and y components:
BDeI(t) = BM sin wt5i.
- [0.5BM sin (wt - 2400)] 5i. + [1"BM sin (wt - 2400
)]y
- [0.5BM sin (wt - I200)] 5i. - [1"BM sin (wt - 1200
)]y
Combining x and y components yields](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-268-320.jpg)
![ACMACHINERYFUNDAMENTALS 245
Hnem = [BMsin wt - O.5BMsin (wt - 240°) - 0.5BM sin(WI' - 1200jx
+ ['7BMsin (WI' - 240°) - '7BM sin (wt - l200)]y
By the angle-addition trigonometric identities,
S nelt) = [BMsin WI' + iBM sin wt -1BM cos WI' + iBM sin wt + IBM cos wt]x
+ [-1BMsinwt + ~BM COS Wt + ~BM sinwt + ~BMCOSWt]S
IS nell) = (1.5BMsin wt)J1 + (1.5BMcos WI')Y I (4- 35)
This time the magnetic fie ld has the same magnitude but rotates in a clock-
wise direction. 1l1erefore, switching the currents in two stator phases reverses the
direction ofmagneticfield rotation in an ac machine.
EXllmple 4-1. Create a MATLAB program that models the behavior of a rotating
magnetic field in the three-phase stator shown in Figure 4-9.
Solutioll
The geometry of the loops in this stator is fIXed as shown in Figure 4-9. The currents in the
loops are
i"",(t) = 1M sin wt A
iw(t) = 1M sin (wt - 120°)
iec,(t) = 1M sin (wt - 240°)
and the resulting magnetic flux densities are
B"",(t)= BM sinwt LO° T
A
A
B"",(t) = BM sin (wt - 120°) L 120°
Bec,(t) = BM sin (wt - 240°) L 240°
<jl = 2rlB = dlB
T
T
(4-2Ia)
(4-21b)
(4-21c)
(4-23a)
(4-23b)
(4-23c)
A simple MATLAB program that plots Boo" BI+" Bee'. and B"", as a ftmction of time is
shown below:
% M-file, mag_ fi e l d.m
% M-file t o ca l c ulate the net magnetic fi e l d produ ced
% by a three-pha se s tat or.
% Set up the bas i c cond ition s
bmax = 1; % Normalize bmax t o 1
fr eq = 60, % 60 Hz
w = 2*p i * fr eq, % a ngular ve l oc ity (rad/ s )
% Fir s t , generate the three component magnetic fi e l ds
t = 0,1 / 6000, 1 / 60,
Baa = s in (w* t ) .* (cos ( O) + j*s in (O)) ,](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-269-320.jpg)
![246 ELECTRIC MACHINERY RJNDAMENTALS
Ebb = s in (w*t - 2*pi /3) * (cos (2*pi /3) + j*s in (2*pi /3)) ;
Ecc = s in (w*t +2*pi /3) * (cos (- 2*pi /3) + j*s in (- 2*pi /3)) ;
!l; Ca l c ulate Enet
Enet = Baa + Ebb + Ecc;
!l; Ca l c ulate a c irc l e representing the expected maximum
!l; va lue o f Enet
c irc l e = 1.5 * (cos (w*t ) + j *s in (w*t ) ) ;
!l; Plo t the magnitude and d irection of the r esulting magne ti c
!l; fi e l ds. No te that Baa i s b l ack, Bbb i s b lue, Bcc i s
!l; magenta , and Enet i s red.
f o r ii = l,length (t )
!l; Plo t the reference c irc l e
p l o t (c irc l e, 'k' ) ;
h o l d o n;
!l; Plo t the f o ur magneti c fi e l ds
p l o t ( [0 real (Baa (il )) ] , [0 i mag (Baa (il )) ] , 'k', 'LineWi dth' ,2);
p l o t ( [0 real (Ebb (il ) ) ] , [0 i mag (Bbb (il ) ) ] , 'b' , 'LineWi dth ' ,2) ;
p l o t ( [0 real (Bec (il ) ) ] , [0 i mag (Bec (il ) ) ] , 'm' , 'LineWi dth ' ,2) ;
p l o t ( [O real (Enet (il )) ] , [0 i mag (Bne t (il )) ] ,' r' ,' LineWi dth ',3 ) ;
axi s square;
axi s( [- 2 2 - 2 2 ] ) ;
drawnow;
ho l d o ff ;
ond
When this program is executed, it draws lines corresponding to the three component mag-
netic fields as well as a line corresponding to the net magnetic field. Execute this program
and observe the behavior of B.....
4.3 MAGNETOMOTIVE FORCE AND FLUX
DISTRIBUTION ON AC MACHINES
In Section 4.2, the flux produced inside an ac machine was treated as if it were in
free space. TIle direction of the flux density produced by a coil of wire was as-
sumed to be perpendicular to the plane of the coil, with the direction of the flux
given by the right-hand rule.
TIle flux in a real mnchine does not behave in the simple manner assumed
above, since there is a ferromagnetic rotor in the center of the machine, with a
small air gap between the rotor and the stator. TIle rotor can be cylindrical, like the
one shown in Figure 4-1 2a, or it can have pole faces projecting out from its
surface, as shown in Figure 4-12b. If the rotor is cylindrical, the machine is said
to have nonsalient poles; if the rotor has pole faces projecting out from it, the](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-270-320.jpg)
![252 ELECTRIC MACHINERY RJNDAMENTALS
However, this equation was derived for the case of a moving wire in a stationnry
mngnetie field. In this case, the wire is stationary and the magnetic field is mov-
ing, so the equation does not directly apply. To use it, we must be in a frame of
reference where the magnetic field appears to be stationary. Ifwe "sit on the mag-
netic field" so that the field appears to be stationary, the sides of the coil will ap-
pear to go by at an apparent velocity vre[, and the equation can be applied. Figure
4- 15b shows the vector magnetic field and velocities from the point of view of a
stationary magnetic field and a moving wire.
TIle total voltage induced in the coil will be the sum ofthe voltages induced
in each of its four sides. These voltages are detennined below:
I. Segment abo For segment ab, a = 180°. Assuming that B is directed radially
outward from the rotor, the angle between v and B in segment ab is 90°,
while the quantity v x B is in the direction of I, so
e"", = (v x B) · 1
= vBI directed out of the page
= - V[BM cos (w",t - 180°)]1
= - vBtJ cos (w",t - 180°) (4- 38)
where the minus sign comes from the fact that the voltage is built up with a
polarity opposite to the assumed polarity.
2. Segment be. The voltage on segment be is zero, since the vector quantity
v x B is perpendicular to I, so
ecb = (v x B) -I = 0 (4- 39)
3. Segment ed. For segment ed, the angle a = 0°. Assuming that B is directed
radially outward from the rotor, the angle between v and B in segment ed is
90°, while the quantity v x B is in the direction ofl, so
eJc = (v x B) - I
= vBI directed out of the page
= v(BM cos wn,l)l
= vBtJ cos w.,/ (4-40)
4. Segment da. The voltage on segment da is zero, since the vector quantity
v x B is perpendicular to I, so
ead = (v x B) - I = 0
Therefore, the total voltage on the coil wi ll be
eind = e"" + edc
= - vBtJ cos(w",t - 180°) + vBtJ cos w",t
Since cos () = - cos «
() - 180°),
(4-41)
(4-42)](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-276-320.jpg)
![ACMACHINERYFUNDAMENTALS 255
where d is the diameter and I is the length of the coil. Therefore, the flux in the machine is
given by
<p = (0.5 mXO.3 m)(0.2 T) = 0.03 Vb
The speed of the rotor is given by
w = (3600 r/minX27T radXI minl60 s) = 377 radls
(a) The magnitudes of the peak. phase voltages are thus
Emu = Nc<Pw
= (15 turnsXO.03 Wb)(377 radls) = 169.7 V
and the three phase voltages are
e"".(t) = 169.7 sin 377t V
ebb.(t) = 169.7 sin (377t -1200) V
e,At) = 169.7 sin (377t - 240°) V
(b) The nns phase voltage of this generator is
Ell = E~x = 16~V = 120V
(c) Since the generator is V-connected,
VT = v'5EIl = 0(120 V) = 208 V
4.5 INDUCED TORQUE IN AN AC MACHINE
In ac machines under nonnaI operating conditions, there are two magnetic fields
present--.:1. magnetic field from the rotor circuit and another magnetic field from
the stator circuit. The interaction of these two magnetic fields produces the torque
in the machine, just as two pennanent magnets near each other will experience a
torque which causes them to line up.
Figure 4-1 7 shows a simplified ac machine with a sinusoidal stator flux dis-
tribution that peaks in the upward direction and a single coil of wire mounted on
the rotor. TIle stator flux distribution in this machine is
Bs<,.a ) = Bs sin a (4- 51)
where Bs is the magnitude of the peak flux density; B:!..a ) is positive when the flux
density vector points radially outward from the rotor surface to the stator surface.
How much torque is produced in the rotor of this simplified ac machine? To find
out, we will analyze the force and torque on each ofthe two conductors separately.
The induced force on conductor I is
F = i(l x B)
= ilBs sin a
The torque on the conductor is
"TiDd.] = (r x F)
= rilBs sin a
with direction as shown
counterclockwise
( 1-43)](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-279-320.jpg)
![o
E '
",IOIX I
,
,
B,
,,'
E '
",IDiX I
,
B, ,
,
o
o ,, 0
,
,
,
,
,
FIGURE 5-8
,
, D,
•
'e'
".
'.•
'0 E,.,
•
'.
SYNCHRONOUS GENERATORS 275
o
o
,
D,
w
,
,
,
,
,
,b,
,
,
,
,
, v,
,
I I... JII>X
, '
o
.' I 'AJIIH
, , '
--:::::]'0 ""
,
,
D,
,
,
,
,
,d,
,
o
~ E""
•
The development of a model for armature reaction: (a) A rotating magnetic field produces the
internal generated voltage EA' (b) The resulting voltage produces a lagging currentflow when
connected to a lagging load. (e) The stator current produces its own magnetic field BS' which
produces its own voltage E_ in the stator windings of the machine. (d) The field Us adds to "I/"
distorting it into H.... The voltage E... adds to EA. producing v
.at the output of the phase.
hand rule to be as shown in Figure 5--8c. The stator magnetic field Bs produces a
voltage of its own in the stator, and this voltage is called E...., on the figure.
With two voltages present in the stator windings, the total voltage in a phase
is just the sum of the internal generated voltage EA. and the annature reaction volt-
age E"a,:
(5-4)
The net magnetic field 8 ..., is just the sum of the rotor and stator magnetic fields:
(5- 5)](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-299-320.jpg)
![SYNCHRONOUS GENERATORS 277
I"
+
jXs R,
+
EA]
""' 'f]
I,
+
I"
R.,
+
R,
jXs R,
v, +
EA2
""' 'f2
(&)
L,
FIGURE 5-10
The full equivalent circuit of a three-phase synchronous generator.
II is now possible 1
0 sketch the equivalent circuit of a three-phase synchro-
nous generator. The full equivalent circuit of such a generator is shown in Fig-
ure 5- 10. This figure shows a dc power source supplying the rotor field circuit,
which is modeled by the coil 's inductance and resistance in series. In series with
RF is an adjustable resistor R adj which controls the flow of field current. The rest
of the equivalent circuit consists of the models for each phase. E:1.ch phase has an
internal generated voltage with a series inductance Xs (consisting of the sum of
the armature reactance and the coil 's self-inductance) and a series resistance RA.
TIle voltages and currents of the three phases are 120° apart in angle, but other-
wise the three phases are identical.
TIlese three phases can be either Y- or Ii-connected as shown in Figure
5- 11. If they are Y-connected, then the tenninal voltage VT is related to the phase
voltage by
(5- 12)](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-301-320.jpg)
![296 ELECTRIC MACHINERY RJNDA MENTALS
60 L 53.13°
v,
,b
,
v,
",
""GURE 5-14
Generator phasor diagrams for Example 5- 3. (a) Lagging power factor; (b) unity power factor;
(c) leading power factor.
(277 vi = [V. + (1 .0 0)(60 A) sin 36.87°]2 + [( 1.0 n X60 A) cos 36.87°]2
76,729 = (V. + 36)1 + 2304
74,425 = (V. + 36)2
272.8 = V,., + 36
V,., = 236.8 V
Since the generator is Y-colUlected, Vr = V3V. = 410 V.
2. If the generator is loaded with the rated current at unity power factor, then
the phasor diagram wi11look like Figure 5- 24h.To find V
. here the right tri-
angle is](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-320-320.jpg)
![SYNCHRONOUS GENERATORS 297
£1 = vi + (XsIA)2
(277V)2 = vi + [(1.00)(60A)]2
76,729 = Vi + 3600
Vi = 73,129
V. = 270.4 V
Therefore, Vr = V3"V. = 46S.4 V.
3. When the generator is loaded with the rated current at O.S PF leading, the re-
suiting phasor diagram is the one shown in Figure 5- 24c. To find V. in this
situation, we construct the triangle OAB shown in the figure. The resulting
equation is
£1 = (V. - XsIA)2 + (XsI./t cos (/)2
Therefore, the phase voltage at the rated load and O.S PF leading is
(277 V)2 = [V. - (l.OnX60A) sin 36.S7°f + [(1.0 0)(60 A) cos 36.S7o]2
76,729 = (V. - 36)2 + 2304
74,425 = (V. - 36)2
272.S = V. - 36
V. = 30S.S V
Since the generator is Y-cOIUlected, Vr = V3"V. = 535 V.
(c) The output power of this generator at 60 A and O.S PF lagging is
POU,=3V.IAcos ()
= 3(236.S VX60AXO.S) = 34.1 kW
The mechanical input power is given by
= 34.1 kW + a + 1.0kW + 1.5kW = 36.6kW
The efficiency of the generator is thus
Pout 34.1 kW
7f = P, x 100% = 366 kW x 100% = 93.2%
rn .
(d) The input torque to this generator is given by the equation
T = Pin =
- W.
36.6 kW
125.7 radls = 291.2 N · m
The induced cOlUltertorque is given by
P.:oov =
Tind = Wv
34.1 kW
125.7 radls = 271.3 N · m
(e) The voltage regulation of a generator is defined as](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-321-320.jpg)
![SYNCHRONOUS GENERATORS 299
% Now initia liz e a ll other va lues
v-ph ase = zeros( 1 ,2 1 ) ;
e_a = 277.0;
x_s = 1. 0;
theta = 36 .S7 .. (p i / 1 SO ) ; % Converted t o radians
% Now ca l c ulate v-ph ase f or each c urrent l eve l
f or ii = 1: 2 1
e od
(x_s .. i _a( ii ) .. cos( the ta ))" 2 )
(x_s .. i _a( ii ) .. s in (the ta )) ;
% Ca l c ulate terminal vo lt age from the phase vo lt age
v_t = v-ph ase .. sqrt (3) ;
% Plot the terminal ch aracteri s ti c, remembering the
% the line c urre nt i s the same as i _a
p l ot ( i _a, v_t , 'Col or' , 'k' , 'Linewi dth ' ,2.0) ;
x l abe l ( 'Line CUrre nt (A) ' , 'Fontwei ght ' , 'Bo l d ' ) ;
y l abe l ( 'Te rmina l Volt age (V) ' , 'Fontwei ght' , 'Bo l d ' ) ;
title ( 'Te rmina l Ch a r acter i s tic f o r O.S PF l agg ing l oad ' , ...
'Fontwei ght' , 'Bo l d ' ) ;
gr i d on ;
ax i s( [ O 60 400 550 ] ) ;
The plot resulting when this M-file is executed is shown in Figure 5- 25a.
(b) If the generator is loaded with a 0.8 PF leading current, the resulting phasor di-
agram looks like the one shown in Figure 5- 24c. To fmd V.' the easiest ap-
proach is to construct a right triangle on the phasor diagram, as shown in the fig-
ure. From Figure 5- 24c, the right triangle gives
E1 = (V. - XsfA sin 9)2 + (XsfAcos 9)2
This equation can be used to solve for V", as a ftmction of the current fA:
V. = JE
l (XsfA cos (J)l + XsfA sin 9
This equation can be used to calculate and plot the terminal characteristic in a
manner similar to that in part a above. The resulting tenninal characteristic is
shown in Figure 5- 25b.
5.9 PARALLEL OPERATION OF
AC GENERATORS
In today's world, an isolated synchrono us generator supplying its own load inde-
pendently of other generators is very rare. Such a situation is found in only a few
out-of-the-way applications such as emergency generators. For all usual genera-
tor applications, there is more than one generator operating in parallel to supply
the power demanded by the loads.An extreme example ofthis situation is the U.S.
power grid, in which literally thousands of generators share the load on the
system.](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-323-320.jpg)
![304 ELECTRIC MACHINERY RJNDAMENTALS
..._
S
C
,_
"_
,h
,;,'C
'"C""
_ J
FIGURE 5-28
A synchrosrope.
at the top and 1800
at the bottom. Since the frequencies of the two systems are
slightly different, the phase angle on the meter changes slowly. If the oncoming
generator or system is faster than the running system (the desired situation), then
the phase angle advances and the synchroscope needle rotates clockwise. If the
oncoming machine is slower, the needle rotates counterclockwise. When the syn-
chroscope needle is in the vertical position, the voltages are in phase, and the
switch can be shut to connect the systems.
Notice, though, that a synchroscope checks the relationships on only one
phase. It gives no infonnation about phase sequence.
In large generators belonging to power systems, this whole process of par-
alleling a new generator to the line is automated, and a computer does this job. For
smaller generators, though, the operator manually goes through the paralleling
steps just described.
Frequency-Power and Voltage-Reactive Power
Characteristics of a Synchronolls Generator
All generators are driven by a prime mover, which is the generator's source of
mechanical power. TIle most common type of prime mover is a steam turbine, but
other types include diesel engines, gas turbines, water turbines, and even wind
turbines.
Regardless of the original power source, all prime movers tend to behave in
a similar fashion--.:1.s the power drawn from them increases, the speed at which
they turn decreases. The decrease in speed is in general nonlinear, but some form
of governor mechanism is usually included to make the decrease in speed linear
with an increase in power demand.
Whatever governor mechanism is present on a prime mover, it will always
be adjusted to provide a slight drooping characteristic with increasing load. The
speed droop (SD) ofa prime mover is defined by the equation
I SO = nnl nn nil x 100% I (5- 27)
where n o] is the no-load prime-mover speed and no is the full-load prime-mover
speed. Most generator prime movers have a speed droop of 2 to 4 percent, as de-
fined in Equation (5- 27). In addition, most governors have some type of set point](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-328-320.jpg)
![306 ELECTRIC MACHINERY RJNDAMENTALS
VTo]
Q, 0
kYAR consumed
""GURE 5-30
Qn Q (reactive power).
kYAR supplied
The curve of terminal voltage (Vr) versus reactive power (Q) for a synchronous generator.
generator, its tenninal voltage drops. Likewise, when a leading load is added to a
synchronous generator, its tenninal voltage increases. It is possible to make a plot
oftenninal voltage versus reactive power. and such a plot has a drooping charac-
teristic like the one shown in Figure 5-30. This characteristic is not intrinsically
linear, but many generator voltage regulators include a feature to make it so. The
characteristic curve can be moved up and down by changing the no-load tenninal
voltage set point on the voltage regulator. As with the frequency-power character-
istic, this curve plays an important role in the parallel operation of synchronous
generators.
TIle relationship between the terminal voltage and reactive power can be
expressed by an equation similar to the frequency-power relationship [Equation
(5-28)] if the voltage regulator produces an output that is linear with changes in
reactive power.
It is important to realize that when a single generator is operating alone, the
real power P and reactive power Q supplied by the generator will be the amount
demanded by the load attached to the generator- the P and Q supplied cannot be
controlled by the generator's controls. Therefore, for any given real power, the
governor set points control the generator's operating frequencyIe and for any given
reactive power, the field current controls the generator's tenninal voltage VT.
Example 5-5. Figure 5-31 shows a generator supplying a load. A second load is to
be connected in parallel with the first one. The generator has a no-load frequency of 61.0
Hz and a slope sp of I MWlHz. Load I consumes a real power of I()(x) kW at 0.8 PF lag-
ging. while load 2 consrunes a real power of 800 kW at 0.707 PF lagging.
(a) Before the switch is closed. what is the operating frequency of the system?
(b) After load 2 is cOIUlected. what is the operating frequency of the system?
(c) After load 2 is cOIUlected. what action could an operator take to restore the sys-
tem frequency to 60 Hz?](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-330-320.jpg)
![SYNCHRONOUS GENERATORS 321
120
11Xl -- -- - -- -- -- -- -- -- --
,
E 80
-
0
fin>'aDtaooow
'-
•, (j(]
!
•
,40
.l
"
f
/ [V "
V
~
20
0.5 LO
Time,s
FIGURE 5-43
The dynamic response when an applied torque equal to 50% of Tmu is suddenly added to a
synchronous generator.
increase until an angle [) is reached at which T;Dd is equal and opposite to T opp' TIlis is
the steady-state operating point ofthe generator with the new load. However, the ro-
tor of the generator has a great deal of inertia, so its torque angle /) actually over-
shoots the steady-state position, and gradually settles out in a damped oscillation, as
shown in Figure 5-43. TIle exact shape ofthis damped oscillation can be detennined
by solving a nonlinear differential equation, which is beyond the scope of this book.
For more information, see Reference 4, p. 345.
The important point about Figure 5-43 is that ifat any point in the transient
response the instantaneous torque exceeds T"""" the synchronous generator will be
unstable. The size of the oscillations depends on how suddenly the additional
torque is applied to the synchronous generator. If it is added very gradually, the
machine should be able to almost reach the static stability limit. On the other
hand, if the load is added sharply, the machine will be stable only up to a much
lower limit, which is very complicated to calculate. For very abrupt changes in
torque or load, the dynamic stability limit may be less than half of the static sta-
bility limit.
Short-Circuit Transients
in Synchronous Generators
By far the severest transient condition that can occur in a synchronous generator
is the situation where the three terminals of the generator are suddenly shorted
out. Such a short on a power system is called afaull. There are several compo-
nents of current present in a shorted synchronous generator, which will be de-
scribed below. TIle same effects occur in less severe transients like load changes,
but they are much more obvious in the extreme case of a short circuit.](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-345-320.jpg)
![350 ELECTRIC MACHINERY RJNDAMENTALS
torque's direction reverses and becomes counterclockwise. In other words, the ma-
chine's torque is now in the direction of motion, and the machine is acting as a mo-
tor. The increasing torque angle 8 results in a larger and larger torque in the direc-
tion of rotation, until eventually the motor's induced torque equals the load torque
on its shaft. At that point, the machine will be operating at steady state and syn-
chronous speed again, but now as a motor.
TIle phasor diagram corresponding to generator operation is shown in Fig-
ure 6-3a, and the phasor diagram corresponding to motor operation is shown in
Figure 6-4a. TIle reason that the quantity jXsI), points from Vo/>, to E), in the gen-
erator and from E), to Vo/> in the motor is that the reference direction of I), was re-
versed in the definition of the motor equi valent circuit. The basic difference be-
tween motor and generator operation in synchronous machines can be seen either
in the magnetic field diagram or in the phasor diagram. In a generator, E), lies
ahead of V
o/>, and BR lies ahead of 8 0 ... In a motor, E), lies behind Vo/>' and BR lies
behind Boe, . In a motor the induced torque is in the direction of motion, and in a
generator the induced torque is a countertorque opposing the direction of motion.
6.2 STEADY-STATE SYNCHRONOUS
MOTOR OPERATION
TIlis section explores the behavior of synchronous motors under varying condi-
tions of load and field current as well as the question of power-factor correction
with synchronous motors. The following discussions will generally ignore the ar-
mature resistance of the motors for simplicity. However, R), will be considered in
some of the worked numerical calculations.
The Synchronous Motor Torque-Speed
Characteristic Curve
Synchronous motors supply power to loads that are basically constant-speed de-
vices. They are usually connected to power systems very much larger than the in-
dividual motors, so the power systems appear as infinite buses to the motors. TIlis
means that the terminal voltage and the system frequency will be constant regard-
less of the amount of power drawn by the motor. 1lle speed of rotation of the mo-
tor is locked to the applied electrical frequency, so the speed of the motor will be
constant regardless of the load. The resulting torque-speed characteristic curve is
shown in Figure 6- 5. The steady-state speed of the motor is constant from no load
all the way up to the maximum torque that the motor can supply (called the pull-
out torque), so the speed regulation of this motor [Equation (4-68)] is 0 percent.
1lle torque equation is
(4-<>1 )
(5- 22)](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-374-320.jpg)
![SYNCHRONOUS MOTORS 353
jXSIA has to increase to reach from the tip of EA to Vo/>, and therefore the annature
current IAalso increases. Notice that the power-factor angle () changes too, becom-
ing Jess and less leading and then more and more lagging.
Example 6-1. A 20S-V, 4S-kVA, O.S-PF-Ieading, a-connected, 60-Hz synchro-
nous machine has a synchronous reactance of 2.5 0 and a negligible armature resistance.
Its friction and windage losses are 1.5 kW, and its core losses are 1.0 kW. Initially, the shaft
is supplying a IS-hp load, and the motor's power factor is O.SO leading.
(a) Sketch the phasor diagram of this motor, and find the values of lA, fL' and EA.
(b) Assume that the shaft load is now increased to 30 hp. Sketch the behavior of the
phasor diagram in response to this change.
(c) Find lA, fL' and EA after the load change. What is the new motor power factor?
Solutioll
(a) Initially, the motor's output power is 15 hp. This corresponds to an output of
POOl = (15 hp)(0.746 KWlhp) = 11.19 kW
Therefore, the electric power supplied to the machine is
Pin = P
out + P""",blo<s + P CO/IeJo.. + Pel""l.,..
= 11.I9kW+ I.5kW+ 1.0kW+OkW = 13.69kW
Since the motor's power factor is O.SO leading, the resulting line current flow is
I - ccciP
-","CC-o
L - v'3VTcos 0
13.69 kW
= V3"(20S VXO.SO) = 47.5 A
and the annature current is h/V'!, with O.Sleading power factor, which gives
the result
IA = 27.4 L 36.S7° A
To find EA
, apply Kirchhoff's voltage law [Equation (6-2)]:
EA = Vo/> - jXsIA
= 20S L 0° V - (j2.5 0)(27.4 L 36.S7° A)
= 20S L 0° V - 6S.5 L 126.S7° V
=249.I -jS4.SV =2SSL - 12.4° V
The resulting phasor diagram is shown in Figure 6-7a.
(b) As the power on the shaft is increased to 30 hp, the shaft slows momentarily, and
the internal generated voltage EA swings out to a larger angle /j while maintain-
ing a constant magnitude. The resulting phasor diagram is shown in Figure 6-7b.
(c) After the load changes, the electric input power of the machine becomes
Pin = P
out + Pmoc.b lo<s + PCO/IeJo.. + P el""l.,..
= (30 hpXO.746 kWlhp) + 1.5 kW + 1.0 kW + 0 kW
= 24.SSkW](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-377-320.jpg)
![354 ELECTRIC MACHINERY RJNDAMENTALS
,
,
,
1
..1. '" 27.4 L 36.87° A
8
""GURE 6-7
"J
(b,
,,,,,
V. '" 208 L r:f' V
jXSIA '" 68.5 L 126.87°
EA"'255L - 12.4°Y
,,,,
V.",208Lr:f'V
E;" '" 255 L _23° V
(a) The motor phasor diagram for Example 6-la. (b) The motor phasor diagram for Example 6- lb.
From the equation for power in tenns of torque angle [Equation (5-20)], it is pos-
sible to find the magnitude of the angle /j (remember that the magnitude of EA is
constant):
'0
p = 3VI/!EA sin Ii
X,
_ . _] XsP
ii -sill 3VE
.,
_ . _] (2.5 flX24.SS kW)
- Sill 3(20S V)(255 V)
= sin- ] 0.391 = 23°
(5- 20)
The internal generated voltage thus becomes EA = 355 L _23° V. Therefore, IA
will be given by
_ VI/! - EA
IA - 'X
J ,
20SLooY-255L-23°Y
=
j2.5fl](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-378-320.jpg)
![SYNCHRONOUS MOTORS 357
FIGURE 6-10
(a) The phasor diagram of an underexcited synchronous motor. (b) The phasor diagram of an
overexcited synchronous motor.
motor has a leading current and supplies Q to the power system. Since the field
current is large in this situation, the motor is said to be overexcited. Phasor dia-
grams illustrating these concepts are shown in Figure 6-10.
EXllmple 6-2. The 20S-V, 45-kVA, O.S-PF-Ieading, 8-cOIUlected, 60-Hz synchro-
nous motor of the previous example is supplying a 15-hp load with an initial power factor
of 0.85 PF lagging. The field current I" at these conditions is 4.0 A.
(a) Sketch the initial phasor diagram of this motor, and fmd the values IAand EA.
(b) If the motor's flux is increased by 25 percent, sketch the new phasor diagram of
the motor. What are EA
, lA, and the power factor of the motor now?
(c) Assume that the flux in the motor varies linearly with the field current I". Make
a plot of 1..1 versus I" for the synchronous motor with a IS-hp load.
Solutioll
(a) From the previous example, the electric input power with all the losses included
is p~ = 13.69 kW. Since the motor's power factor is 0.85 lagging, the resulting
annature current flow is
I - n,R",~"::-;;
A - 3VoIIcos(J
13.69 kW
= 3(20S V)(0.S5) = 25.8 A
The angle (J is cos-1 0.85 = 31.8°, so the phasor current 1..1 is equal to
1..1 = 25.8 L -31.SoA
To find EA
, apply Kirchhoff's voltage law [Equation (6--2)]:
EA = VoII - jXSIA
= 20S L 0° V - (j2.5 0)(25.8 L - 31.So A )
=20SLOo V - 64.5L5S.2° V
= 182L - 17.5° V
The resulting phasor diagram is shown in Figure 6- 11, together with the results
for part b.](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-381-320.jpg)
![358 ELECTRIC MACHINERY RJNDAMENTALS
,
,
,
I;' I
,
I" ,
,
,
fV~"'208LOOV
EA",182L - 17.5° Y.J '- E;' '" 227.5 L _ 13.9° Y
""GURE 6-11
The phasor diagram of the motor in Example 6--2.
(b) Ifthe flux cp is increased by 25 percent, then EA = Kcpw will increase by 25 per-
cent too:
EA2 = 1.25 EAI = 1.25(182 V) = 227.5 V
However, the power supplied to the load must remain constant. Since the dis-
tance EA sin /) is proportional to the power, that distance on the phasor diagram
must be constant from the original flux level to the new flux level. Therefore,
EA]sin 8] = EA2sin ~
~ = sin- t(EAt sin 8])
E"
The annature current can now be found from Kirchhoff's voltage law:
_ VI/! - EA2
1..1.2 - ·X
J ,
I _ 208 LO° V - 227.5 L - 13.9° V
..1. - j2.50
= 56.2~.~OA2° V = 22.5 L 13.2° A
Finally, the motor's power factor is now
PF = cos (13.2°) = 0.974 leading
The resulting phasor diagram is also shown in Figure 6-11.
(e) Because the flux is assumed to vary linearly with field current, EA will also vary
linearly with field current. We know that EA is 182 V for a field current of 4.0A,
so EA for any given field current can be fOlUld from the ratio
~ - ~
182V -4.0A
(6-5)](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-382-320.jpg)
![p,.
-
Infinite bus
Transmission line
-
Q,.
Plant
FIGURE 6-13
SYNCHRONOUS MOTORS 361
,----------------,
P,
-
-
Q,
P,
-
-
'"
P,
-
-Q,
Ind.
motor
Ind.
motor
Synchr.
motor
lOO kW
0.78 PF
lagging
200 kW
0.8 PF
lagging
150 kW
PF = ?
L ________________ ~
A simple power system consisting of an infinite bus supplying an industrial plant through a
transmission line.
Solutioll
(a) In the first case, the real power of load I is 100 kW, and the reactive power of
load I is
Ql = Pt tan ()
= (1 00 kW) tan (cos-l 0.7S) = (100 kW) tan 3S.7°
= SO.2 kVAR
The real power of load 2 is 200 kW, and the reactive power of load 2 is
Q2= P2tan()
= (200 kW) tan (cos-l O.SO) = (200 kW) tan 36.S7°
= 150 kVAR
The real power load 3 is 150 kW. and the reactive power of load 3 is
Q]= p] tan()
= (150 kW) tan (cos-l 0.S5) = (150 kW) tan 3 1.So
= 93 kVAR
Thus, the total real load is
P'o< = Pt + P2 + p]
= lOO kW + 200 kW + 150 kW = 450 kW
and the total reactive load is
Q,o<=Qt+ Q2+Q]
= SO.2 kVAR + 150 kVAR + 93 kVAR = 323.2 kVAR
The equivalent system power factor is thus
PF = cos (J = cos (tan- I .2)= cos (tan- l 323.2 kVAR)
P 450 kW
= cos 35.7° = 0.Sl2 lagging](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-385-320.jpg)
![370 ELECTRIC MACHINERY RJNDAMENTALS
the resulting torque on the bars (and the rotor) is counterclockwise.
Figure 6-1 9b shows the situation at t = 11240 s. Here, the stator magnetic
field has rotated 90° while the rotor has barely moved (it simply cannot speed up
in so short a time). At this point, the voltage induced in the amortisseur windings
is zero, because v is parallel to B. With no induced voltage, there is no current in
the windings, and the induced torque is zero.
Figure 6-1 9c shows the situation at t = 11120 s. Now the stator magnetic
field has rotated 900, and the rotor still has not moved yet. TIle induced voltage
[given by Equation (1-45)] in the amortisseur windings is out of the page in the
bottom bars and into the page in the top bars. The resulting current flow is out of
the page in the bottom bars and into the page in the top bars, causing a magnetic
field Bwto point to the left.1lle resulting induced torque, given by
T;Dd = k Bw x Bs
is counterclockwise.
Finally, Figure 6-1 9d shows the situation at time t = 31240 s. Here, as at
t = 11240 s, the induced torque is zero.
Notice that sometimes the torque is counterclockwise and sometimes it is
essentially zero, but it is always unidirectional. Since there is a net torque in a sin-
gle direction, the motor's rotor speeds up. (1llis is entirely different from starting
a synchronous motor with its normal field current, since in that case torque is first
clockwise and then counterclockwise, averaging out to zero. In this case, torque is
always in the same direction, so there is a nonzero average torque.)
Although the motor's rotor will speed up, it can never quite reach synchro-
nous speed. This is easy to understand. Suppose that a rotor is turning at synchro-
nous speed. Then the speed of the stator magnetic field Bs is the same as the ro-
tor's speed, and there is no relative motion between Bs and the rotor. If there is no
relative motion, the induced voltage in the windings will be zero, the resulting
current fl ow will be zero, and the winding magnetic field will be zero. Therefore,
there will be no torque on the rotor to keep it turning. Even though a rotor cannot
speed up all the way to synchronous speed, it can get close. It gets close enough
to n'YD< that the regular field current can be turned on, and the rotor will pull into
step with the stator magnetic fields.
In a real machine, the field windings are not open-circuited during the start-
ing procedure. If the field windings were open-circuited, then very high voltages
would be produced in them during starting. If the field winding is short-circuited
during starting, no dangerous voltages are produced, and the induced field current
actually contributes extra starting torque to the motor.
To summarize, if a machine has amortisseur windings, it can be started by
the following procedure:
I. Disconnect the field windings from their dc power source and short them out.](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-394-320.jpg)
![INDUCTION MOTORS 387
but unlike a transfonner, the secondary frequency is not necessarily the same as
the primary frequency.
If the rotor of a motor is locked so that it cannot move, then the rotor will
have the same frequency as the stator. On the other hand, if the rotor turns at syn-
chronous speed, the frequency on the rotor will be zero. What will the rotor fre-
quency be for any arbitrary rate of rotor rotation?
At nm= 0 rlmin, the rotor frequencyfr = Jr, and the slip s = I. At nm = n,ync'
the rotor frequencyfr = 0 Hz, and the slip s = O. For any speed in between, the ro-
tor frequency is directly proportional to the difference between the speed of the mag-
netic field n.ync and the speed of the rotor nm. Since the slip of the rotor is defined as
(7-4)
the rotor frequency can be expressed as
(7-8)
Several alternative fonns of this expression exist that are sometimes useful. One
of the more common expressions is deri ved by substituting Equation (7-4) for the
slip into Equation (7--8) and then substituting for n,ync in the denominator of the
expression:
But n,yDC = 120fr I P [from Equation (7- 1)], so
Therefore,
(7- 9)
Example 7-1. A 20S-V, lO-hp, four-pole, 60-Hz, V-connected induction motor has
a full-load slip of 5 percent.
(a) What is the synchronous speed of this motor?
(b) What is the rotor speed of this motor at the rated load?
(c) What is the rotor frequency of this motor at the rated load?
(d) What is the shaft torque of this motor at the rated load?
Solution
(a) The synchronous speed of this motor is
_ 120f,.
n,ync - - p
-
= 120(60 Hz) _
4 poles - ISOOr/ min
(7- 1)](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-411-320.jpg)
![400 ELECTRIC MACHINERY RJNDAMENTALS
or w". = (1 - s)w.ry1tC
= (1 - 0.022XI88.5rad/s) = 184.4rad/s
(b) To find the stator clUTent, get the equivalent impedance of the circuit. The first
step is to combine the referred rotor impedance in parallel with the magnetiza-
tion branch, and then to add the stator impedance to that combination in series.
The referred rotor impedance is
R,
z, = - +jX2
,
= 0.332 + '0 464
0.022 J.
= 15.00 + jO.464 [! = 15.IOLI.76° [!
The combined magnetization plus rotor impedance is given by
1
Z/ = I/jXM
+ 1!L;
= ~=~~1=~~
jO.038 + 0.0662L 1.76°
1
= 0.0773L 31.1 0 = 12.94L31.I O[!
Therefore, the total impedance is
z..,. = z....,+ Z/
= 0.641 + j1.106 + 12.94L31.1° [!
= 11.72 + j7.79 = 14.07L33.6° [!
The resulting stator current is
V
I - ~
,- z..
_ 266LO° V _
- 14.07L33.6° [! -
(c) The power motor power factor is
PF = cos 33.6° = 0.833
(d) The input power to this motor is
Pin = V3"VT h cos ()
18.88L - 33.6° A
lagging
= V3"(460 VX18.88 A)(0.833) = 12,530 W
The stator copper losses in this machine are
PSCL = 3/f R]
= 3(18.88 A)2(0.64 I [!) = 685 W
The air-gap power is given by
PAG = P;n - PSCL = 12,530 W - 685 W = 11,845 W
Therefore, the power converted is
(7- 25)](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-424-320.jpg)
![INDUCTION MOTORS 401
P.:oov = (1 - S)PAG = (I - 0.022X11,845 W) = 11,585 W
The power P"", is given by
Pout = P.:oov - Prot = 11,585W - llOOW = 1O,485W
(
1hp )
= 10,485 W 746 W = 14.1 hp
(e) The induced torque is given by
PAG
Tind = --
w' )'''''
11,845 W
= 18S.5rad/s = 62.8 N
o
m
and the output torque is given by
p-,
Tload =
10,485 W
= 184.4 radls = 56.9 Nom
(In English lUlits, these torques are 46.3 and 41.9 Ib-ft, respectively.)
(jJ The motor's efficiency at this operating condition is
p
~
7f = R x 100%
,.
10,485 W
12530 W x 100% = 83.7%
7.5 INDUCTION MOTOR TORQUE-SPEED
CHARACTERISTICS
How does the torque of an induction motor change as the load changes? How
much torque can an induction motor supply at starting conditions? How much
does the speed of an induction motor drop as its shaflload increases? To find out
the answers to these and similar questions, it is necessary to clearly understand the
relationships among the motor's torque, speed, and power.
In the following material, the torque- speed relationship will be examined
first from the physical viewpoint of the motor's magnetic field behavior. TIlen, a
general equation for torque as a fu nction of slip wilI be derived from the induction
motor equivalent circuit (Figure 7- 12).
Induced Torque from a Physical Standpoint
Figure 7-1 5a shows a cage rotor induction motor that is initially operating at no
load and therefore very nearly at synchronous speed.llle net magnetic field BDeI in
this machine is produced by the magnetization current 1
Mflowing in the motor's
equivalent circuit (see Figure 7-1 2). The magnitude of the magnetization current
and hence of BDeI is directly proportional to the voltage E). If E] is constant, then the](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-425-320.jpg)
![INDUCTION MOTORS 403
tor magnetic fields in the machine. Greater relative motion produces a stronger ro-
tor voltage ER which in turn produces a larger rotor current IR. With a larger rotor
current, the rotor magnetic field DR also increases. However, the angle of the ro-
tor current and DR changes as well. Since the rotor slip is larger, the rotor fre-
quency rises (f, = st ), and the rotor's reactance increases (WLR). Therefore, the
rotor current now lags further behind the rotor voltage, and the rotor magnetic
field shifts with the current. Figure 7-15b shows the induction motor operating at
a fairly high load. Notice that the rotor current has increased and that the angle 8
has increased. The increase in BR tends to increase the torque, while the increase
in angle 8 tends to decrease the torque (TiDd is proportional to sin 8, and 8 > 90°).
Since the first effect is larger than the second one, the overall induced torque in-
creases to supply the motor's increased load.
When does an induction motor reach pullout torque? This happens when the
point is reached where, as the load on the shaft is increased, the sin 8 tenn de-
creases more than the BR tenn increases. At that point, a further increase in load
decreases "TiDd, and the motor stops.
It is possible to use a knowledge of the machine's magnetic fields to approx-
imately derive the output torque-versus-speed characteristic of an induction motor.
Remember that the magnitude of the induced torque in the machine is given by
(4-61)
Each tenn in this expression can be considered separately to derive the overall
machine behavior. The individual tenns are
L BR . TIle rotor magnetic field is directly proportional to the current fl owing in
the rotor, as long as the rotor is unsaturated. TIle current flow in the rotor in-
creases with increasing slip (decreasing speed) according to Equation (7-1 3).
This current fl ow was plotted in Figure 7-11 and is shown again in Fig-
ure 7-16a.
2. B"",. The net magnetic field in the motor is proportional to E] and therefore is
approximately constant (E ] actually decreases with increasing current flow,
but this effect is small compared to the other two, and it will be ignored in
this graphical development). TIle curve for B"", versus speed is shown in Fig-
ure 7-16b.
3. sin 8. TIle angle 8 between the net and rotor magnetic fields can be expressed
in a very useful way. Look at Figure 7-15b. In this figure, it is clear that the
angle 8 is just equal to the power-factor angle ofthe rotor plus 90°:
(7-38)
TIlerefore, sin 8 = sin (OR + 90°) = cos OR. TIlis tenn is the power factor of
the rotor. The rotor power-factor angle can be calculated from the equation
(7-39)](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-427-320.jpg)
![406 ELECTRIC MACHINERY RJNDAMENTALS
+
v
•
-
I,
-
""GURE 7-17
,R, I,
-
+
jXM E,
-
Per-phase equivalent circuit of an induction motor.
~
Refer to the equivalent circuit given in Figure 7-17. In this figure, the air-
gap power supplied to one phase of the motor can be seen to be
, R,
P
AG.t4> = 12s
Therefore, the total air-gap power is
_ 2 R2
P
AG -312 ,
If /l can be determined, then the air-gap power and the induced torque will be
known.
Although there are several ways to solve the circuit in Figure 7-17 for the cur-
rent ll, perhaps the easiest one is to detennine the lllevenin equivalent of the por-
tion of the circuit to the left of the X's in the figure. Thevenin's theorem states that
any linear circuit that can be separated by two tenninals from the rest of the system
can be replaced by a single voltage source in series with an equivalent impedance.
If this were done to the induction motor equivalent circuit, the resulting circuit
would be a simple series combination of elements as shown in Figure 7-1 8c.
To calculate the lllevenin equivalent of the input side of the induction motor
equivalent circuit, first open-circuit the terminals at the X's and fmd the resulting
open-circuit voltage present there. lllen, to find the Thevenin impedance, kill
(short-circuit) the phase voltage and find the Zeq seen "looking" into the tenninals.
Figure 7-1 8a shows the open tenninals used to find the Thevenin voltage.
By the voltage divider rule,
ZM
VTH = V4> Z + Z
M ,
= V jXM
4>Rt + jX] + jXM
llle magnitude of the Thevenin voltage Vru is
(7-4 I a)](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-430-320.jpg)
![INDUCTION MOTORS 407
jX, : ,
v '" jX/oI V
TH R]+jX]+jX/oI •
(-t)V, jX/oI Vrn XM
VTH '" V.
~R]2+(X] +X/oI)2
('J
jX, R,
(bJ
jXrn
(oj
FIGURE 7-18
(a) The Thevenin equivalent voltage of an induction ntotor input circuit. (b) The Thevenin equivalent
impedance of the input circuit. (c) The resulting simplified equivalent circuit of an induction motor.
Since the magnetization reactance XM » X] and XM » RJ, the magnitude of the
Thevenin voltage is approximately
(7-4 (b)
to quite good accuracy.
Figure 7-1 8b shows the input circuit with the input voltage source killed.
The two impedances are in parallel, and the TIlevenin impedance is given by
ZIZM
ZTH = Z l + ZM (7-42)
This impedance reduces to](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-431-320.jpg)
![410 ELECTRIC MACHINERY RJNDAMENTALS
400
]
= 200
e
'0 Braking
If region
;
§
] - 200
~
-400
""GURE 7-10
Tmn --........~ Pullout torque
Motor region
nsyDC..-/ Mechanical speed
Generator region
Induction motor torque-speed characteristic curve. showing the extended operating ranges (braking
region and generator region).
7. If the motor is turning backward relative to the direction of the magnetic fields,
the induced torque in the machine will stop the machine very rapidly and will
try to rotate it in the other direction. Since reversing the direction of magnetic
field rotation is simply a matter of switching any two stator phases, this fact
can be used as a way to very rapidly stop an induction motor. The act of
switching two phases in order to stop the motor very rapidly is called plugging.
TIle power converted to mechanical fonn in an induction motor is equal to
and is shown plotted in Figure 7- 21. Notice that the peak power supplied by the
induction motor occurs at a different speed than the maximum torque; and, of
course, no power is converted to mechanical fonn when the rotor is at zero speed.
Maximum (pullout) Torque in an Induction Motor
Since the induced torque is equal to PAG/w'Y"'" the maximum possible torque oc-
curs when the air-gap power is maximum. Since the air-gap power is equal to the
power consumed in the resistor R2 /s, the maximum induced torque will occur
when the power consumed by that resistor is maximum.](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-434-320.jpg)
![41 2 ELECTRIC MACHINERY RJNDAMENTALS
800 ,--------------------------------,
R,
I
700
600 R.
;""
z
•
~400
"I
~ 300
200
100
R, R,
o~~~~~~~~~~~~~~~~
o 250 500 750 1000 1250 1500 1750 2000
Mechanical speed. r/min
""GURE 7- 22
The effect of varying rotor resistance on the torque-.peed characteristic of a wound-rotor induction
motor.
TIle value of the maximum torque can be found by inserting the expression
for the slip at maximum torque into the torque equation [Equation (7- 50)]. TIle re-
sulting equation for the maximum or pullo ut torque is
(7- 54)
TIlis torque is proportional to the square of the supply voltage and is also in-
versely related to the size of the stator impedances and the rotor reactance. The
smaller a machine's reactances, the larger the maximum torque it is capable of
achieving. Note that slip at which the maximum torque occurs is directly propor-
tional to rotor resistance [Equation (7- 53)], but the value of the maximum torque
is independent of the value of rotor resistance [Equation (7- 54)].
TIle torque-speed characteristic for a wound-rotor induction motor is shown
in Figure 7- 22. Recall that it is possible 1.0 insert resistance into the rotor circuit
of a wound rotor because the rotor circuit is brought out to the stator through slip
rings. Notice on the figure that as the rotor resistance is increased, the pullout
speed of the motor decreases, but the maximum torque remains constant.](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-436-320.jpg)
![414 ELECTRIC MACHINERY RJNDAMENTALS
Example 7-5. A460-V. 25-hp. 60-Hz. four-pole. Y-cOIlllected wOlUld-rotor induc-
tion motor has the following impedances in ohms per phase referred to the stator circuit:
Rl = 0.641 0
Xl = 1.106 0
R2 = 0.3320
X2 = 0.4640 XM = 26.3 0
(a) What is the maximrun torque of this motor? At what speed and slip does it
occur?
(b) What is the starting torque of this motor?
(c) When the rotor resistance is doubled. what is the speed at which the maximum
torque now occurs? What is the new starting torque of the motor?
(d) Calculate and plot the torque-speed characteristics of this motor both with the
original rotor resistance and with the rotor resistance doubled.
Solutioll
The Thevenin voltage of this motor is
(7-4la)
The Thevenin resistance is
(7-44)
J 26.3 n )'
- (0.641 0-'1.1060 + 26.3 0 = 0.590 0
The Thevenin reactance is
XTH - Xl = 1.106 0
(a) The slip at which maximum torque occurs is given by Equation (7- 53):
R,
Smax = ~VijRijfu
=cc+
c=i
(X~rn
"=,,
+=x~
~' (7- 53)
0.3320
= =0.198
Y(0.590 0)1 + (1.106 0 + 0.464 0)2
This corresponds to a mechanical speed of
nm = (I - S)n,yDC = (I - 0.198)(l800r/min) = 1444r/ min
The torque at this speed is
~--,"--c-ce3,Vfl"~cc==C=~"
Trrw; = 2W'YDC[RTH + YRfu + (XTH + X;l2 ]
= 3(255.2 V)2
(7- 54)
2(188.5 rad/s)[O.590 0 + Y(0.590 O)l + (1.106 0 + 0.464 0)2]
= 229N om](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-438-320.jpg)
![INDUCTION MOTORS 415
(b) The starting torque of this motor is found by setting s = I in Equation (7- 50):
3ViHRl
T,tan= W I(R +R)'+(X +X)'l
'YDC TH 2 rn 2
_ 3(255.2 V)1:0.3320)
- (188.5 rad/s)[(0.590 0 + 0.3320)2 + (1.I()5 0 + 0.464 0)2]
= I04N om
(c) If the rotor resistance is doubled, then the slip at maximwn torque doubles, too.
Therefore,
Smax = 0.396
and the speed at maximum torque is
nm = (I - s)n.yDC = (1 - 0.396)(1800 r /min) = 1087 r/min
The maximum torque is still
Trrw; = 229 Nom
The starting torque is now
_ 3(255.2 V)2(0.664 0)
Tot"" - (188.5 rad/s)[(0.590.n + 0.6640)2 + (1.106 0 + 0.4640)2]
= 170 Nom
(d) We will create a MATLAB M-file to calculate and plot the torque-speed char-
acteristic of the motor both with the original rotor resistance and with the dou-
bled rotor resistance. The M-file will calculate the Thevenin impedance using
the exact equations for V
lll and Zrn [Equations (7-4la) and (7-43)] instead of
the approximate equations, because the computer can easily perfonn the exact
calculations. It will then calculate the induced torque using Equation (7- 50) and
plot the results. The resulting M-file is shown below:
% M-file, t orque_speed_curve.m
% M-file c reate a p l ot o f the t orque- speed c urve o f the
% induc tion mot or o f Exampl e 7- 5.
% Fir s t , initia liz e the va lues needed in thi s program.
rl = 0.641; % Stat or res i s tance
xl = 1.10 6; % Stat or r eactance
r2 = 0.332; % Rotor r es i s tance
x2 = 0.464,
xm = 26.3,
v-ph ase = 460 / sqrt (3) ,
n_sync = 1800,
w_sync = 188.5;
% Ca l c ulate the Thevenin
% 7- 41a a nd 7- 43.
v_th 0 v-ph ase • (
= I
% Rotor r eactance
% Magne tizati on branch r eactance
% Phase volt age
% Syn chronou s speed (r / min )
% Sy nchronou s speed (rad /s)
vo lt age and impedance from Equat i ons
sqrt (rl " 2
• (xl
• xm )" 2 ) I ;
,
" 0 ( ( j*xm ) • (d •
- j*xl ) ) I (r1 • j * (xl
• xm ) ) ,
r -" 0 real ( z_ th ) ,
x-" 0 imag ( z_ th ) ,](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-439-320.jpg)
![418 ELECTRIC MACHINERY RJNDAMENTALS
Loo"
like
h.igh.
R,
------------
Desired
curve
Looks like
lowR2
'---------------------------------'---- '.
""GURE 7-14
A torque-speed characteristic curve combining high-resistance effects at low speeds (high slip) with
low-resistance effects at high speed (low slip).
general, the farther away from the stator a rotor bar or part of a bar is, the greater
its leakage reactance, since a smaller percentage of the bar's flux will reach the sta-
tor. Therefore, if the bars of a cage rotor are placed near the surface of the rotor,
they will have only a small leakage flux and the reactance Xl will be small in the
equivalent circuit. On the other hand, ifthe rotor bars are placed deeper into the ro-
tor surface, there will be more leakage and the rotor reactance X2 will be larger.
For example, Figure 7- 25a is a photograph of a rotor lamination showing
the cross section of the bars in the rotor. The rotor bars in the figure are quite large
and are placed near the surface of the rotor. Such a design will have a low resis-
tance (due to its large cross section) and a low leakage reactance and Xl (due to
the bar's location near the stator). Because of the low rotor resistance, the pullout
torque will be quite near synchronous speed [see Equation (7- 53)], and the motor
will be quite efficient. Remember that
P.,oo¥ = (I - s)PAG (7- 33)
so very little of the air-gap power is lost in the rotor resistance. However, since Rl
is small, the motor's starting torque will be small, and its starting current will be
high. TIlis type of design is called the National Electrical Manufacturers Associa-
tion (NEMA) design class A. It is more or less a typical induction motor, and its
characteristics are basically the same as those of a wound-rotor motor with no ex-
tra resistance inserted. Its torque-speed characteristic is shown in Figure 7- 26.
Figure 7- 25d, however, shows the cross section of an induction motor rotor
with smnll bars placed near the surface of the rotor. Since the cross-sectional area
of the bars is small, the rotor resistance is relatively high. Since the bars are lo-
cated near the stator, the rotor leakage reactance is still small. This motor is very
much like a wound-rotor induction motor with extra resistance inserted into the
rotor. Because of the large rotor resistance, this motor has a pullout torque occur-](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-442-320.jpg)
![426 ELECTRIC MACHINERY RJNDAMENTALS
300
250
~> . -.,
200
E
•
Z 150
]
./"
~
/
;? 1
- 100
50 Single-cage design
.- . Double-cage design
o
o 200 400 600 800 1000 1200 1400 1600 1800
n",. rlmin
""GURE 7-29
Comparison of torque-speed characteristics for the single- and double-cage rotors of Example 7-6.
The resulting torque-speed characteristics are shown in Figure 7-29. Note Ihal the double-
cage design has a slightly higher slip in the normal operating range, a smaller maximum
torque and a higher starting torque compared to Ihe corresponding single-cage rotor design.
This behavior matches our theoretical discussions in this section.
7.7 TRENDS IN INDUCTION
MOTOR DESIGN
TIle fundamental ideas behind the induction motor were developed during the late
1880s by Nicola Tesla, who received a patent on his ideas in 1888. At that time,
he presented a paper before the American lnstitute of Electrical Engineers [AlEE,
predecessor of today's Institute of Electrical and Electronics Engineers (IEEE)] in
which he described the basic principles of the wound-rotor induction motor, along
with ideas for two other important ac motors-the synchronous motor and the re-
luctance motor.
Although the basic idea of the induction motor was described in 1888, the
motor itself did not spring forth in full-fledged fonn. There was an initial period
of rapid development, followed by a series of slow, evolutionary improvements
which have continued to this day.
TIle induction motor assumed recognizable modem form between 1888 and
1895. During that period, two- and three-phase power sources were developed to
produce the rotating magnetic fields within the motor, distributed stator windings
were developed, and the cage rotor was introduced. By 1896, fully functional and
recognizable three-phase induction motors were commercially available.
Between then and the early 1970s, there was continual improvement in the
quality of the steels, the casting techniques, the insulation, and the construction](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-450-320.jpg)
![INDUCTION MOTORS 435
Overload
F M] heaters
/ ~I r~~~~
----- - l..L....l..I- 1r--- Resis]or
3TD
Resis]or
lTD 2TD 3TD
Resistor
lTD 2TD 3TD
S<w
Stop
I
OL
lID
lID 2ID
2ID 3ID
FIGURE 7-38
A three-step resistive staner for an induction motor.
Induction
motor
is confined to less than 5 percent slip, and the speed variation over that range is
more or less directly proportional to the load on the shaft of the motor. Even if the
slip could be made larger, the effi ciency of the motor would become very poor,
since the rotor copper losses are directly proportional to the slip on the motor
(remember that PRCL = sPAG).
There are really only two techniques by which the speed of an induction
motor can be controlled. One is to vary the synchronous speed, which is the speed
of the stator and rotor magnetic fields, since the rotor speed always remains near
n,ync. The other technique is to vary the slip of the motor for a given load. E:1.ch of
these approaches will be taken up in more detail.
The synchronous speed of an induction motor is given by](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-459-320.jpg)
![INDUCTION MOTORS 437
simple two-pole induction motor stator suitable for pole changing. Notice that the
individual coils are of very short pitch (60 to 90°). Figure 7-40 shows phase a of
these windings separately for more clarity of detail.
Figure 7-40a shows the current now in phase a of the stator windings at an
instant of time during nonnal operation. Note that the magnetic field leaves the sta-
tor in the upper phase group (a north pole) and enters the stator in the lower phase
group (a south pole). This winding is thus pnxlucing two stator magnetic poles.
a',
I(t) I
,,'
) NtS ) N t S
a] d] a2 d 2 S
B
-
,b,
FIGURE 7-40
Connections
at far end
of stator
,-
",
,
""
,-
",
",
,
""
"
B '
B /,
"
"
B '
N a',
B
S
B
A close-up view of one phase of a pole-changing winding. (a) In the two-pole configuration. one coil
is a north pole and the other one is a south pole. (b) When the connection on one of the two coils is
reversed. they are both nonh poles. and the magnetic flux returns to the stator at points halfway
between the two coils. The south poles are called consequent poles. and the winding is now a four-
pole winding.](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-461-320.jpg)
![INDUCTION MOTORS 441
800,-------------------------------------,
700
600
,
•
Z500
,
~ 400
] 300
• 200P-------
o
o 500
FIGURE 7-42 (roncluded)
1000 1500 2000
Mechanical speed. r/min
I"
(c) The torque-speed characteristic curves for all frequencies.
2500 3000 3500
stator frequency. This process is called derating. If it is not done, the steel in the
core of the induction motor will saturate and excessive magnetization currents
will flow in the machine.
To understand the necessity for derating, recall that an induction motor is
basically a rotating transfonner. As with any transfonner, the flux in the core of an
induction motor can be found from Faraday's law:
vet) = -N~
dl
(1-36)
If a voltage vet) = VM sin wt is applied to the core, the resulting flux ~ is
'Wi ~ J h l)dl
p
= ~ !VM sinwtdt
p
I ~t) = -~coswtl (7- 57)
Note that the electrical frequency appears in the denominator of this expression.
Therefore, if the electrical frequency applied to the stator decreases by 10 percent
while the magnitude of the voltage applied to the stator remains constant, the flux
in the core of the motor wi II increase by about 10 percent and the magnetization
current of the motor will increase. In the unsaturated region of the motor's](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-465-320.jpg)
![444 ELECTRIC MACHINERY RJNDAMENTALS
800
700
R, R, R,
(j()()
E
500
Z
•
,
~ 400
g
"
,
,
~ 300
R] '" 2Ro
200 R2", 3Ro
RJ ",4Ro
RJ '" 5Ro
100 R3", 6Ro
0
0 250 500 750 1000 1250 1500 17'" 2000
Mechanical speed. r/min
fo'IGURE 7-44
Speed control by varying the rotor resistance of a wound-rotor induction motor.
If the torque-speed curve of the load is as shown in the figure, then changing the
rotor resistance will change the operating speed of the motor. However, inserting
extra resistances into the rotor circuit of an induction motor seriously red uces the
efficiency of the machine. Such a method of speed control is nonnally used only
for short periods because of this efficiency problem.
7.10 SOLID·STATE INDUCTION
MOTOR DRIVES
As mentioned in the previous section, the method of choice today for induction
motor speed control is the solid-state variable-frequency induction motor drive. A
typical drive of this sort is shown in Figure 7--45. TIle drive is very flexible: its in-
put power can be either single-phase or three-phase, either 50 or 60 Hz, and any-
where from 208 to 230 V. The output from this drive is a three-phase set of volt-
ages whose frequency can be varied from 0 up to 120 Hz and whose voltage can
be varied from 0 V up to the rated voltage of the motor.
TIle output voltage and frequency control is achieved by using the pulse-
width modulation (PWM) techniques described in Chapter 3. Both output frequency
and output voltage can be controlled independently by pulse-width modulation. fig-
ure 7--46 illustrates the manner in which the PWM drive can control the output fre-
quency while maintaining a constant nns voltage level, while Figure 7--47 illustrates](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-468-320.jpg)
![Initial
Variable
voltage,
variable
frequency,
three-phase
power
00.=
I,
-
+
equivalent
V. (
cin:uit:
I,
Since -
R2('~S)>>R2
+
v
.(
,,'
R2 (' - s)>>X2,
,
this cin:uit
reduces to:
+
Combining
V. (
RF&w and
Reyields:
FIGURE 7-52
R, jXt
R,
,
R,
INDUCTION MOTORS 453
I,
-
P, A
I,
-
A No load
I,
-
P, A lit + IB+ Ie
IL = 3
,.,
12= 0
- jX2 R,
"
I. j
R, jXM ~ R2(';S)
~ Rfri<:tiOll, win<b!O,
~ &""'" »XM
'h
'
The no-load test of an induction motor: (a) test circuit; (b) the resulting motor equivalent cin:uit.
Note that at no load the motor's impedance is essentially the series combination of RI,jXj • andJXM .
the output resistor is in parallel with the magnetization reactance XM and the core
losses Re.
In this motor at no-load conditions, the input power measured by the meters
must equal the losses in the motor. The rotor copper losses are negligible because
the current / l is extremely small [because of the large load resistance R2(1 - s)/s],
so they may be neglected. The stator copper losses are given by](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-477-320.jpg)
![456 ELECTRIC MACHINERY RJNDAMENTALS
I,
-
Adjustable- V
voltage. I,
adjustable- b -
A
frequency.
three-phase
power source I,
, -
A
,,'
I,
R, I,
-
-
v,
'h'
I<'IGURE 7-54
p
p
/,=/,=It..,
XM »IR2+jX21
Rc »IR2 +jX21
So neglect Rc and XM
The locked-rotor test for an induction motor: (a) test circuit: (b) motor equivalem circuit.
small value). Since R2and X2are so small, almost all the input current will flow
through them, instead of through the much larger magnetizing reactance XM .
Therefore, the circuit under these conditions looks like a series combination ofX ],
R], X2 , and Rl .
nlere is one problem with this test, however. In nonnal operation, the stator
frequency is the line frequency of tile power system (50 or 60 Hz). At starting con-
ditions, the rotor is also at line frequency. However, at nonnal operating conditions,
the slip of most motors is only 2 to 4 percent, and the resulting rotor frequency is
in the range of I to 3 Hz. nlis creates a problem in that the line frequency does not
represent the nonnal operating conditions ofthe rotor. Since effective rotor resis-
tance is a strong function of frequency for design class Band C motors, the incor-
rect rotor frequency can lead to misleading results in this test. A typical compro-
mise is to use a frequency 25 percent or less of the rated frequency. While this
approach is acceptable for essentially constant resistance rotors (design classes A
and D), it leaves a lot to be desired when one is trying to find the nonnal rotor re-
sistance of a variable-resistance rotor. Because of these and similar problems, a
great deal of care must be exercised in taking measurements for these tests.](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-480-320.jpg)
![460 ELECTRIC MACHINERY RJNDAMENTALS
R,
""GURE 7-56
jO.67 fl.
,
,
,
,
Rc -s:
(unknown) <-
,
,
,
,
jXM=j14.03fl.
Motor per-phase equivalent circuit for Example 7-8.
= 0.1510 =0.111 = 11.1%
V(0.243 0)2 + (0.67 0 + 0.67 0)2
The maximum torque of this motor is given by
Tmox = 2 W'YDC[Rll{ + VRfu + (Xll{ + X2)]
_ 3(114.6 V)l
(7- 54)
- 2(188.5 rad/s)[O.22 1 0 + V(0.221 Of + (0.670 + 0.67 Of]
= 66.2N o m
7.12 THE INDUCTION GENERATOR
TIle torque-speed characteristic curve in Figure 7- 20 shows that if an induction
motor is driven at a speed greater than n.y"" by an external prime mover, the di-
rection of its inducted torque will reverse and it will act as a generator. As the
torque applied to its shaft by the prime mover increases, the amount of power pro-
duced by the induction generator increases.As Figure 7- 57 shows, there is a max-
imum possible induced torque in the generator mode of operation. This torque is
known as the pushover torque of the generator. If a prime mover applies a torque
greater than the pushover torque to the shaft of an induction generator, the gener-
ator wil I overspeed.
As a generator, an induction machine has severe limitations. Because it
lacks a separate field circuit, an inducti on generator cannot produce reactive
power. In fact, it consumes reactive power, and an external source of reactive
power must be connected to it at all times to maintain its stator magnetic field.
1llis external source of reactive power must also control the terminal voltage of
the generator-with no field current, an induction generator cannot control its
own output voltage. Normally, the generator's voltage is maintained by the exter-
nal power system to which it is connected .
1lle one great advantage of an induction generator is its simplicity. An in-
duction generator does not need a separate field circuit and does not have to be
driven continuously at a fi xed speed. As long as the machine's speed is some](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-484-320.jpg)
![468 ELECTRIC MACHINERY RJNDAMENTALS
(c) Rotor induced voltage
(d) Rotor current
(e) Rotor frequency
(j) PRCL
(g) Synchronous speed
PROBLEMS
7-1. A dc test is performed on a 460-V. ~-connected. lOO-hp induction motor. If Voc =
24 V and foc = 80A. what is the stator resistance R]? Why is this so?
7-2. A 220-V, three-phase. two-pole. 50-Hz induction motor is ruooing at a slip of 5 per-
cent. Find:
(a) The speed of the magnetic fields in revolutions per minute
(b) The speed of the rotor in revolutions per minute
(c) The slip speed of the rotor
(d) The rotor frequency in hertz
7-3. Answer the questions in Problem 7- 2 for a 480-V. three-phase. four-pole. 60-Hz in-
duction motor running at a slip of 0.035.
7-4. A three-phase. 60-Hz induction motor runs at 890 rhnin at no load and at 840 rlmin
at full load.
(a) How many poles does this motor have?
(b) What is the slip at rated load?
(c) What is the speed at one-quarter of the rated load?
(d) What is the rotor's electrical frequency at one-quarter of the rated load?
7-5, ASO-kW, 440-V, 50-Hz, six-pole induction motor has a slip of 6 percent when op-
erating at full-load conditions. At full-load conditions, the friction and windage
losses are 300 W, and the core losses are 600 W. Find the following values for full-
load conditions:
(a) The shaft speed n..
(b) The output power in watts
(c) The load torque 1lood in newton-meters
(d) The induced torque 11... in newton-meters
(e) The rotor frequency in hertz
7-6. A three-phase, 60-Hz, four-pole induction motor runs at a no-load speed of 1790
rlmin and a full-load speed of 1720 r/min. Calculate the slip and the electrical fre-
quency of the rotor at no-load and full-load conditions. What is the speed regulation
of this motor [Equation (4-68)]?
7-7, A 208-V, two-pole, 60-Hz, V-connected woood-rotor induction motor is rated at
IS hp. Its equivalent circuit components are
Rl = 0.200 n
Xl = O.4lOn
Pmoc.b = 250 W
For a slip of 0.05, find
(a) The line ClUTent h
R2 = 0.120 n
X2 = 0.410 n
P.u.c""O
(b) The stator copper losses PSCL
(c) The air-gap power PAG
XM = IS.On
P<Ote = 180W](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-492-320.jpg)
![INDUCTION MOTORS 469
(d) The power converted from electrical to mechanical fonn P<oov
(e) The induced torque Tm
(f) The load torque Tload
(g) The overall machine efficiency
(h) The motor speed in revolutions per minute and radians per second
7-8. For the motor in Problem 7- 7. what is the slip at the pullout torque? What is the
pullout torque of this motor?
7-9. (a) Calculate and plot the torque-speed characteristic ofthe motor in Problem 7- 7.
(b) Calculate and plot the output power versus speed curve of the motor in Prob-
lem 7- 7.
7-10. For the motor of Problem 7- 7. how much additional resistance (referred to the sta-
tor circuit) would it be necessary to add to the rotor circuit to make the maximum
torque occur at starting conditions (when the shaft is not moving)? Plot the
torque-speed characteristic of this motor with the additional resistance inserted.
7-11. If the motor in Problem 7- 7 is to be operated on a 50-Hz power system. what must
be done to its supply voltage? Why? What will the equivalent circuit component
values be at 50 Hz? Answer the questions in Problem 7- 7 for operation at 50 Hz
with a slip of 0.05 and the proper voltage for this machine.
7-12. Figure 7-1 8a shows a simple circuit consisting of a voltage source. a resistor. and
two reactances. Find the Thevenin equi valent voltage and impedance of this circuit
at the terminals. Then derive the expressions for the magnitude of Vrn and for Rrn
given in Equations (7-4lb) and (7-44).
7-13. Figure P7-1 shows a simple circuit consisting ofa voltage source. two resistors. and
two reactances in series with each other. If the resistor RL is allowed to vary but all
the other components are constant. at what value of RL will the maximum possible
power be supplied to it? Prove your answer. (Hint: Derive an expression for load
power in terms of V. Rs. Xs. RL• and XL and take the partial derivative of that ex-
pression with respect to Rd Use this result to derive the expression for the pullout
torque [Equation (7- 54)].
jXs
v(Z)
FlGURE 1'7-1
Circuit for Problem 7- 13.
7-14. A 440-V. 50-Hz, two-pole, V-connected induction motor is rated at 75 kW. The
equivalent circuit parameters are
R[ = 0.075 0
X[ = 0.170
PFAW = 1.0 kW
For a slip of 0.04, find
R2 = 0.065 n
X2 = O.170
Pmioc = 150W
XM = 7.20
Paxe = 1.1 kW](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-493-320.jpg)
![IX: MACHINERY FUNDAMENTALS 483
(d) Suppose the machine is again unloaded, and a torque of 7.5 N • m is applied to
the shaft in the direction of rotation. What is the new steady-state speed? Is this
machine now a motor or a generator?
(e) Suppose the machine is rulUling unloaded. What would the final steady-state
speed of the rotor be if the flux density were reduced to 0.20 T?
Solution
(a) When the switch in Figure 8-6 is closed, a current will flow in the loop. Since
the loop is initially stationary, e jod = O. Therefore, the current will be given by
VB - eiod VB
i = =
R R
This current flows through the rotor loop, producing a torque
2.;
7iod = - 'I"
~
ccw
This induced torque produces an angular acceleration in a counterclockwise di-
rection, so the rotor of the machine begins to turn. But as the rotor begins to
tum, an induced voltage is produced in the motor, given by
so the current i falls. As the current falls, 7ind = (2hr)cpi.t.. decreases, and the ma-
chine winds up in steady state with 7iod = 0, and the battery voltage VB = eind'
This is the same sort of starting behavior seen earlier in the linear dc machine.
(b) At starting conditions, the machine's current is
. VB l20V
I = J[ = 0.3fi = 400 A
At no-load steady-state conditions, the induced torque 7ind must be zero. But
7ind = 0 implies that current i must equal zero, since 7ind = (2hr)cpi, and the flux
is nonzero. The fact that i = 0 A means that the battery voltage VB = eind' There-
fore, the speed of the rotor is
_ VB _ ~
W - (2i1r)cp - 2rlB
l20V
= 2(0.5 mXI.O mXO.25 T) = 480 rad/s
(c) If a load torque of 10 N om is applied to the shaft of the machine, it will begin
to slow down. But as w decreases, eind = (2hr)cpwJ. decreases and the rotor cur-
rent increases [i = (VB - eind.t.. )/R]. As the rotor current increases, ITIOdI in-
creases too, until I 7;",,1 = 171Nd1 at a lower speed w.
At steady state, 171....1 = 17indl = (2hr)cpi. Therefore,
. 7jod 7ind
1= = --
(2hr)cp 2rlB
ION om
= (2XO.5 m)(1.0 mXO.25 T) = 40 A](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-507-320.jpg)
![486 ELECTRIC MACHINERY RJNDAMENTALS
Back side of
coil I
Back side of
coil 4
3 I' 42'
S
Pol,
faces
Comrnutator _
segments -~-
Brushes
""GURE 8-7 (roncluded)
3' 24'
N
I ' 2
,
b
43'
(bl
Back side ofcoil 2
•,
E=4~
Back side of coil 3
3 ] ' 42'
S
3' 24'
N
("
(b) The voltages on the rotor conductors at this time, (c) A winding diagram of this machine showing
the interconnections of the rotor loops.
At the instant shown in Figure 8- 7, the ,2,3', and 4' ends of the loops are
under the north pole face, while the ',2',3, and 4 ends of the loops are underthe
south pole face. TIle voltage in each of the 1,2,3', and 4' ends of the loops is
given by
eind - (v x B) • I
positive out of page
( 1-45)
(8-1 7)
TIle voltage in each of the 1" 2', 3, and 4 ends of the ends of the loops is given by
- vBI positive into the page
( 1-45)
(8-1 8)](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-510-320.jpg)
![IX: MACHINERY FUNDAMENTALS 513
Stator
~~~i"" - L---~s------'>.~ L---~N------'>.~
j ! !
Rotor ~~~ClIT0J:]0~·~0c·IT0J:]6
~.~"
[]"'iJ,,~®~~,,~,,
~~
- Motionof
::f,A· turns Pole magnetomotive force
Compensating
,,(!"inding /~
/
r-~ Rotor
genel1ltor
- Motion of
mmm
magnetomotiveforce '-_____-':1..., =::fpole +::fR +::f""
::f. A • turns
FIGURE 8-3 1
I
Neutral
plane
00'
shifted
:1..., =::fpole
The flux and magnetomotive forces in a de machine with compensating windings.
compensating windings is equal and opposite to the magnetomotive force due to
the rotor at every point under the pole faces. The resulting net magnetomotive
force is just the magnetomotive force due to the poles, so the flux in the machine
is unchanged regardless of the load on the machine. The stator of a large dc ma-
chine with compensating windings is shown in Figure 8- 32.
The major disadvantage of compensating windings is that they are expen-
sive, since they must be machined into the faces of the poles. Any motor that uses
them must also have interpoles, since compensating windings do not cancel
L dildt effects. The interpoles do not have to be as strong, though, since they are
canceling only L dildt voltages in the windings, and not the voltages due to
neutral-plane shifting. Because of the expense of having both compensating wind-
ings and interpoles on such a machine, these windings are used only where the ex-
tremely severe nature of a motor's duty demands them.](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-537-320.jpg)
![IX: MACHINERY FUNDAMENTALS 529
B = 1.0T in air gap
1= 0.3 m (length of coil sides)
r = 0.08 m (radius of coils)
n = 1700 rlmin ccw
The resistance of each rotor coil is 0.04 n.
(a) Is the armature winding shown a progressive or retrogressive winding?
(b) How many current paths are there through the armature of this machine?
(c) What are the magnitude and the polarity of the voltage at the brushes in this
machine?
(d) What is the annature resistance RA of this machine?
(e) If a 1O~ resistor is connected to the tenninals of this machine, how much cur-
rent flows in the machine? Consider the internal resistance of the machine in de-
tennining the current flow.
(f) What are the magnitude and the direction of the resulting induced torque?
(g) Assuming that the speed of rotation and magnetic flux density are constant, plot
the terminal voltage of this machine as a flUlction of the current drawn from it.
8-3. Prove that the equation for the induced voltage of a single simple rotating loop
2
, . , = - "'w
In 7T '+' (8-6)
is just a special case of the general equation for induced voltage in a dc machine
(8--38)
8-4. A dc machine has eight poles and a rated current of 100 A. How much current will
flow in each path at rated conditions if the armature is (a) simplex lap-wound,
(b) duplex lap-wound, (c) simplex wave-wound?
8-5. How many parallel current paths will there be in the armature of a 12-pole machine
if the armature is (a) simplex lap-wolUld, (b) duplex wave-wound, (c) triplex lap-
wound, (d) quadruplex wave-wolUld?
8-6. The power converted from one fonn to another within a dc motor was given by
Use the equations for EA and "Tiod [Equations (8--38) and (8-49)] to prove that
EAtA = "Tiodw..; that is. prove that the electric power disappearing at the point of
power conversion is exactly equal to the mechanical power appearing at that point.
8-7. An eight-pole, 25-kW, 120-V dc generator has a duplex lap-wound armature which
has 64 coils with 16 turns per coil. Its rated speed is 2400 r/min.
(a) How much flux per pole is required to produce the rated voltage in this genera-
tor at no-load conditions?
(b) What is the current per path in the annature of this generator at the rated load?
(c) What is the induced torque in this machine at the rated load?
(d) How many brushes must this motor have? How wide must each one be?
(e) If the resistance of this winding is 0.011 0: per turn, what is the armature resis-
tance RA of this machine?
8-8. Figure PS- 2 shows a small two-pole dc motor with eight rotor coils and four turns
per coil. The flux per pole in this machine is 0.0125 Wh.](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-553-320.jpg)
![rx: MmDRS AND GENERATORS 537
~.Wb
'--- - - - - - - - - - - - - - 3'. A· turns
FIGURE 9-3
The magnetization curve of a ferromagnetic material (4) versus 3').
"'="'0
n ="0 (constant)
' - - - - - - - - - - - - - - - - IF [=~; ]
FIGURE 9-4
The magnetization curve of a dc machine expressed as a plot of E,t versus IF. for a fixed speed ."..
The field current in a dc machine produces a field magnetomotive force
given by '?} = NFIF. nlis magnetomotive force produces a flux in the machine in
accordance with its magnetization curve (Figure 9- 3). Since the field current is di-
rectly proportional to the magnetomoti ve force and since EA is directly propor-
tional to the flux, it is customary to present the magnetization curve as a plot of EA
versus field current for a given speed Wo (Figure 9--4).
It is worth noting here that, to get the maximum possible power per pound
of weight out of a machine, most motors and generators are designed to operate
near the saturation point on the magnetization curve (at the knee of the curve).
This implies that a fairly large increase in field current is often necessary to get a
small increase in EAwhen operation is near full load.
The dc machine magnetization curves used in this book are also available in
electronic form to simplify the solution of problems by MATLAB. Each magneti-
zation curve is stored in a separate MAT file. Each MAT file contains three](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-561-320.jpg)
![rx: MmDRS AND GENERATORS 539
constant-voltage power supply, while a shunt dc motor is a motor whose field
circuit gets its power directly across the armature terminals of the motor. When
the supply voltage to a motor is assumed constant, there is no practical difference
in behavior between these two machines. Unless otherwise specified, whenever
the behavior of a shunt motor is described, the separately excited motor is
included, too.
The Kirchhoff's voltage law (KVL) equation for the armature circuit of
these motors is
(9- 3)
The Terminal Characteristic of a Shunt DC Motor
A tenninal characteristic of a machine is a plot of the machine's output quantities
versus each other. For a motor, the output quantities are shaft torque and speed, so
the terminal characteristic of a motor is a plot of its output torque versus speed.
How does a shunt dc motor respond to a load? Suppose that the load on the
shaft of a shunt motor is increased. Then the load torque "Tlood will exceed the in-
duced torque "TiOO in the machine, and the motor wil I start to slow down. When the
motor slows down, its internal generated voltage drops (EA = K4>wJ.), so the ar-
mature current in the motor IA = (VT - EAJ.)/RA increases. As the annature current
rises, the induced torque in the motor increases (1]00 = K4> IAi ), and finally the in-
duced torque will equal the load torque at a lower mechanical speed of rotation w.
TIle output characteristic of a shunt dc motor can be derived from the in-
duced voltage and torque equations of the motor plus Kirchhoff's voltage law.
(KVL) TIle KVL equation for a shunt motor is
VT = EA + lARA
The induced voltage EA= K4>w, so
VT = K¢w + lARA
Since "Tind = K4>IA, current IA can be expressed as
"Tind
IA = K4>
Combining Equations (9-4) and (9-5) prOOuces
Finally, solving for the motor's speed yields
(9- 3)
(9-4)
(9- 5)
(9-6)
(9- 7)
This equation is just a straight line with a negative slope. TIle resulting
torque- speed characteristic of a shunt dc motor is shown in Figure 9--6a.](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-563-320.jpg)
![rx: MmDRS AND GENERATORS 547
c- f 0 50 ;
•Fi e l d r es i s t a nce (ohms)
c-"
0 0 . 06 ;
•Arma tur e r es i s t a nce (ohms)
i 1 0 1 0, 1 0,300 ;
•Li ne curre nt s (A )
0
- f 0 1200 ;
•Number of turns 0 0 fi e l d
f -" c O 0 84 0 ;
•Arma tur e r eact i on , 200 A
% Ca l c u l a t e the a rma ture c urre nt f or each l oad .
i _a = i _ l - v_t I r _ f ;
% Now ca l c u l a t e the i nt e rna l gene r a t ed vo l t age f or
% each a rma ture c urre nt.
e_a = v_t - i _a * r _a;
(A- t i m)
% Ca l c u l a t e the a rma ture r eact i on MMF f or each a rma ture
% c urre nt.
f _ar = ( i _a I 200) * f _ar O:
% Ca l c u l a t e the e ff ect i ve fi e l d c urre nt.
i f = v_t I r _ f - f _ar I n_ f :
% Ca l c u l a t e the r esu l t i ng i nt e rna l gene r a t ed vo l t age a t
% 1200 r / mi n by i nt e r po l a t i ng the mot or 's magne t i za t i on
% c urve .
e_aO = i nt e r p l ( if_va l ues, ea_va l ue s, i _ f , ' sp line ' ) ;
% Ca l c u l a t e the r esult i ng speed f r om Equa t i on (9 - 13 ) .
n = ( e_a . 1 e_aO ) .. n_O:
% Ca l c u l a t e the i nduced t or que co rrespond i ng t o each
% speed f r om Equa t i on s ( 8- 55 ) a nd (8- 56 ) .
t _ i nd = e_a . * i _a .1 (n .. 2 .. p i I 60):
% Pl ot the t or que - speed c urve
p l ot (t _ i nd, n , 'Col or' , 'k' , 'LineWi dth' ,2 . 0 ) ;
h o l d on ;
x l abe l ( ' t au_( i nd} (N- m) ' , ' Fontwei ght' , 'Bo l d ' ) ;
y l abe l ( ' i tn_( m} rm b f (r / mi n ) ' , 'Fontwei ght' , 'Bo l d ' ) :
( ' b f Shunt OC mot or t or que - speed c h a r act e r i s t i c ' )
ax i s( [ 0 600 1100 1300 ] ) ;
g r i d on ;
h o l d o ff ;
The resulting torque-speed characteristic is shown in Figure 9-10. Note that for any given
load. the speed of the motor with armature reaction is higher than the speed of the motor
without armature reaction.
Speed Control of Shunt DC Motors
How can the speed of a shunt dc motor be controlled? There are two common
methods and one less common method in use.1lle common methods have already
been seen in the simple linear machine in Chapter 1 and the simple rotating loop
in Chapter 8. The two common ways in which the speed of a shunt dc machine
can be controlled are by](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-571-320.jpg)
![548 ELECTRIC MACHINERY RJNDAMENTALS
1300 ,------,--,----,------,--,------,
1280
1260
1240
1220 t--~---'---:
~ 1200
./"80
]]60
]]40
]]20
]]00 '-----c'cC_-~-__C'c---c'cC_-~-~
o 100 200 300 400 500 600
'tiD<! N·m
""GURE 9- 10
The torque--speed characteristic of the motor with armature reaction in Example 9--2.
I. Adjusting the field resistance RF (and thus the field nux)
2. Adjusting the tenninal voltage applied to the annature.
The less common method of speed control is by
3. Inserting a resistor in series with the armature circuit.
E:1.ch of these methods is described in detail below.
CHANGING THE FIELD RESISTANCE. To understand what happens when the
field resistor of a dc motor is changed. assume that the field resistor increases and
observe the response. If the field resistance increases, then the field current de-
creases (IF = Vr lRF i ), and as the field current decreases, the nux <P decreases
with it. A decrease in nux causes an instantaneous decrease in the internal gener-
ated voltage EA(= K<p-tw), which causes a large increase in the machine's anna-
ture current, since
1lle induced torque in a motor is given by "TiDd = K<pfA' Since the nux <p in
this machine decreases while the current fA increases, which way does the induced
torque change? 1lle easiest way to answer this question is to look at an example.
Figure 9- 1I shows a shunt dc motor with an internal resistance of 0.25 O. lt
is currently operating with a tenninal voltage of 250 V and an internal generated
voltage of 245 V. 1llerefore, the annature current now is fA = (250 V -](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-572-320.jpg)
![552 ELECTRIC MACHINERY RJNDAMENTALS
v"
' - - - - - - - - - - - - - - - - f ind
""GURE 9-14
The effect of armature voltage speed control on a shunt motor's torque-speed characteristic.
Bul as the speed w increases, the internal generaled voltage EA(= K4>wi)
increases, causing the armature current to decrease. This decrease in JA decreases
the induced torque, causing Tind to equal Ttoad at a higher rotational speed w.
To summarize the cause-and-effect behavior in this method of speed
control:
I . An increase in VAincreases JA [= (VA i - EA)/RA].
2. Increasing JA increases Tind (= K4>JAi ).
3. Increasing Tind makes TiDd >TJoad increasing w.
4. Increasing w increases EA(= K4>wi).
5. Increasing EA decreases JA [= (VA i - EA)/RAl
6. Decreasing JA decreases Tind until Tind = TJoad at a higher w.
TIle effect of an increase in VA on the torque-speed characteristic of a sepa-
rately excited motor is shown in Figure 9- 14. Notice that the no-load speed of the
motor is shifted by this method of speed control, but the slope of the curve re-
mains constant.
INSERTING A RESISTOR IN SERIES WITH THE ARll ATURE c m.CUIT. If a
resistor is inserted in series with the annature circuit, the effect is to drastically in-
crease the slope of the motor's torque-speed characteristic, making it operate
more slowly if loaded (Figure 9- 15). lllis fact can easily be seen from Equation
(9- 7). The insertion of a resistor is a very wasteful method of speed control, since
the losses in the inserted resistor are very large. For this reason, it is rarely used.
It will be found only in applications in which the motor spends almost all its time
operating at fuJI speed or in applications too inexpensive to justify a better form
of speed control.
TIle two most common methods of shunt motor speed control- field resis-
tance variation and armature voltage variation- have different safe ranges of
operation.](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-576-320.jpg)
![rx: M mDRS AND GENERATORS 557
% Ca l c ula t e the int e rna l gen e r a t ed vo lt age a t 1200 r / min
% f or the r e f e r e nce fi e l d c urre nt (5 A) by int e r po l a ting
% the mot or 's magne tiz a tion c urve. The r e f e r e nce speed
% correspond ing t o thi s fi e l d c urre nt i s 1103 r / min .
e_aO_r e f = int e r p l ( if_va lues, ea_va lues, 5, ' sp line ' ) ;
n_ r e f = 1103;
% Ca l c ula t e the fi e l d c urre nt f or each va lue o f fi e l d
% r es i s t a nce.
i _ f = v_t . / r _ f ;
% Ca l c ula t e the E_aO f or eac h fi e l d c urre nt by
% int e r po l a ting the mot or 's magne ti zation c urve.
e_aO = int e r p l ( if_va lues, ea_va lues, i _ f , ' sp line ' ) ;
% Ca l c ula t e the r esulting speed from Equa tion (9 -1 7):
% n2 = (phil / phi2 ) .. nl = (e_aO_l / e_aO_2 ) .. nl
n2 = ( e_aO_r e f . / e_aO ) .. n_ r e f ;
% Plot the speed ver s u s r _ f c urve.
p l ot (r _ f , n2, 'Col or' , 'k' , 'LineWi dth' ,2.0 ) ;
h o l d on ;
x l abe l ( 'Fi e l d r es i s t a nce, Omega ' , 'Fontwei ght' , 'Bo l d ' ) ;
y l abe l ( ' itn_( m} rm b f (r / min ) ' , 'Fontwei ght' , 'Bo l d ' ) ;
title ( ' Speed vs itR_( F ) rm b f f or a Shunt IX: M
ot or' ,
'Fontwei ght' , 'Bo l d ' ) ;
ax i s( [40 70 0 1400 ] ) ;
g rid on ;
h o l d o ff ;
The resulting plot is shown in Figure 9-1 8.
1400
1200 , ,
IlXXl
.§ 800
~
•600
0
400
200
0
40 45 60 65 70
Field resistance. n
FIGURE 9- 18
Plot of speed versus field resistance for the shunt dc motor of Example 9-3.](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-581-320.jpg)
![rx: MmDRS AND GENERATORS 559
The results of an open field circuit can be quite spectacular. When the au-
thor was an undergraduate, his laboratory group once made a mistake of this sort.
The group was working with a small motor-generator set being driven by a 3-hp
shunt dc motor. The motor was connected and ready to go, but there was just one
little mistake- when the field circuit was connected, it was fused with a O.3-A
fuse instead of the 3-A fu se that was supposed to be used.
When the motor was started, it ran nonnally for about 3 s, and then sud-
denly there was a flash from the fuse. ]mmediately, the motor's speed skyrock-
eted. Someone turned the main circuit breaker off within a few seconds, but by
that time the tachometer attached to the motor had pegged at 4000 r/min. TIle mo-
tor itself was only rated for 800 rimin.
Needless to say, that experience scared everyone present very badly and
taught them to be most careful about fie ld circuit protection. In dc motor starting
and protection circuits, afield loss relay is nonnally included to disconnect the
motor from the line in the event of a loss of field current.
A similar effect can occur in ordinary shunt dc motors operating with light
fields if their annature reaction effects are severe enough. If the annature reaction
on a dc motor is severe, an increase in load can weaken its flux enough to actually
cause the motor's speed to rise. However, most loads have torque-speed curves
whose torque increases with speed, so the increased speed of the motor increases
its load, which increases its annature reaction, weakening its flux again. TIle
weaker flux causes a further increase in speed, further increasing load, etc., until
the motor overspeeds. nlis condition is known as runaway.
In motors operating with very severe load changes and duty cycles, this flux-
weakening problem can be solved by installing compensating windings.
Unfortunately, compensating windings are too expensive for use on ordinary run-of-
the-mill motors. TIle solution to the runaway problem employed for less-expensive,
less-severe duty motors is to provide a turn or two of cumulative compounding to
the motor's poles. As the load increases, the magnetomotive force from the series
turns increases, which counteracts the demagnetizing magnetomotive force of the
annature reaction. A shunt motor equipped with just a few series turns like this is
calJed a stabilized shunt motor.
9.5 THE PERMANENT-MAGNET DC MOTOR
A permanent-magnet de (PMDC) motor is a dc motor whose poles are made of
pennanent magnets. Permanent-magnet dc motors offer a number of benefits
compared with shunt dc motors in some applications. Since these motors do not
require an external field circuit, they do not have the field circuit copper losses as-
sociated with shunt dc motors. Because no field windings are required, they can
be smaller than corresponding shunt dc motors. PMDC motors are especially
common in smaller fractional- and subfractional-horsepower sizes, where the ex-
pense and space of a separate field circuit cannot be justified.
However, PMDC motors also have disadvantages. Pennanent magnets can-
not produce as high a flux density as an externally supplied shunt field, so a](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-583-320.jpg)
![560 ELECTRIC MACHINERY RJNDAMENTALS
Residual flux
density n...
----
--------------.f~t-----------------H ~r~)
""GURE 9-19
Coercive
Magnetizing
intensity He
-----
(.j
(a) The magnetization curve of a typical ferromagnetic material. Note the hysteresis loop. After a
large ntagnetizing intensity H is applied to the core and then removed. a residual flux density n...
rentains behind in the core. This flux can be brought to zero if a coercive magnetizing intensity He is
applied to the core with the opposite polarity. In this case. a relatively sntall value of it will
demagnetize the core.
PMDC motor will have a lower induced torque "rind per ampere of annature cur-
rent lit than a shunt motor of the same size and construction. In addition, PMDC
motors run the risk of demagnetization. As mentioned in Chapler 8, the annature
current lit in a dc machine produces an annature magnetic field of its own. The ar-
mature mlllf subtracts from the mmf of the poles under some portions of the pole
faces and adds to the rnrnfofthe poles under other portions of the pole faces (see
Figures 8- 23 and 8- 25), reducing the overall net nux in the machine. This is the
armature reaction effect. In a PMDC machine, the pole nux is just the residual
flux in the pennanent magnets. If the armature current becomes very large, there
is some risk that the armature mmf may demagnetize the poles, pennanenlly re-
ducing and reorienting the residual flux in them. Demagnetizalion may also be
caused by the excessive heating which can occur during prolonged periods of
overload.
Figure 9- 19a shows a magnetization curve for a typical ferromagnetic ma-
terial. II is a plot of flux density B versus magnelizing intensity H (or equivalently,
a plot of flux <p versus mlllf ?]i). When a strong external magnetomotive force is
applied 10 this material and then removed, a residual flux B.... will remain in the
material. To force the residual flux to zero, it is necessary to apply a coercive mag-
netizing inlensity H e with a polarity opposite 10 the polarity of the magnetizing in-
lensity H that originally established the magnetic field. For normal machine](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-584-320.jpg)
![rx: MmDRS AND GENERATORS 567
% Firs t , i n itia liz e
v_t = 250;
the values needed in thi s program.
% Te rminal vo lt age (V)
r _0
i 0
-
n_ o
0 0.08;
0 1 0, 1 0,300;
0 25;
% Armature + fi e l d r es i s tance (ohms)
% Armature (line) c urre nt s (A)
% Numbe r of series turn s on fi e l d
% Ca l culat e the MMF f o r each l oad
f=n_s *i_a;
% Ca l culat e the int e rnal generated volt age e_a .
e_a = v_t - i _a * r _a;
% Ca l culat e the r esulting interna l generated volt age at
% 1 200 r / min by int e rpolating the mot or 's mag ne tization
% curve.
e_aO = int e rpl (mmf_va lues,ea_values, f ,'spline');
% Ca l culat e the motor's speed fr om Equat i on (9 -1 3) .
n = (e_a . 1 e_aO) * n_O;
% Ca l culat e the induced t orque corresponding t o each
% speed from Equations (8 - 55 ) and (8 - 56 ) .
t _ ind = e_a .* i _a . 1 (n * 2 * p i I 60);
% Plo t the t o rque - speed curve
p l o t (t _ ind, n , ' Co l or' , 'k' , 'LineWi dth', 2 . 0) ;
ho l d on;
x l abel ( ' tau_ {ind) (N-m) ' , ' Fontwe i ght ' , 'Sol d' ) ;
y l abel ( ' itn_ {m) nnbf (r I min ) , , 'Fontwe i ght ' , 'Sol d' ) ;
titl e ( ' Serie s IX: M
ot or To rque - Speed Chara c t e ri s ti c ' ,
'Fontwe i ght , , 'So l d' ) ;
axi s( [ 0 700 0 5000 ] ) ;
gri d on;
ho l d o ff;
The resulting motor torque-speed characteristic is shown in Figure 9- 23. Notice the
severe overspeeding at very small torques.
Speed Control of Series DC Motors
Unlike with the shunt dc motor, there is only one efficient way to change the
speed of a series dc motor. That method is to change the terminal voltage of the
motor. If the terminal voltage is increased, the first term in Equation (9-23) is in-
creased, resulting in a higher speedfor any given torque.
The speed of series dc motors can also be controlled by the insertion of a se-
ries resistor into the motor circuit, but this technique is very wasteful of power and
is used only for intennittent periods during the start-up of some motors.
Until the last 40 years or so, there was no convenient way to change VT, so
the only method of speed control available was the wasteful series resistance
method. That has all changed today with the introduction of solid-state control
circuits. Techniques of obtaining variable terminal voltages were discussed in
Chapter 3 and will be considered further later in this chapter.](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-591-320.jpg)
![572 ELECTRIC MACHINERY RJNDAMENTALS
+
O.04n
<
L,
NF = ](XXl turns per pole
••
• Cumulatively
compounded
• Differentially
compounded
Vr= 250 V
""GURE 9-27
The compounded dc motor in Example 9--6.
(a) What is the shunt field current in this machine al no load?
(b) If the motor is cumulatively compOlUlded, find its speed when I}, = 200 A.
(c) If the motor is differentially compounded, find its speed when I}, = 200 A.
Solutioll
(a) Al no load, the armature ClUTent is zero, so the internal generated vollage of the
motor must equal Vr, which means that it must be 250 V. From the magnetiza-
tion curve, a field current of 5 A will produce a vollage E}, of 250 V at 1200
r/min. Therefore, the shlUll field ClUTent must be 5 A.
(b) When an armature ClUTent of 200 A flows in the motor, the machine's internal
generated vollage is
E}, = Vr - I},(R}, + Rs)
= 250 V - (200 A)(0.04fi) = 242 V
The effective field clUTenl of this cumulatively compolUlded motor is
• A;E ~AR
IF = IF + HI}, - N (9- 28)
F ,
3
= 5A + I()(X) 200 A = 5.6A
From the magnetization curve, E},o = 262 V at speed no = 1200 r/min. There-
fore, the motor's speed will be
E,
n = - n"
E"
242 V .
= 262 V 1200 rlnnn = 1108 rlmin
(c) If the machine is differentially compounded, the effective field current is
• NSE '3'AR
IF = IF - NF I}, - NF (9- 28)
3
= 5A - 1000 200 A = 4.4 A](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-596-320.jpg)
![578 ELECTRIC MACHINERY RJNDAMENTALS
When EA = 187.5 V, fA has fallen to 350 A and it is time to cut out the second
starting resistor R2. When it is cut out, the current should jump back to 700 A.
Therefore,
R +R = VT -EA =250V-187.5V =008930
A 1 I 700 A .
~
(9- 38)
This total resistance will be in the circuit until fA again falls to 350 A. This
occurs when EA reaches
EA = VT - fAR,o< = 250V - (350A)(0.08930) = 218.75V
When EA = 218.75 V, fA has fallen to 350 A and it is time to cut out the third
starting resistor R1. When it is cut out, only the internal resistance of the motor
is left. By now, though, RA alone can limit the motor's current to
VT - EA 250 V - 218.75 V
lAo = R = 0.050
,
= 625 A (less than allowed maximrun)
From this point on, the motor can speed up by itself.
From Equations (9- 34) to (9- 36), the required resistor values can be
calculated:
RJ = R'OI.3 - RA = 0.08930 - 0.05 0 = 0.03930
R2 = R'0I2 - RJ - RA = 0.1786 0 - 0.0393 0 - 0.05 0 = 0.0893 0
R] = R"".l - R2- R3 - RA = 0.357 0 - 0.1786 0 - 0.0393 0 - 0.05 0 = 0.1786 0
And RJ, Rl., and RJ are cut out when EA reaches 125, 187.5, and 218.75 V,
respectively.
DC Motor Starting Circuits
Once the starting resistances have been selected, how can their shorting contacts
be controlled to ensure that they shut at exactly the correct moment? Several dif-
ferent schemes are used to accomplish this switching, and two of the most com-
mon approaches will be examined in this section. Before that is done, though, it is
necessary to intrrxluce some of the components used in motor-starting circuits.
Figure 9- 30 illustrates some of the devices commonly used in motor-
control circuits. 1lle devices illustrated are fuses, push button switches, relays,
time delay relays, and overloads.
Figure 9- 30a shows a symbol for a fuse. The fuses in a motor-control cir-
cuit serve to protect the motor against the danger of short circuits. They are placed
in the power supply lines leading to motors. If a motor develops a short circuit, the
fuses in the line leading to it will burn out, opening the circuit before any damage
has been done to the motor itself.
Figure 9-30b shows spring-type push button switches. There are two basic
types of such switches-normally open and nonnally shut. Nonnally open con-
tacts are open when the button is resting and closed when the button has been](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-602-320.jpg)
![rSCR]
Th~
r
ph= SCR.,
input
FIGURE 9-35
lrSCR2
l,-
SCR~
II
I,
Operation
"0'
possible
lrSCRJ
V-SC~
(.,
(b,
rx: MmDRS AND GENERATORS 585
,Free-
wheeling
diode)
D,
v,
K.
Operation
"oj
possible
+
I
v, E,C)
-
(a) A two-quadrant solid-state dc motor controller. Since current cannot flow out of the positive
terminals of the armature. this motor cannot act as a generator. returning power to the system.
(b) The possible operating quadrants of this motor controller.
a Ward-Leonard system pennits rotation and regeneration in either direction, it is
called afour-quadrant control system.
The disadvantages of a Ward-Leonard system should be obvious. One is that
the user is forced to buy three full machines of essentially equal ratings, which is
quite expensive. Another is that three machines will be much less efficient than
one. Because of its expense and relatively low efficiency, the Ward-Leonard sys-
tem has been replaced in new applications by SCR-based controller circuits.
A simple dc armature voltage controller circuit is shown in Figure 9- 35.
The average voltage applied to the armature of the motor, and therefore the aver-
age speed of the motor, depends on the fraction of the time the supply voltage is
applied to the armature. This in turn depends on the relative phase at which the](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-609-320.jpg)
![594 ELECTRIC MACHINERY RJNDAMENTALS
P
F = (SA:Y(500) = 12S0W
The brush losses at full load are given by
Pbtuob = VBofA = (2 V)(170A) = 340W
The rotational losses at full load are essentially equivalent to the rotational losses at no
load, since the no-load and full-load speeds of the motor do not differ too greatly. These
losses may be ascertained by detennining the input power to the armature circuit at no load
and assuming that the annature copper and brush drop losses are negligible, meaning that
the no-load armature input power is equal to the rotationa1losses:
P,ot. = POOle + P_ = (240 VX13.2 A) = 3168W
(a) The input power of this motor at the rated load is given by
P~ = VrlL = (250VXI7SA) = 43,7S0W
Its output power is given by
P00' = Pm - Pb<wb - P<:U - P<Ote - P1DOC.b - P !MaY
= 43,7S0W - 340 W - 1734 W - 1250W - 3168W - (O.OIX43,750W)
= 36,82oW
where the stray losses are taken to be I percent of the input power.
(b) The efficiency of this motor at full10ad is
T] = ~ou, x 100%
~
_ 36,820W
- 43,7SoW x 100% = 84.2%
9.11 INTRODUCTION TO DC GENERATORS
DC generators are dc machines used as generators. As previously pointed out,
there is no real difference between a generator and a motor except for the direc-
tion of power fl ow. There are five major types of dc generators, classified accord-
ing to the manner in which their field flux is produced:
I. Separately excited generator. In a separately excited generator, the field flux
is derived from a separate power source independent of the generator itself.
2. Shunt generator. In a shunt generator, the field flux is derived by connecting
the field circuit directly across the tenninals of the generator.
3. Series generator. In a series generator, the field flux is produced by connect-
ing the field circuit in series with the armature of the generator.
4. Cumulatively compounded generator. In a cumulatively compounded gener-
ator, both a shunt and a series field are present, and their effects are additive.
S. Differentially compounded generator. In a differentially compounded genera-
tor, both a shunt and a series field are present, but their effects are subtracti ve.
TIlese various types of dc generators differ in their terminal (voJtage--current)
characteristics, and therefore in the applications to which they are suited.](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-618-320.jpg)
![616 ELECTRIC MACHINERY RJNDAMENTALS
v,
L-_ _ _ _ _ _ _ _ _ ~
'------------ 1,
""GURE 9-64
Graphical derivation of the terminal characteristic of a cumulatively compounded dc generator.
I, I,
" - • +
R, R,
L, I, I
v.,
/- I" = IL+I,,
+ V,
E, • V, 1,,=][;
,
L,
Vr = E" - I" (R" + Rsl
""GURE 9-65
The equivalent cin:uit of a differentially compounded dc generator with a long-shunt connection.
equivalenl circuit of a differentially compounded dc generator is shown in Figure
9--65. Notice that the armature current is now fl owing out of a dotted coil end,
while the shunt field current is fl owing into a dotted coil end. In this machine, the
net magnetornotive force is
(9- 55)
~AR I (9- 56)](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-640-320.jpg)
![>
J
•
~
•
"
~
•
•
,
,
a
1!
~
160
150
140
130
120
110
100
90
80
70
60
50
/
40
30
20
/
/
/
10
/
o
o 10
/
/
rx: MmDRS AND GENERATORS 625
5,.., 1200r~
/"
/
/
/
/
/
20 30 40 60 70
Series field current. A
FIGURE 1'9-5
The magnetization curve for the series moior in Problem 9--13. This curve was taken al a constant
speed of 1200 r/min.
What is this motor's efficiency at the rated conditions? [Note:Assrune that ( 1) the
brush voltage drop is 2 V, (2) the core loss is to be determined at an armature volt-
age equal to the armature voltage lUlder full load, and (3) stray load losses are 1 per-
cent of full load.]
Problems 9- 16 to 9--19 refer to a 240-V, IOO-A de motor which has both shunt and series
windings. Its characteristics are
RA =O.14f.!
Rs = 0.04 n
R" = 200 n
Radj = 0 to 300 n,currently set to 120 n
N" = 1500 turns
Nsf', = 12 turns
nO. = 1200 r/min](https://image.slidesharecdn.com/electrical-machinery-fundamentals-220905101643-f02b16d0/85/Electrical-Machinery-Fundamentals-pdf-649-320.jpg)
Electrical-Machinery-Fundamentals.pdf
- 5. ELECTRIC MACHINERY FUNDAMENTALS FOURTH EDITION Stephen J. Chapman BAE SYSTEMS Australia Higher Education Boston Burr Ridge, IL Dubuque, IA Madison, WI New York San Francisco SI. l ouis Bangkok Bogota Caracas Kuala l umpur Lisbon London Madrid Mexico City Milan Montreal New Delhi Santiago Seoul Singapore Sydney Taipei Toronto
- 6. • Higher Education ELECTRIC M ACHINERY RJNDAMENTALS. FOURTH EDITION Published by McGraw-Hill. a business unit of The McGraw-Hill Companies. Inc., 1221 Avenue of the Americas, New Yort. NY 10020. Copyright 0 2005, 1999. 1991. 1985 by The McGraw,Hill Companies. Inc. All rights reserved. No part of this publication may be reproduced or distributed in any form or by any means. or stored in a database or retrieval system. without the prior written con' sent of The McGraw-Hill Companies. Inc., including. but not limited to, in any network or other electronic storage or transmission. or broadcast for distance learning. Some ancillaries. including electronic and prim components. may not be available to customers out, side the United States. This book is printed on acid'free paper. 1234567890DOaDOC09876543 ISBN 0--07- 246523--9 Publisher: Elizabeth A. Jones Senior sponsoring editor: Carlise Paulson Managing developmental editor: EmilyJ. Lupash Marketing manager: Val''"" R. Bercier Senior project manager: Sheila M. Frank Senior production supervisor: Laura Fuller Senior media project manager: Tammy Juran Senior designer: Da·id W. Hash Lead photo research coordinator: Carrie K. Burger Compositor: GAC- Indianapolis Typeface: /0//2 Times Rotnlln Printer: R. R. Donnelley Crawfordsville. IN Libmry of Co n ~ress Gltalo~in~-in-l'ublic:ltion Data Chapman. Stephen J. Electric machinery fundamentals / Stephen Chapman. - 4th ed. p. em. Includes index. ISBN 0-07- 246523--9 I. Electric machinery. I. Title. TK2000.C46 2005 621.31 ·042---dc22 www.mhhe.oom 2003065174 CIP
- 7. THIS WORK IS DEDICATED WITH LOVE TO MY MOTHER, LOUISE G. CHAPMAN, ON THE OCCASION OF HER EIGHTY-RFfH BIRTHDAY.
- 9. ABOUT THE AUTHOR Stephen J. Chapman received a B.S. in Electrical Engineering from Louisiana State University (1975) and an M.S.E. in Electrical Engineering from the Univer- sity of Central Florida ( 1979), and pursued further graduate studies at Rice University. From 1975 to 1980, he served as an officer in the U.S. Navy, assigned to teach electrical engineering at the U.S. Naval Nuclear Power School in Orlando, Florida. From 1980 to 1982, he was affiliated with the University of Houston, where he ran the power systems program in the College of Technology. From 1982 to 1988 and from 1991 to 1995, he served as a member of the technical stafT oftile Massachusetts Institute of Technology's Lincoln Laboratory, both at the main facility in Lexington, Massachusetts, and at the field site on Kwa- jalein Atoll in the Republic of the Marshall Islands. While there, he did research in radar signal processing systems. He ultimately became the leader of four large operational range instrumentation radars at the Kwajalein field site (TRADEX, ALTAIR, ALCOR, and MMW). From 1988 to 1991 , Mr. Chapman was a research engineer in Shell Devel- opment Company in Houston, Texas, where he did seismic signal processing re- search. He was also affiliated with the University of Houston, where he continued to teach on a part-time basis. Mr. Chapman is currently manager of systems modeling and operational analysis for BAE SYSTEMS Australia, in Melbourne. Mr. Chapman is a senior member of the Institute of Electrical and Elec- tronic Engineers (and several of its component societies). He is also a member of the Association for Computing Machinery and the Institution of Engineers (Australia). vu
- 11. BRIEF CONTENTS Chapter 1 Introduction to Machinery Principles Chapter 2 Transformers 65 Chapter 3 Introduction to Power Electronics 152 Chapter 4 AC Machinery Fundamentals 230 Chapter 5 Synchronolls Generators 267 Chapter 6 Synchronolls Motors 346 Chapter 7 Induction Motors 380 Chapter 8 DC Machinery Fundamentals 473 Chapter 9 DC Motors and Generators 533 Chapter 10 Single-Phase and Special-Purpose Motors 633 Appendix A Three-Phase Circuits 681 Appendix B Coil Pitch and Distributed Windings 707 Appendix C Salient-Pole Theory ofSynchronolls Machines 727 Appendix D Tables of Constants and Conversion Factors 737 "
- 13. TABLE OF CONTENTS Chapter 1 Introduction to Machinery Principles 1.1 Electrical Machines, Transformers, and Daily Life 1.2 A Note on Units and Notation Notation 2 1.3 Rotational Motion, Newton's Law, and Power Relationships 3 Angular Position (J I Angular Velocity w / Angular Acceleration a / Torque T / Newton 's Law o/Rotation I W ork WPower P I.. The Magnetic Field 8 Production ofa Magnetic Field / Magnetic Circuits / Magnetic Behavior 01 Ferromagnetic Materials I Energy Losses in a Ferromagnetic Core 1.5 Faraday's Law-Induced Voltage from a Time-Changing Magnetic Field 28 1.6 Production of Induced Force on a Wire 32 1.7 Induced Voltage on a Conductor Moving in a Magnetic Field 34 I." The Linear OC Machine- A Simple Example 36 Starting the Linear DC Machine / The linear DC Machine as a Motor I The Linear DC Machine as a Generator I Starting Problems with the Linear Machine I.. Real, Reactive, and Apparent Power in AC Circuits 47 Alternative Fonns ofthe Power Equations I Complex Power I The Relationships beflt'een Impedance Angle, Current Angle, and Power I The Power Triangle 1.10 Summary 53 Questions 54 Problems 55 References 64 "
- 14. XII TABLE OF CONTENTS Chapter 2 Transformers 65 2.1 Why Transfonners Are Important to Modern Life 66 2.2 Types and Construction of Transformers 66 2.3 The Ideal Transfonner 68 Power in an Ideal Transformer I Impedance TransfornUltion through a Transfornler I Analysis of Circuits Containing Ideal Transformers 2.4 Theory of Operation of Real Single-Phase Transformers 76 The Voltage Ratio across a Transformer I The Magnetization Current in a Real Transformer I The Current Ratio on a Transformer and the Dot Conrention 2.5 The Equivalent Circuit of a Transformer 86 The Exact Equivalent Circuit ofa Real Transformer I ApproxinUlte Equivalent Circuits ofa Transformer I Determining the Values of Components in the Transfonner Model 2.6 The Per-Unit System of Measurements 94 2.7 Transfonner Voltage Regulation and Efficiency 100 The Transformer Phasor Diagram I Transfonner Efficiency 2.8 Transfonner Taps and Voltage Regulation 108 2.9 The Autotransfonner 109 Voltage and Current Relationships in an Autotransformer I The Apparent Power Rating Advantage ofAutotransfornlers I The Internal Impedance ofan Autotransformer 2.10 Three-Phase Transfonners 11 6 Three-Phase Transformer Connections I The Per-Unit System for Three-Phase Transformers 2.11 Three-Phase Transfonnation Using Two Transformers 126 The Open-il (or V-V) Connection I The Open-"3'e-Open- Delta Connection I The Scott-T Connection I The Three- Phase T Connection 2.12 Transfonner Ratings and Related Problems 134 The Voltage and Frequency Ratings ofa Transformer I The Apparent Power Rating ofa Transfornler I The Problem ofCurrent Inrnsh I The Transformer Nameplate 2.13 Instnunent Transformers 140 2.14 Swnmary 142 Questions 143 Problems 144 References 15 1
- 15. Chapter 3 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.' 3.' Chapter 4 4.1 TABLE OF CONTENTS XlU Introduction to Power Electronics Power Electronic Components The Diode / The Two- Wire Thyristor or PNPN Diode / The Three-Wire Thyristor ofSCR / The Gate Turnoff Thyristor / The DlAC / The TRIA C / The Power Transistor / The Insulated-Gate Bipolar Transistor / Power atui Speed Comparison ofPower Electronic Components Basic Rectifier Circuits The Half-Wave Rectifier / The Full-Wave Rectifier / The Three-Phase Half-Wave Rectifier / The Three-Phase Full- Wave Rectifier / Filtering Rectifier Output Pulse Circuits A Relaxation Oscillator Using a PNPN Diode / Pulse Synchronization Voltage Variation by AC Phase Control AC Phase Controlfora DC Load Drivenfrom an AC Source / AC Phase Angle Control for an AC Load / The Effect ofInductive Loads on Phase Angle Control DC-to-DC Power Control-Choppers Forced Commutation in Chopper Circuits / Series- Capacitor Commutation Circuits / Parallel-Capacitor Commutation Circuits Inverters The Rectifier / External Commutation lnverters / Self- Commutation Inverters / A Single-Phase Current Source Inverter / A Three-Phase Current Source lnverter / A Three-Phase Voltage Source Inverter / Pulse-Width Modulation lnverters Cycloconverters Basic Concepts / Noncirculating Current Cycloconverters / Circulating Current Cycloconverters Hannonic Problems Summary Questions Problems References AC Machinery Fundamentals A Simple Loop in a Uniform Magnetic Field The Voltage Induced in a Simple Rotating Loop / The Torque lnduced in a Current-Cart}'ing Loop 152 152 163 171 177 186 193 209 218 221 223 223 229 230 230
- 16. XIV TABLEOF CONTENTS 4.2 The Rotating Magnetic Field 238 Proofofthe Rotating Magnetic Field Concept I The Relationship between Electrical Frequency and the Speed ofMagnetic Field Rotation I Reversing the Direction of Magnetic Field Rotation 4.3 Magnetomotive Force and Flux Distribution on AC Machines 246 4.4 Induced Voltage in AC Machines 250 The Induced Voltage in a Coil on a Two-Pole Stator I The Induced Voltage in a Three-Phase Set ofCoils I The RMS Voltage in a Three-Phase Stator 4.5 Induced Torque in an AC Machine 255 4.• Wmding Insulation in an AC Machine 258 4.7 AC Machine Power Flows and Losses 26 1 The Losses in AC Machines I The Power-Flow Diagram 4.S Voltage Regulation and Speed Regulation 262 4.9 Swnmary 264 Questions 265 Problems 265 References 266 Chapter 5 Synchronous Generators 267 5. 1 Synchronous Generator Construction 267 5.2 The Speed of Rotation of a Synchronous Generator 272 5.3 The Internal Generated Voltage of a Synchronous Generator 273 5.4 The Equivalent Circuit of a Synchronous Generator 274 5.5 The Phasor Diagram of a Synchronous Generator 279 5.• Power and Torque in Synchronous Generators 280 5.7 Measuring Synchronous Generator Model Parameters 283 The Short-Circuit Ratio 5.8 The Synchronous Generator Operating Alone 288 The Effect ofLoad Changes on Synchronous Generator Operating Alone I Example Problems 5.9 Parallel Operation of AC Generators 299 The Conditions Requiredfor Paralleling I The General Procedure for Paralleling Generators I Frequency-Power and Voltage-Reactive Power Characteristics ofa Synchronous Generator I Operation of Generators in Parallel with Large Power Systems I Operation ofGenerators in Parallel with Other Generators ofthe Same Size 5.10 Synchronous Generator Transients 319 Transient Stability ofSynchronous Generators I Short-Circuit Transients in Synchronous Generators
- 17. 5.11 5. 12 Chapter 6 6.1 6.2 6.3 6.4 6.5 6.6 Chapter 7 7.1 7.2 7.3 TABLE OF CONTENTS XV Synchronous Generator Ratings The Voltage, Speed, and Frequency Ratings / Apparent Power atui Power-Factor Ratings / Synchronous Generator Capability CUf1Jes / Short-Time Operation and Sef1Jice Factor Summary Questions Problems References Synchronous Motors Basic Principles of Motor Operation The Equiralent Circuit ofa Synchronous Motor / The Synchronous Motorfrom a Magnetic Field Perspective Steady-State Synchronous Motor Operation The Synchronous Motor Torque-Speed Characteristic CUf1Je / The Effect ofLoad Changes on a Synchronous Motor / The Effect of Field Changes on a Synchronous Motor / The Synchronous Motor atui Power,Factor Correction / The Synchronous Capacitor or Synchronous Condenser Starting Synchronous Motors Motor Starting by Reduced Electrical Frequency / Motor Starting with an utemal Prime Mover / Motor Starting by Using Amortisseur Windings / The Effect of Amortisseur Windings on Motor Stability Synchronous Generators and Synchronous Motors Synchronous Motor Ratings Summary Questions Problems References Induction Motors Induction Motor Construction Basic Induction Motor Concepts The Development ofInduced Torque in an ltuiuction Motor / The Concept ofRotor Slip / The Electrical Frequency on the Rotor The Equivalent Circuit of an Induction Motor The Transformer Model ofan Induction Motor / The Rotor Circuit Model/The Final Equiralent Circuit 326 336 337 338 345 346 346 350 364 37 1 372 373 374 374 379 380 380 384 388
- 18. XVI TABLE OF CONTENTS 7.4 Power and Torque in Induction Motors 394 Losses and the Pml'er-Flow Diagram I Power atui Torque in an Induction Motor I Separating the Rotor Copper Losses and the Pmwr Converted in an lnduction Motor S Equivalent Cirr:uit 7.5 Induction Motor Torque-Speed Characteristics 401 lnduced Torquefrom a Physical Statuipoint IThe Derivation ofthe lnduction Motor ltuiuced-Torque Equation I Comments on the Induction Motor Torque-Speed Cun'e I Maximum (Pullout) Torque in an ltuiuction Motor 7.• Variations in Induction Motor Torque-Speed Characteristics 416 Control ofMotor Characteristics by Cage Rotor Design I Deep-Bar and Double-Cage Rotor Designs I lnduction Motor Design Classes 7.7 Trends in Induction Motor Design 426 7.8 Starting Induction Motors 430 lnduction Motor Starting Circuits 7.9 Speed Control of Induction Motors 434 lnduction Motor Speed Control by Pole Changing I Speed Control by Changing the Line Frequency I Speed Control by Changing the Line Voltage I Speed Control by Changing the Rotor Resistance 7.10 Solid-State Induction Motor Drives 444 Frequency (Speed)Adjustment I A Choice of Voltage and Frequency Patterns I Independently Adjustable Acceleration atui Deceleration Ramps I Motor Protection 7.11 Detennining Circuit Model Parameters 452 The No-Load Test I The DC Testfor Stator Resistance I The Locked-Rotor Test 7.12 The Induction Generator 460 The lnduction Generator Operating Alone I lnduction Generator Applications 7.13 Induction Motor Ratings 464 7.14 Swnmary 466 Questions 467 Problems 468 Rererences 472 Chapter 8 DC Machinery Fundamentals 473 8.1 A Simple Rotating Loop between Curved Pole Faces 473
- 19. TABLE OF CONTENTS XVU The lliltage lnduced in a Rotating Loop / Getting DC Voltage out ofthe Rotating Loop / The Induced Torque in the Rotating Loop 8.2 Commutation in a Simple Four-Loop IX Machine 485 8.3 Commutation and Armature Construction in Real DC Machines 490 The Rotor Coils / Connections to the Commutator Segments / The Lap Winding / The Wave Winding / The Frog-Leg Winding 8.4 Problems with Conunutation in Real Machines 502 Armature Reaction / L dildt Voltages / Solutions to the Problems with Commutation 8.5 The Internal Generated Voltage and Induced Torque Equations of Real DC Machines 514 8.6 The Construction of DC Machines 518 Pole and Frame Construction / Rotor or Armature Constrnction / Commutator and Brushes / Winding Insulation 8.7 Power Flow and Losses in DC Machines 524 The Losses in DC Machines / The Power-Flow Diagram 8.8 Summary 527 Questions 527 Problems 527 References 530 Chapter 9 DC Motors and Generators 533 9.1 Introduction to DC Motors 533 9.2 The Equivalent Circuit of a IX Motor 535 9.3 The Magnetization Curve of a DC Machine 536 9.4 Separately Excited and Shunt IX Motors 538 The Tenninal Characteristic ofa Shunt DC Motor / Nonlinear Analysis ofa Shunt DC Motor / Speed Control ofShunt DC Motors / The Effect ofan Open Field Circuit 9.5 The Pennanent-Magnet DC Motor 559 9.6 The Series IX Motor 562 Induced Torque in a Series DC Motor / The Terminal Characteristic ofa Series DC Motor / Speed Control of Series DC Motors 9.7 The Compounded DC Motor 568 The Torque-Speed Characteristic ofa Cumulatively Compounded DC Motor / The Torque- Speed
- 20. XVIII TABLE OF CONTENTS Characteristic ofa Differentially Compoutuied DC Motor / The Nonlinear Analysis ofCompounded DC Motors / Speed Control in the Cumulatively Compoutuied DC Motor 9.8 DC Motor Starters 9.9 9.10 9.11 9.12 9.13 9.14 9.15 9.16 9.17 Chapter 10 10.1 DC Motor Problems on Starting / DC Motor Starting Circuits The Ward-Leonard System and Solid-State Speed Controllers Protection Circuit Section / StartlStop Circuit Section / High.Power Electronics Section / Low-Power Electronics Section DC Motor Efficiency Calculations Introduction to IX Generators The Separately Excited Generator The Terminal Characteristic ofa Separately Excited DC Generator / Control ofTerminal Voltage / Nonlinear Analysis ofa Separately Excited DC Generator The Shunt DC Generator Voltage Buildup in a Shunt Generator / The Tenninal Characteristic ofa Shunt DC Generator / Voltage Control fo r a Shunt DC Generator / The Analysis ofShunt DC Generators The Series DC Generator The Terminal Characteristic ofa Series Generator The Crunul atively Compounded DC Generator The Terminal Characteristic ofa Cumulatively Compounded DC Generator / Voltage Control of Cumulatively Compounded DC Generators / Analysis of Cumulatively Compounded DC Generators The Differentially CompolUlded DC Generator The Terminal Characteristic ofa Differentially Compounded DC Generator / Voltage Control of Differentially Compounded DC Generators / Graphical Analysis ofa Differentially Compounded DC Generator Srunmary Questions Problems References Single-Phase and Special-Purpose Motors The Universal Motor Applications of Universal Motors / Speed Control of Universal Motors 573 582 592 594 596 602 608 611 615 619 620 621 631 633 634
- 21. TA BLE OF CONTENTS XIX 10.2 Introduction to Single-Phase Induction Motors 637 The Double.Rerolving-Field Theory ofSingle.Phase Induction Motors / The Cross·Field Theory ofSingle. Phase Induction Motors 10.3 Starting Single-Phase Induction Motors 646 Split-Phase Windings / Capacitor.Start Motors / Pennanent Split-Capacitor and Capacitor.Start, Capacitor.Run Motors / Shaded-Pole Motors / Comparison ofSingle.Phase Induction Motors 10.4 Speed Control of Single-Phase Induction Motors 656 10.5 The Circuit Model of a Single-Phase Induction Motor 658 Circuit Analysis with the Single-Phase Induction Motor Equiralent Circuit 10.6 Other Types of Motors 665 Reluctance Motors / Hysteresis Motors / Stepper Motors / Brushless DC Motors 10.7 Summary 677 Questions 678 Problems 679 References 680 Appendix A Three-Phase Circuits 68 1 A.I Generation of Three-Phase Voltages and Currents 68 1 A.2 Voltages and Currents in a Three-Phase Circuit 685 Voltages and Currents in the ~~'e (Y) Connection / Voltages and Currents in the Delta (8) Connection A.3 Power Relationships in lbree-Phase Circuits 690 Three-Phase Po....er Equations Involving Phase Quantities / Three-Phase Po....er Equations Involving Line Quantities A.4 Analysis of Balanced Three-Phase Systems 693 A.5 One-Line Diagrams 700 A.6 Using the Power Triangle 700 Qnestions 703 Problems 704 References 706 Appendix B Coil Pitch and Distributed Windings 707 8.1 The Effect of Coil Pitch on AC Machines 707 The Pitch ofa Coil / The Induced Voltage ofa Fractional· Pitch Coil / Harmonic Problems and Fractional-Pitch Windings
- 22. XX TABLE OF CONTENTS 8.2 Distributed Windings in AC Machines 716 The Breadth or Distribution Factor I The Generated Voltage Including Distribution Effects / Tooth or Slot Harmonics 8.3 Swnmary 724 Questions 725 Problems 725 References 726 Appendix C Salient-Pole Theory of Synchronous Machines 727 C. I Development of the Equivalent Circuit of a Salient-Pole Synchronous Generator 728 C.2 Torque and Power Equations of Salient-Pole Machine 734 Problems 735 Appendix D Tables of Constants and Conversion Factors 737
- 23. PREFACE In the years since the first edition of Electric Machinery Fundamentals was published, there has been rapid advance in the development of larger and more sophisticated solid-state motor drive packages. The first edition of this book stated that de motors were the method of choice for demanding variable-speed applica- tions.11131 statement is no longer true today. Now, the system of choice for speed control applications is most often an ac induction motor with a solid-state motor drive. DC motors have been largely relegated to special-purpose applications where a de power source is readily available, such as in automotive electrical systems. The third edition orthe book was extensively restructured to reflect these changes.1lle material on ac motors and generators is now covered in Chapters 4 through 7, before the material on dc machines. In addition, the dc machinery cov- erage was reduced compared to earlier editions.1lle fourth edition continues with this same basic structure. Chapter I provides an introduction to basic machinery concepts, and con- cludes by applying those concepts to a linear dc machine, which is the simplest possible example of a machine. Glapter 2 covers transformers, and Chapter 3 is an introduction to solid-state power electronic circuits. The material in Chapter 3 is optional, but it supports ac and dc motor control discussions in Chapters 7, 9, and 10. After Chapter 3, an instructor may choose to teach either dc or ac machin- ery first. Chapters 4 through 9 cover ac machinery, and Chapters 8 and 9 cover dc machinery. 1llese chapter sequences have been made completely independent of each other, so that instructors can cover the material in the order that best suits their needs. For example, a one-semester course with a primary concentration in ac machinery might consist of parts of Chapters I to 7, with any remaining time devoted to dc machinery. A one-semester course with a primary concentration in dc machinery might consist of parts of Chapters I, 3, 8, and 9, with any remain- ing time devoted to ac machinery. Chapter 0 is devoted to single-phase and special-purpose motors, such as universal motors, stepper motors, brushless dc motors, and shaded-pole motors. XXI
- 24. XXII PREFACE TIle homework problems and the ends of chapters have been revised and corrected, and more than 70 percent of the problems are either new or modified since the last edition. In recent years, there have been major changes in the methods used to teach machinery to electrical engineering and electrical technology students. Excellent analytical tools such as MATLAB have become widely available in university en- gineering curricula. TIlese tools make very complex calculations simple to per- form, and allow students to explore the behavior of problems interactively. This edition of Electric Machinery Fundamentals makes sclected use of MATLAB to enhance a student's learning experience where appropriate. For example, students use MATLAB in Chapter 7 to calculate the torque-speed characteristics of induc- tion motors and to explore the properties of double-cage induction motors. TIlis text does not teach MATLAB; it assumes that the student is familiar with it through previous work. Also, the book does not depend on a student hav- ing MATLAB. MATLAB provides an enhancement to the learning experience if it is available, but if it is not, the examples involving MATLAB can simply be skipped, and the remainder of the text still makes sensc. Supplemental materials supporting the book are available from the book's website, at www.mhhe.com/engcslelectricallchapman. The materials available at that address include MATLAB source code, pointers to sites of interest to ma- chinery students, a list of errata in the text, some supplemental topics that are not covered in the main text, and supplemental MATLAB tools. TIlis book would never have been possible without the help of dozens of people over the past 18 years. I am not able to acknowledge them al l here, but I would especially like to thank Charles P. LeMone, Teruo Nakawaga, and Tadeo Mose of Toshiba International Corporation for their invaluable help with the solid-state machinery control material in Chapter 3. I would also like to thank Jeffrey Kostecki, Jim Wright, and others at Marathon Electric Company for sup- piying measured data from some of the real generators that the company builds. TIleir material has enhanced this revision. Finally, I would like to thank my wife Rosa and our children Avi, David, Rachel, Aaron, Sarah, Naomi, Shira, and Devorah for their forbearance during the revision process. I couldn't imagine a better incentive to write! Stepllell J. Chapman Metboume, Victoria, Australia
- 25. 1.1 ELECTRICAL MACHINES, TRANSFORMERS, AND DAILY LIFE CHAPTER 1 INTRODUCTION TO MACHINERY PRINCIPLES An electrical machine is a device that can convert either mechanical energy to electrical energy or electrical energy to mechanical energy. When such a device is used to convert mechanical energy to e lectrical energy, it is called a generator. When it converts electrical energy to mechanical energy, it is called a motor. Since any given electrical machine can convert power in either direction, any machine can be used as either a generator or a motor. Almost all practical motors and gen- erators convert energy from one form to another through the action of a magnetic fie ld, and only machines using magnetic fields to perform such conversions are considered in this book. The transformer is an electrical device that is closely related to electrical machines. It converts ac electrical energy at one voltage level to ac electrical en- ergy at another voltage level. Since transfonners operate on the same principles as generators and motors, depending on the action ofa magnetic field to accomplish the change in voltage level, they are usually studied together with generators and motors. These three types of electric devices are ubiquitous in modern daily life. Electric motors in the home run refrigerators, freezers, vacuum cleaners, blenders, air conditioners, fans, and many similar appliances. In the workplace, motors pro- vide the motive power for almost all tools. Of course, generators are necessary to supply the power used by alJ these motors. I
- 26. 2 ELECTRIC MACHINERY FUNDAMENTALS Why are electric motors and generators so common? The answer is very simple: Electric power is a clean and efficient energy source that is easy to trans- mit over long distances, and easy to control. An electric motor does not require constant ventilation and fuel the way that an internal-combustion engine does, so the motor is very well suited for use in environments where the pollutants associ- ated with combustion are not desirable. Instead, heat or mechanical energy can be converted to electrical fonn at a distant location, the energy can be transmitted over long distances to the place where it is to be used, and it can be used cleanly in any horne, office, or factory. Transfonners aid this process by reducing the en- ergy loss between the point of electric power generation and the point of its use. 1.2 A NOTE ON UNITS AND NOTATION TIle design and study of electric machines and power systems are among the old- est areas of electrical engineering. Study began in the latter part of the nineteenth century. At that time, electrical units were being standardized internationally, and these units came to be universally used by engineers. Volts, amperes, ohms, watts, and similar units, which are part of the metric system of units, have long been used to describe electrical quantities in machines. In English-speaking countries, though, mechanical quantities had long been measured with the English system of units (inches, feet, pounds, etc.). This prac- tice was followed in the study of machines. TIlerefore, for many years the electri- cal and mechanical quantities of machines have been measured with different sys- tems of units. In 1954, a comprehensive system of units based on the metric system was adopted as an international standard. This system of units became known as the Systeme International (SI) and has been adopted throughout most of the world. The United States is practically the sole holdout--even Britain and Canada have switched over to Sl. TIle SI units will inevitably become standard in the United States as time goes by, and professional societies such as the Institute of Electrical and Elec- tronics Engineers (IEEE) have standardized on metric units for all work. How- ever, many people have grown up using English units, and this system will remain in daily use for a long time. Engineering students and working engineers in the United States today must be familiar with both sets of units, since they will en- counter both throughout their professional lives. Therefore, this book includes problems and examples using both SI and English units. TIle emphasis in the ex- amples is on SI units, but the older system is not entirely neglected. Notation In this book, vectors, electrical phasors, and other complex values are shown in bold face (e.g., F), while scalars are shown in italic face (e.g., R). In addition, a special font is used to represent magnetic quantities such as magnetomotive force (e.g., 'iJ).
- 27. INTRODUCTION TO MACHINERY PRINCIPLES 3 1.3 ROTATIONAL MOTION, NEWTON'S LAW, AND POWER RELATIONSHIPS Almost all electric machines rotate about an axis, called the shaft of the machine. Because of the rotational nature of machinery, it is important to have a basic un- derstanding of rotational motion. This section contains a brief review of the con- cepts ofdistance, velocity, acceleration, Newton's law, and power as they apply to rotating machinery. For a more detailed discussion of the concepts of rotational dynamics, see References 2, 4, and 5. In general, a three-dimensional vector is required to completely describe the rotation of an object in space. However, machines nonnally turn on a fixed shaft, so their rotation is restricted to one angu lar dimension. Relative to a given end of the machine's shaft, the direction of rotation can be described as either clockwise (CW) or counterclockwise (CCW). For the purpose of this volume, a counter- clockwise angle of rotation is assumed to be positive, and a clockwise one is as- sumed to be negative. For rotation about a fixed shaft, all the concepts in this sec- tion reduce to scalars. Each major concept of rotational motion is defined below and is related to the corresponding idea from linear motion. Angular Position 0 The angular position () of an object is the angle at which it is oriented, measured from some arbitrary reference point. Angular position is usually measured in radians or degrees. It corresponds to the linear concept of distance along a line. Angular Velocity w Angular velocity (or speed) is the rate of change in angular position with respect to time. It is assumed positive if the rotation is in a counterclockwise direction. Angular velocity is the rotational analog of the concept of velocity on a line. One- dimensional linear velocity along a line is defmed as the rate of change of the dis- placement along the line (r) with respect to time d, v~ - dt (I-I ) Similarly, angular velocity w is defined as the rate of change of the angular dis- placement () with respect to time. w = de dt (1- 2) If the units of angular position are radians, then angular velocity is measured in ra- dians per second. In dealing with ordinary electric machines, engineers often use units other than radians per second to describe shaft speed. Frequently, the speed is given in
- 28. 4 ELECTRIC MACHINERY FUNDAMENTALS revolutions per second or revolutions per minute. Because speed is such an im- portant quantity in the study of machines, it is customary to use different symbols for speed when it is expressed in different units. By using these different symbols, any possible confusion as to the units intended is minimized. TIle following sym- bols are used in this book to describe angular velocity: Wm angular velocity expressed in radians per second f.. angular velocity expressed in revolutions per second nm angular velocity expressed in revolutions per minute TIle subscript m on these symbols indicates a mechanical quantity, as opposed to an electrical quantity. If there is no possibility of confusion between mechanical and electrical quantities, the subscript is often left out. TIlese measures of shaft speed are related to each other by the following equations: (l- 3a) ( 1- 3b) Angular Acceleration a Angular acceleration is the rate of change in angular velocity with respect to time. It is assumed positive if the angular velocity is increasing in an algebraic sense. Angular acceleration is the rotational analog of the concept of acceleration on a line. Just as one-dimensional linear acceleration is defined by the equation angular acceleration is defined by d, a~ - dt a = dw dt (1-4) (1- 5) If the units of angular velocity are radians per second, then angular acceleration is measured in radians per second squared. Torque "T In linear motion, aforce applied to an object causes its velocity to change. In the absence of a net force on the object, its velocity is constant. TIle greater the force applied to the object, the more rapidly its velocity changes. TIlere exists a similar concept for rotation. When an object is rotating, its angular velocity is constant unless a torque is present on it. The greater the torque on the object, the more rapidly the angular velocity of the object changes. What is torque? It can loosely be called the "twisting force" on an object. Intuitively, torque is fairly easy to understand. Imagine a cylinder that is free to
- 29. FIGURE I- I , , , , • (a) r=O Torque is zero INTRODUCTION TO MACHINERY PRINCIPLES 5 • F , Torque is counterclockwise 'bJ (a) A force applied to a cylinder so that it passes through the axis of rotation. T = O. (b) A force applied to a cylinder so that its line of action misses the axis of rotation. Here Tis counterclockwise. rotate aboul its axis. If a force is applied to Ihe cylinder in such a way thai its line of action passes through the axis (Figure I-Ia), then the cylinder will not rotate. However, if the same force is placed so that its line of action passes 1 0 Ihe righl of Ihe axis (Figure I-I b), then Ihe cylinder will lend 1 0 rotate in a counterclockwise direction. The torque or twisting action on the cylinder depends on ( I) the magni- tude of the applied force and (2) the distance between the axis of rotation and the line of action of the force. The torque on an object is defined as the product of the force applied 1 0 the object and the smallest distance between the line of action ofthe force and the ob- ject's axis of rotation. If r is a vector pointing from the axis of rotation to the poinl of applicalion of the force, and if F is the applied force, then the torque can be de- scribed as 7" = (force applied)(perpendicular distance) = (F) (r sin ()) = rF sin () (1-6) where () is the angle between the vector r and the vector F. The direction of the lorque is clockwise if it would tend 10 cause a clockwise rotation and counler- clockwise if it wou Id tend to cause a counterclockwise rotalion (Figure 1-2). The units of torque are newton-meters in SI units and pound-feel in lhe Eng- lish system.
- 30. 6 ELECTRIC MACHINERY FUNDAMENTALS rsin(1800 - 1I)=rsinll ~ , __ _J , , , , , , I 1800 _ II T = (perpendicular distance) (force) T = (r sin 9)F. cou nterclockwise Newton's Law of Rotation ,,,, F ' FIGURE 1-1 Derivation of the equation for the torque on an object. Newton's law for objects moving along a straight line describes the relationship between the force applied to an object and its resulting acceleration. This rela- tionship is given by the equation where F = nuJ F = net force applied to an object m = mass of the object a = resulting acceleration (1- 7) In SI units, force is measured in newtons, mass in kilograms, and acceleration in meters per second squared. In the English system. force is measured in pounds, mass in slugs, and acceleration in feet per second squared. A similar equation describes the relationship between the torque applied to an object and its resulting angular acceleration. This relationship, called Newton S law ofrotation, is given by the equation 7" = Ja (1- 8) where 7" is the net applied torque in newton-meters or pound-feet and a is the re- sulting angular acceleration in radians per second squared. 1lle tenn J serves the same purpose as an object's mass in linear motion. It is called the moment of ineT1ia of the object and is measured in kilogram-meters squared or slug-feet squared. Calculation of the moment of inertia of an object is beyond the scope of this book. For infonnation about it see Ref. 2.
- 31. INTRODUCTION TO MACHINERY PRINCIPLES 7 Work W For linear motion, work is defined as the application of aforce through a distance. In equation fonn, W = fFdr (1- 9) where it is assumed that the force is coil inear with the direction of motion. For the special case of a constant force applied collinearly with the direction of motion, this equation becomes just W = Fr (1 -1 0) The units of work are joules in SI and foot-pounds in the English system. For rotational motion, work is the application of a torque through an angle. Here the equation for work is w ~ f,dO (I -II ) and if the torque is constant, W = TO ( 1-12) Power P Power is the rate of doing work, or the increase in work per unit time. The equa- tion for power is p = dW dt ( 1-13) It is usually measured in joules per second (watts), but also can be measured in foot-pounds per second or in horsepower. By this defmition, and assuming that force is constant and collinear with the direction of motion, power is given by p = dd~ = :r(Fr) = F(~;) = Fv ( 1-1 4) Similarly, assuming constant torque, power in rotational motion is given by p = dd~ = :r(TO) = T(~~) = TW p = TW ( 1-15) Equation (1-1 5) is very important in the study of electric machinery, because it can describe the mechanical power on the shaft of a motor or generator. Equation (1-I5) is the correct relationship runong power, torque, and speed if power is measured in watts, torque in newton-meters, and speed in radians per sec- ond. If other units are used to measure any of the above quantities, then a constant
- 32. 8 ELECTRIC MACHINERY FUNDAMENTALS must be intnx:luced into the equation for unit conversion factors. It is still common in U.S. engineering practice to measure torque in pound-feet, speed in revolutions per minute, and power in either watts or horsepower. If the appropriate conversion factors are included in each tenn, then Equation (I-IS) becomes T (lb-ft) n (r/min) P (watts) = 7.04 ( 1-16) P (h ) _ T (lb-ft) n (r/min) orsepower - 5252 ( 1-17) where torque is measured in pound-feet and speed is measured in revolutions per minute. 1.4 THE MAGNETIC FIELD As previously stated, magnetic fields are the fundrunental mechanism by which en- ergy is converted from one fonn to another in motors, generators, and transfonn- ers. Four basic principles describe how magnetic fields are used in these devices: I. A current-carrying wire produces a magnetic field in the area around it. 2. A time-changing magnetic field induces a voltage in a coil of wire ifit passes through that coil. (This is the basis of transfonner action.) 3. A current-carrying wire in the presence of a magnetic field has a force in- duced on it. (This is the basis of motor action.) 4. A moving wire in the presence of a magnetic field has a voltage induced in it. (This is the basis of generator action.) TIlis section describes and elaborates on the production of a magnetic field by a current-carrying wire, while later sections of this chapter explain the remaining three principles. Production of a Magnetic Field TIle basic law governing the production of a magnetic field by a current is Ampere's law: ( 1-18) where H is the magnetic field intensity produced by the current I""" and dl is a dif- ferential element of length along the path of integration. In SI units, I is measured in amperes and H is measured in ampere-turns per meter. To better understand the meaning of this equation, it is helpful to apply it to the simple example in Figure -3. Figure 1-3 shows a rectangular core with a winding of N turns of wire wrapped about one leg of the core. If the core is composed of iron or certain other similar metals (collectively calledferromagnefic mnterials), essentially all the magnetic field produced by the current will remain inside the core, so the path of integration in Ampere's law is the mean path length of the core (. TIle current
- 33. INTRODUCTION TO MACHINERY PRINCIPLES 9 Mean path length Ie FIGURE 1-3 A simple magnetic core. Nturns It~'1I--- Cross-sectional =.A passing within the path of integration I""" is then Ni, since the coil of wire cuts the path of integration Ntimes while carrying current i. Ampere's law thus becomes H( = Ni ( 1-19) Here H is the magnitude of the magnetic field intensity vector H. Therefore, the magnitude or the magnetic field intensity in the core due to the applied current is H =Ni Ie ( 1-20) The magnetic field intensity H is in a sense a measure of the "effort" that a current is putting into the establishment of a magnetic field. The strength of the magnetic field nux prOOuced in the core also depends on the material of the core. The relationship between the magnetic field intensity H and the resulting mag- netic flux density B produced within a material is given by ( 1-21) where H = magnetic field intensity /L = magnetic penneabi/ity of material B = resulting magnetic flux density prOOuced TIle actual magnetic flux density produced in a piece of material is thus given by a product of two tenns: H, representing the effort exerted by the current to establish a magnetic field /L, representing the relative ease of establishing a magnetic field in a given material
- 34. 10 ELECIRIC MACHINERY FUNDAMENTALS The units of magnetic field intensity are ampere-turns per meter, the units of per- meability are henrys per meter, and the units of the resulting flux density are webers per square meter, known as teslas (T). TIle penneability of free space is called J.Lo, and its value is /J.o = 47T X 10- 7 Him ( 1-22) TIle penneability of any other material compared to the penneability of free space is called its relative permeability: - " /L, - J.Lo ( 1-23) Relative penneability is a convenient way to compare the magnetizability of materials. For example, the steels used in modern machines have relative penne- abilities of 2000 to 6000 or even more. This means that, for a given amount of current, 2000 to 6000 times more flux is established in a piece of steel than in a corresponding area of air. (The penneability of air is essentially the same as the penneability of free space.) Obviously, the metals in a transformer or motor core play an extremely important part in increasing and concentrating the magnetic flux in the device. Also, because the permeability of iron is so much higher than that of air, the great majority of the flux in an iron core like that in Figure 1-3 remains inside the core instead of traveling through the surrounding air, which has much lower per- meability. The small leakage flux that does leave the iron core is very important in detennining the flux linkages between coils and the self-inductances of coils in transformers and motors. In a core such as the one shown in Figure 1-3, the magnitude of the flux density is given by ( 1-24) Now the total flux in a given area is given by (1-25a) where dA is the differential unit of area. If the flux density vector is perpendicu- lar to a plane of areaA, and if the fl ux density is constant throughout the area, then this equation reduces to (I - 25b) TIlUS, the total flux in the core in Figure 1-3 due to the current i in the wind- ing is ( 1-26) where A is the cross-sectional area of the core.
- 35. INTRODUCTION TO MACHINERY PRINCIPLES II - - I " v + ) - R + g= Ni - I v = - R ", ,b, FIGURE 1-4 (3) A simple electric cin:uit. (b) The magnetic circuit analog to a transformer COTe. Magnetic Circuits In Equation ( 1- 26) we see that the current in a coil ofwire wrapped around a core produces a magnetic nux in the core. This is in some sense analogous to a voltage in an electric circuit producing a current now. It is possible to define a "magnetic circuit" whose behavior is governed by equations analogous to those for an elec- tric circuit. The magnetic circuit model of magnetic behavior is often used in the design of electric machines and transfonners to simplify the otherwise quite com- plex design process. In a simple electric circuit such as the one shown in Figure 1-4a, the volt- age source V drives a current J around the circuit through a resistance R. The rela- tionship between these quantities is given by Ohm's law: V = JR In the electric circuit, it is the voltage or electromotive force that drives the cur- rent now. By analogy, the corresponding quantity in the magnetic circuit is called the magnetomotiveforce (mmI). The rnagnetomotive force of the magnetic circuit is equal to the effective current now applied to the core, or g=Ni ( 1- 27) where gis the symbol for magnetomotive force, measured in ampere-turns. Like the voltage source in the electric circuit, the magnetomotive force in the magnetic circuit has a polarity associated with it. The positive end of the mmf source is the end from which the nux exits, and the negative end of the mmf source is the end at which the nux reenters. The polarity of the mmf from a coil of wire can be detennined from a modification of the right-hand rule: If the fillgers of the right hand curl in the direction of the current now in a coil of wire, then the thumb will point in the direction of the positive rnrnf (see Figure 1- 5). In an electric circuit, the applied voltage causes a current J to flow. Simi- larly, in a magnetic circuit, the applied magnetomotive force causes nux <p to be produced. The relationship between voltage and current in an electric circuit is
- 36. 12 ELECIRIC MACHINERY FUNDAMENTALS / / ; rr='- N II ""GURE l-S Determining the polarity of a magnetomotive force source in a magnetic cirwit. Ohm's law (V = IR); similarly, the relationship between magnetomotive force and flux is where g = magnetomotive force of circuit <p = flux of circuit CR = reluctance of circuit ( 1- 28) 1lle reluctance of a magnetic circuit is the counterpart ofelectrical resistance, and its units are ampere-turns per weber. 1llere is also a magnetic analog of conductance. Just as the conductance of an electric circuit is the reciprocal of its resistance, the permeance cP of a magnetic circuit is the reciprocal of its rei uctance: ( 1- 29) The relationship belween rnagnetomotive force and flux can lhus be expressed as ( 1- 30) Under some circumstances, it is easier to work with the penneance of a magnetic circuit than with its reluctance.
- 37. INTRODUcnONTO MACHINERY PRINCIPLES 13 What is the reluctance of the core in Figure 1-3? The resulting flux in this core is given by Equation (1-26): (1-26) ( 1-31) By comparing Equation ( 1-31) with Equation ( 1-28), we see that the reluctance of the core is (1-32) Reluctances in a magnetic circuit obey the same rules as resistances in an electric circuit. TIle equivalent reluctance of a number of reluctances in series is just the sum of the individual reluctances: ( 1-33) Similarly, reluctances in parallel combine according to the equation (1-34) Penneances in series and parallel obey the same rules as electrical conductances. Calculations of the flux in a core performed by using the magnetic circuit concepts are always approximations-at best, they are accurate to within about 5 percent of the real answer. lllere are a number of reasons for this inherent inaccuracy: I. TIle magnetic circuit concept assumes that all flux is confilled within a mag- netic core. Unfortunately, this is not quite true. The permeability of a ferro- magnetic core is 2(x)() to 6()(x) times that of air, but a small fraction of the flux escapes from the core into the surrounding low-permeability air. This flux outside the core is called leakage flux, and it plays a very important role in electric machine design. 2. TIle calculation of reluctance assumes a certain mean path length and cross- sectional area for the core. These assumptions are not really very good, espe- cially at corners. 3. In ferromagnetic materials, the permeabi lity varies with the amount of flux already in the material. This nonJinear effect is described in detail. It adds yet another source oferror to magnetic circuit analysis, since the reluctances used in magnetic circuit calculations depend on the penneability of the material.
- 38. 14 ELECIRIC MACHINERY FUNDAMENTALS N s ""GURE 1-6 The fringing effect of a magnetic field at an air gap. Note the increased cross-sectional area of the air gap compared with the cross-sectional area of the metal. 4. If there are air gaps in the flux path in a core, the effective cross-sectional area of the air gap will be larger than the cross-sectional area of the iron core on either side. The extra effective area is caused by the "fringing effect" of the magnetic field at the air gap (Figure 1-6). It is possible to partially offset these inherent sources of error by using a "cor- rected" or "effective" mean path length and the cross-sectional area instead of the actual physical length and area in the calculations. TIlere are many inherent limitations to the concept of a magnetic circuit, but it is still the easiest design tool avai lable for calculating fluxes in practical ma- chinery design. Exact calculations using Maxwell's equations are just too diffi- cult, and they are not needed anyway, since satisfactory results may be achieved with this approximate method. TIle following examples illustrate basic magnetic circuit calculations. Note that in these examples the answers are given to three significant digits. Example I-t. A ferromagnetic core is shown in Figure 1-7a. Three sides of this core are of unifonn width. while the fourth side is somewhat thinner. The depth of the core (into the page) is 10 cm. and the other dimensions are shown in the figure. There is a 2()()'" turn coil wrapped around the left side of the core. Assruning relative penneability I-Lr of 2500. how much flux will be produced by a I-A input current? Solutio" We will solve this problem twice. once by hand and once by a MATLAB program. and show that both approaches yield the same answer. Three sides of the core have the same cross-sectional areas. while the fourth side has a different area. Thus. the core can be divided into two regions: (I) the single thiImer side and (2) the other three sides taken together. The magnetic circuit corresponding to this core is shown in Figure 1-7b.
- 39. INTRODUcnONTO MACHINERY PRINCIPLES 15 , , f--- 15 cm-_!-____ 30 em -----.,_10 cm_ , , , , , , -----t---------t-------;------t---T- 15 cm ----- ---- -----i------i--- -- ---- --- ; --- - N", 200 turns 30 em ---- ---- ----- ---- ----+-------~-- -- ---- --- '-----i---- I, -----i----' _____+-----+_ _ _-+----+____L 15 cm , , , , , , f--- 15 cm-_i-____ 30 = - - - ---i- 1O cm_ , , , (.j Depth = IOcm ; - + 'iJ( '" NO (b j FIGURE 1-7 (a) The ferromagnetic core of Example I- I. (b) The magnetic circuit corresponding to (a).
- 40. 16 ELECIRIC MACHINERY FUNDAMENTALS The mean path length of region I is 45 cm, and the cross-sectional area is 10 X 10 cm = 100 cm2. Therefore, the reluctance in the first region is cc~~-, O~ .4~ 5~ m~~c-cc = (25DOX47T X 10 7)(0.01m2) = 14,300 A ° turns/Wb (1- 32) The mean path length of region 2 is 130 cm, and the cross-sectional area is 15 X 10 cm = 150 cm2. Therefore, the reluctance in the second region is l2 l2 "" = _ = c:-'-, J.Ul.2 14 ~2 ccccccc-_lc·3 ~m~cccc-c~ = (25DOX47T X 10 7)(0.015 m2) = 27,600 A ° turns/Wb Therefore, the total reluctance in the core is ~ = 'l:!l + 'l:!2 = 14,3DO A ° tumslWb + 27,600 A ° tlUlls/Wb = 41,900 A ° tumslWb The total magnetomotive force is g = Ni = (2DO turnsXI.OA) = 200 A ° turns The total flux in the core is given by g 200 A ° tlUllS cp = CR = 4 1,900 A otlUllsl Wb = 0.0048 Vb (1- 32) This calculation can be perfonned by using a MATLAB script file, if desired. A sim- ple script to calculate the flux in the core is shown below. !l; M -file : exl _ 1.m !l; M -file t o ca l c ula t e the fl ux 11 0 .4 5; i n Exa mp l e 1-1 . 12 1. 3; "' 0 . 01 ; "' 0 . 01 5 ; or 2500; 0 0 4* p i * l E- 7; 0 200; i ~ " !l; Ca l c ula t e the fi r s t r e l u c t a n ce rl = 11 I (ur * u O * a l ) ; d i sp ( ['rl = ' num2str (rl ) I ) ; !l; Ca l c ula t e the second r e l u c t a n ce r 2 = 12 I (ur * u O * a2 ) ; d i sp ( ['r 2 = ' num2str (r 2 ) I ) ; % Le ng t h of r egi on 1 • • • • • • • Le ng t h of r egi on 2 Ar ea of r egi on 1 Ar ea of r egi on 2 Re l a t i ve permeabili t y Pe rmeabili t y of f r ee space Number of turns on cor e CUrre nt in amps
- 41. INTRODUCTION TO MACHINERY PRINCIPLES 17 % Ca l c u l a t e the t ot a l r e l uc t a nce rt ot = rl + r 2; % Ca l c u l a t e the rumf rumf= n * i ; % Fina lly, get the fl ux i n the co r e f l ux = rumf I rt ot ; % Di sp l ay r esult d i sp ( [ 'Flux = ' num2str ( fl ux ) I ) ; When this program is executed, the results are: ,. e1:1_ 1 rl = 14323.9 44 9 r 2 = 27586 , 8568 Flux = 0 , 00 4772 This program produces the same answer as our hand calculations to the number of signifi- cant digits in the problem. Example 1-2. Figure 1- 8a shows a ferromagnetic core whose mean path length is 40 cm. There is a small gap of 0.05 cm in the structure of the otherwise whole core. The cross-sectional area of the core is 12 cm2, the relative permeability of the core is 4(x)(), and the coil of wire on the core has 400 turns. Assmne that fringing in the air gap increases the effective cross-sectional area of the air gap by 5 percent. Given this infonnation, find (a) the total reluctance of the flux path (iron plus air gap) and (b) the current required to produce a flux density of 0.5 T in the air gap. Solutioll The magnetic circuit corresponding to this core is shown in Figure 1-8b. (a) The reluctance of the core is I, 1 < "" = - = J.Ul.< /.tr IJ.ty'I.< (1- 32) cccccc--cO c·4 ~m~=ccc-cc = (4000X47T X 10 7XO.OO2 m2) = 66,300 A • turns/Wb The effecti ve area of the air gap is 1.05 X 12 cm2 = 12.6 cm2, so the reluctance of the air gap is I. "'" = - wi, 0.0005 m = C (4C~--C X~1~Oa'X~O~.OO ~1~ 276~ m "') = 316,OOO A · turns/Wb (1- 32)
- 42. 18 ELECIRIC MACHINERY FUNDAMENTALS N=400 turns g(=Ni) fo'IGURE 1-8 + ,b, " J 1-0.05 em T A=12cm2 CJ;>.. (Reluctance of core) CJ:.>., (Reluctance of air gap) (a) The ferromagnetic core of Example 1- 2. (b) The magnetic circuit corresponding to (a). Therefore, the total reluctance of the flux path is 1l!... = CJ:l" + CQ" = 66,300 A· turns/Wb + 3 16,OOO A· turnsIWb = 382,300 A • tumslWb Note that the air gap contributes most of the reluctance even though it is 800 times shorter than the core. (b) Equation (1 - 28) states that Since the flux cp = BA and 'if = Ni, this equation becomes Ni = BACl! (1 - 28)
- 43. INTRODUcnONTO MACHINERY PRINCIPLES 19 . BAct! ,=-- N (0.5 D(0.00126 m2)(383,200 A ° tlU1lsi Wb) = 400 turns = 0.602 A Notice that, since the air-gap flux was required, the effective air-gap area was used in the above equation. EXllmple 1-3. Figure 1-9a shows a simplified rotor and stator for a dc motor. The mean path length of the stator is 50 cm, and its cross-sectional area is 12 cm2. The mean path length of the rotor is 5 em, and its cross-sectional area also may be assruned to be 12 cm2. Each air gap between the rotor and the stator is 0.05 cm wide, and the cross- sectional area of each air gap (including fringing) is 14 cm2 . The iron of the core has a rel- ative penneability of 2()(x), and there are 200 turns of wire on the core. If the current in the wire is adjusted to be I A, what will the resulting flux density in the air gaps be? Solutioll To detennine the flux density in the air gap, it is necessary to first calculate the magneto- motive force applied to the core and the total reluctance of the flux path. With this infor- mation, the total flux in the core can be found. Finally, knowing the cross-sectional area of the air gaps enables the flux density to be calculated. The reluctance of the stator is ---"" c- 'll,= /-tr IJ.QII ccccccc-~OC·50m",ccc~o-" = (2000X47T X 10 7)(0.0012 m2) = 166,oooA o tlUllsIWb The reluctance of the rotor is ---""~ 'll,= /-tr IJ.QIr O.05m = C(2;;;()()()=X:C4C-~-;X""'lo;O'i'C;;)(;;CO.;;:OO;;;1;;:2C:m:h') = 16,600A o tumslWb The reluctance of the air gaps is '. "'" = ---'"c- /-tr IJ.QI. 0.0005 m = C(1~ )( ~4- ~~X~10~ 'X~O ~.OO~1~4- m" ) = 284,000 A ° tlUllsIWb The magnetic circuit corresponding to this machine is shown in Figure 1-9b. The total re- luctance of the flux path is thus
- 44. 20 ELECIRIC MACHINERY FUNDAMENTALS ~I~I,="m Ic=50cm ,., Stator reluctance ,b, ""GURE 1-9 (a) A simplified diagram of a rotor and stator for a de motor. (b) The magnetic circuit corresponding to (a). CJ:l..q = Ci:!, + CJ:!"t + Ci:!, + Ci:!~2 = 166,000 + 284,000 + 16,600 + 284,000 A • tumslWb = 75 1,ooo A · turns/Wb The net magnetomotive force applied to the core is g = Ni = (200 turnsXI.OA) = 200 A • turns Therefore, the total fl ux in the core is
- 45. INTRODUCTION TO MACHINERY PRINCIPLES 21 g 200 A • turns cp = CR = 751.0CXl A • turnsJ Vb = 0.cXl266 Wb Finally, the magnetic flux density in the motor's air gap is B = cp = 0.000266 Wb A 0.0014 m2 0.19T Magnetic Behavior of Ferromagnetic Materials Earlier in this section, magnetic permeability was defined by the equation ( 1-21) It was explained that the penneability of ferromagnetic materials is very high, up to 6(X)Q times the penneability of free space. In that discussion and in the examples that followed, the penneability was assumed to be constant regardless of the mag- netomotive force applied to the material. Although permeability is constant in free space, this most certainly is not true for iron and other ferromagnetic materials. To illustrate the behavior of magnetic penneability in a ferromagnetic ma- terial, apply a direct current to the core shown in Figure 1-3, starting with 0 A and slowly working up to the maximum permissible current. When the flux prOOuced in the core is plotted versus the magnetomotive force producing it, the resulting plot looks like Figure I-lOa. lllis type of plot is called a saturation curoe or a magnetization culVe. At first , a small increase in the magnetomotive force pro- duces a huge increase in the resulting flux. After a certain point, though, further increases in the magnetomotive force produce relatively smaller increases in the flux. Finally, an increase in the magnetomotive force produces almost no change at all. The region of this figure in which the curve flattens out is called the satu- ration region, and the core is said to be saturated. In contrast, the region where the flux changes very rapidly is calJed the unsaturated region of the curve, and the core is said to be unsaturated. The transition region between the unsaturated re- gion and the saturated region is sometimes called the knee of the curve. Note that the flux produced in the core is linearly related to the applied magnetomotive force in the unsaturated region, and approaches a constant value regardless of magnetomotive force in the saturated region. Another closely related plot is shown in Figure I-lOb. Figure I-lOb is a plot of magnetic flux density B versus magnetizing intensity H. From Equations (1- 20) and (I- 25b), Ni g H ~ - ~ - ( Ie '" ~ BA ( 1-20) (I- 25b) it is easy to see that magnetizing intensity is directly proP011io1UJi to magnetomotive force and magnetic flux density is directly propoT1ional to flux for any given core. Therefore, the relationship between B and H has the same shape as the relationship
- 46. 22 ELECIRIC MACHINERY FUNDAMENTALS ..p.Wb B. T 2.8 2.6 2.4 2.2 2.0 E 1.8 • .~ 1.6 ~ 1.4 1.0 0.8 0.6 0.4 0.2 o 0 (a) 20 30 40 50 F. A · turns v / 100 200 300 500 1000 Magnetizing iotensity H. A · turnslm "I ""GURE - 10 H.A · turnslm ,b, 2000 5000 (a) Sketch of a dc magnetization curve for a ferromagnetic core. (b) The magnetization curve expressed in terms of flux density and magnetizing intensity. (c) A detailed magnetization curve for a typical piece of steel. (d) A plot of relative permeability /J., as a function of magnetizing intensity H for a typical piece of steel.
- 47. 7(XX) 2(xx) ](XX) o 10 FIGURE - 0 (continued) INTRODUCTION TO MACHINERY PRINCIPLES 23 / ~ ' ' ' 20 30 40 50 ]00 200 Magnetizing intensity H (A ' turnslm) I" "- 300 '" i'- 500 between flux and magnetomotive force. The slope of the curve of flux density ver- sus magnetizing intensity at any value of H in Figure I- I Db is by definition the per- meability ofthe core at that magnetizing intensity. The curve shows that the penne- ability is large and relatively constant in the unsaturated region and then gradually drops to a very low value as the core becomes heavily saturated. 1000 Figure I- IDe is a magnetization curve for a typical piece of steel shown in more detail and with the magnetizing intensity on a logarithmic scale. Only with the magnetizing intensity shown logarithmically can the huge saturation region of the curve fit onto the graph. The advantage of using a ferromagnetic material for cores in electric ma- chines and transfonners is that one gets many times more flux for a given magne- tomotive force with iron than with air. However, if the resulting flux has to be pro- portional, or nearly so, to the applied magnetomotive force, then the core must be operated in the unsaturated region of the magnetization curve. Since real generators and motors depend on magnetic flux to produce volt- age and torque, they are designed to produce as much flux as possible. As a result, most real machines operate near the knee of the magnetization curve, and the flux in their cores is not linearly related to the magnetomotive force producing it. This
- 48. 24 ELECIRIC MACHINERY FUNDAMENTALS nonlinearity accounts for many of the peculiar behaviors of machines that will be explained in future chapters. We will use MATLAB to calculate solutions to prob- lems involving the nonlinear behavior of real machines. Example 1-4. Find the relative penneability of the typical ferromagnetic material whose magnetization curve is shown in Figure l-lOc at (a) H = 50. (b) H = 100. (c) H = 500. and (d) H = 1000 A ° turns/m. Solutio" The penneability of a material is given by and the relative permeability is given by IJ- = B H (1- 23) Thus. it is easy to detennine the penneability at any given magnetizing intensity. (a) AtH = 50A otums/m.B = 0.25T. so B 0.25 T IJ- = H = 50 A ° turns/m = O.OO5Q Him = ~ = 0.0050 HIm = 3980 IL, f.1.O 47T X 10 7Hhn (b) At H = lOOA ° turns/m. B = 0.72 T. so B 0.72 T IJ- = H = 100 A ° turns/m = 0.0072 Him = ~ = 0.0072 HIm = 5730 IJ-, f.1.O 47T X 10 7Hhn (c) AtH = 500 A ° turns/m.B = I.40 T, so B 1.40 T IJ- = H = 500 A ° turns/m = 0.0028 Him = ~ = 0.0028 HIm = 2230 IJ-, f.1.O 47T X 10 7Hhn (d) AtH = lOOOA oturns/m,B = 1.51 T, so B 1.51 T IJ- = H = 1000 A ° turns/m = 0.00151 Him = ~ = 0.00151 HIm = 1200 IJ-, f.1.O 47T X 10 7Hhn
- 49. INTRODUCTION TO MACHINERY PRINCIPLES 25 Notice that as the magnetizing intensity is increased, the relative penne- ability first increases and then starts to drop off. The relative permeability of a typ- ical ferromagnetic material as a function of the magnetizing intensity is shown in Figure 1-lOd. This shape is fairly typical of all ferromagnetic materials. It can easi ly be seen from the curve for /L. versus H that the assumption of constant rel- ative penneability made in Examples 1-1 to -3 is valid only over a relatively narrow range of magnetizing intensities (or magnetomotive forces). In the following example, the relative penneability is not assumed constant. Instead, the relationship between Band H is given by a graph. Example 1-5. A square magnetic core has a mean path length of 55 cm and a cross- sectional area of 150 cm2. A2()()...tum coil of wire is wrapped arOlUld one leg of the core. The core is made of a material having the magnetization curve shown in Figure l-lOc. (a) How much current is required to produce 0.012 Wb of flux in the core? (b) What is the core's relative permeability at that current level? (c) What is its reluctance? Solutioll (a) The required flux density in the core is B = q, = 1.012 Wb = 0.8T A 0.015 m2 From Figure l-lOc, the required magnetizing intensity is H = 115 A" turns/m From Equation (1-20), the magnetomotive force needed to produce this magnetizing in- tensity is q= Ni = Hlc = (115 A" turns/mXD.55 m) = 63.25 A" turns so the required current is i = q = 63.25 A" turns = 0.316A N 200 turns (b) The core's permeability at this current is B 0.8T I.L = H = liS A "turnshn = 0.00696 Him Therefore, the relative permeability is I.L 0.00696 Him I.Lr = 1.1.0 = 47T X 10 7 Him 5540 (c) The reluctance of the core is tT'I = q = 63.25 A" turns = 5270 A. _..~ -'" q, 0.012Wb turn", .. u
- 50. 26 ELECIRIC MACHINERY FUNDAMENTALS i(t) Residual flux -------= ,., ,,' ¢ (or 8) b --- --------.,f-J~ "'I_-------- 'J(or H) ,b, ""GURE I- II The hysteresis loop traced out by the flux in a core when the current i{l) is applied to it. Energy Losses in a Ferromagnetic Core Instead of applying a direct current to the windings on Ihe core, let us now apply an alternating currenl and observe what happens. TIle currenl to be applied is shown in Figure I- ila. Assume that the flux in the core is initially zero. As the current increases for the first time, the flux in the core traces out path ab in Figure I- lib. lllis is basically the saturation curve shown in Figure 1- 10. However, when Ihe current falls again, thef1ux traces out a different pathfrom the one itfol- lowed when the current increased. As the current decreases, the flux in the core traces out path bcd, and later when the current increases again, the flux traces out path deb. Notice that the amount of flux present in Ihe core depends nol only on Ihe amount of current applied to the windings of the core, but also on the previous history of the flux in Ihe core.lllis dependence on the preceding flux history and the resulting failure to retrace flux paths is called hysteresis. Path bcdeb traced out in Figure I- II b as the applied current changes is called a hysteresis loop.
- 51. INTRODUCTION TO MACHINERY PRINCIPLES 27 - /' '- I - - - - - - - - I X I - X - I - - / ' - --- - - I - /' "- X /' - - - - - ' - - "- /' - " - - --- " - / X / I • • - - -.. - /' --- - 1 I I " t / - - - -.. - " ,b, FIGURE 1-12 (a) Magnetic domains oriented randomly. (b) Magnetic domains lined up in the presence of an external magnetic field. - - /' - --- -.. Notice that if a large magnetomolive force is first applied to the core and then removed, the flux path in the core will be abc, When the magnetomotive force is removed, the flux in the core does not go to zero. Instead, a magnetic field is left in the core. This magnetic field is called the residual flux in the core. It is in precisely this manner that pennanent magnets are produced. To force the flux to zero, an amount of magnetomotive force known as the coercive magnetomotive force '(tc must be applied to the core in the opposite direction. Why does hysteresis occur? To understand the behavior of ferromagnetic materials, it is necessary to know something about their structure. TIle atoms of iron and similar metals (cobalt, nickel, and some of their alloys) tend to have their magnetic fields closely aligned with each other. Within the metal, there are many small regions called domnins, In each domain, all the atoms are aligned with their magnetic fields pointing in the same direction, so each domain within the material acts as a small permanent magnet. The reason that a whole block of iron can ap- pear to have no flux is that these numerous tiny domains are oriented randomly within the material. An example of the domain structure within a piece of iron is shown in Figure 1- 12. When an external magnetic field is applied to this block of iron, it causes do- mains that happen to point in the direction of the field to grow at the expense of domains pointed in other directions. Domains pointing in the direction of the mag- netic field grow because the atoms at their boundaries physically switch orientation to align themselves with the applied magnetic field. The extra atoms aligned with the field increase the magnetic nux in the iron, which in turn causes more atoms to switch orientation, further increasing the strength of the magnetic field. It is this pos- itive feedback effect that causes iron to have a penneabiJity much higher than air. As the strength of the external magnetic field continues to increase, whole domains that are aligned in the wrong direction eventually reorient themselves as
- 52. 28 ELECIRIC MACHINERY FUNDAMENTALS a unit to line up with the field. Finally, when nearly all the atoms and domains in the iron are lined up with the external field, any further increase in the magneto- motive force can cause only the same flux increase that it would in free space. (Once everything is aligned, there can be no more feedback effect to strengthen the field.) At this point, the iron is saturated with flux. This is the situation in the saturated region of the magnetization curve in Figure 1-10. TIle key to hysteresis is that when the external magnetic field is removed, the domains do not completely randomize again. Why do the domains remain lined up? Because turning the atoms in them requires energy. Originally, energy was provided by the external magnetic field to accomplish the alignment; when the field is removed, there is no source of energy to cause all the domains to rotate back. The piece of iron is now a pennanent magnet. Once the domains are aligned, some of them wi ll remain aligned until a source of external energy is supplied to change them. Examples of sources of ex- ternal energy that can change the boundaries between domains and/or the align- ment of domains are magnetomotive force applied in another direction, a large mechanical shock, and heating. Any of these events can impart energy to the do- mains and enable them to change alignment. (It is for this reason that a permanent magnet can lose its magnetism if it is dropped, hit with a hammer, or heated.) TIle fact that turning domains in the iron requires energy leads to a common type ofenergy loss in all machines and transfonners. The hysteresis loss in an iron core is the energy required to accomplish the reorientation of domains during each cycle of the alternating current applied to the core. It can be shown that the area enclosed in the hysteresis loop formed by applying an alternating current to the core is directly proportional to the energy lost in a given ac cycle. The smaller the applied magnetomotive force excursions on the core, the smaller the area of the resulting hysteresis loop and so the smaller the resulting losses. Figure 1-1 3 illustrates this point. Another type of loss should be mentioned at this point, since it is also caused by varying magnetic fields in an iron core. This loss is the eddy current loss. The mechanism of eddy current losses is explained later after Faraday's law has been introduced. Both hysteresis and eddy current losses cause heating in the core material, and both losses must be considered in the design of any machine or transformer. Since both losses occur within the metal of the core, they are usually lumped together and called core losses. 1.5 FARADAY'S LAW-INDUCED VOLTAGE FROM A TIME-CHANGING MAGNETIC FIELD So far, attention has been focused on the pnxluction of a magnetic field and on its properties. It is now time to examine the various ways in which an existing mag- netic field can affect its surroundings. TIle first major effect to be considered is called Faraday slaw. It is the ba- sis of transfonner operation. Faraday's law states that if a flux passes through a
- 53. INTRODUCTION TO MACHINERY PRINCIPLES 29 ¢ (or 8) , , -' " -----,Hf,',fr-!,f-f------- 'J(or H) , FIGURE -13 , , Area c< hysteresis loss The elTect of the size of magoetomotive force excursions on the magnitude of the hysteresis loss. turn of a coil of wire, a voltage will be induced in the turn of wire that is directly proportional to the rate of change in the flux with respect to time. In equation fonn, (1-35) where eind is the voltage induced in the turn of the coil and <P is the flux passing through the turn. If a coil has N turns and if the same flux passes through all of them, then the voltage induced across the whole coil is given by where eioo = voltage induced in the coil N = number of turns of wire in coil <p = nux passing through coil (1-36) The minus sign in the equations is an expression of Lenz slaw. Lenz's law states that the direction of the voltage buildup in the coil is such that if the coil ends were short circuited, it would produce current that would cause a flux opposing the original nux change. Since the induced voltage opposes the change that causes it, a minus sign is included in Equation ( 1-36). To understand this concept clearly,
- 54. 30 ELECIRIC MACHINERY FUNDAMENTALS Direction of j required ; + '00 ( N turns I. (. ) ""GURE 1- 14 , , • I· (b) Direction of opposing flux ¢ increasing The meaning of Lenz's law: (a) Acoil enclosing an increasing magnetic flux: (b) determining the resulting voltage polarity. examine Figure 1- 14. If the nux shown in the figure is increasing in strength, then the voltage built up in the coil will tend to establish a flux that will oppose the in- crease. A current flowing as shown in Figure 1-1 4b would produce a nux oppos- ing the iflcrease, so the voltage Ofl the coil must be built up with the polarity re- quired to drive that current through the external circuit. 1l1erefore, the voltage must be buill up with Ihe polarity shown in the figure. Since the polarity of the re- sulting voltage can be detennined from physical considerations, Ihe minus sign in Equalions (1- 35) and (1- 36) is often left out. It is left out of Faraday's law in the remainder of this book. 1l1ere is one major difficulIy involved in using Equation (1- 36) in practical problems. That equation assumes that exactly Ihe same flux is present in each tum of the coil. Unfortunately, the flux leaking oUI of the core inlo the surrounding air prevents this from being lrue. If the windings are tightly coupled, so that the vast majority of the flux passing through one turn of the coil does indeed pass through all of them, then Equation ( 1- 36) will give valid answers. Bul if leakage is quite high or if extreme accuracy is required, a different expression that does not make that assumption will be needed. The magnitude of the voliage in the ith tum of the coil is always given by If there are N turns in the coiI of wire, the total vollage on the coil is N eind = ~ ei i - I ( 1- 37) ( 1- 38)
- 55. INTRODUcnONTO MACHINERY PRINCIPLES 31 (1-39) ( 1-40) The term in parentheses in Equation (1--40) is called the flux linkage Aof the coil, and Faraday's law can be rewritten in terms of flux linkage as ( 1-41) where ( 1-42) The units of flux linkage are weber-turns. Faraday's law is the fundamental property of magnetic fields involved in transformer operation. The effect of Lenz's law in transformers is to predict the polarity of the voltages induced in transformer windings. Faraday's law also explains the eddy current losses mentioned previously. A time-changing flux induces voltage within a ferromagnetic core in just the same manner as it would in a wire wrapped around that core. TIlese voltages cause swirls of current to fl ow within the core, much like the eddies seen at the edges of a river. It is the shape of these currents that gives rise to the name eddy currents. These eddy currents are fl owing in a resistive material (the iron of the core), so energy is dissipated by them. The lost energy goes into heating the iron core. The amount of energy lost to eddy currents is proportional to the size of the paths they follow within the core. For this reason, it is customary to break up any ferromagnetic core that may be subject to alternating fluxes into many small strips, or laminntions, and to build the core up out of these strips. An insulating oxide or resin is used between the strips. so that the current paths for eddy currents are limited to very small areas. Because the insulating layers are extremely thin, this action reduces eddy current losses with very little effect on the core's mag- netic properties. Actual eddy current losses are proportional to the square of the lamination thickness, so there is a strong incentive to make the laminations as thin as economically possible. EXllmple 1-6. Figure 1-15 shows a coil of wire wrapped around an iron core. If the flux in the core is given by the equation cp = 0.05 sin 377t Wb If there are 100 turns on the core. what voltage is produced at the terminals of the coil? Of what polarity is the voltage during the time when flux is increasing in the reference
- 56. 32 ELECIRIC MACHINERY FUNDAMENTALS /' / Required direction of i ; - + N= 100 turns Opposing ~ .- H, t I / ~ _ 0.05 Sin 3771 Wb ""GURE 1-15 The core of Example 1--6. Determination of the voltage polarity at the terminals is shown. direction shown in the figure? Asswne that all the magnetic flux stays within the core (i.e., assume that the flux leakage is zero). Solutioll By the same reasoning as in the discussion on pages 29- 30, the direction of the voltage while the flux is increasing in the reference direction must be positive to negative, as shown in Figure 1- 15. The magnitude of the voltage is given by or alternatively, d~ e;"" = NYt = (100 turns) :, (0.05 sin 377t) = 1885 cos 377t eiod = 1885 sin(377t + goO) V 1.6 PRODUCTION OF INDUCED FORCE ONAWIRE A second major effect of a magnetic field on its surroundings is that it induces a force on a current-carrying wire within the field. The basic concept involved is il- lustrated in Figure 1- 16. The figure shows a conductor present in a unifonn mag- netic field of flux density D, pointing into the page. 1lle conductor itself is I me- ters long and contains a current of i amperes. The force induced on the conductor is given by F = i(I X D) ( 1-43)
- 57. INTRODUCTION TO MACHINERY PRINCIPLES 33 , , - , , " J , , , , , , , , , , I , , , , " , I , , J , , FlGURE 1- 16 h Acurrent-carrying wire in the presence of3. , , - , , magnetic field. where i = magnitude of current in wire I = length of wire, with direction of I defined to be in the direction of current flow B = magnetic flux density vector The direction of the force is given by the right-hand rule: If the index finger of the right hand points in the direction of the vector I and the middle finger points in the direction of the flux density vector B, then the thumb points in the direction of the resultant force on the wire. TIle magnitude of the force is given by the equation F = UR sin () ( 1-44) where () is the angle between the wire and the flux density vector. Example 1-7. Figure 1- 16 shows a wire carrying a current in the presence ofa magnetic field. The magnetic flux density is 0.25 T. directed into the page. If the wire is 1.0 m long and carries 0.5 A ofcurrent in the direction from the top of the page to the bot- tom of the page. what are the magnitude and direction of the force induced on the wire? Solutioll The direction of the force is given by the right-hand rule as being to the right. The magni- tude is given by F = ilB sin (J (1-44) = (0.5 AXI.O m)(0.25 T) sin 90° = 0.125 N Therefore. F = 0.125 N. directed to the right The induction of a force in a wire by a current in the presence of a magnetic fie ld is the basis of motor action. Almost every type of motor depends on this basic principle for the forces and torques which make it move.
- 58. 34 ELECIRIC MACHINERY FUNDAMENTALS 1.7 INDUCED VOLTAGE ON A CONDUCTOR MOVING IN A MAGNETIC FIELD There is a third major way in which a magnetic field interacts with its surround- ings. If a wire with the proper orientation moves through a magnetic field, a volt- age is induced in it. TIlis idea is shown in Figure -7. TIle voltage induced in the wire is given by eiD<! = (v X B) • I ( 1-45) where v = velocity of the wire B = magnetic nux density vector I = length of conductor in the magnetic field Vector I points along the direction of the wire toward the end making the smallest angle with respect to the vector v X B. The voltage in the wire will be built up so that the positive end is in the direction of the vector v X B. TIle following exam- ples illustrate this concept. Example 1-8. Figure 1-17 shows a conductor moving with a velocity of 5.0 m1s to the right in the presence of a magnetic field. The flux density is 0.5 T into the page, and the wire is 1.0 m in length, oriented as shown. What are the magnitude and polarity of the resulting induced voltage? Solutio" The direction of the quantity v X B in this example is up. Therefore, the voltage on the con- ductor will be built up positive at the top with respect to the bottom of the wire. The direc- tion of vector I is up, so that it makes the smallest angle with respect to the vector ' X B. Since ' is perpendicular to B and since v X B is parallel to I, the magnitude of the induced voltage reduces to ejmd = (' X B) ·I = (vB sin 90°) I cos 0° = vBI = (5.0 mls)(0.5 TXI.O m) = 2.5 V Thus the induced voltage is 2.5 V, positive at the top of the wire. (1---45) Example 1-9. Figure 1-18 shows a conductor moving with a velocity of 10 m1s to the right in a magnetic field. The flux density is 0.5 T, out of the page, and the wire is 1.0 m in length, oriented as shown. What are the magnitude and polarity of the resulting induced voltage? Solutioll The direction of the quantity v X B is down. The wire is not oriented on an up-down line, so choose the direction of I as shown to make the smallest possible angle with the direction
- 59. INTRODUCTION TO MACHINERY PRINCIPLES 35 x x + ~ ,:XB x • ++ x x x x ,~ x X I X X , X X X X Jo'IGURE 1- 17 X X - '" X X A conductor moving in the presence of a magnetic field. • • • • II • • • • 3<1' • I • • vX II • • • Jo'IGURE 1- 18 The conductor of Example 1--9. of ' X B. The voltage is positive at the bottom of the wire with respect to the top of the wire.The magnitude of the voltage is eiod = (' X B) ' I = (vB sin 90°) l cos 30° = (10.0 m1sX0.51)(I.Om) cos 30° = 4.33 V (1--45) The induction of voltages in a wire moving in a magnetic field is funda- mental to the operation of all types of generators. For this reason, it is called gen- erator action.
- 60. 36 ELECIRIC MACHINERY FUNDAMENTALS Switch Magnetic field into page R 1 X X X + -=- VB e ;00 X X X ""GURE 1-19 A linear dc machine. The magnetic field points into the page. 1.8 THE LINEAR DC MACHINE- A SIMPLE EXAMPLE • A linear dc machine is about the simplest and easiest-to-understand version of a dc machine, yet it operates according to the same principles and exhibits the same behavior as real generators and motors. It thus serves as a good starting point in the study of machines. A linear dc machine is shown in Figure - 9. It consists of a battery and a resistance connected through a switch to a pair of smooth, frictionless rails. Along the bed of this "railroad track" is a constant, uniform-density magnetic field di- rected into the page. A bar of conducting metal is lying across the tracks. How does such a strange device behave? Its behavior can be determined from an application of four basic equations to the machine. These equations are I. The equation for the force on a wire in the presence ofa magnetic field: F i(I X B) I ( 1- 43) where F = force on wire i = magnitude of currenl in wire I = length of wire, with direction of! defined to be in the direction of current now B = magnetic nux density vector 2. The equation for the voltage induced on a wire moving in a magnelic field: l e;Dd (vXB) -1 where e;Dd = voltage induced in wire v = velocity of the wire B = magnetic nux density vector I = length of conductor in the magnelic field ( 1- 45)
- 61. INTRODUCTION TO MACHINERY PRINCIPLES 37 ~o R • , x x X i (/) + - Find -=- VB e;!!d ,. X X X FIGURE 1-10 Starting a linear dc machine. 3. Kirchhoff's voltage law for Ihis machine. From Figure J- J9lhis law gives VB - iR - ei!!d = 0 ( 1- 46) 4. Newton's law for the bar across the tracks: (1- 7) We will now explore the fundamental behavior of this simple dc machine using these four equations as lools. Starting the Linear DC Machine Figure 1- 20 shows the linear dc machine under starting conditions. To start this machine, simply close the switch. Now a currenl fl ows in the bar, which is given by Kirchhoff's voltage law: ( 1- 47) Since the bar is initially at rest, e;Dd = 0, so i = VBIR. The current flows down through the bar across the tracks. But from Equation (1- 43), a currenl flowing Ihrough a wire in the presence ofa magnetic field induces a force on Ihe wire. Be- cause of the geometry of the machine, this force is Find = ilB to the right ( 1- 48) Therefore, the bar will accelerale to the right (by Newton's law). However, when Ihe velocity of the bar begins 10 increase, a voltage appears across the bar. The voltage is given by Equation (1 - 45), which reduces for this geometry to e;Dd = vBI positive upward ( 1- 49) The voliage now reduces the current fl owing in the bar, since by Kirch- hoff's voliage law
- 62. 38 ELECIRIC MACHINERY FUNDAMENTALS v(t) V, BI o e;oo (t) V, o i (t) V, R o F;oo (t) VBIB R o (a) 'hI "I ,dl ""GURE 1-21 The linear de machine on starting. (a) Velocity v(t) as a function of time; (b) induced voltage e....(/); (c) currem i(t); (d) induced force Fu.il). ( 1-47) As eioo increases, the current i decreases. TIle result of this action is that eventually the bar wi ll reach a constant steady-state speed where the net force on the bar is zero. TIlis will occur when eioo has risen all the way up to equal the voltage VB. At that time, the bar will be mov- ing at a speed given by VB = e;Dd = v" BI V, v'" = BI ( I-50) TIle bar will continue to coast along at this no-load speed forever unless some ex- ternal force disturbs it. When the motor is started, the velocity v, induced voltage eiDd , current i, and induced force Find are as sketched in Figure 1-21. To summarize, at starting, the linear dc machine behaves as follows: I, Closing the switch produces a current flow i = VB /R. 2, The current flow produces a force on the bar given by F = ilB.
- 63. INTRODUCTION TO MACHINERY PRINCIPLES 39 R " i (/) X X X + - F;md _ fo'_ e;md I , X X X FIGURE 1-22 The linear dc machine as a motor. 3. The bar accelerales to the right, producing an induced voltage e;md as it speeds up. 4. lllis induced voltage reduces the current flow i = (VB - e;Dd t)!R. 5. The induced force is thus decreased (F = i J.IB) until eventually F = o. At that point, e;Dd = VB, i = 0, and the bar moves at a constant no-load speed v" = VB ! BI. This is precisely the behavior observed in real motors on starting. The Linear DC Machine as a Molor Assume that the linear machine is initial ly running at the no-load steady-state con- ditions described above. What will happen to this machine if an external load is applied to it? To find out, let's examine Figure 1- 22. Here, a force Fto.d is applied to the bar opposite the direction of motion. Since the bar was initially at steady state, application of the force Ftoad will result in a net force on the bar in the direc- tion opposite the direction of motion (FDel = Fto.d - Find). The effect of this force will be to slow the bar. But just as soon as the bar begins to slow down, the in- duced voltage on the bar drops (e;Dd = vJ.BI). As the induced voltage decreases, the current flow in the bar rises: ( 1- 47) 1l1erefore, the induced force rises too (Find = itIB). The overall result of this chain of events is that the induced force rises until it is equal and opposite to the load force, and the bar again travels in steady state, but at a lower speed. When a load is attached to the bar, the velocity v, induced voltage eind, current i, and in- duced force Find are as sketched in Figure 1- 23. 1l1ere is now an induced force in the direction of motion of the bar, and power is being convenedJrom electricalfonn to mechanicalJonn to keep the bar moving. The power being converted is
- 64. 40 ELECIRIC MACHINERY FUNDAMENTALS v(I) V, BI o eiOO (I) V, o j (I) F BI o FiOO (I) F~ o '" 'hI '<I ,dl ""GURE 1- 23 The linear de machine operating at no-load conditions and then loaded as a motor. (a) Velocity '01(1) as a function of time; (b) induced voltage e..JI); (c) current i(I); (d) induced force Fu.il). ( I-51 ) An amount of electric power equal to eindi is consumed in the bar and is replaced by mechanical power equal to Findv, Since power is converted from electrical 1 0 mechanical form, this bar is operating as a motor. To summarize this behavior: I. A force ~oad is applied opposite to the direction of motion, which causes a net force F..., opposite 1 0 the direction of motion. 2. The resulting acceleration a = Fo.. /m is negalive, so the bar slows down (vJ.) . 3. The voltage eiod = vJ.Bl falls, and so i = (VB - eiodJ.YR increases. 4. The induced force Fiod = itlB increases until IFiod I = IFtoad I at a lower speed v. 5. An amount of electric power equal to eiodi is now being converted to me- chanical power equal to fiodV, and the machine is acting as a motor. A real dc molor behaves in a precisely analogous fashion when it is loaded: As a load is added to its shaft, the motor begins to slow down, which reduces its in- ternal voltage, increasing its current now. The increased currenl flow increases its induced torque, and the induced lorque wi ll equal the load torque of the motor at a new, slower speed.
- 65. INTRODUcnONTO MACHINERY PRINCIPLES 41 R " - x x X i (t) ~=- V + F -- , Fjmd '00 I ~ - , x x x ""GURE 1-24 The linear de machine as a generator. Note that the power converted from electrical form 1 0 mechanical form by this linear motor was given by the equation Pcoo¥ = F ;"dV, The power converted from electrical form to mechanical form in a real rotaling motor is given by the equation ( I- 52) where the induced torque "TjDd is the rotational analog ofthe induced force FjDd, and Ihe angu lar velocity w is the rotational analog of the linear velocity v. The Linear DC Machine as a Generator Suppose that the linear machine is again operating under no-load steady-state con- ditions. This time, apply a force in the direction ofmotion and see what happens. Figure - 24 shows the linear machine with an applied force Fapp in the di- rection of motion. Now the applied force will cause the bar to accelerate in the direction of motion, and Ihe velocity v of the bar will increase. As Ihe velocity increases, ejmd = vtBI will increase and will be larger than Ihe ballery voltage VB' With eind > VB, the currenl reverses direction and is now given by the equation . eiDd - VB ,~ R ( I- 53) Since this current now flows up through the bar, it induces a force in the bar given by Find = ilB to Ihe left ( I- 54) TIle direction of the induced force is given by the right-hand rule. TIlis induced force opposes the applied force on the bar. Finally, the induced force will be equal and opposite to the applied force, and the bar wi ll be moving at a higher speed than before. Notice Ihat now the bat- tery is charging. The linear machine is now serving as a generator, converting me- chanical power Findv into electric power ejDdi . To summarize this behavior:
- 66. 42 ELECIRIC MACHINERY FUNDAMENTALS I. A force F app is applied in the direction of motion; F oet is in the direction of motion. 2. Acceleration a = F"",/m is positive, so the bar speeds up (vt). 3. The voltage eiod = vtBl increases, and so i = (eiod t-VBYR increases. 4. The induced force F ;od = itlB increases until IFind I = IFload I at a higher speed v. 5. An amount of mechanical power equal to F ;odv is now being converted to electric power e;odi, and the machine is acting as a generator. Again, a real dc generator behaves in precisely this manner: A torque is ap- plied to the shaft in the direction ofmotion, the speed ofthe shaft increases, the in- ternal voltage increases, and current fl ows out of the generator to the loads. The amount of mechanical power converted to electrical form in the real rotating gen- erator is again given by Equation ( I-52): ( I-52) It is interesting that the same machine acts as both motor and generator. The only difference between the two is whether the externally applied forces are in the direction of motion (generator) or opposite to the direction of motion (motor). Electrically, when eind > VB, the machine acts as a generator, and when e iod < VB, the machine acts as a motor. Whether the machine is a motor or a generator, both induced force (motor action) and induced voltage (generator action) are present at all times. nlis is generally true of all machines- both actions are present, and it is only the relative directions of the external forces with respect to the direction of motion that determine whether the overall machine behaves as a motor or as a generator. Another very interesting fact should be noted: This machine was a genera- tor when it moved rapidly and a motor when it moved more slowly, but whether it was a motor or a generator, it always moved in the same direction. Many begin- ning machinery students expect a machine to turn one way as a generator and the other way as a motor. This does not occur. Instead, there is merely a small change in operating speed and a reversal of current fl ow. Starting Problems with the Linear Machine A linear machine is shown in Figure 1-25. This machine is supplied by a 250-V dc source, and its internal resistance R is given as about 0. 10 n. (1lle resistor R models the internal resistance of a real dc machine, and this is a fairly reasonable internal resistance for a medium-size dc motor.) Providing actual numbers in this figure highlights a major problem with ma- chines (and their simple linear model). At starting conditions, the speed of the bar is zero, so eind = O. TIle current flow at starting is . VB 250 V Isu.n= R" = 0.1 n = 2500 A
- 67. FIGURE 1- 15 o.lOn " - i (I) INTRODUCTION TO MACHINERY PRINCIPLES 43 U =0.5 T. directed into the page x x X 0.5 m X X X The linear dc machine with componem values illustrating the problem of excessive starting currem. fa o.lOn " i (t) -=- VB =250V FIGURE 1- 16 R",., " X X X 0.5 m X X X A linear dc machine with an extra series resistor inserted to control the starting currem. This current is very high, often in excess of 10 times the rated current of the ma- chine. Such currents can cause severe damage to a motor. Both real ac and real dc machines suffer from similar high-currenl problems on starting. How can such damage be prevented? TIle easiest method for this simple lin- ear machine is 1 0 insert an extra resistance into Ihe circuit during starting 1 0 limit the current fl ow until ej!>d builds up enough to limit it. Figure -26 shows a start- ing resistance inserted into the machine circuitry. The same problem exists in real dc machines, and it is handled in precisely Ihe same fashion-a resistor is inserted into Ihe motor annature circuit during starting. TIle control of high starting current in real ac machines is handled in a different fashion, which will be described in Chapter 8. EXllmple 1-10. The linear dc machine shown in Figure 1-27a has a battery volt- age of 120 V. an internal resistance of 0.3 n. and a magnetic flux density of 0.1 T. (a) What is this machine's maximum starting current? What is its steady-state velocity at no load? (b) Suppose that a 30-N force pointing to the right were applied to the bar. What would the steady-state speed be? How much power would the bar be producing or consruning? How much power would the battery be producing or consuming?
- 68. 44 ELECIRIC MACHINERY FUNDAMENTALS -=- 120V -::~ 120V -::~ 120V ""GURE -27 0.30 " 0.30 0.30 - x X "J X F oo 30 N X 'bJ X F toad =30 N X ,<J U =O.1 T . directed into the JX1ge X X 10m X X U =O.1 T . directed into the JX1ge X X + - F 'PP- 3O N e;nd - , X X U =O.1 T . directed into the JX1ge X X + - F;nd- 3O N e;nd - , X X The linear de machine of Example 1- 10. (a) Starting conditions; (b) operating as a generator; (e) operating as a motor. Explain the difference between these two figures. Is this machine acting as a motor or as a generator? (c) Now suppose a 30-N force pointing to the left were applied to the bar. What would the new steady-state speed be? Is this machine a motor or a generntor now? (d) Assrune that a force pointing to the left is applied to the bar. Calculate speed of the bar as a flUlction of the force for values from 0 N to 50 N in IO-N steps. Plot the velocity of the bar versus the applied force. (e) Assume that the bar is lUlloaded and that it suddenly nms into a region where the magnetic field is weakened to 0.08 T. How fast will the bar go now? Solutioll (a) At starting conditions, the velocity of the bar is 0, so em = O. Therefore, i = VB - eiAd = l20Y - OV = 400 A R 0.3 0
- 69. INTRODUCTION TO MACHINERY PRINCIPLES 45 When the machine reaches steady state, Find = 0 and i = O. Therefore, VB = eind = v,J31 V, v... = BI 120 V = (0.1 TXlOm) = 120mls (b) Refer to Figure 1-27b. If a 30-N force to the right is applied to the bar, the final steady state will occur when the induced force Find is equal and opposite to the applied force FOPP' so that the net force on the bar is zero: F.pp = Find = ilB Therefore, . Find 30 N 1 = IB = (IOmXo. l T) = 30 A flowing up through the bar The induced voltage eind on the bar must be eind = VB+iR = 120 V + (30AX0.3 0 ) = 129 V and the final steady-state speed must be ,~ v" =m 129 V = (0.1 TXlOm) = 129m1s The bar is producing P = (129 VX30 A) = 3870 W of power, and the battery is consuming P = (120 VX30 A) = 3600 W. The difference between these two num- bers is the 270 W of losses in the resistor. This machine is acting as a generator. (c) Refer to Figure 1-25c.This time, the force is applied to the left, and the induced force is to the right. At steady state, Fopp = Find = ilB . Find 30 N 1 = IB = (IO mXO.IT) = 30 A flowing down through the bar The induced voltage eind on the bar must be eind= VB - iR = 120 V - (30 AX0.3 0 ) = III V and the final speed must be ,~ v.. = m 11I V = (0.1 TXIO m) = 111 mls This machine is now acting as a motor, converting electric energy from the bat- tery into mechanical energy of motion on the bar.
- 70. 46 ELECIRIC MACHINERY FUNDAMENTALS (d) This task is ideally suited for MATLAB. We can take advantage ofMATLAB's vectoriled calculations to detennine the velocity of the bar for each value of force. The MATLAB code to perform this calculation is just a version of the steps that were performed by hand in part c. The program shown below calcu- lates the current, induced voltage, and velocity in that order, and then plots the velocity versus the force on the bar. % M - f il e, exl _ 10 .m % M - f il e to ca l cu l ate and p l ot the vel oci t y of f unc t i on of l oad . % a linear motor as a VB = 120; r = 0 . 3; % Battery vol tage (V) % Res i s tance (ohms ) 1 = 1 ; B = 0 . 6; % Sel ect the F = 0, 10,50; •Ca l cu l ate 1 0 , ./ (1 •Ca l cu l ate e i nd = VB - •Ca l cu l ate v_ba r 0 e i nd % Bar l ength (m) % Flux density (T) for ces to appl y t o • the bar Force (N) 'he current s f l owi ng 1 0 'he mot or. • B) ; •CUrrent (A ) 'ho i nduced vol tages 0 0 'ho bar. i • c, •I nduced vol tage 'ho vel oci t i es of the bar. . / (1 • B) ; % Vel oci t y (m/ s) % Pl ot the vel oci t y of the bar ver s u s f or ce . p l ot (F,v_ba r ) ; t i t l e ('Pl ot of Vel oci t y ve r s u s Appli ed For ce'); x l abel (' Force (N) ' ) ; y l abel ('Vel oci t y (m/ s)'); axi s ( [0 500 200 ] ) ; (V) The resulting plot is shown in Figure 1- 28. Note that the bar slows down more and more as load increases. (e) If the bar is initially unloaded, then eind = VB. If the bar suddenly hits a region of weaker magnetic field, a transient will occur. Once the transient is over, though, eind will again equal VB. This fact can be used to determine the final speed of the bar. The initial speed was 120 rnls. The final speed is VB = eind = vllBl V, v.. = Bl 120 V = (0.08 TXIO m) = 150 mls Thus, when the flux in the linear motor weakens, the bar speeds up. The same behavior oc- curs in real dc motors: When the field flux of a dc motor weakens, it turns faster. Here, again, the linear machine behaves in much the same way as a real dc motor.
- 71. INTRODUCTION TO MACHINERY PRINCIPLES 47 200 ISO 160 140 ~ 120 60 40 20 0 o FIGURE 1-28 , 0 15 20 25 Force (N) Plot of velocity versus force for a linear de machine. ~ , 30 1.9 REAL, REACTlVE,AND APPARENT POWER IN AC CIRCUITS , 40 45 50 In a dc circuit such as Ihe one shown in Figure l- 29a, the power supplied to the dc load is simply Ihe product of the voltage across the load and the current flow- ing through it. p = VI ( I- 55) Unfortunately, the situation in sinusoidal ac circuits is more complex, be- cause there can be a phase difference between the ac voltage and the ac currenl supplied to Ihe load. TIle instantaneous power supplied to an ac load will still be Ihe product of the instantaneous voltage and the instantaneous currenl, but the av- erage power supplied 1 0 the load wi ll be affected by the phase angle between the voltage and the current. We will now explore the effects of this phase difference on the average power supplied to an ac load. Figure l- 29b shows a single-phase voltage source supplying power 10 a single-phase load with impedance Z = ZL OO. If we assume that the load is in- ductive, then the impedance angle 0 of the load will be positive, and the currenl will lag the voltage by 0 degrees. The voltage applied to this load is vet) = yI1V cos wi ( I- 56)
- 72. 48 ELECIRIC MACHINERY FUNDAMENTALS 1 v + ) - 1 1 + v(t) '" - 1 I (a) 1- / L 0° - - V = VLO" 'hI I R I I Z I z = Z L 0 ""GURE 1-29 (a) A de voltage source supplying a load with resistance R. (b) An ac voltage source supplying a load with impedance Z = Z L (J fl. where V is the nns value of the voltage applied to the load, and the resulting cur- rent flow is i(t) = V2I COS(wl - ()) where I is the rms value of the current fl owing through the load. 1lle instantaneous power supplied to this load at any time t is pet) = v(t)i(t) = 2VI cos wt COS(wl - ()) ( I-57) ( I-58) 1lle angle () in this equation is the impedance angle of the load. For inductive loads, the impedance angle is positive, and the current waveform lags the voltage waveform by () degrees. Ifwe apply trigonometric identities to Equation (1-58), it can be manipu- lated into an expression of the form pet) = VI cos () (1 + cos 2wt) + VI sin () sin 2wt ( I-59) 1lle first tenn of this equation represents the power supplied to the load by the component of current that is in phase with the voltage, while the second tenn rep- resents the power supplied to the load by the component of current that is 90° out ofphase with the voltage. The components of this equation are plotted in Figure 1-30. Note that thefirst term of the instantaneous power expression is always pos- itive, but it produces pulses of power instead of a constant value. The average value of this term is p = Vl cos () ( 1-60) which is the average or real power (P) supplied to the load by term 1of the Equa- tion (I- 59). The units of real power are watts (W), where 1 W = 1 V X 1A.
- 73. INTRODUCTION TO MACHINERY PRINCIPLES 49 p('1 Component I // f f f , , " , , , , , , , , , , , , , , , , , , , , , , , , , Component 2 , , , 0 , , , , , , , ,~2 I 0 ,2 , 4 , 6' • , 10 w o. , , , , , , " ""GURE 1-30 The components of power supplied to a single-phase load versus time. The first component represents the power supplied by the component of current j" phase with the voltage. while the second term represents the power supplied by the component of current 90° OUI Ofphase with the voltage. Nole that Ihe second tenn of the instantaneous power expression is positive half of the time and negative half of the time, so Ihal the average power supplied by this term is zero. This tenn represents power that is first transferred from the source 10 Ihe load, and then returned from Ihe load to the source. The power that continually bounces back and forth between the source and the load is known as re- active power (Q). Reactive power represents the energy thai is first stored and then released in the magnelic field of an inductor, or in the electric field of a capacitor. The reactive power of a load is given by Q = v/ sin() ( 1-61) where () is the impedance angle of the load. By convention, Q is positive for in- ductive loads and negative for capacitive loads, because Ihe impedance angle () is positive for inductive loads and negative for capacitive loads. TIle units of reac- tive power are voH-amperes reactive (var), where I var = 1 V X 1A. Even though the dimensional units are the same as for watts, reactive power is traditionally given a unique name to distinguish it from power actually supplied 10 a load. TIle apparent power (S) supplied to a load is defined as the product of the voHage across the load and the current Ihrough the load. TIlis is the power thai "appears" to be supplied to the load if the phase angle differences between volt- age and current are ignored. Therefore, the apparenl power of a load is given by
- 74. 50 ELECIRIC MACHINERY FUNDAMENTALS S = V[ ( 1-62) 1lle units of apparent power are volt-amperes (VA), where I VA = I V X 1A. As with reactive power, apparent power is given a distinctive set of units to avoid confusing it with real and reactive power. Alternative Forms of the Power Equations If a load has a constant impedance, then Ohm's law can be used to derive alterna- tive expressions for the real, reactive, and apparent powers supplied to the load. Since the magnitude of the voltage across the load is given by V = JZ ( 1-63) substituting Equation ( 1--63) into Equations (1--60) to (1--62) produces equations for real, reactive, and apparent power expressed in tenns of current and impedance: P = [lZcos () Q = [ lZ sin () S = [ lZ where Izi is the magnitude of the load impedance Z. Since the impedance of the load Z can be expressed as Z ~ R + jX ~ Izl co, 0 +j Izl ' in 0 ( 1-64) ( 1-65) ( 1-66) we see from this equation that R = Izi cos () and X = Izi sin (), so the real and reactive powers of a load can also be expressed as where R is the resistance and X is the reactance of load Z. Complex Power ( 1-67) ( 1-68) For simplicity in computer calculations, real and reactive power are sometimes represented together as a complex power S, where s ~ P + jQ ( 1-69) 1lle complex power S supplied to a load can be calculated from the equation S = VI* ( 1-70) where the asterisk represents the complex conjugate operator. To understand this equation, let's suppose that the voltage applied to a load is V = V L a and the current through the load is I = [ L {3. 1llen the complex power supplied to the load is
- 75. INTRODUcnONTO MACHINERY PRINCIPLES 51 p - - 1 - + 1 Q (~ v z Z = IziLon T -I FIGURE 1-31 An inductive load has a posilil'e impedance angle (J. This load produces a lngging current. and it consumes both real power P and reactive power Q from the source. S = VI* = (VL a )(JL-f3) = VI L(a - (3) = VI cos(a - (3) +jVI sin(a - (3) The impedance angle () is the difference between the angle of the voltage and the angle of the current (() = a - /3), so this equation reduces to S = VI cos () +jVI sin () ~ P +jQ The Relationships between Impedance Angle, Current Angle, and Power As we know from basic circuit theory, an inductive load (Figure 1- 31) has a pos- itive impedance angle (), since the reactance of an inductor is positive. If the im- pedance angle () of a load is positive. the phase angle of the current flowing through the load will lag the phase angle of the voltage across the load by (). I = V = VLoo= ~ L_ () Z IzlL6 Izl Also, if the impedance angle () of a load is positive, the reactive power consumed by the load wi ll be positive (Equation 1-65), and the load is said to be consuming both real and reactive power from the source. In contrast, a capacitive load (Figure 1- 32) has a negative impedance angle (), since the reactance of a capacitor is negative. If the impedance angle () of a load is negative, the phase angle of the current flowing through the load wi ll lead the phase angle of the voltage across the load by (). Also, if the impedance an- gie () of a load is negative, the reactive power Q consumed by the load will be negative (Equation 1-65). In this case, we say that the load is consuming real power from the source and supplying reactive power to the source. The Power Triangle The real, reactive, and apparent powers supplied to a load are related by the power triangle. A power triangle is shown in Figure 1- 33. The angle in the lower left
- 76. 52 ELECIRIC MACHINERY FUNDAMENTALS p - - 1 - + I Q + rv v z Z = 121Lon - 1 -I ""GURE 1-32 A capacitive loo.d has a nega/il'e impedance angle (j, This load produces a leading current, and it consumes real pO'er P from the source and while supplying reactive power Q to the source, s Q = SsinO o P = ScosO p cosO =- S sinO = SJ S tanO = ~ FI GURE 1-33 The power triangle, corner is the impedance angle (), The adjacent side of Ihis triangle is Ihe real power P supplied to the load, the opposite side ofthe triangle is the reactive power Q supplied to the load, and the hypotenuse of the triangle is the apparent power S of the load, 1lle quantity cos () is usually known as the power factor of a load, The power factor is defined as the fraction of the apparent power S that is actually sup- plying real power to a load, TIlUS, PF = cos () ( 1- 71) where () is the impedance angle of the load, Note that cos () = cos (- ()), so the power factor produced by an impedance angle of +30° is exactly the same as the power factor produced by an impedance angle of -30°, Because we can't tell whether a load is inductive or capacitive from the power factor alone, it is customary to state whether the current is leading or lagging the voltage whenever a power factor is quoted, TIle power triangle makes the relationships among real power, reactive power, apparent power, and the power factor clear, and provides a convenient way to calculate various power-related quantities if some of them are known, Example I- II. Figure 1- 34 shows an ac voltage source supplying power to a load with impedance Z = 20L - 30° n. Calculate the current I supplied to the load, the power factor of the load, and the real, reactive, apparent, and complex power supplied to the load,
- 77. INTRODUCTION TO MACHINERY PRINCIPLES 53 - 1 I + '" ' = 120LO"V Z Z = 20L - 30"n - T I FIGURE 1-34 The circuit of Example I- II. Solutioll The current supplied to this load is I = V = 120LO° V = 6L300 A Z 20L 300 n The power factor of the load is PF = cos (J = cos (-30°) = 0.866 leading (1 - 71) (Note that this is a capacitive load, so the impedance angle (J is negative, and the current leads the voltage.) The real power supplied to the load is P = Vlcos (J P = (120 VX6A) cos (-30°) = 623.5 W The reactive power supplied to the load is Q=Vlsin(J Q = (120 V)(6A) sin (-30°) = -360 VAR The apparent power supplied to the load is S = VI Q = (120 V)(6A) = 720 VA The complex power supplied to the load is S = VI* 1.10 SUMMARY = (l20LOOV)(6L-30° A)* = (l20LO° V)(6L30° A) = 720L30° VA = 623.5 - j360 VA (1- 60) (1- 61) (1- 62) (1- 70) This chapter has reviewed briefly the mechanics of systems rotating about a sin- gle axis and introduced the sources and effects of magnetic fields important in the understanding of transformers, motors, and generators. Historically, the English system of units has been used to measure the mechanical quantities associated with machines in English-speaking countries.
- 78. 54 ELECIRIC MACHINERY FUNDAMENTALS Recently, the 51 units have superseded the English system almost everywhere in the world except in the United States, but rapid progress is being made even there. Since 51is becoming almost universal, most (but not all) of the examples in this book use this system of units for mechanical measurements. Electrical quantities are always measured in 51units. In the section on mechanics, the concepts of angular position, angular veloc- ity, angular acceleration, torque, Newton's law, work, and power were explained for the special case of rotation about a single axis. Some fundamental relationships (such as the power and speed equations) were given in both 51and English units. TIle prOOuction of a magnetic field by a current was explained, and the spe- cial properties of ferromagnetic materials were explored in detail. The shape ofthe magnetization curve and the concept of hysteresis were explained in terms of the domain theory of ferromagnetic materials, and eddy current losses were discussed. Faraday's law states that a voltage will be generated in a coil of wire that is proportional to the time rate of change in the flux passing through it. Faraday's law is the basis oftransfonner action, which is explored in detail in Chapter 3. A current-carrying wire present in a magnetic field, if it is oriented properly, will have a force induced on it. This behavior is the basis of motor action in all real machines. A wire moving through a magnetic field with the proper orientation will have a voltage induced in it. TIlis behavior is the basis of generator action in all real machines. A simple linear dc machine consisting of a bar moving in a magnetic field illustrates many of the features of real motors and generators. When a load is at- tached to it, it slows down and operates as a motor, converting electric energy into mechanical energy. When a force pulls the bar faster than its no-load steady-state speed, it acts as a generator, converting mechanical energy into electric energy. In ac circuits, the real power P is the average power supplied by a source to a load. TIle reactive power Q is the component of power that is exchanged back and forth between a source and a load. By convention, positive reactive power is consumed by inductive loads (+ 0) and negative reactive power is consumed (or positive reactive power is supplied) by capacitive loads (- 0). TIle apparent power S is the power that "appears" to be supplied to the load if only the magnitudes of the voltages and currents are considered. QUESTIONS I-I. What is torque? What role does torque play in the rotational motion of machines? 1-2. What is Ampere's law? 1-3. What is magnetizing intensity? What is magnetic flux density? How are they related? 1-4. How does the magnetic circuit concept aid in the design of transformer and machine cores? 1-5. What is reluctance? 1-6. What is a ferromagnetic material? Why is the permeability of ferromagnetic mate- rials so high?
- 79. INTRODUCTION TO MACHINERY PRINCIPLES 55 1-7. How does the relative penneability of a ferromagnetic material vary with magneto- motive force? 1-8. What is hysteresis? Explain hysteresis in tenns of magnetic domain theory. 1-9. What are eddy current losses? What can be done to minimize eddy current losses in a core? 1- 10. Why are all cores exposed to ac flux variations laminated? I- II. What is Faraday's law? 1- 12. What conditions are necessary for a magnetic field to produce a force on a wire? 1- 13. What conditions are necessary for a magnetic field to produce a voltage in a wire? 1- 14. Why is the linear machine a good example of the behavior observed in real dc machines? 1- 15, The linear machine in Figure 1- 19 is running at steady state. What would ha~n to the bar if the voltage in the battery were increased? Explain in detail. 1- 16. Just how does a decrease in flux produce an increase in speed in a linear machine? 1- 17, Will current be leading or lagging voltage in an inductive load? Will the reactive power of the load be positive or negative? 1- 18. What are real, reactive, and apparent power? What lUlits are they measured in? How are they related? 1- 19. What is power factor? PROBLEMS 1- 1. A motor's shaft is spinning at a speed of 3000 r/min. What is the shaft speed in radians per second? 1- 2. A flywheel with a moment of inertia of 2 kg 0 m2 is initially at rest. If a torque of 5 N o m (cOlUlterc1ockwise) is suddenly applied to the flywheel, what will be the speed of the flywheel after 5 s? Express that speed in both radians per second and revolutions per minute. 1-3. A force of 10 N is applied to a cylinder, as shown in Figure PI- I. What are the mag- nitude and direction of the torque produced on the cylinder? What is the angular ac- celeration a of the cylinder? 3D' , r= 0.25 m J=5k:s o m2 F = ION fo'IGURE 1 '1-1 The cylinder of Problem - 3.
- 80. 56 ELECIRIC MACHINERY FUNDAMENTALS 1-4. A motor is supplying 60 N · m of torque to its load. If the motor's shaft is turning at 1800 r/min. what is the mechanical power supplied to the load in watts? In horse- power? 1-5. A ferromagnetic core is shown in Figure PI- 2. The depth of the core is 5 cm. The other dimensions of the core are as shown in the figure. Find the value of the current that will produce a flux of 0.005 Wb. With this current. what is the flux density at the top of the core? What is the flux density at the right side of the core? Assrune that the relative permeability of the core is I()(x). I. I' ,m 1 r lOem - i--- 20 cm - --+ - =~- T ",m ; [. + - I-e- -- - -- -- - - -- - - 400 turns ",m - - -- - - - I-e- [. ",m 1 Core depth - 5 em fo'IGURE PI- 2 The core of Problems 1- 5 and 1- 16. 1-6. A ferromagnetic core with a relative permeability of 1500 is shown in Figure PI- 3. The dimensions are as shown in the diagram. and the depth of the core is 7 cm. The air gaps on the left and right sides of the core are 0.070 and 0.050 cm. respectively. Because of fringing effects. the effective area of the air gaps is 5 percent larger than their physical size. If there are 400 IlU1lS in the coil wrapped arOlUld the center leg of the core and if the current in the coil is 1.0 A. what is the flux in each of the left. center. and right legs of the core? What is the flux density in each air gap? 1-7. A two-legged core is shown in Figure PI-4. The winding on the left leg of the core (Nt) has 400 turns. and the winding on the right (N2) has 300 turns. The coils are wound in the directions shown in the figure. If the dimensions are as shown. then what flux would be produced by currents il = 0.5 A and i2 = 0.75 A? Assume I-L, = J(XXl and constant. 1-8. A core with three legs is shown in Figure PI- 5. Its depth is 5 cm. and there are 200 IlU1lSon the leftmost leg. The relative penneability of the core can be assruned to be 1500 and constant. What flux exists in each of the three legs of the core? What is the flux density in each of the legs? Assume a 4 percent increase in the effective area of the air gap due to fringing effects.
- 81. ---r 'om - f- ; - - -- - 400 turns JOcm - O.07cm 0.05 cm - - - - - - f- 'om ---.L Core depth = 7 em H GURE )'1-3 The core of Problem 1-6. r--15 cm-+------- 50'm-------+-15 Cffi---1 T 15 em ;, ;, -- - - -- - - - - - - - - - 50 om - -- 400 turns 300 turns -- - - - - - - - - N, N, - - - -- -- - - -- 15 em ~ Core depth'" 15 em FIGURE PI- 4 The core of Problems - 7 and - 12. 57
- 82. 58 ELECIRIC MACHINERY FUNDAMENTALS ~r- 25cm----t- 15cm-+--- 25'm---r l C 9 C,"m'-ll T ',m ; - - 2A - - - 200 turns - O.04cm - - - - - t 25cm + ',m ~ Core depth = 5 cm ""GURE PI-S The core of Problem 1--8. 1-9. The wire shown in Figure PI--6 is carrying 5.0 A in the presence of a magnetic field. Calculate the magnitude and direction of the force induced on the wire. ---- 1= 1 m I - b n =O.25 T. - - - - - to the right i=5.0A _ ""GURE 1'1-6 A current-carrying wire in a magnetic field (Problem 1- 9). 1- 10. The wire shown in Figure PI- 7 is moving in the presence of a magnetic field. With the information given in the figure. detennine the magnitude and direction of the in- duced voltage in the wire. I- II. Repeat Problem 1- 10 for the wire in Figure PI-8. 1- 12. The core shown in Figure PI-4 is made of a steel whose magnetization curve is shown in Figure PI-9. Repeat Problem 1- 7. but this time do not asswne a constant value of Pro How much flux is produced in the core by the currents specified? What is the relative permeability of this core under these conditions? Was the asswnption
- 83. x x x X X X X X X X X '_5m1s X X X X , {:O.Sm INTRODUCTION TO MACHINERY PRINCIPLES 59 x x x x ~ 45' ~ X ~ X ~ ~ ~ X ~ X X , ~ ~ • / ' X x X I=O.SOm X X X X X X ') X X HGURE 1 '1-7 A wire moving in a magnetic field (Problem 0 ", 0.25 T. into the page 1- 10). lV=,mlS U =O.5 T HGURE " 1-8 A wire moving in a magnetic field (Problem I- II). in Problem 1- 7 that the relative penneability was equal to 1()(x) a good assumption for these conditions? Is it a good assumption in general? 1-13. A core with three legs is shown in Fig ure Pi- IO. Its depth is 8 em, and there are 400 turns on the center leg. The remaining dimensions are shown in the figure. The core is composed of a steel having the magnetization curve shown in Figure I- Uk . An- swer the following questions about this core: (a) What current is required to produce a flux density of 0.5 T in the central1eg of the core? (b) What current is required to produce a flux density of 1.0 T in the central leg of the core? Is it twice the current in part (a)? (c) What are the reluctances of the central and right legs of the core under the con- ditions in part (a)? (d) What are the reluctances of the central and right legs of the core under the con- ditions in part (b)? (e) What conclusion can you make about reluctances in real magnetic cores? 1- 14. A two-legged magnetic core with an air gap is shown in Figure PI - II. The depth of the core is 5 cm. the length of the air gap in the core is 0.06 cm. and the munber of turns on the coil is I(x)(). The magnetization curve of the core material is shown in
- 84. 60 ELECIRIC MACHINERY FUNDAMENTALS 1.00 E • ~ 0.75 ." , ~ " B 0.50 " 0.25 0.0 100 1000 Magnetizing intensity H (A ' turns/m) ""GURE 1'1-9 The magnetization curve for the core material of Problems 1- 12 and 1- 14. ; - N 400 turns ~gCm1- 16cm-----1-8cm+- 16C1l1-+8cm-1 Depth '" 8 em ""GURE Pi- IO The core of Problem 1- 13. T 8cm t 16 em 1 I 80m -.L Figure PI-9. Assume a 5 percent increase in effective air-gap area to account for fringing. How much current is required to produce an air-gap flux density of 0.5 T? What are the flux densities of the four sides of the core at that current ? What is the total flux present in the air gap?
- 85. INTRODUcnONTO MACHINERY PRINCIPLES 61 T Oem ; 1)- - --- ----- -- - --- --- N: ,O(X) turns O06om1= JOem - - ------ --- - - - - I-f- Oem ~ L 0 em~_---_ 30 em------ll -;-:--ll r l 'om Depth : 5 em ""GURE Pl-ll The core ofProbJem - 4. 1- 15. A transformer core with an effective mean path length of JO in has a 300-tW1l coil wrapped arOlUld one leg. Its cross-sectional area is 0.25 inl. and its magnetization curve is shown in Figure 1- IOc. If current of 0.25 A is flowing in the coil. what is the total flux in the core? What is the flux density? 1- 16. The core shown in Figure PI- 2 has the flux cp shown in Figure PI- l2. Sketch the voltage present at the terminals of the coil. 1- 17. Figure PI- 13 shows the core ofa simple dc motor. The magnetization curve for the metal in this core is given by Figure I- JOc and d. Assume that the cross-sectional area of each air gap is 18 cm2 and that the width of each air gap is 0.05 em. The ef- fective diameter of the rotor core is 4 em. (a) It is desired to build a machine with as great a flux density as possible while avoiding excessive saturation in the core. What would be a reasonable maxi- mum flux density for this core? (b) What would be the total flux in the core at the flux density of part (a)? (c) The maximum possible field current for this machine is I A. Select a reasonable nwnber of turns of wire to provide the desired flux density while not exceeding the maximum available current. 1- 18. Asswne that the voltage applied to a load is V = 208L -30° V and the current flow- ing through the load is I = 5L 15° A. (a) Calculate the complex power S consruned by this load. (b) Is this load inductive or capacitive? (c) Calculate the power factor of this load.
- 86. 62 ELECIRIC MACHINERY FUNDAMENTALS omo -------- ------------------------- 0.005 O~---}----' 2 ----'3-'~"4----~ 5----" 6C--->C---.8 C---- f(ms) - 0.005 - 0.010 --------------------- ------------- FIGURE 1'1- 12 Plot of flux 4> as a function of time for Problem 1- 16. 4,m N=? Ntums 4,m Depth = 4cm FIGURE 1 '1- 13 The core of Problem 1- 17. (d) Calculate the reactive power consmned or supplied by this load. Does the load consume reactive power from the source or supply it to the source? 1- 19. Figure PI - 14 shows a simple single-phase ac power system with three loads. The voltage source is V = l20LO° V. and the impedances of the three loads are 2:J = 5L _90° n Answer the following questions about this power system. (a) Assrune that the switch shown in the figure is open. and calculate the current I. the power factor. and the real. reactive. and apparent power being supplied by the load.
- 87. INTRODUCTION TO MACHINERY PRINCIPLES 63 (b) Assume that the switch shown in the figure is closed, and calculate the current I, the power factor, and the real, reactive, and apparent power being supplied by the load. (c) What happened to the current flowing from the source when the switch closed? Why? - 1 +1 +1 +1 + '"., V z, Z, Z, - T 1 1 1 FIGURE PI- 14 The circuit of Problem 1- 9. 1-20. Demonstrate that Equation (I- 59) can be derived from Equation (I- 58) using the simple trigonometric identities: pet) = v(t)i(t) = 2VI cos wt cos(wt - (J) pet) = VI cos (J (I + cos 2wt) + VI sin esin 2wt (I- 58) (I- 59) 1-2 1. The linear machine shown in Figure PI- IS has a magnetic flux density of 0.5 T directed into the page, a resistance of 0.25 n,a bar length I = 1.0 m, and a battery voltage of 100 V. (a) What is the initial force on the bar at starting? What is the initial current flow? (b) What is the no-load steady-state speed of the bar? (c) If the bar is loaded with a force of 25 N opposite to the direction of motion, what is the new steady-state speed? What is the efficiency of the machine under these circrunstances? 1=0 1 0.25 n i " - X VB = 00 V -=- X FIGURE PI- IS The linear machine in Problem - 21. 1-22. A linear machine has the following characteristics: B = 0.33 T into page 1= 0.5 m R = 0.50 n VB = 120V H =0.5 T x X X 1m X X X
- 88. 64 ELECIRIC MACHINERY FUNDAMENTALS (a) If this bar has a load of 10 N attached to it opposite to the direction of motion, what is the steady-state speed of the bar? (b) If the bar nms off into a region where the flux density falls to 0.30 T, what hap- pens to the bar? What is its fmal steady-state speed? (c) Suppose VB is now decreased to 80 V with everything else remaining as in part b. What is the new steady-state speed of the bar? (d) From the results for parts band c, what are two methods of controlling the speed of a linear machine (or a real dc motor)? REFERENCES I. Alexander. Charles K., and Matthew N. O. Sadiku: Fundamentals ofElectric Circuits, McGraw- Hill. 2CXXl. 2. Beer. E , and E. Johnston. Jr.: Vector Mechanicsfor Engineers: Dynamics, 6th ed., McGraw-Hill. New Yort.I 997. 3. Hayt. William H.: Engineering Electromllgnetics, 5th ed.. McGraw-Hill. New Yort. 1989. 4. Mulligan. J. E : Introductory College Physics, 2nd ed.. McGraw-HilL New York. 1991. 5. Sears. Francis W., Mark W. Zemansky. and Hugh D. Young: University Physics, Addison-Wesley. Reading. Mass., 1982.
- 89. CHAPTER 2 TRANSFORMERS A transformer is a device that changes ac electric power at one voltage level to ac electric power at another voltage level through the action of a magnetic field. It consists of two or more coils of wire wrapped around a common ferromagnetic core. These coils are (usually) not directly connected. The only connection be- twecn the coils is the common magnetic nux present within the core. FIGURE 2-1 The fi rst practical modern transformer. built by William Stanley in 1885. Note that the core is made up of individual sheets of metal (laminations). (Courtesy ofGeneml Electric Company.) 65
- 90. 66 ELECIRIC MACHINERY FUNDAMENTALS One of the transfonner windings is connected to a source of ac electric power, and the second (and perhaps third) transformer winding supplies electric power to loads.1lle transfonner winding connected to the power source is called the primary winding or input winding, and the winding connected to the loads is called the secondnry winding or output winding. If there is a third winding on the transformer, it is called the tertiary winding. 2.1 WHY TRANSFORMERS ARE IMPORTANT TO MODERN LIFE TIle first power distribution system in the United States was a 120-V dc system in- vented by Thomas A. Edison to supply power for incandescent light bulbs. Edi- son's first central power station went into operation in New York City in Septem- ber 1882. Unfortunately, his power system generated and transmitted power at such low voltages that very large currents were necessary to supply significant amounts of power. These high currents caused huge voltage drops and power losses in the transmission lines, severely restricting the service area of a generat- ing station. In the 1880s, central power stations were located every few city blocks to overcome this problem. The fact that power could not be transmitted far with low-voltage dc power systems meant that generating stations had to be small and localized and so were relatively inefficient. TIle invention of the transfonner and the concurrent development of ac power sources eliminated forever these restrictions on the range and power level of power systems. A transfonner ideally changes one ac voltage level to another voltage level without affecting the actual power supplied. If a transfonner steps up the voltage level of a circuit, it must decrease the current to keep the power into the device equal to the power out of it. 1llcrefore, ac electric power can be gener- ated at one central location, its voltage stepped up for transmission over long dis- tances at very low losses, and its voltage stepped down again for fmal use. Since the transmission losses in the lines of a power system are proportional to the square of the current in the lines, raising the transmission voltage and reducing the resulting transmission currents by a factor of 10 with transformers reduces power transmission losses by a factor of lOll Without the transfonner, it would simply not be possible to use electric power in many of the ways it is used today. In a rmx:lern power system, electric power is generated at voltages of 12 to 25 kV. Transfonners step up the voltage to between 110 kV and nearly 1000 kV for transmission over long distances at very low losses. Transfonners then step down the voltage to the 12- to 34.5-kV range for local distribution and fmally pennit the power to be used safely in homes, offices, and factories at voltages as low as 120 V. 2.2 TYPES AND CONSTRUCTION OF TRANSFORMERS The principal purpose of a transformer is to convert ac power at one voltage level to ac power of the same frequency at another voltage level. Transfonners are also
- 91. TRANSFORMERS 67 i, (I) - - + / + ") N, N, ) v, (t . HGURE 2-2 Core-foml transfomler construction. used for a variety of other purposes (e.g., voltage sampling, current sampling, and impedance transformation), but this chapter is primarily devoted to the power transformer. Power transfonners are constructed on one of two types of cores. One type of construction consists of a simple rectangular laminated piece of steel with the transformer windings wrapped around two sides of the rectangle. This type of construction is known as corefonn and is illustrated in Figure 2- 2. The other type consists of a three-legged laminated core with the windings wrapped around the center leg. nlis type of construction is known as shell form and is illustrated in Figure 2- 3. In either case, the core is constructed of thin laminations electrically isolated from each other in order to minimize eddy currents. The primary and secondary windings in a physical transformer are wrapped one on top of the other with the low-voltage winding innermost. Such an arrange- ment serves two purposes: I . It simplifies the problem of insulating the high-voltage winding from the core. 2. It results in much less leakage nux than would be the case if the two windings were separated by a distance on the core. Power transformers are given a variety of different names, depending on their use in power systems. A transformer connected to the output of a generator and used to step its voltage up to transmission levels (110+ kV) is sometimes called a unit transformer. The transfonner at the other end of the transmission line, which steps the voltage down from transmission levels to distribution levels (from 2.3 to 34.5 kV), is called a substation transfonner. Finally, the transformer that takes the distribution voltage and steps it down to the final voltage at which the power is actually used (110, 208, 220 V, etc.) is called a distribution transformer. All these devices are essentially the same- the only difference among them is their intended use.
- 92. 68 ELECIRIC MACHINERY FUNDAMENTALS (a) ""GURE 2-3 (a) Shell-form transformer construction. (b) A typical shell-form transformer. (Courtesy ofGeneml Electric Company.) In addition to the various power transfonners, two special-purpose trans- fonners are used with electric machinery and power systems. TIle first of these special transformers is a device specially designed to sample a high voltage and produce a low secondary voltage directly proportional to it. Such a transfonner is called a potential transfonner. A power transformer also produces a secondary voltage directly proportional to its primary voltage; the difference between a po- tential transfonner and a power transfonner is that the potential transformer is de- signed to handle only a very small current. The second type of special transfonner is a device designed to provide a secondary current much smaller than but directly proportional to its primary current. This device is called a current transformer. Both special-purpose transformers are discussed in a later section of this chapter. 2.3 THE IDEAL TRANSFORMER An ideal transformer is a lossless device with an input winding and an output winding. The relationships between the input voltage and the output voltage, and betwccn the input current and the output current, are given by two simple equa- tions. Figure 2- 4 shows an ideal transfonner. TIle transformer shown in Figure 2- 4 has Np turns of wire on its primary side and Ns turns of wire on its secondary side. 1lle relationship betwccn the volt-
- 93. TRANSFORMERS 69 ip (t) i, (t) - - • • + + ( N, N, "'p(1) ",,(t) J ,., - - ,b, H GURE 2-4 (a) Sketch of an ideal transformer. (b) Schematic symbols of a transformer. age vp(t) applied to the primary side of the transformer and the voltage vsCt) pro- duced on the secondary side is ~ 'p"(t")-!iJ'~-, vsCt) = Ns = a where a is defined to be the turns ratio of the transformer: Np a~ N, (2- 1) (2- 2) llle relationship between the current il...t) flowing into the primary side ofthe trans- fonner and the current isCt) flowing out of the secondary side of the transfonner is ;"I,IL 1 isCt) - a (2- 3a) (2- 3b)
- 94. 70 ELECIRIC MACHINERY FUNDAMENTALS In tenns of phasor quantities, these equations are ,nd ~ ~ I"~11 I, a (2- 4) (2- 5) Notice that the phase angle of Vp is the same as the angle of Vs and the phase an- gie of Ip is the same as the phase angle of Is. TIle turns ratio of the ideal trans- fonner affects the magnitudes of the voltages and currents, but not their angles. Equations (2-1 ) to (2- 5) describe the relationships between the magnitudes and angles of the voltages and currents on the primary and secondary sides of the transformer, but they leave one question unanswered: Given that the primary cir- cuit's voltage is positive at a specific end of the coil, what would the polarity of the secondary circuit's voltage be? In real transformers, it would be possible to tell the secondary's polarity only if the transformer were opened and its windings ex- amined. To avoid this necessity, transfonners utilize the dot convention. The dots appearing at one end of each winding in Figure 2-4 tell the polarity of the voltage and current on the secondary side of the transformer. TIle relationship is as follows: I. If the primary voltage is positive at the dotted end of the winding with respect to the undotted end, then the secondary voltage will be positive at the dotted end also. Voltage polarities are the same with respect to the dots on each side of the core. 2. If the primary current of the transformer fl ows into the dotted end of the pri- mary winding, the secondary current wi ll flow out of the dotted end of the secondary winding. TIle physical meaning of the dot convention and the reason polarities work out this way wi ll be explained in Section 2.4, which deals with the real transfonner. Power in an Ideal Transformer TIle power supplied to the transformer by the primary circuit is given by the equation (2-6) where ()p is the angle between the primary voltage and the primary current. The power supplied by the transformer secondary circuit to its loads is given by the equation (2- 7)
- 95. TRANSFORMERS 71 where ()s is the angle between the secondary voltage and the secondary current. Since voltage and current angles are unaffected by an ideal transfonner, ()p - ()s = (). The primary and secondary windings of an ideal transfonner have the same power factor. How does the power going into the primary circuit of the ideal transformer compare to the power coming out of the other side? lt is possible to find out through a simple application ofthe voltage and current equations [Equations (2-4) and (2- 5)]. 1lle power out of a transformer is Poot = Vsls cos () (2-8) Applying the turns-ratio equations gives Vs = Vp /a and Is = alp, so -"" P out - a (alp) cos () I Poot - VpIp cos () - P;n (2- 9) Thus, the output power ofan ideal transfonner is equal to its input power. The same relationship applies to reactive power Q and apparent power S: IQ ;" VpIp sin () VsIs sin () Qoo, I (2- 10) and Is;, ~ Vplp = VsIs = Soot I (2- 11 ) Impedance Transformation through a Transformer The impedance of a device or an element is defined as the ratio of the phasor volt- age across it to the phasor current flowing through it: V, ZL = 1; (2- 12) One of the interesting properties of a transfonner is that, since it changes voltage and current levels, it changes the ratio between voltage and current and hence the apparent impedance of an element. To understand this idea, refer to Figure 2- 5. If the secondary current is calJed Is and the secondary voltage Vs, then the imped- ance of the load is given by V, ZL = 1; The apparent impedance of the primary circuit of the transfonner is V Z' = --.f.. , Ip Since the primary voltage can be expressed as Vp = aVs (2- 13) (2- 14)
- 96. 72 ELECIRIC MACHINERY FUNDAMENTALS - + v, Z, v, z,= - I, ,,' - + + • • V, - ,b, ""GURE 2--.5 (a) Definition of impedance. (b) Impedance scaling through a transformer. and the primary current can be expressed as " Ip = - a the apparent impedance of the primary is V aV V Z' - ..:...t!. - ~ -a2.:...s. L - Ip - Isla - Is I Z~ =a2ZL I Z, (2- 15) With a transfonner, it is possible to match the magnitude of a load imped- ance to a source impedance simply by picking the proper turns ratio. Analysis of Circuits Containing Ideal Transformers If a circuit contains an ideal transformer, then the easiest way to analyze the cir- cuit for its voltages and currents is to replace the portion of the circuit on one side of the transfonner by an equivalent circuit with the same tenninal characteristics. After the equivalent circuit has been substituted for one side, then the new circuit (without a transformer present) can be solved for its voltages and currents. In the portion of the circuit that was not replaced, the solutions obtained will be the COf-
- 97. - ,I Ztime + V =4S0LO"V - ,., T, I tiDe O.ISo. 1 ,I • :10 - • Zlime + - V =4S0LO"V ,b, FIGURE 2-6 j 0.24 0. j 0.24 0. TRANSFORMERS 73 + V~~ - T, 10: 1 • • I' z~ 4+j3o. 1 ~ - Z + 4+j ~- V ~ The power system of Example 2- 1 (a) without and (b) with transformers at the ends of the transmission line. rect values of voltage and current for the original circuit. 1llen the turns ratio of the transfonner can be used to detennine the voltages and currents on the other side of the transfonner. TIle process of replacing one side of a transformer by its equivalent at the other side's voltage level is known as referring the first side of the transfonner to the second side. How is the equivalent circuit fonned? Its shape is exactly the same as the shape of the original circuit. TIle values of voltages on the side being replaced are scaled by Equation (2-4), and the values of the impedances are scaled by Equa- tion (2- 15). TIle polarities of voltage sources in the equivalent circuit will be re- versed from their direction in the original circuit if the dots on one side of the transformer windings are reversed compared to the dots on the other side of the transformer windings. The solution for circuits containing ideal transformers is illustrated in the following example. Example 2-1. A single-phase power system consists of a 4SO-V 60-Hz gen- erator supplying a load Z_ = 4 + )3 0 through a transmission line of impedance Ztm. = O.IS + jO.24 O. Answer the following questions about this system. (a) If the power system is exactly as described above (Figure 2-6a), what will the voltage at the load be? What will the transmission line losses be?
- 98. 74 ELECIRIC MACHINERY FUNDAMENTALS (b) Suppose a I: 10 step-up transformer is placed at the generator end of the trans- mission line and a 10:I step-down transfonner is placed at the load end of the line (Figure 2- 6b). What will the load voltage be now? What will the transmis- sion line losses be now? Solutioll (a) Figure 2-6a shows the power system without transfonners. Here IG = IIu.. = Ilood' The line current in this system is given by IIu.. = ,----'V-,-_ ZIi". + Zioad 480 L O ° V = " (O".1"' 8'"~ +' jio.'!! 24'f!i"~ ) "; +-'("4'"~ +' j" 3"'ll ) 480 LO° = = 4.18 + j3.24 = 9O.8L-37.8° A Therefore the load voltage is V10ad = Iline~ 480 LO° 5.29L37.8° = (90.8 L -37.8° A)(4 n + j3 n ) = (90.8 L -37.8° A)(5 L36.9° fl) = 454 L - 0.9° V and the line losses are Pim• = (tUDe)' R line = (90.8 A)' (0.1 8 n) = 1484 W (b) Figure 2-6b shows the power system with the transfonners.To analyze this sys- tem, it is necessary to convert it to a common voltage level. This is done in two steps: I. Eliminate transfonner T2by referring the load over to the transmission line's voltage level. 2. Eliminate transformer TI by referring the transmission line's elements and the equivalent load at the transmission line's voltage over to the source side. The value of the load's impedance when reflected to the transmission system's voltage is Z · - , ' Z load - load = (1f)4n + j3 n) = 4000 + j300n The total impedance at the transmission line level is now Zeq = ~ine + Zio.t = 400.18 + j300.24 n = 500.3 L36.88° n
- 99. TRANSFORMERS 75 V =480LO"V 0.18 n jO.24 n - :10 " , , • • , ZHDO , Z'_= : , + 4OO+j300n , , , , , ~ ,,' :Equivalent cin:uit 0.0018 fi jO.0024fi I ,I , " + Z'line ( V =480LO"V Z ' _=4+j3fi - . Equivalent cin:uit ,b , FIGURE 2-7 (a) System with the load referred to the transmission system voltage level. (b) System with the load and transmission line referred to the generator's voltage level. This equi valent circuit is shown in Figure 2- 7a. The total impedance at the transmission line level (4". + Zl~ is now reflected across Tl to the source's voltage level: Z ' = a2 Z ~ ~ = a2 (l1ine + Z k...i) = ( lb)0.18 0 + jO.24 0 + 400 0 + j300f.l) = (0.0018 0 + jO.0024 0 + 4 n + j3 f.!) = 5.003 L36.88° n Notice that Z'k-d = 4 + j3 0 and:!.time = 0.0018 +jllOO24 n. The resulting equi valent cir- cuit is shown in Figure 2- 7b. The generator's current is _ 480LOo V _ ° 10 - 5.003 L36.88° n - 95 .94L-36.88 A Knowing the current Ie. we can now work back and find II;'" and 1_ . Working back through Tl , we get
- 100. 76 ELECIRIC MACHINERY FUNDAMENTALS = 1~(95.94 L-36.88° A) = 9.594L-36.88° A Working back through T2gives NnIline = NSlIload "n Ilood = N IUDe n = 0 (9.594 L-36.880 A) = 95.94L-36.88° A It is now possible to answer the questions originally asked. The load voltage is given by V10ad = Ibdl10ad = (95.94 L-36.88° A)(5 L36.87° 0 ) = 479.7 L-O.Olo V and the line losses are given by PIo!;. = (/n...)lRnne = (9.594 A)l (0.18 n) = 16.7 W Notice that raising the transmission voltage of the power system reduced transmission losses by a factor of nearly 90! Also, the voltage at the load dropped much less in the system with transformers compared to the system without trans- fonners. This simple example dramatically illustrates the advantages of using higher-voltage transmission lines as well as the extreme importance of transform- ers in modern power systems. 2.4 THEORY OF OPERATION OF REAL SINGLE-PHASE TRANSFORMERS TIle ideal transformers described in Section 2.3 can of course never actually be made. What can be produced are real transformers-two or more coils of wire physically wrapped around a ferromagnetic core. The characteristics of a real transformer approximate the characteristics of an ideal transfonner, but only to a degree. This section deals with the behavior of real transformers. To understand the operation of a real transfonner, refer to Figure 2--8. Fig- ure 2- 8 shows a transfonner consisting of two coils of wire wrapped around a transfonner core. 1lle primary of the transfonner is connected to an ac power source, and the secondary winding is open-circuited. TIle hysteresis curve of the transformer is shown in Figure 2- 9.
- 101. TRANSFORMERS 77 iP(t) - + + l' 0- N, vP(t) FIGURE l-8 Sketch of a real transformer with no load attached to its secondary. q, Rux --------ttt--------- Magnetomotive force FIGURE 2-9 The hysteresis curve of the transformer. The basis of transfonner operation can be derived from Faraday's law: dA e iod = dt ( 1-41 ) where Ais the flux linkage in the coil across which the voltage is being induced. The flux linkage A is the sum of the flux passing through each turn in the coil added over all the turns of the coil: N A ~ '2:</>; ( 1-42) ; = 1
- 102. 78 ELECIRIC MACHINERY FUNDAMENTALS The total flux linkage through a coil is notjust N<p, where N is the number of turns in the coil, because the flux passing through each turn of a coil is slightly differ- ent from the flux in the other turns, depending on the position of the turn within the coil. However, it is possible to define an average flux per turn in a coil. If the total flux linkage in all the turns of the coils is Aand if there are N turns, then the averageflux per turn is given by and Faraday's law can be written as - A q,~ N - <& eind - N dt The Voltage Ratio across a Transformer (2-1 6) (2-1 7) Ifthe voltage of the source in Figure 2--8 is vp(t), then that voltage is placed di- rectly across the coils of the primary winding of the transformer. How will the transformer react to this applied voltage? Faraday's law explains what will hap- pen. When Equation (2- 17) is solved for the average flux present in the primary winding of the transfonner, the result is - 1 I <P = N Vp(t) dt p (2- 18) TIlis equation states that the average flux in the winding is proportional to the in- tegral of the voltage applied to the winding, and the constant of proportionality is the reciprocal of the number of turns in the primary winding IINp. TIlis flux is present in the primary coil of the transformer. What effect does it have on the secondary coil of the transfonner? TIle effect depends on how much of the flux reaches the secondary coil. Not all the flux produced in the primary coil also passes through the secondary coil-some of the flux lines leave the iron core and pass through the air instead (see Figure 2- 10). TIle portion of the flux that goes through one of the transfonner coils but not the other one is called leak- age flux. The flux in the primary coil of the transformer can thus be divided into two components: a mutualflux, which remains in the core and links both wind- ings, and a small leakageflux, which passes through the primary winding but re- turns through the air, bypassing the secondary winding: I <pp = <PM + fu where <pp = total average primary flux (2-19) <PM = flux component linking both primary and secondary coils <hp = primary leakage flux
- 103. TRANSFORMERS 79 -- -- - , - - , - , , , • cPM , , , ~ , , '"" , , , , , , I, , , , , , , I, - 0 , , , , 0 - , , +1 , , , , + , , , , , , , , , , , , , , , , , , I , ;cPLP I I 0., ' I , , I I , I " I I " S l I I , , I I I I I I , , i i I I I I I I , , I I , , , , , , I I , , , , , , , , , , , , , , , , , • - - , , , , , , , , , , , , , , , , , , " , , , , , , , / '-, , ./1 " / , , / -- OM , -- , FIGURE l-IO Mutual and leakage fluxes in a transformer core. There is a similar division of flux in the secondary winding between mutual flux and leakage flux which passes through the secondary winding but returns through the air, bypassing the primary winding: where l 4>s = 4>M + 4>LS 4>s = total average secondary flux (2- 20) 4>M = flux component linking both primary and secondary coils fu = secondary leakage flux With the division of the average primary flux into mutual and leakage com- ponents, Faraday's law for the primary circuit can be reexpressed as (2- 21) The first term of this expression can be called ep(l), and the second term can be called eLP(l). If this is done, then Equation (2- 21) can be rewritten as (2- 22)
- 104. 80 ELECIRIC MACHINERY FUNDAMENTALS TIle voltage on the secondary coil of the transfonner can also be expressed in terms of Faraday's law as d<l>s vs<.t) = NSdt _ d<l>M dfu - Ns dt + Ns dt = esCt) + eL'>(t) TIle primary voltage due to the mutualflux is given by _ d4>M ep(t) - Np dt and the secondary voltage due to the mutualflux is given by _ d4>M es(t) - Ns dt Notice from these two relationships that TIlerefore, ep(t) _ d4>M _ edt) Np - dt - Ns (2- 23) (2- 24) (2- 25) (2- 26) (2- 27) TIlis equation means that the ratio ofthe primary voltage caused by the mutual flux to the secondary voltage caused by the mutualflux is equal to the turns ratio of the transformer. Since in a well-designed transfonner 4>M » <hp and 4>M » 4>u" the ratio of the total voltage on the primary of a transformer to the to- tal voltage on the secondary of a transfonner is approximately vp(t) !i.E. vs<.t) = Ns = a (2- 28) TIle smaller the leakage fluxes of the transfonner are, the closer the total trans- fonner voltage ratio approximates that of the ideal transfonner discussed in Sec- tion 2.3. The Magnetization Current in a Real Transformer When an ac power source is connected to a transformer as shown in Figure 2--8, a current flows in its primary circuit, even when the secondary circuit is open- circuited. TIlis current is the current required to produce flux in a real ferromag- netic core, as explained in Chapter I. It consists of two components:
- 105. TRANSFORMERS 81 I. The magnetization current iM , which is the current required to produce the flux in the transformer core 2. llle core-loss current iH " which is the current required to make up for hys- teresis and eddy current losses Figure 2-11 shows the magnetization curve of a typical transformer core. If the fl ux in the transformer core is known, then the magnitude of the magnetization current can be found directly from Figure 2-11. Ignoring for the moment the effects of leakage flux, we see that the average flux in the core is given by - 1 f cf> = Np vp(t)dt (2-1 8) If the primary voltage is given by the expression vp(t) = VM cos wt V, then the re- sulting flux must be cf> = ~ fVM cos wtdt p VM . ~ -- smwt Wb wNp (2- 29) If the values of current required to produce a given flux (Figure 2-11 a) are com- pared to the flux in the core at different times, it is possible to construct a sketch of the magnetization current in the winding on the core. Such a sketch is shown in Figure 2-11 b. Notice the following points about the magnetization current: I. The magnetization current in the transfonner is not sinusoidal. The higher- frequency components in the magnetization current are due to magnetic sat- uration in the transfonner core. 2. Once the peak flux reaches the saturation point in the core, a small increase in peak flux requires a very large increase in the peak magnetization current. 3. The fundamental component of the magnetization current lags the voltage ap- plied to the core by 90°. 4. llle higher-frequency components in the magnetization current can be quite large compared to the fundamental component. In general, the further a trans- fonner core is driven into saturation, the larger the hannonic components wil I become. The other component of the no-load current in the transformer is the current required to supply power to make up the hysteresis and eddy current losses in the core. lllis is the core-loss current. Assume that the flux in the core is sinusoidal. Since the eddy currents in the core are proportional to d<PIdt, the eddy currents are largest when the flux in the core is passing through 0 Wb. lllerefore, the core-loss current is greatest as the flux passes through zero. The total current required to make up for core losses is shown in Figure 2-1 2.
- 106. ,.Wb - - - - - - - - , / - - - - - - - - - - ~,A · turns (a) " --- 7'C---------------------t---~ , , , , , , " , " , , , , , , ~~ 'c----_r'--'"--",C_-'--- , , , , , , , " , " , " , , , ------~I---+---- ~, '...../ '-'---------' , ------------c:01' v. ifi(l) '" - N sin WI w, (bj -------t----"'--+-----'m , H GURE 2-11 (a) The magnetization curve of the transformer core. (b) The magnetization current caused by the flux in the transformer core. 82
- 107. FIGURE 2-12 The core-loss current in a transformer. f 1 ; FIGURE 2-13 f -, " , , , , , , , , , v ,,,,, The total excitation current in a transformer. , , '- , , , , v TRANSFORMERS 83 Notice the following points about. the core-loss current: I. The core-loss current is nonlinear because ofthe nonline.1.r effects of hysteresis. 2. 1lle fundamental component of the core-loss current is in phase with the volt- age applied to the core. The total no-load current in the core is called the excitation current of the transfonner. It is just the sum of the magnetization current and the core-loss cur- rent in the core: (2- 30) The total excitation current in a typical transfonner core is shown in Figure 2-1 3.
- 108. 84 ELECIRIC MACHINERY FUNDAMENTALS I, - • + V, N, ""GURE 2-14 A real transformer with a load connected to its secondary. The Current Ratio on a Transformer and the Dot Convention I, • - + V, N, '"''' Now suppose that a load is connected to the secondary of the transformer. The re- sulting circuit is shown in Figure 2- 14. Notice the dots on the windings of the transformer. As in the ideal transfonner previously described, the dots help deter- mine the polarity of the voltages and currents in the core without having physi- cally to examine its windings. The physical significance of the dot convention is that a current flowing into the doffed end ofa winding produces a positive mng- netomotive force '?J', while a current flowing into the undotted end of a winding produces a negative rnagnetomotive force. Therefore, two currents fl owing into the dotted ends of their respective windings produce rnagnetomotive forces that add. If one current flows into a dotted end of a winding and one flows out ofa dot- ted end, then the magnetornotive forces will subtract from each other. In the situation shown in Figure 2- 14, the primary current produces a posi- tive magnetornotive force '?J'p = Npip, and the secondary current produces a neg- ative rnagnetomotive force:lis = - Nsis. Therefore, the net rnagnetomotive force on the core rnust be (2- 31) lllis net magnetornotive force must produce the net flux in the core, so the net magnetornotive force must be equal to (2- 32)
- 109. TRANSFORMERS 85 ;.Wb -------I-------- ~.A .turns FIGURE 2- 15 The magnetization curve of an ideal transformer. where mis the reluctance of the transfonner core. Because the reluctance of a well- designed transfonner core is very small (nearly zero) until the core is saturated, the relationship between the primary and secondary currents is approximately 2i'Det = Npip - Nsis "'" 0 as long as the core is unsaturated. TIlerefore, INpip "'" Nsis I (2- 33) (2- 34) (2- 35) It is the fact that the magnetomotive force in the core is nearly zero which gives the dot convention the meaning in Section 2.3. In order for the magnetomotive force to be nearly zero, current must flow into one dotted end and out ofthe other dotted end. The voltages must be built up in the same way with respect to the dots on each winding in order to drive the currents in the direction required. (TIle po- larity of the voltages can also be determined by Lenz' law if the construction of the transfonner coils is visible.) What assumptions are required to convert a real transformer into the ideal transfonner described previously? 1lley are as follows: I. 1lle core must have no hysteresis or eddy currents. 2. 1lle magnetization curve must have the shape shown in Figure 2- 15. Notice that for an unsaturated core the net magnetomotive force 2i'nel = 0, implying that Npip= Nsis. 3. The leakage flux in the core must be zero, implying that all the flux in the core couples both windings. 4. 1lle resistance of the transfonner windings must be zero.
- 110. 86 ELECIRIC MACHINERY FUNDAMENTALS While these conditions are never exactly met, well-designed power transformers can come quite close. 2.5 THE EQUIVALENT CIRCUIT OF A TRANSFORMER TIle losses that occur in real transformers have to be accounted for in any accurate model oftransforrner behavior. The major items to be considered in the construc- tion of such a model are I. Copper cPR) losses. Copper losses are the resistive heating losses in the pri- mary and secondary windings ofthe transformer. They are proportional to the square of the current in the windings. 2. Eddy current losses. Eddy current losses are resistive heating losses in the core of the transformer. They are proportional to the square of the voltage ap- plied to the transformer. 3. Hysteresis losses. Hysteresis losses are associated with the rearrangement of the magnetic domains in the core during each half-cycle, as explained in Chapter 1. They are a complex, nonlinear function of the voltage applied to the transformer. 4. Leakagef1ux. TIle fluxes <PLP and 4>u. which escape the core and pass through only one of the transformer windings are leakage fluxes. These escaped fluxes produce a self-inductance in the primary and secondary coils, and the effects of this inductance must be accounted for. The Exact Equivalent Circuit of a Real Transformer lt is possible to construct an equivalent circuit that takes into account all the ma- jor imperfections in real transformers. E:1.ch major imperfection is considered in turn, and its effect is included in the transformer model. TIle easiest effect to model is the copper losses. Copper losses are resistive losses in the primary and secondary windings of the transformer core. They are modeled by placing a resistor Rp in the primary circuit of the transformer and a re- sistor Rs in the secondary circuit. As explained in Section 2.4, the leakage flux in the primary windings <PLP produces a voltage <PLP given by (2- 36a) and the leakage flux in the secondary windings 4>u. produces a voltage eLS given by (2- 36b)
- 111. TRANSFORMERS 87 Since much of the leakage flux path is through air, and since air has a constant re- luctance much higher than the core reluctance, the flux fu is directly proportional to the primary circuit current ip and the flux <hs is directly proportional to the sec- ondary current is: q,LP = (IlPNp)ip q,LS = (IlPNs)is where IlP = penneance of fl ux path Np = number of turns on primary coil Ns = number of turns on secondary coil Substitute Equations (2- 37) into Equations (2- 36). TIle result is eLP(t) = Np :r(raWp)ip = NJ, IlP ~{ d . dis eLS(!) = Ns d/!PNs)IS = ~ rJ> dt The constants in these equations can be lumped together. Then _ dip eLP(t) - Lp dt (2- 37a) (2- 37b) (2- 38a) (2- 38b) (2- 39a) (2-39b) where Lp = N}1lP is the self-inductance of the primary coil and Ls = NIIlP is the self-inductance of the secondary coil. Therefore, the leakage flux will be modeled by primary and secondary inductors. How can the core excitation effects be modeled? TIle magnetization current im is a current proportional (in the unsaturated region) to the voltage applied to the core and lagging the applied voltage by 900, so it can be modeled by a reactance XM connected across the primary voltage source. TIle core-loss current i H< is a current proportional to the voltage applied to the core that is in phase with the ap- plied voltage, so it can be modeled by a resistance Re connected across the pri- mary voltage source. (Remember that both these currents are really nonlinear, so the inductance XM and the resistance Re are, at best, approximations of the real ex- citation effects.) The resulting equivalent circuit is shown in Figure 2- 16. Notice that the ele- IllCnts forming the excitation branch are placed inside the primary resistance Rp and the primary inductance Lp. lllis is because the voltage actually applied to the core is really equal to the input voltage less the internal voltage drops of the winding. Although Figure 2- 16 is an accurate model ofa transfonner, it is not a very useful one. To analyze practical circuits containing transformers, it is normally necessary to convert the entire circuit to an equivalent circuit at a single voltage level. (Such a conversion was done in Example 2- 1.) Therefore, the equivalent
- 112. 88 ELECIRIC MACHINERY FUNDAMENTALS I, - + v, ~ R,~ - ""GURE 2-16 The model of a real transformer. I, R, j Xp + V, R, ~ R, . X, " I, ? J ? - + R, ~ ? - ""GURE 2-17 [ • • JX. N, N, d'R, [ " j XM j ,., R, I .X. J e> " [ ,b, Ideal transformer ja2X, JX, * I, - j Xs I, - + aV, + V, (a) The transformer model referred to its primary voltage level. (b) The transfonner model referred to its secondary voltage level. + v, - circuit must be referred either to its primary side or to its secondary side in problem solutions. Figure 2- 17a is the equivalent circuit of the transfonner re- ferred to its primary side, and Figure 2-17b is the equivalent circuit referred to its secondary side.
- 113. TRANSFORMERS 89 Approximate Equivalent Circuits of a Transformer The transfonner models shown before are often more complex than necessary in order to get good results in practical engineering applications. One of the princi- pal complaints about them is that the excitation branch of the model adds another node to the circuit being analyzed, making the circuit solution more complex than necessary. The excitation branch has a very small current compared to the load current of the transfonners. In fact, it is so small that under nonnal circumstances it causes a completely negligible voltage drop in Rp and Xp. Because this is true, a simplified equivalent circuit can be produced that works almost as well as the original model. The excitation branch is simply moved to the front of the trans- fonner, and the primary and secondary impedances are left in series with each other. TIlese impedances are just added, creating the approximate equivalent cir- cuits in Figure 2- 18a and b. In some applications, the excitation branch may be neglected entirely with- out causing serious error. In these cases, the equivalent circuit of the transformer reduces to the simple circuits in Figure 2- 18c and d. I, I, " ,I, I, - )X"lP - - jXeq. - + j v, < R, J - ,,' I, - R.~ v, jXM + aV, Reqp '" Rp + rrR, xeqp '" xp + rrx, .., - jXeqp - - ,~---------~, - ,,' FIGURE 2-18 + V , , - ,b, V, , ,I, - : j R, .x. if )-;;> J I, - V, - ,~---------~, - ,d, Approximate transformer models. (a) Referred to the primary side; (b) referred to the secondary side; (c) with no excitation branch. referred to the primary side; (d) with no excitation branch. referred to the secondary side. + V, -
- 114. 90 ELECIRIC MACHINERY FUNDAMENTALS r:0 at me er ip (t) - J + • • w, + v (t) ""' V vp (t) - - Transformer -0-Ammeter --0-Voltmeter ""GURE 2-19 Connection for transformer open-cirwit test. Detennining the Values of Components in the Transformer Model It is possible to experimentally detennine the values of the inductances and resis- tances in the transfonner model. An adequate approximation of these values can be obtained with only two tests, the open-circuit test and the short-circuit test. In the open-circuit test, a transfonner's secondary winding is open- circuited, and its primary winding is connected to a full-rated line voltage. Look at the equivalent circuit in Figure 2- 17. Under the conditions described, all the in- put current must be fl owing through the excitation branch of the transfonner. The series elements Rp and Xp are too small in comparison to Rcand XM to cause a sig- nificant voltage drop, so essentially all the input voltage is dropped across the ex- citation branch. TIle open-circuit test connections are shown in Figure 2- 19. Full line volt- age is applied to the primary of the transfonner, and the input voltage, input cur- rent, and input power to the transfonner are measured. From this information, it is possible to detennine the power factor of the input current and therefore both the mngnitude and the angle of the excitation impedance. TIle easiest way to calculate the values of Rc and XM is to look first at the admittance of the excitation branch. TIle conductance of the core-loss resistor is given by 1 GC=R c and the susceptance of the magnetizing inductor is given by 1 BM = - XM (2- 40) (2- 41) Since these two elements are in parallel, their admittances add, and the total exci- tation admittance is
- 115. r:0 a meter ip (t) - J + • • w" + v (t) '" V vp (t) - - Transformer FIGURE 2-10 Connection for transformer shon-circuit test. YE = Gc - JBM ~ _I _j _' Rc XM TRANSFORMERS 91 i, (t) - (2- 42) (2- 43) The magnitude of the excitation admittance (referred to the primary circuit) can be found from the open-circuit test voltage and current: IYEI ~ ~oe DC (2- 44) The angle of the admittance can be found from a knowledge of the circuit power factor. 1lle open-circuit power factor (PF) is given by and the power-factor angle () is given by Poe () = cos- 1,,-'7~ YclC10c (2- 45) (2- 46) The power factor is always lagging for a real transfonner, so the angle ofthe current always lags the angle of the voltage by () degrees. 1llCrefore, the admittance YE is I YE = V OC L-() oe loe ~ -- L-cos- 1PF Voe (2- 47) By comparing Equations (2-43) and (2-47), it is possible to determine the values of Rc and XM directly from the open-circuit test data. In the shott-circuit test, the secondary tenninals of the transformer are short- circuited, and the primary tenninals are connected to a fairly low-voltage source, as shown in Figure 2- 20. The input voltage is adjusted until the current in the short- circuited windings is equal to its rated value. (Be sure to keep the primary voltage at a saJe level. It would not be a good idea to burn out the transformer's windings while trying to test it.)1lle input voltage, current, and power are again measured.
- 116. 92 ELECIRIC MACHINERY FUNDAMENTALS Since the input voltage is so low during the short-circuit test, negligible cur- rent flows through the excitation branch. If the excitation current is ignored, then all the voltage drop in the transformer can be attributed to the series elements in the circuit. The magnitude of the series impedances referred to the primary side of the transformer is TIle power factor of the current is given by P,c PF = cos (J = u-''i- VscIsc (2- 48) (2- 49) and is lagging. The current angle is thus negative, and the overall impedance an- gle (J is positive: (2- 50) TIlerefore, VscLO° = Vsc L (J 0 ZSE = lsc L (J 0 lsc (2- 51) TIle series impedance ZSE is equal to ZSE = Req + jXeq = (Rp + a2RS) + j(Xp + a2Xs) (2- 52) It is possible to detennine the total series impedance referred to the primary side by using this technique, but there is no easy way to split the series impedance into primary and secondary components. Fortunately, such separation is not nec- essary to solve nonnal problems. TIlese same tests may also be perfonned on the secondary side of the trans- fonner if it is more convenient to do so because of voltage levels or other reasons. If the tests are performed on the secondary side, the results will naturally yield the equivalent circuit impedances referred to the secondary side of the transfonner in- stead of to the primary side. EXllmple 2-2. The equivalent circuit impedances ofa 20-kVA, 800CV240-V, 6O-Hz transformer are to be determined. The open-circuit test and the short-circuit test were perfonned on the primary side of the transfonner, and the following data were taken: Open-circuit tcst (on prinmry) Voc = 8000 V loc = O.214A Voc = 400W Short-circuit tcst (on prinmry) Vsc = 489V Isc = 2.5 A Psc = 240 W
- 117. TRANSFORMERS 93 Find the impedances of the approximate equivalent circuit referred to the primary side, and sketch that circuit. Solution The power factor during the open,circuit test is Poc PF = cos (J = oi-~ Voc loc 400 W = cos (J = (8000 V)(0.214 A) = 0.234 lagging The excitation admittance is given by = 0.214 A L- - , 0 23' 8(x)() V cos . = 0.cX)OO268 L -76.5° n = 0.0000063 - j O.OOOO261 = i -j i C M Therefore, 1 Rc = 0.0000Cl63 = 159 kO 1 XM = 0.000026 1 = 38.4 k!l The power factor during the short-circuit test is P", PF = cos (J = oi--~ Vsclsc = cos (J = (489~~(~5 A ) = 0.196 lagging The series impedance is given by V ZsE = .....K L -cos- l PF I", = i~; X L78.7° = 195.6 L78.7° = 38.4 + j l92 0 Therefore, the equivalent resistance and reactance are Req = 38.4 0 Xeq= 192 0 The resulting simplified equivalent circuit is shown in Figure 2- 21. (2- 45) (2- 47) (2- 49)
- 118. 94 ELECIRIC MACHINERY FUNDAMENTALS jXeq - - + I ;8.40 + j1920 '... I I" v ~ R, jX.. , j38.4 kD. 159kD. ,v, I - ""GURE 2-21 The equivalent cin:uit of Example 2- 2. 2.6 THE PER-UNIT SYSTEM OF MEASUREMENTS As the relatively simple Example 2- 1 showed, solving circuits containing trans- fonners can be quite a tedious operation because of the need 1 0 refer all the dif- ferent voltage levels on differenl sides of the transfonners in the system to a com- mon level. Only after this step has been taken can the system be solved for its voltages and currents. TIlere is another approach 1 0 solving circuits containing transfonners which eliminates the need for explicit voltage-level conversions at every transformer in the system. Instead, the required conversions are handled automatically by the method itself, without ever requiring the user to worry about impedance transfor- mations. Because such impedance transfonnations can be avoided, circuits con- taining many transfonners can be solved easily with less chance of error. This method of calculation is known as the per-unit (pu) system of measurements. There is yet another advantage to the per-unit system that is quite significant for electric machinery and transfonners. As the size of a machine or transfonner varies, its internal impedances vary widely. Thus, a primary circuit reactance of O. I n might be an atrociously high number for one transfonner and a ridiculously low number for another- it all depends on the device's voltage and power ratings. However, it turns out that in a per-unit system related to the device's ratings, ma- chine and transformer impednncesfall withinfairly nanvw ranges for each type and construction of device. This fact can serve as a usefuIcheck in problem solutions. In the per-unit system, the voltages, currents, powers, impedances, and other electrical quantities are not measured in their usual SI units (volts, amperes, watts, ohms, etc.). Instead, each electrical quantity is measured as a decimal fraction of some base level. Any quantity can be expressed on a per-unit basis by the equation Quantit r unit = Actual value. y pe base value of quantity (2- 53) where "actual value" is a value in volts, amperes, ohms, etc.
- 119. TRANSFORMERS 95 It is customary to select two base quantities to define a given per-unit sys- tem. The ones usually selected are voltage and power (or apparent power). Once these base quantities have been selected, all the other base values are related to them by the usual electrical laws. In a single-phase system, these relationships are (2- 54) (2- 55) (2- 56) and (2- 57) Once the base values of S (or P) and V have been selected, all other base values can be computed easily from Equations (2- 54) to (2- 57). In a power system, a base apparent power and voltage are selected at a spe- cific point in the system. A transfonner has no effect on the base apparent power of the system, since the apparent power into a transfonner equals the apparent power out of the transfonner [Equation (2-11 )] . On the other hand, voltage changes when it goes through a transformer, so the value of VI>a.. changes at every transformer in the system according to its turns ratio. Because the base quantities change in passing through a transfonner, the process of referring quantities to a common voltage level is automatically taken care of during per-unit conversion. EXllmple 2-3. A simple power system is shown in Figure 2- 22. This system con- tains a 480-V generator connected to an ideal I: 10 step-up transfonner, a transmission line, an ideal 20: I step-down transformer, and a load. The impedance of the transmission line is 20 +j60 n, and the impedance of the load is IOL30on. The base values for this system are chosen to be 480 V and 10 kVA at the generator. (a) Find the base voltage, current, impedance, and apparent power at every point in the power system. (b) Convert this system to its per-unit equivalent circuit. (c) Find the power supplied to the load in this system. (d) Find the power lost in the transmission line. YG 480LOo y '-''-' Region I RegIOn 2 FIGURE 2-22 The power system of Example 2- 3. RegIOn 3
- 120. 96 ELECIRIC MACHINERY FUNDAMENTALS Solutio" (a) In the generator region. Vbo.. = 480 V and 5_ = 10 kVA, so s~. lbase I = - ~ -- = ~, 10,000 VA = 2083A 480 V . Vbase I 480 V II z."... I = l ba .. I = 20.83 A = 23.04 The turns ratio oftransfonner Tl is a = 1110 = 0.1, so the base voltage in the transmission line region is v. = V base1 = 480V = 4800 V bo.•• 2 a 0.1 The other base quantities are Sbasel = 10 kVA km~ = 10,000 VA = 2083A .,....~ 4800 V . 4800 V Z basel = 2.083 A = 2304 n The turns ratio of transfonner Tl is a = 2011 = 20, so the base voltage in the load region is _ ~ = 4800 V = 240 V Vbase ) - a 20 The other base quantities are 5110..,)= IOkVA lbo.se) = lOi~;A = 41.67 A 240 V Z~ 1 = 41.67 A = 5.76 n (b) To convert a power system to a per-lUlit system, each component must be di- vided by its base value in its region of the system. The generator sper-lUlit volt- age is its actual value divided by its base value: _ 480LOoV _ ° Vo."" - 480V - 1.0LO pu The transmission line sper-unit impedance is its actual value divided by its base value: 20+j60n . ~iDe."" = 2304 n = 0.0087 + )0.0260 pu The loads per-lUlit impedance is also given by actual value divided by base value: The per-unit equivalent circuit of the power system is shown in Figure 2- 23.
- 121. TRANSFORMERS 97 ,~ IHme 0.0087 pu jO.0260 pu I , ~ - - I - 1 I I I + I "G ",lLOO ~ '" 1.736 L 30° per unit - I IG .I""' '" llino•"" '" IJo.d. I""' '" I"" Jo'IGURE 2-23 The per-unit equivalent circuit for Example 2- 3. (c) The current flowing in this per-unit power system is V I = ~ I""' z.."."" I LO° = " CO '".OOmoS' 7"+ C- J "'·OO.O" 26 fiO;C )'; +CCC"".7'36'L 7i0 30 WO ") I I I I 1LO° = " CO ,".OOmoS' 7 ~+- J"'·OO.O~ 26f,O~ ) ~+~C"'<.5"m,"+' j"O."S6as" ) 1.512 + jO.894 1.757 L30.6° = 0.569 L -30.6° pu Therefore, the per-unit power of the load is pload.1""' = PI""'Rpu = (0.569)2(1.503) = 0.487 and the actual power supplied to the load is PIo.! = flo.l.I""'Sbo>e = (0.487)( 10,000 VA) = 4870W (d) The per-unit power lost in the transmission line is pu....1""' = PI""'R1ine.pu = (0.569)2(0'(XJ87) = 0'(xl282 and the actual power lost in the transmission line is fline = fli....""St-. = (0.00282)(10,000 VA) = 28.2 W When only one device (transfonner or motor) is being analyzed, its own rat- ings are usually used as the base for the per-unit system. If a per-unit system based on the transfonner's own ratings is used, a power or distribution transformer's characteristics will not vary much over a wide range of voltage and power ratings. For example, the series resistance of a transfonner is usually about 0.01 per unit,
- 122. 98 ELECIRIC MACHINERY FUNDAMENTALS .. 2 1' ,.J (a) ,b, ""GURE 2-14 (a) A typical 13.2---kY to 1201240-Y distribution transformer. (Courtesy ofGeneml Electric Company.) (b) A cutaway view of the distribution transformer showing the shell-form transfonner inside it. (Courtesy ofGeneml Electric Company. ) and the series reactance is usually between 0.02 and 0.10 per unit. In general, the larger the transformer, the smaller the series impedances. 1lle magnetizing reac- tance is usually between about 10 and 40 per unit, while the core-loss resistance is usually between about 50 and 200 per unit. Because per-unit values provide a con- venient and meaningful way to compare transformer characteristics when they are of different sizes, transformer impedances are normally given in per-unit or as a percentage on the transformer's nameplate (see Figure 2- 46, later in this chapter). 1lle same idea applies to synchronous and induction machines as well: Their per-unit impedances fall within relatively narrow ranges over quite large size ranges. If more than one machine and one transformer are included in a single power system, the system base voltage and power may be chosen arbitrarily, but the entire system must have the same base. One common procedure is to choose the system base quantities to be equal to the base of the largest component in the system. Per-unit values given to another base can be converted to the new base by converting them to their actual values (volts, amperes, ohms, etc.) as an in- between step. Alternati vely, they can be converted directly by the equations
- 123. TRANSFORMERS 99 Ip,po R., jXoq I" po - - + j + , 0.012 fJ·06 I~H1 I'm V R, JXm V"po ,,~ 49.7 jl2 FIGURE 2-15 The per-unit equivalent circuit ofExample 2-4. Sba.se t (P, Q, S)poon base 2 = (P, Q, S)poon base ]- S-- "=, v.: v.: Vbase ] P" on base 2 = po on base ] Vbase2 (R, X, Z)P" 00 base 2 = (Vbase t?(Sbase 2) (R, X, Z)pu on base t(l< )'(S ) base 2 base ] (2- 58) (2- 59) (2-60) Example 2-4. Sketch the approximate per-unit equivalent circuit for the trans- fonner in Example 2- 2. Use the transformer's ratings as the system base. Solutioll The transfonner in Example 2- 2 is rated at 20 kVA, 8()(x)/240 V. The approximate equiva- lent circuit (Figure 2- 21) developed in the example was referred to the high-voltage side of the transfonner, so to convert it to per-unit, the primary circuit base impedance must be fOlUld. On the primary, Therefore, V!>Me I = 80CXl V Sbo>e I = 20,(XXl VA (Vb... t)l (8()(x) V)2 z....•• t= S = 20 00Cl VA =3200 0 ~, ' _ 38.4 + jl92 0 _ . ZsE,po - 3200 0 - 0.012 + jO.06 pu 159 ill Rc.pu = 3200 0 = 49.7 pu 38.4 kO ZM.pu = 3200 0 = 12 pu The per-unit approximate equivalent circuit, expressed to the transfonner's own base, is shown in Figure 2- 25.
- 124. 100 ELECTRIC MACHINERY RJNDAMENTALS 2.7 TRANSFORMER VOLTAGE REGULATION AND EFFICIENCY Because a real transformer has series impedances within it, the output voltage of a transfonner varies with the load even if the input voltage remains constant. To conveniently compare transfonners in this respect, it is customary to define a quantity called voltage regulation (VR). Full-load voltage regulation is a quantity that compares the output voltage of the transformer at no load with the output voltage at full load. lt is defined by the equation I VR = V S .nlV: n V S . fl x 100% I (2-6 1) Since at no load, Vs = Vp /a, the voltage regulation can also be expressed as (2-62) If the transformer equivalent circuit is in the per-unit system, then voltage regula- tion can be expressed as VR = ~ - ~ p.P" S.fl.p" ~ X S.fl.pu 100% (2-63) Usually it is a good practice to have as small a voltage regulation as possible. For an ideal transfonner, VR = 0 percent. lt is not always a good idea to have a low-voltage regulation, though-sometimes high-impedance and high-voltage reg- ulation transfonners are deliberately used to reduce the fault currents in a circuit. How can the voltage regulation of a transfonner be detennined? The Transformer Phasor Diagram To detennine the voltage regulation of a transfonner, it is necessary to understand the voltage drops within it. Consider the simplified transfonner equivalent circuit in Figure 2-1 Sb. TIle effects of the excitation branch on transformer voltage reg- ulation can be ignored, so only the series impedances need be considered. The voltage regulation of a transfonner depends both on the magnitude of these series impedances and on the phase angle of the current fl owing through the transformer. TIle easiest way to detennine the effect of the impedances and the current phase angles on the transformer voltage regulation is to examine a phasor diagram, a sketch of the phasor voltages and currents in the transformer. In all the following phasor diagrams, the phasor voltage Vs is assumed to be at an angle of 0°, and all other voltages and currents are compared to that refer- ence. By applying Kirchhoff's voltage law to the equivalent circuit in Figure 2-I Sb, the primary voltage can be found as
- 125. TRANSFORMERS 101 (2- 64) A transfonner phasor diagram is just a visual representation of this equation. Figure 2- 26 shows a phasor diagram of a transformer operating at a lagging power factor. It is easy to see that Vp la > ~ for lagging loads, so the voltage reg- ulation of a transformer with lagging loads must be greater than zero. A phasor diagram at unity power factor is shown in Figure 2- 27a. Here again, the voltage at the secondary is lower than the voltage at the primary, so VR > O. However, this time the voltage regulation is a smaller number than it was with a lag- ging current. If the secondary current is leading, the secondary voltage can actually be higher than the referred primary voltage. If this happens, the transformer actually has a negative voltage regulation (see Figure 2- 27b). FIGURE 2-26 Phasor diagram of a traruformer operating at a lagging power factor. (a) I, ,b, FIGURE 2-27 v, , v, v, , Phasor diagram of a transformer operating at (a) unity and (b) teading power factor.
- 126. 102 ELECTRIC MACHINERY RJNDAMENTALS , v, I, , v -" I , I jX"jI, I I -----t-- Vp ... V,+Roq l,cos0 +Xoql.Si~O I " ""GURE 2-28 Derivation of the approximate equation for Vpla. Transformer Efficiency Transformers are also compared and judged on their efficiencies. The efficiency of a device is defined by the equation Pout " ~ - X 100% flo Pout 1/ = x 100% ~ut+ ~oss (2-65) (2-66) TIlese equations apply to motors and generators as well as to transfonners. TIle transformer equivalent circuits make efficiency calculations easy. There are three types of losses present in transfonners: I. Copper (PR) losses. These losses are accounted for by the series resistance in the equivalent circuit. 2. Hysteresis losses. These losses were explained in Chapter I and are ac- counted for by resistor Re. 3. Eddy current losses. lllese losses were explained in Chapter I and are ac- counted for by resistor Re. To calculate the efficiency of a transfonner at a given load, just add the losses from each resistor and apply Equation (2-67). Since the output power is given by (2- 7) the efficiency of the transfonner can be expressed by (2-67)
- 127. TRANSFORMERS 103 Example 2-5. A 15-kVA, 23001230-V transformer is to be tested to detennine its excitation branch components, its series impedances, and its voltage regulation. The fol- lowing test data have been taken from the primary side of the transformer: Open-ciITuit tcsl Voc = 2300 V loc = 0.21 A Poc = SOW Short-circuillesl Vsc = 47V Isc = 6.0A Psc = I60W The data have been taken by using the connections shown in Figures 2- 19 and 2- 20. (a) Find the equivalent circuit of this transformer referred to the high-voltage side. (b) Find the equivalent circuit of this transformer referred to the low-voltage side. (c) Calculate the full-load voltage regulation at O.S lagging power factor, 1.0 power factor, and at O.Sleading power factor. (d) Plot the voltage regulation as load is increased from no load to full load at power factors of O.S lagging, 1.0, and O.S leading. (e) What is the efficiency of the transformer at full load with a power factor of O.S lagging? Solutioll (a) The excitation branch values of the transformer equivalent circuit can be calcu- lated from the open-circuit test data, and the series elements can be calculated from the short-circuit test data. From the open-circuit test data, the open-circuit impedance angle is _ - t Poe 60c - cos V oc:ioc _ - t SOW _ 84" - cos (2300 VXO.21 A) - The excitation admittance is thus = 0.21 A L -S40 2300 V = 9.13 x 1O - ~ L-84°0 = 0.0000095 - jO.OOOO9OS0 The elements of the excitation branch referred to the primary are 1 Rc = 0.0000095 = 105 kO 1 XM = O.()()(X)9()S = II kf! From the short-circuit test data, the short-circuit impedance angle is
- 128. 104 ELECTRIC MACHINERY RJNDA MENTALS V, " , - I Psc sc=cos V. J sc sc _ - I l60 W _ 554" - cos (47 V)(6 A) - . The equi valent series impedance is thus V " ZsE = -[- L ' oc " = ~: L55.4° n = 7.833L55.4° = 4.45 + j6.45 The series elements referred to the primary are Xeq = 6.45 n This equivalent circuit is shown in Figure 2- 29a. (b) To find the equivalent circuit referred to the low-voltage side, it is simply neces- sary to divide the impedance by il.Since a = NpiNs = 10, the resulting values are " R.., jXoq I, " , - " - + + I 4.450 j6.45 0 IhH j j'm V Rc ~ jXm aV, PJ05 k fl +jll k fl I ,,' "', R"" jXoq, ,, - - + j + 0.04450 .fl.0645 fl al~ + ~ I"'m ~= J0500 V, ~=jIJOO " I ,b , fo'IGURE 2- 29 The lransfer equivatent circuit for Example 2- 5 referred 10 (a) its primary side and (b) its secondary side.
- 129. TRANSFORMERS 105 Rc = 1050 n XM = lIOn Roq = 0.0445 n Xoq = 0.0645 n The resulting equivalent circuit is shown in Figure 2- 29b. (c) The full-load current on the secondary side of this transfonner is _ ~ _ 15,OCXl VA _ IS,med - V. - 230 V - 65.2 A S,med To calculate Vpla, use Equation (2-64): V, . a = V s + RoqIs + JXoqIs At PF = 0.8 lagging, current Is = 65.2 L - 36.9° A. Therefore, (2-64) ~ = 230L O ° V + (0.0445 fl)(65.2L-36.9° A) + j(0.0645 OX65.2L -36.9° A) = 230 LO° V + 2.90L -36.9° V + 4.21L53.10 V = 230 + 2.32 - j l.74 + 2.52 + j3.36 = 234.84 + j l.62 = 234.85 L0.40° V The resulting voltage regulation is Vp/a - VS() VR = x 100% V s.() = 234.85;0~230 V x 100% = 2.1% At PF = 1.0, current Is = 65.2 L 0° A. Therefore, (2-62) ':; = 230 LO° V + (0.0445 OX65.2 LO° A) + j(0.0645 ll)(65.2 LO° A) = 230 LOo V + 2.90L OoV + 4.2 I L90o V = 230 + 2.90 + j4.2 1 = 232.9 + j4.2 l = 232.94 L 1.04° V The resulting voltage regulation is VR = 232.9ijo~230 V x 100% = 1.28% At PF = 0.8 leading, current Is = 65.2 L36.9° A. Therefore, V : = 230 LO° V + (0.0445 n X65.2 ":::::36.9° A) + j(0.0645 0 )(65.2 L36.9° A) = 230 LO° V + 2.90 L36.9° V + 4.21L 126.9° V = 230 + 2.32 + j l.74 - 2.52 + j3.36 = 229.80 + j5.10 = 229.85 L 1.27° V The resulting voltage regulation is
- 130. 106 ELECTRIC MACHINERY RJNDA MENTALS v -!- '" 234.9 L 0.4° Y V,,,,230LOoy jXoql, '" 4.21 L 53.1° Y Roql, '" 2.9 L - 36.9° Y I,'" 65.2 L - 36.9° A V -.l'",2329L 104°Y , . . 1.1'~6~5~.2~L~O'~A~==:::::::==:::=~2:30~L~o:,~v~::JJ)4.21L 90'V 2.9LOoy ,hI I, '" 65.2 L 36.9° A V -.l'",2298L 127°Y " . . L /==:::=========:}IL 1269'V ~ a.:L36.90 Y 230LOoy "I fo'IGURE 2- 30 Transformer phasor diagrams for Example 2- 5. VR = 229.852~0~230 V x 100% = -0.062% Each of these three phasor diagrams is shown in Figure 2- 30. (d) The best way to plot the voltage regulation as a function of load is to repeat the calculations in part c for many different loads using MATLAB. A program to do this is shown below. % M-fil e : tra n s_vr.m % M-fil e t o ca l cula t e a nd p l ot the volt age r egul a tion % o f a tra n s f o rme r as a fun c ti on of l oad f or power % f act o r s o f 0 . 8 l agging, 1. 0, a nd 0 . 8 l eading . VS = 230; % Secondary voltage (V) a mps = 0: 6.52: 65 . 2; % CUrrent va lues (A)
- 131. TRANSFORMERS 107 'oq xoq 0.0445; 0.0645; !l; Equ i va l ent R (ohms) !l; Equ i va l ent X (ohms) !l; Ca l c u l ate the c urrent va l u es for the three !l; power f actors. The fi r s t row of I contains !l; the l agg i ng c urrent s, the second row contains !l; the uni t y c urrent s, and the third row contains , "0 l eadi ng c urrent s. I (1 , : ) ~P' • ( 0.8 j*0.6) ; I (2, : ) ~P' • ( 1. 0 I , I (3, : ) amps • ( 0.8 + j* 0.6) ; !l; Ca l c u l ate VP/ a. VPa = VS + Req. * I + j. *Xeq. * I ; !l; Ca l c u l ate vo l tage regul at i on VR = (abs (VPa ) - VS) . / VS .* 1 00; !l; Pl ot the vo l tage regu l at i on p l ot (amps, VR( l ,:), 'b- ' ) ; h o l d on; p l ot (amps, VR(2,:), 'k- ' ) ; p l ot (amps, VR (3, : ), 'r- .' ) ; • • • t i t l e ('Vo l tage Regu l at i on Ver s u s Load' ) ; x l abe l ('Load (A) ' ) ; y l abe l ('Voltage Regu l at i o n (%) ' ) ; Lagg i ng Unity Leadi ng l egend (' O.8 PF l agg i ng' , 'l .O PF' ,'0 . 8 PF l eadi ng' ) ; h o l d o ff ; The plot produced by this program is shown in Figure 2- 31. (e) To find the efficiency of the transformer. first calculate its losses. The copper losses are Peu = (ls)2Req = (65.2 A)2(0.0445ll) = 189 W The core losses are given by (Vp/a)2 (234.85 V)l Pcore = Rc = 1050 n = 52.5 W The output power of the transformer at this power factor is = (230 VX65.2 A) cos 36.9° = 12.()(X) W Therefore. the efficiency of the transformer at this condition is Vslscos () 7f = x 100% Peu + P.:.... + Vslscos () = 189W + 52.5 W + 12.()(X) W x = 98.03% 100% (2- 68)
- 132. 108 ELECTRIC MACHINERY RJNDAMENTALS Voltage regulation versus load 2.5~~Fln I I 0.8 PF lagging ---- 1.0 PF _._.- 0.8 PF leading 2 / .........:--- o .-._._ ._ .••._ ._ .__ ._ ._. __ ._. _•._._ .-'-. .--- _.- . --O.50 !--..J IOc---2 eO~-C 3± O--C 4"O---!50~--60 ±---! 70 Load (A) fo'IGURE 2- 31 Plot of voltage regulation versus load for the transformer of Example 2--5. 2.8 TRANSFORMER TAPS AND VOLTAGE REGULATION In previous sections of this chapter, transformers were described by their turns ra- tios or by their primary-to-secondary-voltage ratios. Throughout those sections, the turns ratio of a given transformer was treated as though it were completely fixed. In almost all real distribution transfonners. this is not quite true. Distribution trans- fonners have a series of taps in the windings to pennit small changes in the turns ratio ortile transfonner after it has left the factory. A typical installation might have four taps in addition to the nominal setting with spacings of 2.5 percent of full-load voltage between them. Such an arrangement provides for adjustments up to 5 per- cent above or below the nominal voltage rating of the transfonner. Example 2-6. A 500-kVA, 13,200/480-V distribution transfonner has four 2.5 percent taps on its primary winding. What are the voltage ratios of this transfonner at each tap setting? Solutioll The five possible voltage ratings of this transfonner are +5.0% tap +2.5% tap Nominal rating -2.5% tap -5.0% tap 13,8601480 V 13,5301480 V 13,2001480 V 12,8701480 V 12,540/480 V
- 133. TRANSFORMERS 109 The taps on a transfonner permit the transfonner to be adjusted in the field to accommodate variations in local voltages. However, these taps nonnally can- not be changed while power is being applied to the transformer. They must be set once and left alone. Sometimes a transfonner is used on a power line whose voltage varies widely with the load. Such voltage variations might be due to a high line imped- ance between the generators on the power system and that particular load (perhaps it is located far out in the country). Nonnal loads need to be supplied an essen- tially constant voltage. How can a power company supply a controlled voltage through high-impedance lines to loads which are constantly changing? One solution to this problem is to use a special transformer called a tap changing under load (TCUL) transformer or voltage regulator. Basically, a TCUL transformer is a transformer with the ability to change taps while power is con- nected to it. A voltage regulator is a TCUL transfonner with built-in voltage sens- ing circuitry that automatically changes taps to keep the system voltage constant. Such special transformers are very common in modem power systems. 2.9 THE AUTOTRANSFORMER On some occasions it is desirable to change voltage levels by only a small runount. For example, it may be necessary to increase a voltage from 110 to 120 V or from 13.2 to 13.8 kV These small rises may be made necessary by voltage drops that occur in power systems a long way from the generators. In such circumstances, it is wasteful and excessively expensive to wind a transfonner with two full wind- ings, each rated at about the same voltage. A special-purpose transformer, called an autotransformer. is used instead. A diagram of a step-up autotransfonner is shown in Figure 2- 32. In Figure 2- 32a, the two coils of the transformer are shown in the conventional manner. In Figure 2- 32b, the first winding is shown connected in an additive manner to the second winding. Now, the relationship between the voltage on the first winding and the voltage on the second winding is given by the turns ratio of the trans- former. However, the voltage at the output ofthe whole transformer is the sum of the voltage on the first winding and the voltage on the second winding. TIle first winding here is called the common winding, because its voltage appears on both sides of the transfonner. The smaller winding is called the series winding, because it is connected in series with the common winding. A diagram of a step-down autotransformer is shown in Figure 2- 33. Here the voltage at the input is the sum of the voltages on the series winding and the common winding, while the voltage at the output is just the voltage on the com- mon winding. Because the transformer coils are physically connected, a different tenni- nology is used for the autotransformer than for other types of transformers. The voltage on the common coil is called the common voltage Vc, and the current in that coil is called the common current [c. The voltage on the series coil is called the series voltage VSE, and the current in that coil is called the series current IsE.
- 134. 110 ELECTRIC MACHINERY RJNDAMENTALS (a) 'bJ ""GURE 2-32 A transfomler with its windings (a) connected in the conventional manner and (b) reconnected as an autotransformer. 'H - IH", ISE • IL"' ISE+ Ic I~ I N~ I, VH - • ) V, Ic N, ""GURE 2-.33 A step-down autotransformer connection. TIle voltage and current on the low-voltage side of the transfonner are called VL and IL , respectively, while the corresponding quantities on the high-voltage side of the transformer are called VH and IH . The primary side of the autotransfonner (the side with power into it) can be either the high-voltage side or the low-voltage side, depending on whether the autotransfonner is acting as a step-down or a step- up transfonner. From Figure 2- 32b the voltages and currents in the coils are re- lated by the equations Vc _ Nc VSE - NSE Nc Ic = NSE IsE (2-68) (2-69)
- 135. TRANSFORMERS III The voltages in the coils are related to the voltages at the tenninals by the equations YL = Ye YH = Ye + VSE (2- 70) (2- 71) and the currents in the coils are related to the currents at the terminals by the equations IL= l e+ lsE IH = I SE Voltage and Current Relationships in an Autotransformer (2- 72) (2- 73) What is the voltage relationship between the two sides of an autotransfonner? It is quite easy to determine the relationship between YHand Vv The voltage on the high side of the autotransfonner is given by (2- 71 ) (2- 74) Finally, noting that YL = Ve, we get (2- 75) (2- 76) The current relationship between the two sides of the transformer can be found by noting that IL = Ie + ISE From Equation (2--69), Ie = (NSEINc) lsE' so N>E IL = N ISE + ISE C Finally, noting that Iy = ISE' we find (2- 72) (2- 77)
- 136. 11 2 ELECTRIC MACHINERY RJNDAMENTALS NSE + Nc - N IH C "' I, NSE + Nc IH - Nc The Apparent Power Rating Advantage of Autotransformers (2- 78) (2- 79) It is interesting to note that not all the power traveling from the primary to the sec- ondary in the autotransformer goes through the windings. As a result, if a con- ventional transformer is reconnected as an autotransfonner, it can handle much more power than it was originally rated for. To understand this idea, refer again to Figure 2- 32b. Notice that the input apparent power to the autotransformer is given by Sin = VLIL and the output apparent power is given by (2-80) (2-81) It is easy to show, by using the voltage and current equations [Equations (2- 76) and (2- 79)], that the input apparent power is again equal to the output apparent power: (2-82) where SIO is defined to be the input and output apparent powers ofthe transformer. However, the apparent power in the transfonner windings is (2-83) 1lle relationship between the power going into the primary (and out the sec- ondary) of the transformer and the power in the transfonner's actual windings can be found as follows: Sw = Vcl c Using Equation (2- 79), we get = VL(JL - IH) = VLIL - ~IH -s - 10NsE + Nc (2-84) (2-85)
- 137. TRANSFORMERS 113 Therefore, the ratio of the apparent power in the primary and secondary ofthe autotransfonner to the app.:1.rent power actually traveling through its windings is (2-86) Equation (2--86) describes the apparent power rating advantage of an auto- transformer over a conventional transfonner. Here 5[0 is the apparent power enter- ing the primary and leaving the secondary of the transformer, while Sw is the ap- parent power actually traveling through the transfonner's windings (the rest passes from primary to secondary without being coupled through the transfonner 's wind- ings). Note that the smaller the series winding, the greater the advantage. For example, a SOOO-kVA autotransfonner connecting a 11 O-kV system to a 138-kV system would have an NelNsE turns ratio of 110:28. Such an autotrans- fonner would actually have windings rated at N" Sw= SION + N. " c 28 (2-85) The autotransfonner would have windings rated at on ly about lOIS kVA, while a conventional transformer doing the same job would need windings rated at S(x)() kVA. The autotransfonner could be S times smaller than the conventional trans- fonner and also would be much less expensive. For this reason, it is very advanta- geous to build transfonners between two nearly equal voltages as autotransfonners. The following example illustrates autotransformer analysis and the rating advantage of autotransformers. Example 2-7. A 100-VA 120/12-V transformer is to be connected so as to form a step-up autotransfonner (see Figure 2- 34). A primary voltage of 120 V is applied to the transformer. (a) What is the secondary voltage of the transfonner? (b) What is its maximum voltampere rating in this mode of operation? (e) Calculate the rating advantage of this autotransfonner connection over the trans- fonner 's rating in conventional 120112- V operation. Solutioll To accomplish a step-up transfonnation with a 120-V primary, the ratio of the HUllSon the common winding Ne to the turns on the series winding NSE in this transfonner must be 120:12 (or 10:1). (a) This transfonner is being used as a step-up transformer. The secondary voltage is VH, and from Equation (2- 75), NSE + Ne VH = Ne VL = 12 + 120 120 V = 132 V 120 (2- 75)
- 138. 114 ELECTRIC MACHINERY RJNDAMENTALS - ,-------~+ • - +~------+ Nd - 120) V'=1 20LOO V( • ""GURE 2-34 The autotransformer of Example 2--7. (b) The maximwn voltampere rating in either winding of this transfonner is 100 VA. How much input or output apparent power can this provide? To fmd out, examine the series winding. The voltage VSE on the winding is 12 V, and the voltampere rat- ing of the winding is 100 VA. Therefore, the maximum series winding current is 100 VA = 8.33A 12V Since ISE is equal to the secondary current Is (or IH) and since the secondary voltage Vs = VH= 132 V, the secondary apparent power is SOUl = V s Is = VHIH = (132 V)(S.33 A) = 1100 VA = Sin (e) The rating advantage can be calculated from part (b) or separately from Equa- tion (2-86). From part b, From Equation (2-86), 1100 VA = II 100 VA S[o N SE + Ne = Sw N SE = 12+120 = 132 = 11 12 12 By either equation, the apparent power rating is increased by a factor of II. (2-86) It is not nonnaJly possible to just reconnect an ordinary transformer as an autotransfonner and use it in the manner of Example 2- 7, because the insulation on the low-voltage side of the ordinary transfonner may not be strong enough to withstand the full output voltage of the autotransfonner connection. In transform-
- 139. TRANSFORMERS 115 FIGURE 2-35 (a) A variable-voltage autotransformer. (b) Cutaway view of the autotransformer. (Courtesy of Superior Electric Company.) ers built specifically as autotransfonners, the insulation on the smaller coil (the se- ries winding) is made just as strong as the insulation on the larger coil. It is common practice in power systems to use autotransfonners whenever two voltages fairly close to each other in level need to be transfonned, because the closer the two voltages are, the greater the autotransformer power advantage be- comes. 1lley are also used as variable transfonners, where the low-voltage tap moves up and down the winding. This is a very convenient way to get a variable ac voltage. Such a variable autotransfonner is shown in Figure 2- 35. The principal disadvantage of autotransformers is that, unlike ordinary transformers, there is a direct physical connection between the primary and the secondary circuits, so the electrical isolation of the two sides is lost. If a particu- lar application does not require electrical isolation, then the autotransfonner is a convenient and inexpensive way to tie nearly equal voltages together. The Internal Impedance of an Autotransformer Autotransformers have one additional disadvantage compared to conventional transformers. It turns out that, compared to a given transformer connected in the conventional manner, the effective per-unit impedance of an autotransformer is smaller by a factor equal to the reciprocal of the power advantage of the auto- transfonner connection. The proof of this statement is left as a problem at the end of the chapter. The reduced internal impedance of an autotransfonner compared to a con- ventional two-winding transformer can be a serious problem in some applications where the series impedance is needed to limit current flows during power system faults (short circuits). The effect of the smaller internal impedance provided by an autotransformer must be taken into account in practical applications before auto- transfonners are selected.
- 140. 116 ELECTRIC MACHINERY RJNDAMENTALS Example 2-8. A transfonner is rated at 1000 kVA, 1211.2 kY, 60 Hz when it is op- erated as a conventional two-winding transformer. Under these conditions, its series resis- tance and reactance are given as I and 8 percent per unit, respectively. This transfonner is to be used as a 13.2I12-kV step-down autotransformer in a power distribution system. In the autotransformer connection, (a) what is the transformer's rating when used in this man- ner and (b) what is the transformer's series impedance in per-unit? Solutio" (a) The NclNsE turns ratio must be 12:1.2 or 10:1. The voltage rating of this trans- former will be 13.2112 kV, and the apparent power (voltampere) rating will be NSF. + Nc 5[0 = NSF. Sw = 1+ IO I()(X)kVA = IIOOOkVA 1 ' (b) The transfonner's impedance in a per-unit system when cOIUlected in the con- ventional manner is Zoq = 0.01 + jO.08 pu separate windings The apparent power advantage of this autotransfonner is II, so the per-unit im- pedance of the autotransfonner connected as described is 0.01 + jO.08 Zoq= II = 0.00091 + jOJ'lJ727 pu autotransformer 2,10 THREE-PHASE TRANSFORMERS Almost all the major power generation and distribution systems in the world today are three-phase ac systems. Since three-phase systems play such an important role in modern life, it is necessary to understand how transformers are used in them. Transformers for three-phase circuits can be constructed in one of two ways. One approach is simply to take three single-phase transformers and connect them in a three-phase bank. An alternative approach is to make a three-phase transfonner consisting of three sets of windings wrapped on a common core. TIlese two possible types of transfonner construction are shown in Figures 2- 36 and 2- 37. The construction of a single three-phase transformer is the preferred practice today, since it is lighter, smaller, cheaper, and slightly more efficient. The older construction approach was to use three separate transfonners. That approach had the advantage that each unit in the bank could be replaced individually in the event of trouble, but that does not outweigh the ad vantages of a combined three- phase unit for most applications. However, there are still a great many installa- tions consisting of three single-phase units in service. A discussion of three-phase circuits is included in Appendix A. Sorne read- ers may wish to refer to it before studying the following material.
- 141. TRANSFORMERS 117 N" N" N~ ~ f---< ~ f No; N ~ f---< " FIGURE 2-36 A three-phase transformer bank composed of independent transformers. N " N~ No; ~ ~ ~ N " N" N" ~ ~ ~ FIGURE 2-37 A three-phase transformer wound on a single three-legged COTe.
- 142. 118 ELECTRIC MACHINERY RJNDAMENTALS Three-Phase Transformer Connections A three-phase transfonner consists of three transformers, either separate or com- bined on one core. The primaries and secondaries of any three-phase transfonner can be independently connected in either a wye (Y) or a delta (d ). This gives a to- tal of four possible connections for a three-phase transfonner bank: I. Wye-wye (Y-Y) 2. Wye-delta (Y-d) 3. Oelta-wye (d-Y) 4. Oelta-delta (d--&) 1llese connections are shown in Figure 2- 38. 1lle key to analyzing any three-phase transformer bank is to look at a single transfonner in the bank. Any single transfonner in the bank behaves exactly like the single-phase transformers already studied. 1lle impedance, voltage regula- tion, efficiency, and similar caJculations for three-phase transfonners are done on a per-phase basis, using exactly the same techniques already developed for single-phase transfonners. 1lle advantages and disadvantages of each type of three-phase transfonner connection are discussed below. WYE-WYE CONNECTION. TIle Y-Y connection of three-phase transformers is shown in Figure 2- 38a. In a Y-Y connection, the primary voltage on each phase of the transformer is given by V4>P = V LP / G. The primary-phase voltage is re- lated to the secondary-phase voltage by the turns ratio of the transformer. The phase voltage on the secondary is then related to the line voltage on the secondary by Vu; = GV4>S.1llerefore, overall the voltage ratio on the transformer is y - y (2-87) 1lle Y-Y connection has two very serious problems: I. If loads on the transfonner circuit are unbalanced, then the voltages on the phases of the transfonner can become severely unbalanced. 2. Third-harmonic voltages can be large. If a three-phase set of voltages is applied to a Y- Y transfonner, the voltages in any phase will be 1200 apart from the voltages in any other phase. However, the third-hannonic components ofeach ofthe three phases will be in phase with each other, since there are three cycles in the third hannonic for each cycle of the fun- damental frequency. There are always some third-harmonic components in a transfonner because of the nonlinearity of the core, and these components add up.
- 143. TRANSFORMERS 119 " " • • N" N" b + +b' I v .:( • Nn • )+v " N" - - , - " • • Nn N" " " (.j FIGURE 2-38 Three-phase transfonner connections and wiring diagrams: (a) Y- V: (b) y-~: (e) ~Y; (d) 6.~. The result is a very large third-harmonic component of voltage on top of the 50- ar 6O-Hz fundamental voltage. This third-harmonic voltage can be larger than the fundamental voltage itself. Both the unbalance problem and the third-harmonic problem can be solved using one of two techniques: I. Solidly ground the neutrals ofthe transfonners, especially the primary wind- ing's neutral. nlis connection permits the additive third-hannonic components to cause a current flow in the neutral instead of building up large voltages. The neutral also provides a return path for any current imbalances in the load. 2. Add a third (tel1iary) winding connected in 11 to the transfonner bank. Ifa third l1-connected winding is added to the transfonner, then the third-hannonic
- 144. 120 ELECTRIC MACHINERY RJNDAMENTALS components of voltage in the.1. will add up, causing a circulating current flow within the winding. nlis suppresses the third-hannonic components of voltage in the same manner as grounding the transfonner neutrals. The .1.-connected tertiary windings need not even be brought out of the transformer case, but they often are used to supply lights and auxiliary power within the substation where it is located. The tertiary windings must be large enough to handle the circulating currents, so they are usually made about one-third the power rating of the two main windings. One or the other ofthese correction techniques must be used any time a Y-Y transfonner is installed. In practice, very few Y-Y transfonners are used, since the same jobs can be done by one of the other types of three-phase transformers. WYE-DELTA CONNECTION. TIle Y--d connection of three-phase transformers is shown in Figure 2- 38b. In this connection, the primary line voltage is related to the primary phase voltage by VLP = V3V4>p, while the secondary line voltage is equal to the secondary phase voltage VLS = V<!>S' The voltage ratio of each phase is V ~ =a V., so the overall relationship betwccn the line voltage on the primary side of the bank and the line voltage on the secondary side of the bank is VLP _ V3Vp.p VL'> - V4>S I~~ = V3a (2-88) TIle Y-.6. connection has no problem with third-hannonic components in its voltages, since they are consumed in a circulating current on the.1. side. nlis con- nection is also more stable with respect to unbalanced loads, since the .1. partially redistributes any imbalance that occurs. TIlis arrangement does have one problem, though. Because of the connec- tion, the secondary voltage is shifted 30" relative to the primary voltage of the transformer.llle fact that a phase shift has occurred can cause problems in paral- leling the secondaries of two transformer banks together. TIle phase angles of transformer secondaries must be equal if they are to be paralleled, which means that attention must be paid to the direction of the 3~'' phase shift occurring in each transformer bank to be paralleled together. In the United States, it is customary to make the secondary voltage lag the primary voltage by 30°. Although this is the standard, it has not always been ob- served, and older installations must be checked very carefully before a new trans- fonner is paralleled with them, to make sure that their phase angles match.
- 145. TRANSFORMERS 121 V[J> r " b • • N" ::{ • Nn • Nn b' N" N" -:A.S V~ • , " " v ., ( N" • "LP • N" ) V ., ,----' b j b' • • N" Nn , ~ " • • N" Nn ,b , FIGURE 2-38 (b) Y-b. (continued) The connection shown in Figure 2- 38b will cause the secondary voltage to be lagging if the system phase sequence is abc. Ifthe system phase sequence is acb, then the connection shown in Figure 2- 38b will cause the secondary voltage to be leading the primary voltage by 30°. DELTA-WYE CONNECTION. A !:J..- Y connection of three-phase transformers is shown in Figure 2- 38c. In a !:J..- Y connection, the primary line voltage is equal to the primary-phase voltage VLP = Vo/>p, while the secondary voltages are related by VLS = V3V¢S' TIlerefore, the line-to-line voltage ratio of this transformer con- nection is
- 146. 122 ELECTRIC MACHINERY RJNDAMENTALS lj: .' + • " V LP [ V" "..-::;0+ N~ N" N" V~ b • Nn N" Nn • • 0 b' " " + + • V ,,( N" - - • N" } " + b ~ I • • Nn N~ , - b' • • N" N" (0) ""GURE 2-38 (e) d.- Y (continued) VLP _ V4>P VLS - V3"V¢>S (2-89) TIlis connection has the same advantages and the same phase shift as the Y- .d transformer. TIle connection shown in Figure 2- 38c makes the secondary voltage lag the primary voltage by 30°, as before.
- 147. TRANSFORMERS 123 Nn • +d " + + • )"LS v~[ V ii Nn • N" Nn , - b' N" • v ., • b + • c " + + • v ,,[ N" • 1 v " N" d - b _ I I b' • • N" Nn , ~ • • Nn Nn ,d , FIGURE 2-38 (d) ~6. (concluded) DELTA-DELTA CONNECTION. 1lle .6.---d connection is shown in Figure 2- 38d. In a 11- 11 connection, VLP = Vq.p and VLS = V.jtS, so the relationship between pri- mary and secondary line voltages is (2- 90) This transformer has no phase shift associated with it and no problems with unbalanced loads or hannonics. The Per-Unit System for Three-Phase Transformers The per-unit system of measurements applies just as well to three-phase trans- fonners as to single-phase transformers. TIle single-phase base equations (2- 53)
- 148. 124 ELECTRIC MACHINERY RJNDAMENTALS to (2- 56) apply to three-phase systems on a per-phase basis. If the total base voltampere value of the transfonner bank is called Sb..., then the base voltampere value of one of the transfonners SI4>.I>o.. is S"'. SI4>.hase = - 3 - (2- 9) and the base phase current and impedance of the transfonner are (2- 92a) (2- 92b) (2- 93a) (2- 93b) Line quantities on three-phase transformer banks can also be represented in the per-unit system. 1lle relationship between the base line voltage and the base phase voltage of the transformer depends on the connection of windings. If the windings are connected in delta, VL.l>ose = V••b. .. , while if the windings are con- nected in wye, VL hase = V3"V4>.ba... 1lle base line current in a three-phase trans- fonner bank is given by (2- 94) 1lle application of the per-unit system to three-phase transformer problems is similar to its application in the single-phase examples already given. Example 2-9. A 50-kVA 13.S0CV20S-V 6.-Y distribution transformer has a resis- tance of I percent and a reactance of 7 percent per lUlit. (a) What is the transfonner's phase impedance referred to the high-voltage side? (b) Calculate this transfonner's voltage regulation at full load and O.S PF lagging, using the calculated high-side impedance. (c) Calculate this transformer's voltage regulation under the same conditions, using the per-unit system. Solutioll (a) The high-voltage side of this transfonner has a base line voltage of 13,800 V and a base apparent power of 50 kVA. Since the primary is 6.-connected, its phase voltage is equal to its line voltage. Therefore, its base impedance is (2-93b)
- 149. TRANSFORMERS 125 = 3(13,SOOV)2 = II 426ll 50,DOOVA ' The per-unit impedance of the transfonner is Zoq = 0.0 I + jJ.07 pu so the high-side impedance in ohms is Zoq = Zoq.~ = (0.01 + jJ.07 pu)(11,426 ll) = 114.2 + jSOOll (b) To calculate the voltage regulation of a three-phase transfonner bank, determine the voltage regulation of any single transformer in the bank. The voltages on a single transfonner are phase voltages, so V - aV VR = #' V # x 100% " " The rated transformer phase voltage on the primary is 13,SOO V, so the rated phase current on the primary is given by S 14> = 3V • The rated apparent power S = 50 kVA, so _ 50,OOOVA _ 14> - 3(13,SOO V) - 1.20S A The rated ph~ voltage on the secondary ofthe transfonner is 20S VI v'1 = 120 V. When referred to the high-voltage side of the transformer, this voltage becomes V~ = aV otS = 13,SOO V. Assume that the transfonner secondary is operating at the rated voltage and current, and find the resulting primary phase voltage: V4>P = aV# + Roq l4> + jXoql4> = 13,SOOLO° V + (114.2 llX1.20SL -36.S7° A) +(iSOO llX1.20SL-36.S7° A) = 13,SOO + 13SL -36.S7° + 966.4L53.13° = 13,SOO + 110.4 - jS2.S + 579.S + )773.1 = 14,490 + j690.3 = 14,506 L2.73° V Therefore, VR = V#, ~ aV# x 100% "" = 14,5~3~~3,SOO x 100% = 5.1% (c) In the per-unit system, the output voltage is I L 0", and the current is I L - 36.S7°. Therefore, the input voltage is Vp = I LO° + (O.OIXI L -36.S7°) + (i0.07XI L-36.S7°) = I + O.OOS - jJ.OO6 + 0.042 + jO.056 = 1.05 + jO.05 = 1.051 L2.73°
- 150. 126 ELECTRIC MACHINERY RJNDAMENTALS The voltage regulation is VR = 1.05~.~ 1.0 x 100% = 5.1% Of course, the voltage regulation of the transfonner bank is the same whether the calculations are done in actual ohms or in the per-unit system. 2.11 THREE-PHASE TRANSFORMATION USING TWO TRANSFORMERS In addition to the standard three-phase transfonner connections, there are ways to perform three-phase transformation with only two transformers. All techniques that do so involve a reduction in the power-handling capability of the transform- ers, but they may be justified by certain economic situations. Some of the more important two-transfonner connections are I. The open-.6. (or V- V) connection 2. The open-Y-open-.6. connection 3. The Scott-T connection 4. The three-phase T connection E:1.ch of these transfonner connections is described below. The Open-.6. (or V-V) Connection In some situations a full transformer bank may not be used to accomplish three- phase transformation. For example, suppose that a .6.-.6. transformer bank com- posed of separate transformers has a damaged phase that must be removed for re- pair. 1lle resulting situation is shown in Figure 2- 39. If the two remaining secondary voltages are VA = V L 0° and VA = V L 120° V, then the voltage across the gap where the third transfonner used 1.0 be is given by Vc= - VA - VB - - V LO° - V L-120° - - V - (-0.5V - jO.866V) - -0.5 V + jO.866V = V L 120° V 1llis is exactly the same voltage that wou ld be present if the third transformer were still there. Phase C is sometimes called a ghost phase. Thus, the open-delta connection lets a transfonner bank get by with only two transfonners, allowing some power flow to continue even with a damaged phase removed. How much apparent power can the bank supply with one of its three trans- fonners removed? At first, it seems that it could supply two-thirds of its rated
- 151. TRANSFORMERS 127 , ~----------------~ HGURE 2-39 The open-a or v- v transformer connection. v, ------:--+ v, VA ",VLOO V VB '" V L 120" V '---~ b' apparent power, since two-thirds of the transfonners are still present. Things are not quite that simple, though. To understand what happens when a transfonner is removed, see Figure 2-40. Figure 2-40a shows the transformer bank in nonnal operation connected to a resistive load. If the rated voltage of one transformer in the bank is Vol> and the rated current is It/» then the maximum power that can be supplied to the load is P = 3V4 ,!4> cos (J The angle between the voltage Vol> and the current 101> in each phase is 0° so the to- tal power supplied by the transformer is P = 3V4>I4> cos (J = 3V4>I4> (2- 95) The open-delta transfonner is shown in Figure 2-40b. It is important to note the angles on the voltages and currents in this transfonner bank. Because one of the transformer phases is missing, the transmission line current is now equal to the phase current in each transformer, and the currents and voltages in the transfonner bank differ in angle by 30°. Since the current and voltage angles differ in each of the two transfonners, it is necessary to examine each transfonner individually to determine the maximum power it can supply. For transfonner I, the voltage is at an angle of 150° and the current is at an angle of 120°, so the expression for the maximum power in transformer I is PI = 3V4> 14> cos (150° - 120°) = 3V4>I4> cos 30° V1 = 2 V4>I4> (2- 96)
- 152. 128 ELECTRIC MACHINERY RJNDAMENTALS ..[f /. LOOA Nn I. L300A • - + • V.L300y N" + N" • Nn V. L - 90oy V. LI50o y • I. L --SXf' A ..[3 /. L 120° A • - Nn ..[f l. L _ 120° A I. L 150° A - (., I. LOo A I. L60oA - Nn • + V.L300y N" V. LI50o y • ~ L 120° A I. L 120° A • Nn • I. L _ 120° A - (b' ""GURE 2-40 (a) Voltages and currents in a ~--a transformer banlc. (b) Voltages and currents in an open-~ transformer banlc. R , • • , , 0 • d R , • • , , 0 • d For transfonner 2, the voltage is at an angle of 30° and the current is at an angle of 60°, so its maximum power is P2 = 3V4 ,!4> cos (30° - 60°) = 3V4> J4> cos (-30°) v:l =""2 ~J4> TIlerefore, the total maximum power of tile open-delta bank is given by P = V3V4>J4> (2- 97) (2- 98) TIle rated current is the same in each transfonner whether there are two or three of them, and the voltage is the same on each transfonner; so the ratio of the output power available from the open-delta bank. to the output power available from the normal three-phase bank is
- 153. TRANSFORMERS 129 Popen<1 _ V1"V,f,!1> _ _,_ _ P - 31:[ - V3 - 0.577 3 phase 1> 1> (2- 99) The available power out of the open-delta bank is only 57.7 percent of the origi- nal bank's rating. A good question that could be asked is: What happens to the rest of the open-delta bank's rating? After all, the total power that the two transformers to- gether can produce is two-thirds that of the original bank's rating. To find out, ex- amine the reactive power of the open-delta bank. The reactive power of trans- fonner I is Ql = 3V1>J1> sin (150° - 120°) = 3V1>J1> sin 30° = Yl V1>[1> The reactive power of transfonner 2 is Q2 = 3V1> [1> sin (30° - 60°) = 3 ~J1>sin (-300) = --1; V1>J1> Thus one transfonner is producing reactive power which the other one is con- suming. It is this exchange of energy between the two transformers that limits the power output to 57.7 percent ofthe original bank srating instead ofthe otherwise expected 66.7 percent. An alternative way to look at the rating of the open-delta connection is that 86.6 percent of the rating ofthe two remaining transformers can be used. Open-delta connections are used occasionally when it is desired to supply a small amount of three-phase power to an otherwise single-phase load. In such a case, the connection in Figure 2-4 1 can be used, where transformer Tl is much larger than transfonner Tt . c -----------------, "--, • Th_· T, T, T, ) Single- phase pJu~ power • power • T, • b - - - -y FIGURE 2-41 Using an open-<1 lransformer connection to supply a small amount of lhree-phase power along with a lot of single-phase power. Transformer Tl is much larger than transformer h
- 154. 130 ELECTRIC MACHINERY RJNDAMENTALS V~C , 0 • Missing ph= ""GURE 2-42 • N" :;." • N" . ~ '" • • ,-----,-~o ' b ~--+, r-'--i-~ b' • • co- ---- I--+-~,' " b' V~ 0' The open-Y-open-~ transfonner connection and wiring diagram. Note that this connection is identical to the y....a connection in Figure 3- 38b. except for the absence of the thinl transformer and the presence of the neutral lead. The Open-Wye-Open-Delta Connection TIle open-wye-open-delta connection is very similar to the open-delta connection except that the primary voltages are derived from two phases and the neutral. This type of connection is shown in Figure 2-42. It is used to serve small commercial customers needing three-phase service in rural areas where all three phases are not yet present on the power poles. With this connection, a customer can get three- phase service in a makeshift fashion until demand requires installation of the third phase on the power poles.
- 155. TRANSFORMERS 131 A major disadvantage of this connection is that a very large return current must flow in the neutral of the primary circuit. The Scott-T Connection The Scott-T connection is a way to derive two phases 90° apart from a three-phase power supply. In the early history of ac power transmission, two-phase and three- phase power systems were quite common. In those days, it was routinely neces- sary to interconnect two- and three-phase power systems, and the Scott-T trans- fonner connection was developed for that purpose. Today, two-phase power is primarily limited to certain control applications, but the Scott T is still used to produce the power needed to operate them. The Scott T consists of two single-phase transformers with identical ratings. One has a tap on its primary winding at 86.6 percent of full-load voltage. 1lley are connected as shown in Figure 2-43a. TIle 86.6 percent tap of transfonner T2 is connected to the center tap of transformer Tl . The voltages applied to the primary winding are shown in Figure 2-43b, and the resulting voltages applied to the pri- maries of the two transformers are shown in Figure 2- 43c. Since these voltages are 90° apart, they result in a two-phase output. It is also possible to convert two-phase power into three-phase power with this connection, but since there are very few two-phase generators in use, this is rarely done. The Three-Phase T Connection The Scott-T connection uses two transformers to convert three-phase power to two-phase power at a different voltage level. By a simple modification of that connection, the same two transfonners can also convert three-phase power to three-phase power at a different voltage level. Such a connection is shown in Fig- ure 2- 44. Here both the primary and the secondary windings of transformer Tl are tapped at the 86.6 percent point, and the taps are connected to the center taps of the corresponding windings on transformer Tl . In this connection T] is called the main transfonner and Tl is called the teaser transfonner. As in the Scott T, the three-phase input voltage produces two voltages 90° apart on the primary windings of the transformers. TIlese primary voltages pro- duce secondary voltages which are also 90° apart. Unlike the Scott T, though, the secondary voltages are recombined into a three-phase output. One major advantage of the three-phase T connection over the other three- phase two-transfonner connections (the open-delta and open-wye-open-delta) is that a neutral can be connected to both the primary side and the secondary side of the transfonner bank. This connection is sometimes used in self-contained three- phase distribution transformers, since its construction costs are lower than those of a full three-phase transformer bank. Since the bottom parts of the teaser transfonner windings are not used on ei- ther the primary or the secondary sides, they could be left off with no change in perfonnance. 1llis is, in fact, typically done in distribution transformers.
- 156. 132 ELECTRIC MACHINERY RJNDAMENTALS T, + b 86.6% / tap .<----- V" Center mp V :(, + - O-----~--~ T, " ab'" V L 120" " /r<-",VLO" "c~"'VL - 12O" (a) T, ~-----~. + • V" T, " p2 '" 0.866 V L 90" )-- - - V. (b' V " S2"' - L'Xf' , (d' ""GURE 2-43 " St '" -...t. L 0" , (,' The Scolt-T transformer conneeliOll. (a) Wiring diagram; (b) the three-phase input voltages; (e) the vollages on the transformer primary windings; (d) the two-phase secondary voltages.
- 157. + V oo V ro V :( " T, ( N, 86.6% ''P V " Center b N, "p " + '-----'" e Vah ", V L 1200 Vbc",VLOo V ...,'" V L _ 1200 >--__ V~ ,b , N, a"' - N, ~t T, '--_---'-__.>V" Vct ",.!':L - 1200 " Note: FIGURE 2-44 V VSt '" VBC '" -Lr:f' " Va ", - VSt - VS2 ,d , ,,' r---------------- -Q n , , T, A , + , 57.7% N, , "p +" , , , l _________ ) V n 86.6% V~ "p N, + B " J,e ------- V " T, C + ,e, }----- VBC'" *L 0 0 Va ",.!':L - 1200 " ,., The three-phase T tr:I.nsformer connection. (a) Wiring diagram; (b) the three-phase input voltages; (e) the voltages on the transfonner primary windings; (d) the voltages on the transformer secondary windings; (e) the resulting three-phase secondary voltages. 133 V a
- 158. 134 ELECTRIC MACHINERY RJNDAMENTALS 2.12 TRANSFORMER RATINGS AND RELATED PROBLEMS Transfonners have four major ratings: apparent power, voltage, current, and fre- quency. This section examines the ratings of a transfonner and explains why they are chosen the way they are. It also considers the related question of the current inrush that occurs when a transformer is first connected to the line. The Voltage and Frequency Ratings of a Transformer TIle voltage rating of a transformer serves two functions. One is to protect the winding insulation from breakdown due to an excessive voltage applied to it. TIlis is not the most serious limitation in practical transfonners. TIle second function is related to the magnetization curve and magnetization current of the transformer. Figure 2-11 shows a magnetization curve for a transformer. If a steady-state voltage vet) = V M sin wi V is applied to a transfonner's primary winding, the nux ofthe transfonner is given by 4>(t) = ~p fv(t) dt = ~ f VM sinwtdt p VM - -- cos wt wNp ('-1 00) If the applied voltage v(t) is increased by 10 percent, the resulting maximum nux in the core also increases by 10 percent. Above a certain point on the magne- tization curve, though, a 10 percent increase in nux requires an increase in mag- netization current much larger than 10 percent. TIlis concept is illustrated in Fig- ure 2- 45. As the voltage increases, the high-magnetization currents soon become unacceptable. TIle maximum applied voltage (and therefore the rated voltage) is set by the maximum acceptable magnetization current in the core. Notice that voltage and frequency are related in a reciprocal fashion if the maximum nux is to be held constant: V_ q,max = wN p (2- 101) TIlUS, ifa 60-Hz transfonner is to be operated on 50 Hz, its applied voltage must also be reduced by one-sixth or the peak nux in the core will be too high. TIlis re- duction in applied voltage with frequency is called derating. Similarly, a 50-Hz transfonner may be operated at a 20 percent higher voltage on 60 Hz if this action does not cause insulation problems.
- 159. TRANSFORMERS 135 ------+++-----1- 3' ('"N/). A • turns I 2 3 FIGURE 2-45 The elTect of the peak flux in a tnlnsformer core upon the required magnetization current. Example 2-10. A l-kVA. 230/115-V. 60-Hz single-phase transfonner has 850 turns on the primary winding and 425 llU1lS on the secondary winding. The magnetization curve for this transfonner is shown in Figure 2-46. (a) Calculate and plot the magnetization current of this transfonner when it is run at 230 V on a 60-Hz power source. What is the rms value of the magnetization clUTent ? (b) Calculate and plot the magnetization ClUTent of this transfonner when it is nUl at 230 V on a 50-Hz power source. What is the rms value of the magnetization cur- rent ? How does this ClUTent compare to the magnetization current at 60 Hz? Solutioll The best way to solve this problem is to calculate the flux as a function of time for this core. and then use the magnetization curve to transform each flux value to a corresponding magnetomotive force. The magnetizing current can then be determined from the equation
- 160. 136 ELECTRIC MACHINERY RJNDA MENTALS Magnetization curve for 2301115·V transfonner 1.4 ~~=~=~:CC:;==;:=-';-'=;==-,-~ 1.2 ~ 0.8 " , ti: 0.6 0.4 0.2 °0 ~-C 2~ OO ~-~ ~c-0600 ~-C8~ OO ~-I" OOO ~-c 12 000 OCC1C~ ~-C I60000-c l~ 800 MMF. A - turns HGURE 2-46 Magnetization curve for the 2301115·V transfonner of Example 2- 10. i=~ N, (2- 102) Assruning that the voltage applied to the core is v(t) = VM sin w t volts, the flux in the core as a function of time is given by Equation (2-101): ~(t) = VM - -- coswt wN, (2-1 00) The magnetization curve for this transfonner is available electronically in a file called mag_curve_ l . dat . This file can be used by MATLAB to translate these flux values into corresponding mmf values, and Equation (2- 102) can be used to fmd the required magnetization current values. Finally, the nns value of the magnetization current can be calculated from the equation I = l ITPdt - T O A MATLAB program to perform these calculations is shown below: ~ M-file, mag_curre nt.m ~ M-file t o ca l c ula t e a nd p l o t the magn e tiza tio n ~ c urre nt o f a 230 / 115 tra n s f o rme r oper a ting a t ~ 230 vo lt s a nd 50 /60 Hz. Thi s p r ogr a m a l so ~ ca l c ula t es the rms va lu e o f the mag . current . ~ Load the magne tiza tio n cu rve. It i s in t wo ~ co lumns, with the first column being mmf and ~ the second co lumn be ing flux . l oad mag_curve_l. dat ; mmf_dat a = mag_curve_l ( : , l ) ; (2-103)
- 161. TRANSFORMERS 137 % Initia liz e va lues VM 325; NP = 850; % Max imum vo lt age (V) % Prima r y turn s % Ca l c ula t e a ngula r ve l oc ity f or 60 Hz fr eq = 60; % Freq (Hz ) w = 2 * p i .. fr eq; % Ca l c ula t e flu x ver s u s time time 0, 1 /300 0, 1 /3 0; % 0 t o 1 /3 0 sec flu x = - VM /(w*NP ) * cos(w .* time ) ; % Ca l c ula t e the mmf correspo nd ing t o a g i ven flu x % u s ing the flu x ' s int e r po l a tion func tion. mmf = int e r p l ( flu x_dat a, mmf_dat a, flu x ) ; % Ca l c ula t e the magne tiz a tio n c urre nt im = mmf 1 NP ; % Ca l c ula t e the rms va lue o f the c urre nt irms = sqrt (sum ( im. A 2 )/ l e ngth ( im )) ; d i sp( ['The rms c urre nt a t 60 Hz i s " num2str (inns) ] ) ; % Plot the magne tiz a tion c urre nt. fi gure ( l ) s ubp l ot (2, 1 , 1 ) ; p l ot (time, im ) ; title ('b fMagne tiz a tion Curre nt a t 60 Hz' ) ; x l abe l ('b fTime (s) ' ) ; y l abe l ('b f itI_( m) rm (A) ' ) ; ax i s( [ O 0.04 - 2 2 ] ) ; g rid on ; % Ca l c ula t e a ngula r ve l oc ity f or 50 Hz fr eq = 50; % Freq (Hz ) w = 2 * p i .. fr eq; % Ca l c ula t e flu x ver s u s time time 0, 1 / 2500, 1 / 25; % 0 t o 1 / 25 sec flu x = - VM /(w*NP ) * cos(w . * time ) ; % Ca l c ula t e the mmf correspo nd ing t o a g i ven flu x % u s ing the flu x ' s int e r po l a tion func tion. mmf = int e r p l ( flu x_dat a, mmf_dat a, flu x ) ; % Ca l c ula t e the magne tiz a tio n c urre nt im = mmf 1 NP ; % Ca l c ula t e the rms va lue o f the c urre nt irms = sqrt (sum ( im. A 2 )/ l e ngth ( im )) ; d i sp( ['The rms c urre nt a t 50 Hz i s " num2str (inns) ] ) ;
- 162. 138 ELECTRIC MACHINERY RJNDAMENTALS ~ '" -' 1.414 0.707 0 -0.7(17 - 1.414 0 0.005 om om5 om 0.025 0.Q3 0.Q35 0.04 Time (s) ,,' 1.414, - , - -"'C---,--,--,--"'__---,----, 0.707 50Hz o -0.7(17 -1.414:'---o~~~~cc!~--c'~--o+.co-~o-cc!=~ o 0.005 om om5 om 0.025 0.Q3 0.Q35 0.04 Time (s) ,b, ""GURE 2-47 (a) Magnetization current for the transformer operating at 60 Hz. (b) Magnetization current for the transformer operating at 50 Hz. % Pl ot the magnet i zat i on curre nt. s ubpl ot (2, 1 ,2); p l ot (t i me , i m) ; tit l e (' b f Magnet i zat i on Curren t at 50 Hz' ) ; x l abe l (' b fTime (s) ' ) ; y l abe l (' b f it I _{m) nn (A) ' ) ; axi s( [O 0 . 04 - 2 2 ] ) ; gri d on; When this program executes, the results are ,.. mag_ current The nns c urrent at 60 Hz i s 0 .4 894 The nns c urrent at 50 Hz i s 0 . 79252 The resulting magnetization currents are shown in Figure 2-47. Note that the nns magne- tization current increases by more than 60 percent when the frequency changes from 60 Hz to 50 Hz. The Apparent Power Rating of a Transformer TIle principal purpose of the apparent power rating of a transfonner is that, to- gether with the voltage rating, it sets the current flow through the transformer windings. TIle current now is important because it controls the jlR losses in trans- fonner, which in turn control the heating of the transformer coils. It is the heating
- 163. TRANSFORMERS 139 that is critical, since overheating the coiIs of a transformer drastically shortens the life of its insulation. The actual voltampere rating of a transfonner may be more than a single value. In real transfonners, there may be a voltampere rating for the transformer by itself, and another (higher) rating for the transformer with forced cooling. The key idea behind the power rating is that the hot-spot temperature in the trans- fonner windings must be limited to protect the life of the transfonner. If a transfonner's voltage is reduced for any reason (e.g., if it is operated at a lower frequency than normal), then the transfonner's voltrunpere rating must be re- duced by an equal amount. If this is not done, then the current in the transfonner's windings will exceed the maximum pennissible level and cause overheating. The Problem of Current Inrush A problem related to the voltage level in the transfonner is the problem of current inrush at starting. Suppose that the voltage v(t) = VM sin (wi + 6) v (2-104) is applied at the moment the transfonner is first connected to the power line. The maximum flux height reached on the first half-cycle of the applied voltage depends on the phase of the voltage at the time the voltage is applied. If the initial voltage is vet) = VM sin (wi + 90°) = V M cos wt v (2-1 05) and if the initial flux in the core is zero, then the maximum flux during the first half-cycle wi ll just equal the maximum flux at steady state: V=, q,max = wNp (2-1 01) This flux level is just the steady-state flux, so it causes no special problems. But if the applied voltage happens to be vet) = VM sin wi V the maximum flux during the first half-cycle is given by I f"'· q,(t)= N p 0 VM sinwidt V M I"'· ~ - -- cos wt wNp 0 VM ~ - - [(- 1) - (1)[ wNp (2-106) This maximum flux is twice as high as the normal steady-state flux. If the magnetization curve in Figure 2-11 is examined, it is easy to see that doubling the
- 164. 140 ELECTRIC MACHINERY RJNDAMENTALS Rated current ""GURE 2-48 v(I)=Vm sinrot f The current inrush due to a transformer's magnetization current on starting. maximum flux in the core results in an enonnous magnetization current. In fact, for part of the cycle, the transformer looks like a short circuit, and a very large current fl ows (see Figure 2-48). For any other phase angle of the applied voltage between 90°, which is no problem, and 0°, which is the worst case, there is some excess current flow. The applied phase angle of the voltage is not Ilonnally controlled on starting, so there can be huge inrush currents during the first several cycles after the transfonner is connected to the line. The transfonner and the power system to which it is con- nected must be able to withstand these currents. The Transformer Nameplate A typical nameplate from a distribution transformer is shown in Figure 2-49. TIle infonnation on such a nameplate includes rated voltage, rated kilovoltamperes, rated frequency, and the transfonner per-unit series impedance. lt also shows the voltage ratings for each tap on the transfonner and the wiring schematic of the transfonner. Nameplates such as the one shown also typically include the transformer type designation and references to its operating instructions. 2.13 INSTRUMENT TRANSFORMERS Two special-purpose transfonners are used with power systems for taking mea- surements. One is the potential transfonner, and the other is the current transfonner.
- 165. TRANSFORMERS 141 I ~ ! , ." j ~~ 3PHASI: ClASS 0 A iASIC 1Mf'UlSl UYEl II'IWHtDHtG l VI'll NIlItIG KV Wl:IGHTSI~ POINUS 1m:IIIOR .. NPlDOIC(WC U.UDVOlTS ~ xI "0 "l ~ ••••,r;::;::;::;::;: ~ "0 x , "i "i • , • ~ :., .1:;-""1 I I I ! HI~I HO~ x, xl "l ~ 00GROUND STlW' 11! = I DISTRIBUTION TRANSFORMER COIITAiNSHON-PCaATnr.(J' a~ ,-cccccccccc~cccccccc~______-''''''' __..~·"~ ··"'""" .. '·"·"·····"' .. " ,-~ • f-a ~ o ~ FIGURE 2-49 , , ! , j A sample distribution transfonner nameplate. Note the ratings li5ted: voltage, frequen>;y, apparem power. and tap settings. (Courtesy o/General Electric Company.) A potential transformer is a specially wound transformer with a high- voltage primary and a low-voltage secondary. It has a very low power rating, and its sole purpose is to provide a sample of the power system's voltage to the in- struments monitoring it. Since the principal purpose of the transfonner is voltage sampling, it must be very accurate so as not to distort the true voltage values too badly. Potential transfonners of several accuracy classes may be purchased de- pending on how accurate the readings must be for a given application. Current transformers sample the current in a line and reduce it to a safe and measurable level. A diagram of a typical current transformer is shown in Figure 2- 50. The current transfonner consists of a secondary winding wrapped around a ferromagnetic ring, with the single primary line running through the center of the ring. 1lle ferromagnetic ring holds and concentrates a small sample of the flux from the primary line. That flux then induces a voltage and current in the sec- ondary winding. A current transformer differs from the other transformers described in this chapter in that its windings are loosely coupled. Unlike all the other transfonners, the mutual flux <PM in the current transfonner is smaller than the leakage flux <PL- Because of the loose coupling, the voltage and current ratios of Equations (2- 1) to (2- 5) do not apply to a current transfonner. Nevertheless, the secondary current
- 166. 142 ELECTRIC MACHINERY RJNDAMENTALS - Instruments ""GURE 2-50 Sketch of a current transformer. in a current transfonner is directly proportional to the much larger primary cur- rent, and the device can provide an accurate sample of a line's current for mea- surement purposes. Current transformer ratings are given as ratios of primary to secondary cur- rent. A typical current transfonner ratio might be 600:5,800:5, or 1000:5. A 5-A rating is standard on the secondary of a current transfonner. It is imponant to keep a current transfonner short-circuited at all times, since extremely high voltages can appear across its open secondary terminals. In fact, most relays and other devices using the current from a current transformer have a shoning interlock which must be shut before the relay can be removed for inspection or adjustment. Without this interlock, very dangerous high voltages wi ll appear at the secondary terminals as the relay is removed from its socket. 2.14 SUMMARY A transfonner is a device for converting electric energy at one voltage level to electric energy at another voltage level through the action of a magnetic field. It plays an extremely important role in modern life by making possible the econom- ical long-distance transmission of electric power. When a voltage is applied to the primary ofa transfonner, a flux is produced in the core as given by Faraday's law. The changing flux in the core then induces a voltage in the secondary winding of the transformer. Because transfonner cores have very high penneability, the net magnetomotive force required in the core to produce its flux is very small. Since the net magnetomotive force is very small, the primary circuit's magnetomotive force must be approximately equal and op- posite to the secondary circuit's magnetomotive force. nlis fact yields the trans- fonner current ratio. A real transfonner has leakage nuxes that pass through either the primary or the secondary winding, but not both. In addition there are hysteresis, eddy current, and copper losses. These effects are accounted for in the equivalent circuit of the
- 167. TRANSFORMERS 143 transfonner. Transfonner imperfections are measured in a real transfonner by its voltage regulation and its efficiency. The per-unit system of measurement is a convenient way to study systems containing transformers, because in this system the different system voltage lev- els disappear. In addition, the per-unit impedances of a transfonner expressed to its own ratings base fall within a relatively narrow range, providing a convenient check for reasonableness in problem solutions. An autotransformer differs from a regular transfonner in that the two wind- ings of the autotransformer are connected. The voltage on one side of the trans- fonner is the voltage across a single winding, while the voltage on the other side of the transformer is the sum of the voltages across both windings. Because only a portion of the power in an autotransformer actually passes through the windings, an autotransfonner has a power rating advantage compared to a regular trans- fonner ofequal size. However, the connection destroys the electrical isolation be- tween a transformer's primary and secondary sides. The voltage levels of three-phase circuits can be transfonned by a proper combination of two or three transfonners. Potential transfonners and current transformers can sample the voltages and currents present in a circuit. Both de- vices are very common in large power distribution systems. QUESTIONS 2- 1. Is the ttU1lSratio of a transfonner the same as the ratio of voltages across the trans- fonner? Why or why not? 2-2. Why does the magnetization current impose an upper limit on the voltage applied to a transformer core? 2-3. What components compose the excitation current of a transfonner? How are they modeled in the transformer 's equivalent circuit? 2-4. What is the leakage flux in a transfonner? Why is it modeled in a transformer equivalent circuit as an inductor? 2-5, List and describe the types of losses that occur in a transfonner. 2-6. Why does the power factor of a load affect the voltage regulation of a transfonner? 2-7. Why does the short-circuit test essentially show only PR losses and not excitation losses in a transfonner? 2-8. Why does the open-circuit test essentially show only excitation losses and not PR losses? 2-9. How does the per-lUlit system of measurement eliminate the problem of different voltage levels in a power system? 2-10. Why can autotransformers handle more power than conventional transformers of the same size? 2-11, What are transfonner taps? Why are they used? 2-12. What are the problems associated with the Y- Ythree-phase transformer cOlUlection? 2-13. What is a TCUL transformer? 2-14. How can three-phase transformation be accomplished using only two transformers? What types ofconnections can be used?What are their advantages and disadvantages?
- 168. 144 ELECTRIC MACHINERY RJNDAMENTALS 2- 15. Explain why the open-a transformer connection is limited to supplying 57.7 percent of a normal a - a transfonner bank's load. 2-16. Can a 60-Hz transfonner be operated on a 50-Hz system? What actions are neces- sary to enable this operation? 2-17. What happens to a transformer when it is first cotulected to a power line? Can any- thing be done to mitigate this problem? 2-18. What is a potential transformer? How is it used? 2-19. What is a current transfonner? How is it used? 2-20. A distribution transfonner is rated at 18 kVA, 20,000/480 V, and 60 Hz. Can this transformer safely supply 15 kVA to a 4 15-V load at 50 Hz? Why or why not? 2-21. Why does one hear a hwn when standing near a large power transformer? PROBLEMS 2-1. The secondary winding of a transfonner has a terminal voltage of vi-t) = 282.8 sin 377t V. The turns ratio of the transformer is 100:200 (a = 0.50). If the secondary current of the transfonner is ii-t) = 7.rn sin (377t - 36.87°) A, what is the primary current of this transfonner? What are its voltage regulation and efficiency? The im- pedances of this transfonner referred to the primary side are ~=0.20 0 Xoq = 0.750 0 Rc = 300 0 XM = 80 0 2-2. A 20-kVA 8(x)()/480-V distribution transformer has the following resistances and reactances: Rp= 32 0 Xp = 45 0 Rc = 250 kfl Rs = 0.05 0 Rs= 0.06 0 XM=30 kO The excitation branch impedances are given referred to the high-voltage side of the transformer. (a) Find the equi valent circuit of this transfonner referred to the high-voltage side. (b) Find the per-unit equivalent circuit of this transformer. (c) Assume that this transformer is supplying rated load at 480 V and 0.8 PF lag- ging. What is this transformer's input voltage? What is its voltage regulation? (d) What is the transfonner's efficiency under the conditions of part (c)? 2-3, A 1000-VA 2301115-V transformer has been tested to determine its equivalent cir- cuit. The results of the tests are shown below. Open-circuit test Voc = 230V loc = O.4S A Poc = 30 W Short-circuit tcst Vsc = 19.1 V Isc = 8.7 A Psc = 42.3 W All data given were taken from the primary side of the transformer.
- 169. TRANSFORMERS 145 (a) Find the equivalent circuit of this transformer referred to the low-voltage side of the transfonner. (b) Find the transformer's voltage regulation at rated conditions and ( I) 0.8 PF lag- ging, (2) 1.0 PF, (3) 0.8 PF leading. (c) Detennine the transfonner's efficiency at rated conditions and 0.8 PF lagging. 2-4. A single-phase power system is shown in Figure P2-1. The power source feeds a lOO-kVA 14/2.4-kV transformer through a feeder impedance of 40.0 +j l50 n. The transformer's equivalent series impedance referred to its low-voltage side is 0. 12 + jO.5 n. The load on the transfonner is 90 kW at 0.80 PF lagging and 2300 V. (a) What is the voltage at the power source of the system? (b) What is the voltage regulation of the transfonner? (c) How efficient is the overall power system? 40fl jl50fl 0.12fl jO.5!l • • ( + Lood 90 kW V, ,-,_-, 0.g5 PF lagging ~~~~~~~~----~~-~~ Source Feeder Transformer Load (transmission line) FIGURE P2-1 The circuit of Problem 2-4. 2-5. When travelers from the United States and Canada visit Europe, they encounter a different power distribution system. Wall voltages in North America are 120 V nns at 60 Hz, while typical wall voltages in Europe are 220 to 240 V at 50 Hz. Many travelers carry small step-uplstep-down transfonners so that they can use their ap- pliances in the countries that they are visiting. A typical transformer might be rated at I kVA and 120/240 V. It has 500 turns of wire on the 120-V side and I()(X) turns of wire on the 240-V side. The magnetization curve for this transfonner is shown in Figure P2- 2, and can be found in file p22 .mag at this book's website. (a) Suppose that this transfonner is connected to a 120-V, 60-Hz power source with no load connected to the 240-V side. Sketch the magnetization current that would flow in the transformer. (Use MATLAB to plot the current accurately, if it is available.) What is the nns amplitude of the magnetization current? What percentage of full-load current is the magnetization clUTent? (b) Now suppose that this transformer is connected to a 240-V, 50-Hz power source with no load connected to the 120-Vside. Sketch the magnetization current that would flow in the transformer. (Use MATLAB to plot the current accurately, if it is available.) What is the nns amplitude of the magnetization current? What percentage of full-load current is the magnetization clUTent? (c) In which case is the magnetization current a higher percentage of full-load cur- rent? Why?
- 170. 146 ELECTRIC MACHINERY RJNDAMENTALS 0.001 2 0 ./' V 8 / 0.001 0.000 ~ .. 0.(XX)6 o / " 0.0004 V 2 v 0 o 100 150 200 250 300 350 400 0.000 M:MF. A ' turns H GURE 1'2- 2 Magnetization curve for the transformer of Problem 2- 5. 2-6. A 15-kVA 800CV230-V distribution transformer has an impedance referred to the pri- mary of 80 + j300 il. The components of the excitation branch referred to the pri- mary side are Rc = 350 ill and XM = 70 kil. (a) If the primary voltage is 7967 V and the load impedance is ZL = 3.0 + j 1.5 n. what is the secondary voltage of the transfonner? What is the voltage regulation of the transfonner? (b) If the load is discOIUlected and a capacitor of -j4.0 il is connected in its place. what is the secondary voltage of the transfonner? What is its voltage regulation lUlder these conditions? 2-7. A 5OCXl-kVA 2301l3.8-kV single-phase power transfonner has a per-unit resistance of I percent and a per-unit reactance of 5 percent (data taken from the transfonner's nameplate). The open-circuit test performed on the low-voltage side of the trans- fonner yielded the following data: Voc = 13.8 kV loc = 15.1 A Poc = 44.9kW (a) Find the equi valent circuit referred to the low-voltage side of this transfonner. (b) If the voltage on the secondary side is 13.8 kV and the power supplied is 4000 kW at 0.8 PF lagging. find the voltage regulation of the transfonner. Find its efficiency. 450
- 171. TRANSFORMERS 147 2-8. A 200-MVA. 1512()()"'kV single-phase power transfonner has a per-unit resistance of 1.2 percent and a per-lUlit reactance of 5 percent (data taken from the transformer's nameplate). The magnetizing impedance isjSO per unit. (a) Find the equivalent circuit referred to the low-voltage side of this transformer. (b) Calculate the voltage regulation of this transfonner for a full-load current at power factor of 0.8 lagging. (c) Assrune that the primary voltage of this transformer is a constant 15 kV. and plot the secondary voltage as a function of load current for currents from no load to full load. Repeat this process for power factors of O.S lagging. 1.0. and 0.8 leading. 2-9. A three-phase transfonner bank is to handle 600 kVA and have a 34.5113.S-kV volt- age ratio. Find the rating of each individual transformer in the bank (high voltage. low voltage. turns ratio. and apparent power) if the transfonner bank is connected to (a) Y- Y, (b) Y- /1, (c) /1- Y, (d) /1-11, (e) open 11, (j) open Y-open 11. 2-10. A 13,S00I4S0-V three-phase Y- I1-connected transfonner bank consists of three identical lOO-kVA 7967/4S0-V transfonners. It is supplied with power directly from a large constant-voltage bus. In the short-circuit test, the recorded values on the high-voltage side for one of these transformers are Vsc =560V h e = 12.6 A Psc = 3300W (a) If this bank delivers a rated load at 0.85 PF lagging and rated voltage, what is the line-to-line voltage on the high-voltage side of the transformer bank? (b) What is the voltage regulation under these conditions? (c) Assume that the primary voltage of this transformer is a constant 13.S kV, and plot the secondary voltage as a function of load current for currents from no- load to full-load. Repeat this process for power factors ofO.S5 lagging, 1.0, and 0.85 leading. (d) Plot the voltage regulation of this transfonner as a function of load current for currents from no-load to full-load. Repeat this process for power factors ofO.S5 lagging, 1.0, and 0.85 leading. 2-11. A lOO,()(X)-kVA, 23CVI15-kV /1- 11 three-phase power transformer has a resistance of 0.02 pu and a reactance of0.055 pu. The excitation branch elements are Re = 110 pu and XM = 20 pu. (a) If this transfonner supplies a load of SO MVA at 0.85 PF lagging, draw the pha- sor diagram of one phase of the transformer. (b) What is the voltage regulation of the transfonner bank under these conditions? (c) Sketch the equivalent circuit referred to the low-voltage side of one phase of this transformer. Calculate all the transfonner impedances referred to the low-voltage side. 2-12. An autotransformer is used to connect a 13.2-kV distribution line to a 13.S-kV dis- tribution line. It must be capable of handling 2CXXl kVA. There are three phases, con- nected Y- Y with their neutrals solidly grounded. (a) What must the Ne/NSF. IlU1lS ratio be to accomplish this cotulection? (b) How much apparent power must the windings ofeach autotransfonner handle? (c) If one of the autotransfonners were recOIUlected as an ordinary transformer, what would its ratings be? 2-13. Two phases of a 13.S-kV three-phase distribution line serve a remote rural road (the neutral is also available). A fanner along the road has a 480-V feeder supplying
- 172. 148 ELECTRIC MACHINERY RJNDAMENTALS 120 kW at 0.8 PF lagging of three-phase loads, plus 50 kW at 0.9 PF lagging of single-phase loads. The single-phase loads are distributed evenly among the three phases. Assuming that the ~n- Y-open-6. connection is used to supply power to his fann, find the voltages and currents in each of the two transformers. Also find the real and reactive powers supplied by each transfonner. Asswne the transformers are ideal. 2- 14. A 13.2-kV single-phase generator supplies power to a load through a transmission line. The load's impedance is L10ad = 500 L 36.870 n, and the transmission line's impedance is 21m. = 60 L 53.10 n. 6O L 531°n z,~ (~)Ve"'13.2LOO kV 500L 36.87°n z~ "J 6O L 53 Ion 1:10 10: I 500L 36.87°n • • • • ( ~) ve'" 13.2 L OO kV z~ 'bJ FIGURE 1 '2-3 Circuits for Problem 2-14: (a) without transformers and (b) with transformers. (a) If the generator is directly connected to the load (Figure P2- 3a), what is the ra- tio of the load voltage to the generated voltage? What are the transmission losses of the system? (b) If a 1:10 step-up transfonner is placed at the output of the generator and a 10: I transformer is placed at the load end of the transmission line, what is the new ratio of the load voltage to the generated voltage? What are the transmission losses of the system now? (Note:The transfonners may be assumed to be ideal.) 2-15. A 5000-VA, 4801120-Vconventional transfonner is to be used to supply power from a 600-V source to a 120-V load. Consider the transfonner to be ideal, and assrune that all insulation can handle 600 V. (a) Sketch the transfonner cOIUlection that will do the required job. (b) Find the kilovoltampere rating of the transfonner in the configuration. (c) Find the maximum primary and secondary currents lUlder these conditions.
- 173. TRANSFORMERS 149 2-16. A 5000-VA. 4801120-V conventional transformer is to be used to supply power from a 6()()'"V source to a 480-V load. Consider the transfonner to be ideal. and assrune that all insulation can handle 600 V. Answer the questions of Problem 2-15 for this transformer. 2-17. Prove the following statement: If a transfonner having a series impedance Zeq is con- nected as an autotransfonner. its per-unit series impedance Z~ as an autotransfonner will be Note that this expression is the reciprocal of the autotransformer power advantage. 2-18. Three 25-kVA. 24.000/277-V distribution transformers are connected in /1-y. The open-circuit test was performed on the low-voltage side of this transformer bank. and the following data were recorded: V~"".OC = 480 V I~"".oc = 4.10 A PJ.,OC = 945 W The short-circuit test was performed on the high-voltage side of this transformer bank., and the following data were recorded: V~"".sc = 1600 V Ir",e.sc = 2.00 A P~sc = 1150W (a) Find the per-lUlit equivalent circuit of this transformer bank. (b) Find the voltage regulation of this transfonner bank. at the rated load and 0.90 PF lagging. (c) What is the transformer bank's efficiency under these conditions? 2-19. A 20-kVA. 20,OOO/480-V, 60-Hz distribution transformer is tested with the follow- ing results: Open-circuit test ( mca s u~d from st.'eondary side) Voc = 480V loc = I.60A Poc = 305 W Short-circuit test (measun.>d from primary side) Vsc = 1130 V Isc = UX) A Psc =260W (a) Find the per-lUlit equivalent circuit for this transformer at 60 Hz. (b) What would the rating of this transfonner be if it were operated on a 50-Hz power system? (c) Sketch the per-lUlit equivalent circuit ofthis transfonner referred to the primary side ifit is operating at 50 Hz. 2-20. Prove that the three-phase system of voltages on the secondary of the Y-/1 trans- fonner shown in Figure 2- 38b lags the three-phase system of voltages on the pri- mary of the transformer by 30°. 2-21. Prove that the three-phase system of voltages on the secondary of the /1-Y trans- fonner shown in Figure 2- 38c lags the three-phase system of voltages on the pri- mary of the transformer by 30°. 2-22. A single-phase lO-kVA, 4801120-V transformer is to be used as an autotransfonner tying a 600-V distribution line to a 480-V load. When it is tested as a conventional
- 174. 150 ELECTRIC MACHINERY RJNDAMENTALS transfonner, the following values are measured on the primary (480-V) side of the transformer: Open,circuit tcst Voe = 480 V loe = 0.41 A Poe = 38W Short,circuit test Vsc = 10.0 V Isc = 1O.6A Psc = 26W (a) Find the per-unit equivalent circuit of this transfonner when it is connected in the conventional ma/Uler. What is the efficiency of the transfonner at rated condi- tions and lUlity power factor? What is the voltage regulation at those conditions? (b) Sketch the transfonner connections when it is used as a 600/480-V step-down autotransfonner. (c) What is the kilovoltampere rating of this transformer when it is used in the au- totransformer connection? (d) Answer the questions in a for the autotransformer connection. 2-23. Figure P2-4 shows a power system consisting of a three-phase 480-V 60-Hz gener- ator supplying two loads through a transmission line with a pair of transfonners at either end. Generator 41lO V T, ~~~~~_ _ _ T, 480114.400 V ](XXl kVA R "'O.OlOpu X ",0.040pu ""GURE "2-4 Line ZL",1.5+jIOO 14.4001480 V 500 kVA R", 0.020 pu X =0.085 pu Lood I ZLood t- 0.45L36.87°n Y- connected Lood2 ZLood 2 '" -jO.8ll Y- connected A one-tine diagram of the power system of Problem 2- 23. Note that some impedance values are given in the per-unit system. while others are given in ohms. (a) Sketch the per-phase equivalent circuit of this power system. (b) With the switch opened, find the real power P, reactive power Q, and apparent power S supplied by the generator. What is the power factor of the generator? (c) With the switch closed, find the real power P, reactive power Q, and apparent power S supplied by the generator. What is the power factor of the generator? (d) What are the transmission losses (transformer plus transmission line losses) in this system with the switch open? With the switch closed? What is the effect of adding load 2 to the system?
- 175. TRANSFORMERS 151 REFERENCES 1. Beeman. Donald. Industrial Po....er Systems Hmwbook. New York: McGraw-Hill. 1955. 2. Del TOTO. V. Eloctric Machines and Po·....er Systems. Englewood ClilTs. N.J.: Prentice-Hall. 1985. 3. Feinberg. R. Modern Po....er Transformer PractiCl!. New York: Wiley. 1979. 4. Fitzgerald. A. E.• C. Kingsley. Jr.• and S. D. Umaos. Eloctric Machinery. 5th ed. New Yort: McGraw-Hill. 1990. 5. McPherson. George. An Introduction 10 Electrical Machines and Transformers. New York: Wiley. 1981. 6. M.l.T. Staff. Magnetic Circuits and Transformers. New York: Wiley. 1943. 7. Siemon. G. R.• and A. Stra.ughen. Electric Machines. Reading. Mass.: Addison-Wesley. 1980. 8. Electrical Transmission and Distribution Reference Book. East Pittsburgh: Westinghouse Ele(;tric Corpora.tion.1964.
- 176. CHAPTER 3 INTRODUCTION TO POWER ELECTRONICS Over the last 40 years, a revolution has occurred in the application of electric motors.1lle development of solid-state motor drive packages has progressed to the point where practically any power control problem can be solved by using them. With such solid-state drives, it is possible to run de motors from ac power supplies or ac motors from de power supplies. It is even possible to change ac power at one frequency to ac power at another frequency. Furthermore, the costs of solid-state drive systems have decreased dramati- cally, while their reliability has increased . TIle versatility and the relatively low cost of solid-state controls and drives have resulted in many new applications for ac motors in which they are doing jobs formerly done by dc machines. DC motors have also gained flexibility from the application of solid-state drives. nlis major change has resulted from the development and improvement of a series of high-power solid-state devices. Although the detailed study of such power electronic circuits and components would requi re a book in itself, some familiarity with them is important to an understanding of modem motor applications. nlis chapter is a brief introduction to high-power electronic components and to the circuits in which they are employed. It is placed at this point in the book because the material contained in it is used in the discussions of both ac motor controllers and dc motor controllers. 3.1 POWER ELECTRONIC COMPONENTS Several major types of semiconductor devices are used in motor-control circuits. Among the more important are 152
- 177. INTRODUCTION TO POWER ELECTRONICS 153 I. 1lle diode 2. 1lle two-wire thyristor (or PNPN diode) 3. 1lle three-wire thyristor [or silicon controlled rectifier (SCR)] 4. 1lle gate turnoff (GTO) thyristor 5. 1lle DlAC 6. 1lle TRIAC 7. 1lle power transistor (PTR) 8. 1lle insulated-gate bipolar transistor (IGBT) Circuits containing these eight devices are studied in this chapter. Before the cir- cuits are examined, though, it is necessary to understand what each device does. The Diode A diode is a semiconductor device designed to conduct current in one direction only. The symbol for this device is shown in Figure 3-1. A diode is designed to conduct current from its anode to its cathode, but not in the opposite direction. The voltage-current characteristic of a diode is shown in Figure 3- 2. When a voltage is applied to the diode in the forward direction, a large current flow re- sults. When a voltage is applied to the diode in the reverse direction, the current flow is limited to a very small value (on the order of microamperes or less). If a large enough reverse voltage is applied to the diode, eventually the diode will break down and allow current to flow in the reverse direction. 1llese three regions of diode operation are shown on the characteristic in Figure 3- 2. Diodes are rated by the amount of power they can safely dissipate and by the maximum reverse voltage that they can take before breaking down. The power v, Anode PIV 'D Cathode FIGURE 3-1 FIGURE 3-2 The symbol of a diode. Voltage-current characteristic of a diode.
- 178. 154 ELECTRIC MACHINERY RJNDAMENTALS + FIGURE 3-3 The symbol of a two-wire thyristor or PNPN diode. dissipaled by a diode during forward operation is equal to the forward voltage drop across the diode times the current flowing through it. 1l1is power must be limited to protect the diode from overheating. llle maximum reverse voltage of a diode is known as its peak inverse voltage (PlY). It must be high enough to ensure that the diode does not break down in a circuit and conduct in the reverse direction. Diodes are also rated by their switching time, that is, by the time it takes to go from the off state to the on state, and vice versa. Because power diodes are large, high-power devices with a lot of stored charge in their junctions, they switch states much more slowly than the diodes found in electronic circuits. Es- sentially all power diodes can switch states fast enough to be used as rectifiers in 50- or 60-Hz circuits. However, some applications such as pulse-width modu- lation (PWM) can require power diodes to switch states at rates higher than 10,000 Hz. For these very fast switching applications, special diodes called fast- recovery high-speed diodes are employed. The Two-Wire Thyristor or PNPN Diode Thyristor is the generic name given to a fami ly of semiconductor devices which are made up of four semiconductor layers. One member of this fmnily is the two-wire thyristor, also known as the PNPN diode or trigger diode.1l1is device's name in the Institute of Electrical and Electronics Engineers (IEEE) standard for graphic sym- bols is reverse-blocking diode-f}pe thyristor. Its symbol is shown in Figure 3- 3. llle PNPN diode is a rectifier or diode with an unusual voltage-current characteristic in the forward-biased region. Its voltage-current characteristic is shown in Figure 3-4.llle characteristic curve consists of three regions: I. The reverse-blocking region 2. The forward-blocking region ), The conducting region In the reverse-blocking region, the PNPN diode behaves as an ordinary diode and blocks all current fl ow until the reverse breakdown voltage is reached. In the conducting region, the PNPN diode again behaves as an ordinary diode, al- lowing large amounts of current to flow with very little voltage drop. It is the forward-blocking region that distinguishes a PNPN diode from an ordinary diode.
- 179. v, ;D I Aooo. + ;G )D Gate Cathode INTRODUCTION TO POWER ELECTRONICS 155 ------- ""GURE3-S ""GURE 3-4 Voltage-current characteristic of a PNPN diode. The syntbol of a three-wire thyristor or SCR. When a PNPN diode is forward-biased, no current flows until the forward voltage drop exceeds a certain value called the breakover voltage VBO. When the forward voltage across the PNPN diode exceeds VBO' the PNPN diode turns on and remnins on until the current flowing through it falls below a certain minimum value (typically a few milliamperes). Ifthe current is reduced to a value below this minimum value (called the holding current lu), the PNPN diode turns off and will not conduct until the forward voltage drop again exceeds VBO . In summary, a PNPN diode I. Turns on when the applied voltage vD exceeds VBO 2. Turns off when the current iD drops below lu 3. Blocks all current fl ow in the reverse direction until the maximum reverse voltage is exceeded The Three-Wire Thyristor or SCR The most important member of the thyristor frunily is the three-wire thyristor, also known as the silicon controlled rectifier or SCR. This device was developed and given the name SCR by the General Electric Company in 1958. 1lle name thyristor was adopted later by the Int.ernational Electrotechnical Commission (IEC). 1lle symbol for a three-wire thyristor or SCR is shown in Figure 3- 5.
- 180. 156 ELECTRIC MACHINERY RJNDAMENTALS ""GURE 3-6 Voltage-current characteristics of an SCR. As the name suggests, the SCR is a controlled rectifier or diode. Its voltage- current characteristic with the gate lead open is the same as that of a PNPN diode. What makes an SCR especially useful in motor-control applications is that the breakover or turn-on voltage ofthe device can be adjusted by a current flow- ing into its gate lead. TIle largerthe gate current, the lower VBO becomes (see Fig- ure 3--6). If an SCR is chosen so that its breakover voltage with no gate signal is larger than the highest voltage in the circuit, then it can only be turned on by the application of a gate current. Once it is on, the device stays on until its current falls below IH' Therefore, once an SCR is triggered, its gate current may be re- moved without affecting the on state of the device. In the on state, the forward voltage drop across the SCR is about 1.21.0 1.5 times larger than the voltage drop across an ordinary forward-biased diode. TIll"Ce-wire thyristors or SCRs are by far the most common devices used in power-control circuits. TIley are widely used for switching or rectification appli- cations and are currently available in ratings ranging from a few amperes up to a maximum of about 3()(x) A. In summary, an SCR I. Turns on when the voltage vD applied to it exceeds VBO 2. Has a breakover voltage VBO whose level is controlled by the amount of gate current io present in the SCR
- 181. INTRODUCTION TO POWER ELECTRONICS 157 Anode Cathode (,' Several amperes to I r;:;.. II 2~s - 30p,s several tens of _L_l<.,'==:::::: I===---J,,---, (+, amperes _ () 20 Alp,s - 50 Alp,s-----' (b' FIGURE3-7 One-fourth to one-sixth of the -·00" currem (a) The symbol of a gate turn-off thyristor (GTO). (b) The gate current waveform required to turn a GTO thyristor on aod off. 3. Turns off when the current iD flowing through it drops below IH 4. Blocks all current fl ow in the reverse direction until the maximum reverse voltage is exceeded The Cate ThrnofT Thyristor Among the recent improvements to the thyristor is the gate turnoff (Cm ) thyris- tor. A cm thyristor is an SCR that can be turned off by a large enough negative pulse at its gate lead even if the current iDexceeds IH. Although GTO thyristors have been around since the 1960s, they only became practical for motor-control applications in the late 1970s. Such devices are becoming more and more com- mon in motor-control packages, since they eliminate the need for external com- ponents to turn off SCRs in dc circuits (see Section 3.5). The symbol for a GTO thyristor is shown in Figure 3- 7a. Figure 3- 7b shows a typical gate current waveform for a high-power GTO thyristor. A GTO thyristor typically requires a larger gate current for turn-on than an ordinary SCR. For large high-power devices, gate currents on the order of 0 A or more are necessary. To turn off the device, a large negative current pulse of 20- to 30-iJ.s duration is required. TIle magnitude of the negative current pulse must be one-fourth to one-sixth that of the current flowing through the device.
- 182. 158 ELECTRIC MACHINERY RJNDAMENTALS + FIGURE 3-8 The symbol of a DIAC. -------- ---<~------------r-------------~-- 'D ------__ ~ ~o ""GURE 3-9 Voltage-current characteristic of a DiAC. The DIA C A DIAC is a device containing fi ve semiconductor layers (PNPNP) that behaves like two PNPN diodes connected back to back. It can conduct in either direction once the breakover voltage is exceeded. The symbol for a DIAC is shown in Fig- ure 3--8, and its current-voltage characteristic is shown in Figure 3- 9. It turns on when the applied voltage in either direction exceeds VBO . Once it is turned on, a DIAC remains on until its current falls below ly. The TRIAC A TRIAC is a device that behaves like two SCRs connected back to back with a common gate lead. It can conduct in either direction once its breakover voltage
- 183. INTRODUCTION TO POWER ELECTRONICS 159 + G ""GURE 3-10 The symbol of a lRIAC. FIGURE 3-11 Voltage-current characteristic of a lRIAC. is exceeded. The symbol for a TRIAC is shown in Figure 3- 10, and its current- voltage characteristic is shown in Figure 3-11. The breakover voltage in a TRIAC decreases with increasing gate current in just the same manner as it does in an SCR, except that a TRIAC responds to either positive or negative pulses at its gate. Once it is turned on, a TRIAC remains on until its current fall s below [y. Because a single TRIAC can conduct in both directions, it can replace a more complex pair of back-to-back SCRs in many ac control circuits. However, TRIACs generally switch more slowly than SCRs, and are available only at lower power ratings. As a result, their use is largely restricted to low- to medium-power applications in 50- or 6O-Hz circuits, such as simple lighting circuits.
- 184. 160 ELECTRIC MACHINERY RJNDAMENTALS Collector 'cI < .~ E " }c, § 0 - - Base Q )i 0 U Emitter ,., Emitter-collector voltage veE> V ,b, ""GURE 3-12 (a) The symbol ofa power transistor. (b) The voltage-current characteristic of a power transistor. The Power Transistor TIle symbol for a transistor is shown in Figure 3- 12a, and the collector-to-emitter voltage versus collector current characteristic for the device is shown in Figure 3- 12b. As can be seen from the characteristic in Figure 3- 12b, the transistor is a device whose collector current ie is directly proportional 1 0 its base current iBover a very wide range of collector-to-emitter vollages (VCE) . Power transistors (PTRs) are commonly used in machinery-control applica- tions to switch a current on or off. A transistor with a resistive load is shown in Figure 3- 13a, and its ie-VcE characteristic is shown in Figure 3- 13b with the load line of the resistive load. Transistors are nonnally used in machinery-control ap- plications as switches; as such they should be either completely on or completely off. As shown in Figure 3- 13b, a base current of iB4 would completely turn on this transistor, and a base current of zero would completely turn off the transistor. If the base current of this transistor were equal to iB3 , then the transistor would be neither fully on nor fully off. This is a very undesirable condition, since a large collector current will fl ow across a large collector-to-emitter vollage vcr, dissipati ng a lot of power in the transistor. To ensure that the transistor conducts without wasting a lot of power, it is necessary to have a base current high enough to completely saturate it. Power transistors are most often used in inverter circuits. Their m~or draw- back in switching applications is that large power transistors are relatively slow in changing from the on to the off state and vice versa, since a relatively large base current has to be applied or removed when they are turned on or off.
- 185. +v (a) FIGURE 3-13 .~ • ! , " • • u INTRODUCTION TO POWER ELECTRONICS 161 0, Off Emitter-collector voltage VCE ,b, (a) A transistor with a resistive load. (b) The voltage-current characteristic of this transistor and load. Collector GateQ_____~ fo'IGURE 3-14 Emitter The symbol of an IGBT. The Insulated-Gate Bipolar Transistor The insulated-gate bipolar transistor (IGBT) is a relatively recent development. It is similar to the power transistor, except that it is controlled by the voltage applied to a gate rather than the current flowing into the base as in the power transistor. The impedance of the control gate is very high in an IGBT, so the amount of current flowing in the gate is extremely small. The device is essentially equivalent to the combination of a rnetal-oxide-semiconductor field-effect transistor (MOSFET) and a power transistor. llle symbol of an IGBT is shown in Figure 3-14.
- 186. 162 ELECTRIC MACHINERY RJNDAMENTALS Since the IGBT is controlled by a gate voltage with very little current flow, it can switch much more rapidly than a conventional power transistor can. IGBTs are therefore being used in high-power high-frequency applications. Power and Speed Comparison of Power Electronic Components Figure 3- 15 shows a comparison of the relative speeds and power-handling capa- bilities of SCRs, GTO thyristors, and power transistors. Clearly SCRs are capable of higher-power operation than any of the other devices. GTO thyristors can op- erate at almost as high a power and much faster than SCRs. Finally, power tran- sistors can handle less power than either type of thyristor, but they can switch more than 10 times faster. 10' 10' -, ---1..-- < > , ~GTO ! 10' , " 0 x a ~ 10' , ~ / , , & SCR i 10' , ) 0 I " I I 10' I I ITR I I 10' I 10' 10' 10' 10' 10' 10' Operating frequency. Hz ""GURE 3-15 A comparison of the relative speeds and power-handling capabilities of SCRs. GTO thyristors. and power transistors.
- 187. INTRODUCTION TO POWER ELECTRONICS 163 3.2 BASIC RECTIFIER CIRCUITS A rectifier circuit is a circuit that converts ac power to dc power. There are many different rectifier circuits which produce varying degrees of smoothing in their dc output. TIle four most common rectifier circuits are I. TIle half-wave rectifier 2. TIle full-wave bridge rectifier 3. TIle three-phase half-wave rectifier 4. TIle three-phase full-wave rectifier A good measure of the smoothness of the dc voltage out of a rectifier circuit is the ripple factor of the dc output. The percentage ofripple in a dc power sup- ply is defined as the ratio of the nns value of the ac components in the supply's voltage to the dc value of the voltage 100% I (3-1 ) where Vac.rTD. is the nns value of the ac components of the output voltage and Voc is the dc component of voltage in the output. The smaller the ripple factor in a power supply, the smoother the resulting dc waveform. The dc component of the output voltage Voc is quite easy to calculate, since it is just the average of the output voltage of the rectifier: (3- 2) The rms value of the ac part of the output voltage is harder to calculate, though, since the dc component ofthe voltage must be subtracted first. However, the ripple factor r can be calculated from a different but equivalent formula which does not require the rms value of the ac component of the voltage. This formula for ripple is I, ~ J(%::f-1 x 100% 1 (3- 3) where VlDl• is the nns value of the total output voltage from the rectifier and Voc is the dc or average output voltage from the rectifier. In the following discussion of rectifier circuits, the input ac frequency is as- sumed to be 60 Hz. The Half-Wave Rectifier A half-wave rectifier is shown in Figure 3-1 6a, and its output is shown in Figure 3-J 6b. TIle diode conducts on the positive half-cycle and blocks current flow on
- 188. 164 ELECTRIC MACHINERY RJNDAMENTALS (a) ~--~----~----~---.-- ' I I I _, ,b, I I I '-' FIGURE 3-16 (a) A luIf-wave rectifier ci['(;uit. (b) The output voltage of the rectifier ci['(;uit. the negative half-cycle. A simple half-wave rectifier of this sort is an extremely poor approximation to a constant dc waveform- it contains ac frequency compo- nents at 60 Hz and all its hamlOnics. A half-wave rectifier such as the one shown has a ripple factor r = 121 percent, which means it has more ac voltage compo- nents in its output than dc voltage components. Clearly, the half-wave rectifier is a very poor way to produce a dc voltage from an ac source. Example 3-1. Calculate the ripple factor for the half-wave rectifier shown in Fig- ure 3- 16, both analytically and using MATLAB. Solutioll In Figure 3- 16, the ac source voltage is vjt) = VM sin wtvolts. The output voltage of the rectifier is O<wt<Tr Tr S wts2Tr 80th the average voltage and the rms voltage must be calculated in order to calculate the ripple factor analytically. The average voltage out of the rectifier is W f" . =2TrO VMsinwtdt = ~ - ....!!! cos wt ( ~ )1"· 2Tr W 0
- 189. INTRODUCTION TO POWER ELECTRONICS 165 = The nns value of the total voltage out of the rectifier is = cos2wt d 2 t = v M J(f;t- f,; Sin 2wt)I:/~ = v M J(±- 8~ sin 27T) - (0 - 8~ sin o) VM = 2 Therefore, the ripple factor of this rectifier circuit is I, - 121% I The ripple factor can be calculated with MATLAB by implementing the average and rms voltage calculations in a MATLAB function, and then calculating the ripple from Equation (3- 3). The first part of the function shown below calculates the average of an in- put wavefonn, while the second part of the function calculates the nns value of the input waveform. Finally, the ripple factor is calculated directly from Equation (3- 3). fun c tion r = ripp l e (wave f orm ) % Func tion t o ca l c ula t e the rippl e on a n input wavef o rm. % Ca l c ula t e the aver age va lu e o f the wave f orm nva l s = s iz e (wave f orm ,2); t emp = 0; f or ii = l:nva l s t emp = t emp + wave f orm (ii ) ; e od aver age = t emp / nva l s; % Ca l c ula t e rms va lue o f wave f orm t emp = 0;
- 190. 166 ELECTRIC MACHINERY RJNDAMENTALS f or ii = l:nv a l s t emp = t e mp + wavef o rm (ii )A 2; end rms = sqrt (t emp / nva l s) ; ~ Ca l c ula t e rippl e f act or r = sqrt (( rms / aver age )A2 - 1 ) * 100 ; FlUlction ripp l e can be tested by writing an m-file to create a half-wave rectified wave- fonn and supply that wavefonn to the function. The appropriate M-file is shownbelow: ~ M-file: t est _ ha lfwave .m ~ M-file t o ca l c ula t e the rippl e on the output o f a h a lf-wave ~ wave r ectifie r. ~ Firs t , gene r a t e the out put o f a ha lf-wave r ectifi e r wave f orm = ze r os( 1 , 128 ) ; f or ii = 1 : 128 wave f orm (ii ) = ha lfwave (ii*pi / 64 ) ; end ~ Now ca l c ula t e the rippl e f a c t o r r = rippl e (wave f orm ) ; ~ Print out the r esult s tring = ['The ripp l e i s ' num2str (r ) '%.'] ; d i sp(string ) ; The output of the half-wave rectifier is simulated by flUlction h a l f wave . func tion vo lt s = ha lfwave(wt ) ~ Func tion t o s imula t e the output o f a ha lf-wave r ectifi e r. ~ wt = Phase in r adi a n s (=omega x time) ~ Convert input t o the r a nge 0 <= wt < 2*p i while wt >= 2*p i " 0 " 2*pi ; end while " < 0 " 0 " • 2*pi ; end ~ Simula t e the out put o f the h a lf-wave r ectifi e r if wt >= 0 & wt <= p i vo lt s = s in (wt ) ; e l se vo lt s = 0 ; end When t es t _ h a l f wave is executed, the results are: ,. te s t ha1fwave The rippl e i s 121.1772% . This answer agrees with the analytic solutioncalculated above.
- 191. INTRODUCTION TO POWER ELECTRONICS 167 The Full-Wave Rectifier A full-wave bridge rectifier circuit is shown in Figure 3- 17a, and its output volt- age is shown in Figure 3- 17c. In this circuit, diodes DJ and D3conduct on the positive half-cycle of the ac input, and diodes Dl and D4conduct on the negative half-cycle. TIle output voltage from this circuit is smoother than the output volt- age from the half-wave rectifier, but it still contains ac frequency components at 120 Hz and its hannonics. The ripple factor of a fu II-wave rectifier of this sort is r = 48.2 percent- it is clearly much better than that of a half-wave circuit. + vif) "- + "- - FIGURE 3-17 D, D, •• "'I I I , / - ,., - ,b, 'e ' D, + Lo., »,~'" D, D, vlood(tl + Lo" D, I I , / - (a) A full-wave bridge rectifier circuit. (bl The output voltage of the rectifier circuit. (el An alternative full-wave rectifier circuit using two diodes and a center-tapped transfonner.
- 192. 168 ELECTRIC MACHINERY RJNDAMENTALS "~ / vIII) ~-*-+--, r-~7'~"~~+-~'--T-t-+---, , , i > / > / / ,- ....../ ,_/ '......./ (a) ,b, ~------------------------- , '0' HGURE 3- 18 (a) A three-phase half-wave rectifier cin:uit. (b) The three-phase input voltages to the rectifier cin:uit. (c) The output voltage of the rectifier cin:uit. Another possible full-wave rectifier circuit is shown in Figure 3- l7b. In this circuit, diode Dt conducts on the positive half-cycle of the ac input with the cur- rent returning through the center tap of the transfonner, and diode Dl conducts on the negative half-cycle of the ac input with the current returning through the cen- ter tap of the transfonner. The output waveform is identical to the one shown in Figure 3- 17c. The Three-Phase Half-Wave Rectifier A three-phase half-wave rectifier is shown in Figure 3- 18a. The effect of having three diodes with their cathodes connected to a common point is that at any in- stant the diode with the largest voltage applied to it will conduct, and the other two diodes will be reverse-biased. 1lle three phase voltages applied to the rectifier circuit are shown in Figure 3- 18b, and the resulting output voltage is shown in Figure 3- 18c. Notice that the voltage at the output of the rectifier at any time is just the highest of the three input voltages at that moment.
- 193. '" ""I VB(I) r Vdt)r (a) • 1 Lo'" (b) Lood INTRODUCTION TO POWER ELECTRONICS 169 ." _, "" GURE3-19 (a) A three-phase full-wave rectifier circuit. (b) This circuit places the lowes/ of its three input voltages at its output. This output voltage is even smoother than that of a fuJI-wave bridge recti- fier circuit. It contains ac voltage components at 180 Hz and its harmonics. The ripple factor for a rectifier of this sort is 18.3 percent. The Three-Phase Full-Wave Rectifier A three-phase full-wave rectifier is shown in Figure 3- 19a. Basically, a circuit of this sort can be divided into two component parts. One part of the circuit looks just like the three-phase half-wave rectifier in Figure 3- 18, and it serves to con- nect the highest of the three phase voltages at any given instant to the load. The other part of the circuit consists of three diodes oriented with their an- odes connected to the load and their cathodes connected to the supply voltages (Figure 3- I9b). This arrangement connects the lowest of the three supply voltages to the load at any given time. Therefore, the three-phase fuJI-wave rectifier at aJl times connects the high- est of the three voltages to one end of the load and always connects the lowest of the three voltages to the other end of the load. The result of such a connection is shown in Figure 3- 20.
- 194. 170 ELECTRIC MACHINERY RJNDAMENTALS v(l) ,, I I I I ,,, , I , / I / I ,, ,,, I I I I I--------T- - - - - - - I (a) v(l) , , , , , , , , , , , , , , , , , , , , , / / / V V V V V V / / , / / , / / / / / / I 1 1 / / / / / / / " " 1 " " " " " , , / / 1 / / 1 ~~~__~-L~L-~~__L-~~__~-+__L- ' I--------T- - - - - - - I ,b, ""GURE 3-10 (a) The highest and lowest voltages in the three-phase full-wave rectifier. (b) The resulting output voltage. TIle output of a three-phase fuJi-wave rectifier is even smoother than the output of a three-phase half-wave rectifier. The lowest ac frequency component present in it is 360 Hz, and the ripple factor is only 4.2 percent. Filtering Rectifier Output The output of any of these rectifier circuits may be further smoothed by the use of low-pass filters to remove more of the ac frequency components from the output. Two types of elements are commonly used to smooth the rectifier's output: I. Capacitors connected across the lines to smooth ac voltage changes 2. Inductors connected in series with the line to smooth ac current changes A common filter in rectifier circuits used with machines is a single series induc- tor, or choke. A three-phase fuJi-wave rectifier with a choke filter is shown in Fig- ure 3- 21.
- 195. INTRODUCTION TO POWER ELECTRONICS 171 L ;~ I + '.~ " ~ '<~ Lood - FIGURE 3-21 A three-phase full-wave bridge circuit with an inductive filter for reducing output ripple. 3.3 PULSE CIRCUITS The SCRs, GTO thyristors, and TRIACs described in Section 3.1 are turned on by the application of a pulse of current to their gating circuits. To build power con- trollers, it is necessary to provide some methoo of producing and applying pulses to the gates of these devices at the proper time to turn them on. (In addition, it is necessary to provide some methoo of producing and applying negative pulses to the gates of GTO thyristors at the proper time to turn them off.) Many techniques are available to produce voltage and current pulses. They may be divided into two broad categories: analog and digital. Analog pulse gen- eration circuits have been used since the earliest days of solid-state machinery controls. They typically rely on devices such as PNPN diodes that have voltage- current characteristics with discrete nonconducting and conducting regions. The transition from the nonconducting to the conducting region of the device (or vice versa) is used to generate a voltage and current pulse. Some simple analog pulse generation circuits are described in this section. These circuits are collectively known as relaxation oscillators. Digital pulse generation circuits are becoming very common in modern solid-state motor drives. They typically contain a microcomputer that executes a program stored in read-only memory (ROM). The computer progrrun may consider many different inputs in deciding the proper time to generate firing pulses. For ex- ample, it may considerthe desired speed of the motor, the actual speed of the mo- tor, the rate at which it is accelerating or decelerating, and any specified voltage or current limits in detennining the time to generate the firing pulses. The inputs that it considers and the relative weighting applied to those inputs can usually be changed by setting switches on the microcomputer's circuit board, making solid- state motor drives with digital pulse generation circuits very flexible. A typical dig- ital pulse generation circuit board from a pulse-width-modulated induction motor drive is shown in Figure 3- 22. Examples of solid-state ac and dc motor drives con- taining such digital firing circuits are described in Chapters 7 and 9, respectively. The production of pulses for triggering SCRs, GTOs, and TRIACs is one of the most complex aspects of solid-state power control. The simple analog circuits
- 196. 172 ELECTRIC MACHINERY RJNDAMENTALS +~---, c ""GURE 3-23 ""GURE 3-22 A typical digital pulse generation circuit board from a pulse-width- modulated (PWM) induction motor drive. (Courtesy ofMagneTek Drives and Systems.) A relaxation oscillator (or pulse generator) using a PNPN diode. shown here are examples of only the most primitive lypes of pulse-producing cir- cuits-more advanced ones are beyond the scope of this book. A Relaxation Oscillator Using a PNPN Diode Figure 3- 23 shows a relaxation oscillator or pulse-generating circuit built with a PNPN diode. In order for this circuit to work, the following conditions must be true: I. The power supply voltage Voc must exceed VBO for the PNPN diode. 2. VodRI must be less than /H for the PNPN diode. 3. RI must be much larger than R2 . When the switch in the circuit is first closed, capacitor C will charge through resistor RI with time constant 7" = RIC. As the voltage on the capacitor builds up, it will eventually exceed VBO and the PNPN diode will turn on. Once
- 197. INTRODUCTION TO POWER ELECTRONICS 173 Voc 1--------------- l'o(l) (b' l'o(l) (" ""GURE 3-14 (a) The voltage across the capacitor in the relaxation oscillator. (b) The output voltage of the relaxation oscillator. (c) The output voltage of the oscillator after R[ is decreased. the PNPN diode turns on, the capacitor will discharge through it. 1lle discharge will be very rapid because R2is very small compared to RI . Once the capacitor is discharged, the PNPN diode will turn off, since the steady-state current corning through RI is less than the current /y of the PNPN diode. The voltage across the capacitor and the resuIting output voltage and current are shown in Figure 3- 24a and b, respectively. The timing of these pulses can be changed by varying RI . Suppose that re- sistor RI is decreased. Then the capacitor will charge more quickly, and the PNPN diode will be triggered sooner. 1lle pulses wi ll thus occur closer together (see Figure 3- 24c).
- 198. 174 ELECTRIC MACHINERY RJNDAMENTALS +~-r---1 iD j +) SCR vct.tl iG - R, R ,-, .. c c (a) (b' +~-r-----, R ,------[Q c (,' ""GURE J-15 (a) Using a pulse generator to directly trigger an SCR. (b) Coupling a pulse generator to an SCR through a transformer. (c) Connecting a pulse generntor to an SCR through a transistor amplifier to increase the strength of the pulse. nlis circuit can be used to trigger an SCR directly by removing R2and con- necting the SCR gate lead in its place (see Figure 3- 25a). Alternatively, the pulse circuit can be coupled to the SCR through a transfonner, as shown in Figure 3- 25b. If more gate current is needed to drive the SCR or TRIAC, then the pulse can be amplified by an extra transistor stage, as shown in Figure 3- 25c. 1lle same basic circuit can also be built by using a DIAC in place of the PNPN diode (see Figure 3- 26). It will function in exactly the same fashion as pre- viously described. In general, the quantitative analysis of pulse generation circuits is very com- plex and beyond the scope of this book. However, one simple example using a re- laxation oscillator follows. It may be skipped with no loss of continuity, if desired.
- 199. INTRODUCTIONTO POWER ELECTRONICS 175 + ~-----, R, .----T--~ + vr:f,t) FIGURE 3-26 A relaxation oscillator using a DIAC instead of a PNPN diode. + ~--, Voc=120V .-------,--~ + )'"'" FIGURE 3-27 The relaxation oscillator of ExampJe 3- 2. Example 3-2. Figure 3- 27 shows a simple relaxation oscillator using a PNPN diode. In this circuit, Voc = 120 V RI = 100 ill C = I J.tF R2 = I kO VBO = 75 V IH = 0 rnA (a) Delennine the firing frequency of this circuit. (b) Detennine the firing frequency of this circuit if RI is increased to 150 ill. Solutioll (a) When the PNPN diode is turned off, capacitor C charges through resistor RI with a time constant T = RIC, and when the PNPN diode turns on, capacitor C discharges through resistor R2with time constant T = R2C. (Actually, the dis- charge rate is controlled by the parallel combination of RI and R2, bul since RI » R2' the parallel combination is essentially the same as R2itself.) From elementary circuit theory, the equation for the voltage on the capacitor as a function of time during the charging portion of the cycle is vc(t) = A + B e- ·IR,C
- 200. 176 ELECTRIC MACHINERY RJNDAMENTALS where A and B are constants depending on the initial conditions in the circuit. Since vC<O) = 0 V and vC<D<» = Yoc, it is possible to solve for A and B: A = vc(D<» = Yoc A + B = vdO) = 0 => B = - Yoc Therefore, (3--4) The time at which the capacitor will reach the breakover voltage is found by solving for time t in Equation (3--4) : (3- 5) In this case, 120V-75V tl = - (1ookOXI/.!F)ln 120V = 98ms Similarly, the equation for the voltage on the capacitor as a flUlction of time dur- ing the discharge portion of the cycle turns out to be vc(t) = VBQe- ·,R,C so the current flow through the PNPN diode becomes V,a i(t) = -- e- tlR,c R, (3-6) (3-7) If we ignore the continued trickle of current through R], the time at which i(t) reaches IH and the PNPN diode turns off is Therefore, the total period of the relaxation oscillator is T = tl + tl = 98 ms + 2 ms = lOOms and the frequency of the relaxation oscillator is /= 1 = 10Hz T (b) If R] is increased to 150 kO, the capacitor charging time becomes Yoc - ¥ BO t] = - RIC in V oc = -(150 kO)( 1 F) I 120 V - 75 V /.! n 120V = 147 ms (3-8)
- 201. INTRODUCTION TO POWER ELECTRONICS 177 The capacitor discharging time remains unchanged at 1,/1, t2 = - RlC ln -V = 2 ms '0 Therefore, the total period of the relaxation oscillator is T = t] + t2 = 147ms + 2ms = 149ms and the frequency of the relaxation oscillator is 1 f= 0.149 s = 6.71 Hz Pulse Synchronization In ac applications, it is important that the triggering pulse be applied to the con- trolling SCRs at the same point in each ac cycle. TIle way this is nonnally done is to synchronize the pulse circuit to the ac power line supplying power to the SCRs. This can easily be accomplished by making the power supply to the triggering cir- cuit the same as the power supply to the SCRs. If the triggering circuit is supplied from a half-cycle of the ac power line, the RC circuit will always begin to charge at exactly the beginning of the cycle, so the pulse wi ll always occur at a fIXed time with respect to the beginning of the cycle. Pulse synchronization in three-phase circuits and inverters is much more complex and is beyond the scope of this book. 3.4 VOLTAGE VARIATION BY AC PHASE CONTROL The level of voltage applied to a motor is one of the most common variables in motor-control applications. The SCR and the TRIAC provide a convenient tech- nique for controlling the average voltage applied to a load by changing the phase angle at which the source voltage is applied to it. AC Phase Control for a DC Load Driven from an AC Source Figure 3- 28 illustrates the concept of phase angle power control. The figure shows a voltage-phase-control circuit with a resistive dc load supplied by an ac source. TIle SCR in the circuit has a breakover voltage for iG = 0 A that is greater than the highest voltage in the circuit, while the PNPN diode has a very low breakover voltage, perhaps 10 V or so. The full-wave bridge circuit ensures that the voltage applied to the SCR and the load will always be dc. If the switch SI in the picture is open, then the voltage VI at the tenninals of the rectifier will just be a full-wave rectified version of the input voltage (see Fig- ure 3- 29). If switch SI is shut but switch S2 is left open, then the SCR will always be off. TIlis is true because the voltage out of the rectifier will never exceed VBO for
- 202. 178 ELECTRIC MACHINERY RJNDAMENTALS s, S, 1.0'" + ),~ + vit) "-- R + + 'D (,p, ;, - + vc(t) C ""GURE 3-18 A circuit controlling the voltage to a dc load by phase angle control. ""GURE 3-29 The voltage at the output of the bridge circuit with switch S[ open. the SCR. Since the SCR is always an open circuit, the current through it and the load, and hence the voltage on the load, will still be zero. Now suppose that switch S2 is closed.1llen, at the beginning of the first half- cycle after the switch is closed, a voltage builds up across the RC network, and the capacitor begins to charge. During the time the capacitor is charging, the SCR is off, since the voltage applied to it has not exceeded VBO' As time passes, the ca- pacitor charges up to the breakover voltage of the PNPN diode, and the PNPN diode conducts. 1lle current fl ow from the capacitor and the PNPN diode flows through the gate of the SCR, lowering VBO for the SCR and turning it on. When the SCR turns on, current flows through it and the 10ad.1llis current flow continues for the rest of the half-cycle, even after the capacitor has discharged, since the SCR turns off only when its current falls below the holding current (since IH is a few milliamperes, this does not occur until the extreme end of the half-cycle). At the beginning of the next half-cycle, the SCR is again off. The RC circuit again charges up over a finite period and triggers the PNPN diode. The PNPN diode once more sends a current to the gate ofthe SCR, turning it on. Once on, the SCR remains on for the rest of the cycle again. The voltage and current wave- forms for this circuit are shown in Figure 3- 30. Now for the critical question: How can the power supplied to this load be changed? Suppose the value of R is decreased. Then at the beginning ofeach half-
- 203. I I , I , PNPN diode fires I I , I , I I , I , INTRODUCTION TO POWER ELECTRONICS 179 FIGURE 3-30 The voltages across the capacitor. SCR. and load. and the current f---'L--''---"---'-...L--'---- t through the load. when switches 51 and 51 are closed. fo'lGURE 3-31 The effect of decreasing R on the output voltage applied to the load in the cin:uit of Figure 3-28. cycle, the capacitor will charge more quickly, and the SCR will fire sooner. Since the SCR will be on for longer in the half-cycle, more power will be supplied to the load (see Figure 3- 31). The resistor R in this circuit controls the power flow to the load in the circuit. The power supplied to the load is a function of the time that the SCR fires; the earlier that it fires, the more power will be supplied. The firing time of the SCR is customarily expressed as afiring angle, where the firing angle is the angle of the applied sinusoidal voltage at the time of firing. The relationship between the firing angle and the supplied power wi ll be derived in Example 3- 3.
- 204. 180 ELECTRIC MACHINERY RJNDAMENTALS Lo'" R ....(1) c ", ,b, ""GURE 3-32 (a) A circuit controlling the ...oltage to an ac load by phase angle control. (b) Voltages on the source, the load, and the SCR in this controller. AC Phase Angle Control for an AC Load II is possible to modify the circuit in Figure 3- 28 to control an ac load simply by moving the load from the dc side of the circuit to a point before the rectifiers. The resulting circuit is shown in Figure 3- 32a, and its voltage and circuit wavefonns are shown in Figure 3- 32b. However, there is a much easier way to make an ac power controller. If the same basic circuit is used with a DIAC in place of the PNPN diode and a TRIAC in place of the SCR, then the diode bridge circuit can be completely taken out of the circuit. Because both the DIAC and the TRIAC are two-way devices, they op-
- 205. ;~ - + Lo'" /- Y. ,n DIAC == C - ;~ + - + Lo'" ," l~ circuit INTRODUCTIONTO POWER ELECTRONICS 181 - - + - - / t D ) TRIAC FIGURE 3-33 An ac phase angle controller using a DIAC and lRIAC. FIGURE 3-34 An ac phase angle controller using a TRIAC triggered by a digital pulse circuit. erate equally well on either half-cycle of the ac source. An ac phase power con- troller with a DIAC and a TRIAC is shown in Figure 3- 33. Example 3-3. Figure 3- 34 shows an ac phase angle controller supplying power to a resistive load. The circuit uses a TRIAC triggered by a digital pulse circuit that can pro- vide firing pulses at any point in each half-cycle of the applied voltage vit).Assume that the supply voltage is 120 V nns at 60 Hz. (a) Determine the rms voltage applied to the load as a function of the firing angle of the pulse circuit. and plot the relationship between firing angle and the supplied voltage. (b) What firing angle would be required to supply a voltage of 75 V rms to the load? Solutioll (a) This problem is ideally suited to solution using MATLAB because it involves a repetitive calculation of the rms voltage applied to the load at many different fir- ing angles. We will solve the problem by calculating the waveform produced by firing the TRIAC at each angle from 10 to 1790 • and calculating the rms voltage of the resulting waveform. (Note that only the positive half cycle is considered. since the negative half cycle is symmetrical.) The first step in the solution process is to produce a MATLAB function that mimics the load voltage for any given wt and firing angle. Function
- 206. 182 ELECTRIC MACHINERY RJNDA MENTALS ac_ p h ase_ contro ll e r does this. It accepts two input arguments, a nor- malized time wt in radians and a firing angle in degrees. If the time wt is earlier than the firing angle, the load voltage at that time will be 0 V. If the time wt is after the firing angle, the load voltage will be the same as the source voltage for that time. fun c tion volt s = ac_phase_controll e r (wt ,deg) % Func tion t o s imula t e the output of the pos itive h a lf % cycl e o f a n ac phase a ng l e controll e r with a peak % voltage o f 1 20 .. SQRT (2 ) = 1 70 V. % wt = Phase in r adi a n s (=omega x time) % deg = Firing a ng l e in degr ees % Degr ees t o r adi a n s conver s i on f act or deg2 r ad = p i / 1 80 ; % Simula t e the output o f the ph ase a ng l e controll e r. if wt > deg .. deg2 r ad; vo lt s = 1 70 .. s in (wt ) ; e l se vo lt s = 0; ond The next step is to write an m-file that creates the load waveform for each pos- sible firing angle, and calculates and plots the resulting nns voltage. The m-file shown below uses ftmction aC....Jlh ase_ con tro ll e r to calculate the load voltage waveform for each firing angle, and then calculates the nns voltage of that waveform. % M-fi1 e, vo1ts_vs-ph ase_ a ng l e .m % M-fi1 e t o cal cul a t e the rms vo lt age app lied t o a l oad as % a fun c ti on of the ph ase a ng l e firin g c irc uit , a nd t o % p l ot the r esulting r e l a tion ship . % Loop over a ll firing a ng l es ( 1 t o 179 degr ees ) deg = zer os( 1 , 1 79 ) ; rms = zer os( 1 , 1 79 ) ; f or ii = 1 , 1 79 ond % Save firing a ng l e deg (ii ) = ii ; % Firs t , gen e r a t e the wavef orm t o a na l yze . wavef o rm = z e r os (1 , 180); f o r jj = 1 , 1 80 wavef o rm ( j j ) = ac-ph as e_contro11e r (j j "pi / 1 80, ii ) ; end % Now ca l cula t e the rms volt age of the wave f orm t emp = s um (wavef o rm. " 2 ) ; rms (ii ) = sqrt (t emp / 1 80) ;
- 207. 180 160 140 120 • 100 ~ 80 60 40 20 00 180 160 140 120 • 100 ~ 80 60 40 20 o o INTRODUCTION TO POWER ELECTRONICS 183 % Pl o t rms vol t age o f the l oad as a fun c t i on of f i r i ng a ng l e p l o t (deg, rms); t i t l e( ' Load Vo lt age vs . Fi r i ng Ang l e ' ) ; x l abel ( ' Fir i ng a ng l e (deg ) ' ) ; y l abel ( 'RMS vo lt age (V) ' ) ; g r i d on ; Two examples of the waveform generated by this function are shown in Figure 3- 35. 0.' 0.' Load ,·ol!aJ!e for a 45° firi nJ! anJ!le V / 1.0 1.0 1.5 2.0 w i (radians) (a) ----- 1.5 2.0 WI (radians) ,b , 1 2.5 3.0 1 2.5 3.0 3.5 3.5 FIGURE 3-35 Waveform produced by vo l t s _ vs...Jlh ase_ a n g 1 e for a firing angle of (a) 45°; (b) 90°.
- 208. 184 ELECTRIC MACHINERY RJNDAMENTALS 140 120 ~ 100 ~ 80 ~ 60 ~ • " 40 20 ------ '""- "- " w ~ ro 80 100 120 140 lro IW Firing angle (deg) ""GURE 3-36 Plot of rms load voltage versus TRIAC firing angle. When this m-fIle is executed, the plot shown in Figure 3- 36 results. Note that the earlier the firing angle, the greater the rms voltage supplied to the load. However, the relationship between firing angle and the resulting voltage is not linear, so it is not easy to predict the required firing angle to achieve a given load voltage. (b) The firing angle required to supply 75 V to the load can be fOlUld from Figure 3- 36. It is about 99°. The Effect of Inductive Loads on Phase Angle Control If the load attached to a phase angle controller is inductive (as real machines are), then new complications are introduced to the operation of the controller. By the nature of inductance, the current in an inductive load cannot change instanta- neously. nlis means that the current to the load will not rise immediately on firing the SCR (or TRIAC) and that the current will not stop flowing at exactly the end of the half-cycle. At the end of the half-cycle, the inductive voltage on the load will keep the device turned on for some time into the next half-cycle, until the cur- re nt flowing through the load and the SCR finally fall s below ly. Fig ure 3- 37 shows the effect of this delay in the voltage and current wavefonns for the circuit in Fig ure 3- 32. A large inductance in the load can cause two potentially serio us proble ms with a phase controller: I. The inductance can cause the current buildup to be so slow when the SCR is switched on that it does not exceed the holding current before the gate current disappears. If this happens, the SCR will not remain on, because its current is less than [y.
- 209. INTRODUCTION TO POWER ELECTRONICS 185 , ,, , ,, / / / / / / / / / FIGURE 3-37 , ,, , , , ,, / / The elTect of an inductive load on the current and voltage waveforms of the cin:uit shown in Figure 3- 32. 2. If the current continues long enough before decaying to IH after the end of a given cycle, the applied voltage could build up high enough in the next cycle to keep the current going, and the SCR wi II never switch off. The nonnal solution to the first problem is to use a special circuit to provide a longer gating current pulse to the SCR. nlis longer pulse allows plenty of tirne
- 210. 186 ELECTRIC MACHINERY RJNDAMENTALS Freewheeling diod ~---r--;='I "d"t:::;"i"I) R2 load SCR c R, ""GURE 3-38 A phase angle controller illustrating the use of a free-wheeling diode with an inductive load. for the current through the SCR to rise above IH, pennitting the device to remain on for the rest of the half-cycle. A solution to the second problem is to add afree-wheeling diode. A free- wheeling diode is a diode placed across a load and oriented so that it does not con- duct during nonnal current flow. Such a diode is shown in Figure 3- 38.At the end ofa half-cycle, the current in the inductive load will attempt to kccp fl owing in the same direction as it was going. A voltage will be built up on the load with the po- larity required to keep the current flowing. This voltage will forward-bias the free- wheeling diode, and it will supply a path for the discharge current from the load. In that manner, the SCR can turn off without requiring the current of the inductor to instantly drop to zero. 3.5 DC-TO-DC POWER CONTROL- CHOPPERS Sometimes it is desirable to vary the voltage available from a dc source before ap- plying it to a load. TIle circuits which vary the voltage of a dc source are called de- to-de conveners or choppers. In a chopper circuit, the input voltage is a constant dc voltage source, and the output voltage is varied by varying thefmerion ofthe time that the dc source is connected to its load. Figure 3- 39 shows the basic prin- ciple of a chopper circuit. When the SCR is triggered, it turns on and power is sup- plied to the load. When it turns off, the dc source is disconnected from the load. In the circuit shown in Figure 3- 39, the load is a resistor, and the voltage on the load is either Voc or O. Similarly, the current in the load is either VoclR or O. lt is possible to smooth out the load voltage and current by adding a series inductor to filter out some of the ac components in the waveform. Figure 3-40 shows a chopper circuit with an inductive filter. The current through the inductor increases exponentially when the SCR is on and decreases exponentially when the SCR is off. If the inductor is large, the time constant of the current changes (T = LIR) will
- 211. INTRODUCTION TO POWER ELECTRONICS 187 SCR + L~ + Voc ,~( R~ ,,' Voc ,b, ,,' FIGURE 3-39 (a) The basic principte of a chopper circuit. (b) The input voltage to the circuit. (c) The resulting voltage on the load. be long relative to the on/off cycle of the SCR and the load voltage and current will be almost constant at some average value. In the case of ac phase controllers, the SCRs automatically turn off at the end of each half-cycle when their currents go to 7..ero. For dc circuits, there is no point at which the current naturally falls below IH, so once an SCR is turned on, it never turns off. To turn the SCR off again at the end of a pulse, it is necessary to apply a reverse voltage to it for a short time. TIlis reverse voltage stops the current flow and turns off the SCR. Once it is off, it will not turn on again until another pulse enters the gate of the SCR. TIle process of forcing an SCR to turn off at a desired time is known as forced commutation. GTO thyristors are ideally suited for use in chopper circuits, since they are self-commutating. In contrast to SCRs, GTOs can be turned off by a negative cur- rent pulse applied to their gates. Therefore, the extra circuitry needed in an SCR
- 212. 188 ELECTRIC MACHINERY RJNDAMENTALS SCR + "j I;~ + L + Voc vI(t) ,~( D, R~ ,,' Voc e------------------------------ "------------------------------ , ,b, , --- vI(t) , --- --- vlood(l) -- -- - - - - , / , / , / / / / , , , ", ""GURE 3-40 A chopper circuit with an inductive filter 10 smooth oUllhe load voltage and current. chopper circuit to turn off the SCR can be eliminated from a GTO thyristor chop- per circuit (Figure 3---4 1a). Power transistors are also self-commutating and are used in chopper circuits that fall within their power limits (Figure 3---41b). Chopper circuits are used with dc power systems to vary the speed of dc motors. TIleir greatest advantage for dc speed control compared to conventional methods is that they are more efficient than the systems (such as the Ward- Leonard system described in Chapter 6) that they replace. Forced Commutation in Chopper Circuits When SCRs are used in choppers, a forced-commutation circuit must be included to turn off the SCRs at the desired time. Most such forced-commutation circuits
- 213. + VOC INTRODUCTION TO POWER ELECTRONICS 189 L Lo,d ;~ Voc - I, I +}~ ;J Lood I ;, f/ - " (a) (b' FIGURE 3-41 (a) A chopper cin:uit made with a GTO thyristor. (b) A chopper cin:uit made with a transistor. + c ~----------------------, I D FIGURE 3-42 I SCR ~-T +,---f--, R L - '--+---' A series-capacitor forced-commutation chopper cin:uit. Lo,d + depend for their turnoff voltage on a charged capacitor. Two basic versions of ca- pacitor commutation are examined in this brief overview: I . Series-capacitor commutation circuits 2. Parallel-capacitor commulalion circuits Series-Capacitor Commutation Circuits Figure 3-42 shows a simple dc chopper circuit with series-capacitor commuta- lion. Ii consists of an SCR, a capacitor, and a load, all in series with each other.
- 214. 190 ELECTRIC MACHINERY RJNDAMENTALS Voc ------- - Discharge 1" '" RC L--1______~~______~________ , Voc --- --------- --------- -------- ""GURE 3-43 The capacitor and load voltages in the series chopper circuit. TIle capacitor has a shunt discharging resistor across it, and the load has a free- wheeling diode across it. TIle SCR is initially turned on by a pulse applied to its gate. When the SCR turns on, a voltage is applied to the load and a current starts flowing through it. But this current flows through the series capacitor on the way to the load, and the capacitor gradually charges up. When the capacitor's voltage nearly reaches Voc. the current through the SCR drops below iH and the SCR turns off. Once the capacitor has turned off the SCR, it gradually discharges through resistor R. When it is totally discharged, the SCR is ready to be fired by another pulse at its gate. The voltage and current wavefonns for this circuit are shown in Figure 3-43. Unfortunately, this type of circuit is limited in tenns of duty cycle, since the SCR cannot be fired again until the capacitor has discharged. The discharge time depends on the time constant 1" = RC, and C must be made large in order to let a lot of current flow to the load before it turns off the SCR. But R must be large, since the current leaking through the resistor has to be less than the holding cur- rent of the SCR. These two facts taken together mean that the SCR cannot be re- fired quickly after it turns off. It has a long recovery time. An improved series-capacitor commutation circuit with a shortened recov- ery time is shown in Figure 3-44. TIlis circuit is similar to the previous one except that the resistor has been replaced by an inductor and SCR in series. When SCR is fired, current will flow to the load and the capacitor will charge up, cutting off SCRI. Once it is cut off, SCR2 can be fired, discharging the capacitor much more
- 215. INTRODUCTION TO POWER ELECTRONICS 191 ;" [ n n n , ;" [ n n n , i i!l SCR I vc(l) i!2 L + co ~ vc(l) - SCR, ~ - + Inductive D load I ~~-~ - ~ .= -!--~~--~- , Ready '" - fire (a) (bj FIGURE 3-44 (a) A series-capacitor forced-commutation chopper cin:uit with improved capacitor recovery time. (b) The resulting capacitor and load voltage waveforms. Note that the capacitor discharges much more rapidly, so SCR[ could be refired sooner than before. quickly than the resistor would. TIle inductor in series with SCR1 protects SCR1 from instantaneous current surges that exceed its ratings. Once the capacitor dis- charges, SCR1 turns off and SCR] is ready to fi re again. Parallel-Capacitor Commutation Circuits The other common way to achieve forced commutation is via the parallel- capacitor commutation scheme. A simple example of the parallel-capacitor scheme is shown in Figure 3-45. In this scheme, SCR] is the main SCR, supply- ing power to the load, and SCR2 controls the operation of the commutating capac- itor. To apply power to the load, SCRI is fired. When this occurs, a current flows through the SCR to the load, supplying power to it. Also, capacitor C charges up through resistor R to a voltage equal to the supply voltage Voc. When the time comes to turn off the power to the load, SCR2 is fired. When SCR1 is fired, the voltage across it drops to zero. Since the voltage across a
- 216. 192 ELECTRIC MACHINERY RJNDAMENTALS • R D ,~ vOC L " - - C. SCR] ~ ~ ""GURE 3-45 A parallel-<:apacitor forced-commuta.tion chopper circuit. • R L D ,~ L oc I o---J v _ Vc + " SCR] ~ ~ ""GURE 3-46 A parallel-<:apacitor forced-commuta.tion chopper circuit with improved capacitor charging time. SCRJ permits the load power to be turned off more quickly thaD it could be with the basic parallel- capacitor circu it. capacitor cannot change instantaneously, the voltage on the left side of the capaci- tor must instantly drop to -Voc volts. This turns off SCRb and the capacitor charges through the load and SCR2 to a voltage of Voc volts positive on its left side. Once capacitor C is charged, SCRl turns off, and the cycle is ready to begin again. Again, resistor R] must be large in order for the current through it to be less than the holding current of SCR2. But a large resistor Rt means that the capacitor will charge only slowly after SCR] fi res. This limits how soon SCR] can be turned off after it fires, setting a lower limit on the on time of the chopped waveform. A circuit with a reduced capacitor charging time is shown in Figure 3-46. In this circuit SCR1 is triggered at the same time as SCR] is, and the capacitor can
- 217. INTRODUCTION TO POWER ELECTRONICS 193 charge much more rapidly. This allows the current to be turned off much more rapidly if it is desired to do so. In any circuit of this sort, the free-wheeling diode is extremely important. When SCR[ is forced off, the current through the inductive load must have an- other path available to it, or it could possibly damage the SCR. 3.6 INVERTERS Perhaps the most rapidly growing area in modern power electronics is static fre- quency conversion, the conversion of ac power at one frequency to ac power at another frequency by means of solid-state electronics. Traditionally there have been two approaches to static ac frequency conversion: the cycloconverter and the rectifier-inverter. The cycJoconverter is a device for directly converting ac power at one frequency to ac power at another frequency, while the rectifier-inverter first converts ac power to dc power and then converts the dc power to ac power again at a different frequency. lllis section deals with the operation of rectifier-inverter circuits, and Section 3.7 deals with the cycJoconverter. A rectifier-inverter is divided into two parts: I. A rectifier to produce dc power 2. An inveT1er to produce ac power from the dc power. Each part is treated separately. The Rectifier The basic rectifier circuits for converting ac power to dc power are described in Section 3.2. These circuits have one problem from a motor-control point of view-their output voltage is fixed for a given input voltage. This problem can be overcome by replacing the diodes in these circuits with SCRs. Figure 3-47 shows a three-phase full-wave rectifier circuit with the diodes in the circuits replaced by SCRs. The average dc output voltage from this circuit depends on when the SCRs are triggered during their positive half-cycles. If they are triggered at the beginning of the half-cycle, this circuit will be the same as that of a three-phase full-wave rectifier with diodes. Ifthe SCRs are never triggered, the output voltage wi ll be 0 V. For any other firing angle between 0° and 1800 on the wavefonn, the dc output voltage will be somewhere between the maximum value and 0 V. When SCRs are used instead of diodes in the rectifier circuit to get control of the dc voltage output, this output voltage wi ll have more harmonic content than a simple rectifier would, and some fonn of filter on its output is important. Figure 3-47 shows an inductor and capacitor filter placed at the output of the rectifier to help smooth the dc output.
- 218. 194 ELECTRIC MACHINERY RJNDAMENTALS • L , , -, vB<t) vc<t) c , , -, ""GURE 3-47 A three-phase rectifier circuit using SCRs to provide control of the dc output voltage level. L, I, J - "-' S~ SCR2 SCR1 ;C C·l r{ '-'I Rectifier Synchronous motor SCR4 <>-'-' SCR~ o-C-J SCR6 o-C-J ""GURE 3-48 An external commutation inverter. External Commutation Inverters Inverters are classified into two basic types by the commutation technique used: external commutation and self-commutation. Extenwl commutation inverters are inverters in which the energy required to turn off the SCRs is provided by an ex- ternal motor or power supply. An example of an external commutation inverter is shown in Figure 3-48. The inverter is connected to a three-phase synchronous motor, which provides the countervoltage necessary to turn off one SCR when its companion is fired. llle SCRs in this circuit are triggered in the following order: SCRj, SC~, SCR2, SCR4, SCR1, SCR5. When SCRt fi res, the internal generated voltage in the synchronous motor provides the voltage necessary to turn off SCRJ . Note that if the load were not connected to the inverter, the SCRs would never be turned off and after ~ cycle a short circuit would develop through SCRt and SC~. lllis inverter is also called a load-commutated inverter.
- 219. INTRODUCTION TO POWER ELECTRONICS 195 Self-Commutation Inverters Ifit is not possible to guarantee that a load will always provide the proper coun- tervoltage for commutation, then a self-commutation inverter must be used. A self-commutation inverter is an inverter in which the active SCRs are turned off by energy stored in a capacitor when another SCR is switched on. It is also possi- ble to design self-commutation inverters using GTOs or power transistors, in which case commutation capacitors are not required. There are three m~or types of self-commutation inverters:current source in- verters (CSls), voltage source inverters (VSls), and pulse-width modulation (PWM) inverters. Current source inverters and voltage source inverters are simpler than PWM inverters and have been used for a longer time. PWM inverters require more complex control circuitry and faster switching components than CSls and VSls. CSls and VSls are discussed first. Current source inverters and voltage source inverters are compared in Figure 3-49. In the current source inverter, a rectifier is connected to an inverter through a large series inductor Ls.The inductance of Ls is sufficiently large that the direct current is constrained to be almost constant. TIle SCR current output waveform will be roughly a square wave, since the current flow Is is constrained to be nearly constant. The line-to-Iine voltage will be approximately triangular. It is easy to limit overcurrent conditions in this design, but the output voltage can swing widely in response to changes in load. In the voltage source inverter, a rectifier is connected to an inverter through a series inductor Ls and a parallel capacitor C. The capacitance of C is sufficiently large that the voltage is constrained to be almost constant. The SCR line-to-line voltage output wavefonn will be roughly a square wave, since the voltage Vc is constrained to be nearly constant. TIle output current now wi ll be approximately triangular. Voltage variations are small in this circuit, but currents can vary wildly with variations in load, and overcurrent protection is difficult to implement. TIle frequency of both current and voltage source inverters can be easily changed by changing the firing pulses on the gates of the SCRs, so both inverters can be used to drive ac motors at variable speeds (see Chapter 10). A Single-Phase Current Source Inverter A single-phase current source inverter circuit with capacitor commutation is shown in Figure 3- 50. It contains two SCRs, a capacitor, and an output trans- former. To understand the operation of this circuit, assume initially that both SCRs are off. If SCR[ is now turned on by a gate current, voltage Voc will be applied to the upper half of the transfonner in the circuit. This voltage induces a voltage Voc in the lower half of the transfonner as well, causing a voltage of 2Voc to be built up across the capacitor. The voltages and currents in the circuit at this time are shown in Figure 3- 50b. Now SCRl is turned on. When SCR2 is turned on, the voltage at the cathode of the SCR will be Voc. Since the voltage across a capacitor cannot change
- 220. 196 ELECTRIC MACHINERY RJNDAMENTALS Current source invener Voltage source invener L, I, L, - I I r + I Main circuit ~ ~ ~ y,( f--o configuration ~ ~ ~ c f--o ~ ~ ~ f--o I I I - I Rectifier Inverter Rectifier Inverter Type of source Current source - Is almost constant Voltage source - Vs almost constam Output impedance High Low Line /"""-... , ', ' Vi r- Line voltage / voltage lO'J"ilf a If 21f (1800 conduction) Output waveform =t1 , CUrrent OJ /tv Currem (1200 conduction) I. Easy to control overcurrent I. Difficult to limit current Characteristics conditions with this design because of capacitor 2. Output voltage varies widely 2. Output voltage variations with changes in load small because of capacitor ""GURE 3-49 Comparison of current source inveners and voltage source inverters. instantaneously, this forces the voltage at the top of the capacitor to instantly be- come 3Voc, turning off SCRt . At this point, the voltage on the bottom half of the transfonner is built up positive at the bottom to negative at the top of the winding, and its magnitude is Voc. The voltage in the bottom half induces a voltage Voc in the upper half of the transformer, charging the capacitor C up to a voltage of 2Voc, oriented positive at the bottom with respect to the top of the capacitor. The condi- tion of the circuit at this time is shown in Figure 3- 5Oc. When SCR] is fired again, the capacitor voltage cuts off SCR2, and this process repeats indefinitely. The resulting voltage and current wavefonns are shown in Figure 3- 51.
- 221. + (a) + (b' + «, oc L ' ",- ~- FIGURE 3-50 INTRODUCTION TO POWER ELECTRONICS 197 + Ji...(t) C ", - SCR 2 _4----' SCRl - c-':= ;, Voc + + ;, if. ;~ Lo,' • - • ;~ (a) A simple single-phase inverter circuit. (b) The voltages and currems in the circuil when SCR[ is triggered. (c) The voltages and currents in the circuit when SCR1is lriggered. A Three-Phase Current Source Inverter Figure 3- 52 shows a three-phase current source inverter. In this circ uit, the six SCRs fire in the order SCR], SC~, SCR2, SC~, SCR1, SCR5. Capacitors Cl through C6 provide the commutation required by the SCRs.
- 222. 198 ELECTRIC MACHINERY RJNDA MENTALS 3Voc SCR2 turned SCR] turned ~ off ~] I~ " 2Voc " I - - - - I I I I I ; I I o - ,----" vif), ;,.(1) 2Voc Voc o I - H GURE 3-51 I / ~ , SCR2 turned SCR] tu ~ off off I~ " I I I - I I I I I ; I I , ----" ~ I, I, I No< t ~d - - - "'SCRI cathode - - - "'SCR2 cathode _ _ v)/) - - - jefl) Plots of the voltages and current in the inverter circuit: VI is the voltage al the cathode of seRlo and V1 is the voltage at the cathode of SeRlo Since the voltage at their anodes is Voc. any time VI or V1 exceeds Voc. that SCR is turned off. il.ood is the current supplied to the invener's load.
- 223. INTRODUCTION TO POWER ELECTRONICS 199 Three- ph~ input ~ ~ ~ I, I L, Rectifier FIGURE 3-52 SCR[ ~ D, D, SCI<, ~ A three-phase current source inverter. SCR2 ~ C, " " D, D, C, " " SCR~ ~ SCRJ ~ C, C, D, 0 b f ' Motor , D, C, C, SCR,; ~ To understand the operation of this circuit, examine Figure 3- 53. Assume that initially SCR[ and SCR~ are conducting, as shown in Figure 3- 53a. Then a voltage will build up across capacitors Ct , C3, C4, and Cs as shown on the dia- gram. Now assume that SCR6 is gated on. When SC~ is turned on, the voltage at point 6 drops to zero (see Figure 3- 53b). Since the voltage across capacitor Cs cannot change instantaneously, the anode of SCRsis biased negati ve, and SCRs is turned off. Once SC~ is on, all the capacitors charge up as shown in Figure 3- 53c, and the circuit is ready to turn off SCR6 whenever SCR4 is turned on. This same commutation process applies to the upper SCR bank as well. The output phase and line current from this circuit are shown in Figure 3- 53d. A Three-Phase Voltage Source Inverter Figure 3- 54 shows a three-phase voltage source inverter using power transistors as the active elements. Since power transistors are self-commutating, no special commutation components are included in this circuit.
- 224. N Q Q Three- p,,", input - I, I L, Rectifier FlG URE J-..5 3 ---. SCR] J D, L- D, SCR4 J SCR2 SCR3 C, J + ,;- C, J " + - D, C] D3 " - b Motor I - ./ c "- D, D. C, C, " ,,~ + "- -" + " C. " SCR, ~ SCRo -' ,,' T _~ p"'~ input I, I L, Rectifier SCR ] J D, D, SCR., J SCR2 SCR1 C, J + ,,'- C, ~ " + - D, " C1 D3 D, D. C, ~,' + - v~ _11 + ~,' " SCR, SCRo -' ~ ,b, " b Motor c ./ '.WhenSC~fires. "'6 ---+ O. Therefore the anode voltage of SCR~ ('~ ) becomes negative. and SC R~ turns off. 1be operation of the three-phase CSI. (3) Initially. SCRI and SCRI are conducting. Note how the commutating capacitocs have charged up. (b) 1be situation when SC~ fires. 1be voltage at the aoode of SCR6 falls almost instantaneously to zero. Since the voltage across capacitor Cl cannot change instantaneously. the voltage at the anode ofSeRl will become negative. arxI SCRI will tum off.
- 225. Three- p,,", input I, I L, Rectifier ---. SCR] J D, I D, SCR4 J N FlGURE 3-S3 (concluded) - SCR2 SCR3 C, J C, -' + ,,- " + - D, ~3 D3 D, D. C, C, " + " " C. + " SCI<, ~ J SCR, ..i ," " - b MOlOr , I - V Gale pulses SCR 1 1 6 2 4 3 S 6 conducting I SCR t SCR2 SCRJ SCR t nlerva/s SCI<, SCR, SCR, SCI<, SCR, , - , '.(1'1 I I , I, I I I - I, ]1----,--------'---'----,-------,---- "(:~II------,-----,-------'----'--------,----- - I, ,d, 8: (c) Now SCR[ and SC~ are conducting, and the commutating capacitors charge up as shown, (d) The gating pulses, SCR conducting intervals, and the output current from this invener,
- 226. 202 ELECTRIC MACHINERY RJNDAMENTALS +~~ Th~ ph~ invene ~ ~ ~ Rectifier - ""GURE 3-54 - ~ + /V, T + T, D, V, = =C T, D, - ,,' T, T, (a) A three-phase voltage source inverter using power transistors. T, D, D, T. D, D. In this circuit, the transistors are made to conduct in the order Tb T6 , T2, T4, T1, T5. The output phase and line voltage from this circuit are shown in Fig- ure 3- 54b. Pulse-Width Modulation Inverters Pulse-width modulation is the process of modifying the width of the pulses in a pulse train in direct proportion to a small control signal; the greater the control voltage, the wider the resulting pulses become. By using a sinusoid of the desired frequency as the control voltage for a PWM circuit, it is possible to produce a high-power wavefonn whose average voltage varies sinusoidally in a manner suitable for driving ac motors. 1lle basic concepts of pulse-width modulation are illustrated in Figure 3- 55. Figure 3- 55a shows a single-phase PWM inverter circuit using IGBTs. The states of IGSTt through IGBT4 in this circuit are controlled by the two comparators shown in Figure 3- 55b. A comparator is a device that compares the input voltage Vinet) to a refer- ence signal and turns transistors on or off depending on the results of the test. ComparatorA compares VinCt) to the reference voltage v..(t) and controls IGBTs Tt and Tl based on the results of the comparison. Comparator B compares Vinet) to the reference voltage v,(t) and controls IGBTs Tl and T4 based on the results of the comparison. If VinCt) is greater than v..(t) at any given time t, then comparator A will turn on Tt and turn off Tl . Otherwise, it will turn off Tt and turn on T2. Simi- larly, if Vinet) is greater than vy(t) at any given time t, then comparator B will turn b ,
- 227. T, T, T, , , -v, '-------' "'c(l) Vs f-----, -v, "'w,(/) 2V, v, o -v, - 2V, "'bc(1) 2V, v, o -v, - 2V, "'c,.(1) 2V, v, o -v, - 2V, FIGURE 3-54 (concluded) INTRODUCTION TO POWER ELECTRON ICS 203 T, T, T, T, T, T. T, T. L ,b , (b) The output phase and line voltages from the inverter.
- 228. 204 ELECTRIC MACHINERY RJNDAMENTALS + VBI~ T, VB3~ T v_(I) , , + - Lo,d + + VB2~ T, ) 'J" ('in VB4~ T, - - (., ""GURE J-S5 The basic concepts of pulse-width modulation. (a) A single-phase PWM cin:uil using IGBTs. off TJ and tum on T4. Otherwise, it will turn on T) and turn off T4 . The reference voltages vit) and vy(!) are shown in Figure 3-55c. To understand the overall operation of this PWM inverter circuit, see what happens when different control voltages are applied to it. First, assume that the control voltage is av. 1llen voltages vit) and v.(t) are identical, and the load volt- age out of the circuit V1oad(t) is zero (see Figure 3- 56). Next, assume that a constant positive control voltage equal to one-half of the peak reference voltage is applied to the circuit. 1lle resulting output voltage is a train of pulses with a 50 percent duty cycle, as shown in Figure 3- 57. Finally, assume that a sinusoidal control voltage is applied to the circuit as shown in Figure 3- 58. 1lle width of the resulting pulse train varies sinusoidally with the control voltage. 1lle result is a high-power output wavefonn whose aver- age voltage over any small region is directly proportional to the average voltage of the control signal in that region. 1lle fundamental frequency of the output waveform is the same as the frequency of the input control voltage. Of course, there are hannonic components in the output voltage, but they are not usually a concern in motor-control applications. 1lle hannonic components may cause ad- ditional heating in the motor being driven by the inverter, but the extra heating can be compensated for either by buying a specially designed motor or by derating an ordinary motor (running it at less than its full rated power). A complete three-phase PWM inverter would consist of three of the single- phase inverters described above with control voltages consisting of sinusoids
- 229. INTRODUCTION TO POWER ELECTRONICS 205 Comparator A '" '" I'in > I'x 0" Off viI) l'in < I'x Off 0" Comparator B '~ '" I'in > I'y Off 0" vft) l'in < I'y 0" Off ,b, ,,' FIGURE 3-55 (col/eluded) (b) The comparators used to control the on and off states ofttle transistors. (c) The reference voltages used in the comparators. shifted by 1200 between phases. Frequency control in a PWM inverter of this sort is accomplished by changing the frequency of the input control voltage. A PWM inverter switches states many times during a single cycle of the re- sulting output voltage. At the time ofthis writing, reference voltages with frequen- cies as high as 12 kHz are used in PWM inverter designs, so the components in a PWM inverter must change states up to 24,(X)Q times per second. This rapid switch- ing means that PWM inverters require faster components than CSls or YSls. PWM inverters need high-power high-frequency components such as GTO thyristors,
- 230. 206 ELECTRIC MACHINERY RJNDAMENTALS • " , , , , , , • " , , , , , , V.jt) "'." , , , , , , • " , , , , , , • " , , , , , , Vio '" 0 f-*--~'----'*--~-~I--*---il--*--lf--*-- , , , , , r- r- , , , , , , " • , , , , , , " • , , , , , , " • , , , , , , " • ,,,,, ,-- ,-- , , Vmd(l) ", 0 r-------------------- , ""GURE 3-56 The output of the PWM circuit with an input voltage of 0 V. Note that v,(t) '" v,(t). so v~t) '" o.
- 231. ) , , , , , " , , FIGURE 3-57 , , , , , , " • ,- , , , , , , " • ,- INTRODUCTION TO POWER ELECTRONICS 207 , , , , , , " • ,- , , , , , , " • , , , , , , " • r , The output of the PWM circuit with an input voltage equal to one-half of the peak comparator voltage.
- 233. INTRODUCTION TO POWER ELECTRONICS 209 IGBTs, and/or power transistors for proper operation. (At the time of this writing, IGBTs have the advantage for high-speed, high-power switching, so they are the preferred component for building PWM inverters.) The control voltage fed to the comparator circuits is usually implemented digitally by means of a microcomputer mounted on a circuit board within the PWM motor controller. The control voltage (and therefore the output pulse width) can be controlled by the microcomputer in a manner much more sophisticated than that described here. It is possible for the microcomputer to vary the control voltage to achieve different frequencies and voltage levels in any desired manner. For example, the microcomputer could im- plement various acceleration and deceleration ramps, current limits, and voltage- versus-frequency curves by simply changing options in software. A real PWM-bascd induction motor drive circuit is described in Section 7.10. 3.7 CYCLOCONVERTERS The cyc1oconverter is a device for directly converting ac power at one frequency to ac power at another frequency. Compared to rectifier-inverter schemes, cyc1o- converters have many more SCRs and much more complex gating circuitry. De- spite these disadvantages, cyc1oconverters can be less expensive than rectifier- inverters at higher power ratings. Cyc1oconverters are now available in constant-frequency and variable- frequency versions. A constant-frequency cyc1oconverter is used to supply power at one frequency from a source at another frequency (e.g., to supply 50-Hz loads from a 6O-Hz source). Variable-frequency cyc1oconverters are used to provide a variable output voltage and frequency from a constant-voltage and constant- frequency source. They are often used as ac induction motor drives. Although the details of a cyc1oconverter can become very complex, the ba- sic idea behind the device is simple. TIle input to a cyc1oconverter is a three-phase source which consists ofthree voltages equal in magnitude and phase-shifted from each other by 120°. TIle desired output voltage is some specified wavefonn, usu- ally a sinusoid at a different frequency. The cycloconverter generates its desired output waveform by selecting the combination of the three input phases which most closely approximates the desired output voltage at each instant oftime. There are two major categories of cycloconverters, noncirculating current cycloconverters and circulating current cycloconverters. These types are distin- guished by whether or not a current circulates internally within the cyclocon- verter; they have different characteristics. The two types of cycloconverters are described following an introduction to basic cycloconverter concepts. Basic Concepts A good way to begin the study of cycloconverters is to take a closer look at the three-phase fu ll-wave bridge rectifier ci rcuit described in Section 3.2. TIlis circuit is shown in Figure 3- 59 attached to a resistive load. In that figure, the diodes are divided into two halves, a positive half and a negative half. In the positive half, the
- 234. 210 ELECTRIC MACHINERY RJNDAMENTALS D, D, VA(I) '" VMsin WI V VB(I) '" VMsin (ro/- 120") V Vc(l) '" VMsin (&1- 240°) V fo'IGURE 3-59 D, D, Lo.d D, D, + }~(') - A three-phase full-wave diode bridge ci['(;uit connected to a resistive load. diode with the highest voltage applied to it at any given time will conduct, and it will reverse-bias the other two diodes in the section. In the negative half, the diode with the lowest voltage applied to it at any given time will conduct, and it will reverse-bias the other two diodes in the section. TIle resulting output voltage is shown in Figure 3--60. Now suppose that the six diodes in the bridge circuit are replaced by six SCRs as shown in Figure 3--61. Assume that initially SCR] is conducting as shown in Figure 3--61b. nlis SCR will continue to conduct until the current through it falls below IH. Ifno other SCR in the positive halfis triggered, then SCR[ will be turned ofT when voltage VA goes to 7..ero and reverses polarity at point 2. However, if SCRl is triggered at any time after point I, then SCR[ will be instantly reverse-biased and turned off. TIle process in which SCR2 forces SCR[ to turn off is calJed/orced com- mutation; it can be seen that forced commutation is possible only for the phase an- gles between points I and 2. 1lle SCRs in the negative half behave in a similar manner, as shown in Figure 3--61c. Note that ifeach of the SCRs is fired as soon as commutation is possible, then the output of this bridge circuit will be the same as the output of the full-wave diode bridge rectifier shown in Figure 3- 59. Now suppose that it is desired to produce a linearly decreasing output volt- age with this circuit, as shown in Figure 3--62. To produce such an output, the con- ducting SCR in the positive half of the bridge circuit must be turned off whenever its voltage falls too far below the desired value. 1llis is done by triggering another SCR voltage above the desired value. Similarly, the conducting SCR in the nega- tive half of the bridge circuit must be turned ofT whenever its voltage rises too far above the desired value. By triggering the SCRs in the positive and negative halves at the right time, it is possible to produce an output voltage which de- creases in a manner roughly corresponding to the desired wavefonn. It is obvious from examining Figure 3--62 that many harmonic components are present in the resulting output voltage.
- 235. (,) (q) (.) 'Pro1 ;KJ] 0] P"!1ddil "iil'HOA 1C10] ;KU (:l) 's<PpOfP J111Q-JAnl1ii"U ;KJ] WOJJ "iiC1IOA mdlnO "IlL (q) 's"POfP JI11Ij-JAmsod JI/I UJO.IJ Jiil1l[OA ]ndlnO JIlL (e) 09-£ :nm:: >1.'1 , , " , ' ,,- - ~.... ,,'--..... ,--......... ,,-- ~ .... /' " /" ...., /" "v" ...., / ',// ...., /" ~ )- /' X /, < , /~ / ' / ' / /~ I , ~ / , / ' / ' I~ I , /~ ' / , 1 ' ,/~/,I, ' 1 , 1 , 1 1 1 1 ,I ,I 1 '/ 1 , I ,I ' I .1 'I vI , Y A v .~ f , / " 1/ / / / / 1 I , 1 ' 1 1 I 1 , 1 1 1 I " " , , " ' , 1 ' 1 1 , I 1 ' I , ' I , I " 1 I , 1 ' / ' I 'I , I ,I I '( ) I~ I )1 ~ , I 1 I 1 1 1 " / I I / I I // '// ,/ 'x'/ ',,/ ';1// 1 o , , I ' / ' 1 I ' I , 1 1 , I , I ' 1 I ' f I , I~/ 1 , 1 ' , , , , III'I/~/ I /1 ./ ',/ / ~./ , , , iiu!I:lnpuOJ ta iiunJnpUOJ 9 a l!unJnpUO:l 'a /.... " /'. / ' / , /, / ' " ~ / , '" / ~ I , I / '. / • / , ~ / 1 t , / r t , / 1 ~ , / / .... 1 ' / ~/ / , / ' / , / / , I / ' / / , " , / ," , .( A /, .... / , / ' / ' , / , / , / , , / " .... / .... ',,-----' ',,-----''' ',------''' ' (/j-J~ (lfI~ (I)V~ ..... ..... -- '-, /"....... ",,- -......... ,,- --- ~ ..... , / / . . . " '" / .... / ....". , / ' / , , / , / ' " " / /<. X ',,~ / ~, / , / .... / / , / , / ' / ~ / , I / '. / , / , / 1 t ( . 1 , ~ , / t / , / , /1 ~ / , / ' / , / ' / ; ; / l!unJnpUO:l 'a (lj-J~ , / ' / , ; ~ (1)11... l!unJnpUO:l Za l!ufpnpUOJ la II Z S:JINOlli.:J3"l3 M3MOd 0.1 NOIl.JOamUNI , , ; , ; , ; " (!)V~ " A- " A IVA £L'I (!)JlOO'I~ " A (11"", " A (!)rod~
- 236. 212 ELECTRIC MACHINERY RJNDAMENTALS Posit holf Negal half ,,,!lrSCRI ,,,!lrSCR! SCR I conducting , ", ~ , ~ , ~ , .-,-'/ '" V-SCR2 lrSCR~ ,,,,, ----~ -- - -~ , ""GURE 3-61 ,,,, lrSCR3 lrSC~ (a) ,b, ", + Lo'" }~(" - (a) A three-phase full-wave SCR bridge cirwit connected to a resistive load. (b) The operation of the positive half of the SCRs. Assume that initially SCRI is conducting. If SCR1is triggered at any time after point I. then SCRI will be reverse-biased and shut off. (c) The operation of the negative half of the SCRs. Assume that initially SC~ is conducting. If SCR! is triggered at any time after point I. then SC~ will be reverse-biased and shut off.
- 237. , , , , , , , , v " , , , , , , , , , , " ,,,, , ,,, , , , , INTRODUCTION TO POWER ELECTRONICS ,,,,, , , , , , , - , , , , , , ,,, , , , , , , , , ' , 'i " , " ," ' , , , " " ' " / ' , " , " f ' , ' " , 1 " I ' 1 ' / " " , / ' f / , I / ' " ' " 't' ,I '/ 213 , , , , " I, /' I, I' , I ' ' " ' " " f , ' " " 1 , ' I " ' 1 ' I, I , " I " , ' I " , , ' ' ' I / 1/ / / 1/ / ' I ,/ ! ' , " ," l , " 1 " , " , ' I ' " , " I " , , I I , I 1 ' 1 ' / 1 1 / " 1 , " ' I , I , I 1 ' I , ) / / ' I ,I " / , I , , / ' X /, >( >, / ,/" ", ,/ ~" / " / / '" / " / " ~- ," "~ - _/ ,,- - /" "~ - -" ,,~" '~ - -~ FIGURE 3-62 Approximating a linearly decreasing voltage with the three-phase full-wave SCR bridge circuit.
- 238. 214 ELECTRIC MACHINERY RJNDAMENTALS Positive group Negative group j;;<) , SCRI r SCR2 r SCR1 r SCR7 SCR, SCR, -' -' -' + ,,(') ( Lood - SCR4 r SCR ~ r SC~ r SCRIO SCRll SCRl2 J J J ""GURE 3-63 One phase of a noncin:ulating current cycloconvener cin:ui1. If two of these SCR bridge circuits are connected in parallel with opposite polarities, the result is a noncirculating current cycloconverter. Noncirculating Current Cycloconverters One phase of a typical noncirculating current cycJoconverter is shown in Figure 3--63. A fuJi three-phase cycloconverter consists of three identical units of this type. Each unit consists of two three-phase full-wave SCR bridge circuits, one conducting current in the positive direction (the positive group) and one conduct- ing current in the negative direction (the negative group).llle SCRs in these cir- cuits are triggered so as to approximate a sinusoidal output voltage, with the SCRs in the positive group being triggered when the current fl ow is in the positive di- rection and the SCRs in the negative group being triggered when the current flow is in the negative direction. The resulting output voltage is shown in Figure 3--64. As can be seen from Figure 3--64, noncirculating current cycJoconverters produce an output voltage with a fairly large harmonic component. TIlese high harmonics limit the output frequency of the cycloconverter to a value less than about one-third of the input frequency. In addition, note that current flow must switch from the positive group to the negative group or vice versa as the load current reverses direction. The cycJo- converter pulse-control circuits must detect this current transition with a current polarity detector and switch from triggering one group of SCRs to triggering the other group. There is generally a brief period during the transition in which nei- ther the positive nor the negative group is conducting. This current pause causes additional glitches in the output waveform. TIle high harmonic content, low maximum frequency, and current glitches associated with noncirculating current cycloconverters combine to limit their use.
- 239. INTRODUCTION TO POWER ELECTRONICS 215 c--- Negative + Positive - - - - - - - - - - - - - - - - - - - group group FIGURE 3-64 The output voltage and current from a noncin:ulating current cycJoconvener connected to an inductive load. Note the switch from the operation of the negative group to the operation of the positive group at the time the current changes direction. In any practical noncirculating current cyc1oconverter, a fi Iter (usually a series in- ductor or a transformer) is placed between the output of the cyc1oconverter and the load, to suppress some of the output hannonics. Circulating Current Cyclocollver ters One phase of a typical circulating current cyc1oconverter is shown in Figure 3- 65. It differs from the noncirculating current cyc1oconverter in that the positive and negative groups are connected through two large inductors, and the load is sup- plied from center taps on the two inductors. Unlike the noncirculating current cy- c1oconverter, both the positive and the negative groups are conducting at the same time, and a circulating current fl ows around the loop fonned by the two groups and the series inductors. The series inductors must be quite large in a circuit ofthis sort to limit the circulating current to a safe value. The output voltage from the circulating current cyc1oconverter has a smaller hannonic content than the output voltage from the noncirculating current cyc1o- converter, and its maximum frequency can be much higher. It has a low power factor due to the large series inductors, so a capacitor is often used for power- factor compensation. The reason that the circulating current cyc1oconverter has a lower hannonic content is shown in Figure 3--66. Figure 3--66a shows the output voltage of the positive group, and Figure 3--66b shows the output voltage of the negative group. The output voltage V1oad(t) across the center taps of the inductors is _ vpo,(t) - vrw:s(t) Vtoait) - 2 (3- 9)
- 240. N - ~ v, v, Vc ~ ~ ~ isolation transformer HGURE 3-65 SCR[ V, VC/ SCR4 + "~ lr SCR2 lr SCR3 lr lr SCR~ lr SCR6 lr "'_(1) '" Vp:>«t) - '>'..g(t) o - One phase of a six-pulse type of cin:ulaling current cycloconverter. L, "., + J SCR7 J SCR8 J SCI<, v, + V/ Lo., J seRlO J seRll .J SCRl2 > .. - - L,
- 241. INTRODUCTION TO POWER ELECTRONICS 217 Many of the high-frequency harmonic components which appear when the posi- tive and negative groups are examined separately are common to both groups. As such, they cancel during the subtraction and do not appear at the tenninals of the cycloconverter. Some recirculating current cycloconverters are more complex than the one shown in Figure 3- 65. With more sophisticated designs, it is possible to make cycloconverters whose maximum output frequency can be even higher than their input frequency. These more complex devices are beyond the scope of this book. , , , , , , , , , , , , , , / , , , , , ,, , , , , , 1 , 1 1 , " " , , 1 ' 1 1 ' " " ' , , , > ' 1 ' , I' 1 ' , 1 " 1 1 , 1 ' " I I I , / ~/ ~I I ,/ ) ' i< >( / / ' .... /".... /', / ' / , " .... _-" '" ,. ,"--' ,"--,,/ .... , , , , , , , , v " , , , , , , , , , , , , " , " .... _-" FIGURE 3-66 , " , , , , , , , , , (a) / -' ,b, , , " , , , , , , '-' Voltages in the six-pulse circulating current cycloconverter. (a) The voltage out of the positive group; (b) the voltage out of the negative group.
- 242. 218 ELECTRIC MACHINERY RJNDAMENTALS , ,-~, ,--, ,-- ,'-' , ,' - ' ,.- , ',,' ' ,/' '/' '" , , " ' "- A P;:- ' / /' / / / I//~/ / / '~//'/ , " * 1 'I '/ 1 ~ 'I '( I' , ' / ' , , " I / ' , I ' 1 " ' 1 ' I ' / / / / 1 / 1 / 1 , 1 / ' 1 ' / 1 " 1 ' II ' I ' I 1 " ' I 1 1 1 " 1 " II I, I' I ,II I I I I " I , / 1 / / / / , ' / ' 1 / ' I , I ' I , ' ' " ' " " '-/ ,I I'- , > 'I I' , / I, I' ,I / ' I, , " /, /' /' / ' /' , / , 1 , 1 ' / ' , ' " , / ' I " ' I , , / ~I , I ~I V ,/ )/ )l -( ( /~, ,A, ,', /' /, /, " '--" '--" ,-,- ,~.," '- - ," '- .'/ HGURE 3--66 (concluded) (e) the resulting load voltage. 3.8 HARMONIC PROBLEMS ") Power electronic components and circuits are so flexible and useful that equip- menl controlled by them now makes up 50 to 60 percent of the total load on most power systems in the developed world. As a result, the behavior of these power electronic circuits strongly inn uences the overall operation of the power systems that they are connected to. TIle principal problem associated with power electronics is the harmonic components of voltage and current induced in the power system by the switching transients in power electronic controllers. TIlese hannonics increase the total cur- rent flows in the lines (especially in the neutral of a three-phase power system). The extra currents cause increased losses and increased heating in power system components, requiring larger components to supply the same total load. In addi- tion, the high neutral currents can trip protective relays, shutting down portions of a power system. As an example of this problem, consider a balanced three-phase motor with a wye connection that draws 10 A at full load. When this motor is connected to a power system, the currents flowing in each phase will be equal in magnitude and 1200 out of phase with each other, and the return current in the neutral will be 0 (see Figure 3--67). Now consider the same motor supplied with the same total power through a rectifier-inverter that pn:x:luces pulses of current. TIle currents in the power line now are shown in Figure 3--68. Note that the nns current of each line is still lOA, but the neutral also has an rms current of 15 A!The current in the neutral consists entirely of hannonic components.
- 243. Ie. /" '-- o o I u<t.un:J r- '-- o o IU<llIlQ / ::> o "1 o , v ) o "1 o , o o Ie-- /' 1 "--- o o o , ~ ') o , 219
- 244. N N o 2') , ) 5 , ~ 0 5 - 10 ) - I , - 2') 20 , ) , • ~ ) 5 - 10 ) - I 5 - 20 Time ,,' Time ,b, 20 1 }, 1,,1 ' I 'I' I' 'I I I 10 1 I III I III ! :11 I r I II 1 II II 1 I u -51 I II I I 1 I I - 10 I II I I I I I - 15 1 I" " I i I -20 1 I /1 Time 'e' 20 1 11 It h 11 1/ 111 ' I I I III II I I I II I II II 10 1 I I III II I I I II I II II ! :111 II II II I III u -51 I II I II II I II I I I II I - 10 II I II I I I I I I II I II I I I - 15 II I II I I I I I I II I II I I I - 20 II I I I I I " I Time ,d , Jo'IGURE 3-68 Current flow for a balanced three- phase. wye<onnected mOlOr connected to the power line through a power electronic controller that produces current pulses: (a) phase a; (b) phaseb; (e) phasec; (d) neutral The nlls current flow in phases a. b, and c is 0 A. wh.ile the nns current flow in the neutral is 15 A.
- 245. INTRODUCTION TO POWER ELECTRONICS 221 The spectra of the currents in the three phases and in the neutral are shown in Figure 3--69. For the motor connected directly to the line, only the fundamen- tal frequency is present in the phases, and nothing at all is present in the neutral. For the motor connected through the power controller, the current in the phases in- cludes both the fundamental frequency and all of the odd hannonics. TIle current in the neutral consists principally of the third, ninth, and fifteenth harmonics. Since power electronic circuits are such a large fraction of the total load on a modern power system, their high hannonic content causes significant problems for the power system as a whole. New standards* have been created to limit the amount of harmonics produced by power electronic circuits, and new controllers are designed to minimize the hannonics that they produce. 3.9 SUMMARY Power electronic components and circuits have produced a m~or revolution in the area of motor controls during the last 35 years or so. Power electronics provide a convenient way to convert ac power to dc power, to change the average voltage level of a dc power system, to convert dc power to ac power, and to change the frequency of an ac power system. The conversion of ac to dc power is accomplished by rectifier circuits, and the resulting dc output voltage level can be controlled by changing the firing times of the devices (SCRs, TRIACs, GTO thyristors, etc.) in the rectifier circuit. Adjustment of the average dc voltage level on a load is accomplished by chopper circuits, which control the fraction of time for which a fixed dc voltage is applied to a load. Static frequency conversion is accomplished by either rectifier-inverters or cycloconverters. Inverters are of two basic types: externally cornrnutated and self- commutated. Externally commutated inverters rely on the attached load for com- mutation voltages; self-commutated inverters either use capacitors to produce the required commutation voltages or use self-commutating devices such as GTO thyristors. Self-commutated inverters include current source inverters, voltage source inverters, and pulse-width modu lation inverters. Cycloconverters are used to directly convert ac power at one frequency to ac power at another frequency. There are two basic types of cycloconverters: non- circulating current and circulating current. Noncirculating current cycloconverters have large harmonic components and are restricted to relatively low frequencies. In addition, they can suffer from glitches during current direction changes. Circu- lating current cycloconverters have lower hannonic components and are capable of operating at higher frequencies. TIley require large series inductors to limit the circulating current to a safe value, and so they are bulkier than noncirculating cur- rent cycloconverters of the same rating. *See IEC l00Q.3-2. EMC: Part 3. Section 2. "Limits for harmonic current emission (equipment input current s 16 A per phase)," and ANSI/IEEE Standard 519-1992, "IEEE recommended practices and requiremems for harmonic control in power systems."
- 246. N N N 30.000 J 25.000 J 20.000 i J .• '0.. 15.000 ~ • J lOilOO 0 '.000 " 45.000 40.000 35.000 30.000 .a 25.000 o ~ ~ 20.000 15ilOO lOilOO '.000 J o 2 4 o - 2 4 6 8 10 12 14 Harrnonicnumber ,,' I - 6 - 8 0 12 14 Harmonic number 'e' 16 16 25.000 J 20.000 • " 15.000 , .• f ) 10000 ) 5.000 J o FIGURE 3-69 - 2 • 4 I • 6 8 10 12 14 16 Harmonic number ,b, (a) The spectrum of the phase current in the balanced three-phase. wye-connected motor connected directly to the power line. Only the fundamental frequency is present. (b) The spectrum of the phase current in the balanced three-phase. wye-<:onnected motor connected through a power electronic controller that produces current pulses. T he fundamental frequency and all odd harmonics are pre.-;ent. (c) The neutral current for the motor connected through a electronic power controller. The third. ninth. and fifteenth harmonics are present in the current
- 247. INTRODUCTION TO POWER ELECTRONICS 223 QUESTIONS 3-1. Explain the operation and sketch the output characteristic of a diode. 3-2. Explain the operation and sketch the output characteristic of a PNPN diode. 3-3. How does an SCR differ from a PNPN diode? When does an SCR conduct? 3-4. What is a GTO thyristor? How does it differ from an ordinary three-wire thyristor (SCR)? 3-5. What is an IG8T? What are its advantages compared to other power electronic devices? 3-6. What is a DIAC? ATRIAC? 3-7. Does a single-phase full-wave rectifier produce a better or worse dc output than a three-phase half-wave rectifier? Why? 3-8. Why are pulse-generating circuits needed in motor controllers? 3-9. What are the advantages of digital pulse-generating circuits compared to analog pUlse-generating circuits? 3-10. What is the effect of changing resistor R in Figure 3- 32? Explain why this effect occurs. 3-11. What is forced conunutation? Why is it necessary in dc-to-dc power-control circuits? 3-12. What device(s) could be used to build dc-to-dc power-control circuits without forced conunutation? 3-13. What is the purpose of a free-wheeling diode in a control circuit with an inductive load? 3-14. What is the effect of an inductive load on the operation of a phase angle controller? 3-15. Can the on time of a chopper with series-capacitor commutation be madearbitrarily long? Why or why not? 3-16. Can the on time of a chopper with parallel-capacitor conunutation be made arbitrar- ily long? Why or why not? 3-17. What is a rectifier-inverter? What is it used for? 3-18. What is a current-source inverter? 3-19. What is a voltage-source inverter? Contrast the characteristics of a VSI with those of a CSI. 3-20. What is pulse-width modulation? How do PWM inverters compare to CSI and VSI inverters? 3-21. Are power transistors more likely to be used in PWM inverters or in CSI inverters? Why? PROBLEMS 3-1. Calculate the ripple factor of a three-phase half-wave rectifier circuit. both analyti- cally and using MATLAB. 3-2. Calculate the ripple factor of a three-phase full-wave rectifier circuit. both analyti- cally and using MATLAB. 3-3. Explain the operation of the circuit shown in Figure P3-1. What would happen in this circuit if switch S, were closed?
- 248. 224 ELECTRIC M ACHINERY RJNDA MENTALS + 21 D, • • I + )'~(') ") Y. Lood S, /- (-~I SCR 0, v,...(l) '" 339 sin 3771 V C2 C] FlGURE P3-1 The cin:uit of Problems 3- 3 through 3--6. 3-4. What would the nns voltage on the load in the circuit in Figure P3-1 be if the firing angle of the SCR were (a) 0°, (b) 30°, (c) 90°? '3-5. For the circuit in Figure P3-1, assume that VBO for the DlAC is 30 V, Ct is I p.F, R is adjustable in the range I to 20 ill, and switch SI is open. What is the firing angle of the circuit when R is 10 kO? What is the rillS voltage on the load lUlder these con- ditions? (Caution: This problem is hard to solve analytically because the voltage charging the capacitor varies as a function of time.) 3-6. One problem with the circuit shown in Figure P3-1 is that it is very sensitive to vari- ations in the input voltage v.At). For example, suppose the peak. value of the input voltage were to decrease. Then the time that it takes capacitor C] to charge up to the breakover voltage of the DIAC will increase, and the SCR will be triggered later in each half-cycle. Therefore, the rillS voltage supplied to the load will be reduced both by the lower peak voltage and by the later firing. This same effect happens in the opposite direction if voc(t) increases. How could this circuit be modi fied to reduce its sensitivity to variations in input voltage? 3-7. Explain the operation of the circuit shown in Figure P3- 2, and sketch the output voltage from the circuit. 3-8. Figure P3- 3 shows a relaxation oscillator with the following parameters: R] = variable C= IJ.tF VBO = 30V R2 = 1500 0 Voc = 100 V lH = 0.5 mA (a) Sketch the voltages vc(t), vo(t), and rrJt) for this circuit. (b) If R, is currently set to 500 kO, calculate the period of this relaxation oscillator. 3-9. In the circuit in Figure P3-4, T] is an autotransformer with the tap exactly in the center of its winding. Explain the operation of this circuit. Assruning that the load is inductive, sketch the voltage and current applied to the load. What is the purpose of SCR2? What is the purpose of D2? (This chopper circuit arrangement is known as a Jones circuit.) *The asterisk in front of a problem number indicates that it is a more difficult problem.
- 249. INTRODUCTION TO POWER ELECTRONICS 225 • "~---T ~ • • c, Lood -- .,'----.L T, T, FIGURE 1'3-2 The inverter circuit of Problem 3- 7. +~-----; Voc'" lOOV " J + + Vc (I) ( '>'0 (/) R2", 1500 n C'" 1.0~ FIGURE 1'3-3 The relaxation oscillator circuit of Problem 3--8. + J SCR] ,icC ~CR2 D, • T] (Autotnlnsformer) D, Lo,d FlGURE P3-4 The chopper ci["(;uit of Problem 3- 9.
- 250. 226 ELECTRIC MACHINERY RJNDAMENTALS 3-10. A series-capacitor forced commutation chopper circuit supplying a purely resistive load is shown in Figure P3- 5. Voc = 120 V lH = 8 rnA VBO = 200 V Rl =20 kll Rlood = 250 0 C = 150 /lF (a) When SCRl is turned on. how long will it remain on? What causes it to tum off? (b) When SCRl turns off. how long will it be until the SCR can be turned on again? (Assume that 3 time constants must pass before the capacitor is discharged.) (c) What problem or problems do these calculations reveal about this simple series- capacitor forced-commutation chopper circuit ? (d) How can the problem(s) described in part c be eliminated? + SCR 0---/ + R C )" - voc / D ,~ ~ Rwm Lo,' ""GURE 1'3-5 The simple series-capacitor forced-commulation cin:uit of Problem 3-10. 3-11. A parallel-capacitor forced-conunutation chopper circuit supplying a purely resis- tive load is shown in Figure P3-6. Voc = 120 V lH = 5 rnA VBO = 250V R] =20kfi Rlood = 250 0 C= 15 /lF (a) When SCR] is turned on, how long will it remain on? What causes it to IlUll off? (b) What is the earliest time that SCR] can be turned off after it is turned on? (Assume that 3 time constants must pass before the capacitor is charged.) (c) When SCR] turns off, how long will it be until the SCR can be tlUlled on again? (d) What problem or problems do these calculations reveal about this simple parallel- capacitor forced-commutation chopper circuit? (e) How can the problem(s) described in part d be eliminated? 3-12. Figure P3- 7 shows a single-phase rectifier-inverter circuit. Explain how this circuit ftmctions. What are the purposes of C] and C2? What controls the output frequency of the inverter?
- 251. INTRODUCTION TO POWER ELECTRONICS 227 + + , D RLOAD " ~ R, " - C - + 0--/ SCRl 0--/ FIGURE P3-6 The simple parallel-capacitor forced commutation circuit of Problem 3- 11. FIGURE P3-7 The single-phase rectifier-inverter circuil of Problem 3- 12. '3- 13. A simple full-wave ac phase angle voltage controller is shown in Figure P3-8. The component values in this circuit are R = 20 to 300 kf.!. currently set to 80 kf.! C = 0.1 5 p.,F VBO = 40 V (for PNPN diode DJ) VBO = 250 V (for SCRl ) rs(t) = VMsin wt V where VM = 169.7 V and w = 377 radls (a) At what phase angle do the PNPN diode and the SCR tum on? (b) What is the rms voltage supplied to the load under these circwnstances? ' 3- 14. Figure P3-9 shows a three-phase full -wave rectifier circuit supplying power to a dc load. The circuit uses SCRs instead of diodes as the rectifying elements.
- 252. 228 ELECTRIC MACHINERY RJNDAMENTALS + R SCR I ) vD (t) v,m C D, ""GURE I'J-S The full-wave phase angle voltage controller of Problem 3- t3. lr r lr SCRI SCR2 SCR3 + Lo., ir' r Ir SCR, SCR3 SC", ""GURE PJ-9 The lhree-phase full-wave reclifier circuit of Problem 3- t4. (a) What will the nns load voltage and ripple be if each SCR is triggered as soon as it becomes forward-biased? At what phase angle should the SCRs be triggered in order to operate this way? Sketch or plot the output voltage for this case. (b) What will the rms load voltage and ripple be if each SCR is triggered at a phase angle of 90° (that is, halfway through the half-cycle in which it is forward bi- ased)? Sketch or plot the output voltage for this case. '3-15. Write a MATLAB program that imitates the operation of the pulse-width modula- tion circuit shown in Figure 3-55, and answer the following questions. (a) Assume that the comparison voltages vjt) and vI') have peak amplitudes of 10 V and a frequency of 500 Hz. Plot the output voltage when the input voltage is Vinet) = 10 sin 2'lT ft V, andf = 60 Hz . (b) What does the spectrwn of the output voltage look like? What could be done to reduce the hannonic content of the output voltage? (c) Now assume that the frequency of the comparison voltages is increased to 1000 Hz. Plot the output voltage when the input voltage is Vinet) = 10 sin 2'lTft V and/ = 60 Hz. (d) What does the spectrum of the output voltage in c look like? (e) What is the advantage of using a higher comparison frequency and more rapid switching in a PWM modulator?
- 253. INTRODUCTION TO POWER ELECTRONICS 229 REFERENCES I. Dewan. S. 8.. G. R. Siemon. and A. Straughen. Power Semiconductor Drives. New York: Wiley- Interscience.1984. 2. IEEE. Graphic Symbols for Electrical and Electronics Diagrams. IEEE Standard 315-19751ANSI Standard Y32.2-1975. 3. Millman. Jacob. and Christos C. Halkias. Integrated Electronics: Analog and Digital Circuits and Systems. New York: McGraw-Hill. 1972. 4. Vithayathil. Joseph. Pov.·er Electronic:;: Principles andApplications. New York: McGraw-Hill. 1995. 5. Werninck. E. H. (ed.). Electric Motor Handbook. London: McGraw-Hill. 1978.
- 254. CHAPTER 4 AC MACHINERY FUNDAMENTALS ACmachines are generators that convert mechanical energy to ac electrical energy and motors that convert ac electrical energy to mechanical en- ergy. The fundamental principles of ac machines are very simple, but unfortu- nately, they are somewhat obscured by the complicated construction of real ma- chines. This chapter will first explain the principles of ac machine operation using simple examples, and then consider some of the complications that occur in real ac machines. TIlcre are two major classes of ac rnachines-synchronous machines and in- duction machines. Synchronous machines are motors and generators whose mag- netic field current is supplied by a separate de power source, while induction ma- chines are motors and generators whose field current is supplied by magnetic induction (transformer action) into their fi eld windings. The field circuits of most synchronous and induction machi nes are located on their rotors. nlis chapter cov- ers some of the fundamentals common to both types of three-phase ac machines. Synchronous machines will be covered in detail in Chapters 5 and 6, and induc- tion machines will be covered in Chapter 7. 4.1 A SIMPLE LOOP IN A UNIFORM MAGNETIC FIELD We wil l start our study of ac machines with a simple loop of wire rotating within a uniform magnetic field. A loop of wire in a uniform magnetic field is the sim- plest possible machine that produces a sinusoidal ac voltage. nlis case is not rep- resentative of real ac machines, since the flux in real ac machines is not constant in either magnitude or direction. However, the factors that control the voltage and torque on the loop will be the same as the factors that control the voltage and torque in real ac machines. 230
- 255. ACMACHINERYFUNDAMENTALS 231 rom -------= ----- -----~---~---- ~~~ VcJ a/ -----L -----------T ---_ ' .... _-_ .... N s -~ ---------------------ii- "---- " is a uniform magnetic field, aligned as shown. ,., FIGURE 4- 1 , + d I , I , I , I , I , I 'h' b , + r- " A simple rotating loop in a uniform magnetic field. (a) Front view; (b) view of coil. Figure 4- 1shows a simple machi ne consisting of a large stationary magnet producing an essentially constant and uniform magnetic field and a rotating loop of wire within that field. The rotating part of the machine is called the rotor, and the stationary part of the machine is called the stator. We will now deterrnine the voltages present in the rotor as it rotates within the magnetic field. The Voltage Induced in a Simple Rotating Loop If the rotor of this machine is rotated, a voltage will be induced in the wire loop. To detennine the magnitude and shape of the voltage, examine Figure 4- 2. 1lle loop of wire shown is rectangular, with sides ab and cd perpendicular to the plane of the page and with sides be and da parallel to the plane of the page. The mag- netic field is constant and unifonn, pointing from left to right across the page. To determine the total voltage e,OI on the loop, we will examine each seg- ment of the loop separately and sum all the resulting voltages. The voltage on each segment is given by Equation (1-45): eind = (v x H) -' ( 1-45) I. Segment abo In this segment, the velocity of the wire is tangential to the path of rotation, while the magnetic field B points to the right, as shown in Figure 4- 2b. The quantity v x B points into the page, which is the same direction as segment abo Therefore, the induced voltage on this segment of the wire is eoo = (v x H) ·' = vBI sin (Jab into the page (4-1) 2. Segment be. In the first half of this segment, the quantity v x B points into the page, and in the second half of this segment, the quantity v x B points out of
- 256. 232 ELECTRIC MACHINERY RJNDAMENTALS 8 (a) (b) (,) ""GURE 4-2 (a) Velocities and oriemations of the sides of the loop with respect to the magnetic field. (b) The direction of motion with respect to the magnetic field for side abo (c) The direction of motion with respect to the magnetic field for side cd. the page. Since the length I is in the plane of the page, v x B is perpendicular to I for both portions of the segment. lllerefore the voltage in segment be will be zero: (4-2) ), Segment ed, In this segment, the velocity of the wire is tangential to the path of rotation, while the magnetic field B points to the right, as shown in Figure 4- 2c. The quantity v x B points into the page, which is the same direction as segment ed. TIlerefore, the induced voltage on this segment of the wire is edc = (v x B) ·1 = vBI sin (Jed out of the page (4-3) 4. Segment da. Just as in segment be, v x B is perpendicular to I. TIlerefore the voltage in this segment will be zero too: ead = 0 (4-4) 1lle total induced voltage on the loop ei!>d is the sum of the voltages on each of its sides: = vBI sin (Jab + vBI sin (Jed (4-5) Note that (Jab = 1800 - (Jed, and recall the trigonometric identity sin (J = sin (1800 - (J). Therefore, the induced voltage becomes eind = 2vBL sin (J (4-6) 1lle resulting voltage eind is shown as a function of time in Figure 4- 3. TIlere is an alternative way to express Equation (4-6), which clearly relates the behavior of the single loop to the behavior of larger, real ac machines. To de- rive this alternative expression, examine Figure 4- 2 again. If the loop is rotating at a constant angular velocity w, then angle (J of the loop will increase linearly with time. In other words,
- 257. ACMACHINERYFUNDAMENTALS 233 • 2 FIGURE 4-3 Plot of e... versus a. 3• -,- (J = wt e. radians Also, the tangential velocity v of the edges of the loop can be expressed as v = rw (4-7) where r is the radius from axis of rotation out to the edge of the loop and w is the an- gular velocity of the loop. Substituting these expressions into Equation (4-6) gives e;nd = 2rwBI sin wi (4-!l) Notice also from Figure 4-1 b that the area A of the loop is just equal to 2rl. Therefore, e;nd = ABw sin wt (4-9) Finally, note that the maximum flux through the loop occurs when the loop is per- pendicular to the magnetic flux density lines. This flux isjust the product of the loop's surface area and the flux density through the loop. q,max = AB Therefore, the final fonn of the voltage equation is Ie;nd q,rmu.w sin wt I (4-1 0) (4-11 ) Thus, the voltage generated in the loop is a sinusoid whose magnitude is equal to the product oftheJ1ux inside the machine and the speed ofrotation ofthe machine. This is also true of real ac machines. In general, the voltage in any real machine will depend on three factors: I. TIle flux in the machine 2. TIle speed of rotation 3. A constant representing the construction of the machine (the number of loops, etc.)
- 258. 234 ELECTRIC MACHINERY RJNDAMENTALS ------------------------ ,,---~ ---~R---~...--- J~: ""'""-- (jI ----~--------!"-~--- , -- ~--- ------------------------ II nis a uniform magnetic field. aligned as shown. The xin a wire indicates current flowing into the page. and the • in a wire indicates current flowing out of the page. (a) HGURE 4-4 , , I , I I , I , d - I , I ,b, A current-carrying loop in a unifonn magnetic field. (a) Front view; (b) view of coil. I into page ~' .~ , F ,., l out of page ,,' ,"'IGURE 4-5 ~ r. ,"' into page n roc'" 0 ,b, r. ,"' out of page • ,d, b , I, " (a) Derivation of force and torque on segment ab. (b) Derivation of force and torque on segment bc. (c) Derivation of force and torque on segment cd. (d) Derivation of force and torque on segment da. The Torque Induced in a Current-Carrying Loop Now assume that the rotor loop is at some arbitrary angle () with respect to the magnetic field, and that a current i is fl owing in the loop, as shown in Figure 4--4. If a current flows in the loop, then a torque will be induced on the wire loop. To detennine the magnitude and direction of the torque, examine Figure 4- 5. The force on each segment of the loop will be given by Equation (1--43), F = i(l x B) ( 1-43)
- 259. where ACMACHINERYFUNDAMENTALS 235 i = magnitude of current in the segment I = length of the segment, with direction of I defined to be in the direction of current flow B = magnetic flux density vector The torque on that segment will then be given by 7" = (force applied)(perpendicular distance) = (F) (r sin (j) = rF sin (j (1-6) where (J is the angle between the vector r and the vector F. The direction of the torque is clockwise if it would tend to cause a clockwise rotation and counter- clockwise if it wou Id tend to cause a counterclockwise rotation. I. Segment abo In this segment, the direction of the current is into the page, while the magnetic field B points to the right, as shown in Figure 4- 5a. The quantity I x B points down. Therefore, the induced force on this segment of the wire is TIle resulting torque is F =i(lxB) = ilB down 7"ab = (F) (r sin (jab) = rilB sin (jab clockwise (4-1 2) 2. Segment be. In this segment, the direction of the current is in the plane of the page, while the magnetic field B points to the right, as shown in Figure 4- 5b. TIle quantity I x B points into the page. Therefore, the induced force on this segment of the wire is F =i(lxB) = ilB into the page For this segment, the resulting torque is 0, since vectors r and I are parallel (both point into the page), and the angle (jbc is O. 7"bc = (F) (r sin (jab) ~O (4-1 3) 3. Segment ed. In this segment, the direction of the current is out of the page, while the magnetic field B points to the right, as shown in Figure 4- 5c. The quantity I x B points up. Therefore, the induced force on this segment of the wire is F =i(lxB) = ilB up
- 260. 236 ELECTRIC MACHINERY RJNDAMENTALS The resulting torque is Ted = (F) (r sin Oed) = rilB sin Oed clockwise (4-1 4) 4. Segmentda. In this segment. the direction of the current is in the plane of the page, while the magnetic field B points to the right, as shown in Figure 4- 5d. The quantity I x B points out of the page. 1llerefore, the induced force on this segment of the wire is F = i(l x B) = ilB out of the page For this segment, the resulting torque is 0, since vectors r and I are parallel (both point out of the page), and the angle 000 is O. Too = (F) (r sin O "J ~O (4- 15) 1lle total induced torque on the loop Tind is the sum ofthe torques on each of its sides: = rilB sin O ab + rilB sin O ed Note that O ab = O c.t, so the induced torque becomes TiDd = 2rilB sin 0 (4-1 6) (4-1 7) TIle resulting torque TiDd is shown as a function of angle in Figure 4-6. Note that the torque is maximum when the plane ofthe loop is parallel to the magnetic field, and the torque is zero when the plane of the loop is perpendicular to the mag- netic field. TIlere is an alternative way to express Equation (4-17), which clearly re- lates the behavior of the single loop to the behavior of larger, real ac machines. To derive this alternative expression, examine Figure 4-7. If the current in the loop is as shown in the figure, that current will generate a magnetic flux density Bloop with the direction shown. The magnitude of Bloop will be _ 1!i.. Bloop - G where G is a factor that depends on the geometry of the loop.* Also, note that the area of the loop A is just equal to 2rl. Substituting these two equations into Equa- tion (4-17) yields the result (4- 18) *If the loop were a cirde. then G", 2r. where r is the radius of the circle. so B""" '" lJ.inr. For a rec- tangular loop. the value of G will vary depending on the exact length-to-width ratio of the loop.
- 261. FIGURE 4-6 Plot of 1'... versus (J. .... (.) (b) ACMACHINERYFUNDAMENTALS 237 e. radians Jo'IGURE4-7 Derivation of the induced torque equation. (a) The current in the loop produces a magnetic flul( density "loop perpendicular to the plane of the loop; (b) geometric relationship between Dioop and Os. (4-1 9) where k = AGIJ1 is a factor depending on the construction of the machine, Bs is used for the stator magnetic field to distinguish it from the magnetic field gener- ated by the rotor, and () is the angle between B loop and Bs.The angle between B loop and Bs can be seen by trigonometric identities to be the same as the angle () in Equation (4-1 7). Both the magnitude and the direction of the induced torque can be deter- mined by expressing Equation (4--19) as a cross product: (4- 20) Applying this equation to the loop in Fig ure 4- 7 produces a torque vector into the page, indicating that the torque is clockwise, with the magnitude given by Equa- tion (4-1 9). Thus, the torque induced in the loop is proportional to the strength of the loop's magnetic field, the strength ofthe external magnetic field, and the sine of the angle between them. This is also true of real ac machines. In general, the torque in any real machine will depend on four factors:
- 262. 238 ELECTRIC MACHINERY RJNDAMENTALS I. The strength of the rotor magnetic field 2. The strength of the external magnetic field 3. The sine of the angle between them 4. A constant representing the construction of the machine (geometry. etc.) 4.2 THE ROTATING MAGNETIC FIELD In Section 4.1, we showed that if two magnetic fields are present in a machine, then a torque will be created which will tend to line up the two magnetic fields. If one magnetic field is produced by the stator of an ac machine and the other one is produced by the rotor of the machine, the n a torque will be induced in the rotor which will cause the rotor to turn and align itself with the stator magnetic field. If there were some way to make the stator magnetic field rotate, then the in- duced torque in the rotor would cause it to constantly "chase" the stator magnetic field around in a circle. lllis, in a nutshell, is the basic principle of all ac motor operation. How can the stator magnetic field be made to rotate? llle fundamental prin- ciple of ac machine operation is that ifa three-phase set ofcurrents, each ofequal mngnitude and differing in phase by 120°,flows in a three-phase winding, then it will produce a rotating mngnetic field of constant mngnitude. The three-phase winding consists of three separate wi ndi ngs spaced 120 electrical degrees apart around the surface of the machine. llle rotating magnetic field concept is illustrated in the simplest case by an empty stator contai ning just three caiIs, each 1200 apart (see Figure 4-8a). Since such a winding produces only one north and one south magnetic pole, it is a two- pole winding. To understand the concept of the rotating magnetic field, we will apply a set of currents to the stator of Figure 4--8 and see what happens at specific instants of time. Assume that the currents in the three coils are given by the equations iaa' (1) = 1M sin wt A ibb' (1) = 1M sin (wt - 120°) icc' (1) = 1M sin (wt - 240°) A A (4- 2I a) (4- 21 b) (4- 2Ic) llle current in coil aa' flows into the a end of the coil and out the a' end of the coil. It produces the magnetic field inte nsity A ·turns/ m (4- 22a) where 0° is the spatial angle of the magnetic field intensity vector, as shown in Figure 4-8b. llle direction of the magnetic field intensity vector Had(t) is given by the right-hand rule: If the fingers of the right hand curl in the direction of the current flow in the coil, then the resulting magnetic field is in the direction that the thumb points. Notice that the magnitude of the magnetic field intensity vector H"..,(l) varies sinusoidally in time, but the direction of Had(l) is always constant. Similarly, the magnetic field intensity vectors Hb/;o,(l) and H«(l) are
- 263. ACMACHINERYFUNDAMENTALS 239 ° , °b' FIGURE 4- 8 11",, -(/) 11",-(/) " ",,(0 0" "I o a' (a) A simple three-phase stator. Currents in this stator are assumed positive if they flow into the unprimed end and out the primed end of the coils. The magnetizing intensities produced by each coil are also shown. (b) The magnetizing intensity vector H.... (/) produced by a current flowing in coil 00'. A · turns/ m (4-22b) A·turns/ m (4- 22c) The flux densities resulting from these magnetic field intensities are give n by Equation ( 1-2 1): They are Baa' (1) = BM sin wl L 0° T Bbb' (1) = BM sin (wl- 120°) L 120° BC<", (1) = BM sin (wl- 240°) L 240° T T ( 1-21) (4- 23a) (4-23b) (4- 23c) where BM = J1HM . 1lle currents and their corresponding flux densities can be ex- amined at specific times to detennine the resulting net magnetic field in the stator. For example, at time wt = 0°, the magnetic field from coil ad will be B"",,= 0 The magnetic field from coil bb' will be BbI>' = BM sin (_1 20°) L 1200 and the magnetic field from coil ee' wi] I be BC<", = BM sin (_240°) L 240° (4- 24a) (4-24b) (4- 24c)
- 264. 240 ELECTRIC MACHINERY RJNDAMENTALS , '"b' ""GURE 4-9 0, WI = 0" (a) o b o , o '"b' ncr. ~R~ "w '" , c '" , WI =90° (b) (a) The vector magnetic field in a stator at time WI = 0°. (b) The vector magnetic field in a sta.tor a.t time WI = 90°. TIle total magnetic field from all three coils added together will be Bne! = Baa' + Bw + B ee' = 0 + (-f BM) L 120° + (fBM) L240° = 1.5BML-9Q0 TIle resulting net magnetic field is shown in Figure 4- 9a. b As another example, look at the magnetic field at time wt = 90°. At that time, the currents are i",,' = 1M sin 90° A i"",= IM sin (- 300) A iec,= IM sin (-1 500) A and the magnetic fields are B"",= BM LO° B"", = -0.5 BML 1200 Bec' = -0.5 BML 2400 The resulting net magnetic field is Bn.. = Baa' + B"". + B..... = BML 0° + (-O.5BM) L 120° + (-O.5BM) L 240° = 1.5 BM LO°
- 265. ACMACHINERYFUNDAMENTALS 241 The resulting magnetic field is shown in Figure 4- 9b. Notice that although the di- rection of the magnetic field has changed, the magnitude is constant. TIle mag- netic field is maintaining a constant magnitude while rotating in a counterclock- wise direction. Proof of the Rotating Magnetic Field Concept At any time t, the magnetic field wi ll have the same magnitude I.5BM, and it wi ll continue to rotate at angular velocity w. A proof of this statement for all time t is now given. Refer again to the stator shown in Figure 4-8. In the coordinate system shown in the fi gure, the x direction is to the right and the y direction is upward. TIle vector:l1 is the unit vector in the horizontal direction, and the vector S' is the unit vector in the vertical direction. To find the total magnetic flux density in the stator, simply add vectorially the three component magnetic fields and detennine their sum. The net magnetic nux density in the stator is given by B••(I) ~ B_, (I) + B~, (I) + B~, (I) = BM sin wt LO° + BMsin (wt - 120°) L 120° +BM sin (wi_ 240°) L 2400T Each of the three component magnetic fields can now be broken down into its x and y components. Bnet(t) = BMsin wt x - [O.5BMsin (wt - 1200)]x + ['] BMsin (wt - 1200)]y - [O.5BMsin (wt - 2400)]x - ['] BM sin (wt - 2400)]y Combining x and y components yields Dnet(t) = [BMsin wi - O.SBMsin (wt - 120°) - O.5BMsin (wt - 240°)] x + ['7BMsin(wt - 120°) - '7BMsin (wt - 2400)]y By the angle-addition trigonometric ide ntities, BnetCt) = [BMsin wi + iBM sin wt + 1BM cos wi + i BMsin wt -1BMcos wt]x + [-1BMsinwt - ~BMCOSWt + ~BM sinwt - ~BMCOSWt]S' IBneln = (1.5BMsin wt):I1 - (1.5BMcos wI)y I (4- 25) Equation (4- 25) is the final expression for the net magnetic flux density. Notice that the magnitude of the field is a constant I.SBMand that the angle changes con- tinually in a counterclockwise direction at angular velocity w. Notice also that at
- 266. 242 ELECTRIC MACHINERY RJNDAMENTALS N - 0 . b ""GURE 4-10 The rotating magnetic field in a stator represented as moving north and south stator poles. wi = 0°, BDeI = I.SBM L _90° and that at wt = 90°, Bne, = 1.58M L 0°. 1l1ese re- sults agree with the specific examples examined previously. The Relationship between Electrical Frequency and the Speed of Magnetic Field Rotation Figure 4-1 0 shows that the rotating magnetic field in this stator can be represented as a north pole (where the flux leaves the stator) and a south pole (where the flux enters the stator). These magnetic poles complete one mechanical rotation around the stator surface for each electrical cycle of the applied current. 1l1erefore, the mechanical speed of rotation of the magnetic field in revolutions per second is equal to the electric frequency in hertz: two poles two poles (4- 26) (4- 27) Here1m and w,., are the mechanical speed in revolutions per second and radians per second, while!. and W e are the electrical speed in hertz and radians per second. Notice that the windings on the two-pole stator in Figure 4- 10 occur in the order (taken counterclockwise) a-c'-b-a '-c-b' What would happen in a stator if this pattern were repeated twice within it? Fig- ure 4-ll a shows such a stator. There, the pattern of windings (taken counter- clockwise) is a-c '-b-a'-c-b '-a-c '-b-a '-c-b' which is just the pattern of the previous stator repeated twice. When a three-phase set of currents is applied to this stator, two north poles and two south poles are pro- duced in the stator winding, as shown in Figure 4-11 b. In this winding, a pole
- 267. ACMACHINERYFUNDAMENTALS 243 b, b,~ ~ @ ~ s @ " ai / II " "; • 0 0 0 w. w. 0 "i / oi " i!1 @ ~i!1 b, b, b' 'll (a) (b) , " b , " b Back: 1 '"' of stator coils • • 8 • X • X • II ,s, II ,N , I-I ,s, II ,N , , , , , , , , , ! ! ! I I I ", '; b, "i " bi ", 'i b, "; C2 bi ) j j I , b , Counterclock:wise b' f') FIGURE 4- 11 (a) A simple four-pole stator winding. (b) The resulting stator magnetic poles. Notice that there are moving poles of alternating polarity every 90° around the stator surface. (c) A winding diagram of the stator as seen from its inner surface, showing how the stator currents produce north and south magnetic poles. moves only halfway around the stator surface in one electrical cycle. Since one electrical cycle is 360 electrical degrees, and since the mechanical motion is 180 mechanical degrees, the relationship between the electrical angle O e and the me- chanical angle 0", in this stator is (4- 28) Thus for the four-pole winding, the electrical frequency of the current is twice the mechanical frequency of rotation:
- 268. 244 ELECTRIC MACHINERY RJNDAMENTALS fe = 2fm four poles W e = 2wm four poles (4- 29) (4- 30) Ingeneral, if the number of magnetic poles on an ac machine stator is P, then there are PI2 repetitions of the winding sequence a-c '-b-a '-e-b' around its inner surface, and the electrical and mechanical quantities on the stator are related by le, ~ iem l (4- 3 1) (4- 32) (4- 33) Also, noting that fm = n,,/60, it is possible to relate the electrical frequency in hertz to the resulting mechanical speed of the magnetic fields in revolutions per minute. nlis relationship is Reversing the Direction of Magnetic Field Rotation (4- 34) Another interesting fact can be observed about the resulting magnetic field.lfthe current in any two of the three coils is swapped, the direction of the mngnetie field's rotation will be reversed. This means that it is possible to reverse the direc- tion of rotation of an ac motor just by switching the connections on any two of the three coils. lllis result is verified below. To prove that the direction of rotation is reversed, phases bb' and ee' in Fig- ure 4-8 are switched and the resulting flux density Bn .. is calculated. llle net magnetic flux density in the stator is given by B...,(t) = B"".(t) + Bw(t) + BeAt) = BM sin wi L 0° + BM sin (wi- 240°) L 120° + BM sin (wI- 120°) L 240° T Each of the three component magnetic fie lds can now be broken down into its x and y components: BDeI(t) = BM sin wt5i. - [0.5BM sin (wt - 2400)] 5i. + [1"BM sin (wt - 2400 )]y - [0.5BM sin (wt - I200)] 5i. - [1"BM sin (wt - 1200 )]y Combining x and y components yields
- 269. ACMACHINERYFUNDAMENTALS 245 Hnem = [BMsin wt - O.5BMsin (wt - 240°) - 0.5BM sin(WI' - 1200jx + ['7BMsin (WI' - 240°) - '7BM sin (wt - l200)]y By the angle-addition trigonometric identities, S nelt) = [BMsin WI' + iBM sin wt -1BM cos WI' + iBM sin wt + IBM cos wt]x + [-1BMsinwt + ~BM COS Wt + ~BM sinwt + ~BMCOSWt]S IS nell) = (1.5BMsin wt)J1 + (1.5BMcos WI')Y I (4- 35) This time the magnetic fie ld has the same magnitude but rotates in a clock- wise direction. 1l1erefore, switching the currents in two stator phases reverses the direction ofmagneticfield rotation in an ac machine. EXllmple 4-1. Create a MATLAB program that models the behavior of a rotating magnetic field in the three-phase stator shown in Figure 4-9. Solutioll The geometry of the loops in this stator is fIXed as shown in Figure 4-9. The currents in the loops are i"",(t) = 1M sin wt A iw(t) = 1M sin (wt - 120°) iec,(t) = 1M sin (wt - 240°) and the resulting magnetic flux densities are B"",(t)= BM sinwt LO° T A A B"",(t) = BM sin (wt - 120°) L 120° Bec,(t) = BM sin (wt - 240°) L 240° <jl = 2rlB = dlB T T (4-2Ia) (4-21b) (4-21c) (4-23a) (4-23b) (4-23c) A simple MATLAB program that plots Boo" BI+" Bee'. and B"", as a ftmction of time is shown below: % M-file, mag_ fi e l d.m % M-file t o ca l c ulate the net magnetic fi e l d produ ced % by a three-pha se s tat or. % Set up the bas i c cond ition s bmax = 1; % Normalize bmax t o 1 fr eq = 60, % 60 Hz w = 2*p i * fr eq, % a ngular ve l oc ity (rad/ s ) % Fir s t , generate the three component magnetic fi e l ds t = 0,1 / 6000, 1 / 60, Baa = s in (w* t ) .* (cos ( O) + j*s in (O)) ,
- 270. 246 ELECTRIC MACHINERY RJNDAMENTALS Ebb = s in (w*t - 2*pi /3) * (cos (2*pi /3) + j*s in (2*pi /3)) ; Ecc = s in (w*t +2*pi /3) * (cos (- 2*pi /3) + j*s in (- 2*pi /3)) ; !l; Ca l c ulate Enet Enet = Baa + Ebb + Ecc; !l; Ca l c ulate a c irc l e representing the expected maximum !l; va lue o f Enet c irc l e = 1.5 * (cos (w*t ) + j *s in (w*t ) ) ; !l; Plo t the magnitude and d irection of the r esulting magne ti c !l; fi e l ds. No te that Baa i s b l ack, Bbb i s b lue, Bcc i s !l; magenta , and Enet i s red. f o r ii = l,length (t ) !l; Plo t the reference c irc l e p l o t (c irc l e, 'k' ) ; h o l d o n; !l; Plo t the f o ur magneti c fi e l ds p l o t ( [0 real (Baa (il )) ] , [0 i mag (Baa (il )) ] , 'k', 'LineWi dth' ,2); p l o t ( [0 real (Ebb (il ) ) ] , [0 i mag (Bbb (il ) ) ] , 'b' , 'LineWi dth ' ,2) ; p l o t ( [0 real (Bec (il ) ) ] , [0 i mag (Bec (il ) ) ] , 'm' , 'LineWi dth ' ,2) ; p l o t ( [O real (Enet (il )) ] , [0 i mag (Bne t (il )) ] ,' r' ,' LineWi dth ',3 ) ; axi s square; axi s( [- 2 2 - 2 2 ] ) ; drawnow; ho l d o ff ; ond When this program is executed, it draws lines corresponding to the three component mag- netic fields as well as a line corresponding to the net magnetic field. Execute this program and observe the behavior of B..... 4.3 MAGNETOMOTIVE FORCE AND FLUX DISTRIBUTION ON AC MACHINES In Section 4.2, the flux produced inside an ac machine was treated as if it were in free space. TIle direction of the flux density produced by a coil of wire was as- sumed to be perpendicular to the plane of the coil, with the direction of the flux given by the right-hand rule. TIle flux in a real mnchine does not behave in the simple manner assumed above, since there is a ferromagnetic rotor in the center of the machine, with a small air gap between the rotor and the stator. TIle rotor can be cylindrical, like the one shown in Figure 4-1 2a, or it can have pole faces projecting out from its surface, as shown in Figure 4-12b. If the rotor is cylindrical, the machine is said to have nonsalient poles; if the rotor has pole faces projecting out from it, the
- 271. ACMACHINERYFUNDAMENTALS 247 o o o o ,,' ,b, FIGURE 4- 12 (a) An ac machine with a cylindrical or nonsalient-pole rotor. (b) An ac machine with a salient-pole rotor. machine is said to have salient poles. Cylindrical rotor or nonsalient-pole ma- chines are easier to understand and analyze than salient-pole machines, and this discussion will be restricted to machines with cylindrical rotors. Machines with salient poles are discussed briefly in Appendix C and more extensively in Refer- ences I and 2. Refer to the cylindrical-rotor machine in Figure 4-1 2a. The reluctance of the air gap in this machine is much higher than the reluctances of either the rotor or the stator, so the flux density vector B takes the shortest possible path across the air gap and jumps perpendicularly between the rotor and the stator. To produce a sinusoidal voltage in a machine like this, the magnitude ofthe flux density vector B must vary in a sinusoidal manner along the surface of the air gap. TIle flux density will vary sinusoidally only if the magnetizing intensity H (and magnetomotive force ?f) varies in a sinusoidal manner along the surface of the air gap (see Figure 4-1 3). The most straightforward way to achieve a sinusoidal variation of magneto- motive force along the surface of the air gap is to distribute the turns of the wind- ing that produces the magnetomotive force in closely spaced slots around the surface of the machine and to vary the number of conductors in each slot in a sinusoidal manner. Figure 4-14a shows such a winding, and Figure 4-1 4b shows the magnetomotive force resulting from the winding. TIle number of conductors in each slot is given by the equation nc = Necos a (4- 36) where Ne is the number of conductors at an angle of 0°. As Figure 4-1 4b shows, this distribution of conductors produces a close approximation to a sinusoidal dis- tribution of magnetomotive force. Furthermore, the more slots there are around the surface of the machine and the more closely spaced the slots are, the better this approximation becomes.
- 272. 248 ELECTRIC MACHINERY RJNDA MENTALS B=BMsina Stator Air gap Rotor -I--f--L a ,,) • or (lH.)-I) ~__L-__L-__L-__ -__~__~__~__+--a (b) '". ~__L-__L-__L-__ -__~__~__~__+--a 'e) ""GURE 4- 1J (a) A cylindrical rotor with sinusoidally varying air-gap flux density. (b) The magnetomotive force or magnetizing imensity as a function of angle a in the air gap. (c) The flux density as a function of angle a in the air gap.
- 273. ACMACHINERYFUNDAMENTALS 249 3 3 7 7 • 10 • -- ill! 10 -- -- a 10 0 @IO • Ell 7 7 • 0 Assume Nc '" 10 3 3 ,., ~ 20 ,3 'L , , 10 -!- --', , , , , 0 a 60 120 180 240 300 "60 , , , - 10 ";- ---t , , - 20 "" ,b, FIGURE 4- 14 (a) An ac machine with a distributed stator winding designed to produce a sinusoidally varying air- gap flux density. The number of conductors in each slot is indicated on the diagram. (b) The magnetomotive force distribution resulting from the winding. compared to an ideal distribution. In practice, it is not possible to distribute windings exactly in accordance with Equation (4- 36), since there are only a finite number of slots in a real ma- chine and since only integral numbers of conductors can be included in each slot. The resulting magnetomotive force distribution is only approximately sinusoidal, and higher-order harmonic components will be present. Fractional-pitch windings are used to suppress these unwanted harmonic components, as explained in Ap- pendix S.l.
- 274. 250 ELECTRIC MACHINERY RJNDAMENTALS Furthennore, it is often convenient for the machine designer to include equal numbers of conductors in each slot instead of varying the number in accor- dance with Equation (4--36). Windings ofthis type are described in Appendix B.2; they have stronger high-order harmonic components than windings designed in accordance with Equation (4- 36). The harmonic-suppression techniques of Ap- pendix B.I are especially important for such windings. 4.4 INDUCED VOLTAGE IN AC MACHINES Just as a three-phase set of currents in a stator can produce a rotating magnetic field, a rotating magnetic field can produce a three-phase set of voltages in the coils of a stator. 1lle equations governing the induced voltage in a three-phase sta- tor will be developed in this section. To make the development easier, we will be- gin by looking at just one single-turn coil and then expand the results to a more general three-phase stator. The Induced Voltage in a Coil on a Two-Pole Stator Figure 4-1 5 shows a rotating rotor with a sinusoidally distributed magnetic field in the center of a stationary coil. Notice that this is the reverse of the situation studied in Section 4.1, which involved a stationary magnetic field and a rotating loop. We will assume that the magnitude of the flux density vector B in the air gap between the rotor and the stator varies sinusoidally with mechanical angle, while the direction of B is always radially outward. 1llis sort of flux distribution is the ideal to which machine designers aspire. (What happens when they don't achieve it is described in Appendix B.2. ) If (l is the angle measured from the direction of the peak rotor flux density, then the magnitude of the nux density vector B at a point around the rotor is given by B = BM cos (l (4- 37a) Note that at some locations around the air gap, the nux density vector will really point in toward the rotor; in those locations, the sign of Equation (4- 37a) is nega- tive. Since the rotor is itself rotating within the stator at an angular velocity W m, the magnitude of the nux density vector B at any angle a around the stator is given by IB - BMcos(wt - a) I 1lle equation for the induced voltage in a wire is e= (v x B) ·1 where v = velocity of the wire relative to the magneticfield B = magnetic flux density vector I = length of conductor in the magnetic field (4- 37b) ( 1-45)
- 275. AC MACHINERY FUNDAMENTALS 251 I d ,,) " Air-gap flux density: Air gap t B(a ) =BM cos(oo",l - a ) Stator Rotor 0-- vrel o c-d , a, , : "M , o (J - b Voltage is really into the page. since B is negative here. (b) FIGURE 4- 15 B e b 'e) (a) A rotating rotor magnetic field inside a stationary stator coil. Detail of coil. (b) The vector magnetic flux densities and velocities on the sides of the coil. The velocities shown are from a frame of reference in which the magnetic field is stationary. (c) The flux density distribution in the air gap.
- 276. 252 ELECTRIC MACHINERY RJNDAMENTALS However, this equation was derived for the case of a moving wire in a stationnry mngnetie field. In this case, the wire is stationary and the magnetic field is mov- ing, so the equation does not directly apply. To use it, we must be in a frame of reference where the magnetic field appears to be stationary. Ifwe "sit on the mag- netic field" so that the field appears to be stationary, the sides of the coil will ap- pear to go by at an apparent velocity vre[, and the equation can be applied. Figure 4- 15b shows the vector magnetic field and velocities from the point of view of a stationary magnetic field and a moving wire. TIle total voltage induced in the coil will be the sum ofthe voltages induced in each of its four sides. These voltages are detennined below: I. Segment abo For segment ab, a = 180°. Assuming that B is directed radially outward from the rotor, the angle between v and B in segment ab is 90°, while the quantity v x B is in the direction of I, so e"", = (v x B) · 1 = vBI directed out of the page = - V[BM cos (w",t - 180°)]1 = - vBtJ cos (w",t - 180°) (4- 38) where the minus sign comes from the fact that the voltage is built up with a polarity opposite to the assumed polarity. 2. Segment be. The voltage on segment be is zero, since the vector quantity v x B is perpendicular to I, so ecb = (v x B) -I = 0 (4- 39) 3. Segment ed. For segment ed, the angle a = 0°. Assuming that B is directed radially outward from the rotor, the angle between v and B in segment ed is 90°, while the quantity v x B is in the direction ofl, so eJc = (v x B) - I = vBI directed out of the page = v(BM cos wn,l)l = vBtJ cos w.,/ (4-40) 4. Segment da. The voltage on segment da is zero, since the vector quantity v x B is perpendicular to I, so ead = (v x B) - I = 0 Therefore, the total voltage on the coil wi ll be eind = e"" + edc = - vBtJ cos(w",t - 180°) + vBtJ cos w",t Since cos () = - cos « () - 180°), (4-41) (4-42)
- 277. ACMACHINERYFUNDAMENTALS 253 ei!>d = vB&I cos wmt + vB&I cos wmt = 2vB&I cos wmt (4--43) Since the velocity of the end conductors is given by v = rwm , Equation (4--43) can be rewritten as ei!>d = 2(rwm )B&I cos wmt = 2rlB~m cos wmt Finally, the flux passing through the coil can be expressed as <p = 2rlBm (see Problem 4- 7), while W m = W e = W for a two-pole stator, so the induced voltage can be expressed as I eind - ¢w cos wi I (4-44) Equation (4--44) describes the voltage induced in a single-turn coil. Ifthe coil in the stator has Nc turns of wire, then the total induced voltage of the coil will be I eind - Nc¢w cos wt I (4-45) Notice that the voltage produced in stator of this simple ac machine wind- ing is sinusoidal with an amplitude which depends on the flux <p in the machine, the angular velocity w of the rotor, and a constant depending on the construction of the machine (Nc in this simple case). This is the same as the result that we ob- tained for the simple rotating loop in Section 4.1. Note that Equation (4--45) contains the term cos wt instead of the sin wt found in some of the other equations in this chapter. 1lle cosine tenn has no spe- cial significance compared to the sine- it resulted from our choice of reference direction for 0: in this derivation. If the reference direction for 0: had been rotated by 90° we would have had a sin wt tenn. The Induced Voltage in a Three-Phase Set of Coils If three coils, each of Nc turns, are placed around the rotor magnetic field as shown in Figure 4-1 6, then the voltages induced in each of them will be the same in magnitude but wi II differ in phase by 120°.1lle resulting voltages in each of the three coils are e.....(t) = Nc <Pw sin wt V ew(r) = Nc <Pw sin (wt - 120°) ee,,,(t) = Nc <pw sin (wt- 2400) v V (4--46a) (4-46b) (4--46c) Therefore, a three-phase sct of currents can generate a unifonn rotating magnetic field in a machine stator, and a uniform rotating magnetic field can gen- erate a three-phase sct of voltages in such a stator.
- 278. 254 ELECTRIC MACHINERY RJNDAMENTALS ~.M FIGURE 4- 16 The production of three-phase voltages from three coils spaced 120° apan. The RMS Voltage in a Three-Phase Stator TIle peak voltage in any phase of a three-phase stator of this sort is Since w = 2nf, this equation can also be written as Enuu = 27rNc <pf TIlerefore, the nns voltage of any phase of this three-phase stator is 2,,- EA = lfNc<pf IE,- !2,,-Nc<l>f I (4-47) (4-48) (4-49) (4- 50) TIle nns voltage at the terminnls of the machine will depend on whether the stator is Y- or .1.-connected. Ifthe machine is V-connected, then the tenninal voltage will be V3 times EA; if the machine is .1.-connected, then the tenninal voltage will just be equal to EA. Example 4-2. The following information is known about the simple two-pole generator in Figure 4--16. The peak flux density of the rotor magnetic field is 0.2 T, and the mechanical rate of rotation of the shaft is 3600 r/min. The stator diameter of the machine is 0.5 m, its coil length is 0.3 m, and there are 151lU1ls per coil. The machine is V-connected. (a) What are the three phase voltages of the generator as a ftmction of time? (b) What is the nns phase voltage of this generator? (c) What is the nns tenninal voltage of this generator? Solutioll The flux in this machine is given by <P = 2rlB = dlB
- 279. ACMACHINERYFUNDAMENTALS 255 where d is the diameter and I is the length of the coil. Therefore, the flux in the machine is given by <p = (0.5 mXO.3 m)(0.2 T) = 0.03 Vb The speed of the rotor is given by w = (3600 r/minX27T radXI minl60 s) = 377 radls (a) The magnitudes of the peak. phase voltages are thus Emu = Nc<Pw = (15 turnsXO.03 Wb)(377 radls) = 169.7 V and the three phase voltages are e"".(t) = 169.7 sin 377t V ebb.(t) = 169.7 sin (377t -1200) V e,At) = 169.7 sin (377t - 240°) V (b) The nns phase voltage of this generator is Ell = E~x = 16~V = 120V (c) Since the generator is V-connected, VT = v'5EIl = 0(120 V) = 208 V 4.5 INDUCED TORQUE IN AN AC MACHINE In ac machines under nonnaI operating conditions, there are two magnetic fields present--.:1. magnetic field from the rotor circuit and another magnetic field from the stator circuit. The interaction of these two magnetic fields produces the torque in the machine, just as two pennanent magnets near each other will experience a torque which causes them to line up. Figure 4-1 7 shows a simplified ac machine with a sinusoidal stator flux dis- tribution that peaks in the upward direction and a single coil of wire mounted on the rotor. TIle stator flux distribution in this machine is Bs<,.a ) = Bs sin a (4- 51) where Bs is the magnitude of the peak flux density; B:!..a ) is positive when the flux density vector points radially outward from the rotor surface to the stator surface. How much torque is produced in the rotor of this simplified ac machine? To find out, we will analyze the force and torque on each ofthe two conductors separately. The induced force on conductor I is F = i(l x B) = ilBs sin a The torque on the conductor is "TiDd.] = (r x F) = rilBs sin a with direction as shown counterclockwise ( 1-43)
- 280. 256 ELECTRIC MACHINERY RJNDAMENTALS a IUia)1 '" Bs sin a ""GURE4- 17 A simplified ac machine with a sinusoidal sta.tor flux distribution a.nd a single coil of wire mounted in the rotor. TIle induced force on conductor 2 is F = i(lxB) = ilBs sin a TIle torque on the conductor is 1";oo.t = (r x F) = rilBs sin a with direction as shown counterclockwise TIlerefore, the torque on the rotor loop is l1"ind = 2rilBs sin a counterclockwise I ( 1-43) (4- 52) Equation (4- 52) can be expressed in a more convenient fonn by examining Figure 4-1 8 and noting two facts: I. The current i flowing in the rotor coil produces a magnetic field of its own. The direction of the peak of this magnetic field is given by the right-hand rule, and the magnitude of its magnetizing intensity HR is directly propor- tional to the current flowing in the rotor:
- 281. ACMACHINERYFUNDAMENTALS 257 FIGURE 4- 18 B, , ,,,,,, II.., '-j<" _ f a _ "'-- ---l__ _ _____ , , , , , , I I1~ HR r= 180" - a The components magnetic flux density inside Ihe machine of Figure 4--17. HR = Ci where C is a constant of proportionality. (4- 53) 2. 1lle angle between the peak of the stator flux density Bs and the peak of the rotor magnetizing intensity HR is y. Furthennore, y=180o-a sin y= sin ( 180° _a) = sin 0: (4- 54) (4- 55) By combining these two observations, the torque on the loop can be expressed as "Tioo = KHI13ssin a counterclockwise (4- 56) where K is a constant dependent on the construction of the machine. Note that both the magnitude and the direction of the torque can be expressed by the equation I"Tind - KHR X Bs I (4- 57) Finally, since B R = /LHR, this equation can be reexpressed as (4- 58) where k = KIp. Note that in general k will not be constant, since the magnetic per- meability p varies with the amount of magnetic saturation in the machine. Equation (4--58) is just the same as Equation (4--20), which we derived for the case of a single loop in a unifonn magnetic field. It can apply to any ac machine, not
- 282. 258 ELECTRIC MACHINERY RJNDAMENTALS just to the simple one-loop rotor just described. Only the constant k will differ from machine to machine. This equation will be used only for a qualitative study of torque in ac machines, so the actual val ue of k is unimportant for our purposes. TIle net magnetic field in this machine is the vector sum of the rotor and sta- tor fields (assuming no saturation): B..., = BR + Bs (4- 59) TIlis fact can be used to produce an equivalent (and sometimes more useful) ex- pression for the induced torque in the machine. From Equation (4- 58) "TiDd = kB R X Bs But from Equation (4- 59), Bs = BDe, - BR, so "Tind = kBR X (BDe, - BR) = k(BR x B"",) - k(BR x BR) Since the cross prOOuct of any vector with itself is zero, this reduces to (4- 58) (4-60) so the induced torque can also be expressed as a cross product of BR and BDe, with the same constant k as before. The magnitude of this expression is (4-6 1) where /j is the angle between BR and B...,. Equations (4-58) to (4-6 1) will be used to help develop a qualitative un- derstanding of the torque in ac machines. For example, look at the simple syn- chronous machine in Figure 4-1 9. Its magnetic fields are rotating in a counter- clockwise direction. What is the direction of the torque on the shaft of the machine's rotor? By applying the right-hand rule to Equation (4- 58) or (4-60), the induced torque is found to be clockwise, or opposite the direction of rotation of the rotor. Therefore, this machine must be acting as a generator. 4.6 WINDING INSULATION IN AN ACMACHINE One of the most critical parts of an ac machine design is the insulation of its wind- ings. If the insulation of a motor or generator breaks down, the machine shorts out. The repair ofa machine with shorted insulation is quite expensive, ifit is even possible. To prevent the winding insulation from breaking down as a result of overheating, it is necessary to limit the temperature of the windings. TIlis can be partially done by providing a cooling air circulation over them, but ultimately the maximum winding temperature limits the maximum power that can be supplied continuously by the machine.
- 283. w <8> 0, , , , , , , r ! B, , , , , ACMACHINERYFUNDAMENTALS 259 FlGURE 4- 19 A simplified synchronous machine showing its rotor and stator magnetic fields. Insulation rarely fails from immediate breakdown at some critical tempera- ture. Instead, the increase in temperature produces a gradual degradation of the in- sulation, making it subject to failure from another cause such as shock, vibration, or electrical stress. 1l1ere was an old rule of thumb that said that the life ex- pectancy of a motor with a given type of insulation is halved for each 10 percent rise in temperature above the rated temperature of the winding. This rule still ap- plies to some extent today. To standardize the temperature limits of machine insulation, the National Electrical Manufacturers Association (NEMA) in the United States has defined a series of insulation system classes. Each insulation system class specifies the maximum temperature rise pennissible for that class of insulation. 1l1ere are three common NEMA insulation classes for integral-horsepower ac motors: 8, F, and H. Each class represents a higher pennissible winding temperature than the one before it. For example, the annature winding temperature rise above ambient tem- perature in one type of continuously operating ac induction motor must be limited to 80°C for class 8, 105°C for class F, and 125°C for class H insulation. The effect of operating temperatu re on insulation life for a typical machine can be quite dramatic. A typical curve is shown in Figure 4-20. This curve shows the mean life of a machine in thousands of hours versus the temperature of the windings, for several different insulation classes. The specific temperature specifications for each type of ac motor and gen- erator are set out in great detail in NEMA Standard MG 1-1993, Motors and Gen- erators. Similar standards have been defined by the International Electrotechnical Commission (IEC) and by various national standards organizations in other countries.
- 284. ~ < B u g ~ '--... i'- ~ "" 8 ------"'- 1 0 ~ '--... i'- 1 ~ 0 , N , • 0 ~ M • , ij ~ "" ~ ------ 1 ~ ~ g 0 00 ~ 8 0 ~ spuemoljlll! SJIloH 260
- 285. AC MACHINERY FUNDAMENTALS 261 4.7 AC MACHINE POWER FLOWS AND LOSSES AC generators take in mechanical power and produce electric power, while ac motors take in electric power and produce mechanical power. In either case, not all the power input to the machine appears in useful form at the other end- there is always some loss associated with the process. The efficiency of an ac machine is defined by the equation ?"UI " ~ - X 100% P in (4-62) The difference between the input power and the output power of a machine is the losses that occur inside it. lllerefore, (4-63) The Losses in AC Machines The losses that occur in ac machines can be divided into four basic categories: I. Electrical or copper losses (/ 2R losses) 2. Core losses 3. Mechanical losses 4. Stray load losses ELECTRICAL OR COPPER LOSSES. Copper losses are the resistive heating losses that occur in the stator (annature) and rotor (field) windings of the machine. TIle sta- tor copper losses (SCL) in a three-phase ac machine are given by the equation (4-64) where IAis the current flowing in each annature phase and Rio. is the resistance of each armature phase. The rotor copper losses (RCL) of a synchronous ac machine (induction ma- chines will be considered separately in Chapter 7) are given by (4-65) where IFis the current flowing in the field winding on the rotor and RF is the re- sistance of the field winding. The resistance used in these calculations is usually the winding resistance at nonnal operating temperature. CORE LOSSES. The core losses are the hysteresis losses and eddy current losses occurring in the metal of the motor. These losses were described in Chapter I.
- 286. 262 ELECTRIC MACHINERY RJNDAMENTALS TIlese losses vary as the square of the flux density (8 2 ) and, for the stator, as the l.5th power of the speed of rotation of the magnetic fields (nI.5) . MECHANICAL LOSSES. The mechanical losses in an ac machine are the losses associated with mechanical effects. There are two basic types of mechanical losses:friction and windage. Friction losses are losses caused by the friction of the bearings in the machine, while windage losses are caused by the friction between the moving parts of the machine and the air inside the motor's casing. TIlese losses vary as the cube of the speed of rotation of the machine. TIle mechanical and core losses ofa machine are often lumped together and called the no-load rotationnlloss of the machine. At no load, all the input power must be used to overcome these losses. Therefore, measuring the input power to the stator of an ac machine acting as a motor at no load will give an approximate value for these losses. STRAY LOSSES (OR MISCELLANEOUS LOSSES). Stray losses are losses that cannot be placed in one ofthe previous categories. No matter how carefully losses are accounted for, some always escape inclusion in one of the above categories. All such losses are lumped into stray losses. For most machines, stray losses are taken by convention to be I percent of full load. The Power-Flow Diagram One of the most convenient techniques for accounting for power losses in a ma- chine is the power-flow diagram. A power-flow diagram for an ac generator is shown in Figure 4-21 a. In this figure, mechanical power is input into the machine, and then the stray losses, mechanical losses, and core loses are subtracted. After they have been subtracted, the remaining power is ideally converted from me- chanical to electrical fonn at the point labeled P<X>f!¥' TIle mechanical power that is converted is given by (4-66) and the same amount of electrical power is produced. However, this is not the power that appears at the machine's terminals. Before the tenninals are reached, the electrical12 R losses must be subtracted. In the case of ac motors, this power-flow diagram is simply reversed. The power-flow diagram for a motor is shown in Figure 4- 21b. Example problems involving the calculation of ac motor and generator effi- ciencies will be given in the next three chapters. 4.8 VOLTAGE REGULATION AND SPEED REGULATION Generators are often compared to each other using a figure of merit called voltage regulation. Voltage regulation (VR) is a measure of the ability of a generator to keep a constant voltage at its terminals as load varies. It is defmed by the equation
- 287. Stray losses losses Pi. '" 3V.JA cos 6 '" ..fjVdL cos 6 I PR losses FIGURE 4- 21 C~ losses (a) Core ACMACHINERYFUNDAMENTALS 263 PR losses P00' ",3V.1.ot cos 60r ..fjVdL cos 6 losses losses ,b, (a) The power-flow diagram of a three-phase ac generator. (b) The power-flow diagram of a three- phase ac motor. IVR = v..l Vfl Vfl X 100% I (4-67) where V.t is the no-load tenninal voltage of the generator and Vfl is the full-load tenninal voltage of the generator. It is a rough measure of the shape of the gener- ator's voltage-current characteristic-a positive voltage regulation means a drooping characteristic, and a negative voltage regulation means a rising charac- teristic. A small VR is "better" in the sense that the voltage at the tenninals of the generator is more constant with variations in load. Similarly, motors are often compared to each other by using a figure of merit called speed regulation. Speed regulation (SR) is a measure of the ability of a motor to keep a constant shaft speed as load varies. It is defined by the equation 100% 1 (4-68) x (4-69)
- 288. 264 ELECTRIC MACHINERY RJNDAMENTALS It is a rough measure of the shape of a motor's torque-speed characteristic- , positive speed regulation means that a motor's speed drops with increasing load, and a negative speed regulation means a motor's speed increases with increasing load. TIle magnitude of the speed regulation tells approximately how steep the slope of the torque-speed curve is. 4.9 SUMMARY There are two major types of ac machines: synchronous machines and induction machines. The principal difference between the two types is that synchronous ma- chines require a dc field current to be supplied to their rotors, while induction ma- chines have the field current induced in their rotors by transfonner action. TIley will be explored in detail in the next three chapters. A three-phase system of currents supplied to a system of three coiIs spaced 120 electrical degrees apart on a stator wi ll produce a unifonn rotating magnetic field within the stator. The direction of rotation of the magnetic field can be re- versed by simply swapping the connections to any two of the three phases. Con- versely, a rotating magnetic field wi II produce a three-phase set of voltages within such a set of coils. In stators of more than two poles, one complete mechanical rotation of the magnetic fields produces more than one complete electrical cycle. For such a sta- tor, one mechanical rotation produces PI2 electrical cycles. Therefore, the electri- cal angle of the voltages and currents in such a machine is related to the mechan- ical angle of the magnetic fields by P (J~ = "2(Jm TIle relationship between the electrical frequency of the stator and the mechanical rate of rotation of the magnetic fields is ".p f~ = 120 TIle types of losses that occur in ac machines are electrical or copper losses (PR losses), core losses, mechanical losses, and stray losses. E.1.ch of these losses was described in this chapter, along with the definition of overall machine effi- ciency. Finally, voltage regulation was defined for generators as 1 VR = Vol V Il V Il x 1 00% 1 and speed regulation was defined for motors as ISR = nal nn nil x 100% 1
- 289. ACMACHINERYFUNDAMENTALS 265 QUESTIONS 4-1. What is the principal difference between a synchronous machine and an induction machine? 4-2. Why does switching the current flows in any two phases reverse the direction of ro- tation of a stator's magnetic field? 4-3, What is the relationship between electrical frequency and magnetic field speed for an ac machine? 4-4. What is the equation for the induced torque in an ac machine? PROBLEMS 4-1. The simple loop rotating in a lUlifonn magnetic field shown in Figure 4--1 has the following characteristics: B =0.5Ttotheright r = O.lm l = 0.5 m w = 103 radls (a) Calculate the voltage elOl(f) induced in this rotating loop. (b) Suppose that a 5-0 resistor is COIlllected as a load across the terminals of the loop. Calculate the current that would flow through the resistor. (c) Calculate the magnitude and direction of the induced torque on the loop for the conditions in b. (d) Calculate the electric power being generated by the loop for the conditions in b. (e) Calculate the mechanical power being consumed by the loop for the conditions in b. How does this nwnber compare to the amount ofelectric power being gen- erated by the loop? 4-2. Develop a table showing the speed of magnetic field rotation in ac machines of 2, 4, 6, 8, 10, 12, and 14 poles operating at frequencies of 50, 60, and 400 Hz. 4-3. A three-phase, four-pole winding is installed in 12 slots on a stator. There are 40 turns of wire in each slot of the windings. All coils in each phase are cOIlllected in series, and the three phases are connected in.6.. The flux per pole in the machine is 0.060 Wh, and the speed of rotation of the magnetic field is 1800 rhnin. (a) What is the frequency of the voltage produced in this winding? (b) What are the resulting phase and tenninal voltages of this stator? 4-4. A three-phase, Y-COIlllected, 50-Hz, two-pole synchronous machine has a stator with 2()(x) turns of wire per phase. What rotor flux would be required to produce a tenni- nal (line-to-line) voltage of 6 kV? 4-5. Modify the MATLAB problem in Example 4--1 by swapping the currents flowing in any two phases. What happens to the resulting net magnetic field? 4-6. If an ac machine has the rotor and stator magnetic fields shown in Figure P4--1, what is the direction of the induced torque in the machine? Is the machine acting as a mo- tor or generator? 4-7. The flux density distribution over the surface of a two-pole stator of radius rand length l is given by B = BMCOs(W.,f-a) Prove that the total flux lUlder each pole face is (4--37b)
- 290. 266 ELECTRIC MACHINERY RJNDAMENTALS H , B•• 0 , 0 , 0 , , , , y , , w B, o ""GURE 1)~- 1 The ac machine of Problem 4--6. 4--8. In the early days of ac motor development. machine designers had great difficulty controlling the core losses (hysteresis and eddy currents) in machines. They had not yet developed steels with low hysteresis. and were not making laminations as thin as the ones used today. To help control these losses. early ac motors in the United States were run from a 25-Hz ac power supply. while lighting systems were run from a separale 60-Hz ac power supply. (a) Develop a table showing the speed of magnetic field rotation in ac machines of 2.4.6.8. 10. 12. and 14 poles operating at 25 Hz. What was the fastest rota- tional speed available to these early motors? (b) For a given motor operating at a constant flux density B, how would the core losses of the motor flmning at 25 Hz compare to the core losses of the motor nmning at 60 Hz? (c) Why did the early engineers provide a separate 60-Hz power system for lighting? REFERENCES I. Del Toro. Vincent. Electric Machines and Po....er Systetru. Englewood Cliffs. N.J.: Prentice-Halt. 1985. 2. Fitzgerald. A. E.. and Charles Kingsley. Electric Machinery. New York: McGraw-Hill. 1952. 3. Fitzgerald. A. E.. Charles Kingsley. and S. D. Umans. Electric Machinery. 5th ed.. New York: McGraw-Hill. 1990. 4. International Electrotechnical Commission. Rotating Electrical Machines Part I: Rating and Perfortnllnce. IEC 34-1 (RI994). 1994. 5. Liwschitz-Garik. Michael. and Clyde Whipple. Altunating-Current Machinery. Princeton. N.J.: Van Nostrand. 1961. 6. McPherson. George. An Introduction to Electrical Machines and Transformers. New Yort: Wiley. 1981. 7. National Electrical Manufacturers Association. Motors and Gl'nerators, Publication MG1-1993. Washington. D.C.. 1993. 8. Werninck. E. H. (ed.). Electric Motor Handbook. London: McGraw-Hill. 1978.
- 291. CHAPTER 5 SYNCHRONOUS GENERATORS Synchronous generators or alternntors are synchronous machines used to con- vert mechanical power to ac electric power. This chapter explores the opera- tion of synchronous generators, both when operating alone and when operating to- gether with other generators. 5.1 SYNCHRONOUS GENERATOR CONSTRUCTION In a synchronous generator, a de current is applied to the rotor winding, which produces a rotor magnetic field. The rotor of the generator is then turned by a prime mover, producing a rotating magnetic field within the machine. This rotat- ing magnetic field induces a three-phase set of voltages within the stator windings of the generator. Two terms commonly used to describe the windings on a machine arefield windings and armature windings. In general, the tenn "field windings" applies to the windings that produce the main magnetic field in a machine, and the term "armature windings" applies to the windings where the main voltage is induced. For synchronous machines, the field windings are on the rotor, so the tenns "rotor windings" and "field windings" are used interchangeably. Similarly, the terms "stator windings" and "annature windings" are used interchangeably. The rotor of a synchronous generator is essentially a large electromagnet. The magnetic poles on the rotor can be of either salient or nonsalient construction. The tenn salient means "protruding" or "sticking out," and a salient pole is a mag- netic pole that sticks out from the surface of the rotor. On the other hand, a 267
- 292. 268 ELECTRIC MACHINERY RJNDAMENTALS o 0, s End View Side View fo'IGURE 5-1 A nonsalient two-pole rotor for a synchronous machine. nonsalient pole is a magnetic pole constructed flush with the surface of the rotor. A nonsalient-pole rotor is shown in Figure 5-1 , while a salient-pole rotor is shown in Figure 5- 2. Nonsalient-pole rotors are nonnally used for two- and four-pole ro- tors, while salient-pole rotors are nonnally used for rotors with four or more poles. Because the rotor is subjected to changing magnetic fields, it is constructed ofthin laminations to reduce eddy current losses. A de current must be supplied to the field circuit on the rotor. Since the ro- tor is rotating, a special arrangement is required to get the de power to its field windings. There are two common approaches to supplying this dc power: I. Supply the dc power from an external dc source to the rotor by means of slip rings and brushes. 2. Supply the dc power from a special de power source mounted directly on the shaft of the synchronous generator. Slip rings are metal rings completely encircling the shaft of a machine but in- sulated from it. One end of the dc rotor winding is tied to each ofthe two slip rings on the shaft of the synchronous machine. and a stationary brush rides on each sli p ring. A "brush" is a block of graphitelike carbon compound that conducts electric- ity freely but has very low friction. so that it doesn't wear down the slip ring. If the positive end of a dc voltage source is connected to one brush and the negative end is connected to the other, then the same dc voltage wi II be applied to the field wind- ing at all times regardless of the angular position or speed of the rotor. Slip rings and brushes create a few problems when they are used to supply dc power to the field windings of a synchronous machine. TIley increase the amount of maintenance required on the machine, since the brushes must be checked for wear regularly. In addition, brush voltage drop can be the cause of significant power losses on machines with larger field currents. Despite these problems, slip rings and brushes are used on all smaller synchronous machines, because no other method of supplying the dc field current is cost-effective. On larger generators and motors, brnshless exciters are used to supply the dc field current to the machine. A brushless exciter is a small ac generator with its
- 293. (a) HGURE 5-2 Slip rings (a) A salient six-pole rotor for a synchronous ntachine. (b) Photograph of a salient eight-pole synchronous ntachine rotor showing the windings on the individual rotor poles. (Courtesy of Geneml Electric Company. ) (e) Photograph of a single S3.lient pole front a rotor with the field SYNCHRONOUS GENERATORS 269 ,b, ,d, windings not yet in place. (Courtesy ofGeneml Electric Company.) (d) A single salient pole shown after the field windings are installed but before it is mounted on the rotor. (Courtesy ofWestinglwuse Electric Company.) field circuit mounted on the stator and its armature circuit mounted on the rotor shaft. The three-phase output of the exciter generator is rectified to direct current by a three-phase rectifier circuit also mounted on the shaft of the generator, and is then fed into the main dc field circuit. By controlling the small dc field current of the exciter generator (located on the stator), it is possible to adjust the field current on the main machine without slip rings and brushes. This arrangement is shown schematically in Figure 5-3, and a synchronous machine rotor with a brushless exciter mounted on the same shaft is shown in Figure 5-4. Since no mechanical contacts ever occur between the rotor and the stator, a brushless exciter requires much less maintenance than slip rings and brushes.
- 294. 270 ELECTRIC MACHINERY RJNDAMENTALS Exciter Exciter armature :L Three-phase rectifier r , , , , I" ---,- Synchronous machine Main Field -----------------~-------------+---------------- N~_h Three-phase input (low current) ""GURE 5-3 fu , citer ,ld fi ,------, Three-phase output Maln annature A brushless exciter circuit. A small thrre-phase current is rectified and used to supply the field circuit of the exciter. which is located on the stator. The output of the armature cirwit of the exciter (on the rotor) is then rectified and used to supply the field current of the main machine. ""GURE 5-4 Photograph of a synchronous machine rotor with a brushless exciter mounted on the same shaft. Notice the rectifying electronics visible next to the armature of the exciter. (Courtesy of Westinghouse Electric Company.)
- 295. Pilot exciter Pilot exciter field Permanent magnets I I I SYNCHRONOUS GENERATORS 271 Exciter , I Synchronous : generator , , Exciter armature ----I- Main field , , , , Th~ , ph~ , , rectifier , , , , --r- , --~-----------~----------~----------r-------------------- I Three-phase" Pllot eXCl1er annature FIGURE 5-5 Threo· ph~ rectifier , rO~"='~P"~'~~==+ 'I~ -4 R, , , -t- ExcIter field Main armature A brushless excitation scheme that includes a pilot exciter. The permanent magnets of the pilot exciter produce the field current of the exciter. which in turn produces the field current of the main machine. To make the excitation of a generator completely independent of any exter- nal power sources, a small pilot exciter is often included in the system. Apilot ex- citer is a small ac generator with permanent magnets mounted on the rotor shaft and a three-phase winding on the stator. It produces the power for the field circuit of the exciter, which in turn controls the field circuit of the main machine. If a pilot exciter is included on the generator shaft, then no external electric power is required to run the generator (see Figure 5-5). Many synchronous generators that include brushless exciters also have slip rings and brushes, so that an auxiliary source of dc field current is available in emergencies. The stator of a synchronous generator has already been described in Chap- ter 4, and more details of stator construction are found in Appendix B. Synchro- nous generator stators are nonnally made of prefonned stator coils in a double- layer winding. The winding itself is distributed and chorded in order to reduce the hannonic content of the output voltages and currents, as described in Appendix B. A cutaway diagram of a complete large synchronous machine is shown in Figure 5-6. This drawing shows an eight-pole salient-pole rotor, a stator with dis- tributed double-layer windings, and a brushless exciter.
- 296. 272 ELECTRIC MACHINERY RJNDAMENTALS ""GURE 5-6 A cutaway diagram of a large synchronous machine. Note the saliem·pole construction and the on· shaft exciter. (Courtesy ofGl'neral Electric Company.) 5.2 THE SPEED OF ROTATION OF A SYNCHRONOUS GENERATOR Synchronous generators are by defmition synchronous, meaning that the electrical frequency produced is locked in or synchronized with the mechanical rate of rotation of the generator. A synchronous generator's rotor consists of an electro- magnet to which direct current is supplied. TIle rotor's magnetic field points in whatever direction the rotor is turned. Now, the rate of rotation of the magnetic fields in the machine is related to the stator electrical frequency by Equation (4-34): (4- 34) where !. = electrical frequency, in Hz nm = mechanical speed of magnetic field, in r/min (equals speed of rotor for synchronous machines) P = number of poles Since the rotor turns at the same speed as the magnetic field, this equation relates the speed ofrotor rotation to the resulting electrical frequency. Electric power is generated at 50 or 60 Hz, so the generator must turn at a fixed speed depending on the number of poles on the machine. For example, to generate 60-Hz power in a two-pole machine, the rotor must turn at 3600 r/min. To generate 50-Hz power in a four-pole machine, the rotor must turn at 1500 rImin. TIle required rate of rota- tion for a given frequency can always be calculated from Equation (4- 34).
- 297. ; ,,' FIGURE 5-7 SYNCHRONOUS GENERATORS 273 '" '" "")'DC (constant) ..-- ,b , (a) Plot of flux versus field current for a synchronous generator. (b) The magnetization curve for the synchronous generator. 5.3 THE INTERNAL GENERATED VOLTAGE OFASYNCHRONOUSGENERATOR In Chapler 4, the magnitude of the voltage induced in a given stator phase was found to be (4- 50) This voltage depends on the flux ~ in the machine, the frequency or speed of ro- lation, and the machine's construction. In solving problems with synchronous ma- chines, this equation is sometimes rewritten in a simpler fonn that emphasizes the quantities that are variable during machine operalion. This simpler form is I EA ~ K<J>w I (5-1 ) where K is a constant representing the construction of the machine. If w is ex- pressed in electrical radians per second, then Nc K ~ V2 (5- 2) while if w is expressed in mechanical radians per second, then Nc P K ~ V2 (5- 3) The internal generated voltage EA is directly proportional to the flux and to the speed, but the flux itself depends on the current fl owing in the rotor field cir- cuit. The field circuit IF is related to the flux ~ in the manner shown in Fig- ure 5- 7a. Since EA is directly proportional to the flux, the internal generated volt- age EA is related to the field current as shown in Figure 5- 7b. lllis plot is called the magnetization cUIYe or the open-circuit characteristic of the machine.
- 298. 274 ELECTRIC MACHINERY RJNDAMENTALS 5.4 THE EQUIVALENT CIRCUIT OF A SYNCHRONOUS GENERATOR The voltage EA is the internal generated voltage produced in one phase ofa syn- chronous generator. However, this voltage EAis not usually the voltage that ap- pears at the terminals of the generator. In fact, the only time the internal voltage EAis the same as the output voltage Vo/> of a phase is when there is no annature current fl owing in the machine. Why is the output voltage Vo/> from a phase not equal to EA , and what is the relationship between the two voltages? TIle answer to these questions yields the model of a synchronous generator. TIlere are a number of factors that cause the difference between EA and Vo/>: I, The distortion of the air-gap magnetic field by the current flowing in the sta- tor, called annature reaction. 2, The self-inductance of the annature coils. ), The resistance of the armature coils. 4, The effect of salient-pole rotor shapes. We will explore the effects of the first three factors and derive a machine model from them. In this chapter, the effects of a salient-pole shape on the operation of a synchronous machine will be ignored; in other words, all the machines in this chapter are assumed to have nonsalient or cylindrical rotors. Making this assump- tion will cause the calculated answers to be slightly inaccurate if a machine does indeed have salient-pole rotors, but the errors are relatively minor. A discussion of the effects of rotor pole saliency is inc1 uded in Appendix C. TIle first effect mentioned, and nonnally the largest one, is armature reac- tion. When a synchronous generator's rotor is spun, a voltage EA is induced in the generator's stator windings. If a load is attached to the terminals of the generator, a current flows. But a three-phase stator current flow will produce a magnetic field of its own in the machine. This stator magnetic field distorts the original ro- tor magnetic field, changing the resulting phase voltage. This effect is called armature reaction because the annature (stator) current affects the magnetic field which produced it in the first place. To understand annature reaction, refer to Figure 5--8. Figure 5--8a shows a two-pole rotor spinning inside a three-phase stator. There is no load connected to the stator. The rotor magnetic field DR produces an internal generated voltage EA whose peak value coincides with the direction of DR. As was shown in the last chapter, the voltage will be positive out of the conductors at the top and negative into the conductors at the bottom of the figure. With no load on the generator, there is no annature current flow, and EAwill be equal to the phase voltage Vo/>. Now suppose that the generator is connected to a lagging load. Because the load is lagging, the peak current wi ll occur at an angle behind the peak voltage. TIlis effect is shown in Figure 5--8b. TIle current flowing in the stator windings produces a magnetic field of its own. This stator magnetic field is called Ds and its direction is given by the right-
- 299. o E ' ",IOIX I , , B, ,,' E ' ",IDiX I , B, , , o o ,, 0 , , , , , FIGURE 5-8 , , D, • 'e' ". '.• '0 E,., • '. SYNCHRONOUS GENERATORS 275 o o , D, w , , , , , ,b, , , , , , v, , I I... JII>X , ' o .' I 'AJIIH , , ' --:::::]'0 "" , , D, , , , , ,d, , o ~ E"" • The development of a model for armature reaction: (a) A rotating magnetic field produces the internal generated voltage EA' (b) The resulting voltage produces a lagging currentflow when connected to a lagging load. (e) The stator current produces its own magnetic field BS' which produces its own voltage E_ in the stator windings of the machine. (d) The field Us adds to "I/" distorting it into H.... The voltage E... adds to EA. producing v .at the output of the phase. hand rule to be as shown in Figure 5--8c. The stator magnetic field Bs produces a voltage of its own in the stator, and this voltage is called E...., on the figure. With two voltages present in the stator windings, the total voltage in a phase is just the sum of the internal generated voltage EA. and the annature reaction volt- age E"a,: (5-4) The net magnetic field 8 ..., is just the sum of the rotor and stator magnetic fields: (5- 5)
- 300. 276 ELECTRIC MACHINERY RJNDAMENTALS v, FlGURES-9 A simple cirwit (see text). Since the angles of EAand BR are the same and the angles of E"a. and Bs, are the same, the resulting magnetic field Boe. will coincide with the net voltage Vo/>. The resulting voltages and currents are shown in Figure 5--8d. How can the effects of armature reaction on the phase voltage be modeled? First, note that the voltage E"a. lies at an angle of 90° behind the plane of maxi- mum current IA . Second, the voltage E."" is directly proportional to the current IA . If X is a constant of proportionality, then the armature reaction voltage can be ex- pressed as E"., = - jXIA (5-6) TIle voltage on a phase is thus ' " I Vc-.-_--o Eo- A -_--c j X ccI" , 1 (5- 7) Look at the circuit shown in Figure 5- 9. The Kirchhoff's voltage law equa- tion for this circuit is (5- 8) TIlis is exactly the same equation as the one describing the annature reaction volt- age. Therefore, the annature reaction voltage can be modeled as an inductor in series with the internal generated voltage. In addition to the effects of armature reaction, the stator coils have a self- inductance and a resistance. If the stator self-inductance is called LA(and its cor- responding reactance is called XA) while the stator resistance is called RA , then the total difference betwccn EAand Vo/> is given by (5- 9) TIle annature reaction effects and the self-inductance in the machine are both rep- resented by reactances, and it is customary to combi ne them into a single reac- tance, called the synchronous reactance of the machine: XS= X + XA Therefore, the final equation describing Vo/> is I V4> - EA- jXS IA - RAIA I (5- 10) (5- 11 )
- 301. SYNCHRONOUS GENERATORS 277 I" + jXs R, + EA] ""' 'f] I, + I" R., + R, jXs R, v, + EA2 ""' 'f2 (&) L, FIGURE 5-10 The full equivalent circuit of a three-phase synchronous generator. II is now possible 1 0 sketch the equivalent circuit of a three-phase synchro- nous generator. The full equivalent circuit of such a generator is shown in Fig- ure 5- 10. This figure shows a dc power source supplying the rotor field circuit, which is modeled by the coil 's inductance and resistance in series. In series with RF is an adjustable resistor R adj which controls the flow of field current. The rest of the equivalent circuit consists of the models for each phase. E:1.ch phase has an internal generated voltage with a series inductance Xs (consisting of the sum of the armature reactance and the coil 's self-inductance) and a series resistance RA. TIle voltages and currents of the three phases are 120° apart in angle, but other- wise the three phases are identical. TIlese three phases can be either Y- or Ii-connected as shown in Figure 5- 11. If they are Y-connected, then the tenninal voltage VT is related to the phase voltage by (5- 12)
- 302. 278 ELECTRIC MACHINERY RJNDAMENTALS E.n + v, v, + jXs (a) + ~______C=~____--Q + jXs v, jXs 'bJ ""GURE 5- 11 The generator equivalent circuit connected in (a) Yand (b) 8. If they are a-connected, then (5- 13) TIle fact that the three phases of a synchronous generator are identical in all respects except for phase angle nonnally leads to the use of a per-phase equiva- lent circuit. The per-phase equivalent circuit of this machine is shown in Fig-
- 303. SYNCHRONOUS GENERATORS 279 v, FIGURE 5-12 The per-phase equivalent circuit of a synchronous generator. The internal field circuit resistance and the external variable resistance have been contbined into a single resistor Rr . FIGURE 5-13 The phasor diagrant of a synchronous generator at unity power factor. ure 5- 12. One important fact must be kept in mind when the per-phase equivalent circuit is used: The three phases have the same voltages and currents only when the loads attached to them are balanced. If the generator's loads are not balanced, more sophisticated techniques of analysis are required. 1l1ese techniques are be- yond the scope of this book. 5.5 THE PHASOR DIAGRAM OF A SYNCHRONOUS GENERATOR Because the voltages in a synchronous generator are ac voltages, they are usually expressed as phasors. Since phasors have both a magnitude and an angle, the re- lationship between them must be expressed by a two-dimensional plot. When the voltages within a phase (E,t, V4n jXSIA, and RAIA) and the current IAin the phase are plotted in such a fashion as to show the relationships among them, the result- ing plot is called a phasor diagram. For example, Figure 5- 13 shows these relationships when the generator is supplying a load at unity power factor (a purely resistive load). From Equation (5- 11 ), the total voltage E,t differs from the tenninal voltage of the phase V4> by the resistive and inductive voltage drops. All voltages and currents are referenced to V4n which is arbitrarily assumed to be at an angle of 0°. This phasor diagram can be compared to the phasor diagrams of generators operating at lagging and leading power factors.1l1ese phasor diagrams are shown
- 304. 280 ELECTRIC MACHINERY RJNDAMENTALS V, jXSIA lARA ,,' E, jXSIA lARA V, ,b, ""GURE 5-14 The phasor diagram of a synchronous generator at (3) lagging and (b) leading power factor. in Figure 5- 14. Notice that, for a given phase voltage and armnture current, a larger internal generated voltage EA is needed for lagging loads than for leading loads. Therefore, a larger field current is needed with lagging loads to get the same tenninal voltage, because (5- 1) and w must be constant to keep a constant frequency. Alternatively, for a given field current and magnitude ofload current, the terminal voltage is lowerfor lagging loads and higherfor leading loads. In real synchronous machines, the synchronous reactance is nonnally much larger than the winding resistance RA, so RA is often neglected in the qualitative study of voltage variations. For accurate numerical results, R A must of course be considered. 5.6 POWERAND TORQUE IN SYNCHRONOUS GENERATORS A synchronous generator is a synchronous machine used as a generator. It con- verts mechanical power to three-phase electrical power. The source of mechanical power, the prime mover, may be a diesel engine, a stearn turbine, a water turbine, or any similar device. Whatever the source, it must have the basic property that its speed is almost constant regardless of the power demand. If that were not so, then the resulting power system's frequency would wander. Not all the mechanical power going into a synchronous generator becomes electrical power out ofthe machine.llle difTerence between input power and output power represents the losses of the machine. A power-flow diagram for a synchro-
- 305. SYNCHRONOUS GENERATORS 281 p~= foppw.. Stray losses FIGURE 5-15 windage losses , , find w.. I , losses , , , (copper losses) The power-flow diagram of a synchronous generntor. nous generator is shown in Figure 5- 15. The input mechanical power is the shaft power in the generator fln= "Tappwm , while the power converted from mechanical to electrical fonn internally is given by = 3E,./1t cos "y (5- 14) (5- 15) where 'Y is the angle between Elt and lit- TIle difference between the input power to the generator and the power converted in the generator represents the mechan- ical, core, and stray losses of the machine. TIle real electrical output power of the synchronous generator can be ex- pressed in line quantities as and in phase quantities as ?"UI = 3'4,IA cos (J The reactive power output can be expressed in line quantities as Q,UI = ~VTIL sin (J or in phase quantities as (5- 16) (5- 17) (5- 18) (5- 19) If the annature resistance RIt is ignored (since Xs» RIt), then a very useful equation can be derived to approximate the output power of the generator. To de- rive this equation, examine the phasor diagram in Figure 5- 16. Figure 5- 16 shows a simplified phasor diagram of a generator with the stator resistance ignored. No- tice that the vertical segment be can be expressed as either Elt sin /j or Xs lit cos (J. Therefore, EAsin /j lA cos (J = X,
- 306. 282 ELECTRIC MACHINERY RJNDAMENTALS , o r ""GURE 5-16 """, " , , , , ............1:;' , , , , , , ", , jXs l,t I , , , V , • ___ L.L " b , Simplified phasor diagram with armature resistance ignored. and substituting this expression into Equation (5- 17) gives E,t sin .s =Xs l,t cos(} (5- 20) Since the resistances are assumed to be zero in Equation (5- 20), there are no elec- trical losses in this generator, and this equation is both PCOII ¥ and Pout. Equation (5- 20) shows that the power produced by a synchronous genera- tor depends on the angle 8 between Vq,and EA. The angle 8 is known as the torque angle of the machine. Notice also that the maximum power that the generator can supply occurs when 8 = 900. At 8 = 90°, sin 8 = I , and (5- 21) TIle maximum power indicated by this equation is called the static stability limit of the generator. Nonnally, real generators never even come close to that limit. Full-load torque angles of 15 to 20° are more typical of real machines. Now take another look at Equations (5- 17), (5- 19), and (5- 20). IfVq, is as- sumed constant, then the real power output is directly prop011ionni to the quanti- ties J,t cos () and E,t sin 8, and the reactive power output is directly proportional to the quantity J,t sin (). These facts are useful in plotting phasor diagrams of syn- chronous generators as loads change. From Chapter 4, the induced torque in this generator can be expressed as (4- 58) or as (4-60)
- 307. SYNCHRONOUS GENERATORS 283 The magnitude of Equation (4--60) can be expressed as Tind = kB,/J"", sin /j (4-6 1) where /j is the angle between the rotor and net magnetic fields (the so-called torque angle). Since BRproduces the voltage E... and BOel produces the voltage Vo/>. the angle /j between E... and V0/> is the same as the angle /j between BR and B_. An alternative expression for the induced torque in a synchronous generator can be derived from Equation (5- 20). Because PC<JO¥ = TiDdWm • the induced torque can be expressed as (5- 22) This expression describes the induced torque in terms of electrical quantities, whereas Equation (4--60) gives the same infonnation in terms of magnetic quantities. 5.7 MEASURING SYNCHRONOUS GENERATOR MODEL PARAMETERS The equivalent circuit of a synchronous generator that has been derived contains three quantities that must be detennined in order to completely describe the be- havior of a real synchronous generator: I. The relationship between field current and nux (and therefore between the field current and E...) 2. 1lle synchronous reactance 3. 1lle annature resistance This section describes a simple technique for determining these quantities in a synchronous generator. The first step in the process is to perfonn the open-circuit test on the gener- ator. To perform this test, the generator is turned at the rated speed, the terminals are disconnected from all loads, and the field current is set to zero. 1llen the field current is gradually increased in steps, and the tenninal voltage is measured at each step along the way. With the tenninals open, I... = 0, so E... is equal to ~. It is thus possible to construct a plot of E... or Vr versus IFfrom this information. This plot is the so-called open-circuit characteristic (OCC) ofa generator. With this character- istic, it is possible to fmd the internal generated voltage of the generator for any given field currenl. A typical open-circuit characteristic is shown in Figure 5-1 7a. Notice that at first the curve is almost perfectly linear, until some saturation is ob- served at high field currents. The unsaturated iron in the frame of the synchronous machine has a reluctance several thousand times lower than the air-gap reluctance, so at first almost all the magnetomotive force is across the air gap, and the result- ing nux increase is linear. When the iron finally saturates, the reluctance of the iron
- 308. 284 ELECTRIC MACHINERY RJNDAMENTALS Air-gap lioe , , , , , , , , , (a) (b, Open-cin:uit characteristic (OCC) Short-cin:uit characteristic (SCC) nGURES- 17 (a) The open-cin:uit characteristic (OCC) of 3. synchronous generator. (b) The short-cin:uit characteristic (SCC) of a synchronous generator. increases dramatically, and the flux increases much more slowly with an increase in magnetomotive force.1lle linear portion of an ace is called the air-gap line of the characteristic. 1lle second step in the process is to conduct the shon-circuit test. To per- form the short-circuit test, adjust the field current to zero again and short-circuit the tenninals of the generator through a set of ammeters. Then the armature cur- rent lit or the line current IL is measured as the field current is increased. Such a plot is called a short-circuit characteristic (SeC) and is shown in Figure 5- 17b. It is essentially a straight line. To understand why this characteristic is a straight line, look at the equivalent circuit in Figure 5- 12 when the terminals of the machine are short-circuited. Such a circuit is shown in Figure 5- 8a. Notice that when the tenninals are short-circuited, the annature current lit is given by EA IA= RA + jXs (5- 23) and its magnitude is just given by
- 309. SYNCHRONOUS GENERATORS 285 (b, ( ., ------------ Bsta! B..., (" FIGURE 5-18 (a) The equivalent circuit of a synchronous generator during the short-circuit test. (b) The resulting phasor diagram. (c) The magnetic fields during the short-circuit test. r~E~ A""" 1 -, A - VRi + xl (5- 24) The resulting phasor diagram is shown in Figure 5-1 8b, and the corresponding magnetic fields are shown in Figure 5-1 8c. Since Bsalmost cancels BR, the net magnetic field BDet is very small (corresponding to internal resistive and inductive drops only). Since the net magnetic field in the machine is so small, the machine is unsaturated and the sec is linear. To understand what infonnation these two characteristics yield, notice that, with Vo/> equal to zero in Figure 5-1 8, the internnl mnchine impedance is given by EA Zs = VRA 2 + X2 = - (5- 25) , IA Since Xs» RIl , this equation reduces to (5- 26) If Ell and III are known for a given situation, then the synchronous reactance Xs can be found. Therefore, an approximate method for detennining the synchronous reac- tance Xs at a given field current is I. Get the internal generated voltage Ell from the ace at that field current. 2. Get the short-circuit current now l,o.,sc at that field current from the Sec. 3. Find Xs by applying Equation (5- 26).
- 310. 286 ELECTRIC MACHINERY RJNDAMENTALS Air-gap line ___- - -ace sec x, o o ""GURE 5- 19 A sketch of the approximate synchronous reacl3.nce of a synchronous generator as a function of the field current in the machine. The constant value of reactance found at low values of field current is the uns(J/umted synchronous reactance of the machine. TIlere is a problem with this approach, however. The internal generated voltage Ell comes from the acc, where the machine is partially saturated for large field currents, while III is taken from the sec, where the machine is unsatu- rated at all field currents. TIlerefore, at higher field currents, the Ell taken from the aec at a given field current is not the same as the Ell at the srune field current un- der short-circuit conditions, and this difference makes the resulting value of Xs only approximate. However, the answer given by this approach is accurate up to the point of saturation, so the unsaturated synchronous reactance Xs.~ of the machine can be found simply by applying Equation (5- 26) at any field current in the linear por- tion (on the air-gap line) of the acc curve. TIle approximate value of synchronous reactance varies with the degree of saturation of the ace, so the val ue of the synchronous reactance to be used in a given problem should be one calculated at the approximate load on the machine. A plot of approximate synchronous reactance as a function of field current is shown in Figure 5- 19. To get a more accurate estimation of the saturated synchronous reactance, refer to Section 5- 3 of Reference 2. If it is important to know a winding's resistance as well as its synchronous reactance, the resistance can be approximated by applying a dc voltage to the windings while the machine is stationary and measuring the resulting current flow. TIle use of dc voltage means that the reactance of the windings will be zero during the measurement process.
- 311. SYNCHRONOUS GENERATORS 287 This technique is not perfectly accurate, since the ac resistance will be slightly larger than the dc resistance (as a result of the skin effect at higher fre- quencies). The measured value of the resistance can even be plugged into Equa- tion (5- 26) to improve the estimate of Xs, if desired. (Such an improvement is not much help in the approximate approach- saturation causes a much larger error in the Xs calculation than ignoring Rio. does.) The Short-Circuit Ratio Another parameter used to describe synchronous generators is the short-circuit ra- tio.1lle short-circuit ratio of a generator is defined as the ratio of thefield current requiredfor the rated voltage at open circuit to the field current required for the rated armature current at short circuit. It can be shown that this quantity is just the reciprocal of the per-unit value of the approximate saturated synchronous re- actance calculated by Equation (5- 26). Although the short-circuit ratio adds no new information about the genera- tor that is not already known from the saturated synchronous reactance, it is im- portant to know what it is, since the tenn is occasionally encountered in industry. Example 5-1. A 2oo-kVA, 480-y' 50-Hz, V-connected synchronous generator with a rated field current of 5 A was tested, and the following data were taken: 1. VT,OC at the rated h was measured to be 540 V. 2. h,se at the rated If was found to be 300 A. 3. When a dc voltage of 10 V was applied to two of the tenninals, a current of 25 A was measured. Find the values of the armature resistance and the approximate synchronous reactance in ohms that would be used in the generator model at the rated conditions. Solutioll The generator described above is V-connected, so the direct current in the resistance test flows through two windings. Therefore, the resistance is given by V 2R -= 10. - loe Voe 10 V Rio. = 2/0e = (2)(25 A) = 0.2 n The internal generated voltage at the rated field current is equal to V, EIo. = V</I.oc = v"J =5~V =3 ll.8V The short-circuit current 110. is just equal to the line current, since the generator is Y- cOIUlected: Ilt,se = 14se = 300 A
- 312. 288 ELECTRIC MACHINERY RJNDAMENTALS I, R, I, + 0.2!1 R, jl.02 !1 + V, EA=312L&Q V, L, ""GURE 5-10 The per-phase equivalent ci["(;uit of the generator in Example 5- 1. Therefore, the synchronous reactance at the rated field current can be calculated from Equation (5- 25): )(0.20)2 + Xi = 3jrio 8 A V )(0.2 0)2 + Xs = 1.039 0 0.04 + xi = 1.08 Xi = 1.04 Xs= 1.020 (5- 25) How much effect did the inclusion of R" have on the estimate of Xs? Not much. If Xs is evaluated by Equation (5- 26), the result is X - E" _ 311.8V - 1.040 s - fA - 300 A - Since the error in Xs due to ignoring R" is much less than the error due to saturation effects, approximate calculations are normally done with Equation (5- 26). The resulting per-phase equivalent circuit is shown in Figure 5- 20. 5.S THE SYNCHRONOUS GENERATOR OPERATING ALONE The behavior of a synchronous generator under load varies greatly depending on the power factor of the load and on whether the generator is operating alone or in parallel with other synchronous generators. In this section, we will study the be- havior of synchronous generators operating alone. We will study the behavior of synchronous generators operating in parallel in Section 5.9. TIuoughout this section, concepts will be illustrated with simplified phasor diagrams ignoring the effect of RA- In some of the numerical examples the resis- tance RAwill be included. Unless otherwise stated in this section, the speed of the generators will be assumed constant, and all terminal characteristics are drawn assuming constant
- 313. SYNCHRONOUS GENERATORS 289 Lo,' FIGURE 5-21 A single generator supplying a load. speed. Also, the rotor nux in the generators is assumed constant unless their field current is explicitly changed. The Effect of Load Changes on a Synchronous Generator Operating Alone To understand the operating characteristics of a synchronous generator operating alone, examine a generator supplying a load. A diagram of a single generator sup- plying a load is shown in Figure 5- 21. What happens when we increase the load on this generator? An increase in the load is an increase in the real and/or reactive power drawn from the generator. Such a load increase increases the load current drawn from the generator. Because the field resistor has not been changed, the field cur- rent is constant, and therefore the flux cp is constant. Since the prime mover also keeps a constant speed w, the magnitude ofthe internal generated voltage Ell = Kcpw is constant. If Ell is constant, just what does vary with a changing load? The way to find out is to construct phasor diagrams showing an increase in the load, keeping the constraints on the generator in mind. First, examine a generator operati ng at a lagging power factor. If more load is added at the same powerfactor, then 11111increases but remains at the same an- gie () with respect to V0/> as before. Therefore, the annature reaction voltage jXs IIl is larger than before but at the same angle. Now since Ell = Vo/> + jXsIIl j Xs III must stretch between Vo/> at an angle of 0° and Ell, which is constrained to be of the same magnitude as before the load increase. If these constraints are plotted on a phasor diagram, there is one and only one point at which the annature reac- tion voltage can be parallel to its original position while increasing in size. The re- sulting plot is shown in Figure 5- 22a. If the constraints are observed, then it is seen that as the load increases, the voltage V0/> decreases rather sharply. Now suppose the generator is loaded with unity-power-factor loads. What happens if new loads are added at the same power factor? With the same con- straints as before, it can be seen that this time Vo/> decreases only slightly (see Fig- ure 5- 22b).
- 314. 290 ELECTRIC MACHINERY RJNDAMENTALS E'A E, jXslA jXSIA , " 0 V' // V , .~ . I, I', ~ ~ ~ ~ , ~ , ~ , ~ , ~ , ~ ',,/'y~ v ,,, (a) " , ~ ~ ~ ~ E'A , ~ ~ I', ~ ~ I, " , ,,' ""GURE 5-11 " 111. I'll. jXSIA E, V. V; 'h' jXs IA V' V • • The elIect ofan increase in generator loads at constant power factor upon its terminal voltage. (a) Lagging power factor; (b) unity power factor; (c) teading power factor. Finally, let the generator be loaded with leading-power-factor loads. If new loads are added at the same power factor this time, the annature reaction voltage lies outside its previous value, and Vo/> actually rises (see Figure 5- 22c). In this last case, an increase in the load in the generator produced an increase in the tenninal voltage. Such a result is not something one would expect on the basis of intuition alone. General conclusions from this discussion of synchronous generator behav- ior are I. If lagging loads (+ Q or inductive reactive power loads) are added to a gen- erator, Vo/> and the terminal voltage Vrdecrease significantly. 2. If unity-power-factor loads (no reactive power) are added to a generator, there is a slight decrease in V0/> and the tenninal voltage. 3. If leading loads (--Q or capacitive reactive power loads) are added to a gener- ator, V0/> and the tenninal voltage will rise. A convenient way to compare the voltage behavior of two generators is by their voltage regulation. The voltage regu lation (VR) of a generator is defined by the equation
- 315. SYNCHRONOUS GENERATORS 291 Vn1 - Va I VR = Va x 100% (4-67) where V ol is the no-load voltage of the generator and Vfl is the full-load voltage of the generator. A synchronous generator operating at a lagging power factor has a fairly large positive voltage regulation, a synchronous generator operating at a unity power factor has a small positive voltage regulation, and a synchronous gen- erator operating at a leading power factor often has a negative voltage regulation. Normally, it is desirable to keep the voltage supplied to a load constant, even though the load itself varies. How can tenninal voltage variations be corrected for? The obvious approach is to vary the magnitude of E), to compensate for changes in the load. Recall that E), = Kcpw. Since the frequency should not be changed in a nonnal system, E), must be controlled by varying the flux in the machine. For example, suppose that a lagging load is added to a generator. Then the terminal voltage will fall, as was previously shown. To restore it to its previous level, decrease the field resistor RF" If RF decreases, the field current wil I increase. An increase in IF increases the flux, which in turn increases E)" and an increase in E), increases the phase and terminal voltage. nlis idea can be summarized as follows: I. Decreasing the field resistance in the generator increases its field current. 2. An increase in the field current increases the flux in the machine. 3. An increase in the flux increases the internal generated voltage E), = Kcpw. 4. An increase in E), increases Vo/> and the terminal voltage of the generator. The process can be reversed to decrease the tenninal voltage. It is possible to regulate the tenninal voltage of a generator throughout a series of load changes simply by adjusting the field current. Example Problems The following three problems illustrate simple calculations involving voltages, currents, and power flows in synchronous generators. The first problem is an ex- ample that includes the armature resistance in its calculations, while the next two ignore R),. Part of the first example problem addresses the question: How must a generator sfield current be adjusted to keep VT constant as the load changes? On the other hand, part of the second example problem asks the question: lfthe load changes and the field is left alone, what happens to the terminnl voltage? You should compare the calculated behavior of the generators in these two problems to see if it agrees with the qualitative arguments of this section. Finally, the third example illustrates the use of a MATLAB program to derive the terminal charac- teristics of synchronous generator. Example 5-2. A 480-V, 6()"Hz, ~ -co lUlected, four-pole SynChroflOUSgeflerator has the OCC shown in Figure 5--23a. This geflerator has a synchronous reactaflce of 0.1 n afld
- 316. 292 ELECTRIC MACHINERY RJNDAMENTALS > • ~ • " ~ .. .§ § ., , •• y 0 ~ 600 500 400 300 200 100 I I o 0.0 V / I / / 1.0 2.0 3.0 I,t '" 692.8 L - 36.87° A ""GURE 5-23 / / ' / / 4.0 5.0 6.0 Field current. A v • ,., ,b, 7.0 8.0 9.0 10.0 (a) Open-drwit characteristic of the generator in Example 5- 2. (b) Phasor diagram of the generator in Example 5- 2. an annature resistance of 0.015 n. At fullload, the machine supplies 1200 A at 0.8 PF lag- ging. Under full-load conditions. the friction and windage losses are 40 kW. and the core losses are 30 kW. Ignore any field circuit losses.
- 317. SYNCHRONOUS GENERATORS 293 (a) What is the speed of rotation of this generator? (b) How much field current must be supplied to the generator to make the terminal voltage 480 V at no load? (c) If the generator is now cOlUlected to a load and the load draws 1200 A at 0.8 PF lagging, how much field current will be required to keep the terminal voltage equal to 480 V? (d) How much power is the generator now supplying? How much power is supplied to the generator by the prime mover? What is this machine's overall efficiency? (e) If the generator's load were suddenly disconnected from the line, what would happen to its terminal voltage? (jJ Finally, suppose that the generator is cOIUlected to a load drawing 1200 A at 0.8 PF leading. How much field current would be required to keep Vrat 480 V? Solutioll This synchronous generator is .d.-connected, so its phase voltage is equal to its line voltage V. = Vr, while its phase current is related to its line current by the equation IL = ~/• . (a) The relationship between the electrical frequency produced by a synchronous generator and the mechanical rate of shaft rotation is given by Equation (4--34): Therefore, ".p fe = 120 12!X60 Hz) _ 4 poles 1800 r/ min (4--34) (b) In this machine, Vr = V• . Since the generator is at no load, IA = 0 and EA = V•. Therefore, Vr = V. = EA = 480 V, and from the open-circuit characteristic, I" = 4.5 A. (c) If the generator is supplying 1200 A. then the armature current in the machine is 1..1 = 12~A = 692.8 A The phasor diagram for this generator is shown in Figure 5- 23b. If the terminal voltage is adjusted to be 480 V, the size of the internal generated voltage EA is given by EA = V. + RAIA + jXsI, = 480 LO° V + (0.015 n X692.8 L -36.87° A ) + (j0.1 0)(692.8 L -36.87° A ) = 480 LO° V + 10.39 L -36.87° V + 69.28 L53.13° V = 529.9 + j49.2 V = 532 L5.JO V To keep the tenninal voltage at 480 V, E, must be adjusted to 532 V. From Fig- ure 5- 23, the required field current is 5.7 A. (d) The power that the generator is now supplying can be found from Equation (5-16): (5--1 6)
- 318. 294 ELECTRIC MACHINERY RJNDAMENTALS = VJ(480 VXI200 A) cos 36.87° = 798 kW To detennine the power input to the generator, use the power-flow diagram (Fig- ure 5-15). From the power-flow diagram, the mechanical input power is given by The stray losses were not specified here, so they will be ignored. In this genera- tor, the electrical losses are P olo< 10.. = 311RA = 3(692.8 A)2(0.015 f.!) = 21.6 kW The core losses are 30 kW, and the friction and windage losses are 40 kW, so the total input power to the generator is P in = 798kW + 21.6kW + 30kW + 40kW = 889.6kW Therefore, the machine's overall efficiency is Pout 798 kW 7f = p x 100% = 8896 kW x 100% = 89.75% rn . (e) If the generator's load were suddenly disconnected from the line, the current IA would drop to zero, making EA = V•. Since the field current has not changed, lEAl has not changed and V. and Vr must rise to equal EA' Therefore, if the load were suddenly dropped, the terminal voltage of the generator would rise to 532 V. (f) If the generator were loaded down with 1200 A at 0.8 PF leading while the ter- minal voltage was 480 V, then the internal generated voltage would have to be EA = V. + RAIA + jXs I, = 480LO° V + (0.015 n)(692.8L36.87° A) + (j0.1 nX692.8L36.87° A) = 480 LO° V + 10.39 L36.87° V + 69.28 L 126.87° V = 446.7 + j61.7 V = 451 L7.10 V Therefore, the internal generated voltage EA must be adjusted to provide 451 V if Vr is to remain 480 V. Using the open-circuit characteristic, the field current would have to be adjusted to 4.1 A. Which type of load (leading or lag ging) needed a larger field current to maintain the rated voltage? Which type of load (leading or lagging) placed more thermal stress on the generator? Why? Example 5-3. A 480-V, 50-Hz, Y-connected, six-pole synchronous genera- tor has a per-phase synchronous reactance of 1.0 n. Its full-load armature current is 60 A at 0.8 PF lagging. This generator has friction and windage losses of 1.5 kW and core losses of 1.0 kW at 60 Hz at full load. Since the armature resistance is being ig- nored, assrune that the j 2R losses are negligible. The field current has been adjusted so that the terminal voltage is 480 V at no load. (a) What is the speed of rotation of this generator? (b) What is the terminal voltage of this generator if the following are true?
- 319. SYNCHRONOUS GENERATORS 295 I. It is loaded with the rated current at 0.8 PF lagging. 2, It is loaded with the rated current at 1.0 PF. 3, It is loaded with the rated current at 0.8 PF leading. (c) What is the efficiency ofthis generator (ignoring the lUlknown electrical losses) when it is operating at the rated current and 0.8 PF lagging? (d) How much shaft torque must be applied by the prime mover at full load? How large is the induced cOlUltertorque? (e) What is the voltage regulation of this generator at 0.8 PF lagging? At 1.0 PF? At 0.8 PF leading? Solution This generator is V-connected, so its phase voltage is given by V. = Vr/ v'J. That means that when Vr is adjusted to 480 V, V. = 277 V. The field current has been adjusted so that Vr... = 480 V, so V. = 277 V. At no load, the armature current is zero, so the armature re- action voltage and the I},R}, drops are zero. Since I}, = 0, the internal generated voltage E}, = V. = 277 V. The internal generated voltage E},( = Kq,w) varies only when the field current changes. Since the problem states that the field current is adjusted initially and then left alone, the magnitude of the internal generated voltage is E}, = 277 V and will not change in this example. (a) The speed of rotation of a synchronous generator in revolutions per minute is given by Equation (4-34): Therefore, 1201. " = - - m p _ limP Ie - 120 = 120(50 H z) _ 1000 rlmin 6 poles Alternatively, the speed expressed in radians per second is Wm = (1000r/min)e6~~n)e~~ad) = 104.7 radl s (4-34) (b) I. If the generator is loaded down with rated current at 0.8 PF lagging, the re- sulting phasor diagram looks like the one shown in Figure 5- 24a. In this phasor diagram, we know that V. is at an angle of 0°, that the magnitude of E}, is 277 V, and that the quantity jXsI}, is jXsI}, = j(1.0 nX60 L - 36.87° A) = 60 L53.13° V The two quantities not known on the voltage diagram are the magnitude of V. and the angle 0 of E},. To find these values, the easiest approach is to con- struct a right triangle on the phasor diagram, as shown in the figure. From Figure 5- 24a, the right triangle gives E1 = (V. + Xsl}, sin 9)2 + (Xsl}, cos 9)2 Therefore, the phase voltage at the rated load and 0.8 PF lagging is
- 320. 296 ELECTRIC MACHINERY RJNDA MENTALS 60 L 53.13° v, ,b , v, ", ""GURE 5-14 Generator phasor diagrams for Example 5- 3. (a) Lagging power factor; (b) unity power factor; (c) leading power factor. (277 vi = [V. + (1 .0 0)(60 A) sin 36.87°]2 + [( 1.0 n X60 A) cos 36.87°]2 76,729 = (V. + 36)1 + 2304 74,425 = (V. + 36)2 272.8 = V,., + 36 V,., = 236.8 V Since the generator is Y-colUlected, Vr = V3V. = 410 V. 2. If the generator is loaded with the rated current at unity power factor, then the phasor diagram wi11look like Figure 5- 24h.To find V . here the right tri- angle is
- 321. SYNCHRONOUS GENERATORS 297 £1 = vi + (XsIA)2 (277V)2 = vi + [(1.00)(60A)]2 76,729 = Vi + 3600 Vi = 73,129 V. = 270.4 V Therefore, Vr = V3"V. = 46S.4 V. 3. When the generator is loaded with the rated current at O.S PF leading, the re- suiting phasor diagram is the one shown in Figure 5- 24c. To find V. in this situation, we construct the triangle OAB shown in the figure. The resulting equation is £1 = (V. - XsIA)2 + (XsI./t cos (/)2 Therefore, the phase voltage at the rated load and O.S PF leading is (277 V)2 = [V. - (l.OnX60A) sin 36.S7°f + [(1.0 0)(60 A) cos 36.S7o]2 76,729 = (V. - 36)2 + 2304 74,425 = (V. - 36)2 272.S = V. - 36 V. = 30S.S V Since the generator is Y-cOIUlected, Vr = V3"V. = 535 V. (c) The output power of this generator at 60 A and O.S PF lagging is POU,=3V.IAcos () = 3(236.S VX60AXO.S) = 34.1 kW The mechanical input power is given by = 34.1 kW + a + 1.0kW + 1.5kW = 36.6kW The efficiency of the generator is thus Pout 34.1 kW 7f = P, x 100% = 366 kW x 100% = 93.2% rn . (d) The input torque to this generator is given by the equation T = Pin = - W. 36.6 kW 125.7 radls = 291.2 N · m The induced cOlUltertorque is given by P.:oov = Tind = Wv 34.1 kW 125.7 radls = 271.3 N · m (e) The voltage regulation of a generator is defined as
- 322. 298 ELECTRIC MACHINERY RJNDA MENTALS v - " VR = nl () X 100% V, (4--67) By this defmition, the voltage regulation for the lagging, lUlity, and leading power-factor cases are 1. Lagging case: VR = 480 ~I~ ~1O V x 100% = 17.1% 2. Unity case: VR = 480 ~6; ~68 V x 100% = 2.6% 3. Leading case: VR = 480 ~3; ~35 V x 100% = -10.3% In Example 5- 3, lagging loads resulted in a drop in terminal voltage, unity- power-factor loads caused little effect on VT, and leading loads resulted in an in- crease in tenninal voltage. Example 5-4. Assume that the generator of Example 5- 3 is operating at no load with a tenninal voltage of 480 V. Plot the tenninal characteristic (terminal voltage versus line current) of this generator as its armature ClUTent varies from no-load to full load at a power factor of (a) 0.8 lagging and (b) 0.8 leading. Assume that the field current remains constant at all times. Solutioll The terminal characteristic of a generator is a plot of its tenninal voltage versus line cur- rent. Since this generator is Y-cOIlllected. its phase voltage is given by V. = VTIV'3 . If Vr is adjusted to 480 V at no-load conditions. then V. = fA = 277 V. Because the field ClUTent remains constant, fA will remain 277 V at all times.The output current h from this gener- ator will be the same as its armature current IA because it is V-connected. (a) If the generator is loaded with a 0.8 PF lagging clUTent. the resulting phasor di- agram looks like the one shown in Figure 5- 24a. In this phasor diagram. we know that V. is at an angle of 0°. that the magnitude of EAis 277 V. and that the quantity jXSIA stretches between V. and EA as shown. The two quantities not known on the phasor diagram are the magnitude of V. and the angle 0 of EA. To find V.. the easiest approach is to construct a right triangle on the phasor dia- gram. as shown in the figure. From Figure 5--24a. the right triangle gives ~ = (V. + XSIAsin (J)2 + (XSIAcos (J)2 This equation can be used to solve for V., as a flUlction of the current IA: V. = JE1 (XiAcos 0)2 - XiA sin 0 A simple MATLAB M-filecan be used to calculate V .(and hence VT) as a func- tion of current. Such an M-file is shown below: ~ M-file : t e rm_cha r _a .m ~ M-file t o p l ot the t e rmina l cha r act e ri s ti cs o f the ~ gene r a t or o f Examp l e 5-4 with a n 0.8 PF l agging l oad . ~ Firs t , initia lize the c urrent a mp litudes (2 1 va lues ~ in the r a nge 0- 60 A) i _a = (0, 1: 20) .. 3;
- 323. SYNCHRONOUS GENERATORS 299 % Now initia liz e a ll other va lues v-ph ase = zeros( 1 ,2 1 ) ; e_a = 277.0; x_s = 1. 0; theta = 36 .S7 .. (p i / 1 SO ) ; % Converted t o radians % Now ca l c ulate v-ph ase f or each c urrent l eve l f or ii = 1: 2 1 e od (x_s .. i _a( ii ) .. cos( the ta ))" 2 ) (x_s .. i _a( ii ) .. s in (the ta )) ; % Ca l c ulate terminal vo lt age from the phase vo lt age v_t = v-ph ase .. sqrt (3) ; % Plot the terminal ch aracteri s ti c, remembering the % the line c urre nt i s the same as i _a p l ot ( i _a, v_t , 'Col or' , 'k' , 'Linewi dth ' ,2.0) ; x l abe l ( 'Line CUrre nt (A) ' , 'Fontwei ght ' , 'Bo l d ' ) ; y l abe l ( 'Te rmina l Volt age (V) ' , 'Fontwei ght' , 'Bo l d ' ) ; title ( 'Te rmina l Ch a r acter i s tic f o r O.S PF l agg ing l oad ' , ... 'Fontwei ght' , 'Bo l d ' ) ; gr i d on ; ax i s( [ O 60 400 550 ] ) ; The plot resulting when this M-file is executed is shown in Figure 5- 25a. (b) If the generator is loaded with a 0.8 PF leading current, the resulting phasor di- agram looks like the one shown in Figure 5- 24c. To fmd V.' the easiest ap- proach is to construct a right triangle on the phasor diagram, as shown in the fig- ure. From Figure 5- 24c, the right triangle gives E1 = (V. - XsfA sin 9)2 + (XsfAcos 9)2 This equation can be used to solve for V", as a ftmction of the current fA: V. = JE l (XsfA cos (J)l + XsfA sin 9 This equation can be used to calculate and plot the terminal characteristic in a manner similar to that in part a above. The resulting tenninal characteristic is shown in Figure 5- 25b. 5.9 PARALLEL OPERATION OF AC GENERATORS In today's world, an isolated synchrono us generator supplying its own load inde- pendently of other generators is very rare. Such a situation is found in only a few out-of-the-way applications such as emergency generators. For all usual genera- tor applications, there is more than one generator operating in parallel to supply the power demanded by the loads.An extreme example ofthis situation is the U.S. power grid, in which literally thousands of generators share the load on the system.
- 324. 300 ELECTRIC MACHINERY RJNDAMENTALS 550 > • 500 00 • r----I---- --------I---- -------- ~ .." ., ~ 450 400 0 0 20 30 40 50 Line current. A ,,' 550 ,----,---,--,---,----,----, > ~ 500 • ~ e- o " § 450 ~ 4OO0·'----" IO c---C 20 ~---" 30 c---- 4O ~---c 50 '---~ 60 Line current. A ,b, ""GURE 5-25 (a) Thrminal characteristic for the generator of Example 5-4 when loaded with an 0.8 PF lagging loo.d. (b) Thrminal characteristic for the generator when loaded with an 0.8 PF leading load. Why are synchronous generators operated in parallel? There are several ma- jor advantages to such operation: I. Several generators can supply a bigger load than one machine by itself. 2. Having many generators increases the reliability of the power system, since the failure of anyone of them does not cause a total power loss to the load. 3. Having many generators operating in parallel allows one or more of them to be removed for shutdown and preventive maintenance.
- 325. SYNCHRONOUS GENERATORS 301 ' Generator I Lood / s, . - Generator 2 HGURE 5-26 A generator being paralleled with a running power system. 4. If only one generator is used and it is not operating at near full load, then it will be relatively inefficient. With several smaller machines in parallel, it is possible to operate only a fraction of them. The ones that do operate are op- erating near full load and thus more efficiently. This section explores the requirements for paralleling ac generators, and then looks at the behavior of synchronous generators operated in parallel. The Conditions Required for Paralleling Figure 5- 26 shows a synchronous generator Gt supplying power to a load, with another generator Gl about to be paralleled with Gt by closing the switch St. What conditions must be met before the switch can be closed and the two generators connected? If the switch is closed arbitrarily at some moment, the generators are liable to be severely damaged, and the load may lose power. If the voltages are not ex- actly the same in each conductor being tied together, there will be a very large cur- rent flow when the switch is closed. To avoid this problem, each of the three phases must have exactly the same voltage magnitude and phase angle as the con- ductor to which it is connected. In other words, the voltage in phase a must be ex- actly the same as the voltage in phase a' , and so forth for phases b-b' and c-c'. To achieve this match, the following paralleling conditions must be met: I. 1lle rms line voltages of the two generators must be equal. 2. 1lle two generators must have the same phase sequence. 3. 1lle phase angles of the two a phases must be equal. 4. 1lle frequency of the new generator, called the oncoming generator, must be slightly higher than the frequency of the running system. These paralleling conditions require some explanation. Condition I is obvi- o",- in order for two sets of voltages to be identical, they must ofcourse have the same rms magnitude of voltage. The voltage in phases a and a' will be completely
- 326. 302 ELECTRIC MACHINERY RJNDAMENTALS v, • v, • v, v, abc phase sequence acb phase sequence ,,' Generator I Lo"" Generator 2 Switch S[ ,b, ""GURE 5-27 (a) The two possible phase sequences of a three-phase system. (b) The three-light-bulb method for checking phase sequence. identical at all times if both their magnitudes and their angles are the same, which explains condition 3. Condition 2 ensures that the sequence in which the phase voltages peak in the two generators is the same. Ifthe phase sequence is different (as shown in Fig- ure 5- 27a), then even though one pair of voltages (the a phases) are in phase, the other two pairs of voltages are 1200 out of phase. If the generators were connected in this manner, there would be no problem with phase a, but huge currents would fl ow in phases band c, damaging both machines. To correct a phase sequence problem, simply swap the connections on any two of the three phases on one of the machines. If the frequencies of the generators are not very nearly equal when they are connected together, large power transients will occur until the generators stabilize at a common frequency. The frequencies of the two machines must be very nearly equal, but they cannot be exactly equal. 1lley must differ by a small amount so
- 327. SYNCHRONOUS GENERATORS 303 that the phase angles of the oncoming machine will change slowly with respect to the phase angles of the running system. In that way, the angles between the volt- ages can be observed and switch SI can be closed when the systems are exactly in phase. The General Procedure for Paralleling Generators Suppose that generator Gl is to be connected to the running system shown in Fig- ure 5- 27. TIle following steps should be taken to accomplish the paralleling. First, using voltmeters, the field current ofthe oncoming generator should be adjusted until its tenninal voltage is equal to the line voltage ofthe running system. Second, the phase sequence of the oncoming generator must be compared to the phase sequence of the running system. TIle phase sequence can be checked in a number of different ways. One way is to alternately connect a small induction motor to the terminals of each of the two generators. If the motor rotates in the same direction each time, then the phase sequence is the same for both generators. If the motor rotates in opposite directions, then the phase sequences differ, and two of the conductors on the incoming generator must be reversed. Another way to check the phase sequence is the three-light-bulb method. In this approach, three light bulbs are stretched across the open terminals of the switch connecting the generator to the system as shown in Figure 5- 27b. As the phase changes between the two systems, the light bulbs first get bright (large phase difference) and then get dim (small phase difference). Ifall three bulbs get bright and dark together, then the systems have the same phase sequence. If the bulbs brighten in succession, then the systems have the opposite phase sequence, and one of the sequences must be reversed. Next, the frequency of the oncoming generator is adjusted to be slightly higher than the frequency of the running system. TIlis is done first by watching a frequency meter until the frequencies are close and then by observing changes in phase between the systems. TIle oncoming generator is adjusted to a slightly higher frequency so that when it is connected, it will come on the line supplying power as a generator, instead of consuming it as a motor would (this point will be explained later). Once the frequencies are very nearly equal, the voltages in the two systems will change phase with respect to each other very slowly. TIle phase changes are observed, and when the phase angles are equal, the switch connecting the two sys- tems together is shut. How can one tell when the two systems are finally in phase? A simple way is to watch the three light bulbs described above in connection with the discussion of phase sequence. When the three light bulbs all go out, the voltage difference across them is zero and the systems are in phase. This simple scheme works, but it is not very accurate. A better approach is to employ a synchroscope. A synchro- scope is a meter that measures the difference in phase angle between the a phases of the two systems. The face of a synchroscope is shown in Figure 5- 28. TIle dial shows the phase difference between the two a phases, with 0 (meaning in phase)
- 328. 304 ELECTRIC MACHINERY RJNDAMENTALS ..._ S C ,_ "_ ,h ,;,'C '"C"" _ J FIGURE 5-28 A synchrosrope. at the top and 1800 at the bottom. Since the frequencies of the two systems are slightly different, the phase angle on the meter changes slowly. If the oncoming generator or system is faster than the running system (the desired situation), then the phase angle advances and the synchroscope needle rotates clockwise. If the oncoming machine is slower, the needle rotates counterclockwise. When the syn- chroscope needle is in the vertical position, the voltages are in phase, and the switch can be shut to connect the systems. Notice, though, that a synchroscope checks the relationships on only one phase. It gives no infonnation about phase sequence. In large generators belonging to power systems, this whole process of par- alleling a new generator to the line is automated, and a computer does this job. For smaller generators, though, the operator manually goes through the paralleling steps just described. Frequency-Power and Voltage-Reactive Power Characteristics of a Synchronolls Generator All generators are driven by a prime mover, which is the generator's source of mechanical power. TIle most common type of prime mover is a steam turbine, but other types include diesel engines, gas turbines, water turbines, and even wind turbines. Regardless of the original power source, all prime movers tend to behave in a similar fashion--.:1.s the power drawn from them increases, the speed at which they turn decreases. The decrease in speed is in general nonlinear, but some form of governor mechanism is usually included to make the decrease in speed linear with an increase in power demand. Whatever governor mechanism is present on a prime mover, it will always be adjusted to provide a slight drooping characteristic with increasing load. The speed droop (SD) ofa prime mover is defined by the equation I SO = nnl nn nil x 100% I (5- 27) where n o] is the no-load prime-mover speed and no is the full-load prime-mover speed. Most generator prime movers have a speed droop of 2 to 4 percent, as de- fined in Equation (5- 27). In addition, most governors have some type of set point
- 329. .5 , I " J o ", o ,b , SYNCHRONOUS GENERATORS 305 Power. kW Power. kW HGURE 5-29 (a) The speed-versus-power curve for a typical prime mover. (b) The resulting frequency-versus-power curve for the generator. adjustment to allow the no-load speed of the turbine to be varied. A typical speed- versus-power plot is shown in Figure 5- 29. Since the shaft speed is related to the resulting electrical frequency by Equa- tion (4- 34), (4- 34) the power output of a synchronous generator is related to its frequency. An exam- ple plot of frequency versus power is shown in Figure 5- 29b. Frequency-power characteristics of this sort play an essential role in the parallel operation of syn- chronous generators. The relationship between frequency and power can be described quantita- tively by the equation where P = power output of the generator Jot = no-load frequency of the generator !.y. = operating frequency of system sp = slope of curve, in kW/Hz or MW/Hz (5- 28) A similar relationship can be derived for the reactive power Q and terminal voltage VT. As previously seen, when a lagging load is added to a synchronous
- 330. 306 ELECTRIC MACHINERY RJNDAMENTALS VTo] Q, 0 kYAR consumed ""GURE 5-30 Qn Q (reactive power). kYAR supplied The curve of terminal voltage (Vr) versus reactive power (Q) for a synchronous generator. generator, its tenninal voltage drops. Likewise, when a leading load is added to a synchronous generator, its tenninal voltage increases. It is possible to make a plot oftenninal voltage versus reactive power. and such a plot has a drooping charac- teristic like the one shown in Figure 5-30. This characteristic is not intrinsically linear, but many generator voltage regulators include a feature to make it so. The characteristic curve can be moved up and down by changing the no-load tenninal voltage set point on the voltage regulator. As with the frequency-power character- istic, this curve plays an important role in the parallel operation of synchronous generators. TIle relationship between the terminal voltage and reactive power can be expressed by an equation similar to the frequency-power relationship [Equation (5-28)] if the voltage regulator produces an output that is linear with changes in reactive power. It is important to realize that when a single generator is operating alone, the real power P and reactive power Q supplied by the generator will be the amount demanded by the load attached to the generator- the P and Q supplied cannot be controlled by the generator's controls. Therefore, for any given real power, the governor set points control the generator's operating frequencyIe and for any given reactive power, the field current controls the generator's tenninal voltage VT. Example 5-5. Figure 5-31 shows a generator supplying a load. A second load is to be connected in parallel with the first one. The generator has a no-load frequency of 61.0 Hz and a slope sp of I MWlHz. Load I consumes a real power of I()(x) kW at 0.8 PF lag- ging. while load 2 consrunes a real power of 800 kW at 0.707 PF lagging. (a) Before the switch is closed. what is the operating frequency of the system? (b) After load 2 is cOIUlected. what is the operating frequency of the system? (c) After load 2 is cOIUlected. what action could an operator take to restore the sys- tem frequency to 60 Hz?
- 331. SYNCHRONOUS GENERATORS 307 y "- Lo'" 1 /' Turbine generator I I Lo'" 2 FIGURE 5-31 The power system in Example 5- 5. Solutioll This problem states that the slope of the generator's characteristic is I MW/Hz and that its no-load frequency is 61 Hz. Therefore, the power produced by the generator is given by P = sl--JnJ - J.y. ) p f.y• = Jo l - sp (a) The initial system frequency is given by = 61 Hz - lOOOkW I MW/Hz = 61 Hz - I Hz = 60 Hz (b) After load 2 is connected, = 61 Hz - 1800 kW I MW/Hz = 61 Hz - 1.8 Hz = 59.2 Hz (5--28) (c) After the load is connected, the system frequency falls to 59.2 Hz. To restore the system to its proper operating frequency, the operator should increase the gov- ernor no-load set points by 0.8 Hz, to 61.8 Hz. This action will restore the sys- tem frequency to 60 Hz. To summarize, whe n a generator is operating by itself supplying the system loads, then I . 1lle real and reactive power supplied by the generator will be the amount de- manded by the attached load. 2. 1lle governor sct points of the generator will control the operating frequency of the power syste m.
- 332. 308 ELECTRIC MACHINERY RJNDAMENTALS /, -p o Consumed ,,' ""GURE 5-32 p. kW supplied v, - Q o Consumed Q. kVAR supplied ,b, Curves for an infinite bus: (a) frequency versus power and (b) tenninal voltage versus reactive power. 3. The field current (or the field regulator set points) control the terminal volt- age of the power system. nlis is the situation found in isolated generators in remote field environments. Operation of Generators in Parallel with Large Power Systems When a synchronous generator is connected to a power system, the power system is often so large that nothing the operator of the generator does will have much of an effect on the power system. An example of this situation is the connection of a single generator to the U.S. power grid. 1lle U.S. power grid is so large that no reasonable action on the part of the one generator can cause an observable change in overall grid frequency. nlis idea is idealized in the concept of an infinite bus. An infinite bus is a power system so large that its voltage and frequency do not vary regardless of how much real and reactive power is drawn from or supplied to it. The power- frequency characteristic of such a system is shown in Figure 5- 32a, and the reac- tive power-voltage characteristic is shown in Figure 5- 32b. To understand the behavior of a generator connected to such a large system, examine a system consisting of a generator and an infinite bus in parallel supply- ing a load. Assume that the generator's prime mover has a governor mechanism, but that the field is controlled manually by a resistor. lt is easier to explain gener- ator operation without considering an automatic field current regulator, so this dis- cussion will ignore the slight differences caused by the field regulator when one is present. Such a system is shown in Figure 5- 33a. When a generator is connected in parallel with another generator or a large system, the frequency and terminnl voltage ofall the mnchines must be the same,
- 333. SYNCHRONOUS GENERATORS 309 Infinite bus ;- Generator )::::::::::::::::::::~U ,,' f. Pjmoo,' kW PiJJ. bw Pc· kW ,b, FIGURE 5-33 (a) A synchronous generator operating in parallet with an infinite bus. (b) The frequency-versus- power diagram (or lwuse diagmm) for a synchronous generator in parallel with an infinite bus. since their output conductors are tied together. Therefore, their real power- frequency and reacti ve power- voltage characteristics can be plotted back to back, with a common vertical axis. Such a sketch, sometimes infonnally called a house diagram, is shown in Figure 5- 33b. Assume that the generator has just been paralleled with the infinite bus ac- cording to the procedure described previously. Then the generator will be essen- tially "floating" on the line, supplying a small amount of real power and little or no reactive power. nlis situation is shown in Figure 5- 34. Suppose the generator had been paralleled to the line but, instead of being at a slightly higher frequency than the running system, it was at a slightly lower fre- quency. In this case, when paralleling is completed, the resulting situation is shown in Figure 5- 35. Notice that here the no-load frequency of the generator is less than the system's operating frequency. At this frequency, the power supplied by the gen- erator is actually negative. In other words, when the generator's no-load frequency is less than the system's operating frequency, the generator actually consumes elec- tric power and runs as a motor. It is to ensure that a generator comes on line sup- plying power instead ofconsuming it that the oncoming machine's frequency is ad- justed higher than the running system's frequency. Many real generators have a
- 334. 310 ELECTRIC MACHINERY RJNDAMENTALS !.. Hz P. k:W P. k:W ""GURE 5-34 The frequency-versus-power diagram at the moment just after paralleling. !.. Hz P. k:W Pc<O P. k:W (consuming) ""GURE 5-35 The frequency-versus-power diagram if the no-load frequency of the generator were slightly less than system frequency before paralleling. reverse-power trip connected to them, so it is imperative that they be paralleled with their frequency higher than that of the running system. Ifsuch a generator ever starts to consume power. it will be automatically disconnectedfrom the line. Once the generator has been connected, what happens when its governor set points are increased? The effect of this increase is to shift the no-load frequency of the generator upward. Since the frequency of the system is unchanged (the fre- quency of an infmite bus cannot change), the power supplied by the generator in- creases. lllis is shown by the house diagram in Figure 5- 36a and by the phasor di- agram in Figure 5- 36b. Notice in the phasor diagram that E). sin /) (which is proportional to the power supplied as long as VT is constant) has increased, while the magnitude of E). (= K$w) remains constant, since both the field current IF and the speed of rotation w are unchanged. As the governor set points are further in- creased, the no-load frequency increases and the power supplied by the generator increases. As the power output increases, E). remains at constant magnitude while E). sin /) is further increased.
- 335. Piofbw P_ '" constant'" PB+PG , I, FIGURE 5-36 , 1' , ,,' ,b, SYNCHRONOUS GENERATORS 311 P. k:W E, -E~-----t"P~ ---} O<P G1 -- --- --- v, The effect of increasing the governor's set points on (a) the house diagram; (b) the phasor diagram What happens in this system if the power output of the generator is in- creased until it exceeds the power consumed by the load? If this occurs, the extra power generated flows back into the infinite bus. The infmite bus, by definition, can supply or consume any amount of power without a change in frequency, so the extra power is consumed. After the real power of the generator has been adjusted to the desired value, the phasor diagram of the generator looks like Figure 5-36b. Notice that at this time the generator is actually operating at a slightly leading power factor, supply- ing negative reactive power. Alternatively, the generator can be said to be con- suming reactive power. How can the generator be adjusted so that it will supply some reactive power Q to the system? This can be done by adjusting the field cur- rent of the machine. To understand why this is true, it is necessary to consider the constraints on the generator's operation under these circumstances. The first constraint on the generator is that the power must remain constant when IF is changed. The power into a generator is given by the equation Pin = 'TiDdWm , Now, the prime mover of a synchronous generator has a fixed torque- speed
- 336. 312 ELECTRIC MACHINERY RJNDAMENTALS ~ , , , , , ~~ , , , I, ~Qr-Q~_-_--i~_" ""GURE 5-37 I, , , I, The effect of increasing the generator's field current on the phasor diagram of the machine. characteristic for any given governor setting. This curve changes only when the governor set points are changed. Since the generator is tied to an infinite bus, its speed cannot change. If the generator's speed does not change and the governor set points have not been changed, the power supplied by the generator must re- main constant. If the power supplied is constant as the field current is changed, then the distances proportional to the power in the phasor diagram (110. cos () and EIo. sin 8) cannot change. When the field current is increased, the nux <P increases, and therefore EIo. (= K<piw) increases. If EIo. increases, but EIo. sin 8 must remain con- stant, then the phasor EIo. must "slide" along the line of constant power, as shown in Figure 5- 37. Since Vo/> is constant, the angle of jXsllo. changes as shown, and therefore the angle and magnitude of 110. change. Notice that as a result the distance proportional to Q (110. sin ()) increases. In other words, increasing the field current in a synchronous generator operating in parallel with an infinite bus increases the reactive power output ofthe generator. To summarize, when a generator is operating in parallel with an infinite bus: I. The frequency and tenninal voltage of the generator are controlled by the sys- tem to which it is connected. 2. The governor set points of the generator control the real power supplied by the generator to the system. 3, The field current in the generator controls the reactive power supplied by the generator to the system. lllis situation is much the way real generators operate when connected to a very large power system. Operation of Generators in Par allel with Other Generators of the Same Size When a single generator operated alone, the real and reactive powers (P and Q) supplied by the generator were fixed, constrained to be equal to the power de- manded by the load, and the frequency and tenninal voltage were varied by the
- 337. SYNCHRONOUS GENERATORS 313 governor set points and the field current. When a generator operated in parallel with an infinite bus, the frequency and tenninal voltage were constrained to be constant by the infmite bus, and the real and reactive powers were varied by the governor set points and the field current. What happens when a synchronous gen- erator is connected in parallel not with an infinite bus, but rather with another gen- erator of the same size? What will be the effect of changing governor set points and field currents? If a generator is connected in parallel with another one of the same size, the resulting system is as shown in Figure 5- 38a. In this system, the basic constraint is that the sum of the real and reactive powers supplied by the two generators must equal the P and Q demanded by the load. The system freq uency is not con- strained to be constant, and neither is the power of a given generator constrained to be constant. The power-frequency diagram for such a system immediately after Gl has been paralleled to the line is shown in Figure 5- 38b. Here, the total power P,,,, (which is equal to PJood) is given by (5- 29a) and the total reactive power is given by (5- 29b) What happens if the governor set points of Gl are increased? When the gov- ernor set points of G2 are increased, the power-frequency curve of G2 shifts up- ward, as shown in Figure 5- 38c. Remember, the total power supplied to the load must not change. At the original frequency fj, the power supplied by GJ and Gl will now be larger than the load demand, so the system cannot continue to oper- ate at the same frequency as before. In fact, there is only one freq uency at which the sum of the powers out of the two generators is equal to PJood ' That frequency fl is higher than the original system operating frequency. At that freq uency, Gl sup- plies more power than before, and GJ supplies less power than before. Therefore, when two generators are operating together, an increase in gov- ernor set points on one of them I. Increases the system frequency. 2. Increases the power supplied by that generator. while reducing the power supplied by the other one. What happens if the field current of G2 is increased? TIle resulting behavior is analogous to the real-power situation and is shown in Figure 5- 38d. When two generators are operating together and the field current of Gl is increased, I. The system terminal voltage is increased. 2. The reactive power Q supplied by that generator is increased, while the re- active power supplied by the other generator is decreased.
- 338. kW kVAR 314 t, 601h P~ ,b, Generator I t, h ------- ----j--------"'~ : II pc. , , , , , , pis) p,. P~ P~ ,,' Generator I V, ----- --r------ , V" , , , , , , Qe. QGI {b Q,. Q. ,d, P" ~ Vn Qm P~ Generntor 2 kW Generator 2 kVAR HGURE 5-38 (a) A generator COIloected in parallel with another machine of the same size. (b) The corresponding house diagram at the moment generator 2 is paralleled with the system. (e) The effect of increasing generator 2's governor set points on the operation of the system. (d) The effect of increasing generator 2's field current on the operation of the system.
- 339. Generator I Slope = I MW/Hz kW P t =1.5 MW 61.5 Hz 60 H, /= 60 Hz SYNCHRONOUS GENERATORS 315 Generator 2 P2= 1.0MW kW FIGURE 5-39 The llOuse dia.gram for the system in Example 5- :5. If the slopes and no-load frequencies of the generator's speed droop (frequency-power) curves are known, then the powers supplied by each generator and the resulting system frequency can be determined quantitatively. Example 5--6 shows how this can be done. EXllmple 5-6. Figure 5- 38a shows two generators supplying a load. Generator I has a no-load frequency of 61.5 Hz and a slope SpI of I MW/Hz. Generator 2 has a no-load frequency of 61.0 Hz and a slope sn of I MWlHz. The two generators are supplying a real load totaling 2.5 MW at 0.8 PF lagging. The resulting system power-frequency or house diagram is shown in Figure 5- 39. (a) At what frequency is this system operating, and how much power is supplied by each of the two generators? (b) Suppose an additional I-MW load were attached to this power system. What would the new system frequency be, and how much power would GI and G2 supply now? (c) With the system in the configuration described in part b, what will the system frequency and generator powers be if the governor set points on G2 are in- creased by 0.5 Hz? Solutioll The power produced by a synchronous generator with a given slope and no-load frequency is given by Equation (5- 28): P I = SPI(foJ.l - l>y.) P2 = sn(foJ2 - l>y.) Since the total power supplied by the generators must equal the power consumed by the loads, P loo.d = P I + P2 These equations can be used to answer all the questions asked.
- 340. 316 ELECTRIC MACHINERY RJNDAMENTALS (a) In the first case, both generators have a slope of I MW/Hz, and GI has a no-load frequency of 61 .5 Hz, while G2 has a no-load frequency of 61.0 Hz. The total load is 2.5 MW. Therefore, the system frequency can be found as follows: P load = P I + P l = Spt(f ol.1 - I.Y' ) + sn(f ol.l - I.Y' ) 2.5 MW = (1 MW/Hz)(61.5 Hz - I.Y ' ) + (1 MW/HzX61 Hz - f.y.) = 61.5 MW - (I MW/Hz)l.y• + 61 MW - (1 MW/Hz)f.ys = 122.5 MW - (2 MW/Hz)f.y• therefore (" = 122.5 MW - 2.5 MW = 60 0 H Jsy. (2MW/Hz) . z The resulting powers supplied by the two generators are P I = spl(fnJ.1 - f.y. ) = (1 MW/HzX61.5 Hz - 60.0 Hz) = 1.5 MW P2 = sn(fnJ.2 - f .y. ) = (1 MW/HzX61.0 Hz - 60.0 Hz) = I MW (b) When the load is increased by I MW, the total load becomes 3.5 MW. The new system frequency is now given by Pload = Spt(f ol.1 - I.Y' ) + sn(f ol.l - f.y.) 3.5 MW = (1 MW/Hz)(61.5 Hz - I.y.) + (1 MW/HzX61 Hz - f.ys) = 61.5 MW - (I MW/Hz)l.y• + 61 MW - (1 MW/Hz)f.y. = 122.5 MW - (2 MW/Hz)f.y• therefore (" = 122.5 MW - 3.5 MW = 595 H Jsy. (2MW/Hz) . z The resulting powers are P I = spl(fnJ.1 - f.y.) = (1 MW/HzX61.5 Hz - 59.5 Hz) = 2.0 MW P2 = sn(fnJ.2 - f.y.) = (1 MW/HzX61.0 Hz - 59.5 Hz) = 1.5MW (c) If the no-load governor set points of Gl are increased by 0.5 Hz, the new system frequency becomes P load = Spt(f ol.1 - I.y.) + sn(f ol.l - I.Y ') 3.5 MW = (1 MW/Hz)(61.5 Hz - f.y.) + (1 MW/HzX61.5 Hz - f.y.) = 123 MW - (2 MW/Hz)f.y• 123 MW - 3.5 MW f.y• = (2MW/Hz) = 59.75 Hz The resulting powers are P I = P l = Spt(fol.l - I.y.) = (1 MW/HzX61.5 Hz - 59.75 Hz) = 1.75 MW
- 341. SYNCHRONOUS GENERATORS 317 Notice that the system frequency rose, the power supplied by G2 rose, and the power supplied by G1 fell. When two generators of similar size are operating in parallel, a change in the governor set points of one of them changes both the system frequency and the power sharing between them. It would nonnally be desired to adjust only one of these quantities at a time. How can the power sharing of the power system be ad- justed independently of the system frequency, and vice versa? The answer is very simple. An increase in governor set points on one gen- erator increases that machine's power and increases system frequency.A decrease in governor set points on the other generator decreases that machine's power and decreases the system frequency. Therefore, to adjust power sharing without changing the system frequency, increase the governor set points ofone generator and simultaneously decrease the governor set points ofthe other generator (see Figure 5-40a). Similarly, to adjust the system frequency without changing the power sharing, simultaneously increase or decrease both governor set points (see Figure 5-40b). Reactive power and tenninal voltage adjustments work in an analogous fashion. To shift the reactive power sharing without changing Vn simultaneously increase the field current on one generator and decrease the field current on the other (see Figure 5-40c). To change the tenninal voltage without affecting the re- active power sharing, simultaneously increase or decrease bothfield currents (see Figure 5-4(kl). To summarize, in the case of two generators operating together: I. TIle system is constrained in that the total power supplied by the two genera- tors together must equal the amount consumed by the load. Neither/.y• nor VT is constrained to be constant. 2. To adjust the real power sharing between generators without changing/.y" simultaneously increase the governor set points on one generator while de- creasing the governor set points on the other. TIle machine whose governor set point was increased will assume more of the load. 3. To adjust /.Y' without changing the real power sharing, simultaneously in- crease or decrease both generators' governor set points. 4. To adjust the reactive power sharing between generators without changing VT, simultaneously increase the field current on one generator while decreas- ing the field current on the other. The machine whose field current was in- creased will assume more of the reactive load. 5. To adjust VTwithout changing the reactive power sharing, simultaneously in- crease or decrease both generators' field currents. It is very important that any synchronous generator intended to operate in par- allel with other machines have a drooping frequency-power characteristic. If two generators have flat or nearly flat characteristics, then the power sharing between
- 342. Genemtor I ,- kW P, P" , kW P, Generator I - kVAR Q, Q, Generator I kVAR Q, H GURE 5-40 /. '" Generator 2 ------ , I=constant l , , , , , , , , , -' P, P' , ,,' I. Hz ,b , v, Generator 2 , VT = con5lant I , ,,' v, ,d , , , , ,- Generator 2 Qi kW kW kVAR kVAR (a) Shifting power sharing without affecting system frequency. (b) Shifting system frequency without affecting power sharing. (c) Shifting reactive power sharing without affecting temJinal voltage. (d) Shifting terminal voltage without affecting reactive power sharing. 318
- 343. SYNCHRONOUS GENERATORS 319 t, p, ------------~==========;---------------- p, FIGURE 5-41 Two synchronous generators with flat frequency-power characteristics. A very tiny change in the no- load frequency of either of these machines could cause huge shifts in the power sharing. Ihem can vary widely with only the tiniest changes in no-load speed.1llis problem is illustraled by Figure 5-41. Notice that even very tiny changes inInJ in one of the generators would cause wild shifts in power sharing. To ensure good control of power sharing between generators, they should have speed droops in the range of 2 to 5 percent. 5.10 SYNCHRONOUS GENERATOR TRANSIENTS When the shaft torque applied to a generator or the output load on a generator changes suddenly, there is always a transient lasting for a finite period of time be- fore the generator returns to steady state . For example, when a synchronous gen- erator is paralleled with a running power system, it is initially turning faster and has a higher frequency than the power system does. Once it is paralleled, there is a transient period before the generator steadies down on the line and runs at line frequency while supplying a small amount of power to the load. To illustrate this situation, refer to Figure 5-42. Figure 5-42a shows the magnetic fields and the phasor diagram of the generator at the moment just before it is paralleled with the power system. Here, the oncoming generator is supplying no load, its stator current is zero, E... = Vq., and RR = Ro... At exactly time t = 0, the switch connecting the generator to the power sys- tem is shut, causing a stator current to fl ow. Since the generator's rotor is still turning faster than the system speed, it continues to move out ahead of the sys- tem's voltage Vo/>. The induced torque on the shaft of the generator is given by (4-60) The direction of this torque is opposite to the direction of motion, and it increases as the phase angle between DR and D...,Cor E... and Vo/» increases.nlis torque opposite
- 344. w 320 ELECTRIC MACHINERY RJNDAMENTALS ", ~jXSIA Ill. ~ ,b, ""GURE 5-42 w Bs Tind '" k BR x".., Tind is clockwise D, (a) The phasor diagram and magnetic fields of a generator at the momem of paralleling with 3. large power system. (b) The phasor diagram and house diagram shortly after a. Here. the rotor has moved on ahead of the net nt3.gnetic fields. producing a clockwise torque. This torque is slowing the rotor down to the synchronous speed of the power system. the direction ofmotion slows down the generator until it finally turns at synchronous speed with the rest of the power system. Similarly, if the generator were turning at a speed lower than synchronous speed when it was paralleled with the power system, then the rotor would fall be- hind the net magnetic fields, and an induced torque in the direction of motion would be induced on the shaft of the machine. This torque would speed up the rotor until it again began turning at synchronous speed. Transient Stability of Synchronolls Generators We learned earlier that the static stability limit of a synchronous generator is the maximum power that the generator can supply under any circumstances. The maximum power that the generator can supply is given by Equation (5- 21): _ 3'"4,EA Pmax - X, and the corresponding maximum torque is _ 3'"4,EA T=- X W , (5- 21) (5- 30) In theory, a generator should be able to supply up to this amount of power and torque before becoming unstable. In practice, however, the maximum load that can be supplied by the generator is limited to a much lower level by its dynamic stability limit. To understand the reason for this limitation, consider the generator in Figure 5-42 again. If the torque applied by the prime mover (Tapp) is suddenly increased, the shaft of the generator will begin to speed up, and the torque angle [) will increase as described. As the angle [) increases, the induced torque Tind of the generator will
- 345. SYNCHRONOUS GENERATORS 321 120 11Xl -- -- - -- -- -- -- -- -- -- , E 80 - 0 fin>'aDtaooow '- •, (j(] ! • ,40 .l " f / [V " V ~ 20 0.5 LO Time,s FIGURE 5-43 The dynamic response when an applied torque equal to 50% of Tmu is suddenly added to a synchronous generator. increase until an angle [) is reached at which T;Dd is equal and opposite to T opp' TIlis is the steady-state operating point ofthe generator with the new load. However, the ro- tor of the generator has a great deal of inertia, so its torque angle /) actually over- shoots the steady-state position, and gradually settles out in a damped oscillation, as shown in Figure 5-43. TIle exact shape ofthis damped oscillation can be detennined by solving a nonlinear differential equation, which is beyond the scope of this book. For more information, see Reference 4, p. 345. The important point about Figure 5-43 is that ifat any point in the transient response the instantaneous torque exceeds T"""" the synchronous generator will be unstable. The size of the oscillations depends on how suddenly the additional torque is applied to the synchronous generator. If it is added very gradually, the machine should be able to almost reach the static stability limit. On the other hand, if the load is added sharply, the machine will be stable only up to a much lower limit, which is very complicated to calculate. For very abrupt changes in torque or load, the dynamic stability limit may be less than half of the static sta- bility limit. Short-Circuit Transients in Synchronous Generators By far the severest transient condition that can occur in a synchronous generator is the situation where the three terminals of the generator are suddenly shorted out. Such a short on a power system is called afaull. There are several compo- nents of current present in a shorted synchronous generator, which will be de- scribed below. TIle same effects occur in less severe transients like load changes, but they are much more obvious in the extreme case of a short circuit.
- 346. 322 ELECTRIC MACHINERY RJNDAMENTALS o Time Phase a DC component -~ o Time Phase b - 0 ~ Time DC component Phase c ""GURE 5-44 The total fault currents as a function of time during a three-phase fault at the terminals of a synchronous generator. When a fault occurs on a synchronous generator, the resulting current flow in the phases of the generator can appear as shown in Figure 5-44. The current in each phase shown in Figure 5-42 can be represented as a dc transient component added on top of a symmetrical ac component. 1lle symmetrical ac component by itself is shown in Figure 5-45. Before the fault, only ac voltages and currents were present within the gen- erator, while after the fault, both ac and dc currents are present. Where did the dc currents come from? Remember that the synchronous generator is basically inductive-it is modeled by an internal generated voltage in series with the syn- chronous reactance. Also, recall that a current cannot change instantaneously in an inductor. When the fault occurs, the ac component of current jumps to a very
- 347. SubtraDsient period Subtransient period ,I Transient period Transient p<riod "n-fit:1F'f1) - • , I Extrapolation of transient envelope I FIGURE 5-45 Extrapolation of steady value The symmetric ac COntponent of the fault current. SYNCHRONOUS GENERATORS 323 Steady-state period Steady-state ",nod l Actual envelope large value, but the total current cannot change at that instant. The dc component of current is just large enough that the sum of the ac and dc components just after the fault equals the ac current flowing just before the fault. Since the instanta- neous values of current at the moment of the fault are different in each phase, the magnitude of the dc component of current wil I be different in each phase. These dc components of current decay fairly quickly, but they initially av- erage about 50 or 60 percent of the ac current flow the instant after the fault occurs. The total initial current is therefore typicalJy 1.5 or 1.6 times the ac com- ponent taken alone. The ac symmetrical component of current is shown in Figure 5-45. It can be divided into roughly three periods. During the first cycle or so after the fault oc- curs, the ac current is very large and falls very rapidly. This period of time is called the subtransient period. After it is over, the current continues to fall at a slower rate, until at last it reaches a steady state.1lle period of time during which it falls at a slower rate is calJed the transient period, and the time after it reaches steady state is known as the steady-state period. If the rms magnitude of the ac component of current is plotted as a function of time on a semilogarithmic scale, it is possible to observe the three periods of fault current. Such a plot is shown in Figure 5-46. It is possible to detennine the time constants of the decays in each period from such a plot. The ac nns current fl owing in the generator during the subtransient period is called the subtransient current and is denoted by the symbol r. 1llis current is caused by the damper windings on synchronous generators (see Chapter 6 for a discussion of damper windings). 1lle time constant of the subtransient current is
- 348. 324 ELECTRIC MACHINERY RJNDAMENTALS I.A (logarithmic scale) Subtransient period ,,,,,, Transiem period Steady- state period -- --"" - -~-'----'- I (linear) ""GURE 5-46 A semilogarithmic plot of the magnitude of the ac component offault current as a function of time. The subtransient and transient time constants of the generator can be determined from such a plot. given the symbol T~, and it can be detennined from the slope of the subtransient current in the plot in Figure 5-46. This current can often be 10 times the size of the steady-state fault current. TIle nns current flowing in the generator during the transient period is called the transient current and is denoted by the symbolf'. It is caused by a dc component of current induced in thefield circuit at the time ofthe short. This field current increases the internal generated voltage and causes an increased fault cur- rent. Since the time constant of the dc field circuit is much longer than the time constant of the damper windings, the transient period lasts much longer than the subtransient period. TIlis time constant is given the symbol T'. TIle average nns current during the transient period is often as much as 5 times the steady-state fault current. After the transient period, the fault current reaches a steady-state condition. TIle steady-state current during a fault is denoted by the symboll". It is given ap- proximately by the fundamental frequency component of the internal generated voltage Ell within the machine divided by its synchronous reactance: EA l.. = X, steady state (5- 3 1) TIle nns magnitude of the ac fault current in a synchronous generator varies continuously as a function of time. If r is the subtransient component of current at the instant of the fault, l' is the transient component of current at the instant of the fault, and I... is the steady-state fault current, then the nns magnitude of the current at any time after a fault occurs at the tenninals of the generator is I(t) = (r _ J')e-tlT" + (I' _1,,)e-tlT' + I... (5- 32)
- 349. SYNCHRONOUS GENERATORS 325 It is customary to define subtransient and transient reactances for a syn- chronous machine as a convenient way to describe the subtransient and transient components of fault current. 1lle subtransient reactance of a synchronous gener- ator is defilled as the ratio of the fundamental component of the internal generated voltage to the subtransient component of current at the beginning of the fault. It is given by X"= ~~ subtransient (5- 33) Similarly, the transient reactance of a synchronous generator is defined as the ra- tio of the fundamental component ofEA to the transient component of current l' at the beginning of the fault. This value of current is found by extrapolating the sub- transient region in Figure 5-46 back to time zero: X' = ~~ transient (5- 34) For the purposes of sizing protective equipment, the subtransient current is often assumed to be E,.,IX", and the transient current is assumed to be EAIX', since these are the maximum values that the respective currents take on. Note that the above discussion of faults assumes that all three phases were shorted out simultaneously. If the fault does not involve all three phases equally, then more complex methods of analysis are required to understand it.1llese meth- ods (known as symmetrical components) are beyond the scope of this book. Example 5-7. A lOO-MVA, 13.5-kV, V-connected, three-phase, 60-Hz synchro- nous generator is operating at the rated voltage and no load when a three-phase fault de- velops at its tenninals. Its reactances per unit to the machine's own base are Xs = 1.0 X' = 0.25 X" = 0.12 and its time constants are T' = 1.1Os T" = O.04s The initial dc component in this machine averages 50 percent of the initial ac component. (a) What is the ac component of current in this generator the instant after the fault occurs? (b) What is the total current (ac plus dc) flowing in the generator right after the fault occurs? (c) What will the ac component of the ClUTent be after two cycles? After 5 s? Solutioll The base current of this generator is given by the equation (2-95) 100 MVA = VJ(13.8 kV) = 4184 A
- 350. 326 ELECTRIC MACHINERY RJNDAMENTALS The subtransient, transient, and steady-state currents, per unit and in amperes, are /" Ell 1.0 8 333 = X,,= 0.12 = . = (8.333X4184A) = 34,900 A I' = ~~ = Ol.~ = 4.00 = (4.00)(4184 A) = 16,700 A Ell 1.0 I" = X' = 1.0 = 1.00 = (1.00)(4184 A) = 4184 A (a) The initial ac component of current is r = 34,900 A. (b) The total current (ac plus dc) at the beginning of the fault is Ita = 1.5r = 52,350 A (c) The ac component of current as a function of time is given by Equation (5--32): l(t) = (r _ l')e-llT" + (I' _I,,)~T' + I" (5- 32) = 18,2()()e-4°·04. + 12,516r'1.l · + 4184A At two cycles, t = 1/30 s, the total current is IUO)= 7910A + 12,142A + 4184A = 24,236 A After two cycles, the transient component ofcurrent is clearly the largest one and this time is in the transient period ofthe short circuit. At 5 s, the current is down to 1(5) = 0 A + 133 A + 4184 A = 4317 A This is part of the steady-state period ofthe short circuit. 5.11 SYNCHRONOUS GENERATOR RATINGS TIlere are certain basic limits to the speed and power that may be obtained from a synchronous generator. 1l1ese limits are expressed as ratings on the machine. The purpose of the ratings is to protect the generator from damage due to improper op- eration. To this end, each machine has a number of ratings listed on a nameplate attached to it. Typical ratings on a synchronous machine are voltage, frequency, speed, ap- parent power (kilovoltamperes), power factor. field current, and service factor. 1l1ese ratings, and the interrelationships among them, will be discussed in the fol- lowing sections. The Voltage, Speed, and Frequency Ratings The rated frequency of a synchronous generator depends on the power system to which it is connected. The commonly used power system frequencies today are
- 351. SYNCHRONOUS GENERATORS 327 50 Hz (in Europe, Asia, etc.), 60 Hz (in the Americas), and 400 Hz (in special- purpose and control applications). Once the operating frequency is known, there is only one possible rotational speed for a given number of poles. The fixed rela- tionship between frequency and speed is given by Equation (4- 34): as previously described. "mP k = 120 (4- 34) Perhaps the most obvious rating is the voltage at which a generator is de- signed to operate. A generator's voltage depends on the flux, the speed of rotation, and the mechanical construction of the machine. For a given mechanical frame size and speed, the higher the desired voltage, the higher the machine's required flux. However, flux cannot be increased forever, since there is always a maximum allowable field current. Another consideration in setting the maximum allowable voltage is the breakdown value of the winding insulation- nonnal operating voltages must not approach breakdown too closely. Is it possible to operate a generator rated for one frequency at a different fre- quency? For example, is it possible to operate a 6O-Hz generator at 50 Hz? 1lle answer is a qualified yes, as long as certain conditions are met. Basically, the problem is that there is a maximum flux achievable in any given machine, and since Ell = K<pw, the maximum allowable E). changes when the speed is changed. Specifically, if a 6O-Hz generator is to be operated at 50 Hz, then the operating voltage must be derated to 50/60, or 83.3 percent, of its original value. Just the opposite effect happens when a 50-Hz generator is operated at 60 Hz. Apparent Power and Power-Factor Ratings There are two factors that detennine the power limits of electric machines. One is the mechanical torque on the shaft of the machine, and the other is the heating of the machine's windings. In all practical synchronous motors and generators, the shaft is strong enough mechanically to handle a much larger steady-state power than the machine is rated for, so the practical steady-state limits are set by heating in the machine's windings. There are two windings in a synchronous generator, and each one must be protected from overheating. These two windings are the annature winding and the field winding. 1lle maximum acceptable annature current sets the apparent power rating for a generator, since the apparent power S is given by S=3Vo/>lll (5- 35) If the rated voltage is known, then the maximum acceptable armature current de- tennines the rated kilovoltamperes of the generator: Srated = 3 '"4..rated IAmax Srated = V3VL.rated l L.max (5- 36) (5- 37)
- 352. 328 ELECTRIC MACHINERY RJNDAMENTALS ""GURE 5-47 How the rotor field current lintit sets the rated power factor of a generator. It is important to realize that, for heating the annature windings, the powerfactor ofthe armature current is irrelevant. The heating effect of the stator copper losses is given by (5- 38) and is independent of the angle of the current with respect to Vo/>. Because the cur- rent angle is irrelevant to the annature heating, these machines are rated in kilo- voltamperes instead of kilowatts. 1lle other wi nding of concern is the field winding. 1lle field copper losses are given by (5- 39) so the maximum allowable heating sets a maximum field current for the machine. Since Ell = K4>w this sets the maximum acceptable size for Ell. 1lle effect of having a maximum IF and a maximum E). translates directly into a restriction on the lowest acceptable power factor of the generator when it is operating at the rated kilovoltamperes. Fig ure 5-47 shows the phasor diagram of a synchronous generator with the rated voltage and armature current. The current can assume many different angles, as shown. The internal generated voltage Ell is the sum of V0/> and jXs Ill. Notice that for some possible current angles the required E). exceeds E A,mu. If the generator were operated at the rated annature current and these power factors, the field winding would burn up. TIle angle of III that requires the maximum possible Ell while V0/> remains at the rated value gives the rated power factor of the generator. It is possible to op- erate the generator at a lower (more lagging) power factor than the rated value, but only by cutting back on the kilovoltamperes supplied by the generator.
- 353. • • FIGURE 5-48 • 3Vl X, v • (a) ,b, 0 Volts kW " .;" , , E, ' B .', , , , , SYNCHRONOUS GENERATORS 329 A Volts B , .', , p 3V. l.ot cosO , , , k:VAR A Q=3V.l.ot sinO Derivation of a synchronous generator capability curve. (a) The generator phasor diagram; (b) the corresponding power units. Synchronous Generator Capability Curves The stator and rotor heat limits, together with any external limits on a synchro- nous generator, can be expressed in graphical fonn by a generator capability dia- gram. A capability diagram is a plot of complex power S = P + j Q. It is derived from the phasor diagram of the generator, assuming that Vo/> is constant at the ma- chine's rated voltage. Figure 5-48a shows the phasor diagram of a synchronous generator operat- ing at a lagging power factor and its rated voltage. An orthogonal set of axes is drawn on the diagram with its origin at the tip of V0/> and with units of volts. On this diagram, vertical segment AB has a length Xs/). cos (), and horizontal segment OA has a length XsI). sin (). The real power output of the generator is given by P = 3'4,IA cos () (5- 17)
- 354. 330 ELECTRIC MACHINERY RJNDAMENTALS the reactive power output is given by Q =3V4 ,lA sine and the apparent power output is given by S=3V4 ,lA (5-1 9) (5- 35) so the vertical and horizontal axes of this figure can be recalibrated in terms of real and reactive power (Figure 5--48b). The conversion factor needed to change the scale of the axes from volts to voJtamperes (power units) is 3~ 1 Xs: ,nd 3 ~ P = 3V4 ,lAcos e= X (XsIA cos e) (5--40) , . 3V.p . Q = 3 ~IA Sin e= X (XsIA Sin e) , (5-41) On the voltage axes, the origin of the phasor diagram is at -Vo/,on the hori- zontal axis, so the origin on the power diagram is at (5-42) TIle field current is proportional to the machine's flux, and the flux is proportional to Elt = Kcf>w. TIle length corresponding to Elt on the power diagram is 3EA V.p X, (5-43) TIle annature current lit is proportional to Xsllt' and the length corresponding to Xsflt on the power diagram is 3Vq,11l.- 1lle final synchronous generator capability curve is shown in Figure 5-49. It is a plot of P versus Q, with real power P on the horizontal axis and reactive power Q on the vertical axis. Lines of constant armature current lit appear as lines of constant S = 3Vq,IIt, which are concentric circles around the origin. Lines of constant field current correspond to lines of constant EIt, which are shown as cir- cles of magnitude 3EItVq,IXs centered on the point 3V ' - "-'-' X, (5-42) TIle armature current limit appears as the circle corresponding to the rated lit or rated kilovoJtarnperes, and the field current limit appears as a circle corre- sponding to the rated IF or Ell.- Any point that lies within both circles is a safe op- erating pointfor the generator. It is also possible to show other constraints on the diagram, such as the max- imum prime-mover power and the static stability limit. A capability curve that also reflects the maximum prime-mover power is shown in Figure 5- 50.
- 355. FIGURE 5-49 Q. k:VAR 3.' • X, SYNCHRONOUS GENERATORS 331 Rotor current limit P. k:W Stator current limit The resulting generator capability curve. FIGURE 5-50 Q. k:VAR P. k:W Prime-mover power limit Origin of rotor current circle: 3.' Q= - ~ ' X, A capability dia.gram showing the prime-mover power limit.
- 356. 332 ELECTRIC MACHINERY RJNDAMENTALS Example 5-8. A 480-V, 50-Hz, Y-connected, six-pole synchronous generator is rated at 50 kVA at 0.8 PF lagging. It has a synchronous reactance of 1.0 n per phase. As- swne that this generator is cOIUlected to a steam turbine capable of supplying up to 45 kW. The friction and windage losses are 1.5 kW, and the core losses are 1.0 kW. (a) Sketch the capability curve for this generator, including the prime-mover power limit. (b) Can this generator supply a line current of 56A at 0.7 PF lagging? Why or why not? (c) What is the maximwn amOlUlt of reactive power this generator can produce? (d) If the generator supplies 30 kW of real power, what is the maximum amount of reactive power that can be simultaneously supplied? Solutioll The maximum current in this generator can be fOlUld from Equation (5--36): s...'ed = 3VoI!.l1IIod IA.max (5- 36) The voltage Vol! of this machine is VT 4S0 V Vol! = V3" = ~ = 277 V so the maximum armature ClUTent is ~ 50kVA (".max = 3Vol! = 3(277 V) = 60 A With this infonnation, it is now possible to answer the questions. (a) The maximum permissible apparent power is 50 kVA, which specifies the max- imum safe armature current. The center of the EA circles is at Q= _ 3Vl X, = _ 3(277V)2 = - 230kVAR 1.0 n The maximum size of EA is given by EA = Vol! + jXslA = 277 LO° V + (i1.0 nX60 L - 36.87° A) = 313 + j48 V = 317 LS.7° V Therefore, the magnitude of the distance proportional to EA is 3EA VoI! DE: = X, = 3(317 VX277 V) = 263 kVAR I.on (5-42) (5-43) The maximum output power available with a prime-mover power of 45 kW is approximately
- 357. SYNCHRONOUS GENERATORS 333 Q. k:VAR Stator currenl " - 50 limit , - -_ _ / Field current limit -- - 25 FIGURE 5-51 " ,, -- 150 75 , P.k:W - 75 - 00 - 125 - 150 - 175 - 200 , Maximum printe- mover power - 225 ~-- - Origin of maximum I =::- rotor current - 250 circle The capability diagram for the generator in Example 5--8. PDIU_ = PIIIU';' - Pmech I.,.. - Paxe ..,.. = 45 kW - 1.5 kW - 1.0 kW = 42.5 kW (This value is approximate because the / 2R loss and the stray load loss were not considered.) The resulting capability diagram is shown in Figure 5--51. (b) A current of 56 A at 0.7 PF lagging produces a real power of P = 3Vo/J,I cos () = 3(277 VX56 AXO.7) = 32.6kW and a reactive power of Q = 3V4>/,Isin () = 3(277 V)(56AXO.714) = 33.2kVAR
- 358. 334 ELECTRIC MACHINERY RJNDAMENTALS " < > " , , & .~ ! " 200 00 0 - 00 - 200 r- - 300 r- , ------ - _________ ,________ L ,;_ __ -': L_ >-- - --;---- - :1.0 PF ---- -'--- -- -------- - ------ -- -_.... - .. ~ .. ~ ~ ~~~~~~~~~~-=~~~~--~~ o 50 100 150 200 250 300 350 400 450 500 Real power. kW ""GURE 5-52 Capability curve for a real synchronous generator rated at 470 kVA. (Courtesy ofMaratlwn Electric Company.) Plotting this point on the capability diagram shows that it is safely within the maximum (It curve but outside the maximrun I" curve. Therefore, this point is not a safe operating condition. (e) When the real power supplied by the generator is zero, the reactive power that the generator can supply will be maximwn. This point is right at the peak of the capability curve. The Q that the generator can supply there is Q = 263 kVAR - 230 kVAR = 33 kVAR (d) If the generator is supplying 30 kW of real power, the maximum reactive power that the generator can supply is 31.5 kVAR. This value can be fOlUld by entering the capability diagram at 30 kW and going up the constant-kilowatt line lUltii a limit is reached. The limiting factor in this case is the field clUTent- the anna- ture will be safe up to 39.8 kVAR. Fig ure 5- 52 shows a typical capability for a real synchronous generator. Note that the capability boundaries are not a perfect circle for a real generator. nlis is true because real synchronous generators with salient poles have additional effects that we have not modeled. These effects are described in Appendix C.
- 359. SYNCHRONOUS GENERATORS 335 Short-Time Operation and Service Factor The most important limit in the steady-state operation ofa synchronous generator is the heating of its armature and field windings. However, the heating limit usu- ally occurs at a point much less than the maximum power that the generator is magnetically and mechanically able to supply. In fact, a typical synchronous gen- erator is often able to supply up to 300 percent of its rated power for a while (un- til its windings burn up). This ability to supply power above the rated amount is used to supply momentary power surges during motor starting and similar load transients. It is also possible to use a generator at powers exceeding the rated values for longer periods of time, as long as the windings do not have time to heat up too much before the excess load is removed. For example, a generator that could supply 1 MW indefinitely might be able to supply 1.5 MW for a couple of minutes without serious hann, and for progressively longer periods at lower power levels. However, the load must finally be removed, or the windings will overheat. The higher the power over the rated value, the shorter the time a machine can tolerate it. Figure 5- 53 illustrates this effect. This figure shows the time in seconds re- quired for an overload to cause thennal damage to a typical electrical machine, whose windings were at nonnal operating temperature before the overload oc- curred.ln this particular machine, a 20 percent overload can be tolerated for JOOO seconds, a 100 percent overload can be tolerated for about 30 seconds, and a 200 percent overload can be tolerated for about 10 seconds before damage occurs. The maximum temperature rise that a machine can stand depends on the in- sulation class of its windings. There are four standard insulation classes: A, B, F, and H. While there is some variation in acceptable temperature depending on a machine's particular construction and the method of temperature measurement, these classes generally correspond to temperature rises of 60, 80, 105, and 125°C, respectively, above ambient temperature. 1lle higher the insulation class of a given machine, the greater the power that can be drawn out of it without over- heating its windings. Overheating of windings is a very serious problem in a motor or generator. It was an old rule of thumb that for each 1DoC temperature rise above the rated windings temperature, the average lifetime of a machine is cut in half (see Figure 4- 20). Modern insulating materials are less susceptible to breakdown than that, but temperature rises still drastically shorten their lives. For this reason, a syn- chronous machine should not be overloaded unless absolutely necessary. A question related to the overheating problem is:Just how well is the power requirement of a machine known? Before installation, there are often only ap- proximate estimates of load. Because of this, general-purpose machines usualJy have a sef>!icefactor. The service factor is defined as the ratio of the actual max- imum power of the machine to its nameplate rating. A generator with a service factor of 1.15 can actually be operated at 115 percent of the rated load indefinitely without harm. The service factor on a machine provides a margin of error in case the loads were improperly estimated.
- 360. 336 ELECTRIC MACHINERY RJNDAMENTALS w' w' w, lei' o , 1.2 ""GURE 5-.53 '" , 1.4 ~ -~ , , 1.6 1.8 2 Per-unit current , , • , , , 2.2 2.4 2.6 2.8 ThemJal damage curve for a typical synchronous machine. assuming that the windings were already at operational temperature when the overload is applied. (Courtesy ofMaratlwn Electric Company.) 5.12 SUMMARY A synchronous generator is a device for converting mechanical power from a prime mover to ac electric power at a specific voltage and frequency. The term synchronous refers to the fact that this machine's electrical frequency is locked in or synchronized with its mechanical rate of shaft rotation. 1lle synchronous gen- erator is used to produce the vast majority of electric power used throughout the world. TIle internal generated voltage of this machine depends on the rate of shaft rotation and on the magnitude of the field nux. The phase voltage of the machine differs from the internal generated voltage by the effects of annature reaction in the generator and also by the internal resistance and reactance of the annature wind- ings. The tenninal voltage of the generator will either equal the phase voltage or be related to it by V3, depending on whether the machine is ,6,- or V-connected. TIle way in which a synchronous generator operates in a real power system depends on the constraints on it. When a generator operates alone, the real and 3
- 361. SYNCHRONOUS GENERATORS 337 reactive powers that must be supplied are detennined by the load attached to it, and the governor set points and field current control the frequency and terminal voltage, respectively. When the generator is connected to an infinite bus, its fre- quency and voltage are fixed, so the governor set points and field current control the real and reactive power flow from the generator. In real systems containing generators of approximately equal size, the governor set points affect both fre- quency and power flow, and the field current affects both tenninal voltage and re- active power flow. A synchronous generator's abilit.y to produce electric power is primarily limited by heating within the machine. When the generator's windings overheat, the life of the machine can be severely shortened. Since here are two different windings (armature and field), there are two separate constraints on the generator. The maximum allowable heating in the armature windings sets the maximum kilovoltamperes allowable from the machine, and the maximum allowable heat- ing in the field windings sets the maximum size of E),- The maximum size of Elt and the maximum size of lit together set the rated power factor of the generator. QUESTIONS 5-1. Why is the frequency of a synchronous generator locked into its rate of shaft rotation? 5-2. Why does an alternator's voltage drop sharply when it is loaded down with a lag- ging load? 5-3. Why does an alternator's voltage rise when it is loaded down with a leading load? 5-4. Sketch the phasor diagrams and magnetic field relationships for a synchronous gen- erator operating at (a) unity power factor, (b) lagging power factor, (c) leading power factor. 5-5. Explain just how the synchronous impedance and annature resistance can be deter- mined in a synchronous generator. 5-6. Why must a 60-Hz generator be derated if it is to be operated at 50 Hz? How much derating must be done? 5-7. Would you expect a 400-Hz generator to be larger or smaller than a 6O-Hz genera- tor of the same power and voltage rating? Why? 5-8. What conditions are necessary for paralleling two synchronous generators? 5-9. Why must the oncoming generator on a power system be paralleled at a higher fre- quency than that of the nmning system? 5-10. What is an infinite bus? What constraints does it impose on a generator paralleled with it? 5-11. How can the real power sharing between two generators be controlled without af- fecting the system's frequency? How can the reactive power sharing between two generators be controlled without affecting the system's terminal voltage? 5-12. How can the system frequency of a large power system be adjusted without affect- ing the power sharing among the system's generators? 5-13. How can the concepts of Section 5.9 be expanded to calculate the system frequency and power sharing among three or more generators operating in parallel? 5-14. Why is overheating such a serious matter for a generator?
- 362. > • 00 • ~ • 0 ., ;; '5 , ., , ~ 338 ELECTRIC MACHINERY RJNDA MENTALS 5-15. Explain in detail the concept behind capability curves. 5-16. What are short-time ratings? Why are they important in regular generator operation? PROBLEMS 5-1. At a location in Europe. it is necessary to supply 300 kW of 60-Hz power. The only power sources available operate at 50 Hz. It is decided to generate the power by means of a motor-generator set consisting of a synchronous motor driving a syn- chronous generator. How many poles should each of the two machines have in or- der to convert 50-Hz power to 60-Hz power? 5-2. A 23OO-V. lOOO-kVA. O.S-PF-Iagging. 60-Hz. two-pole. V-connected synchronous generator has a synchronous reactance of 1.1 0 and an armature resistance of 0.1 5 o. At 60 Hz. its friction and windage losses are 24 kW. and its core losses are IS kW. The field circuit has adc voltage of 200 V. and the maximum I" is IDA. The resistance of the field circuit is adjustable over the range from 20 to 200 O. The OCC of this generator is shown in Figure P5- 1. 3(XX) 2700 2400 2100 1800 1500 1200 900 600 300 / o 0.0 / / 1.0 fo'IGURE " 5- 1 V / 2.0 /' /" V V / 3.0 4.0 5.0 6.0 Field current. A The open-circuit characteristic for the generator in Problem 5- 2. 7.0 8.0 9.0 10.0 (a) How much field current is required to make Vr equal to 2300 V when the gen- erator is mlUling at no load? (b) What is the internal generated voltage of this machine at rated conditions?
- 363. SYNCHRONOUS GENERATORS 339 (c) How much field current is required to make Vr equal to 2300 V when the gen- erator is rulUling at rated conditions? (d) How much power and torque must the generator's prime mover be capable of supplying? (e) Construct a capability curve for this generator. 5-3. Assume that the field current of the generator in Problem 5- 2 has been adjusted to a value of 4.5 A. (a) What will the tenninal voltage of this generator be if it is connected to a 6.-colUlected load with an impedance of 20 L 30° O ? (b) Sketch the phasor diagram of this generator. (c) What is the efficiency of the generator at these conditions? (d) Now asswne that another identical 6.-colUlected load is to be paralleled with the first one. What happens to the phasor diagram for the generator? (e) What is the new tenninal voltage after the load has been added? (f) What must be done to restore the terminal voltage to its original value? 5-4. Assume that the field current of the generator in Problem 5- 2 is adjusted to achieve rated voltage (2300 V) at full-load conditions in each of the questions below. (a) What is the efficiency of the generator at rated load? (b) What is the voltage regulation of the generator if it is loaded to rated kilo- voltamperes with 0.8-PF-Iagging loads? (c) What is the voltage regulation of the generator if it is loaded to rated kilo- voltamperes with 0.8-PF-Ieading loads? (d) What is the voltage regulation of the generator if it is loaded to rated kilo- voltamperes with lUlity power factor loads? (e) Use MATLAB to plot the terminal voltage of the generator as a flUlction of load for all three power factors. 5-5. Assume that the field current of the generator in Problem 5- 2 has been adjusted so that it supplies rated voltage when loaded with rated current at unity power factor. (a) What is the torque angle 0 of the generator when supplying rated current at unity power factor? (b) When this generator is running at full load with lUlity power factor, how close is it to the static stability limit of the machine? 5-6. A 480-V, 4oo-kVA, 0.85-PF-Iagging, 50-Hz, four-pole, 6.-connected generator is driven by a 500-hp diesel engine and is used as a standby or emergency generator. This machine can also be paralleled with the normal power supply (a very large power system) if desired. (a) What are the conditions required for paralleling the emergency generator with the existing power system? What is the generator's rate of shaft rotation after paralleling occurs? (b) If the generator is cOIUlected to the power system and is initially floating on the line, sketch the resulting magnetic fields and phasor diagram. (c) The governor setting on the diesel is now increased. Show both by means of house diagrams and by means of phasor diagrams what happens to the genera- tor. How much reacti ve power does the generator supply now? (d) With the diesel generator now supplying real power to the power system, what happens to the generator as its field current is increased and decreased? Show this behavior both with phasor diagrams and with house diagrams. 5-7. A 13.8-kV, IO-MVA, 0.8-PF-Iagging, 60-Hz, two-pole, Y-COlUlected steam-turbine generator has a synchronous reactance of 12 n per phase and an armature resistance
- 364. 340 ELECTRIC MACHINERY RJNDAMENTALS of 1.5 n per phase.This generator is operating in parallel with a large power system (infinite bus). (a) What is the magnitude of EA at rated conditions? (b) What is the torque angle of the generator at rated conditions? (c) If the field current is constant, what is the maximwn power possible out of this generator? How much reserve power or torque does this generator have at full load? (d) At the absolute maximum power possible, how much reactive power will this generator be supplying or consruning? Sketch the corresponding phasor dia- gram. (Assrune h is still unchanged.) 5-8. A 480-V, lOO-kW, two-pole, three-phase, 60-Hz synchronous generator's prime mover has a no-load speed of3630 rfmin and a full-load speed of3570 rfmin.1t is op- erating in parallel with a 480-V, 75-kW, four-pole, 60-Hz synchronous generator whose prime mover has a no-load speed of 1800 rhnin and a full-load speed of 1785 rfmin. The loads supplied by the two generators consist of 100 kW at 0.85 PF lagging. (a) Calculate the speed droops of generator I and generator 2. (b) Find the operating frequency of the power system. (c) Find the power being supplied by each of the generators in this system. (d) If Vr is 460 V, what must the generator's operators do to correct for the low ter- minal voltage? 5-9. TIrree physically identical synchronous generators are operating in parallel. They are all rated for a full load of 3 MW at 0.8 PF lagging. The no-load frequency ofgen- erator A is 61 Hz, and its speed droop is 3.4 percent. The no-load frequency of generator B is 61.5 Hz, and its speed droop is 3 percent. The no-load frequency ofgenerator C is 60.5 Hz, and its speed droop is 2.6 percent. (a) If a total load consisting of 7 MW is being supplied by this power system, what will the system frequency be and how will the power be shared among the three generators? (b) Create a plot showing the power supplied by each generator as a function of the total power supplied to all loads (you may use MATLAB to create this plot). At what load does one of the generators exceed its ratings? Which generator ex- ceeds its ratings first? (c) Is this power sharing in a acceptable? Why or why not ? (d) What actions could an operator take to improve the real power sharing among these generators? 5-10. A paper mill has installed three steam generators (boilers) to provide process steam and also to use some its waste products as an energy source. Since there is extra ca- pacity, the mill has installed three 5-MW turbine generators to take advantage of the situation. Each generator is a 4160-V, 6250-kVA, 0.85-PF-Iagging, two-pole, Y-cOIlllected synchronous generator with a synchronous reactance of 0.75 n and an annature resistance of 0.04 n. Generators I and 2 have a characteristic power- frequency slope sp of 2.5 MWlHz, and generators 2 and 3 have a slope of3 MW/Hz. (a) If the no-load frequency of each of the three generators is adjusted to 61 Hz, how much power will the three machines be supplying when actual system fre- quency is 60 Hz? (b) What is the maximum power the three generators can supply in this condition without the ratings ofone of them being exceeded? At what frequency does this limit occur? How much power does each generator supply at that point?
- 365. SYNCHRONOUS GENERATORS 341 (c) What would have to be done to get all three generators to supply their rated real and reactive powers at an overall operating frequency of 60 Hz? (d) What would the internal generated voltages of the three generators be under this condition? Problems 5- 11 to 5- 21 refer to a four-pole, Y-connected synchronous generator rated at 470 kVA, 480 V, 60 Hz, and 0.85 PF lagging. Its armature resistance RA is 0.016 n. The core losses of this generator at rated conditions are 7 kW, and the friction and windage losses are 8 kW. The open-circuit and short-circuit characteristics are shown in Figure P5- 2. 5- 11. (a) What is the saturated synchronous reactance of this generator at the rated conditions? (b) What is the unsaturated synchronous reactance of this generator? (c) Plot the saturated synchronous reactance of this generator as a function of load. 5-12. (a) What are the rated current and internal generated voltage of this generator? (b) What field current does this generator require to operate at the rated voltage, current, and power factor? 5-13. What is the voltage regulation of this generator at the rated current and power factor? 5-14. If this generator is operating at the rated conditions and the load is suddenly re- moved, what will the tenninal voltage be? 5-15. What are the electrical losses in this generator at rated conditions? 5-16. If this machine is operating at rated conditions, what input torque must be applied to the shaft of this generator? Express your answer both in newton-meters and in polUld-feet. 5-17. What is the torque angle 0 of this generator at rated conditions? 5-18. Assume that the generator field current is adjusted to supply 480 V under rated con- ditions. What is the static stability limit of this generator? (Note: You may ignore RA to make this calculation easier.) How close is the full-load condition of this genera- tor to the static stability limit? 5-19. Assume that the generator field current is adjusted to supply 480 V under rated con- ditions. Plot the power supplied by the generator as a function of the torque angle o. (Note: You may ignore RA to make this calculation easier.) 5-20. Assume that the generator's field current is adjusted so that the generator supplies rated voltage at the rated load current and power factor. If the field current and the magnitude of the load current are held constant, how will the terminal voltage change as the load power factor varies from 0.85 PF lagging to 0.85 PF leading? Make a plot of the tenninal voltage versus the impedance angle of the load being supplied by this generator. 5-2 1. Assrune that the generator is connected to a 480-V infinite bus, and that its field cur- rent has been adjusted so that it is supplying rated power and power factor to the bus. You may ignore the annature resistance RA when answering the following questions. (a) What would happen to the real and reactive power supplied by this generator if the field flux is reduced by 5 percent? (b) Plot the real power supplied by this generator as a function of the flux cp as the flux is varied from 75 percent to 100 percent of the flux at rated conditions.
- 366. Open Circuit Characteristic 1200 1100 , , , , , , , , , , , , , '- c > ~ • • , '§ , ., , ~ < • ~ a , 3 • ~ lllOO 9lXl c ,/ C /' 800 700 C /' I c 6lXl 500 c / c 400 c / / 300 200 V 100 c °0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Field current. A Shon Circuit Characteristic 16lXl 1400 C 1200 c lllOO c 800 c 6lXl c 400 c 200 V o o / 0.2 / / , , 0.4 0.6 / , 0.8 HGURE )'5- 2 Field current. A ,b, I. I 1.2 1.3 1.4 1.5 /' / / , , , 1.2 1.4 (a) Open-cirwit characteristic curve for the generator in Problems 5- 11 to 5- 21. (b) Short-cin:uit characteristic curve for the generator in Problems 5- 11 to 5- 21. 342
- 367. SYNCHRONOUS GENERATORS 343 (c) Plot the reactive power supplied by this generator as a function of the flux cp as the flux is varied from 75 percent to 100 percent of the flux at rated conditions. (d) Plot the line current supplied by this generator as a function of the flux cp as the flux is varied from 75 percent to lDO percent of the flux at rated conditions. 5-22. A lDO-MVA. 12.S- kV. 0.8S-PF-Iagging. SO-Hz. two-pole. Y-cOlUlected synchronous generator has a per-unit synchronous reactance of 1.1 and a per-lUlit annature resis- tance of 0.012. (a) What are its synchronous reactance and annature resistance in oluns? (b) What is the magnitude of the internal generated voltage E./t at the rated condi- tions? What is its torque angle 0 at these conditions? (c) Ignoring losses in this generator. what torque must be applied to its shaft by the prime mover at full load? 5-23. A three-phase Y-cOIUlected synchronous generator is rated 120 MVA. 13.2 kV. 0.8 PF lagging. and 60 Hz. Its synchronous reactance is 0.9 n.and its resistance may be ignored. (a) What is its voltage regulation? (b) What would the voltage and apparent power rating of this generator be if it were operated at 50 Hz with the same annature and field losses as it had at 60 Hz? (c) What would the voltage regulation of the generator be at 50 Hz? 5-24. Two identical 600-kVA. 480-V synchronous generators are connected in parallel to supply a load. The prime movers of the two generators happen to have different speed droop characteristics. When the field currents of the two generators are equal. one delivers 4DO A at 0.9 PF lagging, while the other delivers 3DO A at 0.72 PF lagging. (a) What are the real power and the reactive power supplied by each generator to the load? (b) What is the overall power factor of the load? (c) In what direction must the field current on each generator be adjusted in order for them to operate at the same power factor? 5-25. A generating station for a power system consists offour 120-MVA, IS-kV, 0.85-PF- lagging synchronous generators with identical speed droop characteristics operating in parallel. The governors on the generators' prime movers are adjusted to produce a 3-Hz drop from no load to full load. TIrree of these generators are each supplying a steady 7S MW at a frequency of 60 Hz, while the fourth generator (called the swing generator) handles all increme ntal load changes on the system while main- taining the system's frequency at 60 Hz. (a) At a given instant, the total system loads are 260 MW at a frequency of 60 Hz. What are the no-load frequencies of each of the system's generators? (b) If the system load rises to 290 MW and the generator's governor set points do not change, what will the new system frequency be? (c) To what frequency must the no-load frequency of the swing generator be ad- justed in order to restore the system frequency to 60 Hz? (d) If the system is operating at the conditions described in part c, what would hap- pen if the swing generator were tripped off the line (disconnected from the power line)? 5-26. Suppose that you were an engineer plaMing a new electric cogeneration facility for a plant with excess process steam. You have a choice of either two IO-MW turbine- generators or a single 20-MW turbine-generator. What would be the advantages and disadvantages of each choice?
- 368. 344 ELECTRIC MACHINERY RJNDAMENTALS 5-27. A 25-MVA. three-phase. 13.8-kV. two-pole. 60-Hz Y-connected synchronous gen- erator was tested by the open-circuit test. and its air-gap voltage was extrapolated with the following results: Open-circuit test Field current. A Line voltage. t V Extrapolated air-gap voltage. tV 320 13.0 15.4 365 13.8 17.5 380 14.1 18.3 475 15.2 22.8 The short-circuit test was then peIfonned with the following results: Short-circuit test Field current. A Affilature current. A 320 040 The armature resistance is 0.24 n per phase. 365 1190 380 1240 475 1550 570 16.0 27.4 570 1885 (a) Find the unsaturated synchronous reactance of this generator in oluns per phase and per unit. (b) Find the approximate saturated synchronous reactance Xs at a field current of 380 A. Express the answer both in ohms per phase and per lUlit. (c) Find the approximate saturated synchronous reactance at a field current of 475 A. Express the answer both in ohms per phase and in per-unit. (d) Find the short-circuit ratio for this generator. 5-28. A 20-MVA, 12.2-kY, 0.8-PF-Iagging, Y-connected synchronous generator has a neg- ligible annature resistance and a synchronous reactance of 1.1 per lUlit. The gener- ator is connected in parallel with a 60-Hz, 12.2-kV infinite bus that is capable of supplying or consuming any amOlUlt of real or reactive power with no change in frequency or tenninal voltage. (a) What is the synchronous reactance of the generator in oluns? (b) What is the internal generated voltage EA of this generntor lUlder rated conditions? (c) What is the annature current IAin this machine at rated conditions? (d) Suppose that the generator is initially operating at rated conditions. If the inter- nal generated voltage EA is decreased by 5 percent, what will the new annature current IA be? (e) Repeat part d for 10, 15, 20, and 25 percent reductions in EA. (j) Plot the magnitude of the annature current 1..1 as a function of EA. (You may wish to use MATLAB to create this plot.)
- 369. SYNCHRONOUS GENERATORS 345 REFERENCES 1. Chaston. A. N. Electric Machinery. Reston. Va.: Reston Publishing. 1986. 2. Del TOTO. V. Electric Machines and Po·....er Systelll!i. Englewood ClilTs. N.J.: Prentice-Hall. 1985. 3. Fitzgerald. A. E., and C. Kingsley. Jr. Electric Machinery. New Yor(: McGraw-Hill. 1952. 4. Fitzgerald. A. E., C. Kingsley, Jr., and S. D. Umans. Electric Machinery. 5th ed., New York: McGraw-Hill. 1990. 5. Kosow. Irving L. Electric Machinery and Transformers. Englewood ClilTs. N.J.: Prentice-Hall. 1972. 6. Liwschitz-Garik. Michael. and Clyde Whipple. AlteflUlting-Current Machinery. Princeton. N.J.: Van Nostrand. 1961. 7. McPherson. George. An Introduction to Electrical Machines and Traruformers. New Yor(: Wiley. 1981. 8. Siemon. G. R., and A. Straughen. Electric Machines. Reading, Mass.: Addison-Wesley. 1980. 9. Werninck. E. H. (ed.). Electric Motor Hatufbook. London: McGraw-Hill. 1978.
- 370. CHAPTER 6 SYNCHRONOUS MOTORS Synchronous motors are synchronous machines used to convert electrical power to mechanical power. This chapter explores the basic operation of synchronous motors and relates their behavior to that of synchronous generators. 6.1 BASIC PRINCIPLES OF MOTOR OPERATION To understand the basic concept of a synchronous motor, look at Figure 6-1 , which shows a two-pole synchronous motor. 1lle field current IF of the motor produces a steady-state magnetic field HR. A three-phase set of voltages is applied to the stator orthe machine, which produces a three-phase current flow in the windings. As was shown in Chapter 4, a three-phase sct of currents in an annature winding produces a uniform rotating magnetic field Bs. Therefore, there are two magnetic fields present in the machine, and the rotorfield will tend to line up with the stator field, just as two bar magnets will tend to line up if placed near each other. Since the stator magnetic field is rotating, the rotor magnetic field (and the rotor itself) will constantly try to catch up. TIle larger the angle between the two magnetic fields (up to a certain maximum), the greater the torque on the rotor of the machine. The basic principle of synchronous motor operation is that the rotor "chases" the rotating stator magnetic field around in a circle, never quite catching up with it. Since a synchronous motor is the same physical machine as a synchronous generator, all of the basic speed, power, and torque equations of Chapters 4 and 5 apply to synchronous motors also. 346
- 371. SYNCHRONOUS MOTORS 347 o / , 11, 0, o Tind ",k HRx nS '" counterclockwise o FIGURE 6-1 A two-pole synchronous motor. The Equivalent Circuit of a Synchronolls Motor A synchronous motor is the same in all respects as a synchronous generator, except that the direction of power flow is reversed. Since the direction of power fl ow in the machine is reversed, the direction of current fl ow in the stator of the motor may be expected to reverse also. Therefore, the equivalent circuit of a synchronous motor is exactly the same as the equivalent circuit of a synchronous generator, except that the reference direction of IA. is reversed. 1lle resulting full equivalent circuit is shown in Figure 6- 2a, and the per-phase equivale nt circuit is shown in Figure 6- 2b. As be- fore, the three phases of the equivalent circuit may be either Y- or d-connected. Because of the change in direction of lA., the Kirchhoff's voltage law equa- tion for the equivalent circuit changes too. Writing a Kirchhoff's voltage law equation for the new equivalent circuit yields I V4> - EA+ jXS IA + RAIA I lEA- V4> - jXS IA - RAIA I (6-1) (6-2) This is exactly the srune as the equation for a generator, except that the sign on the current term has been reversed. The Synchronolls Motor from a Magnetic Field Perspective To begin to understand synchronous motor operation, take another look at a syn- chronous generator connected to an infinite bus. The generator has a prime mover
- 372. 348 ELECTRIC MACHINERY RJNDAMENTALS I" j Xs R, )v., E" I, R.. I" R, j Xs ) V .' R, V, + E" "-' L, I" - j Xs R, )v" E" (a) I, RJ,{ I, - - /' j Xs R, V, L, E, V . ,b, ""GURE 6-2 (a) The full equivalent circuit of a three-phase synchronous motor. (b) The per-phase equivalent circuit. turning its shaft, causing it to rotate. The direction of the applied torque Tapp from the prime mover is in the direction of motion, because the prime mover makes the generator rotate in the first place. The phasor diagram of the generator operating with a large field current is shown in Figure 6-3a, and the corresponding magnetic field diagram is shown in Figure 6- 3b. As described before, RR corresponds to (produces) EA , Rnet corre- sponds to (produces) Vo/>, and Rs corresponds to E"at (= -jXsIA). TIle rotation of both the phasor diagram and magnetic fie ld diagram is counterclockwise in the figure, following the standard mathematical convention of increasing angle. TIle induced torque in the generator can be found from the magnetic field diagram. From Equations (4-60) and (4-6 1) the induced torque is given by
- 373. , FIGURE 6-3 (a) •, SYNCHRONOUS MOTORS 349 B, w.~ -rlf,=-------B~ ,b , (a) Phasor diagram ofa synchronous generator operating at a lagging power factor. (b) The corresponding magnetic field diagram. ~/~ -- ,, 8 , FIGURE 6-4 u, , 'V , ,,~ w.~ "cr,-------'" ,b , B, (a) Phasor diagram ofa synchronous motor. (b) The corresponding magnetic field diagram. (4-60) (4-61) Notice that from the magnetic field diagram the induced torque in this machine is clockwise, opposing the direction of rotation. In other words, the induced torque in the generator is a countertorque, opposing the rotation caused by the external applied torque "Taw Suppose that, instead of turning the shaft in the direction of motion, the prime mover suddenly loses power and starts to drag on the machine's shaft. What happens to the machine now? The rotor slows down because of the drag on its shaft and falls behind the net magnetic field in the machine (see Figure 6-4a). As the ro- tor, and therefore BR, slows down and falls behind Bne, the operation ofthe machine suddenly changes. By Equation (4--60), when BR is behind B..." the induced
- 374. 350 ELECTRIC MACHINERY RJNDAMENTALS torque's direction reverses and becomes counterclockwise. In other words, the ma- chine's torque is now in the direction of motion, and the machine is acting as a mo- tor. The increasing torque angle 8 results in a larger and larger torque in the direc- tion of rotation, until eventually the motor's induced torque equals the load torque on its shaft. At that point, the machine will be operating at steady state and syn- chronous speed again, but now as a motor. TIle phasor diagram corresponding to generator operation is shown in Fig- ure 6-3a, and the phasor diagram corresponding to motor operation is shown in Figure 6-4a. TIle reason that the quantity jXsI), points from Vo/>, to E), in the gen- erator and from E), to Vo/> in the motor is that the reference direction of I), was re- versed in the definition of the motor equi valent circuit. The basic difference be- tween motor and generator operation in synchronous machines can be seen either in the magnetic field diagram or in the phasor diagram. In a generator, E), lies ahead of V o/>, and BR lies ahead of 8 0 ... In a motor, E), lies behind Vo/>' and BR lies behind Boe, . In a motor the induced torque is in the direction of motion, and in a generator the induced torque is a countertorque opposing the direction of motion. 6.2 STEADY-STATE SYNCHRONOUS MOTOR OPERATION TIlis section explores the behavior of synchronous motors under varying condi- tions of load and field current as well as the question of power-factor correction with synchronous motors. The following discussions will generally ignore the ar- mature resistance of the motors for simplicity. However, R), will be considered in some of the worked numerical calculations. The Synchronous Motor Torque-Speed Characteristic Curve Synchronous motors supply power to loads that are basically constant-speed de- vices. They are usually connected to power systems very much larger than the in- dividual motors, so the power systems appear as infinite buses to the motors. TIlis means that the terminal voltage and the system frequency will be constant regard- less of the amount of power drawn by the motor. 1lle speed of rotation of the mo- tor is locked to the applied electrical frequency, so the speed of the motor will be constant regardless of the load. The resulting torque-speed characteristic curve is shown in Figure 6- 5. The steady-state speed of the motor is constant from no load all the way up to the maximum torque that the motor can supply (called the pull- out torque), so the speed regulation of this motor [Equation (4-68)] is 0 percent. 1lle torque equation is (4-<>1 ) (5- 22)
- 375. SYNCHRONOUS MOTORS 351 fpullou1 ----------------- n_. - n, SR= on xlOO% '" SR=O% f",,0<1 ----------------- L-__________________~-------- ,. '.~ FIGURE 6-S The torque-speed characteristic of a synchronous motor. Since the speed of the motor is oonstam. its speed regulation is zero. The maximum or pullout torque occurs when /j = 900. Nonnal full-load torques are much less than that, however. In fact, the pullout torque may typically be 3 times the full-load torque of the machine. When the torque on the shaft of a synchronous motor exceeds the pullout torque, the rotor can no longer remain locked to the stator and net magnetic fields. Instead, the rotor starts to slip behind them. As the rotor slows down, the stator magnetic field "laps" it repeatedly, and the direction of the induced torque in the rotor reverses with each pass. The resulting huge torque surges, first one way and then the other way, cause the whole motor to vibrate severely. The loss of syn- chronization after the pullout torque is exceeded is known as slipping poles. The maximum or pullout torque of the motor is given by (6-3) (6-4) These equations indicate that the larger the field current (and hence E,...), the greater the maximum torque ofthe nwtor. There is therefore a stability advantage in operating the motor with a large field current or a large E,.,. The Effect of Load Changes on a Synchronous Motor If a load is attached to the shaft of a synchronous motor, the motor will develop enough torque to keep the motor and its load turning at a synchronous speed. What happens when the load is changed on a synchronous motor?
- 376. 352 ELECTRIC MACHINERY RJNDAMENTALS (a) , , , , IIA2 ilAl ""GURE 6-6 ,,,,,,, IV, EA4 "j'__CC____________ _ (b) (a) Pltasor diagram of a motor operating at a leading power factor. (b) The effect of an increase in load on the operation of a synchronous motor. To find oul, examine a synchronous motor operating initially with a leading power factor, as shown in Figure 6--6. If the load on the shaft of the motor is in- creased, the rotor will initially slow down. As it does, the torque angle 8 becomes larger, and the induced torque increases. The increase in induced torque eventu- ally speeds the rotor back up, and the motor again turns at synchronous speed but with a larger torque angle 8. What does the phasor diagrrun look like during this process? To find out, ex- amine the constraints on the machine during a load change. Figure 6--6a shows the motor's phasor diagram before the loads are increased. The internal generated volt- age EAis equal to K<pw and so depends on only the field current in the machine and the speed of the machine. The speed is constrained to be constant by the input power supply, and since no one has touched the field circuit, the field current is constant as well. TIlerefore, lEAl must be constant as the load changes. TIle dis- tances proportional to power (EA sin 8 and JAcos ()) will increase, but the magni- tude of EAmust remain constant. As the load increases, EA swings down in the manner shown in Figure 6-6b. As EA swings down further and further, the quantity
- 377. SYNCHRONOUS MOTORS 353 jXSIA has to increase to reach from the tip of EA to Vo/>, and therefore the annature current IAalso increases. Notice that the power-factor angle () changes too, becom- ing Jess and less leading and then more and more lagging. Example 6-1. A 20S-V, 4S-kVA, O.S-PF-Ieading, a-connected, 60-Hz synchro- nous machine has a synchronous reactance of 2.5 0 and a negligible armature resistance. Its friction and windage losses are 1.5 kW, and its core losses are 1.0 kW. Initially, the shaft is supplying a IS-hp load, and the motor's power factor is O.SO leading. (a) Sketch the phasor diagram of this motor, and find the values of lA, fL' and EA. (b) Assume that the shaft load is now increased to 30 hp. Sketch the behavior of the phasor diagram in response to this change. (c) Find lA, fL' and EA after the load change. What is the new motor power factor? Solutioll (a) Initially, the motor's output power is 15 hp. This corresponds to an output of POOl = (15 hp)(0.746 KWlhp) = 11.19 kW Therefore, the electric power supplied to the machine is Pin = P out + P""",blo<s + P CO/IeJo.. + Pel""l.,.. = 11.I9kW+ I.5kW+ 1.0kW+OkW = 13.69kW Since the motor's power factor is O.SO leading, the resulting line current flow is I - ccciP -","CC-o L - v'3VTcos 0 13.69 kW = V3"(20S VXO.SO) = 47.5 A and the annature current is h/V'!, with O.Sleading power factor, which gives the result IA = 27.4 L 36.S7° A To find EA , apply Kirchhoff's voltage law [Equation (6-2)]: EA = Vo/> - jXsIA = 20S L 0° V - (j2.5 0)(27.4 L 36.S7° A) = 20S L 0° V - 6S.5 L 126.S7° V =249.I -jS4.SV =2SSL - 12.4° V The resulting phasor diagram is shown in Figure 6-7a. (b) As the power on the shaft is increased to 30 hp, the shaft slows momentarily, and the internal generated voltage EA swings out to a larger angle /j while maintain- ing a constant magnitude. The resulting phasor diagram is shown in Figure 6-7b. (c) After the load changes, the electric input power of the machine becomes Pin = P out + Pmoc.b lo<s + PCO/IeJo.. + P el""l.,.. = (30 hpXO.746 kWlhp) + 1.5 kW + 1.0 kW + 0 kW = 24.SSkW
- 378. 354 ELECTRIC MACHINERY RJNDAMENTALS , , , 1 ..1. '" 27.4 L 36.87° A 8 ""GURE 6-7 "J (b, ,,,,, V. '" 208 L r:f' V jXSIA '" 68.5 L 126.87° EA"'255L - 12.4°Y ,,,, V.",208Lr:f'V E;" '" 255 L _23° V (a) The motor phasor diagram for Example 6-la. (b) The motor phasor diagram for Example 6- lb. From the equation for power in tenns of torque angle [Equation (5-20)], it is pos- sible to find the magnitude of the angle /j (remember that the magnitude of EA is constant): '0 p = 3VI/!EA sin Ii X, _ . _] XsP ii -sill 3VE ., _ . _] (2.5 flX24.SS kW) - Sill 3(20S V)(255 V) = sin- ] 0.391 = 23° (5- 20) The internal generated voltage thus becomes EA = 355 L _23° V. Therefore, IA will be given by _ VI/! - EA IA - 'X J , 20SLooY-255L-23°Y = j2.5fl
- 379. SYNCHRONOUS MOTORS 355 ,., ,b, FIGURE 6-8 (a) A synchronous motor operating at a Jagging power factor. (b) The effect of an increase in field current on the operation of this motor. and IL will become IL = V3IA = 7 1.4 A The final power factor will be cos (-ISO) or 0.966 leading. The Effect of Field Current Changes on a Synchronous Motor We have seen how a change in shaft load on a synchronous motor affects the motor. There is one other quantity on a synchronous motor that can be readily adjusted- its field current. What effect does a change in field current have on a synchronous motor? To find out, look at Figure 6--8. Figure 6--8a shows a synchronous motor ini- tially operating at a lagging power factor. Now, increase its field current and see what happens to the motor. Note that an increase in field current increases the magnitude of E,t but does not affect the real power supplied by the motor. 1lle power supplied by the motor changes only when the shaft load torque changes. Since a change in IF does not affect the shaft speed nm , and since the load attached
- 380. 356 ELECTRIC MACHINERY RJNDAMENTALS Lagging power factor PF '" 1.0 Leading power factor FIGURE 6-9 IF Synchronous motor V curves. to the shaft is unchanged, the real power supplied is unchanged. Of course, VT is also constant, since it is kept constant by the power source supplying the motor. The distances proportional to power on the phasor diagram (EA sin 8 and IAcos ()) must therefore be constant. When the field current is increased, EAmust increase, but it can only do so by sliding out along the line of constant power. 111is effect is shown in Figure 6--8b. Notice that as the value of EA increases, the magnitude of the annature cur- rent IAfirst decreases and then increases again. At low EA , the armature current is lagging, and the motor is an inductive load. It is acting like an inductor-resistor combination, consuming reactive power Q. As the field current is increased, the annature current eventually lines up with Vo/>, and the motor looks purely resistive. As the field current is increased further, the annature current becomes leading, and the motor becomes a capacitive load. 11 is now acting like a capacitor-resistor combination, consuming negative reactive power -Q or, alternatively, supplying reactive power Q to the system. A plot of IA versus IF for a synchrono us motor is shown in Figure 6- 9. Such a plot is called a synchronous motor V cu",e, for the obvious reason that it is shaped like the letter V. There are several V curves drawn, corresponding to dif- ferent real power levels. For each curve, the minimum armature current occurs at unity power factor, when only real power is being supplied to the motor. At any other point on the curve, some reactive power is being supplied to or by the mo- tor as well. For field currents less than the value giving minimum lA, the annature current is lagging, consuming Q. For field currents greater than the value giving the minimum lA, the annature current is leading, supplying Q to the power system as a capacitor would. 111erefore, by controlling the field current of a synchronous motor, the reactive power supplied to or consumed by the power system can be controlled. When the projection of the phasor EA onto V0/> (EAcos 8) is shoner than V0/> itself, a synchronous motor has a lagging current and consumes Q. Since the field current is small in this situation, the motor is said to be underexcited. On the other hand, when the projection of EAonto Vo/> is longer than Vo/> itself, a synchronous
- 381. SYNCHRONOUS MOTORS 357 FIGURE 6-10 (a) The phasor diagram of an underexcited synchronous motor. (b) The phasor diagram of an overexcited synchronous motor. motor has a leading current and supplies Q to the power system. Since the field current is large in this situation, the motor is said to be overexcited. Phasor dia- grams illustrating these concepts are shown in Figure 6-10. EXllmple 6-2. The 20S-V, 45-kVA, O.S-PF-Ieading, 8-cOIUlected, 60-Hz synchro- nous motor of the previous example is supplying a 15-hp load with an initial power factor of 0.85 PF lagging. The field current I" at these conditions is 4.0 A. (a) Sketch the initial phasor diagram of this motor, and fmd the values IAand EA. (b) If the motor's flux is increased by 25 percent, sketch the new phasor diagram of the motor. What are EA , lA, and the power factor of the motor now? (c) Assume that the flux in the motor varies linearly with the field current I". Make a plot of 1..1 versus I" for the synchronous motor with a IS-hp load. Solutioll (a) From the previous example, the electric input power with all the losses included is p~ = 13.69 kW. Since the motor's power factor is 0.85 lagging, the resulting annature current flow is I - n,R",~"::-;; A - 3VoIIcos(J 13.69 kW = 3(20S V)(0.S5) = 25.8 A The angle (J is cos-1 0.85 = 31.8°, so the phasor current 1..1 is equal to 1..1 = 25.8 L -31.SoA To find EA , apply Kirchhoff's voltage law [Equation (6--2)]: EA = VoII - jXSIA = 20S L 0° V - (j2.5 0)(25.8 L - 31.So A ) =20SLOo V - 64.5L5S.2° V = 182L - 17.5° V The resulting phasor diagram is shown in Figure 6- 11, together with the results for part b.
- 382. 358 ELECTRIC MACHINERY RJNDAMENTALS , , , I;' I , I" , , , fV~"'208LOOV EA",182L - 17.5° Y.J '- E;' '" 227.5 L _ 13.9° Y ""GURE 6-11 The phasor diagram of the motor in Example 6--2. (b) Ifthe flux cp is increased by 25 percent, then EA = Kcpw will increase by 25 per- cent too: EA2 = 1.25 EAI = 1.25(182 V) = 227.5 V However, the power supplied to the load must remain constant. Since the dis- tance EA sin /) is proportional to the power, that distance on the phasor diagram must be constant from the original flux level to the new flux level. Therefore, EA]sin 8] = EA2sin ~ ~ = sin- t(EAt sin 8]) E" The annature current can now be found from Kirchhoff's voltage law: _ VI/! - EA2 1..1.2 - ·X J , I _ 208 LO° V - 227.5 L - 13.9° V ..1. - j2.50 = 56.2~.~OA2° V = 22.5 L 13.2° A Finally, the motor's power factor is now PF = cos (13.2°) = 0.974 leading The resulting phasor diagram is also shown in Figure 6-11. (e) Because the flux is assumed to vary linearly with field current, EA will also vary linearly with field current. We know that EA is 182 V for a field current of 4.0A, so EA for any given field current can be fOlUld from the ratio ~ - ~ 182V -4.0A (6-5)
- 383. SYNCHRONOUS MOTORS 359 The torque angle lj for any given field current can be found from the fact that the power supplied to the load must remain constant: EA I sin 01 = EA2sin ~ ~ = sin- I (EA I sin 01) E" (6-6) These two pieces of infonnation give us the phasor voltage EA. Once EA is avail- able, the new armature current can be calculated from Kirchhoff's voltage law: _ V</I - EAl IAl - 'X J , (6- 7) A MATLAB M-file to calculate and plot IA versus IF using Equations (6- 5) through (6- 7) is shown below: % M-fil e : v_curve. m % M-fil e c reat e a p l o t o f a rmatur e curre nt ver s u s fi e l d % current f o r the syn chro nou s mo t o r of Exampl e 6- 2 . % Firs t , initia liz e the fi e l d curre nt values (2 1 va lues % in the range 3 . S- 5.S A) i _ f = (3S : 1 :5S ) / 1 0; % Now initia liz e a ll o ther values i _a = z e r os( 1 ,2 1 ) ; x_s = 2.5; v_pha se = 20S; del tal = -1 7 .5 .. p i / 1 SO; % Pre - a llocate i _a array % Synchro no u s reac tance % Phase vo lt age at 0 degrees % de lt a 1 in radian s e_a l = l S2 .. (cos (delta l ) + j .. s in (delt a l )) ; % Ca l culat e the armature current f o r each value f or ii = 1: 21 ond % Ca l culat e magnitude of e _a2 e_a2 = 45.5 .. i _ f (ii ) ; % Ca l culat e delta2 delta2 = as in ( abs (e_a l ) / abs (e_a2 ) .. s in (de lt al ) ) ; % Ca l culat e the phasor e_a2 e_a2 = e_a2 " (cos (delt a2 ) + j" s in (delt a2 )) ; % Ca l culat e i _a i _a( ii ) = ( v_pha se % Plot the v - curve p l ot (i _ f. abs (i _a) , ' Co l or ' , 'k' , 'Linewi dth' , 2.0 ) ; x l abel ( 'Fie l d Current (A) ' , 'Fo ntwe i ght' , 'Sol d ' ) ; y l abel ( ' Armature Current (A) ' , 'Fo ntwe i ght ' , 'Sold' ) ; titl e ( ' Sync hro no u s M o tor V- CUrve ' , 'Fo ntwe i ght ' , 'Sol d ' ) ; gr id on; The plot produced by this M-flle is shown in Figure 6-12. Note that for a field current of 4.0 A, the annature current is 25.8 A. This result agrees with part a of this example.
- 384. 360 ELECTRIC MACHINERY RJNDAMENTALS 30 29 28 < 27 " ~ 26 ! 25 ~ 24 23 22 / / / / / / " 21 3.' 4.0 4.5 5.0 ,.5 6.0 Field current. A ""GURE 6- 12 Vcurve for the synchronous motor of Example 6--2. The Synchronolls Motor and Power-Factor Correction Figure 6-13 shows an infinite bus whose OUlpUI is connected through a transmis- sion line 1 0 an industrial plant at a distant point. The industrial plant shown con- sists of three loads. Two of the loads are induction motors with lagging power fac- tors, and the third load is a synchronous motor with a variable power factor. What does the ability to set the power factor of one of the loads do for the power system? To find out, examine the following example problem. (Note: A re- view of the three-phase power equations and their uses is given in Appendix A. Some readers may wish to consult it when studying this problem.) Example 6-3. The infinite bus in Figure 6-13 operates at 480 V. Load I is an in- duction motor consruning 100 kW at 0.78 PF lagging, and load 2 is an induction motor con- sruning 200 kW at 0.8 PF lagging. Load 3 is a synchronous motor whose real power con- srunption is 150 kW. (a) If the synchronous motor is adjusted to operate at 0.85 PF lagging, what is the transmission line current in this system? (b) If the synchronous motor is adjusted to operale at 0.85 PF leading, what is the transmission line current in this system? (c) Assrune thai the transmission line losses are given by line loss where LL stands for line losses. How do the transmission losses compare in the two cases?
- 385. p,. - Infinite bus Transmission line - Q,. Plant FIGURE 6-13 SYNCHRONOUS MOTORS 361 ,----------------, P, - - Q, P, - - '" P, - -Q, Ind. motor Ind. motor Synchr. motor lOO kW 0.78 PF lagging 200 kW 0.8 PF lagging 150 kW PF = ? L ________________ ~ A simple power system consisting of an infinite bus supplying an industrial plant through a transmission line. Solutioll (a) In the first case, the real power of load I is 100 kW, and the reactive power of load I is Ql = Pt tan () = (1 00 kW) tan (cos-l 0.7S) = (100 kW) tan 3S.7° = SO.2 kVAR The real power of load 2 is 200 kW, and the reactive power of load 2 is Q2= P2tan() = (200 kW) tan (cos-l O.SO) = (200 kW) tan 36.S7° = 150 kVAR The real power load 3 is 150 kW. and the reactive power of load 3 is Q]= p] tan() = (150 kW) tan (cos-l 0.S5) = (150 kW) tan 3 1.So = 93 kVAR Thus, the total real load is P'o< = Pt + P2 + p] = lOO kW + 200 kW + 150 kW = 450 kW and the total reactive load is Q,o<=Qt+ Q2+Q] = SO.2 kVAR + 150 kVAR + 93 kVAR = 323.2 kVAR The equivalent system power factor is thus PF = cos (J = cos (tan- I .2)= cos (tan- l 323.2 kVAR) P 450 kW = cos 35.7° = 0.Sl2 lagging
- 386. 362 ELECTRIC MACHINERY RJNDAMENTALS Finally, the line current is given by PtrX 450 kW IL = v'JVLcos 0 = v'J(480V)(0.812) = 667 A (b) The real and reactive powers of loads I and 2 are unchanged, as is the real power of load 3. The reactive power of load 3 is Q3 = PJ tan() = (150 kW) tan (_cos-l 0.85) = (150 kW) tan (_31.8°) = -93 kVAR Thus, the total real load is P'fJI = PI + P2 + PJ = lOOkW + 200kW + 150kW = 450kW and the total reactive load is Q'fJI= QI+Q2+ Q3 = 80.2 kVAR + 150 kVAR - 93 kVAR = 137.2 kVAR The equivalent system power factor is thus PF = cos O= cos (tan- l Q) = cos (tan- l 137.2kVAR) P 450kW = cos 16.96° = 0.957 lagging Finally, the line current is given by P'fJI 450 kW IL = V3VL cos 0 = v'3(480 VXO.957) = 566 A (e) The transmission losses in the first case are The transmission losses in the second case are Notice that in the second case the transmission power losses are 28 percent less than in the first case, while the power supplied to the loads is the same. As seen in the preceding example, the abi lity to adjust the power factor of one or more loads in a power system can significantly affect the operating effi- ciency of the power system. TIle lower the power factor of a system, the greater the losses in the power lines feeding it. Most loads on a typical power system are induction motors, so power syste ms are almost invariably lagging in power factor. Having one or more leading loads (overexcited synchronous motors) on the sys- tem can be useful for the following reasons: I. A leading load can supply some reactive power Q for nearby lagging loads, instead of it coming from the generator. Since the reactive power does not have to travel over the long and fairly high-resistance transmission lines, the
- 387. SYNCHRONOUS MOTORS 363 transmission line current is reduced and the power system losses are much lower. (nlis was shown by the previous example.) 2. Since the transmission lines carry less current, they can be smaller for a given rated power flow. A lower equipment current rating reduces the cost of a power system significantly. 3. In addition, requiring a synchronous motor to operate with a leading power factor means that the motor must be run overexcited. nlis mode of operation increases the motor's maximum torque and reduces the chance of acciden- tally exceeding the pullout torque. The use of synchronous motors or other equipment to increase the overall power factor of a power system is called power-factor correction. Since a syn- chronous motor can provide power-factor correction and lower power system costs, many loads that can accept a constant-speed motor (even though they do not necessarily need one) are driven by synchronous motors. Even though a synchro- nous motor may cost more than an induction motor on an individual basis, the ability to operate a synchronous motor at leading power factors for power-factor correction saves money for industrial plants. This results in the purchase and use of synchronous motors. Any synchronous motor that exists in a plant is run overexcited as a matter of course to achieve power-factor correction and to increase its pullout torque. However, running a synchronous motor overexcited requires a high field current and flux, which causes significant rotor heating. An operator must be careful not to overheat the field windings by exceeding the rated field current. The Synchronolls Capacitor or Synchronous Condenser A synchronous motor purchased to drive a load can be operated overexcited to supply reactive power Q for a power system. In fact, at some times in the past a synchronous motor was purchased and run without a load, simply for power- factor correction. nle phasor diagram of a synchronous motor operating overex- cited at no load is shown in Figure 6- 14. Since there is no power being drawn from the motor, the distances propor- tional to power (Ell sin /j and III cos () ) are zero. Since the Kirchhoff's voltage law equation for a synchronous motor is I, I (6-1) ""GURE 6- 14 The phasor diagram of a synchronous Cllpacitor or synchronous condenser.
- 388. 364 ELECTRIC MACHINERY RJNDAMENTALS Lagging PF (+ Q consumed) Saturation Leading PF (+ QsuppIied) L-________~_________ ~ (a) ""GURE 6-15 ,b, (a) The V curve of a synchronous capacitor. (b) The corresponding machine phasor diagram. the quantity jXSIA. points to the left, and therefore the armature current IA. points straight up. If V4> and IA. are examined, the voltage-current relationship between them looks like that of a capacitor. An overexcited synchronous motor at no load looks just like a large capacitor to the power system. Some synchronous motors used to be sold specifically for power-factor cor- rection. 1llese machines had shafts that did not even come through the frame of the motor- no load could be connected to them even if one wanted to do so. Such special-purpose synchronous motors were often called synchronous condensers or synchronous capacitors. (Condenser is an old name for capacitor.) 1lle V curve for a synchronous capacitor is shown in Figure 6-15a. Since the real power supplied to the machine is zero (except for losses), at unity power factor the current fA. = D. As the field current is increased above that point, the line current (and the reactive power supplied by the motor) increases in a nearly linear fashion until saturation is reached. Figure 6-I5b shows the effect of increasing the field current on the motor's phasor diagram. Today, conventional static capacitors are more economical to buy and use than synchronous capacitors. However, some synchronous capacitors may still be in use in older industrial plants. 6.3 STARTING SYNCHRONOUS MOTORS Section 6.2 explained the behavior of a synchronous motor under steady-state conditions. In that section, the motor was always assumed to be initially turning at synchronous speed. What has not yet been considered is the question: How did the motor get to synchronous speed in the first place? To understand the nature of the starting problem, refer to Figure 6- I6. nlis figure shows a 6D-Hz synchronous motor at the moment power is applied to its stator windings. The rotor of the motor is stationary, and therefore the magnetic
- 389. D , B, f=O S 'find = 0 ,,' FIGURE 6-16 SYNCHRONOUS MOTORS 365 D , D , • f=I1240s w 'find = Counterclockwise '{.u-/I= 111205 'find = 0 B, ,b, ,,' B, B, w 't-t-- B, w 1= 31240 5 1=I/60s 'f;nd = clockwise 'f;nd = 0 ,d, ,., Staning problems in a synchronous motor---the torque alternates rapidly in magnitude and direction. so that the net 5taning torque is zero. field DR is stationary. The stator magnetic field Ds is starting to sweep around the motor at synchronous speed. Figure 6-1 6a shows the machine at time t = 0 s, when DR and Ds are exactly lined up. By the induced-torque equation (4- 58) the induced torque on the shaft of the rotor is zero. Figure 6--16b shows the situa- tion at time t = 11240 s. In such a short time, the rotor has barely moved, but the stator magnetic field now points to the left. By the induced-torque equation, the torque on the shaft of the rotor is now counterclockwise. Figure 6-1 6c shows the situation at time t = 1/120 s. At that point DR and Ds point in opposite direc- tions, and TiDd again equals zero. At t = 1160 s, the stator magnetic field now points to the right, and the resulting torque is clockwise. Finally, at t = 1/60 s, the stator magnetic field is again lined up with the ro- tor magnetic field, and T iDd = O. During one electrical cycle, the torque was first counterclockwise and then clockwisc, and the average torque over the complete
- 390. 366 ELECTRIC MACHINERY RJNDAMENTALS cycle was zero. What happens to the motor is that it vibrates heavily with each electrical cycle and finally overheats. Such an approach to synchronous motor starting is hardly satisfactory- managers tend to frown on employees who burn up their expensive equipment. So just how can a synchronous motor be started? TIuee basic approaches can be used to safely start a synchronous motor: I. Reduce the speed ofthe stator mngneticfield to a low enough value that the rotor can accelerate and lock in with it during one half-cycle of the magnetic field 's rotation. This can be done by reducing the frequency of the applied electric power. 2. Use an extenwl prime mover to accelerate the synchronous motor up to syn- chronous speed, go through the paralleling procedure, and bring the machine on the line as a generator. TIlen, turning off or disconnecting the prime mover wil I make the synchronous machine a motor. 3. Use damper windings or amortisseur windings. The function of damper windings and their use in motor starting will be explained below. Each of these approaches to synchronous motor starting will be described in turn. Motor Starting by Reducing Electrical Frequency If the stator magnetic fields in a synchronous motor rotate at a low enough speed, there will be no problem for the rotor to accelerate and to lock in with the stator magnetic field. TIle speed of the stator magnetic fields can then be increased to operating speed by gradually increasingf.. up to its normal 50- or 6O-Hz value. TIlis approach to starting synchronous motors makes a lot of sense, but it does have one big problem: Where does the variable electrical frequency corne from? Regular power systems are very carefully regulated at 50 or 60 Hz, so un- til recently any variable-frequency voltage source had to come from a dedicated generator. Such a situation was obviously impractical except for very unusual circumstances. Today, things are different. Chapter 3 described the rectifier-inverter and the cycloconverter, which can be used to convert a constant input frequency to any de- sired output frequency. With the development of such modern solid-state variable- frequency drive packages, it is perfectly possible to continuously control the elec- trical frequency applied to the motor all the way from a fraction of a hertz up to and above full rated frequency. If such a variable-frequency drive unit is included in a motor-control circuit to achieve speed control, then starting the synchronous motor is very easy- simply adjust the frequency to a very low value for starting, and then raise it up to the desired operating frequency for normal running. When a synchronous motor is operated at a speed lower than the rated speed, its internal generated voltage Ell = Kcpw will be smaller than normal. If Ell is reduced in magnitude, then the terminal voltage applied to the motor must be
- 391. SYNCHRONOUS MOTORS 367 reduced as well in order to keep the stator current at safe levels. The voltage in any variable-frequency drive or variable-frequency starter circuit must vary roughly linearly with the applied frequency. To learn more about such solid-state motor-drive units, refer to Chapter 3 and Reference 9. Motor Starting with an External Prime Mover The second approach to starting a synchronous motor is to attach an external start- ing motor to it and bring the synchronous machine up to full speed with the ex- ternal motor. 1l1en the synchronous machine can be paralleled with its power sys- tem as a generator, and the starting motor can be detached from the shaft of the machine. Once the starting motor is turned off, the shaft of the machine slows down, the rotor magnetic field BR falls behind B..." and the synchronous machine starts to act as a motor. Once paralleling is completed, the synchronous motor can be loaded down in an ordinary fashion. This whole procedure is not as preposterous as it sounds, since many syn- chronous motors are parts of motor-generator sets, and the synchronous machine in the motor-generator set may be started with the other machine serving as the starting motor. Also, the starting motor only needs to overcome the inertia of the synchronous machine without a load- no load is attached until the motor is par- alleled to the power system. Since only the motor's inertia must be overcome, the starting motor can have a much smaller rating than the synchronous motor it starts. Since most large synchronous motors have brushless excitation systems mounted on their shafts, it is often possible to use these exciters as starting motors. For many medium-size to large synchronous motors, an external starting motor or starting by using the exciter may be the only possible solution, because the power systems they are tied to may not be able to handle the starting currents needed to use the amortisseur winding approach described next. Motor Starting by Using Amortisseur Windings By far the most popular way to start a synchronous motor is to employ anwrtisseur or damper windings. Amortisseur windings are special bars laid into notches carved in the face of a synchronous motor's rotor and then shorted out on each end by a large shoT1ing ring. A pole face with a set of amortisseurwindings is shown in Figure 6-17, and amortisseur windings are visible in Figures 5- 2 and 5-4. To understand what a set of amortisseur windings does in a synchronous motor, examine the stylized salient two-pole rotor shown in Figure 6- 18. This ro- tor shows an amortisseur winding with the shorting bars on the ends ofthe two ro- tor pole faces connected by wires. (This is not quite the way nonnal machines are constructed, but it will serve beautifully to illustrate the point of the windings.) Assume initially that the main rotorfield winding is disconnected and that a three-phase set of voltages is applied to the stator of this machine. When the
- 392. 368 ELECTRIC MACHINERY RJNDAMENTALS o o Shorting "'" Shorting "'" o o o o FIGURE 6-17 A rotor field pole for a synchronous machine showing amortisseur windings in the pole face. (Courtesy ofGeneml Electric Company.) fo'IGURE 6- 18 A simplified diagram of a salient two- pole machine showing amortisseur windings. power is first applied at time t = as, assume that the magnetic field Bs is vertical, as shown in Figure 6- 19a. As the magnetic field Bs sweeps along in a counter- clockwise direction, it induces a voltage in the bars of the amortisseur winding given by Equation (1-45): where ei!>d = (v x B) • I v = velocity of the bar relative to the magnetic field B = magnetic nux density vector I = length of conductor in the magnetic field ( 1-45)
- 393. eind and i out of page ®® w 1-+'-- 8. 00 00 00 eind and i into page I find = counterclockwise Shorting b= (a) 1=05 eind and i into page 00 00 00 '" I find = counterclockwise ".• • d . eind an J ® ® out of page (c) 1= 11120& FIGURE 6-19 SYNCHRONOUS MOTORS 369 "0 0" (b) 1=112405 0 0 00 +t)--u, I find=O 0 0 0 0 (d) 1=312405 The development of a unidirectional torque with synchronous motor amonisseur windings. The bars at the top of the rotor are moving to the right relative to the magnetic field, so the resulting direction of the induced voltage is out of the page. Similarly, the induced voltage is into the page in the bottom bars. These voltages produce a current fl ow out of the top bars and into the bottom bars, resulting in a winding magnetic field Bwpointing to the right. By the induced-torque equation
- 394. 370 ELECTRIC MACHINERY RJNDAMENTALS the resulting torque on the bars (and the rotor) is counterclockwise. Figure 6-1 9b shows the situation at t = 11240 s. Here, the stator magnetic field has rotated 90° while the rotor has barely moved (it simply cannot speed up in so short a time). At this point, the voltage induced in the amortisseur windings is zero, because v is parallel to B. With no induced voltage, there is no current in the windings, and the induced torque is zero. Figure 6-1 9c shows the situation at t = 11120 s. Now the stator magnetic field has rotated 900, and the rotor still has not moved yet. TIle induced voltage [given by Equation (1-45)] in the amortisseur windings is out of the page in the bottom bars and into the page in the top bars. The resulting current flow is out of the page in the bottom bars and into the page in the top bars, causing a magnetic field Bwto point to the left.1lle resulting induced torque, given by T;Dd = k Bw x Bs is counterclockwise. Finally, Figure 6-1 9d shows the situation at time t = 31240 s. Here, as at t = 11240 s, the induced torque is zero. Notice that sometimes the torque is counterclockwise and sometimes it is essentially zero, but it is always unidirectional. Since there is a net torque in a sin- gle direction, the motor's rotor speeds up. (1llis is entirely different from starting a synchronous motor with its normal field current, since in that case torque is first clockwise and then counterclockwise, averaging out to zero. In this case, torque is always in the same direction, so there is a nonzero average torque.) Although the motor's rotor will speed up, it can never quite reach synchro- nous speed. This is easy to understand. Suppose that a rotor is turning at synchro- nous speed. Then the speed of the stator magnetic field Bs is the same as the ro- tor's speed, and there is no relative motion between Bs and the rotor. If there is no relative motion, the induced voltage in the windings will be zero, the resulting current fl ow will be zero, and the winding magnetic field will be zero. Therefore, there will be no torque on the rotor to keep it turning. Even though a rotor cannot speed up all the way to synchronous speed, it can get close. It gets close enough to n'YD< that the regular field current can be turned on, and the rotor will pull into step with the stator magnetic fields. In a real machine, the field windings are not open-circuited during the start- ing procedure. If the field windings were open-circuited, then very high voltages would be produced in them during starting. If the field winding is short-circuited during starting, no dangerous voltages are produced, and the induced field current actually contributes extra starting torque to the motor. To summarize, if a machine has amortisseur windings, it can be started by the following procedure: I. Disconnect the field windings from their dc power source and short them out.
- 395. SYNCHRONOUS MOTORS 371 2. Apply a three-phase voltage to the stator of the motor, and let the rotor accel- erate up to near-synchronous speed. The motor should have no load on its shaft , so that its speed can approach n.ync as closely as possible. 3. Connect the dc field circuit to its power source. After this is done, the motor will lock into step at synchronous speed, and loads may then be added to its shaft. The Effect of Amortisseur Windings on Motor Stability If amortisseur windings are added to a synchronous machine for starting, we get a free bonus-an increase in machine stability. The stator magnetic field rotates at a constant speed n.YD<, which varies only when the system frequency varies. If the rotor turns at n,YD<, then the amortisseur windings have no induced voltage at all. If the rotor turns slower than n,YD<, then there will be relative motion between the rotor and the stator magnetic field and a voltage will be induced in the windings. nlis voltage produces a current fl ow, and the current fl ow produces a magnetic field. The interaction of the two magnetic fields produces a torque that tends to speed the machine up again. On the other hand, if the rotor turns faster than the stator magnetic field, a torque will be produced that tries to slow the rotor down. Thus, the torque produced by the anwrtisseur windings speeds up slow mnchines and slows down fast machines. These windings therefore tend to dampen out the load or other transients on the machine. It is for this reason that amortisseur windings are also called damper windings. Amortisseur windings are also used on synchronous generators, where they serve a similar stabilizing function when a generator is operating in parallel with other generators on an infinite bus. If a variation in shaft torque occurs on the generator, its rotor will momentarily speed up or slow down, and these changes will be opposed by the amortisseur windings. Amortisseur windings improve the overall stability of power systems by reducing the magnitude of power and torque transients. Amortisseur windings are responsible for most of the subtransient current in a faulted synchronous machine. A short circuit at the terminals of a generator is just another fonn of transient, and the amortisseur windings respond very quickly to it. 6.4 SYNCHRONOUS GENERATORS AND SYNCHRONOUS MOTORS A synchronous generator is a synchronous machine that converts mechanical power to electric power, while a synchronous motor is a synchronous machine that converts electric power to mechanical power. In fact, they are both the same physical machine.
- 396. 372 ELECTRIC MACHINERY RJNDAMENTALS Supply Consume reactive power E" cos {j > V6 reactive power E" cos {j < V. Q Q Supply pow~ P E, E, " ~. ~V • , , o • ~ I, E" leads V. Consume pow~ p I, " ~ ~V' ~ E , E" Jags E, V , ""GURE 6-10 Phasor diagrams showing the generation and consumption of real power P and reactive power Q by synchronous generators and motors. A synchronous machine can supply real power to or consume real power from a power system and can supply reactive power to or consume reactive power from a power system. All four combinations of real and reactive power flows are possible, and Figure 6-20 shows the phasor diagrams for these conditions. Notice from the figure that I. The distinguishing characteristic of a synchronous generator (supplying P) is that E" lies ahead o/V", while for a motor E" lies behind V",. 2. The distinguishing characteristic of a machine supplying reactive power Q is that E" cos lj > V", regardless of whether the machine is acting as a generator or as a motor. A machine that is consuming reactive power Q has E" cos lj < V",. 6.5 SYNCHRONOUS MOTOR RATINGS Since synchronous motors are the same physical machines as synchronous genera- tors, the basic machine ratings are the same.The one major difference is that a large
- 397. SYNCHRONOUS MOTORS 373 '" GENERAL@ ELECTRIC " SYNCHRONOUS MOTOR FIGURE 6-21 A typical nameplate for a large synchronous motor. (Courtesy o!General Electric Company.) Ell gives a leading power factor instead of a lagging one, and therefore the effect of the maximum field current limit is expressed as a rating at a leading power factor. Also, since the output of a synchronous motor is mechanical power, a synchronous motor's power rating is usually given in horsepower rather than kilowatts. TIle nameplate of a large synchronous motor is shown in Figure 6-21. In addition to the information shown in the figure, a smaller synchronous motor would have a service factor on its nameplate. In general, synchronous motors are more adaptable to low-speed, high- power applications than induction motors (see Chapter 7). They are therefore commonly used for low-speed, high-power loads. 6.6 SUMMARY A synchronous motor is the same physical machine as a synchronous generator, except that the direction of real power fl ow is reversed. Since synchronous motors are usually connected to power systems containing generators much larger than the motors, the frequency and tenninal voltage of a synchronous motor are fixed (i.e., the power system looks like an infinite bus to the motor). The speed of a synchronous motor is constant from no load to the maximum possible load on the motor. The speed of rotation is _ _ 120..r.: nm - n sync - p The maximum possible power a machine can produce is _ 3V1>EA Pm:u.- X, (5- 21)
- 398. 374 ELECTRIC MACHINERY RJNDAMENTALS If this value is exceeded, the rotor will not be able to stay locked in with the sta- tor magnetic fields, and the motor will slip poles. If the field current of a synchronous motor is varied while its shaft load re- mains constant, then the reactive power supplied or consumed by the motor will vary. If Ell cos 8 > ~, the motor will suppl y reactive power, while if Ell cos 8< Vo/» the motor will consume reactive power. A synchronous motor has no net starting torque and so cannot start by itself. TIlere are three main ways to start a synchronous motor: I. Reduce the stator frequency to a safe starting level. 2. Use an external prime mover. 3. Put amortisseur or damper windings on the motor to accelerate it to near- synchronous speed before a direct current is applied to the field windings. If damper windings are present on a motor, they will also increase the sta- bility of the motor during load transients. QUESTIONS 6-1. What is the difference between a synchronous motor and a synchronous generator? 6-2. What is the speed regulation of a synchronous motor? 6-3. When would a synchronous motor be used even though its constant-speed charac- teristic was not needed? 6-4. Why can't a synchronous motor start by itself? 6-5. What techniques are available to start a synchronous motor? 6-6. What are amortisseur windings?Why is the torque produced by them unidirectional at starting, while the torque produced by the main field winding alternates direction? 6-7. What is a synchronous capacitor? Why would one be used? 6-8. Explain, using phasor diagrams, what happens to a synchronous motor as its field current is varied. Derive a synchronous motor V curve from the phasor diagram. 6-9. Is a synchronous motor's field circuit in more danger of overheating when it is op- erating at a leading or at a lagging power factor? Explain, using phasor diagrams. 6-10. A synchronous motor is operating at a fixed real load, and its field current is in- creased. If the armature current falls, was the motor initially operating at a lagging or a leading power factor? 6-11. Why must the voltage applied to a synchronous motor be derated for operation at frequencies lower than the rated value? PROBLEMS 6-1. A 480-V, 60 Hz four-pole synchronous motor draws 50 A from the line at unity power factor and full load. Assuming that the motor is lossless, answer the follow- ing questions: (a) What is the output torque of this motor? Express the answer both in newton- meters and in pound-feet.
- 399. SYNCHRONOUS MOTORS 375 (b) What must be done to change the power factor to 0.8 leading? Explain your an- swer, using phasor diagrams. (c) What will the magnitude of the line current be if the power factor is adjusted to 0.8 leading? 6-2. A 480-V, 60 Hz 4OO-hp, 0.8-PF-Ieading, six-pole, ~-connected synchronous motor has a synchronous reactance of 1.1 {} and negligible annature resistance. Ignore its friction, windage, and core losses for the purposes of this problem. (a) If this motor is initially supplying 400 hp at 0.8 PF lagging, what are the mag- nitudes and angles of EA and IA? (b) How much torque is this motor producing? What is the torque angle O? How near is this value to the maximum possible induced torque of the motor for this field current setting? (c) If lEAl is increased by IS percent, what is the new magnitude of the armature current? What is the motor's new power factor? (d) Calculate and plot the motor's V curve for this load condition. 6-3. A 2300-V, I()(X)-hp, 0.8-PF leading, 60-Hz, two-pole, Y-cotulected synchronous mo- tor has a synchronous reactance of2.8 n and an annature resistance of0.4 n. At 60 Hz, its friction and windage losses are 24 kW, and its core losses are 18 kW. The field circuit has a dc voltage of2oo V, and the maximwn IF is 10 A. The open-circuit characteristic of this motor is shown in Figure P6-1. Answer the following ques- tions about the motor, assuming that it is being supplied by an infinite bus. (a) How much field current would be required to make this machine operate at tulity power factor when supplying full load? (b) What is the motor's efficiency at full load and unity power factor? (c) If the field current were increased by 5 percent, what would the new value of the annature current be? What would the new power factor be? How much re- active power is being consumed or supplied by the motor? (d) What is the maximrun torque this machine is theoretically capable of supplying at tulity power factor? At 0.8 PF leading? 6-4. Plot the V curves (fA versus IF) for the synchronous motor of Problem 6- 3 at no- load, half-load, and full-load conditions. (Note that an electronic version of the open-circuit characteristics in Figure P6-1 is available at the book's website. It may simplify the calculations required by this problem. Also, you may assrune that RA is negligible for this calculation.) 6-5. If a 60-Hz synchronous motor is to be operated at 50 Hz, will its synchronous reac- tance be the same as at 60 Hz, or will it change? (Hint: Think about the derivation of Xs.) 6-6. A 480-V, lOO-kW, 0.85-PF-Ieading, 50-Hz, six-pole, V-connected synchronous mo- tor has a synchronous reactance of 1.5 n and a negligible annature resistance. The rotational losses are also to be ignored. This motor is to be operated over a continu- ous range of speeds from 300 to 1000 rlmin, where the speed changes are to be ac- complished by controlling the system frequency with a solid-state drive. (a) Over what range must the input frequency be varied to provide this speed con- trol range? (b) How large is EA at the motor's rated conditions? (c) What is the maximwn power that the motor can produce at rated speed with the EA calculated in part (b)? (d) What is the largest EA could be at 300 r/min?
- 400. 376 ELECTRIC MACHINERY RJNDAMENTALS 3(XX) 2750 2500 2250 2(XX) 1750 1500 1250 IlXXl 750 250 / / o 0.0 ""GURE 1'(;-1 / / / 1.0 2.0 /" / / / / 3.0 4.0 5.0 6.0 Field current. A The open-circuit characteristic for the motor in Problems 6-3 and 6-4. 7.0 8.0 9.0 10.0 (e) Assuming that the applied voltage V. is derated by the same amOlUlt as EA. what is the maximwn power the motor could supply at 300 r/min? if) How does the power capability of a synchronous motor relate to its speed? 6-7. A 20S-V. Y-connected synchronous motor is drawing 40 A at unity power factor from a 20S-V power system. The field current flowing under these conditions is 2.7 A. Its synchronous reactance is O.S n. Assume a linear open-circuit characteristic. (a) Find the torque angle o. (b) How much field current would be required to make the motor operate at O.S PF leading? (c) What is the new torque angle in part b? 6-8. A synchronous machine has a synchronous reactance of 2.0 n per phase and an ar- mature resistance of 0.4 n per phase. If EA = 460 L -80 V and V. = 4S0 L 0° V, is this machine a motor or a generator? How much power P is this machine consum- ing from or supplying to the electrical system? How much reactive power Q is this machine consuming from or supplying to the electrical system?
- 401. SYNCHRONOUS MOTORS 377 6-9. Figure P6--2 shows a synchronous motor phasor diagram for a motor operating at a leading power factor with no R". For this motor. the torque angle is given by ~X~,J ~'!;; CO~'~''-co tan 0 = -;- VI/! + Xsl" sin () _, ( Xsl" cos () ) • - tan - V I/! + Xi " sin () Derive an equation for the torque angle of the synchronous motor ifthe amlature re- sistance is included. ,,,, FIGURE P6-2 ,,,, . , Xsl" Sin 0 V . ,"---" --'1 , jXsl" 0: Xsl" cos () , Phasor diagram of a motor at a Jeading power factor. 6-10. A 4S0-V, 375-kVA, O.S-PF-Iagging, V-connected synchronous generator has a syn- chronous reactance of 0.4 n and a negligible armature resistance. This generator is supplying power to a 4S0-V, SO-kW, 0 .8-PF-Ieading, V-connected synchronous mo- tor with a synchronous reactance of 1.1 n and a negligible annature resistance. The synchronous generator is adjusted to have a terminal voltage of 480 V when the mo- tor is drawing the rated power at unit y power factor. (a) Calculate the magnitudes and angles of E" for both machines. (b) If the flux of the motor is increased by 10 percent, what happens to the tenni- nal voltage of the power system? What is its new value? (c) What is the power factor of the motor after the increase in motor flux ? 6- 11, A 4S0-V, lOO-kW, 50-Hz, four-pole, V-connected synchronous motor has a rated power factor of 0.S5 Ieading. At full load, the efficiency is 9 1 percent. The annature resistance is O.OS n, and the synchronous reactance is 1.0 n. Find the following quantities for this machine when it is operating at full load: (a) Output torque (b) Input power (c) n.. (d) E" (e) 1 1,,1 if) P coov (g) Pmocb + P core + P""'Y
- 402. 378 ELECTRIC MACHINERY RJNDAMENTALS 6-12. The V-connected synchronous motor whose nameplate is shown in Figure 6-21 has a per-unit synchronous reactance of 0.9 and a per-unit resistance of 0.02. (a) What is the rated input power of this motor? (b) What is the magnitude of EA at rated conditions? (c) If the input power of this motor is 10 MW. what is the maximum reactive power the motor can simultaneously supply? Is it the annature current or the field current that limits the reactive power output? (d) How much power does the field circuit consume at the rated conditions? (e) What is the efficiency of this motor at full load? if) What is the output torque of the motor at the rated conditions? Express the an- swer both in newton-meters and in pound-feet. 6-13. A 440-V. three-phase. V-connected synchronous motor has a synchronous reactance of 1.5 n per phase. The field ClUTent has been adjusted so that the torque angle 0 is 28° when the power supplied by the generator is 90 kW. (a) What is the magnitude of the internal generated voltage EA in this machine? (b) What are the magnitude and angle of the armature current in the machine? What is the motor 's power factor? (c) If the field current remains constant. what is the absolute maximum power this motor could supply? 6-14. A 460-V, 200-kVA. 0.80-PF-Ieading. 400-Hz. six-pole. V-connected synchronous motor has negligible armature resistance and a synchronous reactance of 0.50 per unit. Ignore all losses. (a) What is the speed of rotation of this motor? (b) What is the output torque of this motor at the rated conditions? (c) What is the internal generated voltage of this motor at the rated conditions? (d) With the field ClUTent remaining at the value present in the motor in part c. what is the maximwn possible output power from the machine? 6-15. A lOO-hp. 440-V. 0.8-PF-Ieading. 6.-cormected synchronous motor has an annature resistance of 0.22 n and a synchronous reactance of 3.0 O. Its efficiency at full load is 89 percent. (a) What is the input power to the motor at rated conditions? (b) What is the line ClUTent of the motor at rated conditions? What is the phase cur- rent of the motor at rated conditions? (c) What is the reactive power consumed by or supplied by the motor at rated conditions? (d) What is the internal generated voltage EA of this motor at rated conditions? (e) What are the stator copper losses in the motor at rated conditions? if) What is POOIIV at rated conditions? (g) If EA is decreased by 10 percent. how much reactive power will be consumed by or supplied by the motor? 6-16. Answer the following questions about the machine of Problem 6-1 5. (a) If EA = 430 L 13.5° V and V. = 440 L 0° V. is this machine consuming real power from or supplying real power to the power system? Is it consuming re- active power from or supplying reactive power to the power system? (b) Calculate the real power P and reactive power Q supplied or consumed by the machine under the conditions in part a. Is the machine operating within its rat- ings nnder these circumstances?
- 403. SYNCHRONOUS MOTORS 379 (c) If E, = 470 L _1 20 V and V. = 440 L 00 V, is this machine consuming real power from or supplying real power to the power system? Is it consuming re- active power from or supplying reactive power to the power system? (d) Calculate the real power P and reactive power Q supplied or consruned by the machine WIder the conditions in part c. Is the machine operating within its rat- ings under these circumstances? REFERENCES 1. Chaston. A. N. Electric Machinery. Reston. Va.: Reston Publishing. 1986. 2. Del Toro. V. Electric Machines atuf Pov.·er Systems. Englewood Cliffs. NJ : Prentice-Hall. 1985. 3. Fitzgerald. A. E.. and C. Kingsley. Jr. Electric Machinery. New York: McGraw-HilL 1952. 4. Fitzgerald. A. E.. C. Kingsley. Jr.. and S. D. Umans. Electric Machinery, 5th ed. New York: McGraw-Hill. 1990. 5. Kosow. lrving L. Control ofElectric Motors. Englewood Cliffs. N.J.: Prentice-Hall. 1972. 6. Liwschitz..Garik. MichaeL and Clyde Whipple. Alternating-Current Machinery. Princeton. N.J.: Van Nostrand. 1961. 7. Nasar. Syed A. (ed.). Handbook ofElectric Machines. New York: McGraw-Hill. 1987. 8. Siemon. G. R., and A. Straughen. Electric Machines. Reading. Mass.: Addison-Wesley. 1980. 9. Vithayathil. Joseph. PO'we, Electronics: Principles and Applications. New YorK: McGraw-Hill. 1995. 10. Werninck. E. H. (ed.). Electric MOlOr Handaook. London: McGraw-Hill. 1978.
- 404. CHAPTER 7 INDUCTION MOTORS In the last chapter, we saw how amortisseur windings on a synchronous motor cauId develop a starting torque without the necessity of supplyi ng an external field current to them. In fact, amortisscur windings work so well that a motor could be built without the synchronous motor's main de field circuit at all. A ma- chine with only amortisseur windings is called an induction machine. Such ma- chines are called induction machines because the rotor voltage (which produces the rotor current and the rotor magnetic field) is induced in the rotor windings rather than being physically connected by wires. The distinguishing feature of an induction motor is that no de field current is required to run the machine. Although it is possible to use an induction machine as either a motor or a generator, it has many disadvantages as a generator and so is rarely used in that manner. For this reason, induction machines are usually referred to as induction motors. 7.1 INDUCTION MOTOR CONSTRUCTION An induction motor has the same physical stator as a synchronous machine, with a different rotor construction. A typical two-pole stator is shown in Figure 7-1. It looks (and is) the same as a synchronous machine stator. TIlere are two different types of induction motor rotors which can be placed inside the stator. One is called a cage rotor, while the other is called a wound rotor. Figures 7- 2 and 7- 3 show cage induction motor rotors. A cage induction motor rotor consists ofa series of conducting bars laid into slots carved in the face of the rotor and shorted at either end by large shoT1ing rings. This design is re- ferred to as a cage rotor because the conductors, if examined by themselves, would look like one of the exercise wheels that squirrels or hamsters run on. 380
- 405. ,ore FIGURE 7-2 Condoctor rings Rotor rotor conductors INDUCTION MOTORS 381 FIGURE 7-1 The stator of a typical induction motor. showing the stator windings. (Courlesy of MagneTek, Inc.) ,., ,b , (a) Sketch of cage rotor. (b) A typical cage rotor. (Courtesy ofGeneml Electric Company.)
- 406. 382 ELECTRIC MACHINERY RJNDAMENTALS ,,' ,b, ""GURE 7-3 (a) Cutaway diagram of a typical small cage rotor induction motor. (Courtesy ofMa8neTek. Inc.) (b) Cutaway diagram of a typical large cage TOIor induction motor. (Counesy ofGeneml Electric Company.) TIle other type of rotor is a wound rotor. A wound rotor has a complete set of three-phase windings that are mirror images of the windings on the stator. The three phases of the rotor windings are usually V-connected, and the ends of the three rotor wires are tied to slip rings on the rotor's shaft. TIle rotor windings are shorted through brushes riding on the slip rings. Wound-rotor induction motors therefore have their rotor currents accessible at the stator brushes, where they can be examined and where extra resistance can be inserted into the rotor circuit. It is possible to take advantage of this feature to modify the torque- speed characteris- tic of the motor. Two wound rotors are shown in Figure 7-4, and a complete wound-rotor induction motor is shown in Figure 7- 5.
- 407. INDUCTION MOTORS 383 (a) ,b, FIGURE 7-4 Typical wound rotors for induction motors. Notice the slip rings and the bars connecting the rotor windings to the slip rings. (Courtel)' ofGeneml Electric Company.) FIGURE 7-5 Cuta.way diagram of a. wound-rotor induction motor. Notice the brushes and slip rings. Also notice that the rotor windings are skewed to eliminate slot h3.ITllonics. (Courtesy ofMagneTe/:. Inc.)
- 408. 384 ELECTRIC MACHINERY RJNDAMENTALS Wou nd-rotor induction motors are more expensive than cage induction mo- tors, and they require much more maintenance because of the wear associated with their brushes and slip rings. As a result, wound-rotor induction motors are rarely used. 7.2 BASIC INDUCTION MOTOR CONCEPTS Induction motor operation is basically the srune as that of amortisseur windings on synchronous motors. That basic operation will now be reviewed, and some im- portant induction motor tenns will be defined. The Development of Induced Torque in an Induction Motor Figure 7--6 shows a cage rotor induction motor. A three-phase set of voltages has been applied to the stator, and a three-phase set of stator currents is flowing. These currents produce a magnetic field Bs, which is rotating in a counterclockwise direction.1lle speed of the magnetic field's rotation is given by (7-1 ) where Ie is the system frequency in hertz and P is the number of poles in the ma- chine. This rotating magnetic field Bs passes over the rotor bars and induces a voltage in them. 1lle voltage induced in a given rotor bar is given by the equation eioo = (v x H) • I where v = velocity of the bar relative to the magneticfield B = magnetic flux density vector I = length of conductor in the magnetic field ( 1-45) It is the relative motion of the rotor compared to the stator magnetic field that produces induced voltage in a rotor bar. The velocity of the upper rotor bars relative to the magnetic field is to the right, so the induced voltage in the upper bars is out of the page, while the induced voltage in the lower bars is into the page. nlis results in a current flow out of the upper bars and into the lower bars. How- ever, since the rotor assembly is inductive, the peak rotor current lags behind the peak rotor voltage (see Figure 7--6b). The rotor current flow produces a rotor mag- netic field HR. Finally, since the induced torque in the machine is given by (4- 58) the resulting torque is counterclockwise. Since the rotor induced torque is coun- terclockwise, the rotor accelerates in that direction.
- 409. Maximum induced voltage ,,, @ @ @ 0 @ • 0 0 0 " " " " ,., II, Net voltage , I ER , , I, , , , , , @ 0 @ • , 0 , , , IIi ', 0 , , , , 0 , " " " ,,' 0 0 " 0 0 INDUCTION MOTORS 385 Maximum induced voltage Maximum , induced current ,, " IR H , , @ 0 @ 0 • , 0 , 0 , , 0 , " , , , 0 , " " " ,b, ""CURE 7-6 The development of induced torque in an induction motor. (a) The rotating stator field lis induces a voUage in the rotor bars; (b) the rotor voltage produces a rotor currem flow. which lags behind the voUage because of the inductance of the rotor; (c) the rotor currem produces a rotor magnetic field liN lagging 90° behind itself. and liN imeracts with II... to produce a counterclockwise torque in the machine. There is a fmite upper limit to the motor's speed, however. If the induction motor's rotor were turning at synchronous speed, then the rotor bars would be sta- tionary relative to the magneticfield and there would be no induced voltage. If eioo were equal to 0, then there would be no rotor current and no rotor magnetic field. With no rotor magnetic field, the induced torque would be zero, and the rotor would slow down as a result of friction losses. An induction motor can thus speed up to near-synchronous speed, but it can never exactly reach synchronous speed. Note that in nonnal operation both the rotor and stator mngnetic fields BR and Bs rotate together at synchronous speed n,yDC' while the rotor itselftums at a slower speed.
- 410. 386 ELECTRIC MACHINERY RJNDAMENTALS The Concept of Rotor Slip TIle voltage induced in a rotor bar of an induction motor depends on the speed of the rotor relative to the magneticfields. Since the behavior of an induction motor depends on the rotor's voltage and current, it is often more logical to talk about this relative speed. Two tenns are commonly used to define the relative motion of the rotor and the magnetic fields. One is slip speed, defined as the difference be- tween synchronous speed and rotor speed: where n.up = slip speed of the machine n,yDC = speed of the magnetic fields nm = mechanical shaft speed of motor (7- 2) TIle other tenn used to describe the relative motion is slip, which is the rela- tive speed expressed on a per-unit or a percentage basis. That is, slip is defined as "" s = ~ (x 100%) n sync (7- 3) " - n s = 'YDC m(x 100%) n.". (7-4) lllis equation can also be expressed in terms of angular velocity w (radians per second) as W - W S = sync m(x 100%) w~oc (7- 5) Notice that if the rotor turns at synchronous speed, s = 0, while if the rotor is sta- tionary, s = 1. All normal motor speeds fall somewhere between those two limits. It is possible to express the mechanical speed of the rotor shaft in tenns of synchronous speed and slip. Solving Equations (7-4) and (7- 5) for mechanical speed yields I nm = ( 1 - s)n,yDC I (7-6) "' I wm~ (I - s)w,ync I (7- 7) lllese equations are useful in the derivation of induction motor torque and power relationships. The Electrical Frequency on the Rotor An induction motor works by inducing voltages and currents in the rotor of the machine, and for that reason it has sometimes been called a rotating transformer. Like a transformer, the primary (stator) induces a voltage in the secondary (rotor),
- 411. INDUCTION MOTORS 387 but unlike a transfonner, the secondary frequency is not necessarily the same as the primary frequency. If the rotor of a motor is locked so that it cannot move, then the rotor will have the same frequency as the stator. On the other hand, if the rotor turns at syn- chronous speed, the frequency on the rotor will be zero. What will the rotor fre- quency be for any arbitrary rate of rotor rotation? At nm= 0 rlmin, the rotor frequencyfr = Jr, and the slip s = I. At nm = n,ync' the rotor frequencyfr = 0 Hz, and the slip s = O. For any speed in between, the ro- tor frequency is directly proportional to the difference between the speed of the mag- netic field n.ync and the speed of the rotor nm. Since the slip of the rotor is defined as (7-4) the rotor frequency can be expressed as (7-8) Several alternative fonns of this expression exist that are sometimes useful. One of the more common expressions is deri ved by substituting Equation (7-4) for the slip into Equation (7--8) and then substituting for n,ync in the denominator of the expression: But n,yDC = 120fr I P [from Equation (7- 1)], so Therefore, (7- 9) Example 7-1. A 20S-V, lO-hp, four-pole, 60-Hz, V-connected induction motor has a full-load slip of 5 percent. (a) What is the synchronous speed of this motor? (b) What is the rotor speed of this motor at the rated load? (c) What is the rotor frequency of this motor at the rated load? (d) What is the shaft torque of this motor at the rated load? Solution (a) The synchronous speed of this motor is _ 120f,. n,ync - - p - = 120(60 Hz) _ 4 poles - ISOOr/ min (7- 1)
- 412. 388 ELECTRIC MACHINERY RJNDAMENTALS (b) The rotor speed of the motor is given by n", = (I - s)n.yDC = (I - 0.95)(l800r/min) = 17lOr/min (c) The rotor frequency of this motor is given by Ir = s/e = (0.05)(60 Hz) = 3 Hz Alternatively, the frequency can be found from Equation (7-9): p /, = 120 (n,ync - nm) = lio(l800r/min - 17IOr/min) = 3 Hz (d) The shaft load torque is given by (10 hpX746 W/hp) = (l7IOr/min)(2'lTrad/rXI min/60s) = 41.7N o m The shaft load torque in English units is given by Equation (1- 17): 5252P (7--6) (7--8) (7-9) where 'Tis in pOlUld-feet, P is in horser.ower, and n.. is in revolutions per minute. Therefore, 5252(10 hp) Tload = 17lOr/min = 30.71b o ft 7.3 THE EQUIVALENT CIRCUIT OF AN INDUCTION MOTOR An induction motor relies for its operation on the induction of voltages and cur- rents in its rotor circuit from the stator circuit (transformer action). Because the in- duction of voltages and curre nts in the rotor circuit of an induction motor is es- sentially a transformer operation, the equivalent circuit of an induction motor will turn o ut to be very similar to the equivalent circuit of a transfonner. An induction motor is called a singly excited machine (as opposed to a doubly excited synchro- nous machine), since power is supplied to only the stator circuit. Because an in- duction motor does not have an independe nt field circ uit, its model will not con- tain an internal voltage source such as the internal generated voltage E,t in a synchronous machine. lt is possible to derive the equivalent circuit of an induction motor from a knowledge of transformers and from what we already know about the variation of rotor frequency with speed in induction motors. TIle induction motor model will be
- 413. v, INDUCTION MOTORS 389 I, R, I, I, - - - I. j + + Rc jXM E, - FIGURE 7-7 The transformer model or an induction motor. with rotor and stator connected by an ideal transformer of turns ratio a,/f" developed by starting with the transformer model in Chapter 2 and then deciding how to take the variable rotor frequency and other similar induction motor effects into account. The Transformer Model of an Induction Motor A transfonner per-phase equivalent circuit, representing the operation of an in- duction motor, is shown in Figure 7- 7. As in any transfonner, there is a certain re- sistance and self-inductance in the primary (stator) windings, which must be rep- resented in the equivalent circuit of the machine. The stator resistance will be called R1• and the stator leakage reactance will be called Xl. These two compo- nents appear right at the input to the machine model. Also, like any transformer with an iron core, the nux in the machine is re- lated to the integral of the applied voltage E l . TIle curve of magnetomotive force versus nux (magnetization curve) for this machine is compared to a similar curve for a power transfonner in Figure 7- 8. Notice that the slope of the induction mo- tor's magnetomotive force-nux curve is much shallower than the curve of a good transformer. TIlis is because there must be an air gap in an induction motor, which greatly increases the reluctance of the nux path and therefore reduces the coupling between primary and secondary windings. TIle higher reluctance caused by the air gap means that a higher magnetizing c urrent is required to obtain a given nux level. Therefore, the magnetizing reactance XM in the equivalent circuit will have a much smaller value (or the susceptance EM will have a much larger value) than it would in an ordinary transformer. The primary internal stator voltage El is coupled to the secondary EN by an ideal transformer with an effective turns ratio a.ff" The effective turns ratio a.ff is fairly easy to detennine for a wound-rotor motor- it is basically the ratio of the conductors per phase on the stator to the conductors per phase on the rotor, modi- fied by any pitch and distribution factor differences. It is rather difficult to see a.ff
- 414. 390 ELECTRIC MACHINERY RJNDAMENTALS Transformer ""GURE 7-8 ;.Wb Induction motor ~, A-turns The magnetization curve of an induction motor compared to that of a transformer, clearly in the cage of a case rotor motor because there are no distinct windings on the cage rotor. In either case, there is an effective turns ratio for the motor, 1lle voltage ER produced in the rotor in turn produces a current fl ow in the shorted rotor (or secondary) circuit of the machine, TIle primary impedances and the magnetiwtion current of the induction mo- tor are very similar to the corresponding components in a transformer equivalent circuit. An induction motor equivalent circuit differs from a transfonner equiva- lent circuit primarily in the effects of varying rotor frequency on the rotor voltage ER and the rotor impedances RR and jXR' The Rotor Circuit Model In an induction motor, when the voltage is applied to the stator windings, a volt- age is induced in the rotor windings of the machine, In general, the greater the relative motion between the rotor and the stator magnetic fields, the greater the resulting rotor voltage and rotor frequency, The largest relative motion occurs when the rotor is stationary, called the locked-rotor or blocked-rotor condition, so the largest voltage and rotor frequency arc induced in the rotor at that condition, TIle smallest voltage (0 V) and frequency (0 Hz) occur when the rotor moves at the same speed as the stator magnetic field, resulting in no relative motion, The magnitude and frequency of the voltage induced in the rotor at any speed between these extremes is directly propoT1ional to the slip ofthe rotor, Therefore, if the magnitude of the induced rotor voltage at locked-rotor conditions is called EIlQ, the magnitude of the induced voltage at any slip will be given by the equation (7-1 0)
- 415. INDUCTION MOTORS 391 + R, FlGURE 7-9 The rotor ci['(;uit model of an induction motor. and the frequency of the induced voltage at any slip will be given by the equation (7-8) This voltage is induced in a rotor containing both resistance and reactance. The rotor resistance RR is a constant (except for the skin effect), independent of slip, while the rotor reactance is affected in a more complicated way by slip. The reactance of an induction motor rotor depends on the inductance of the rotor and the frequency of the voltage and current in the rotor. With a rotor induc- tance of LR, the rotor reactance is given by XR = wrLR = 27rfrLR By Equation (7--8),/, = sf~, so XR - 27rSfeLR - s(27rfeLR) - sXRO (7-11 ) where XRO is the blocked-rotor rotor reactance. The resulting rotor equivalent circuit is shown in Figure 7- 9. The rotor cur- rent flow can be found as E, IR = RR + jsXRO (7-1 2) E", IR = R I + X R S } RO (7-1 3) Notice from Equation (7-1 3) that it is possible to treat all of the rotor effects due to varying rotor speed as being caused by a varying impedance supplied with power from a constant-voltage source ERO. The equivalent rotor impedance from this point of view is (7-1 4) and the rotor equivalent circuit using this convention is shown in Figure 7- 10. In the equivalent circuit in Figure 7-1 0, the rotor voltage is a constant EIiO V and the
- 416. 392 ELECTRIC MACHINERY RJNDAMENTALS o 25 R, , I'IGURE 7-10 The rotor cirwit model with all the frequency (slip) effects concentrated in resistor RR. ~ ~ 100 nm. percentage of synchronous speed ""GURE 7-11 Rotor currem as a function of rotor speed. 125 rotor impedance ZR.•q contains all the effects of varying rotor slip. A plot of the current flow in the rotor as developed in Equations (7-1 2) and (7- 13) is shown in Figure7-11. Notice that at very low slips the resistive tenn RRIs» XRQ, so the rotor re- sistance predominates and the rotor current varies linearly with slip. At high slips,
- 417. INDUCTION MOTORS 393 XRO is much larger than RRIs, and the rotor current approaches a steady-state value as the slip becomes very large. The Final Equivalent Circuit To produce the fmal per-phase equivalent circuit for an induction motor, it is nec- essary to refer the rotor part of the model over to the stator side. 1lle rotor circuit model that will be referred to the stator side is the model shown in Figure 7-1 0, which has all the speed variation effects concentrated in the impedance term. In an ordinary transformer, the voltages, currents, and impedances on the secondary side of the device can be referred to the primary side by means of the turns ratio of the transfonner: (7-1 5) Ip = I , _1: ,- a (7-1 6) and (7-1 7) where the prime refers to the referred values of voltage, current, and impedance. Exactly the same sort of transfonnati on can be done for the induction mo- tor's rotor circuit. If the effective turns ratio of an induction motor is aoff, then the transfonned rotor voltage becomes the rotor current becomes and the rotor impedance becomes Ifwe now make the following definiti ons: R2 = a;ffRR (7-1 8) (7-1 9) (7- 20) (7- 21) (7- 22) then the final per-phase equivalent circ uit of the induction motor is as shown in Figure 7-1 2. The rotor resistance RR and the locked-rotor rotor reactance XIIQ are very dif- ficult or impossible to determine directly on cage rotors, and the effective turns ra- tio aoff is also difficult to obtain for cage rotors. Fortunately, though, it is possible to make measurements that will directly give the referred resistance and reac- tance Rl and Xl, even though RR, XRO and aeff are not known separately. The mea- surement of induction motor parameters will be taken up in Section 7.7.
- 418. 394 ELECTRIC MACHINERY RJNDAMENTALS I, R, I, - - + 1 .1 ~ v • Rc jXM E, - 7 - ""GURE 7-12 The per-phase equivalent ci['(;uit of an induction motor. Air-gap power : , , ::~~ ' "~oow.r Pu,,,,,f3vrhcos6 : : , R,~ PtrictiOll '-L ODdwiD<!ay ""GURE 7-1J p= (Stator copper loss) (C~ losses) (Rotor 00_ loss) The power-flow diagram of an induction motor. 7.4 POWER AND TORQUE IN INDUCTION MOTORS Because induction motors are singly exciled machines, their power and torque re- lationships are considerably different from the relationships in the synchronous machines previously studied. TIlis section reviews the power and torque relation- ships in induction motors. Losses and the Power-Flow Diagram An induction motor can be basically described as a rotating transfonner. Its input is a three-phase system of voltages and currents. For an ordinary transfonner, the output is electric power from the secondary windings. TIle secondary windings in an induction motor (the rotor) are shorted out, so no electrical output exists from normal induction motors. Instead, the output is mechanical. The relationship be- tween the input electric power and the output mechanical power of this motor is shown in the power-flow diagram in Figure 7- 13.
- 419. INDUCTION MOTORS 395 The input powerto an induction motor flnis in the form of three-phase elec- tric voltages and currents. TIle first losses encountered in the machine are [ 2R losses in the stator windings (the stator copper loss PSCL) ' Then some amount of power is lost as hysteresis and eddy currents in the stator (P.:ore). The power re- maining at this point is transferred to the rotor of the machine across the air gap between the stator and rotor. This power is called the air-gap power PAG of the machine. After the power is transferred to the rotor, some of it is lost as / lR losses (the rotor copper loss PRCL), and the res.t is converted from electrical to mechani- cal form (PC<JII¥)' Finally, fri ction and windage losses PF&W and stray losses Pmlsc are subtracted. The remaining power is the output of the motor Pout. The core losses do not always appear in the power-flow diagram at the point shown in Figure 7- 13. Because of the nature of core losses, where they are ac- counted for in the machine is somewhat arbitrary. The core losses of an induction motor come partially from the stator circuit and partially from the rotor circuit. Since an induction motor nonnally operates at a speed near synchronous speed, the relative motion of the magnetic fields over the rotor surface is quite slow, and the rotor core losses are very tiny compared to the stator core losses. Since the largest fraction of the core losses comes from the stator circuit, all the core losses are lumped together at that point on the diagram. These losses are represented in the induction motor equivalent circuit by the resistor Rc (or the conductance Gd. If core losses are just given by a number (X watts) instead of as a circuit element they are often lumped together with the mechanical losses and subtracted at the point on the diagram where the mechanical losses are located. The higher the speed of an induction motor, the higher its friction, windage, and stray losses. On the other hand, the higherthe speed ofthe motor (up to n,ync)' the lower its core losses. Therefore, these three categories of losses are sometimes lumped together and called rotational losses. The total rotational losses of a mo- tor are often considered to be constant with changing speed, since the component losses change in opposite directions with a change in speed. Example 7-2, A 4S0-V, 60-Hz, SO-hp, three-phase induction motor is drawing 60 A at 0.S5 PF lagging. The stator copper losses are 2 kW, and the rotor copper losses are 700 W. The friction and windage losses are 600 W, the core losses are ISOO W, and the stray losses are negligible. Find the following quantities: (a) The air-gap power PAG (b) The power converted P"""" (c) TheoutputpowerPOIIt (d) The efficiency of the motor Solutioll To answer these questions, refer to the power-flow diagram for an induction motor (Fig- ure 7- 13). (a) The air-gap power is just the input power minus the stator j 2R losses. The input power is given by
- 420. 396 ELECTRIC MACHINERY RJNDAMENTALS Pin = V3"VT h cos () = V3"(480 V)(60 A)(O.8S) = 42.4 kW From the power-flow diagram, the air-gap power is given by PAG = Pin - PSCL. - P.:ore = 42.4kW - 2 kW - 1.8kW = 38.6kW (b) From the power-flow diagram, the power converted from electrical to mechan- ical fonn is PC<JiIV = PAG - PRCL = 38.6 kW - 700 W = 37.9 kW (c) From the power-flow diagram, the output power is given by Pout = P C<JiIV - PF&W - P mi«: = 37.9kW - 600W - OW = 37.3kW or, in horsepower, 1hp Pout = (37.3 kW) 0.746 kW = SOhp (d) Therefore, the induction motor's efficiency is Power and Torque in an Induction Motor Figure 7- 12 shows the per-phase equivalent circuit of an induction motor. If the equi valent circuit is examined closely, it can be used to derive the power and torque equations governing the operation of the motor. TIle input current to a phase of the motor can be found by di viding the input voltage by the total equivalent impedance: V. I I - (7- 23) Z~ where Zeq = RI +JXI + . I Gc - )BM + V2/S + jX2 (7- 24) Therefore, the stator copper losses, the core losses, and the rotor copper losses can be found. The stator copper losses in the three phases are given by (7- 25) The core losses are given by (7- 26)
- 421. INDUCTION MOTORS 397 so the air-gap power can be found as ICp -A -G - ~ - R - ;" - - -p- ,c -c - - - P - ,~ -' (7- 27) Look closely at the equivalent circuit of the rotor. The only element in the equivalent circuit where the air-gap power can be consumed is in the resistor Rl/S. Therefore, the air-gap power can also be given by I PAG = 3Ii~1 (7- 28) The actual resistive losses in the rotor circuit are given by the equation PRG- = 31~ RR (7- 29) Since power is unchanged when referred across an ideal transfonner, the rotor copper losses can also be expressed as I'P- R- cc - ~ - 31~lC-R- ,'1 (7- 30) After stator copper losses, core losses, and rotor copper losses are sub- tracted from the input power to the motor, the remaining power is converted from electrical to mechanical form. This power converted, which is sometimes called developed mechanical power, is given by = 3Ii R2 _ 312R , " = 31~ R2(~ - 1 ) (7- 31) Notice from Equations (7- 28) and (7- 30) that the rotor copper losses are equal to the air-gap power times the slip: (7- 32) Therefore, the lower the slip of the motor, the lower the rotor losses in the ma- chine. Note also that if the rotor is not turning, the slip S = 1 and the air-gap power is entirely consumed in the rotor. This is logical, since if the rotor is not turning, the output power Pout (= "Tload w",) must be zero. Since P.:<>D¥ = PAG - PRCL, this also gives another relationship between the air-gap power and the power con- verted from electrical to mechanical fonn: P.:onv = PAG - PRCL (7- 33)
- 422. 398 ELECTRIC MACHINERY RJNDAMENTALS Finally, if the friction and windage losses and the stray losses are known, the output power can be found as ' I"p oo C- , - _ ~ ~ C O- ", ---C p~, - &- W ---C P~ ~ - ." -'1 (7- 34) TIle induced torque rind in a machine was defined as the torque generated by the internal electric-to-rnechanical power conversion. This torque differs from the torque actually available at the tenninals of the motor by an amount equal to the friction and windage torques in the machine. The induced torque is given by the equation (7- 35) TIlis torque is also called the developed torque of the machine. TIle induced torque of an induction motor can be expressed in a different fonn as well. Equation (7- 7) expresses actual speed in terms of synchronous speed and slip, while Equation (7- 33) expresses P"DDY in terms of PAG and slip. Substituting these two equations into Equation (7- 35) yields (1 - s)P A G r ind = ( 1 s)WS ytlC (7- 36) TIle last equation is especially usefu l because It expresses induced torque directly in tenns of air-gap power and synchronous speed, which does not vary. A knowl- edge of PAG thus directly yields r ind . Separating the Rotor Copper Losses and the Power Converted in an Induction Motor 's Equivalent Circuit Part of the power coming across the air gap in an induction motor is consumed in the rotor copper losses, and part of it is converted to mechanical power to drive the motor's shaft. It is possible to separate the two uses of the air-gap power and to indicate them separately on the motor equivalent circuit. Equation (7- 28) gives an expression for the total air-gap power in an in- duction motor, while Equation (7- 30) gives the actual rotor losses in the motor. TIle air-gap power is the power which would be consumed in a resistor of value Ris, while the rotor copper losses are the power which would be consumed in a resistor of value R2. TIle difference between them is P eDDY' which must therefore be the power consumed in a resistor of value Reonv = ~2 - R2 = R2(~ - 1) (7- 37)
- 423. INDUCTION MOTORS 399 I, R, I, - - + (SCL) 1 .1 + (RCL) (Core loss) R, jXM E, J ./ - FIGURE 7-14 The per-phase equivalent circuit with rotor losses and PCO« separated. Per-phase equivalent circuit with the rotor copper losses and the power con- verted to mechanical fonn separated into distinct e lements is shown in Figure 7- 14. Example 7-3. A 460-V. 25-hp. 6()"'Hz. four-pole. V-connected induction motor has the following impedances in ohms per phase referred to the stator circuit: Rt = 0.641 n XI = 1.106 n Rl = 0.332 n Xl = 0.464 n XM = 26.3 n The total rotational losses are 1100 W and are assumed to be constant. The core loss is lumped in with the rotational losses. For a rotor slip of 2.2 percent at the rated voltage and rated frequency. find the motor's (a) Speed (b) Stator current (c) Power factor (d) POO<IV and Pout (e) Tiad and Tiood (jJ Efficiency Solutioll The per-phase equivalent circuit of this motor is shown in Figure 7- 12. and the power-flow diagram is shown in Figure 7- 13. Since the core losses are Iwnped together with the fric- tion and windage losses and the stray losses. they will be treated like the mechanical losses and be subtracted after P coov in the power-flow diagram. (a) The synchronous speed is 120fe n,yDC = - P - = 129(60 Hz) _ 1800 r/ min 4 poles The rotor's mechanical shaft speed is nm = (I - s)n,yDC = (1 - 0.022XI800r/ min) = 1760r/ min
- 424. 400 ELECTRIC MACHINERY RJNDAMENTALS or w". = (1 - s)w.ry1tC = (1 - 0.022XI88.5rad/s) = 184.4rad/s (b) To find the stator clUTent, get the equivalent impedance of the circuit. The first step is to combine the referred rotor impedance in parallel with the magnetiza- tion branch, and then to add the stator impedance to that combination in series. The referred rotor impedance is R, z, = - +jX2 , = 0.332 + '0 464 0.022 J. = 15.00 + jO.464 [! = 15.IOLI.76° [! The combined magnetization plus rotor impedance is given by 1 Z/ = I/jXM + 1!L; = ~=~~1=~~ jO.038 + 0.0662L 1.76° 1 = 0.0773L 31.1 0 = 12.94L31.I O[! Therefore, the total impedance is z..,. = z....,+ Z/ = 0.641 + j1.106 + 12.94L31.1° [! = 11.72 + j7.79 = 14.07L33.6° [! The resulting stator current is V I - ~ ,- z.. _ 266LO° V _ - 14.07L33.6° [! - (c) The power motor power factor is PF = cos 33.6° = 0.833 (d) The input power to this motor is Pin = V3"VT h cos () 18.88L - 33.6° A lagging = V3"(460 VX18.88 A)(0.833) = 12,530 W The stator copper losses in this machine are PSCL = 3/f R] = 3(18.88 A)2(0.64 I [!) = 685 W The air-gap power is given by PAG = P;n - PSCL = 12,530 W - 685 W = 11,845 W Therefore, the power converted is (7- 25)
- 425. INDUCTION MOTORS 401 P.:oov = (1 - S)PAG = (I - 0.022X11,845 W) = 11,585 W The power P"", is given by Pout = P.:oov - Prot = 11,585W - llOOW = 1O,485W ( 1hp ) = 10,485 W 746 W = 14.1 hp (e) The induced torque is given by PAG Tind = -- w' )''''' 11,845 W = 18S.5rad/s = 62.8 N o m and the output torque is given by p-, Tload = 10,485 W = 184.4 radls = 56.9 Nom (In English lUlits, these torques are 46.3 and 41.9 Ib-ft, respectively.) (jJ The motor's efficiency at this operating condition is p ~ 7f = R x 100% ,. 10,485 W 12530 W x 100% = 83.7% 7.5 INDUCTION MOTOR TORQUE-SPEED CHARACTERISTICS How does the torque of an induction motor change as the load changes? How much torque can an induction motor supply at starting conditions? How much does the speed of an induction motor drop as its shaflload increases? To find out the answers to these and similar questions, it is necessary to clearly understand the relationships among the motor's torque, speed, and power. In the following material, the torque- speed relationship will be examined first from the physical viewpoint of the motor's magnetic field behavior. TIlen, a general equation for torque as a fu nction of slip wilI be derived from the induction motor equivalent circuit (Figure 7- 12). Induced Torque from a Physical Standpoint Figure 7-1 5a shows a cage rotor induction motor that is initially operating at no load and therefore very nearly at synchronous speed.llle net magnetic field BDeI in this machine is produced by the magnetization current 1 Mflowing in the motor's equivalent circuit (see Figure 7-1 2). The magnitude of the magnetization current and hence of BDeI is directly proportional to the voltage E). If E] is constant, then the
- 426. 402 ELECTRIC MACHINERY RJNDAMENTALS ER IR E, , I, 0 0 , 0 " 0 0 , " , " 0 ' , 0 0 " 0 I B.... / D~g.~ Rotor , ' , 0 , 0 0 ' e, w ,., 0 , , 0 , 0 , , , , , H, , 0 , 0 , 0 , , , , , 0 0 0 , 0 0 0 (,' (b' ""GURE 7- 15 (a) The magnetic fields in an induction motor under light loods. (b) The magnetic fields in an induction motor under heavy loads. 0 0 0 Rotor net magnetic field in the motor is constant. In an actual machine, El varies as the load changes, because the stator impedances Rl and Xl cause varying voltage drops with varying load. However, these drops in the stator wi ndi ngs are relatively small, so El (and hence 1M and Bo..) is approximalely constanl with changes in load. Figure 7-l 5a shows the induction motor at no load. At no load, the rotor slip is very smaJl, and so the relalive motion between the rotor and the magnetic fields is very smaJi and the rotor frequency is also very small. Since the relative motion is smaJl, the voltage ER induced in the bars of the rotor is very smaJl, and the resulting currenl fl ow IR is small. Also, because the rotor frequency is so very small, the reactance of the rotor is nearly zero, and the maximum rotor currenl IR is almost in phase with the rotor voltage ER. TIle rotor current thus produces a small magnetic field DR at an angle just slighlly greater than 90° behind the net magnetic field Boe, . Notice that the stator current must be quite large even at no load, since it must supply most of Bne,. (TIlis is why induction motors have large no-load currents compared to other types of machines.) TIle induced torque, which keeps the rotor turning, is given by the equation (4-60) Its magnitude is given by "Tind = kBRBoe, sin /j (4-6 1) Since the rotor magnetic field is very smaJl, the induced torque is also quite small- just large enough to overcome the motor's rotational losses. Now suppose the induction motor is loaded down (Figure 7-15b). As the motor's load increases, its slip increases, and the rotor speed falls. Since the rotor speed is slower, there is now more relative motion between the rotor and the sta-
- 427. INDUCTION MOTORS 403 tor magnetic fields in the machine. Greater relative motion produces a stronger ro- tor voltage ER which in turn produces a larger rotor current IR. With a larger rotor current, the rotor magnetic field DR also increases. However, the angle of the ro- tor current and DR changes as well. Since the rotor slip is larger, the rotor fre- quency rises (f, = st ), and the rotor's reactance increases (WLR). Therefore, the rotor current now lags further behind the rotor voltage, and the rotor magnetic field shifts with the current. Figure 7-15b shows the induction motor operating at a fairly high load. Notice that the rotor current has increased and that the angle 8 has increased. The increase in BR tends to increase the torque, while the increase in angle 8 tends to decrease the torque (TiDd is proportional to sin 8, and 8 > 90°). Since the first effect is larger than the second one, the overall induced torque in- creases to supply the motor's increased load. When does an induction motor reach pullout torque? This happens when the point is reached where, as the load on the shaft is increased, the sin 8 tenn de- creases more than the BR tenn increases. At that point, a further increase in load decreases "TiDd, and the motor stops. It is possible to use a knowledge of the machine's magnetic fields to approx- imately derive the output torque-versus-speed characteristic of an induction motor. Remember that the magnitude of the induced torque in the machine is given by (4-61) Each tenn in this expression can be considered separately to derive the overall machine behavior. The individual tenns are L BR . TIle rotor magnetic field is directly proportional to the current fl owing in the rotor, as long as the rotor is unsaturated. TIle current flow in the rotor in- creases with increasing slip (decreasing speed) according to Equation (7-1 3). This current fl ow was plotted in Figure 7-11 and is shown again in Fig- ure 7-16a. 2. B"",. The net magnetic field in the motor is proportional to E] and therefore is approximately constant (E ] actually decreases with increasing current flow, but this effect is small compared to the other two, and it will be ignored in this graphical development). TIle curve for B"", versus speed is shown in Fig- ure 7-16b. 3. sin 8. TIle angle 8 between the net and rotor magnetic fields can be expressed in a very useful way. Look at Figure 7-15b. In this figure, it is clear that the angle 8 is just equal to the power-factor angle ofthe rotor plus 90°: (7-38) TIlerefore, sin 8 = sin (OR + 90°) = cos OR. TIlis tenn is the power factor of the rotor. The rotor power-factor angle can be calculated from the equation (7-39)
- 428. 404 ELECTRIC MACHINERY RJNDAMENTALS J, oe I DRI r---_~ ~---------7~-- '. n,ymc ,., L----------c-~-_ '. '.- ,b, 0 '. ,e, '.- ,~ ,d, '.- '. The resulting rotor power factor is given by PFR = cos (JR FlGURE 7-16 Grapltical development of an induction motor torque-speed cltaT3cteristic. (a) Plot of rotor current (and titus IURI) versus speed for an induction motor; (b) plot of net magnetic field versus speed for the motor; (c) plot of rotor power factor versus speed for the motor; (d) the resulting torque-speed characteristic. PF = cos (tan- 1SXRQ ) , R, (7-40) A plot of rotor power factor versus speed is shown in Figure 7-1 6c. Since the induced torque is proportional to the product of these three tenns, the torque-speed characteristic of an induction motor can be constructed from the
- 429. INDUCTION MOTORS 405 graphical multiplication of the previous three plots (Figure 7-1 6a to c). 1lle torque-speed characteristic of an induction motor derived in this fashion is shown in Figure7-16d. This characteristic curve can be divided roughly into three regions. The first region is the low-slip region of the curve. In the low-slip region, the motor slip in- creases approximately linearly with increased load, and the rotor mechanical speed decreases approximately linearly with load. In this region of operation, the rotor reactance is negligible, so the rotor power factor is approximately unity, while the rotor current increases Ii nearly with slip. The entire normnl steady-state operating range ofan induction motor is included in this linear low-slip region. Thus in normal operation, an induction motor has a linear speed droop. The second region on the induction motor's curve can be called the moderate-slip region. In the moderate-slip region, the rotor freq uency is higher than before, and the rotor reactance is on the same order of magnitude as the rotor resistance. In this region, the rotor current no longer increases as rapidly as before, and the power factor starts to drop. TIle peak torque (the pullout torque) of the motor occurs at the point where, for an incremental increase in load, the increase in the rotor current is exactly balanced by the decrease in the rotor power factor. The third region on the induction motor's curve is called the high-slip re- gion. In the high-slip region, the induced torque actually decreases with increased load, since the increase in rotor current is completely overshadowed by the de- crease in rotor power factor. For a typical induction motor, the pullout torque on the curve will be 200 to 250 percent ofthe rated full-load torque of the machine, and the starting torque (the torque at zero speed) will be 150 percent or so ofthe full-load torque. Unlike a syn- chronous motor, the induction motor can start with a fuJI load attached to its shaft. The Derivation of the Induction Motor Induced-Torque Equation It is possible to use the equivalent circuit of an induction motor and the power- flow diagram for the motor to derive a general expression for induced torque as a function of speed. The induced torque in an induction motor is given by Equation (7- 35) or (7- 36): (7- 35) (7- 36) The latter equation is especially useful, since the synchronous speed is a constant for a given frequency and number of poles. Since w' ync is constant, a knowledge of the air-gap power gives the induced torque of the motor. The air-gap power is the power crossing the gap from the stator circuit to the rotor circuit. It is equal to the power absorbed in the resistance R2! s. How can this power be found?
- 430. 406 ELECTRIC MACHINERY RJNDAMENTALS + v • - I, - ""GURE 7-17 ,R, I, - + jXM E, - Per-phase equivalent circuit of an induction motor. ~ Refer to the equivalent circuit given in Figure 7-17. In this figure, the air- gap power supplied to one phase of the motor can be seen to be , R, P AG.t4> = 12s Therefore, the total air-gap power is _ 2 R2 P AG -312 , If /l can be determined, then the air-gap power and the induced torque will be known. Although there are several ways to solve the circuit in Figure 7-17 for the cur- rent ll, perhaps the easiest one is to detennine the lllevenin equivalent of the por- tion of the circuit to the left of the X's in the figure. Thevenin's theorem states that any linear circuit that can be separated by two tenninals from the rest of the system can be replaced by a single voltage source in series with an equivalent impedance. If this were done to the induction motor equivalent circuit, the resulting circuit would be a simple series combination of elements as shown in Figure 7-1 8c. To calculate the lllevenin equivalent of the input side of the induction motor equivalent circuit, first open-circuit the terminals at the X's and fmd the resulting open-circuit voltage present there. lllen, to find the Thevenin impedance, kill (short-circuit) the phase voltage and find the Zeq seen "looking" into the tenninals. Figure 7-1 8a shows the open tenninals used to find the Thevenin voltage. By the voltage divider rule, ZM VTH = V4> Z + Z M , = V jXM 4>Rt + jX] + jXM llle magnitude of the Thevenin voltage Vru is (7-4 I a)
- 431. INDUCTION MOTORS 407 jX, : , v '" jX/oI V TH R]+jX]+jX/oI • (-t)V, jX/oI Vrn XM VTH '" V. ~R]2+(X] +X/oI)2 ('J jX, R, (bJ jXrn (oj FIGURE 7-18 (a) The Thevenin equivalent voltage of an induction ntotor input circuit. (b) The Thevenin equivalent impedance of the input circuit. (c) The resulting simplified equivalent circuit of an induction motor. Since the magnetization reactance XM » X] and XM » RJ, the magnitude of the Thevenin voltage is approximately (7-4 (b) to quite good accuracy. Figure 7-1 8b shows the input circuit with the input voltage source killed. The two impedances are in parallel, and the TIlevenin impedance is given by ZIZM ZTH = Z l + ZM (7-42) This impedance reduces to
- 432. 408 ELECTRIC MACHINERY RJNDAMENTALS (7-43) Because XM » Xl and XM + Xl »Rb the TIlevenin resistance and reactance are approximately given by (7-44) (7-45) TIle resulting equivalent circuit is shown in Figure 7-1 8c. From this circuit, the current 12is given by VTH 12 = ZTH +2; _ ~~~~V~TH!lL~~~ Rrn + R2/ s + jXTH + jX2 The magnitude of this current is V TH /2 = Y(RTH + R2/sP + (Xrn + X2)2 TIle air-gap power is therefore given by R, P = 3[ 2 - AG 2 S = (Rrn + R2/si + (Xrn + X2)2 and the rotor-induced torque is given by P AG T;nd=w ~oc (7-46) (7-47) (7-48) (7-49) (7- 50) A plot of induction motor torque as a function of speed (and slip) is shown in Figure 7-1 9, and a plot showing speeds both above and below the normal mo- tor range is shown in Figure 7- 20. Comments on the Induction Motor Torque-Speed Curve TIle induction motor torque-speed characteristic curve plotted in Figures 7-1 9 and 7- 20 provides several important pieces ofinfonnation about the operation of induction motors. TIlis infonnation is summarized as follows:
- 433. INDUCTION MOTORS 409 Pullout torque 400% " ~ = • • 300% • Starting • " torque § ( ~ 200% " ~ 100% ____________________~U~l~~~~~:~~ o Mechanical speed FIGURE 7-19 A typical induction motor torque-speed characteristic curve. I. 1lle induced torque of the motor is. zero at synchronous speed. 1llis fact has been discussed previously. 2. 1lle torque- speed curve is nearly linear between no load and full load. In this range, the rotor resistance is much larger than the rotor reactance, so the ro- tor current, the rotor magnetic field, and the induced torque increase linearly with increasing slip. 3. There is a maximum possible torque that cannot be exceeded. nlis torque, called the pullout torque or breakdown torque, is 2 to 3 times the rated full - load torque of the motor. The next section of this chapter contains a method for calculating pullout torque. 4. 1lle starting torque on the motor is slightly larger than its full-load torque, so this motor will start carrying any load that it can supply at fu ll power. 5. Notice that the torque on the motor for a given slip varies as the square of the applied voltage. nlis fact is useful in one fonn of induction motor speed con- trol that will be described later. 6. If the rotor of the induction motor is driven faster than synchronous speed, then the direction of the induced torque in the machine reverses and the ma- chine becomes a generator, converting mechanical power to electric power. 1lle use of induction machines as generators will be described later.
- 434. 410 ELECTRIC MACHINERY RJNDAMENTALS 400 ] = 200 e '0 Braking If region ; § ] - 200 ~ -400 ""GURE 7-10 Tmn --........~ Pullout torque Motor region nsyDC..-/ Mechanical speed Generator region Induction motor torque-speed characteristic curve. showing the extended operating ranges (braking region and generator region). 7. If the motor is turning backward relative to the direction of the magnetic fields, the induced torque in the machine will stop the machine very rapidly and will try to rotate it in the other direction. Since reversing the direction of magnetic field rotation is simply a matter of switching any two stator phases, this fact can be used as a way to very rapidly stop an induction motor. The act of switching two phases in order to stop the motor very rapidly is called plugging. TIle power converted to mechanical fonn in an induction motor is equal to and is shown plotted in Figure 7- 21. Notice that the peak power supplied by the induction motor occurs at a different speed than the maximum torque; and, of course, no power is converted to mechanical fonn when the rotor is at zero speed. Maximum (pullout) Torque in an Induction Motor Since the induced torque is equal to PAG/w'Y"'" the maximum possible torque oc- curs when the air-gap power is maximum. Since the air-gap power is equal to the power consumed in the resistor R2 /s, the maximum induced torque will occur when the power consumed by that resistor is maximum.
- 435. INDUCTION MOTORS 411 800 120 700 105 600 90 , 500 75 • Z ~ , ~ 400 60 c , ~ " , , " 300 45 0 200 30 100 15 0 0 250 500 750 1000 1250 1500 1750 2000 Mechanical speed. r/min FIGURE 7-21 Induced torque and power convened versus motor speed in revolutions per minute for an example four-pole induction motor. When is the power supplied to Rl/s at its maximum? Refer to the simplified equivalent circuit in Figure 7- 18c. In a situation where the angle of the load im- pedance is fixed, the maximum power transfer theorem states that maximum power transfer to the load resistor Rlls wilIoccur when the magnitude of that im- pedance is equal to the magnitude of the source impedance. TIle equivalent source impedance in the circuit is Zsoun:c = RTH + jXTH + jX2 so the maximum power transfer occurs when R -i- = YRfH + (XTH + Xii (7- 51) (7- 52) Solving Equation (7- 52) for slip, we see that the slip at pullout torque is given by (7- 53) Notice that the referred rotor resistance Rl appears only in the numerator, so the slip of the rotor at maximum torque is directly proportional to the rotor resistance.
- 436. 41 2 ELECTRIC MACHINERY RJNDAMENTALS 800 ,--------------------------------, R, I 700 600 R. ;"" z • ~400 "I ~ 300 200 100 R, R, o~~~~~~~~~~~~~~~~ o 250 500 750 1000 1250 1500 1750 2000 Mechanical speed. r/min ""GURE 7- 22 The effect of varying rotor resistance on the torque-.peed characteristic of a wound-rotor induction motor. TIle value of the maximum torque can be found by inserting the expression for the slip at maximum torque into the torque equation [Equation (7- 50)]. TIle re- sulting equation for the maximum or pullo ut torque is (7- 54) TIlis torque is proportional to the square of the supply voltage and is also in- versely related to the size of the stator impedances and the rotor reactance. The smaller a machine's reactances, the larger the maximum torque it is capable of achieving. Note that slip at which the maximum torque occurs is directly propor- tional to rotor resistance [Equation (7- 53)], but the value of the maximum torque is independent of the value of rotor resistance [Equation (7- 54)]. TIle torque-speed characteristic for a wound-rotor induction motor is shown in Figure 7- 22. Recall that it is possible 1.0 insert resistance into the rotor circuit of a wound rotor because the rotor circuit is brought out to the stator through slip rings. Notice on the figure that as the rotor resistance is increased, the pullout speed of the motor decreases, but the maximum torque remains constant.
- 437. INDUCTION MOTORS 413 It is possible to take advantage of this characteristic of wound-rotor induc- tion motors to start very heavy loads. If a resistance is inserted into the rotor cir- cuit, the maximum torque can be adjusted to occur at starting conditions. There- fore, the maximum possible torque would be available to start heavy loads. On the other hand, once the load is turning, the extra resistance can be removed from the circuit, and the maximum torque will move up to near-synchronous speed for reg- ular operation. Example 7-4. A two-pole, 50-Hz induction motor supplies 15 kW to a load at a speed of 2950 r/min. (a) What is the motor's slip? (b) What is the induced torque in the motor in Nom under these conditions? (c) What will the operating speed of the motor be if its torque is doubled? (d) How much power will be supplied by the motor when the torque is doubled? Solution (a) The synchronous speed of this motor is = 120f,. = 120(50 Hz) = 30CXl r/ min n.yDC P 2 poles Therefore, the motor's slip is = 3000r/min - 29.50 r / min(x 100%) 3(x)() rl mm = 0.0167 or 1.67% (7-4) (b) The induced torque in the motor must be assruned equal to the load torque, and POO/IiV must be assumed equal to Pload' since no value was given for mechanical losses. The torque is thus ~oov TiDd =W- m ~~~~~1 5~k~W~c.-~~-c = (2950 r/minX27Trad/rXI min/60 s) = 48.6N o m (c) In the low-slip region, the torque-speed curve is linear, and the induced torque is directly proportional to slip. Therefore, if the torque doubles, then the new slip will be 3.33 percent. The operating speed of the motor is thus 11m = (1 - s)n.yDC = (1 - 0.0333X3000r/min) = 2900 r / min (d) The power supplied by the motor is given by = (97.2 N 0 m)(2900 r/ minX27Trad/ rXI minI 60 s) = 29.5 kW
- 438. 414 ELECTRIC MACHINERY RJNDAMENTALS Example 7-5. A460-V. 25-hp. 60-Hz. four-pole. Y-cOIlllected wOlUld-rotor induc- tion motor has the following impedances in ohms per phase referred to the stator circuit: Rl = 0.641 0 Xl = 1.106 0 R2 = 0.3320 X2 = 0.4640 XM = 26.3 0 (a) What is the maximrun torque of this motor? At what speed and slip does it occur? (b) What is the starting torque of this motor? (c) When the rotor resistance is doubled. what is the speed at which the maximum torque now occurs? What is the new starting torque of the motor? (d) Calculate and plot the torque-speed characteristics of this motor both with the original rotor resistance and with the rotor resistance doubled. Solutioll The Thevenin voltage of this motor is (7-4la) The Thevenin resistance is (7-44) J 26.3 n )' - (0.641 0-'1.1060 + 26.3 0 = 0.590 0 The Thevenin reactance is XTH - Xl = 1.106 0 (a) The slip at which maximum torque occurs is given by Equation (7- 53): R, Smax = ~VijRijfu =cc+ c=i (X~rn "=,, +=x~ ~' (7- 53) 0.3320 = =0.198 Y(0.590 0)1 + (1.106 0 + 0.464 0)2 This corresponds to a mechanical speed of nm = (I - S)n,yDC = (I - 0.198)(l800r/min) = 1444r/ min The torque at this speed is ~--,"--c-ce3,Vfl"~cc==C=~" Trrw; = 2W'YDC[RTH + YRfu + (XTH + X;l2 ] = 3(255.2 V)2 (7- 54) 2(188.5 rad/s)[O.590 0 + Y(0.590 O)l + (1.106 0 + 0.464 0)2] = 229N om
- 439. INDUCTION MOTORS 415 (b) The starting torque of this motor is found by setting s = I in Equation (7- 50): 3ViHRl T,tan= W I(R +R)'+(X +X)'l 'YDC TH 2 rn 2 _ 3(255.2 V)1:0.3320) - (188.5 rad/s)[(0.590 0 + 0.3320)2 + (1.I()5 0 + 0.464 0)2] = I04N om (c) If the rotor resistance is doubled, then the slip at maximwn torque doubles, too. Therefore, Smax = 0.396 and the speed at maximum torque is nm = (I - s)n.yDC = (1 - 0.396)(1800 r /min) = 1087 r/min The maximum torque is still Trrw; = 229 Nom The starting torque is now _ 3(255.2 V)2(0.664 0) Tot"" - (188.5 rad/s)[(0.590.n + 0.6640)2 + (1.106 0 + 0.4640)2] = 170 Nom (d) We will create a MATLAB M-file to calculate and plot the torque-speed char- acteristic of the motor both with the original rotor resistance and with the dou- bled rotor resistance. The M-file will calculate the Thevenin impedance using the exact equations for V lll and Zrn [Equations (7-4la) and (7-43)] instead of the approximate equations, because the computer can easily perfonn the exact calculations. It will then calculate the induced torque using Equation (7- 50) and plot the results. The resulting M-file is shown below: % M-file, t orque_speed_curve.m % M-file c reate a p l ot o f the t orque- speed c urve o f the % induc tion mot or o f Exampl e 7- 5. % Fir s t , initia liz e the va lues needed in thi s program. rl = 0.641; % Stat or res i s tance xl = 1.10 6; % Stat or r eactance r2 = 0.332; % Rotor r es i s tance x2 = 0.464, xm = 26.3, v-ph ase = 460 / sqrt (3) , n_sync = 1800, w_sync = 188.5; % Ca l c ulate the Thevenin % 7- 41a a nd 7- 43. v_th 0 v-ph ase • ( = I % Rotor r eactance % Magne tizati on branch r eactance % Phase volt age % Syn chronou s speed (r / min ) % Sy nchronou s speed (rad /s) vo lt age and impedance from Equat i ons sqrt (rl " 2 • (xl • xm )" 2 ) I ; , " 0 ( ( j*xm ) • (d • - j*xl ) ) I (r1 • j * (xl • xm ) ) , r -" 0 real ( z_ th ) , x-" 0 imag ( z_ th ) ,
- 440. 416 ELECTRIC MACHINERY RJNDAMENTALS ~ Now ca l c ulate the t orqu e- speed chara c teri s ti c f or many ~ s lips between 0 and 1. Not e that the fir s t s lip value ~ i s set t o 0.001 ins tead of exact l y 0 to avoi d d i v i de- ~ by-zero probl ems. s = (0 ,1,50 ) / 50; s( l ) = 0 . 001; % Slip run = (1 - s) * n_sync; % M ech a ni cal speed ~ Ca l c ulate t orque f or original rotor r es i s tance f or ii = 1,51 t _ indl (U ) = (3 * v_th" 2 * r2 / s (U )) / ... (w_sync * (( r _ th + r2 / s (ii )) " 2 + (x_th + x2 ) " 2 ) ) ; ond ~ Ca l c ulate t orque f or doubl ed roto r r es i s tance f or ii = 1,51 t _ ind2 (U ) = (3 * v_th" 2 * (2*r2 ) / s( U )) / ... (w_sync " (( r _ th + (2*r2 )/s( U ))" 2 + (x_th + x2 )" 2 ) ) ; ond ~ Plot the t orque- speed c urve p l ot (run , t _ indl, ' Co l or' , 'k' , 'LineWi dth ' ,2 . 0); h o l d on; p l ot (nm , t _ ind2, ' Co l or', 'k' , 'LineWi dth ' ,2.0, 'LineStyl e ' , '-. ' ) ; x l abe l ( ' itn_( m)' , 'Pontwei ght' , 'Bo l d ' ) ; y l abe l ( ' tau_( ind) ' , 'Pontwei ght' , 'Bo l d ' ) ; title ( 'Induc tion mot or t orque- speed c harac teri s ti c ' , ... 'Pontwe i ght' , 'Bol d ' ) ; l egend ( ' Original R_(2) ' , 'Doubled R_(2} ' ) ; gri d on; h o l d o ff ; The resulting torque-speed characteristics are shown in Figure 7-23. Note that the peak torque and starting torque values on the curves match the calculations of parts (a) through (c). Also. note that the starting torque of the motor rose as R2increased. 7.6 VARIATIONS IN INDUCTION MOTOR TORQUE-SPEED CHARACTERISTICS Section 7.5 contained the derivation of the torque-speed characteristic for an induction motor. In fact, several characteristic curves were shown, depending on the rotor resistance. Example 7- 5 illustrated an induction motor designer's dilemma-if a rotor is designed with high resistance, then the motor's starting torque is quite high, but the slip is also quite high at normal operating conditions. Recall that P C<X1V = (I - s)pAG, so the higher the slip, the smaller the fraction of air-gap power actually converted to mechanicalfonn, and thus the lower the mo- tor's efficiency. A motor with high rotor resistance has a good starting torque but poor efficiency at nonnal operating conditions. On the other hand, a motor with low rotor resistance has a low starting torque and high starting current, but its ef- ficiency at nonnal operating conditions is quite high. An induction motor designer
- 441. INDUCTION MOTORS 41 7 2'0 ,--,---,--,---,--,---,--,---,--, .- 200 .- , I'" • z '0---- , ," ." 100 r- FIGURE 7-23 - - Original R2 . - . Doubled R2 nO. rlmin Torque-speed characteristics for the motor of Example 7- 5. , , is forced to compromise between the conflicting requirements of high starting torque and good efficiency. One possible solution to this difficulty was suggested in passing in Section 7.5: use a wound-rotor induction motor and insert extra resistance into the rotor during starting.1lle extra resistance could be completely removed for better effi- ciency during nonnal operation. Unfortunately, wound-rotor motors are more ex- pensive, need more maintenance, and require a more complex automatic control circuit than cage rotor motors. Also, it is sometimes important to completely seal a motor when it is placed in a hazardous or explosive environment, and this is eas- ier to do with a completely self-contained rotor. It would be nice to figure out some way to add extra rotor resistance at starting and to remove it during normal running without slip rings and without operator or control circuit inteTYention. Figure 7- 24 illustrates the desired motor characteristic. 1llis figure shows two wound-rotor motor characteristics, one with high resistance and one with low resistance. At high slips, the desired motor should behave like the high-resistance wound-rotor motor curve; at low slips, it should behave like the low-resistance wound-rotor motor curve. Fortunately, it is possible to accomplish just this effect by properly taking advantage of leakage reactance in induction motor rotor design. Control of Motor Characteristics by Cage Rotor Design The reactance Xl in an induction motor equivalent circuit represents the referred fonn of the rotor's leakage reactance. Recall that leakage reactance is the reac- tance due to the rotor nux lines that do not also couple with the stator windings. In
- 442. 418 ELECTRIC MACHINERY RJNDAMENTALS Loo" like h.igh. R, ------------ Desired curve Looks like lowR2 '---------------------------------'---- '. ""GURE 7-14 A torque-speed characteristic curve combining high-resistance effects at low speeds (high slip) with low-resistance effects at high speed (low slip). general, the farther away from the stator a rotor bar or part of a bar is, the greater its leakage reactance, since a smaller percentage of the bar's flux will reach the sta- tor. Therefore, if the bars of a cage rotor are placed near the surface of the rotor, they will have only a small leakage flux and the reactance Xl will be small in the equivalent circuit. On the other hand, ifthe rotor bars are placed deeper into the ro- tor surface, there will be more leakage and the rotor reactance X2 will be larger. For example, Figure 7- 25a is a photograph of a rotor lamination showing the cross section of the bars in the rotor. The rotor bars in the figure are quite large and are placed near the surface of the rotor. Such a design will have a low resis- tance (due to its large cross section) and a low leakage reactance and Xl (due to the bar's location near the stator). Because of the low rotor resistance, the pullout torque will be quite near synchronous speed [see Equation (7- 53)], and the motor will be quite efficient. Remember that P.,oo¥ = (I - s)PAG (7- 33) so very little of the air-gap power is lost in the rotor resistance. However, since Rl is small, the motor's starting torque will be small, and its starting current will be high. TIlis type of design is called the National Electrical Manufacturers Associa- tion (NEMA) design class A. It is more or less a typical induction motor, and its characteristics are basically the same as those of a wound-rotor motor with no ex- tra resistance inserted. Its torque-speed characteristic is shown in Figure 7- 26. Figure 7- 25d, however, shows the cross section of an induction motor rotor with smnll bars placed near the surface of the rotor. Since the cross-sectional area of the bars is small, the rotor resistance is relatively high. Since the bars are lo- cated near the stator, the rotor leakage reactance is still small. This motor is very much like a wound-rotor induction motor with extra resistance inserted into the rotor. Because of the large rotor resistance, this motor has a pullout torque occur-
- 443. INDUCTION MOTORS 419 ,,' ,b , 'e' ,d, FIGURE 7-1.5 Laminations from typical cage induction motor rotors. showing the cross section of the rotor bars: (a) NEMA design class A- large bars near the surface; (b) NEMA design class B- large, deep rotor bars; (e) NEMA design class C--double-cage rotor design; (d) NEMA desisn class D-small bars near the surface. (Counesy ofMagneTek, Inc.) ring at a high slip, and its starting torque is quite high. A cage motor with this type of rotor construction is called NEMA design class D. Its torque-speed character- istic is also shown in Figure 7- 26. Deep-Bar and Double-Cage Rotor Designs Both of the previous rotor designs are essentially similar to a wound-rotor motor with a sct rotor resistance. How can a variable rotor resistance be produced to combine the high starting torque and low starting current of a class 0 design with the low nonnal operating slip and high efficiency of a class A design?
- 444. 420 ELECTRIC MACHINERY RJNDAMENTALS o~~~~~~--~--~ o 20 40 60 80 100 Percentage of synchronous speed FIGURE 7-16 Typical torque-speed curves for different rotor designs. Ii is possible to produce a variable rotor resistance by the use of deep rotor bars or double-cage rotors. TIle basic concept is illustrated with a deep-bar rotor in Figure 7-27. Figure 7-27a shows a current flowing through the upper part ofa deep rotor bar. Since current flowing in that area is tightly coupled to the stator, the leakage inductance is small for this region. Figure 7-27b shows current flow- ing deeper in the bar. Here, the leakage inductance is higher. Since all parts of the rotor bar are in parallel electrically, the bar essentially represents a series of par- allel electric circuits, the upper ones having a smaller inductance and the lower ones having a larger inductance (Figure 7- 27c). At low slip, the rotor's frequency is very small, and the reactances of all the parallel paths through the bar are small compared to their resistances. The imped- ances of all parts of the bar are approximately equal, so current flows through all parts of the bar equally. The resulting large cross-sectional area makes the rotor resistance quite small, resulting in goOO efficiency at low slips. At high slip (start- ing conditions), the reactances are large compared to the resistances in the rotor bars, so all the current is forced to flow in the low-reactance part of the bar near the stator. Since the effective cross section is lower, the rotor resistance is higher than before. With a high rotor resistance at starting conditions, the starting torque is relatively higher and the starting current is relatively lower than in a class A de- sign. A typical torque-speed characteristic for this construction is the design class B curve in Figure 7- 26. A cross-sectional view of a double-cage rotor is shown in Figure 7- 25c. Ii consists of a large, low-resistance set of bars buried deeply in the rotor and a small, high-resistance set of bars set at the rotor surface. Ii is similar to the deep- bar rotor, except that the difference between low-slip and high-slip operation is
- 445. Stator a--:::::------ Rotor with deep bars "j FIGURE 7-27 Slip ring ,f R R R ,'j INDUCTION MOTORS 421 Top of bar L L, L, L, Bottom of bar 'bj Slip ring Flux linkage in a deep-bar rotor. (a) For a current flowing in the top of the bar. the flux is tightly linked to the stator. and leakage inductance is small; (b) for a current flowing in the bottom of the bar. the flux is loosely linked to the stator. and leakage inductance is large; (c) resulting equivalent circuit of the rotor bar as a function of depth in the rotor. even more exaggerated. At starting conditions, only the small bar is effective, and the rotor resistance is quite high. This high resistance results in a large starting torque. However, at normal operating speeds, both bars are effective, and the re- sistance is almost as low as in a deep-bar rotor. Double-cage rotors of this sort are used to produce NEMA class 8 and class C characteristics. Possible torque-speed characteristics for a rotor of this design are designated design class 8 and design class C in Figure 7- 26. Double-cage rotors have the disadvantage that they are more expensive than the other types ofcage rotors, but they are cheaper than wound-rotor designs. TIley allow some of the best features possible with wound-rotor motors (high starting torque with a low starting current and go<Xl effi ciency at nonnal operating condi- tions) at a lower cost and without the need of maintaining slip rings and brushes. Induction Motor Design Classes It is possible to produce a large variety oftorque-speed curves by varying the ro- tor characteristics of induction motors. To help industry select appropriate motors for varying applications in the integral-horsepower range, NEMA in the United
- 446. 422 ELECTRIC MACHINERY RJNDAMENTALS States and the International Electrotechnical Commission (IEC) in Europe have defined a series of standard designs with different torque- speed curves. TIlese standard designs are referred to as design classes, and an individual motor may be referred to as a design class X motor. It is these NEMA and IEC design classes that were referred to earlier. Figure 7-26 shows typical torque-speed curves for the four standard NEMA design classes. The characteristic features of each stan- dard design class are given below. DESIGN CLASS A. Design class A motors are the standard motor design, with a normal starting torque, a nonnal starting current, and low slip. TIle full-load slip of design A motors must be less than 5 percent and must be less than that of a de- sign B motor of equivalent rating. TIle pullout torque is 200 to 300 percent of the full-load torque and occurs at a low slip (less than 20 percent). TIle starting torque of this design is at least the rated torque for larger motors and is 200 percent or more of the rated torque for smaller motors. TIle principal problem with this de- sign class is its extremely high inrush current on starting. Current flows at starting are typically 500 to 800 percent of the rated current. In sizes above about 7.5 hp, some form of reduced-voltage starting must be used with these motors to prevent voltage dip problems on starting in the power system they are connected to. In the past, design class A motors were the standard design for most applications below 7.5 hp and above about 200 hp, but they have largely been replaced by design class 8 motors in recent years. Typical applications for these motors are driving fans, blowers, pumps, lathes, and other machine tools. DESIGN CLASS B. Design class 8 motors have a nonnal starting torque, a lower starting current, and low slip. TIlis motor produces about the same starting torque as the class A motor with about 25 percent less current. TIle pullout torque is greater than or equal to 200 percent ofthe rated load torque, but less than that ofthe class A design because of the increased rotor reactance. Rotor slip is still relatively low (less than 5 percent) at full load. Applications are similar to those for design A, but design B is preferred because of its lower starting-current requirements. Design class 8 motors have largely replaced design class A motors in new installations. DESIGN CLASS C. Design class C motors have a high starting torque with low starting currents and low slip (less than 5 percent) at full load. TIle pullout torque is slightly lower than that for class A motors, while the starting torque is up to 250 percent ofthe full-load torque. TIlese motors are built from double-cage rotors, so they are more expensive than motors in the previous classes. They are used for high-starting-torque loads, such as loaded pumps, compressors, and conveyors. DESIGN CLASS D. Design class D motors have a very high starting torque (275 percent or more of the rated torque) and a low starting current, but they also have a high slip at full load. TIley are essentially ordinary class A induction motors, but with the rotor bars made smaller and with a higher-resistance material. The high rotor resistance shifts the peak torque to a very low speed. It is even possible for
- 447. INDUCTION MOTORS 423 FIGURE 7-18 Rotor cross section. showing the construction of the former design class F induction motor. Since the rotor bars are deeply buried. they have a very high leakage reactance. The high leakage reactance reduces the starting torque and current of this motor. so it is called a soft-start design. (Courtesy of MagneTek, Inc.) the highest torque to occur at zero speed ( 100 percent slip). Full-load slip for these motors is quite high because of the high rotor resistance. It is typically7 to II per- cent. but may go as high as 17 percent or more. These motors are used in applica- tions requiring the acceleration of extremely high-inertia-type loads, especially large flywheels used in punch presses or shears. In such applications, these motors gradually accelerate a large flywheel up to full speed, which then drives the punch. After a punching operation, the motor then reaccelerates the flywheel over a fairly long time for the next operation. In addition to these four design classes, NEMA used to recognize design classes E and F, which were called soft-start induction motors (see Figure 7- 28). These designs were distinguished by having very low starting currents and were used for low-starling-torque loads in situations where starting currents were a problem. 1l1ese designs are now obsolete.
- 448. 424 ELECTRIC MACHINERY RJNDAMENTALS Example 7-6. A460-V. 30-hp. 60-Hz. four-pole. V-connected induction motor has two possible rotor designs. a single-cage rotor and a double-cage rotor. (The stator is iden- tical for either rotor design.) The motor with the single-cage rotor may be modeled by the following impedances in ohms per phase referred to the stator circuit: Rl = 0.641 0 Xl = 0.750 0 Rl = 0.3000 Xl = 0.5000 XM = 26.3 0 The motor with the double-cage rotor may be modeled as a tightly coupled. high- resistance outer cage in parallel with a loosely coupled. low-resistance inner cage (similar to the structure of Figure 7- 25c). The stator and magnetization resistance and reactances will be identical with those in the single-cage design. The resistance and reactance of the rotor outer cage are: Ru, = 3.200 0 Xu, = 0.500 0 Note that the resistance is high because the outer bar has a small cross section. while the re- actance is the same as the reactance of the single-cage rotor. since the outer cage is very close to the stator. and the leakage reactance is small. The resistance and reactance of the inner cage are Ru = 0.400 0 Xu = 3.300 0 Here the resistance is low because the bars have a large cross-sectional area. but the leak- age reactance is quite high. Calculate the torque-speed characteristics associated with the two rotor designs. How do they compare? Solutio" The torque-speed characteristic of the motor with the single-cage rotor can be calculated in exactly the same manner as Example 7- 5. The torque-speed characteristic of the motor with the double-cage rotor can also be calculated in the same fashion. except that at each slip the rotor resistance and reactance will be the parallel combination of the impedances of the iIiller and outer cages. At low slips. the rotor reactance will be relatively lUlimportant. and the large inner cage will playa major part in the machine's operation. At high slips. the high reactance of the inner cage almost removes it from the circuit. A MATLAB M-ftle to calculate and plot the two torque-speed characteristics is shown below: ~ M -file: t orque_speed_2. m ~ M -file c reate and p l ot of the torque- speed ~ i nduc t i on mot or wi th a doubl e - cage rotor curve of an des i gn. ~ Fir s t , i nit i a li ze rl = 0.64 1; " 0 0.750; e' 0 0.300; e2i 0 0.400; e' o 0 3.200; " 0 0.500; the va l u es n eeded i n thi s program. % Stato r r es i s tance •Stato r r eactance •Rotor r es i s tance foe s i ngl e - • cage motor •Rotor r es i s tance f oe i nner • cage of doubl e - cage mot or •Rotor r es i s tance foe outer • cage of doubl e - cage mot or •Rotor r eactance foe s i ngl e - • cage motor
- 449. x2 i = 3.300; x20 = 0.500; xm = 26.3; v-ph ase = 460 / sqrt (3) ; n_sync = 1 800; w_sync = 1 88.5; % Ca l c ula t e the Thevenin % 7- 4 1a a nd 7- 43. v_th 0 v-ph ase • ( = I INDUCTION MOTORS 425 ~ Rot or r eact a nce f or inne r ~ c age o f doub l e - cage mo t o r ~ Rot or r eact a nce f or out e r ~ c age o f doub l e - cage mo t o r ~ M agne ti zation br a nc h r eac t a n ce ~ Phase vo lt age ~ Syn c hronou s speed (r / min ) % Syn c hronou s speed (r ad/ s) vo lt age a nd impedance from Equa tion s sqrt (rl " 2 • (x l • xm ) " 2 ) I ; , " 0 ( ( j *xm ) • (d • - j *xl ) ) I (r1 • j * (x l • xm ) ) ; r -" 0 r ea l ( z_ th ) ; x-" 0 imag ( z_ th ) ; % Now ca l c ula t e the mot or speed f or ma ny s lips bet ween % 0 a nd 1. Not e tha t the fir s t s lip va lue i s set t o % 0.00 1 in s t ead o f exactly 0 t o avo i d d i v i de - by- ze r o % p r obl ems. % Slip s = ( 0, 1 ,50 ) / 50; s( l ) = 0.00 1 ; % Avoi d d i v i s i on-by- zer o run = ( 1 - s) * n_sync; % Mecha ni cal speed % Ca l c ula t e t or que f or the s ing l e - cage r ot or . f or ii = 1 ,5 1 t _ ind l ( ii ) = (3 * v_th" 2 * r 2 / s( ii )) / ... (w_sync * (( r _ th + r 2/ s( ii ))" 2 + (x_ th + x2 )" 2 ) ) ; e od % Ca l c ula t e r es i s t a nce a nd r eact a nce o f the doub l e - cage % r ot or a t thi s s lip, a nd the n u se those va lues t o % ca l c ula t e the induced t or que. f or ii = 1 :5 1 L r 0 1 / (r 2 i • j *s ( ii ) *x2 i ) • 1 / (r 20 • j *s (ii ) *x2o ) ; ,-r 0 l / y_r ; •Eff ective r ot or impedan ce r 2e ff 0 r ea l ( z_ r ) ; •Eff ective r ot or r es i s t a n ce x2e ff 0 imag ( z_ r ) ; •Eff ective r ot or r eac t a n ce ~ Ca l c ula t e induced t or que f or doub l e - cage r ot or . t _ ind2 ( ii ) = (3 * v_th" 2 * r 2e ff / s( ii )) / ... (w_sync * (( r _ th + r 2e f f/s (ii ))" 2 + (x_ th + x2eff )" 2 ) ) ; e od % Plot the t or que - speed c urves p l ot (run , t _ ind l , 'Col or' , 'k' , 'LineWi dth' ,2.0 ) ; h o l d on ; p l ot (run , t _ ind2, 'Col or' , 'k' , 'LineWi dth' ,2.0, 'LineSt y l e ' , '- . ' ) ; x l abe l ( ' itn_ (m} ' , 'Fontwei ght' , 'Bo l d ' ) ; y l abe l ( ' t au_ ( ind} ' , 'Fontwei ght' , 'Bo l d ' ) ; title ( 'Indu c ti on mot or t or que - spe ed ch a r act e ri s tics ' , 'Pontwei ght' , 'Bo l d ' ) ; l egend ( ' Sing l e - Cage Des i gn' , 'Doub l e - Cage Des i g n' ) ; g rid on ; h o l d o ff ;
- 450. 426 ELECTRIC MACHINERY RJNDAMENTALS 300 250 ~> . -., 200 E • Z 150 ] ./" ~ / ;? 1 - 100 50 Single-cage design .- . Double-cage design o o 200 400 600 800 1000 1200 1400 1600 1800 n",. rlmin ""GURE 7-29 Comparison of torque-speed characteristics for the single- and double-cage rotors of Example 7-6. The resulting torque-speed characteristics are shown in Figure 7-29. Note Ihal the double- cage design has a slightly higher slip in the normal operating range, a smaller maximum torque and a higher starting torque compared to Ihe corresponding single-cage rotor design. This behavior matches our theoretical discussions in this section. 7.7 TRENDS IN INDUCTION MOTOR DESIGN TIle fundamental ideas behind the induction motor were developed during the late 1880s by Nicola Tesla, who received a patent on his ideas in 1888. At that time, he presented a paper before the American lnstitute of Electrical Engineers [AlEE, predecessor of today's Institute of Electrical and Electronics Engineers (IEEE)] in which he described the basic principles of the wound-rotor induction motor, along with ideas for two other important ac motors-the synchronous motor and the re- luctance motor. Although the basic idea of the induction motor was described in 1888, the motor itself did not spring forth in full-fledged fonn. There was an initial period of rapid development, followed by a series of slow, evolutionary improvements which have continued to this day. TIle induction motor assumed recognizable modem form between 1888 and 1895. During that period, two- and three-phase power sources were developed to produce the rotating magnetic fields within the motor, distributed stator windings were developed, and the cage rotor was introduced. By 1896, fully functional and recognizable three-phase induction motors were commercially available. Between then and the early 1970s, there was continual improvement in the quality of the steels, the casting techniques, the insulation, and the construction
- 451. INDUCTION MOTORS 427 1903 1910 1920 , 1940 19!54 1974 FIGURE 7-30 The evolution of the induction motor. The motors shown in this figure are all rated at 220 V and 15 hp. There has been a dramatic decrease in motor size and material requirements in induction motors since the first practical ones were produced in the 1890s. (Courtesy ofGeneml Electric Company.) FIGURE 7-31 Typical early large induction motors. The motors shown were rated at 2(xx) hp. (Courtesy ofGeneml Electric Company.) features used in induction motors. 1l1ese trends resulted in a smaller motor for a given power output, yielding considerable savings in construction costs. In fact, a modern 100-hp motor is the same physical size as a 7.S-hp motor of 1897. This progression is vividly illustrated by the IS-hp induction motors shown in Figure 7- 30. (See also Figure 7- 31.)
- 452. 428 ELECTRIC MACHINERY RJNDAMENTALS However, these improvements in induction motor design did not necessar- ily lead to improvements in motor operating efficiency. The major design effort was directed toward reducing the initial materials cost ofthe machines, not toward increasing their efficiency. The design effort was oriented in that direction because electricity was so inexpensive, making the up-front cost of a motor the principal criterion used by purchasers in its selection. Since the price of oil began its spectacular climb in 1973, the lifetime oper- ating cost of machines has become more and more important, and the initial in- stallation cost has become relatively less important. As a result of these trends, new emphasis has been placed on motor e fficiency both by designers and by end users of the machines. New lines of high-efficiency induction motors are now being produced by all major manufacturers, and they are fanning an ever-increasing share of the in- duction motor market. Several techniques are used to improve the efficiency of these motors compared to the traditional standard-efficiency designs. Among these techniques are I. More copper is used in the stator windings to red uce copper losses. 2. The rotor and stator core length is increased to reduce the magnetic nux den- sity in the air gap of the machine. This reduces the magnetic saturation of the machine, decreasing core losses. 3. More steel is used in the stator of the machine, allowing a greater amount of heat transfer out of the motor and reducing its operating temperature. The ro- tor's fan is then redesigned to reduce windage losses. 4. The steel used in the stator is a special high-grade electrical steel with low hysteresis losses. 5. The steel is made of an especially thin gauge (i.e., the laminations are very close together), and the steel has a very high internal resistivity. Both effects tend to reduce the eddy current losses in the motor. 6. The rotor is carefull y machined to produce a unifonn air gap, reducing the stray load losses in the motor. In addition to the general techniques described above, each manufacturer has his own unique approaches to improving motor efficiency. A typical high- efficiency induction motor is shown in Figure 7- 32. To aid in the comparison of motor efficiencies, NEMA has adopted a stan- dard technique for measuring motor efficiency based on Method 8 of the IEEE Standard 11 2, Test Procedure for Polyphase Induction Motors and Generators. NEMA has also introduced a rating called NEMA nominal efficiency, which ap- pears on the nameplates of design class A, 8 , and C motors. The nominal effi- ciency identifies the average efficiency of a large number of motors of a given model, and it also guarantees a certain minimum efficiency for that type of motor. The standard NEMA nominal efficiencies are shown in Figure 7- 33.
- 453. INDUCTION MOTORS 429 FIGURE 7-32 A General Electric Energy Saver motor. typical of modem high-efficiency induction motors. (Courtesy ofGeneml Electric Company.) Nomin al Guanmh.'t!d minimum Nominal Guara nte... >d minimum efficiency, o/~ efficiency, % efficiency, % efficiency, '7~ 95.0 94.1 SO.O 77.0 94.5 93.6 78.5 75.5 94.1 93.0 77.0 74.0 93.6 92.4 75.5 72.0 93.0 91.7 74.0 70.0 92.4 91.0 72.0 68.0 91.7 90.2 70.0 66.0 91.0 89.5 68.0 64.0 90.2 88.5 66.0 62.0 89.5 87.5 64.0 59.5 88.5 86.5 62.0 57.5 87.5 85.5 59.5 55.0 86.5 84.0 57.5 52.5 85.5 82.5 55.0 50.' 84.0 81.5 52.5 48.0 82.5 80.0 50.' 46.0 81.5 78.5 FIGURE 7-33 Table of NEMA nominal efficiency standards. The nominal efficiency represents the mean efficiency of a large number of sample motors. and the 8uar:mteed minimum efficiency represents the lowest permissible efficiency for any given motor of the class. (Reproduced by permissionfrom Motors and GenemfOrs, NEMA Publication MG-I. copyright 1987 by NEMA.)
- 454. 430 ELECTRIC MACHINERY RJNDAMENTALS Other standards organizations have also established efficiency standards for induction motors, the most important ofwhich are the 8ritish (8S-269), IEC (IEC 34-2), and Japanese (JEC-37) standards. However, the techniques prescribed for measuring induction motor efficiency are different in each standard and yield dif- ferent results for the same physical machine. If two motors are each rated at 82.5 percent efficiency, but they are measured according to different standards, then they may not be equally efficient. When two motors are compared, it is important to compare efficiencies measured under the same standard. 7.8 STARTING INDUCTION MOTORS Induction motors do not present the types of starting problems that synchronous motors do. In many cases, induction motors can be started by simply connecting them to the power line. However, there are sometimes good reasons for not doing this. For example, the starting current required may cause such a dip in the power system voltage that across-the-line staning is not acceptable. For wound-rotor induction motors, starting can be achieved at relatively low currents by inserting extra resistance in the rotor circuit during starting. lllis extra resistance not only increases the starting torque but also reduces the starting current. For cage induction motors, the starting current can vary widely depending primarily on the motor's rated power and on the effective rotor resistance at start- ing conditions. To estimate the rotor current at starting conditions, all cage motors now have a starting code letter (not to be confused with their design class letter) on their nameplates. The code letter sets limits on the amount of current the mo- tor can draw at starting conditions. 1l1ese limits are expressed in tenns of the starting apparent power of the motor as a function of its horsepower rating. Figure 7- 34 is a table containing the starting kilovoltamperes per horsepower for each code letter. To determine the starting current for an induction motor, read the rated volt- age, horsepower, and code letter from its nameplate. 1l1en the starting apparent power for the motor will be S.1Mt = (rated horsepower)(code letter factor) and the starting current can be found from the equation S.tan IL = V3V T (7- 55) (7- 56) Example 7-7. What is the starting ClUTent of a 15-hp, 208-V, code-Ietter-F, three- phase induction motor? Solutio" According to Figure 7- 34, the maximwn kilovoltamperes per horsepower is 5.6. Therefore, the maximum starting kilovoltamperes of this motor is S,w<. = (15 hp)(5.6) = 84 kVA
- 455. INDUCTION MOTORS 431 Nominal code Lock... >d rotor, Nominal code Locked rotor, letter kYAJhp letter kVAJhp A 0--3.15 L 9.00-10.00 B 3.15--3.55 M 10.00-11.00 C 3.55-4.00 N 11.20-12.50 D 4.00-4.50 P 12.50-14.00 E 4.50-5.00 R 14.00-16.00 F 5.00-5.60 S 16.00-18.00 G 5.60--6.30 T 18.00-20.00 H 6.30--7.10 U 20.00-22.40 J 7.7- 8.00 V 22.40 and up K 8.00-9.00 FIGURE 7-34 Table of NEMA code letters. indicating the starting kilovoltamperes per horsepower of rating for a motor. Each code letter extends up to. but does not include. the lower bound of the next higher class. (Reproduced 17y permission from Motors and Generators. NEMA Publication MG-I. copyright 1987 byNEMA.) The starting current is thus (7- 56) 84kVA = vi3"(208 V) = 233 A If necessary, the starting current of an induction motor may be reduced by a starting circuit. However, if this is done, it will also reduce the starting torque of the motor. One way to reduce the starting current is to insert extra inductors or resis- tors into the power line during starting. While fonnerly common, this approach is rare today. An alternative approach is to reduce the motor's terminal voltage dur- ing starting by using autotransformers to step it down. Figure 7- 35 shows a typi- cal reduced-voltage starting circuit using autotransfonners. During starting, con- tacts 1 and 3 are shut, supplying a lower voltage to the motor. Once the motor is nearly up to speed, those contacts are opened and contacts 2 are shut. These con- tacts put fuJI line voltage across the motor. It is important to realize that while the starting current is reduced in direct proportion to the decrease in terminal voltage, the starting torque decreases as the square of the applied voltage. Therefore, only a certain amount of current reduc- tion can be done if the motor is to start with a shaft load attached.
- 456. 432 ELECTRIC MACHINERY RJNDAMENTALS Line terminals 2 2 " " " " 3 3 Motor terminals Starting sequence: (a) Close I and 3 (b) Open I and 3 (c) Close 2 ""GURE 7-35 An autotransfonner starter for an induction motor. ~ ---1 1 F, M 'I II ~ F, M , ~ F, M, Disronnect switch Start Overload heaters Induction motor Stop OL LL---r---"--;~M ""GURE 7- 36 A typical across-too-line starter for an induction motor. Induction Motor Starting Circuits A typical full-voltage or across-the-Iine magnetic induction motor starter circuit is shown in Figure 7- 36, and the meanings of the symbols used in the figure are ex- plained in Figure 7- 37. This operation of this circuit is very simple. When the start button is pressed, the relay (or contactor) coil M is energized, causing the normally open contacts Mt , M2, and M) to shut. When these contacts shut, power is applied to the induction motor, and the motor starts. Contact M4 also shuts,
- 457. ---11 If- 0 e 0 II )( rx, OL )f FIGURE 7-37 Normally open Normally shut INDUCTION MOTORS 433 Disconnect switch Push button; push to close Push button; push to open Relay coil; contacts change state when the coil energizes Contact open when coil deenergized Contact shut when coil deenergized Overload heater Overload contact; opens when the heater gets too wann Typical components found in induction motor control circuits. which shorts out the starting switch, allowing the operator to release it without re- moving power from the M relay. When the stop button is pressed, the M relay is deenergized, and the M contacts open, stopping the motor. A magnetic motor starter circuit of this sort has several bui It-in protective features: I. Short-circuit protection 2. Overload protection 3. Undervoltage protection Short-circuit protection for the molor is provided by fuses Ft , F2, and Fl. If a sudden short circuit develops within the motor and causes a current flow many limes larger than the rated current, these fuses will blow, disconnecting the motor from the power supply and preventing it from burning up. However, these fuses must not burn up during normal molor starting, so they are designed to require currents many times greater than the full -load current before they open the circuit. This means that short circuits through a high resistance and/or excessive motor loads will not be cleared by the fuses. Overload protection for the motor is provided by the devices labeled OL in the figure. These overload protection devices consist of two parts, an overload
- 458. 434 ELECTRIC MACHINERY RJNDAMENTALS heater element and overload contacts. Under nonnal conditions, the overload con- tacts are shut. However, when the temperature of the heater elements rises far enough, the OL contacts open, deenergizing the M relay, which in turn opens the normally open M contacts and removes power from the motor. When an induction motor is overloaded, it is eventually drunaged by the ex- cessive heating caused by its high currents. However, this damage takes time, and an induction motor will not nonnally be hurt by brief periods of high currents (such as starting currents). Only if the high current is sustained will damage occur. The overload heater elements also depend on heat for their operation, so they wil l not be affected by brief perioos of high current during starting, and yet they wil I operate during long periods of high current, removing power from the motor be- fore it can be damaged. Undefi!oltage protection is provided by the controller as well. Notice from the figure that the control power for the M relay comes from directly across the lines to the motor. If the voltage applied to the motor falls too much, the voltage applied to the M relay will also fall and the relay will deenergize. TIle M contacts then open, removing power from the motor tenninals. An induction motor starting circuit with resistors to reduce the starting cur- rent flow is shown in Figure 7- 38. TIlis circuit is similar to the previous one, ex- cept that there are additional components present to control removal of the start- ing resistor. Relays lID, 2TD, and 3TD in Figure 7- 38 are so-called time-delay relays, meaning that when they are energized there is a set time delay before their contacts shut. When the start button is pushed in this circuit, the M relay energizes and power is applied to the motor as before. Since the 1ID, 2TD, and 3ID contacts are all open, the full starting resistor is in series with the motor, reducing the start- ing current. When the M contacts close, notice that the 1ID relay is energized. How- ever, there is a finite delay before the lTD contacts close. During that time, the motor partially speeds up, and the starting current drops off some. After that time, the 1TO contacts close, cutting out part of the starting resistance and simultane- ously energizing the 2TD relay. After another delay, the 2TD contacts shut, cut- ting out the second part of the resistor and energizing the 3TD relay. Finally, the 3TD contacts close, and the entire starting resistor is out of the circuit. By a judicious choice of resistor values and time delays, this starting circuit can be used to prevent the motor starting current from becoming dangerously large, while still allowing enough current flow to ensure prompt acceleration to normal operating speeds. 7.9 SPEED CONTROL OF INDUCTION MOTORS Until the advent of modern solid-state drives, induction motors in general were not good machines for applications requiring considerable speed control. The normal operating range of a typical induction motor (design classes A, B, and C)
- 459. INDUCTION MOTORS 435 Overload F M] heaters / ~I r~~~~ ----- - l..L....l..I- 1r--- Resis]or 3TD Resis]or lTD 2TD 3TD Resistor lTD 2TD 3TD S<w Stop I OL lID lID 2ID 2ID 3ID FIGURE 7-38 A three-step resistive staner for an induction motor. Induction motor is confined to less than 5 percent slip, and the speed variation over that range is more or less directly proportional to the load on the shaft of the motor. Even if the slip could be made larger, the effi ciency of the motor would become very poor, since the rotor copper losses are directly proportional to the slip on the motor (remember that PRCL = sPAG). There are really only two techniques by which the speed of an induction motor can be controlled. One is to vary the synchronous speed, which is the speed of the stator and rotor magnetic fields, since the rotor speed always remains near n,ync. The other technique is to vary the slip of the motor for a given load. E:1.ch of these approaches will be taken up in more detail. The synchronous speed of an induction motor is given by
- 460. 436 ELECTRIC MACHINERY RJNDAMENTALS 120f e p (7- 1) so the only ways in which the synchronous speed ofthe machine can be varied are (I) by changing the electrical frequency and (2) by changing the number of poles on the machine. Slip control may be accomplished by varying either the rotor re- sistance or the terminal voltage of the motor. Induction Motor Speed Control by Pole Changing TIlere are two major approaches to changing the number of poles in an induction motor: I. The method of consequent poles 2. Multiple stator windings TIle method of consequent poles is quite an old method for speed control, having been originally developed in 1897. It relies on the fact that the number of poles in the stator windings of an induction motor can easily be changed by a fac- tor of 2:I with only simple changes in coil connections. Figure 7- 39 shows a , Winding connections at back end of stator b fo'IGURE 7-39 d, --- b', p~= 60° 0, , "", , / ' , b', "", --- " b " ---- , " /' a, A two-pole stator winding for pole changing. Notice the very small rotor pitch of these windings.
- 461. INDUCTION MOTORS 437 simple two-pole induction motor stator suitable for pole changing. Notice that the individual coils are of very short pitch (60 to 90°). Figure 7-40 shows phase a of these windings separately for more clarity of detail. Figure 7-40a shows the current now in phase a of the stator windings at an instant of time during nonnal operation. Note that the magnetic field leaves the sta- tor in the upper phase group (a north pole) and enters the stator in the lower phase group (a south pole). This winding is thus pnxlucing two stator magnetic poles. a', I(t) I ,,' ) NtS ) N t S a] d] a2 d 2 S B - ,b, FIGURE 7-40 Connections at far end of stator ,- ", , "" ,- ", ", , "" " B ' B /, " " B ' N a', B S B A close-up view of one phase of a pole-changing winding. (a) In the two-pole configuration. one coil is a north pole and the other one is a south pole. (b) When the connection on one of the two coils is reversed. they are both nonh poles. and the magnetic flux returns to the stator at points halfway between the two coils. The south poles are called consequent poles. and the winding is now a four- pole winding.
- 462. 438 ELECTRIC MACHINERY RJNDAMENTALS Now suppose that the direction of current flow in the lower phase group on the stator is reversed (Figure 7--40b).1llen the magnetic field will leave the stator in both the upper phase group and the lower phase group-each one will be a north magnetic pole. The magnetic fl ux in this machine must return to the stator between the two phase groups, producing a pair of consequent south magnetic poles. Notice that now the stator has four magnetic poles-twice as many as before. 1lle rotor in such a motor is of the cage design, since a cage rotor always has as many poles induced in it as there are in the stator and can thus adapt when the number of stator poles changes. When the motor is reconnected from two-pole to four-pole operation, the resulting maximum torque of the induction motor can be the same as before (constant-torque connection), half of its previous value (square-law-torque con- nection, used for fans, etc.), or twice its previous value (constant-output-power connection), depending on how the stator windings are rearranged. Figure 7--4 1 shows the possible stator connections and their effect on the torque-speed curve. 1lle major disadvantage of the consequent-pole method of changing speed is that the speeds must be in a ratio of 2:I. 1lle traditional approach to overcom- ing this limitation was to employ multiple stator windings with different numbers of poles and to energize only one set at a time. For example, a motor might be wound with a four-pole and a six-pole set of stator windings, and its synchronous speed on a 6O-Hz system could be switched from 1800 to 1200 r/min simply by supplying power to the other set of windings. Unfortunately, multiple stator wind- ings increase the expense of the motor and are therefore used only when ab- solutely necessary. By combining the method of consequent poles with multiple stator wind- ings, it is possible to build a four-speed induction motor. For example, with sepa- rate four- and six-pole windings, it is possible to produce a 6O-Hz motor capable of running at 600, 900, 1200, and 1800 r/min. Speed Control by Changing the Line Frequency If the electrical frequency applied to the stator of an induction motor is changed, the rate of rotation of its magnetic fields "')'DC will change in direct proportion to the change in electrical frequency, and the no-load point on the torque-speed characteristic curve will change with it (see Figure 7--42). The synchronous speed ofthe motor at rated conditions is known as the base speed. By using variable fre- quency control, it is possible to adjust the speed of the motor either above or be- low base speed. A properly designed variable-frequency induction motor drive can be very flexible. It can control the speed of an induction motor over a range from as little as 5 percent of base speed up to about twice base speed. However, it is important to maintain certain voltage and torque limits on the motor as the fre- quency is varied, to ensure safe operation. When running at speeds below the base speed of the motor, it is necessary to reduce the terminal voltage applied to the stator for proper operation. 1lle ter- minal voltage applied to the stator should be decreased linearly with decreasing
- 463. INDUCTION MOTORS 439 T, T, T, T, T, T, T, T, T, Lines Lines S","" L, L, L, S"""d L, L, L, Low T, T, T, T4, T~ T6 Low T, T, T, Tt -T2-Tj High T, T, T, S","" L, Low T, High T, FIGURE 7-41 T, (a) T, T, Lines L, T, T, (,) T, T, L, T, T, 0",," T)-TrTJ together T, T4, T." T6 0",," T)-TrTJ together High T, T, T, (b) (b, High speed 1:: (all) ,l""d(:"'::::1 r---' ~ (Cl Speed, rlmin (d) together T4, T~, T6 0",," Possible connections of the stator coils in a pole-changing motor. together with the resulting torque-speed characteristics: (a) Constant-torque connection- the torque capabilities of the motor remain approximately constant in both high-speed and low-speed connections. (b) Constant- lwrsepok'er connection---lhe power capabilities of the motor remain approximately constant in both h.igh-speed and low-speed connections. (el Fan torque connection---lhe torque capabilities of the motor change with speed in the same manner as f.an-type loads.
- 464. 440 ELECTRIC MACHINERY RJNDAMENTALS 800 ,-------------------------------------, 700 600 , • Z 500 • ¥400 ~ I 300 200 lOOe--, 800 700 600 , • Z 500 • ~ 400 " , 300 0 ~ 200 100 0 0 ""GURE 7-42 Mechanical speed. r/min ,., 1000 1500 2000 2500 3000 3500 Mechanical speed. r/min ,b, Variable-frequency speed control in an induction motor: (a) The family of torque-speed characteristic curves for speeds below base speed. assuming that the line voltage is derated linearly with frequency. (b) The family of torque-speed characteristic curves for speeds above base speed. assuming that the line voltage is held constant.
- 465. INDUCTION MOTORS 441 800,-------------------------------------, 700 600 , • Z500 , ~ 400 ] 300 • 200P------- o o 500 FIGURE 7-42 (roncluded) 1000 1500 2000 Mechanical speed. r/min I" (c) The torque-speed characteristic curves for all frequencies. 2500 3000 3500 stator frequency. This process is called derating. If it is not done, the steel in the core of the induction motor will saturate and excessive magnetization currents will flow in the machine. To understand the necessity for derating, recall that an induction motor is basically a rotating transfonner. As with any transfonner, the flux in the core of an induction motor can be found from Faraday's law: vet) = -N~ dl (1-36) If a voltage vet) = VM sin wt is applied to the core, the resulting flux ~ is 'Wi ~ J h l)dl p = ~ !VM sinwtdt p I ~t) = -~coswtl (7- 57) Note that the electrical frequency appears in the denominator of this expression. Therefore, if the electrical frequency applied to the stator decreases by 10 percent while the magnitude of the voltage applied to the stator remains constant, the flux in the core of the motor wi II increase by about 10 percent and the magnetization current of the motor will increase. In the unsaturated region of the motor's
- 466. 442 ELECTRIC MACHINERY RJNDAMENTALS magnetization curve, the increase in magnetization current will also be about 10 percent. However, in the saturated region of the motor's magnetiwtion curve, a 10 percent increase in flux requires a much larger increase in magnetization current. Induction motors are normally designed to operate near the saturation point on their magnetization curves, so the increase in flux due to a decrease in frequency will cause excessive magnetization currents to flow in the motor. (This same prob- lem was observed in transfonners; see Section 2.1 2.) To avoid excessive magnetization currents, it is customary to decrease the applied stator voltage in direct proportion to the decrease in frequency whenever the frequency falls below the rated frequency of the motor. Since the applied volt- age v appears in the numerator of Equation (7-57) and the frequency wappears in the denominator of Equation (7-57), the two effects counteract each other, and the magnetization current is unaffected. When the voltage applied to an induction motor is varied linearly with fre- quency below the base speed, the flux in the motor will remain approximately constant. TIlerefore, the maximum torque which the motor can supply remains fairly high. However, the maximum power rating of the motor must be decreased linearly with decreases in frequency to protect the stator circuit from overheating. TIle power supplied to a three-phase induction motor is given by P = v'JVLILcos (J If the voltage VL is decreased, then the maximum power P must also be decreased, or else the current flowing in the motor wi II become excessive, and the motor will overheat. Figure 7-42a shows a family of induction motor torque-speed characteris- tic curves for speeds below base speed, assuming that the magnitude of the stator voltage varies linearly with frequency. When the electrical frequency applied to the motor exceeds the rated fre- quency of the motor, the stator voltage is held constant at the rated value. Al- though saturation considerations would pennit the voltage to be raised above the rated value under these circumstances, it is limited to the rated voltage to protect the winding insulation of the motor. The higher the electrical frequency above base speed, the larger the denominator of Equation (7-57) becomes. Since the nu- merator tenn is held constant above rated frequency, the resulting flux in the ma- chine decreases and the maximum torque decreases with it. Figure 7-42b shows a family of induction motor torque- speed characteristic curves for speeds above base speed, assuming that the stator voltage is held constant. If the stator voltage is varied linearly with frequency below base speed and is held constant at rated value above base speed, then the resulting family of torque-speed characteristics is as shown in Figure 7-42c. TIle rated speed for the motor shown in Figure 7-42 is 1800 r/min. In the past, the principal disadvantage of electrical frequency control as a method of speed changing was that a dedicated generator or mechanical fre- quency changer was required to make it operate. This problem has disappeared with the development of modern solid-state variable-frequency motor drives. In
- 467. INDUCTION MOTORS 443 800 700 600 E "" Z ; ~ 400 ~ • , , • 300 • / - / Lo,' 200 100 0 -------- 0 250 '00 "" 1000 1250 1500 1750 2000 Mechanical speed. r/min FIGURE 7-43 Variable-line-voltage speed control in an induction motor. fact, changing the line frequency with solid-state motor drives has become the method of choice for induction motor speed control. Note that this method can be used with any induction motor, unlike the pole-changing technique, which re- quires a motor with special stator windings. A typical solid-state variable-frequency induction motor drive will be de- scribed in Section 7.10. Speed Control by Changing the Line Voltage The torque developed by an induction motor is proportional to the square of the applied voltage. Ifa load has a torque-speed characteristic such as the one shown in Figure 7-43, then the speed ofthe motor may be controlled over a limited range by varying the line voltage. This method of speed control is sometimes used on small motors driving fans. Speed Control by Changing the Rotor Resistance In wound-rotor induction motors, it is possible to change the shape of the torque- speed curve by inserting extra resistances into the rotor circuit of the ma- chine. The resulting torque- speed characteristic curves are shown in Figure 7-44.
- 468. 444 ELECTRIC MACHINERY RJNDAMENTALS 800 700 R, R, R, (j()() E 500 Z • , ~ 400 g " , , ~ 300 R] '" 2Ro 200 R2", 3Ro RJ ",4Ro RJ '" 5Ro 100 R3", 6Ro 0 0 250 500 750 1000 1250 1500 17'" 2000 Mechanical speed. r/min fo'IGURE 7-44 Speed control by varying the rotor resistance of a wound-rotor induction motor. If the torque-speed curve of the load is as shown in the figure, then changing the rotor resistance will change the operating speed of the motor. However, inserting extra resistances into the rotor circuit of an induction motor seriously red uces the efficiency of the machine. Such a method of speed control is nonnally used only for short periods because of this efficiency problem. 7.10 SOLID·STATE INDUCTION MOTOR DRIVES As mentioned in the previous section, the method of choice today for induction motor speed control is the solid-state variable-frequency induction motor drive. A typical drive of this sort is shown in Figure 7--45. TIle drive is very flexible: its in- put power can be either single-phase or three-phase, either 50 or 60 Hz, and any- where from 208 to 230 V. The output from this drive is a three-phase set of volt- ages whose frequency can be varied from 0 up to 120 Hz and whose voltage can be varied from 0 V up to the rated voltage of the motor. TIle output voltage and frequency control is achieved by using the pulse- width modulation (PWM) techniques described in Chapter 3. Both output frequency and output voltage can be controlled independently by pulse-width modulation. fig- ure 7--46 illustrates the manner in which the PWM drive can control the output fre- quency while maintaining a constant nns voltage level, while Figure 7--47 illustrates
- 469. • • / Voltage. V Voltage. V 100 20 o 10 - 100 FIGURE 7-46 INDUCTION MOTORS 445 fo'IGURE 7-45 A typical solid-state variable-frequency induction motor drive. (Courtesy ofMagneTek, Inc.) I. ms (a) 30 40 50 I. ms ,bl Variable-frequency control with a PWM waveform: (a) 6O-Hz.. 120-V PWM waveform: (b) 30-Hz. 12()'V PWM waveform.
- 470. 446 ELECTRIC MACHINERY RJNDAMENTALS Voltage, V 100 t , ms - 100 "J Voltage, V 100 10 30 50 O ~ mf t , ms - 100 20 40 'bJ ""GURE 7-47 Variable voltage control with a PWM waveform: (a) 6O-Hz, 120-V PWM waveform: (b) 6().Hz, 6().V PWM waveform. the manner in which the PWM drive can control the nns voltage level while main- taining a constant frequency. As we described in Section 7.9, it is often desirable to vary the output fre- quency and output nns voltage together in a linear fashion. Figure 7-48 shows typical output voltage wavefonns from one phase of the drive for the situation in which frequency and voltage are varied simultaneously in a linear fashion.· Fig- ure 7-48a shows the output voltage adjusted for a frequency of60 Hz and an nns voltage of 120 V. Figure 7-48b shows the output adjusted for a frequency of 30 Hz and an nns voltage of 60 V, and Figure 7-48c shows the output adjusted for a frequency of 20 Hz and an nns voltage of 40 V. Notice that the peak voltage out of the drive remains the same in all three cases; the nns voltage level is controlled by the fraction of time the voltage is switched on, and the frequency is controlled by the rate at which the polarity of the pulses switches from positive to negative and back again. TIle typical induction motor drive shown in Figure 7-45 has many built-in features which contribute to its adjustability and ease of use. Here is a summary of some of these features. *The output waveforms in Figure 7-47 are actually simplified waveforms. The real induction motor drive has a much higher carrier frequency than that shown in the figure.
- 471. INDUCTION MOTORS 447 Voltage. V PVM waveform t. ms ,., Voltage. V PVM waveform 100 20 30 o 10 40 50 t. ms - 100 ,b, Voltage. V PWM waveform 100 o WUlli:lm 30 , 40 t. ms - 100 10 20 ,< , FIGURE 7-48 Simultaneous voltage and frequency control with a PVM wavefonn: (a) 6O-Hz. 120-V PWM waveform: (b) 30-Hz. 60-V PWM waveform: (c) 2O-Hz. 40-V PWM waveform. Frequency (Speed) Adjustment The output frequency of the drive can be controlled manually from a control mounted on the drive cabinet, or it can be controlled remotely by an external volt- age or current signal. The ability to adjust the frequency of the drive in response to some external signal is very important, since it permits an external computer or process controller to control the speed of the motor in accordance with the over- all needs of the plant in which it is installed.
- 472. 448 ELECTRIC MACHINERY RJNDAMENTALS A Choice of Voltage and Frequency Patterns TIle types of mechanical loads which might be attached to an induction motor vary greatly. Some loads such as fans require very little torque when starting (or running at low speeds) and have torques which increase as the square of the speed. Other loads might be harder to start, requiring more than the rated full-load torque of the motor just to get the load moving. TIlis drive provides a variety of voltage- versus-frequency patterns which can be selected to match the torque from the in- duction motor to the torque required by its load. TIlree of these patterns are shown in Figures 7-49 through 7- 5 J. Figure 7-49a shows the standard or general-purpose voltage-versus- frequency pattern, described in the previous section. This pattern changes the out- put voltage linearly with changes in output frequency for speeds below base speed and holds the output voltage constant for speeds above base speed. (The small constant-voltage region at very low frequencies is necessary to ensure that there will be some starting torque at the very lowest speeds.) Figure 7-49b shows the resulting induction motor torque-speed characteristics for several operating fre- quencies below base speed. Figure 7- 50a shows the voltage-versus-frequency pattern used for loads with high starting torques. This pattern also changes the output voltage linearly with changes in output frequency for speeds below base speed, but it has a shal- lower slope at frequencies below 30 Hz. For any given frequency below 30 Hz, the output voltage will be higherthan it was with the previous pattern. This higher voltage will produce a higher torque, but at the cost of increased magnetic satura- tion and higher magnetization currents. The increased saturation and higher cur- rents are often acceptable for the short periods required to start heavy loads. Fig- ure 7- 50b shows the induction motor torque-speed characteristics for several operating frequencies below base speed. Notice the increased torque available at low frequencies compared to Figure 7-49b. Figure 7-51 a shows the voltage-versus-frequency pattern used for loads with low starting torques (called soft-start loads).TIlis pattern changes the output voltage parabolically with changes in output frequency for speeds below base speed. For any given frequency below 60 Hz, the output voltage will be lower than it was with the standard pattern. TIlis lower voltage will produce a lower torque, providing a slow, smooth start for low-torque loads. Figure 7- 5 Ib shows the induction motor torque-speed characteristics for several operating frequencies below base speed. Notice the decreased torque available at low frequencies com- pared to Figure 7-49. Independently Adjustable Acceleration and Deceleration Ramps When the desired operating speed of the motor is changed, the drive controlling it will change frequency to bring the motor to the new operating speed. If the speed change is sudden (e.g., an instantaneous jump from 900 to 1200 rImin), the drive
- 473. v 8Ol 700 roo E • '00 Z • , 4Ol ~ ~ 300 WO 100 0 0 FIGURE 7-49 INDUCTION MOTORS 449 O ~----------~ ffi C----------C1~ W c---- f HZ f_. (a) Torque--speed characteristic 200 4Ol roo 800 1000 12lXl 1400 Speed. rlmin ,b, 1roo 18Ol (a) Possible voltage-versus-frequency patterns for the solid-state variable-frequency induction motor drive: general-purpose paT/ern. This pattern consists of a linear voltage-frequency curve below rated frequency and a constant voltage above rated frequency. (b) The resulting torque-speed characteristic curves for speeds below rated frequency (speeds above rated frequency look like Figure 7-4lb).
- 474. 450 ELECTRIC MACHINERY RJNDA MENTALS , • Z • , ~ ~ 8Ol 700 600 5Ol 2llll 100 v V..,od , , , , , , , , , o !------~ 60 ~----~1~ 2~O-- f. Hz f~. "I Torque-speed characteristic O ~~~~~~~-o~L,,~-e~-+~-..~-.t, o 200 400 600 800 1000 1200 1400 1600 1800 Speed. r/min 'hI fo'IGURE 7-50 (a) Possible voltage-versus-frequency patterns for the solid-state variable-frequency induction motor drive: high-starting-torque patfem. This is a modified voltage-frequency pattern suitable for loads requiring high starting torques. lt is the same as the linear voltage- frequency pattern except at low speeds. The voltage is disproportionately high at very low speeds. which produces extra torque at the cost of a h.igher magnetization current. (b) The resulting torque-speed characteristic curves for speeds below rated frequency (speeds above rated frequency look like Figure 7--41b).
- 475. INDUCTION MOTORS 451 v O ~----------iro '----------'I~W o---- f Hz (a) 8Ol Torque--speed characteristic 700 roo E • '00 Z • , 4Ol ~ ~ 300 WO 100 0 0 200 4Ol 800 1000 12lXl 1400 lroo 18Ol Speed. r/min 'h' FIGURE 7-51 (a) Possible voltage-versus-frequency patterns for the solid-state variable-frequency induction motor drive:fan torque pattern. This is a voltage-frequency pattern suitable for use with motors driving fans and centrifugal pumps. which have a very low starting torque. (b) The resulting torque-speed characteristic curves for speeds below rated frequency (speeds above rated frequency look like Figure 7-4lb).
- 476. 452 ELECTRIC MACHINERY RJNDAMENTALS does not try to make the motor instantaneously jump from the old desired speed to the new desired speed. Instead, the rate of motor acceleration or deceleration is limited to a safe level by special circuits built into the electronics of the drive. TIlese rates can be adjusted independently for accelerations and decelerations. Motor Protection TIle induction motor drive has built into it a variety of features designed to protect the motor attached to the drive. The drive can detect excessive steady-state cur- rents (an overload condition), excessive instantaneous currents, overvoltage con- ditions, or undervoltage conditions. In any of the above cases, it will shut down the motor. Induction motor drives like the one described above are now so flexible and reliable that induction motors with these drives are displacing dc motors in many applications which require a wide range of speed variation. 7.11 DETERMINING CIRCUIT MODEL PARAMETERS TIle equivalent circuit of an induction motor is a very useful tool for detennining the motor's response to changes in load. However, if a model is to be used for a real machine, it is necessary to detennine what the element values are that go into the model. How can Rl> R2, Xl> X2, and XM be detennined for a real motor? These pieces of information may be found by perfonning a series of tests on the induction motor that are analogous to the short-circuit and open-circuit tests in a transfonner. TIle tests must be performed under precisely controlled conditions, since the resistances vary with temperature and the rotor resistance also varies with rotor frequency. The exact details of how each induction motor test must be performed in order to achieve accurate results are described in IEEE Standard 11 2. Although the details of the tests are very complicated, the concepts behind them are relatively straightforward and will be explained here. The No-Load Test TIle no-load test of an induction motor measures the rotational losses of the mo- tor and provides infonnation about its magnetization current. The test circuit for this test is shown in Figure 7- 52a. Wattmeters, a voltmeter, and three ammeters are connected to an induction motor, which is allowed to spin freely. The only load on the motor is the friction and windage losses, so all Peon v in this motor is consumed by mechanical losses, and the slip of the motor is very small (possibly as small as 0.00 1 or less). TIle equivalent circuit of this motor is shown in Figure 7- 52b. With its very small slip, the resistance corresponding to its power con- verted, Rl l - sys, is much much larger than the resistance corresponding to the rotor copper losses R2and much larger than the rotor reactance X2. Inthis case, the equivalent circuit reduces approximately to the last circuit in Figure 7- 52b. There,
- 477. Initial Variable voltage, variable frequency, three-phase power 00.= I, - + equivalent V. ( cin:uit: I, Since - R2('~S)>>R2 + v .( ,,' R2 (' - s)>>X2, , this cin:uit reduces to: + Combining V. ( RF&w and Reyields: FIGURE 7-52 R, jXt R, , R, INDUCTION MOTORS 453 I, - P, A I, - A No load I, - P, A lit + IB+ Ie IL = 3 ,., 12= 0 - jX2 R, " I. j R, jXM ~ R2(';S) ~ Rfri<:tiOll, win<b!O, ~ &""'" »XM 'h ' The no-load test of an induction motor: (a) test circuit; (b) the resulting motor equivalent cin:uit. Note that at no load the motor's impedance is essentially the series combination of RI,jXj • andJXM . the output resistor is in parallel with the magnetization reactance XM and the core losses Re. In this motor at no-load conditions, the input power measured by the meters must equal the losses in the motor. The rotor copper losses are negligible because the current / l is extremely small [because of the large load resistance R2(1 - s)/s], so they may be neglected. The stator copper losses are given by
- 478. 454 ELECTRIC MACHINERY RJNDAMENTALS so the input power must equal P;o = PSCL + P.:orc + PF& W + Pmisc = 3f r RI + Prot where Prot is the rotational losses of the motor: PM = Poore + PF&W + Pmis.c (7- 25) (7- 58) (7- 59) TI1US, given the input power to the motor, the rotational losses of the machine may be detennined. TIle equivalent circuit that describes the motor operating in this condition contains resistors Re and R2( I - sys in parallel with the magnetizing reactance XM . The current needed to establish a magnetic field is quite large in an induction mo- tor, because of the high reluctance of its air gap, so the reactance XM will be much smaller than the resistances in parallel with it and the overall input power factor will be very small. With the large lagging current, most of the voltage drop will be across the inductive components in the circuit. The equivalent input impedance is thus approximately (7-60) and if Xl can be found in some other fashion, the magnetizing impedance XMwill be known for the motor. The DC Test for Stator Resistance TIle rotor resistance R2plays an extremely critical role in the operation of an in- duction motor. Among other things, Rl determines the shape of the torque-speed curve, detennining the speed at which the pullout torque occurs. A standard mo- tor test called the locked-rotor test can be used to detennine the total motor circuit resistance (this test is taken up in the next section). However, this test ftnds only the total resistance. To find the rotor resistance Rl accurately, it is necessary to know RI so that it can be subtracted from the total. TIlere is a test for Rl independent of Rl , Xl and X2. This test is called the dc test. Basically, a dc voltage is applied to the stator windings of an induction mo- tor. Because the current is dc, there is no induced voltage in the rotor circuit and no resulting rotor current now. Also, the reactance of the motor is zero at direct current. TIlerefore, the only quantity limiting current now in the motor is the sta- tor resistance, and that resistance can be detennined. TIle basic circuit for the dc test is shown in Figure 7- 53. This figure shows a dc power supply connected to two of the three terminals of a V-connected in- duction motor. To perfonn the test, the current in the stator windings is adjusted to the rated value, and the voltage between the terminals is measured. TIle current in
- 479. Voc (variable) + '-/ FIGURE 7-53 Current- limiting resistor Test ci['(;uit for a dc resistance test. INDUCTION MOTORS 455 R, the stator windings is adjusted to the rated value in an attempl to heat the wind- ings to the same temperature they would have during nonnal operation (remem- ber, winding resistance is a function of temperature). The currenl in Figure 7- 53 flows through two of the windings, so the total resistance in the current path is 2Rt . Therefore, Voe 2Rt = -J- oe I R, Vee I 2loc (7-6 1) With this value of Rt the stator copper losses at no load may be detennined, and the rotational losses may be found as the difference between the input power at no load and the stator copper losses. The value of Rt calculated in this fashion is not completely accurate, since it neglects the skin effect that occurs when an ac voltage is applied to the wind- ings. More details concerning corrections for temperature and skin effect can be found in IEEE Standard 11 2. The Locked-Rotor Test The third test that can be perfonned on an induction motor to detennine its circuit parameters is called the locked-rotor test, or sometimes the blocked-rotor test. This test corresponds to the short-circuit test on a transformer. In this test, the ro- tor is locked or blocked so that it cannot move, a voltage is applied to the motor, and the resulting voltage, current, and power are measured. Figure 7- 54a shows the connections for the locked-rotor test. To perform the locked-rotor test, an ac voltage is applied to the stator, and the current flow is adjusted to be approximately full-load value. When the current is fu ll-load value, the voltage, current, and power flowing into the motor are measured. TIle equiva- lent circuit for this test is shown in Figure 7- 54b. Notice that since the rotor is not moving, the slip s = 1, and so the rotor resistance Ris is just equal to R2(quite a
- 480. 456 ELECTRIC MACHINERY RJNDAMENTALS I, - Adjustable- V voltage. I, adjustable- b - A frequency. three-phase power source I, , - A ,,' I, R, I, - - v, 'h' I<'IGURE 7-54 p p /,=/,=It.., XM »IR2+jX21 Rc »IR2 +jX21 So neglect Rc and XM The locked-rotor test for an induction motor: (a) test circuit: (b) motor equivalem circuit. small value). Since R2and X2are so small, almost all the input current will flow through them, instead of through the much larger magnetizing reactance XM . Therefore, the circuit under these conditions looks like a series combination ofX ], R], X2 , and Rl . nlere is one problem with this test, however. In nonnal operation, the stator frequency is the line frequency of tile power system (50 or 60 Hz). At starting con- ditions, the rotor is also at line frequency. However, at nonnal operating conditions, the slip of most motors is only 2 to 4 percent, and the resulting rotor frequency is in the range of I to 3 Hz. nlis creates a problem in that the line frequency does not represent the nonnal operating conditions ofthe rotor. Since effective rotor resis- tance is a strong function of frequency for design class Band C motors, the incor- rect rotor frequency can lead to misleading results in this test. A typical compro- mise is to use a frequency 25 percent or less of the rated frequency. While this approach is acceptable for essentially constant resistance rotors (design classes A and D), it leaves a lot to be desired when one is trying to find the nonnal rotor re- sistance of a variable-resistance rotor. Because of these and similar problems, a great deal of care must be exercised in taking measurements for these tests.
- 481. INDUCTION MOTORS 457 After a test voltage and frequency have been set up, the current fl ow in the motor is quickly adjusted to about the rated value, and the input power, voltage, and current are measured before the rotor can heat up too much. 1lle input power to the motor is given by P = V3"VT IL cos (J so the locked-rotor power factor can be found as PF = cos (} = V3V, I " and the impedance angle (J is just equal to cos-l PF. (7-62) The magnitude of the total impedance in the motor circuit at this time is and the angle of the total impedance is O. Therefore, ~ = RLR + jXi.R ~ IZ"loo, e+ jIZ"I'in e The locked-rotor resistance RLR is equal to I R" - R, + R, I while the locked-rotor reactance X~ is equal to XLR = X; + X; (7-63) (7-64) (7-65) (7-66) where X; and X; are the stator and rotor reactances at the test frequency, respectively. The rotor resistance Rl can now be found as (7-67) where Rl was detennined in the dc test. The total rotor reactance referred to the sta- tor can also be found. Since the reactance is directly proportional to the frequency, the total equivalent reactance at the nonnal operating frequency can be found as X LR = ~ra1.edX LR = Xl + X 2 llesl (7-68) Unfortunately, there is no simple way to separate the contributions of the stator and rotor reactances from each other. Over the years, experience has shown that motors of certain design types have certain proportions between the rotor and stator reactances. Figure 7- 55 summarizes this experience. In nonnal practice, it really does not matter just how XLR is broken down, since the reactance appears as the sum XI + X2in all the torque equations.
- 482. 458 ELECTRIC MACHINERY RJNDAMENTALS Xl and Xl as functions of XLR Rotor Dcsi ~n X, X, Wound rotor 0.5 XU! 0.5 XU! Design A 0.5 XU! 0.5 XU! Design B 0.4 XU! 0.6 XU! Design C 0.3 XU! 0.7 XU! Design D 0.5 XU! 0.5 XU! H GURE 7-55 Rules of thumb for dividing rotor and stator ci["(;uit reactance. Example 7-S. The following test data were taken on a 7.S-hp, four-pole, 20S-V, 60-Hz, design A, Y-colUlected induction motor having a rated current of 28 A. DC test: No-load test: Voc = 13.6 V Vr = 20SV IA = S.12A la = S.20A le = S.18A loc = 28.0A f = 60 Hz P",,= 420W Locked-rotor test: Vr = 25 V IA = 28.IA fa = 28.0A Ie = 27.6A f = IS Hz P"" = 920W (a) Sketch the per-phase equivalent circuit for this motor. (b) Find the slip at the pullout torque, and find the value of the pullout torque itself. Solutio" (a) From the dc test, Voc 13.6 V R[ = 2/0c = 2(2S.0A) = 0.2430 From the no-load test, IL.av = S.12A + 8.2~A + 8.ISA = S.17A 20S V V</l,nl = ~ = 1lOV
- 483. INDUCTION MOTORS 459 Therefore, 1 1 120 V ZnJ = 8.17 A = 14.70 = XI + XM When XI is known, XMcan be fOlUld. The stator copper losses are PSCL = 3/ f RI = 3(8.17 A)2(0.243 n) = 48.7 W Therefore, the no-load rotational losses are = 420W - 48.7 W = 371.3 W From the locked-rotor test, I = 28.IA + 28.0A + 27.6A = 279A L.av 3 . The locked-rotor impedance is 1 1 _1'. _--"'- _ 25V - ZLR - 110. - ..;J/1o. - V3"(27.9 A) - 0.517 n and the impedance angle (J is _ _ I P,n () - cos V!VTh _ _ I 920W - cos v'3(25 VX27.9 A) = cos- t 0.762 = 40.40 Therefore, RLR = 0.517 cos 40.4° = 0.394 0 = RI + R2. SinceRI = 0.243 n, R2must be 0.151 n. The reactance at IS Hz is X LR = 0.517 sin 40.40 = 0.335 0 The equivalent reactance at 60 Hz is X LR = ; : X LR = (~~~~) 0.3350 = 1.340 For design class A induction motors, this reactance is assumed to be divided equally between the rotor and stator, so XI = X2 = 0.67 n XM = IZruI- XI = 14.7 n - 0.67 0 = 14.03 n The final per-phase equivalent circuit is shown in Figure 7- 56. (b) For this equivalent circuit, the Thevenin equivalents are fOlUld from Equations (7-41 b), (7-44), and (7-45) to be Vlll = 114.6 V Rlll = 0.221 n Xlll = 0.67 n Therefore, the slip at the pullout torque is given by ,~~~R~, ~~~, ' - mox - v'R~ + (Xlll + X;ll (7- 53)
- 484. 460 ELECTRIC MACHINERY RJNDAMENTALS R, ""GURE 7-56 jO.67 fl. , , , , Rc -s: (unknown) <- , , , , jXM=j14.03fl. Motor per-phase equivalent circuit for Example 7-8. = 0.1510 =0.111 = 11.1% V(0.243 0)2 + (0.67 0 + 0.67 0)2 The maximum torque of this motor is given by Tmox = 2 W'YDC[Rll{ + VRfu + (Xll{ + X2)] _ 3(114.6 V)l (7- 54) - 2(188.5 rad/s)[O.22 1 0 + V(0.221 Of + (0.670 + 0.67 Of] = 66.2N o m 7.12 THE INDUCTION GENERATOR TIle torque-speed characteristic curve in Figure 7- 20 shows that if an induction motor is driven at a speed greater than n.y"" by an external prime mover, the di- rection of its inducted torque will reverse and it will act as a generator. As the torque applied to its shaft by the prime mover increases, the amount of power pro- duced by the induction generator increases.As Figure 7- 57 shows, there is a max- imum possible induced torque in the generator mode of operation. This torque is known as the pushover torque of the generator. If a prime mover applies a torque greater than the pushover torque to the shaft of an induction generator, the gener- ator wil I overspeed. As a generator, an induction machine has severe limitations. Because it lacks a separate field circuit, an inducti on generator cannot produce reactive power. In fact, it consumes reactive power, and an external source of reactive power must be connected to it at all times to maintain its stator magnetic field. 1llis external source of reactive power must also control the terminal voltage of the generator-with no field current, an induction generator cannot control its own output voltage. Normally, the generator's voltage is maintained by the exter- nal power system to which it is connected . 1lle one great advantage of an induction generator is its simplicity. An in- duction generator does not need a separate field circuit and does not have to be driven continuously at a fi xed speed. As long as the machine's speed is some
- 485. , 0 0 • - 1(XX) - 1500 0 FIGURE 7-57 ~ region Generator region ~ 1000 _/2000 "~ Mechanical speed. r/min Pushover torque 3000 INDUCTION MOTORS 461 The torque-speed characteristic of an induction machine. showing the generator region of operation. Note the pushover torque. value greater than n.ync for the power system to which it is connected, it will func- tion as a generator. The greater the torque applied to its shaft (up to a certain point), the greater its resulting output power. TIle fact that no fancy regulation is required makes this generator a gooj choice for windmills, heat recovery systems, and similar supplementary power sources attached to an existing power system. In such applications, power-factor correcti on can be provided by capacitors, and the generator's tenninal voltage can be controlled by the external power system. The Induction Generator Operating Alone It is also possible for an induction machine to function as an isolated generator, in- dependent of any power system, as long as capacitors are available to supply the reactive power req uired by the generator and by any attached loads. Such an iso- lated induction generator is shown in Figure 7- 58. The magnetizing current 1M required by an induction machine as a function of tenninal voltage can be found by running the machine as a motor at no load and measuring its annature current as a function of terrninal voltage. Such a magneti- zation curve is shown in Figure 7- 59a. To achieve a given voltage level in an in- duction generator, external capacitors must supply the magnetization current cor- responding to that level. Since the reactive current that a capacitor can produce is directly proportional to the voltage applied to it, the locus of all possible combinations of voltage and current through a capacitor is a straight line. Such a plot of voltage versus current
- 486. 462 ELECTRIC MACHINERY RJNDAMENTALS Terminals p I, - - Three-phase induction generator - - Q p I Q - Q To loads Capacitor bank ""GURE 7-58 An induction generator operating alone with a capacitor bank to supply reactive power. for a given frequency is shown in Figure 7-59b. If a three-phase set ofcapacitors is connected across the terminnls ofan induction generator, the no-load voltage of the induction generator will be the intersection of the generator's magnetization CUlVe and the capacitor sload line. TIle no-load tenninal voltage of an induction generator for three different sets of capacit.:1.nce is shown in Figure 7-59c. How does the voltage build up in an induction generator when it is first started? When an induction generator first starts to turn, the residual magnetism in its field circuit produces a smal Ivoltage. TImt smal I voltage produces a capacitive current fl ow, which increases the voltage, further increasing the capacitive cur- rent, and so forth until the voltage is fully built up. If no residual flux is present in the induction generator's rotor, then its voltage will not build up, and it must be magnetized by momentarily running it as a motor. TIle most serious problem with an induction generator is that its voltage varies wildly with changes in load, especially reactive load. Typical tenninal char- acteristics of an induction generator operating alone with a constant parallel ca- pacitance are shown in Figure 7-60. Notice that, in the case of inductive loading, the voltage collapses very rapidly. This happens because the fixed capacitors must supply all the reactive power needed by both the load and the generator, and any reactive power diverted to the load moves the generator back along its magneti- zation curve, causing a major drop in generator voltage. It is therefore very diffi- cult to start an induction motor on a power system supplied by an induction gen- erator- special techniques must be employed to increase the effective capacitance during starting and then decrease it during nonnal operation. Because of the nature of the induction machine's torque-speed characteristic, an induction generator's frequency varies with changing loads: but since the torque-speed characteristic is very steep in the nonnal operating range, the total fre- quency variation is usually limited to less than 5 percent. This amount of variation may be quite acceptable in many isolated or emergency generator applications.
- 487. INDUCTION MOTORS 463 Capacitor bank: voltage Vc- V (1M" no-load armature current) (Lagging amperes) (a) Medium C Small C -------------;'---- -----------r --- '0' FIGURE 7-59 Small Medium capacitance C (mediumZd capacitance C / ~/ (large Zcl / ~ / / 1 / / / / ~ Large capacitance C / ~ ~ (Small Zcl 1/ , (Capacitor bank: current) (leading amperes) ,b, I Large C (a) The magnetization curve of an induction machine. It is a plot of the tenninal voltage of the machine as a function of its magnetization current (which lags the phase voltage by approximately 90°). (b) Plot of the voltage-.::urrent characteristic of a capacitor bank:. Note that the larger the capacitance. the greater its current for a given voltage. This current leads the phase voltage by approximately 90°. (c) The no-load terntinal voltage for an isolated induction generator can be found by ploning the generator terminal characteristic and the capacitor voltage-.::urrent characteristic on a single set of axes. The intersection of the two curves is the point at which the reactive power demanded by the generator is exactly supplied by the capacitors. and this point gives the no-load ferminall"Oltage of the generator.
- 488. 464 ELECTRIC MACHINERY RJNDAMENTALS v, ""GURE 7-60 The terminal voItage--<:urrent characteristic of an induction generator for a load with a constant lagging power factor. Induction Generator Applications Induction generators have been used since early in the twentieth century, but by the 1960s and 1970s they had largely disappeared from use. However, the induc- tion generator has made a comeback since the oil price shocks of 1973. With en- ergy costs so high, energy recovery became an important part of the economics of most industrial processes. The induction generator is ideal for such applications because it requires very litt Ie in the way of control systems or maintenance. Because of their simplicity and small size per kilowatt of output power, in- duction generators are also favored very strongly for small windmills. Many com- mercial windmills are designed to operate in parallel with large power systems, supplying a fraction of the customer's total power needs. In such operation, the power system can be relied on for voltage and frequency control, and static ca- pacitors can be used for power-factor correction. 7.13 INDUCTION MOTOR RATINGS A nameplate for a typical high-efficiency integral-horsepower induction motor is shown in Figure 7--6 1. The most important ratings present on the nameplate are L Output power 2, Voltage ), Current 4, Power factor 5, Speed 6, Nominal efficiency
- 489. INDUCTION MOTORS 4 65 I ..... NO " . Hf. LOUIS ALLIS FIGURE 7-61 The nameplate of a typical lIigh-efficiency induction motor. (Courtesy ofMagneTek, Inc.) 7. NEMA design class 8. Starting code A nameplate for a typical standard-efficiency induction motor would be similar, except that it might not show a nominal efficiency. The voltage limit on the motor is based on the maximum acceptable mag- netization current flow, since the higher the voltage gets, the more saturated the motor's iron becomes and the higher its magnetization current becomes. Just as in the case of transformers and synchronous machines, a 60-Hz induction motor may be used on a 50-Hz power system, but only if the voltage rating is decreased by an amount proportional to the decrease in frequency. nlis derating is necessary be- cause the flux in the core of the motor is. proportional to the integral of the applied
- 490. 466 ELECTRIC MACHINERY RJNDAMENTALS voltage. To keep the maximum nux in the core constant while the period of inte- gration is increasing, the average voltage level must decrease. TIle current limit on an induction motor is based on the maximum acceptable heating in the motor's windings, and the power limit is set by the combination of the voltage and current ratings with the machine's power factor and efficiency. NEMA design classes, starting code letters, and nominal efficiencies were discussed in previous sections of this chapter. 7.14 SUMMA RY TIle induction motor is the most popular type of ac motor because of its simplicity and ease of operation. An induction motor does not have a separate field circuit; in- stead, it depends on transfonner action to induce voltages and currents in its field circuit. In fact, an induction motor is basically a rotating transfonner. Its equivalent circuit is similar to that of a transfonner, except for the effects of varying speed. An induction motor nonnally operates at a speed near synchronous speed, but it can never operate at exactly n,yDC. There must always be some relative mo- tion in order to induce a voltage in the induction motor's field circuit. TIle rotor voltage induced by the relative motion between the rotor and the stator magnetic field produces a rotor current, and that rotor current interacts with the stator mag- netic field to produce the induced torque in the motor. In an induction motor, the slip or speed at which the maximum torque oc- curs can be controlled by varying the rotor resistance. The value of that maximum torque is independent of the rotor resistance. A high rotor resistance lowers the speed at which maximum torque occurs and thus increases the starting torque of the motor. However, it pays for this starting torque by having very poor speed reg- ulation in its normal operating range. A low rotor resistance, on the other hand, re- duces the motor's starting torque while improving its speed regulation. Any nor- mal induction motor design must be a compromise between these two conflicting requirements. One way to achieve such a compromise is to employ deep-bar or double- cage rotors. lllese rotors have a high effective resistance at starting and a low ef- fective resistance under normal running conditions, thus yielding both a high starting torque and good speed regulation in the same motor. The same effect can be achieved with a wound-rotor induction motor if the rotor field resistance is varied. Speed control of induction motors can be accomplished by changing the number of poles on the machine, by changing the applied electrical frequency, by changing the applied tenninal voltage, or by changing the rotor resistance in the case of a wound-rotor induction motor. TIle induction machine can also be used as a generator as long as there is some source of reactive power (capacitors or a synchronous machine) available in the power system. An induction generator operating alone has serious voltage reg- ulation problems, but when it operates in parallel with a large power system, the power system can control the machine's voltage. Induction generators are usually
- 491. INDUCTION MOTORS 467 rather small machines and are used principally with alternative e nergy sources, such as windmills, or with energy recovery syste ms. A lmost all the really large generators in use are synchronous generators. QUESTIONS 7-1. What are slip and slip speed in an induction motor? 7-2. How does an induction motor develop torque? 7-3. Why is it impossible for an induction motor to operate at synchronous speed? 7-4. Sketch and explain the shape of a typical induction motor torque-speed characteris- tic curve. 7-5. What equivalent circuit element has the most direct control over the speed at which the pullout torque occurs? 7...(j. What is a deep-bar cage rotor? Why is it used? What NEMA design c1ass(es) can be built with it? 7-7. What is a double-cage cage rotor? Why is it used? What NEMA design class(es) can be built with it? 7-8. Describe the characteristics and uses of wound-rotor induction motors and of each NEMA design class of cage motors. 7-9. Why is the efficiency of an induction motor (wolUld-rotor or cage) so poor at high slips? 7-10. Name and describe four means of controlling the speed of induction motors. 7-11. Why is it necessary to reduce the voltage applied to an induction motor as electrical frequency is reduced? 7-12. Why is tenninal voltage speed control limited in operating range? 7-13. What are starting code factors? What do they say about the starting current of an in- duction motor? 7-14. How does a resistive starter circuit for an induction motor work? 7-15. What infonnation is learned in a locked-rotor test? 7-16. What infonnation is learned in a no-load test? 7-17. What actions are taken to improve the efficiency of modern high-efficiency induc- tion motors? 7-18. What controls the tenninal voltage of an induction generator operating alone? 7-19. For what applications are induction generators typically used? 7-20. How can a wOlUld-rotor induction motor be used as a frequency changer? 7-2 1. How do different voltage-frequency patterns affect the torque-speed characteristics of an induction motor? 7-22. Describe the major features of the solid-state induction motor drive featured in Sec- tion 7.10. 7-23. Two 480-V, lOO-hp induction motors are manufactured. One is designed for 50-Hz operation, and one is designed for 6O-Hz operation, but they are otherwise similar. Which of these machines is larger? 7-24. An induction motor is rlUlning at the rated conditions. If the shaft load is now in- creased, how do the following quantities change? (a) Mechanical speed (b) Slip
- 492. 468 ELECTRIC MACHINERY RJNDAMENTALS (c) Rotor induced voltage (d) Rotor current (e) Rotor frequency (j) PRCL (g) Synchronous speed PROBLEMS 7-1. A dc test is performed on a 460-V. ~-connected. lOO-hp induction motor. If Voc = 24 V and foc = 80A. what is the stator resistance R]? Why is this so? 7-2. A 220-V, three-phase. two-pole. 50-Hz induction motor is ruooing at a slip of 5 per- cent. Find: (a) The speed of the magnetic fields in revolutions per minute (b) The speed of the rotor in revolutions per minute (c) The slip speed of the rotor (d) The rotor frequency in hertz 7-3. Answer the questions in Problem 7- 2 for a 480-V. three-phase. four-pole. 60-Hz in- duction motor running at a slip of 0.035. 7-4. A three-phase. 60-Hz induction motor runs at 890 rhnin at no load and at 840 rlmin at full load. (a) How many poles does this motor have? (b) What is the slip at rated load? (c) What is the speed at one-quarter of the rated load? (d) What is the rotor's electrical frequency at one-quarter of the rated load? 7-5, ASO-kW, 440-V, 50-Hz, six-pole induction motor has a slip of 6 percent when op- erating at full-load conditions. At full-load conditions, the friction and windage losses are 300 W, and the core losses are 600 W. Find the following values for full- load conditions: (a) The shaft speed n.. (b) The output power in watts (c) The load torque 1lood in newton-meters (d) The induced torque 11... in newton-meters (e) The rotor frequency in hertz 7-6. A three-phase, 60-Hz, four-pole induction motor runs at a no-load speed of 1790 rlmin and a full-load speed of 1720 r/min. Calculate the slip and the electrical fre- quency of the rotor at no-load and full-load conditions. What is the speed regulation of this motor [Equation (4-68)]? 7-7, A 208-V, two-pole, 60-Hz, V-connected woood-rotor induction motor is rated at IS hp. Its equivalent circuit components are Rl = 0.200 n Xl = O.4lOn Pmoc.b = 250 W For a slip of 0.05, find (a) The line ClUTent h R2 = 0.120 n X2 = 0.410 n P.u.c""O (b) The stator copper losses PSCL (c) The air-gap power PAG XM = IS.On P<Ote = 180W
- 493. INDUCTION MOTORS 469 (d) The power converted from electrical to mechanical fonn P<oov (e) The induced torque Tm (f) The load torque Tload (g) The overall machine efficiency (h) The motor speed in revolutions per minute and radians per second 7-8. For the motor in Problem 7- 7. what is the slip at the pullout torque? What is the pullout torque of this motor? 7-9. (a) Calculate and plot the torque-speed characteristic ofthe motor in Problem 7- 7. (b) Calculate and plot the output power versus speed curve of the motor in Prob- lem 7- 7. 7-10. For the motor of Problem 7- 7. how much additional resistance (referred to the sta- tor circuit) would it be necessary to add to the rotor circuit to make the maximum torque occur at starting conditions (when the shaft is not moving)? Plot the torque-speed characteristic of this motor with the additional resistance inserted. 7-11. If the motor in Problem 7- 7 is to be operated on a 50-Hz power system. what must be done to its supply voltage? Why? What will the equivalent circuit component values be at 50 Hz? Answer the questions in Problem 7- 7 for operation at 50 Hz with a slip of 0.05 and the proper voltage for this machine. 7-12. Figure 7-1 8a shows a simple circuit consisting of a voltage source. a resistor. and two reactances. Find the Thevenin equi valent voltage and impedance of this circuit at the terminals. Then derive the expressions for the magnitude of Vrn and for Rrn given in Equations (7-4lb) and (7-44). 7-13. Figure P7-1 shows a simple circuit consisting ofa voltage source. two resistors. and two reactances in series with each other. If the resistor RL is allowed to vary but all the other components are constant. at what value of RL will the maximum possible power be supplied to it? Prove your answer. (Hint: Derive an expression for load power in terms of V. Rs. Xs. RL• and XL and take the partial derivative of that ex- pression with respect to Rd Use this result to derive the expression for the pullout torque [Equation (7- 54)]. jXs v(Z) FlGURE 1'7-1 Circuit for Problem 7- 13. 7-14. A 440-V. 50-Hz, two-pole, V-connected induction motor is rated at 75 kW. The equivalent circuit parameters are R[ = 0.075 0 X[ = 0.170 PFAW = 1.0 kW For a slip of 0.04, find R2 = 0.065 n X2 = O.170 Pmioc = 150W XM = 7.20 Paxe = 1.1 kW
- 494. 470 ELECTRIC MACHINERY RJNDAMENTALS (a) The line clUTent h (b) The stator power factor (c) The rotor power factor (d) The stator copper losses PSCL (e) The air-gap power PAG (j) The power converted from electrical to mechanical fonn POO/IiY (g) The induced torque 7;... (h) The load torque Tlood (i) The overall machine efficiency 71 OJ The motor speed in revolutions per minute and radians per second 7-15. For the motor in Problem 7-14, what is the pullout torque? What is the slip at the pullout torque? What is the rotor speed at the pullout torque? 7-16. If the motor in Problem 7-14 is to be driven from a 440-V, 60-Hz power supply, what will the pullout torque be? What will the slip be at pullout? 7-17. Plot the following quantities for the motor in Problem 7-14 as slip varies from 0 to 10 percent: (a) T ind:(b) POOGY: (c) P00': (d) efficiency 71. At what slip does P00i. equal the rated power of the machine? 7-18. A 20S-V, 60 Hz six-pole, V-connected, 25-hp design class B induction motor is tested in the laboratory, with the following results: No load: Locked rotor: OC test: 208 V, 22.0 A, 1200 W, 60 Hz 24.6 V, 64.5 A, 2200 W, 15 Hz 13.5 V, 64 A Find the equivalent circuit of this motor, and plot its torque-speed characteristic curve. 7-19. A 460-V, four-pole, 50-hp, 60-Hz, Y-colUlected, three-phase induction motor devel- ops its full-load induced torque at 3.S percent slip when operating at 60 Hz and 460 V. The per-phase circuit model impedances of the motor are RI = 0.33 n XI = 0.42 n XM =30 n X2 = 0.42 n Mechanical, core, and stray losses may be neglected in this problem. (a) Find the value of the rotor resistance R2. (b) Find TmalI , s"""v and the rotor speed at maximum torque for this motor. (c) Find the starting torque of this motor. (d) What code letter factor should be assigned to this motor? 7-20. Answer the following questions about the motor in Problem 7-19. (a) Ifthis motor is started from a 460-V infinite bus, how much current will flow in the motor at starting? (b) If transmission line with an impedance of 0.35 + jO.25 n per phase is used to connect the induction motor to the infinite bus, what will the starting current of the motor be? What will the motor's tenninal voltage be on starting? (c) If an ideal 1.4: I step-down autotransformer is connected between the transmis- sion line and the motor, what will the ClUTent be in the transmission line during starting? What will the voltage be at the motor end of the transmission line dur- ing starting?
- 495. INDUCTION MOTORS 471 7-21. In this chapter. we learned that a step-down autotransformer could be used to reduce the starting current drawn by an induction motor. While this technique works. an au- totransfonner is relatively expensive. A much less expensive way to reduce the start- ing current is to use a device called y-~ starter. If an induction motor is nonnally ~-cOIUlected. it is possible to reduce its phase voltage V. (and hence its starting cur- rent) by simply reconnecting the swtor windings in Y during starting. and then restoring the cOlUlections to ~ when the motor comes up to s~ed. Answer the fol- lowing questions about this type of starter. (a) How would the phase voltage at starting compare with the phase voltage under normal lUlUling conditions? (b) How would the starting current of the Y-colUlected motor compare to the start- ing current if the motor remained in a ~-connection during starting? 7-22. A 460-V. lOO-hp. four-pole. ~-connected. 60-Hz. three-phase induction motor has a full-load slip of 5 percent. an efficiency of 92 percent. and a power factor of 0.87 lagging. At start-up. the motor develops 1.9 times the full-load torque but draws 7.5 times the rated current at the rated voltage. This motor is to be started with an auto- transformer reduced-volwge starter. (a) What should the output volwge of the starter circuit be to reduce the starting torque until it equals the rated torque of the motor? (b) What will the motor starting ClUTent and the current drawn from the supply be at this voltage? 7-23. A wound-rotor induction motor is operating at rated voltage and frequency with its slip rings shorted and with a load of about 25 percent of the rated value for the ma- chine. If the rotor resistance of this machine is doubled by inserting external resis- tors into the rotor circuit. explain what hap~ns to the following: (a) Slip s (b) Motor speed n.. (c) The induced voltage in the rotor (d) The rotor current (e) "rio<! (f) P out (g) PRCL (h) Overall efficiency 7f 7-24. Answer the following questions about a 460-V, ~-COlUlected. two-pole. 75-hp. 60-Hz. starting-code-Ietter-E induction motor: (a) What is the maximum current starting current that this machine's controller must be designed to handle? (b) If the controller is designed to switch the stator windings from a ~ connection to a Y connection during starting. what is the maximum starting current that the controller must be designed to handle? (c) If a 1.25: I step-down autotransfonner starter is used during starting. what is the maximum starting current that will be drawn from the line? 7-25. When it is necessary to stop an induction motor very rapidly. many induction motor controllers reverse the direction of rotation of the magnetic fields by switching any two stator leads. When the direction of rotation of the magnetic fields is reversed. the motor develops an induced torque opposite to the current direction of rotation. so it quickly stops and tries to start turning in the opposite direction. If power is re- moved from the stator circuit at the moment when the rotor s~ed goes through zero.
- 496. 472 ELECTRIC MACHINERY RJNDAMENTALS then the motor has been stopped very rapidly. This technique for rapidly stopping an induction motor is called plugging. The motor of Problem 7- 19 is running al rated conditions and is to be stopped by plugging. (a) What is the slip s before plugging? (b) What is the frequency of the rotor before plugging? (c) What is the induced torque "ria<! before plugging? (d) What is the slip s inunediately after switching the stator leads? (e) What is the frequency of the rotor immediately after switching the stator leads? (j) What is the induced torque "ria<! immediately after switching the stator leads? REFERENCES I. Alger. Phillip. Induction Machines. 2nd ed. New York: Gordon and Breach. 1970. 2. Del Toro. V. Electric Machines and Power Systems. Englewood Cliffs. N.J.: Prentice-Hall. 1985. 3. Filzgerald. A. E. and C. Kingsley. Jr. Electric Machinery. New York: McGraw-Hilt. 1952. 4. Filzgerald. A. E.• C. Kingsley. Jr.• and S. D. Umans. Electric Machinery. 51h ed. New York: McGraw-Hilt. 1990. 5. Institute of Electrical and Electronics Engineers. Standard Test Procedurefor Pol)phase Induction MOlOrs and Genemtors. IEEE Standaro 112-1996. New York: IEEE. 1996. 6. Kosow. Irving L. Control ofElectric Motors. Englewood Cliffs. N.J.: Prentice-Hall. 1972. 7. McPherson. George. An Introduction 10 Electrical Machines and Tmnsfonners. New York: Wiley. 1981. 8. National Eleclrical Manufaclurers Association. MOlOrs and GenemlOrs. Publication No. MG I- 1993. Washington. D.C.: NEMA.I993. 9. Siemon. G. R.• and A. Siraughen. Electric Machines. Reading. Mass.: Addison-Wesley. 1980. 10. Vithayalhil. Joseph. Pov.'ef Electronics: Principles and Applications. New York: McGraw-Hill. 1995. II. Weminck. E. H. (ed.). Electric MOlOr Handbook. London: McGraw-Hill. 1978.
- 497. CHAPTER 8 DC MACHINERY FUNDAMENTALS DCmachines are generators that convert mechanical energy to de electric energy and motors that convert de electric energy to mechanical energy. Most de machines are like ac machines in that they have ac voltages and currents within thern---dc machines have a de output only because a mechanism exists that converts the internal ac voltages to de voltages at their tenninals. Since this mechanism is called a commutator, de machinery is also known as commutating machinery. The fundamental principles invo lved in the operation of de machines are very simple. Unfortunately, they are usually somewhat obscured by the compli- cated construction of real machines. This chapter will first explain the principles of de machine operation by using simple examples and then consider some of the complications that occur in real dc machines. S.I A SIMPLE ROTATING LOOP BETWEEN CURVED POLE FACES The linear machine studied in Section 1.8 served as an introduction to basic ma- chine behavior. Its response to loading and to changing magnetic fields closely re- sembles the behavior of the real dc generators and motors that we will study in Chapter 9. However, real generators and motors do not move in a straight line- they rotate. The next step toward understanding real dc machines is to study the simplest possible example of a rotating machine. The simplest possible rotating dc machine is shown in Figure 8-1. It consists of a single loop of wire rotating about a fixed axis. The rotating part of this ma- chine is called the rotor, and the stationary part is called the stator. TIle magnetic 473
- 498. 474 ELECTRIC MACHINERY RJNDAMENTALS d N s (. ) N s r-= >s - f:::: = '-- - ~ - ~ - f-- - 0 ' < b ..-( ):. •d " • " ',. (b) «) F" s • , N F " (d) ""GURE 8-1 A simple rotating loop between curved pole faces. (a) Perspective view; (b) view of field lines; (e) top view; (d) front view.
- 499. OCMACHINERYFUNDAMENTALS 475 .. - " • FIGURE 8-2 Derivation of an equation for the voltages induced in the loop. field for the machine is supplied by the magnetic north and south poles shown on the stator in Figure 8- 1. Notice that the loop of rotor wire lies in a slot carved in a ferromagnetic core. 1lle iron rotor, together with the curved shape of the pole faces, provides a constant-width air gap between the rotor and stator. Remember from Chapter I that the reluctance of air is much much higher than the reiuctance of the iron in the ma- chine. To minimize the reluctance of the flux path through the machine, the mag- netic flux must take the shortest possible path through the air between the pole face and the rotor surface. Since the magnetic flux must take the shortest path through the air, it is per- pendicular to the rotor surface everywhere under the pole faces. Also, since the air gap is of unifonn width, the reluctance is the same everywhere under the pole faces. The uniform reluctance means that the magnetic flux density is constant everywhere under the pole faces. The Voltage Induced in a Rotating Loop If the rotor of this machine is rotated, a voltage will be induced in the wire loop. To detennine the magnitude and shape of the voltage, examine Figure 8- 2. 1lle loop of wire shown is rectangular, with sides ab and cd perpendicular to the plane of the page and with sides be and da parallel to the plane of the page. The mag- netic field is constant and perpendicular to the surface of the rotor everywhere un- der the pole faces and rapidly falls to zero beyond the edges of the poles. To determine the total voltage e,OI on the loop, examine each segment of the loop separately and sum all the resulting voltages. TIle voltage on each segment is given by Equation (1-45): eind = (v x B) • I ( 1-45)
- 500. 476 ELECTRIC MACHINERY RJNDAMENTALS I. Segment abo In this segment, the velocity of the wire is tangential to the path of rotation. The magnetic field 8 points out perpendicular to the rotor surface everywhere under the pole face and is. zero beyond the edges of the pole face. Under the pole face, velocity v is perpendicular to 8, and the quantity v x 8 points into the page. TIlerefore, the induced voltage on the segment is positive into page under the pole face beyond the pole edges (8-1 ) 2. Segment be. In this segment, the quantity v x 8 is either into or out of the page, while length 1is in the plane of the page, so v x 8 is perpendicular to I. Therefore the voltage in segment be will be zero: ecb = 0 (8- 2) 3. Segment ed. In this segment, the velocity of the wire is tangential to the path of rotation. The magnetic field 8 points in perpendicular to the rotor surface everywhere under the pole face and is. zero beyond the edges of the pole face. Under the pole face, velocity v is perpendicular to 8, and the quantity v x 8 points out of the page. 1llerefore, the induced voltage on the segment is positive out of page under the pole face beyond the pole edges (8- 3) 4. Segment da. Just as in segment be, v x 8 is perpendicular to I. Therefore the voltage in this segment will be zero too: ead = 0 (8-4) 1lle total induced voltage on the loop ei!>d is given by under the pole faces beyond the pole edges (8- 5) When the loop rotates through 180°, segment ab is under the north pole face in- stead of the south pole face. At that time, the direction of the voltage on the seg- ment reverses, but its magnitude remains constant. The resulting voltage e,,,, is shown as a function of time in Figure 8- 3. 1llere is an alternative way to express Equation (8- 5), which clearly relates the behavior of the single loop to the behavior of larger, real dc machines. To de- rive this alternative expression, examine Figure 8-4. Notice that the tangential ve- locity v of the edges of the loop can be expressed as v = rw
- 501. IX: MACHINERY FUNDAMENTALS 477 2vBI Cr----, ,81 0 r-----~.------r------~------T_------ , - vBI - 2vBI FIGURE 8-3 The output voltage of the loop. - v=rw FIGURE 8-4 w , Pole surface area Ap.. nrl • Rotor surface area A =2nrl Derivation of an alternative form of the induced voltage equation.
- 502. 478 ELECTRIC MACHINERY RJNDAMENTALS where r is the radius from axis of rotation out to the edge of the loop and w is the angular velocity of the loop. Substituting this expression into Equation (8- 5) gives _ [2rWBI eioo - 0 e ioo = {2r~BW under the pole faces beyond the pole edges under the pole faces beyond the pole edges Notice also from Figure 8-4 that the rotor surface is a cylinder, so the area of the rotor surface A is just equal to 2nrl. Since there are two poles, the area of the rotor under each pole (ignoring the small gaps between poles) is Ap = nrl. TIlerefore, under the pole faces beyond the pole edges Since the nux density B is constant everywhere in the air gap under the pole faces, the total flux under each pole is just the area of the pole times its flux density: Therefore, the final form of the voltage equation is e. , ~ {~"'W '" 0 under the pole faces beyond the pole edges (8-6) TIms, the voltage generated in the machine is equnl to the product of the flux inside the mnchine and the speed ofrotation ofthe machine, multiplied by a constant representing the mechanical construction of the machine. In general, the voltage in any real machine will depend on the same three factors: I. The flux in the machine 2. The speed of rotation 3. A constant representing the construction of the machine Getting DC Voltage out of the Rotating Loop Figure 8- 3 is a plot of the voltage e"", generated by the rotating loop. As shown, the voltage out of the loop is alternately a constant positive value and a constant negative value. How can this machine be made to produce a dc voltage instead of the ac voltage it now has? One way to do this is shown in Figure 8- 5a. Here two semicircular conduct- ing segments are added to the end of the loop, and two fixed contacts are set up at an angle such that at the instant when the voltage in the loop is zero, the contacts
- 503. IX: MACHINERY FUNDAMENTALS 479 N Commutator s Brushes (.) - <'W (b) FIGURE 8-S Producing a dc output from the machine with a commutator and brushes. (a) Perspective view; (b) the resulting output voltage. short-circuit the two segments. In this fashion, every time the voltage ofthe loop switches direction, the contacts also switch connections, and the output ofthe con- tacts is always built up in the same way (Figure 8- 5b). This connection-switching process is known as commutation. TIle rotating semicircular segments are called commutator segments, and the fixed contacts are calJed brushes.
- 504. 480 ELECTRIC MACHINERY RJNDAMENTALS N Commutator s (a) Current into cod page N ,...!..... Current out of page FcJ, imd ..........:::Jw , S ~,Z, "-b (h) ""GURE 8-6 Derivation of an equation for the induced torque in the loop. Note that the iron core is not shown in part b for clarity. The Induced Torque in the Rotating Loop Suppose a battery is now connected to the machine in Figure 8- 5. The resulting configuration is shown in Figure 8-6. How much torque will be produced in the loop when the switch is closed and a current is allowed to fl ow into it? To deter- mine the torque, look at the close-up of the loop shown in Figure 8-6b. TIle approach to take in detennining the torque on the loop is to look at one segment of the loop at a time and then sum the effects of all the individual seg- ments. TIle force on a segment of the loop is given by Equation (1-43):
- 505. IX: MACHINERY FUNDAMENTALS 481 F = i(lx8) ( 1-43) and the torque on the segment is given by T = rFsin () (1-6) where () is the angle between rand F. The torque is essentially zero whenever the loop is beyond the pole edges. While the loop is under the pole faces, the torque is I. Segmentab. In segment ab, the current from the battery is directed out of the page. TIle magnetic field under the pole face is pointing radially out of the ro- tor, so the force on the wire is given by Fab- i(lx8) - ilB tangent to direction of motion TIle torque on the rotor caused by this force is "Tab - rF sin () - r(iIB) sin 900 - rilB CCW (8- 7) (8-8) 2. Segment be. In segment be, the current from the battery is flowing from the upper left to the lower right in the picture. The force induced on the wire is given by Fbc - i(lx8) - 0 since I is parallel to 8 (8- 9) TIlerefore, "Tb< = 0 (8-10) 3. Segment ed. In segment ed, the current from the battery is directed into the page. The magnetic field under the pole face is pointing radially into the ro- tor, so the force on the wire is given by Fed - i(l X 8 ) - ilB tangent to direction of motion TIle torque on the rotor caused by this force is "Ted - rF sin () - r(iIB) sin 900 - rilB CCW (8-11 ) (8-1 2) 4. Segment 00. In segment 00, the current from the battery is flowing from the upper left to the lower right in the picture. The force induced on the wire is given by
- 506. 482 ELECTRIC MACHINERY RJNDAMENTALS Fda - i(I XB) - 0 since I is parallel to B Therefore, Tdo = 0 TIle resulting total induced torque on the loop is given by under the pole faces beyond the pole edges (8-13) (8-14) (8-15) By using the facts that Ap ,., rrrl and cp = ApB, the torque expression can be re- duced to { ,~. - ~, Tind = : under the pole faces (8-16) beyond the pole edges TIlUS, the torque produced in the machine is the product ofthe flux in the machine and the current in the machine, times some quantity representing the me- chanical construction of the machine (the percentage of the rotor covered by pole faces). In general, the torque in any real machine will depend on the same three factors: L The flux in the machine 2, The current in the machine ), A constant representing the construction of the machine Example 8-1. Figure 8--6 shows a simple rotating loop between curved pole faces connected to a battery and a resistor through a switch. The resistor shown models the total resistance of the battery and the wire in the machine. The physical dimensions and charac- teristics of this machine are r = O.5m R = 0.3!l VB = 120 V I = 1.0m B = O.25T (a) What happens when the switch is closed? (b) What is the machine's maximum starting current? What is its steady-state angu- lar velocity at no load? (c) Suppose a load is attached to the loop, and the resulting load torque is 10 N· m. What would the new steady-state speed be? How much power is supplied to the shaft of the machine? How much power is being supplied by the battery? Is this machine a motor or a generator?
- 507. IX: MACHINERY FUNDAMENTALS 483 (d) Suppose the machine is again unloaded, and a torque of 7.5 N • m is applied to the shaft in the direction of rotation. What is the new steady-state speed? Is this machine now a motor or a generator? (e) Suppose the machine is rulUling unloaded. What would the final steady-state speed of the rotor be if the flux density were reduced to 0.20 T? Solution (a) When the switch in Figure 8-6 is closed, a current will flow in the loop. Since the loop is initially stationary, e jod = O. Therefore, the current will be given by VB - eiod VB i = = R R This current flows through the rotor loop, producing a torque 2.; 7iod = - 'I" ~ ccw This induced torque produces an angular acceleration in a counterclockwise di- rection, so the rotor of the machine begins to turn. But as the rotor begins to tum, an induced voltage is produced in the motor, given by so the current i falls. As the current falls, 7ind = (2hr)cpi.t.. decreases, and the ma- chine winds up in steady state with 7iod = 0, and the battery voltage VB = eind' This is the same sort of starting behavior seen earlier in the linear dc machine. (b) At starting conditions, the machine's current is . VB l20V I = J[ = 0.3fi = 400 A At no-load steady-state conditions, the induced torque 7ind must be zero. But 7ind = 0 implies that current i must equal zero, since 7ind = (2hr)cpi, and the flux is nonzero. The fact that i = 0 A means that the battery voltage VB = eind' There- fore, the speed of the rotor is _ VB _ ~ W - (2i1r)cp - 2rlB l20V = 2(0.5 mXI.O mXO.25 T) = 480 rad/s (c) If a load torque of 10 N om is applied to the shaft of the machine, it will begin to slow down. But as w decreases, eind = (2hr)cpwJ. decreases and the rotor cur- rent increases [i = (VB - eind.t.. )/R]. As the rotor current increases, ITIOdI in- creases too, until I 7;",,1 = 171Nd1 at a lower speed w. At steady state, 171....1 = 17indl = (2hr)cpi. Therefore, . 7jod 7ind 1= = -- (2hr)cp 2rlB ION om = (2XO.5 m)(1.0 mXO.25 T) = 40 A
- 508. 484 ELECTRIC MACHINERY RJNDAMENTALS By Kirchhoff's voltage law, eind = VB - iR, so eiDd = 120 V - (40 AXO.3 ll) = 108 V Finally, the speed of the shaft is ejDd eiDd w = (2/Tr)q, = 2rlB 108 V = (2)(0.5 mX1.0 m)(0.25 1) = 432 radls The power supplied to the shaft is P = TW = (10 N • mX432 rad/s) = 4320 W The power out of the battery is P = V Bi = (120 V)(40 A) = 4800 W This machine is operating as a motor, converting electric power to mechanical power. (d) If a torque is applied in the direction of motion, the rotor accelerates. As the speed increases, the internal voltage eind increases and exceeds Vs,so the current flows out of the top of the bar and into the battery. This machine is now a gen- emtor. This current causes an induced torque opposite to the direction of mo- tion. The induced torque opposes the external applied torque, and eventually I11NdI = IT UldI at a higher speed w. The current in the rotor will be . 7ind Tim I = (2hr)q, = 2rlB 7.5 N. m = (2)(0.5 mX1.0 mXO.25 T) = 30 A The induced voltage eind is eind - VB + iR - 120 V + (30 AXO.3 !l) - 129 V Finally, the speed of the shaft is eiod eiDd W = (2/Tr)q, = 2rlB 129 V = (2X0.5 m)(1.0 mXO.25 T) = 516 radls (e) Since the machine is initially lUlloaded at the original conditions, the speed w = 480 radls. If the flux decreases, there is a transient. However, after the transient is over, the machine must again have zero torque, since there is still no load on its shaft. If"TIDd = 0, then the current in the rotor must be zero, and VB= eind. The shaft speed is thus eiod eiDd w = (2/Tr)q, = 2rlB
- 509. IX: MACHINERY FUNDAMENTALS 485 120 V = (2)(0.5 mX 1.0 m)(0.20 T) = 600 rad/s Notice that when the flux in the machine is decreased, its speed increases. This is the same behavior seen in the linear machine and the same way that real dc motors behave. 8.2 COMMUTATION IN A SIMPLE FOUR-LOOP DC MACHINE Commutation is the process of converting the ac voltages and currents in the rotor of a dc machine to dc voltages and currents at its tenninals. It is the most critical part of the design and operation of any dc machine. A more detailed study is nec- essary to determine just how this conversion occurs and to discover the problems associated with it. In this section, the technique of commutation will be explained for a machine more complex than the single rotating loop in Section 8. 1 but less complex than a real dc machine. Section 8.3 will continue this development and explain commutation in real dc machines. A simple four-loop, two-pole dc machine is shown in Figure 8- 7.lllis ma- chine has four complete loops buried in slots carved in the laminated steel of its rotor. TIle pole faces of the machine are curved to provide a uniform air-gap width and to give a uniform nux density everywhere under the faces. The four loops of this machine are laid into the slots in a special manner. The "unprimed" end of each loop is the outermost wire in each slot, while the "primed" end of each loop is the innennost wire in the slot directly opposite. The winding's connections to the machine's commutator are shown in Figure 8- 7b. Notice that loop 1 stretches between commutator segments a and b, loop 2 stretches between segments band c, and so forth around the rotor. • N s d (.J FIGURE 8-7 (a) Afour-toop two-pole dc machine shown at time WI = 0°. (continues)
- 510. 486 ELECTRIC MACHINERY RJNDAMENTALS Back side of coil I Back side of coil 4 3 I' 42' S Pol, faces Comrnutator _ segments -~- Brushes ""GURE 8-7 (roncluded) 3' 24' N I ' 2 , b 43' (bl Back side ofcoil 2 •, E=4~ Back side of coil 3 3 ] ' 42' S 3' 24' N (" (b) The voltages on the rotor conductors at this time, (c) A winding diagram of this machine showing the interconnections of the rotor loops. At the instant shown in Figure 8- 7, the ,2,3', and 4' ends of the loops are under the north pole face, while the ',2',3, and 4 ends of the loops are underthe south pole face. TIle voltage in each of the 1,2,3', and 4' ends of the loops is given by eind - (v x B) • I positive out of page ( 1-45) (8-1 7) TIle voltage in each of the 1" 2', 3, and 4 ends of the ends of the loops is given by - vBI positive into the page ( 1-45) (8-1 8)
- 511. IX: MACHINERY FUNDAMENTALS 487 " -----:7 ,,/ ·---,;..3,--__., • .;:"",:: =.::45 ,- '__ I ' b N 3' I" 2 + e - I ' ov ov 4' + e - Ib, d + e - 2' , d + e - 4 4 S 3 ov ov 3' E=2e FIGURE 8-8 The same machine at time WI = 45°. showing the voltages on the conductors. The overall result is shown in Figure 8- 7b. In Figure 8- 7b, each coil represents one side (or conductor) of a loop. Ir the induced voltage on anyone side of a loop is called e = vBI, then the total voltage at the brushes or tile machine is 1£ -4, wt -O' I (8-1 9) Notice that there are two parallel paths for current through the machine. The ex- istence of two or more parallcl paths for rotor current is a common feature of all commutation schemes. What happens to the voltage E of the terminals as the rotor continues to rotate? To fmd out, examine Figure 8- 8. lllis figure shows the machine at time wt = 45°. At that time, loops 1 and 3 have rotated into the gap between the poles, so the voltage across each of them is zero. Notice that at this instant the brushes
- 512. 488 ELECTRIC MACHINERY RJNDAMENTALS w • - wl= 90° " 3 , ~E~ N b ~f-I d s " 2 The same machine at time WI = 90°, showing the voltages on the conductors. of the machine are shorting out commutator segments ab and cd. This happens just at the time when the loops between these segments have 0 V across them, so shorting out the segments creates no problem. At this time, only loops 2 and 4 are under the pole faces, so the terminal voltage E is given by I E ~ " (8- 20) Now let the rotor continue to turn through another 45°. TIle resulting situa- tion is shown in Figure 8-9. Here, the 1',2,3, and 4' ends of the loops are under
- 513. E. volts 5, 4, 3, 2, , o° FIGURE 8-10 '-' 45° 90° ° IX: MACHINERY FUNDAMENTALS 489 ~ 180° 225° 270° ° 315 360° wI The resulting output voltage of the machine in Figure 8- 7. the north pole face, and the I, 2', 3', and 4 ends of the loops are under the south pole face. The voltages are still built up out of the page for the ends under the north pole face and into the page for the ends under the south pole face. 1lle re- sulting voltage diagram is shown in Figure 8-1 8b. There are now four voltage- carrying ends in each parallel path through the machine, so the terminal voltage E is given by wt - 900 1 (8- 21) Compare Figure 8- 7 to Figure 8- 9. Notice that the voltages on loops 1 and 3 have reversed between the two pictures. but since their connections have also reversed, the total voltage is still being built up in the same direction as before. This fact is at the heart of every commutation scheme. Whenever the voltage re- verses in a loop, the connections of the loop are also switched, and the total volt- age is still built up in the original direction. The terminal voltage of this machine as a function of time is shown in Fig- ure 8-1 0. It is a better approximation to a constant dc level than the single rotat- ing loop in Section 8.1 produced. As the number of loops on the rotor increases, the approximation to a perfect dc voltage continues to get better and better. In summary, Commutation is the process of switching the loop COIUlections on the rotor of a dc machine just as the voltage in the loop switches polarity, in order to maintain an es- sentially constant dc output voltage. As in the case of the simple rotating loop, the rotating segments to which the loops are attached are called commutator segments, and the stationary pieces that ride on top of the moving segments are called brushes.The commutator segments
- 514. 490 ELECTRIC MACHINERY RJNDAMENTALS in real machines are typically made of copper bars. The brushes are made of a mix- ture containing graphite, so that they cause very little friction as they rub over the rotating commutator segments. 8.3 COMMUTATION AND ARMATURE CONSTRUCTION IN REAL DC MACHINES In real dc machines, there are several ways in which the loops on the rotor (also called the armature) can be connected to its commutator segments. nlese differ- ent connections affect the number of parallel current paths within the rotor, the output voltage of the rotor, and the number and position of the brushes riding on the commutator segments. We wil l now examine the construction of the coils on a real dc rotor and then look at how they are connected to the commutator to pro- duce a dc voltage. The Rotor Coils Regardless of the way in which the windings are connected to the commutator segments, most of the rotor windings themselves consist of diamond-shaped preformed coils which are inserted into the armature slots as a unit (see Figure 8-11 ). Each coil consists of a number of turns (loops) of wire, each turn taped and insulated from the other turns and from the rotor slot. Each side of a turn is called a conductor. 1lle number of conductors on a machine's annature is given by where I Z ~ 2eNc I Z = number of conductors on rotor C = number of coils on rotor Nc = number of turns per coil (8- 22) Nonnally, a coil spans 180 electrical degrees. nlis means that when one side is under the center of a given magnetic pole, the other side is under the cen- ter of a pole of opposite polarity. The physical poles may not be located 180 me- chanical degrees apart, but the magnetic field has completely reversed its polarity in traveling from under one pole to the next. The relationship between the electri- cal angle and mechanical angle in a given machine is given by where O e = electrical angle, in degrees 0", = mechanical angle, in degrees P = number of magnetic poles on the machine (8- 23) If a coil spans 180 electrical degrees, the voltages in the conductors on either side of the coil will be exactly the same in magnitude and opposite in direction at all times. Such a coil is called afull-pitch coil.
- 515. Nc turns insulated from ,~h other FIGURE 8-11 OCMACHINERYFUNDAMENTALS 491 I", length of conductor (a) (b) (a) The shape of a typical prefornled rotor coil. (b) A typical coil insulation system showing the insulation between turns within a coil. (Courtesy ofGeneml Electric Company.) Sometimes a coil is built that spans less than 180 electrical degrees. Such a coil is called afractional-pitch coil, and a rotor winding wound with fractional- pitch coils is called a chorded winding. 1lle amount of chording in a winding is described by a pitch factor p, which is defined by the equation electrical angle of coil p = 180 0 x 100% (8- 24) Sometimes a small amount of chording will be used in dc rotor windings to im- prove commutation. Most rotor windings are hi!o-layer windings, meaning that sides from two different coils are inserted into each slot. One side of each coil will be at the bot- tom of its slot, and the other side will be at the top of its slot. Such a construction requires the individual coils to be placed in the rotor slots by a very elaborate
- 516. 492 ELECTRIC MACHINERY RJNDAMENTALS , • , --- :.------ ""GURE 8-12 The installation of prefonned rotor coils on a dc machine rotor. (Courtesy ofWestinghouse Electric Company.) procedure (see Figure 8-1 2). One side of each of the coils is placed in the bottom of its slot, and then after all the bottom sides are in place, the other side of each coil is placed in the top of its slot. In this fashion, all the windings are woven to- gether, increasing the mechanical strength and unifonnity of the final structure. Connections to the Commutator Segments Once the windings are installed in the rotor slots, they must be connected to the commutator segments. There are a number of ways in which these connections can be made, and the different winding arrangements which result have different advantages and disadvantages. TIle distance (in number of segments) between the commutator segments to which the two ends of a coil are connected is called the commutatorpitch Ye. If the end of a coil (or a set number of coils, for wave construction) is connected to a commutator segment ahead of the one its beginni ng is connected to, the winding is called a progressive winding. If the end of a coil is connected to a commutator segment behind the one its beginning is connected to, the winding is called a ret- rogressive winding. If everything else is identical, the direction of rotation of a progressive-wound rotor will be oppos.ite to the direction of rotation of a retrogressive-wound rotor. Rotor (armature) windings are further classified according to the plex of their windings. A simplex rotor winding is a single, complete, closed winding wound on a rotor. A duplex rotor winding is a rotor with two complete and inde- pendent sets of rotor wi ndings. If a rotor has a duplex winding, then each of the windings will be associated with every other commutator segment: One winding
- 517. IX: MACHINERY FUNDAMENTALS 493 c+ I c C+I C-I C C+I I" Ib, FIGURE 8-13 (3) A coil in 3 progressive rotor winding. (b) A coil in 3 retrogressive rotor winding. will be connected to segments I, 3, 5, etc., and the other winding will be con- nected to segments 2, 4, 6, etc. Similarly, a triplex winding will have three com- plete and independent sets of windings, each winding connected to every third commutator segment on the rotor. Collectively, all armatures with more than one set of windings are said to have multiplex windings. Finally, annature windings are classified according to the sequence of their connections to the commutator segments. There are two basic sequences of arrna- ture winding connections-iap windings and wave windings. In addition, there is a third type of winding, called a frog-leg winding, which combines lap and wave windings on a single rotor. These windings will be examined individually below, and their advantages and disadvantages will be discussed. The Lap Winding The simplest type of winding construction used in modem dc machines is the sim- plex series or lap winding. A simplex lap winding is a rotor winding consisting of coils containing one or more turns of wire with the two ends of each coil coming out at adjacent commutator segments (Figure 8- 13). If the end of the coil is con- nected to the segment after the segment that the beginning of the coil is connected to, the winding is a progressive lap winding and Yc = I; if the end of the coil is connected to the segment before the segment that the beginning of the coil is con- nected to, the winding is a retrogressive lap winding and Yc = - I. A simple two- pole machine with lap windings is shown in Figure 8- 14. An interesting feature of simplex lap windings is that there are as many par- allel current paths through the machine as there are poles on the mnchine. If C is the number of coils and commutator segments present in the rotor and P is the
- 518. 494 ELECTRIC MACHINERY RJNDAMENTALS 2 N I-oo--J s 8 5 ""GURE 8-14 A simple two-pole lap-wound dc machine. number of poles on the machine, then there will be c/P coils in each of the P par- allel current paths through the machine. The fact thai there are P current paths also requires that there be as many brushes on the machine as there are poles in order to tap all the current paths. This idea is illustrated by the simple four-pole motor in Figure 8-15. Notice that, for this motor, there are four current paths through the rotor, each having an equal voltage. The fact that there are many current paths in a multipole machine makes the lap winding an ideal choice for fairly low-voltage, high-current machines, since the high currents required can be split among the several different current paths. This current splitting pennits the size of individual rotor conductors to remain reasonable even when the total current becomes ex- tremely large. TIle fact that there are many parallel paths through a multipole lap-wound machine can lead to a serious problem, however. To understand the nature of this problem, examine the six-pole machine in Figure 8-1 6. Because of long usage, there has been slight wear on the bearings of this machine, and the lower wires are closer to their pole faces than the upper wires are. As a result, there is a larger volt- age in the current paths involving wires under the lower pole faces than in the paths involving wires under the upper pole faces. Since all the paths are connected in par- allel, the result will be a circulating current flowing out some of the brushes in the machine and back into others, as shown in Figure 8-17. Needless to say, this is not gooj for the machine. Since the winding resistance of a rotor circuit is so small, a very tiny imbalance runong the voltages in the parallel paths will cause large cir- culating currents through the brushes and potentially serious heating problems. TIle problem of circulating currents within the parallel paths of a machine with four or more poles can never be entirely resolved, but it can be reduced somewhat by equalizers or equalizing windings. Equalizers are bars located on the rotor of a lap-wound dc machine that short together points at the same voltage
- 519. N N lit · " " " " " I 13 2 14 3 IS ,16y IXX'>: yP , FIGURE 8-15 16 s 41651 ' 62'7 XXx 6 , d , IX: MACHINERY FUNDAMENTALS 495 s 4 s N IOL-_ _ II 13 12 s ,., ;x ;X N " " " " " " 3' 8 4' 9 5' 10 6' 117' I ;X :xxx :xxx 8 h ,b, s ,- i- " " " N " " I 8' 139' 1410 151116~ 13' ;:xxx :X14· • m '~ (a) A four-pole lap-wound de motor. (b) The rotor winding diagram of this machine. Notice that each winding ends on the commutator segment just after the one it begins at. This is a progressive lap winding.
- 520. 496 ELECTRIC MACHINERY RJNDAMENTALS s N N s s N ""GURE 8-16 A six-pole dc motor showing the effects of bearing wear. Notice that the rotor is slightly closer to the lower poles than it is to the upper poles. level in the different parallel paths. The effect of this shorting is 1 0 cause any cir- culating currents that occur to now inside the small sections of windings thus shorted together and to prevent this circulating current from flowing through the brushes of the machine. TIlese circulating currents even partially correct the flux imbalance that caused them to exist in the first place. An equalizer for the four- pole machine in Figure 8-1 5 is shown in Figure 8-1 8, and an equalizer for a large lap-wound dc machine is shown in Figure 8-1 9. If a lap winding is duplex, then there are two completely independent wind- ings wrapped on the rotor, and every other commutator segment is tied to one of the sets. Therefore, an individual coil ends on the second commutator segment down from where it started, and Yc = ~2 (depending on whether the winding is progressive or retrogressive). Since each set of windings has as many current paths as the machine has poles, there are twice as many current paths as the ma- chine has poles in a duplex lap winding. In general, for an m-plex lap winding, the commutator pitch Yc is lap winding (8- 25) and the number of current paths in a machine is I a -mpl lap winding (8- 26)
- 521. v, IX: MACHINERY FUNDAMENTALS 497 Cirwlating Current + .l. .l. ... + + + + + , , , " " - - - - - + + + + + , , , " " - - - - - + + + + + , , , " " - - - - - + + + + + , , , " " - - - - - + + + + + , , , " " - - - - - T T T - e+ slightly greater voltage e- slightly lower voltage FIGURE 8-17 The voltages on the rotor conductors of the machine in Figure 8--16 are unequal. producing circulating currents flowing through its brushes. where a = number of current paths in the rotor m = plex of the windings ( I, 2, 3, etc.) P = number of poles on the machine The Wave Winding + + + + + The series or wave winding is an alternative way to connect the rotor coils to the commutator segments. Figure 8- 20 shows a simple four-pole machine with a
- 522. 498 ELECTRIC MACHINERY RJNDA MENTALS Equalizer bars 1 ?< > X - - - - - - - - - - - - IiI r------- I " " " N I:I , I:I N ':1 , II N " " " II 1 13 2 14 3 5 4 6 5 I' 6::k 7 3' 8 4' 9 5' 10 6' II 7' 1 8' 13 9' 14 10 15~1612 XXX >< >< X XX< x,;x X x,;x x,;x >x: >< f'""" 'h ' "m"~ + ~ + + 6' , , , - - + + 6 , " , - - + + " , 8 , - - + + , , 8' , v, - - + + 4' , 9 , - - + + 4 , " , - - + + ,- , 10 , - - + + 3 , 10' , - - ~ - ""GURE 8-18 ,,) + '" , - + 14 , - Equalizers + 13' , - + 13 , - + 12' , - + 12 , - Equalizers + 11' , - + 11 , - (b) B=h " W 16 16' , ,' 2 ,- B=h + , + , + , + , + , + , + , + , (a) An equalizer connection for the four-pole machine in Figure 8--15. (b) A voltage diagram for the machine shows the points shoned by the equalizers.
- 523. s 2 " , b + N • .-0 .1.- , h 7 s FIGURE 8-10 A simple four-pole wave-wound dc machine. IX: MACHINERY FUNDAMENTALS 499 FlGURE 8-19 A closeup of the commutator of a large lap-wound de machine. The equalizers are moumed in the smaIl ring just in front of the commutator segments. (Courtesy ofGeneml Electric Company.) N
- 524. 500 ELECTRIC MACHINERY RJNDAMENTALS simplex wave winding. In this simplex wave winding, every other rotor coil con- nects back to a commutator segment adjacent to the beginning of the first coil. TIlerefore, there are two coils in series between the adjacent commutator seg- ments. Furthennore, since each pair ofcoils between adjacent segments has a side under each pole face, all output voltages are the sum of the effects of every pole, and there can be no voltage imbalances. TIle lead from the second coil may be connected to the segment either ahead of or behind the segment at which the first coil begins. If the second coil is con- nected to the segment ahead of the first coil, the winding is progressive; if it is connected to the segment behind the first coil, it is retrogressive. In general, ifthere are P poles on the machine, then there are PI2 coils in se- ries between adjacent commutator segments. If the (PI2)th coiI is connected to the segment ahead of the first coil, the winding is progressive. If the (PI2)th coil is connected to the segment behind the first coil, the winding is retrogressive. In a simplex wave winding, there are only two current paths. There are c/2 or one-half of the windings in each current path. The brushes in such a machine will be located a full pole pitch apart from each other. What is the commutator pitch for a wave winding? Figure 8- 20 shows a progressive nine-coil winding, and the end of a coil occurs five segments down from its starting point. In a retrogressive wave winding, the end of the coil occurs four segments down from its starting point. Therefore, the end of a coil in a four- pole wave winding must be connected just before or just after the point halfway around the circle from its starting point. TIle general expression for commutator pitch in any simplex wave winding is simplex wave (8- 27) where C is the number of coils on the rotor and P is the number of poles on the machine. The plus sign is associated with progressive windings, and the minus sign is associated with retrogressive windings. A simplex wave winding is shown in Figure 8- 21. Since there are only two current paths through a simplex wave-wound rotor, only two brushes are needed to draw off the current. TIlis is because the segments undergoing commutation connect the points with equal voltage under all the pole faces. More brushes can be added at points 180 electrical degrees apart if desired, since they are at the same potential and are connected together by the wires un- dergoing commutation in the machine. Extra brushes are usually added to a wave- wound machine, even though they are not necessary, because they reduce the amount of current that must be drawn through a given brush set. Wave windings are well suited to building higher-voltage dc machines, since the number of coils in series between commutator segments pennits a high voltage to be built up more easily than with lap windings. A multiplex wave winding is a winding with multiple independent sets of wave windings on the rotor. These extra sets of windings have two current paths each, so the number of current paths on a multiplex wave winding is
- 525. IX: MACHINERY FUNDAMENTALS 501 8 9 2 3 4 5 6 7 8 9 2 9 7' 1 8' 2 9' 3 I' 4 2' 5 3' 6 4 7 5' 8 6' 9 7' 1 8' 2 9' d , FIGURE 8-21 The rotor winding diagram for the machine in fi gure 8-20. Notice that the end of every second coil in series connects to the segment after the beginning of the first coil. This is a progressive wave winding. I a - 2m I rnul1iplex wave (8- 28) The Frog-Leg Winding Thefrog-leg winding or self-equaliZing winding gets its name from the shape of its coils, as shown in Figure 8- 22. It consists of a lap winding and a wave wind- ing combined. The equalizers in an ordinary lap winding are connected at points of equal voltage on the windings. Wave windings reach between points ofessentially equal vol1age under successive pole faces of the same polarity, which are the same lo- cations that equalizers tie together. A frog-leg or self-equalizing winding com- bines a lap winding with a wave winding, so that the wave windings can function as equalizers for the lap winding. The number of current paths present in a frog-leg winding is I a - ' Pm,." I frog-leg winding (8- 29) where P is the number of poles on the machine and mJap is the plex of the lap winding. EXIllllpie 8-2. Describe the rotor winding arrangement of the four-loop machine in Section 8.2. Solutioll The machine described in Section 8.2 has four coils. each containing one turn. resulting in a total of eight conductors. It has a progressive lap winding.
- 526. 502 ELECTRIC MACHINERY RJNDAMENTALS Coil ~~r,~ W ·d· / ave Win lOgS Fl GURE 8-22 A frog-leg or self-equalizing winding coil. 8.4 PROBLEMS WITH COMMUTATION IN REAL MACHINES TIle commutation process as described in Sections 8.2 and 8.3 is not as simple in practice as it seems in theory, because two major effects occur in the real world to disturb it: L Annature reaction 2, L dildt voltages TIlis section explores the nature of these problems and the solutions employed to mitigate their effects. Annature Reaction If the magnetic field windings of a dc machine are connected to a power supply and the rotor of the machine is turned by an external source of mechanical power, then a voltage will be induced in the cond uctors of the rotor. This voltage wilI be rectified into a dc output by the action of the machine's commutator. Now connect a load to the tenninaIs of the machine, and a current will flow in its armature windings. TIlis current flow wi ll produce a magnetic field of its own, which will distort the original magnetic field from the machine's poles. TIlis distortion of the flux in a machine as the load is increased is called armature re- action. It causes two serious problems in real dc machines. TIle first problem caused by annature reaction is neutral-plane shift. The mngnetic neutral plane is defined as the plane within the machine where the
- 527. N IX: MACHINERY FUNDAMENTALS 503 Magnetic neutral plane ;:,w-I--_ New neutral plane w Old neutral plane y-t--- (b) 0 " w N 0 " S ~ 0 " 0 " N 0 " S (.) o (,) FIGURE 8-23 The development of annature reaction in a dc gelJerator. (a) Initially the pole flux is unifonnly distributed. and the magnetic neutral plane is vertical; (b) the effect of the air gap on the pole flux distribution; (c) the armature magnetic field resulting when a load is connected to the machine; (d) both rotor and pole fluxes are shown. indicating points where they add and subtract; (e) the resulting flux under the poles. The neutral plane has shifted in the direction of motion. velocity of the rotor wires is exactly parallel to the magnetic nux lines, so that eind in the conductors in the plane is exactly zero. To understand the problem of neutral-plane shift, exrunine Figure 8- 23. Fig- ure 8- 23a shows a two-pole dc machine. Notice that the nux is distributed uni- fonnly under the pole faces. The rotor windings shown have voltages built up out of the page for wires under the north pole face and into the page for wires under the south pole face. The neutral plane in this machine is exactly vertical. Now suppose a load is connected to this machine so that it acts as a genera- tor. Current will now out of the positive terminal of the generator, so current will
- 528. 504 ELECTRIC MACHINERY RJNDAMENTALS be flowing out ofthe page for wires under the north pole face and into the page for wires under the south pole face. This current fl ow produces a magnetic field from the rotor windings, as shown in Figure 8- 23c. This rotor magnetic field affects the original magnetic field from the poles that produced the generator's voltage in the first place. In some places under the pole surfaces, it subtracts from the pole flux, and in other places it adds to the pole flux. The overall result is that the magnetic flux in the air gap of the machine is skewed as shown in Figure 8- 23d and e. No- tice that the place on the rotor where the induced voltage in a conductor would be zero (the neutral plane) has shifted. For the generator shown in Figure 8- 23, the magnetic neutral plane shifted in the direction of rotation. If this machine had been a motor, the current in its ro- tor would be reversed and the nux would bunch up in the opposite corners from the bunches shown in the figure. As a result, the magnetic neutral plane would shift the other way. In general, the neutral-plane shifts in the direction of motion for a generator and opposite to the direction of motion for a motor. Furthennore, the amount ofthe shift depends on the amount of rotor current and hence on the load of the machine. So what's the big deal about neutral-plane shift? It's just this: The commu- tator must short out commutator segments just at the moment when the voltage across them is equal to zero. Ifthe brushes are set to short out conductors in the vertical plane, then the voltage between segments is indeed zero until the machine is loaded. When the machine is loaded, the neutral plane shifts, and the brushes short out commutator segments with a finite voltage across them. The result is a current now circulating between the shorted segments and large sparks at the brushes when the current path is interrupted as the brush leaves a segment. The end result is arcing and sparking at the brushes. TIlis is a very serious problem, since it leads to drastically reduced brush Iife, pitting ofthe commutator segments, and greatly increased maintenance costs. Notice that this problem cannot be fixed even by placing the brushes over the full-load neutral plane, because then they wouId spark at no load. In extreme cases, the neutral-plane shift can even lead to flashover in the commutator segments near the brushes. The air near the brushes in a machine is normally ionized as a result of the sparking on the brushes. Flashover occurs when the voltage of adjacent commutator segments gets large enough to sustain an arc in the ionized air above them. If flashover occurs, the resulting arc can even melt the commutator's surface. TIle second major problem caused by annature reaction is called flux weak- ening. To understand flux weakening, refer to the magnetization curve shown in Figure 8- 24. Most machines operate at flux densities near the saturation point. TIlerefore, at locations on the pole surfaces where the rotor magnetomotive force adds to the pole magnetomotive force, only a small increase in nux occurs. But at locations on the pole surfaces where the rotor magnetomotive force subtracts from the pole magnetomotive force, there is a larger decrease in flux. 1lle net result is that the total averageflux under the entire poleface is decreased (see Figure 8- 25).
- 529. IX: MACHINERY FUNDAMENTALS 505 q,. Wb , , ••,1 , , .., , , L-------------cf---}--~--------------- ~·A . tums Pole mmf ,/ """ - annature Pole mmf Pole mmf + annature mmf mmf l1q,i - flux increase under reinforced sections of poles l1q,d - flux decrease under subtracting sections of poles FIGURE 8-24 A typical magnetization curve shows the effects of pole saturation where armature and pole magnetomotive forces add. Flux weakening causes problems in bolh generators and motors. In genera- tors, the effect of flux weakening is simply to reduce the voltage supplied by the generator for any given load. In motors, the effect can be more serious. As the early examples in this chapter showed, when the flux in a motor is decreased, its speed in- creases. But increasing the speed of a motor can increase its load, resulting in more flux weakening. It is possible for some shunt dc motors to reach a runaway condi- tion as a result offlux weakening, where the speed of the motor just keeps increas- ing until the machine is disconnected from the power line or until it destroys itself. L dildt Voltages The second major problem is the L dildt voltage that occurs in commutator seg- ments being shorted out by the brushes, sometimes called inductive kick. To un- derstand this problem, look at Figure 8- 26. This figure represents a series of com- mutator segments and the conductors connected between them. Assuming that the current in the brush is 400 A, the current in each path is 200 A. Notice that when a commutator segment is shorted out, the current flow through that commutator
- 530. 506 ELECTRIC MACHINERY RJNDAMENTALS Stator Field § windings - S § N / ~__ "'--'>. L-/__ "'--'>. 1 Rotor - Motion of generator M fmotor ::f. A • turns - Oilon 0 Pole - / magnetomotive force -", .............. ............... " -- , ..............., ................ , . , Rotor magnetomotive force ........-- Net 9' '!f. A· turns </>.Wb ... 1 - Note: Saturation at pole tips fo'IGURE 8- 25 --..........IP. Wb / Old neutral point New neutral poim The flux and magoetomotive force under the pole faces in a de machine. At those points where the magnetomotive forces subtract. the flux closely follows the net magoetomotive force in the iron; but at those points where the magnetomotive forces add, saturation limits the IOtal flux present. Note also that the neutral point of the rotor has shifted. segment must reverse. How fast must this reversal occur? Assuming that the ma- chine is turning at 800 r/min and that there are 50 commutator segments (a reason- able number for a typical motor), each commutator segment moves under a brush and clears it again in t = 0.00 15 s.1l1erefore, the rate of change in current with re- spect to time in the shorted loop must average di 400 A dt - 0.00 15 s - 266,667 A ls (8- 30)
- 531. IX: MACHINERY FUNDAMENTALS 507 - 200 A Direction of commutator motion 200 A - 200 A 200 A -? 200 A (a) 1=0.0015 s 200 Ar----" - - 200 A 200 A 200 A 200 A r-------i-~,"i------------ , Brush reaches beginning of segment b ,,,,,, ...-- Spark at trailing edge of brush Brush clears end of segment a ----- - Ideal conunutation 200 A - Actual commutation with inductance taken into account (b) FIGURE 8-26 (a) The revel"S3.1 of current flow in a coil undergoing commutation. Note that the current in the coil between segments a and b must reverse direction while the brush ShOMS together the two commutator segments. (b) The current reversal in the coil undergoing commutation as a function of time for both ideal commutation and real commutation. with the coil inductance taken into account. With even a tiny inductance in the loop, a very significant inductive voltage kick v = Ldildt will be induced in the shorted commutator segment. This high voltage naturally causes sparking at the brushes of the machine, resulting in the same arc- ing problems that the neutral-plane shift causes.
- 532. 508 ELECTRIC MACHINERY RJNDAMENTALS Solutions to the Problems with Commutation TIlree approaches have been developed to partially or completely correct the prob- lems of armature reaction and L dildt voltages: I. Brush shifting 2. Commutating poles or interpoles 3. Compensating windings E:1.ch of these techniques is explained below, together with its advantages and dis- advantages. BRUSH SHIFTING. Historically, the first attempts to improve the process of commutation in real dc machines started with attempts to stop the sparking at the brushes caused by the neutral-plane shifts and L dildt effects. The first approach taken by machine designers was simple: If the neutral plane of the machine shifts, why not shift the brushes with it in order to stop the sparking? It certainly seemed like a good idea, but there are several serious problems associated with it. For one thing, the neutral plane moves with every change in load, and the shift direction reverses when the machine goes from motor operation to generator operation. lllerefore, someone had to adjust the brushes every time the load on the machine changed. In addition, shifting the brushes may have stopped the brush sparking, but it actually aggravated the flux-weakening effect of the armature reaction in the machine. TIlis is true because of two effects: I. The rotor magnetomotive force now has a vector component that opposes the magnetomotive force from the poles (see Figure 8- 27). 2. The change in annature current distribution causes the flux to bunch up even more at the saturated parts of the pole faces. Another slightly different approach sometimes taken was to fix the brushes in a compromise position (say, one that caused no sparking at two-thirds of full load). In this case, the motor sparked at no load and somewhat at full load, but if it spent most of its life operating at about two-thirds of full load, then sparking was minimized. Of course, such a machine could not be used as a generator at ,II-the sparking would have been horrible. By about 1910, the brush-shifting approach to controlling sparki ng was al- ready obsolete. Today, brush shifting is only used in very small machines that al- ways run as motors. TIlis is done because better solutions to the problem are sim- ply not economical in such small motors. COMMUTATING POLES OR INTERPOLES. Because of the disadvantages noted above and especially because of the requirement that a person must adjust the brush positions of machines as their loads change, another solution to the problem of brush sparking was developed. TIle basic idea behind this new approach is that
- 533. New neutral plane Brushes Jc---r::..;O~d neutral plane o N o o FIGURE 8-27 Net magnetomotive force ::f... s Rotor magnetomotive force ::fR (a) IX: MACHINERY FUNDAMENTALS 509 N New neutral plane Old neutral plane ---f=---w o ~, o o New net magnetomotive , f_ " (h) s Original net I magnetomotive f=, 1, : ::fR , (a) The net magnetomotive force in a dc machine with its brushes in the vertical plane. (b) The net magnetomotive force in a dc machine with its brushes over the shifted neutral plane. Notice that now there is a component of armature magnetomotive force directly oppOiling the poles' magnetomotive force. and the net magnetomotive force in the machine is reduced. if the voltage in the wires undergoing commutation can be made zero, then there will be no sparking at the brushes. To accomplish this, small poles, called com- mutating poles or interpoles, are placed midway between the main poles. These commutating poles are located directly over the conductors being commutated. By providing a flux from the commutating poles, the voltage in the coils undergoing commutation can be exactly canceled. If the cancellation is exact, then there will be no sparking at the brushes. The commutating poles do not otherwise change the operation of the ma- chine, because they are so small that they affect only the few conductors about to undergo commutation. Notice that the armature reaction under the main pole faces is unaffected, since the effects ofthe commutating poles do not extend that far. nlis means that the flux weakening in the machine is unaffected by commutating poles. How is cancellation of the voltage in the commutator segments accom- plished for all values of loads? nlis is done by simply connecting the interpole
- 534. 510 ELECTRIC MACHINERY RJNDAMENTALS windings in series with the windings on the rotor, as shown in Figure 8- 28. As the load increases and the rotor current increases, the magnitude of the neutral-plane shift and the size of the L dildt effects increase too. Both these effects increase the voltage in the conductors undergoing commutation. However, the interpole flux increases too, producing a larger voltage in the conductors that opposes the volt- age due to the neutral-plane shift. The net result is that their effects cancel over a broad range of loads. Note that interpoles work for both motor and generator op- eration, since when the machine changes from motor to generator, the current both in its rotor and in its interpoles reverses direction. Therefore, the voltage effects from them still cancel. What polarity must the flux in the interpoles be? The interpoles must induce a voltage in the conductors undergoing commutation that is opposite to the voltage caused by neutral-plane shift and L dildt effects. In the case of a generator, the neu- tral plane shifts in the direction of rotation, meaning that the conductors undergoing commutation have the same polarity of voltage as the pole they just left (see Figure 8- 29). To oppose this voltage, the interpolcs must have the opposite flux, which is the flux ofthe upcoming pole. In a motor, however, the neutral plane shifts opposite to the direction of rotation, and the conductors undergoing commutation have the same flux as the pole they are approaching. In order to oppose this voltage, the in- terpoles must have the same polarity as the previous main pole. Therefore, I. The interpoles must be of the same polarity as the next upcoming main pole in a generator. - v, N s R - - ---d+ -I, ""GURE 8-28 A de machine with imerpoles.
- 535. IX: MACHINERY FUNDAMENTALS 511 2. The interpoles must be of the same polarity as the previous main pole in a motor. The use of commutating poles or interpoles is very common, because they correct the sparking problems of dc machines at a fairly low cost. TIley are almost always found in any dc machine of I hp or larger. It is important to realize, though, that they do nothing for the flux distribution under the pole faces, so the flux-weakening problem is still present. Most medium-size, general-purpose mo- tors correct for sparking problems with interpoles and just live with the flux- weakening effects. N New neutral plane Now neutral plane FIGURE 8-29 u s n(a) (b) Determining the required polarity of an interpole. The flux from the interpole must produce a voltage that opposes the existing voltage in the conductor.
- 536. 512 ELECTRIC MACHINERY RJNDAMENTALS - - Rotor (amlature) flux - - - Aux from compensating windings o ,, " , N' , , ,, IS , , I o ,,' N " ""GURE 8-30 (,, o ,b, Neutral plane no/ shifted with load The effect of compensating windings in a dc machine. (a) The pole flux in the machine; (b) the fluxes from the armature and compensating windings. Notice that they are equal and opposite; (c) the net flux in the machine. which is just the original pole flux. , , , , COMPENSATING WINDINGS. For very heavy, severe duty cycle motors, the flux-weakening problem can be very serious. To completely cancel armature re- action and thus eliminate both neutral-plane shift and flux weakening, a different technique was developed. This third technique involves placing compensating windings in slots carved in the faces of the poles parallel to the rotor conductors, to cancel the distorting effect of annature reaction. These windings are connected in series with the rotor windings, so that whenever the load changes in the rotor, the current in the compensating windings changes, too. Figure 8- 30 shows the ba- sic concept. In Figure 8- 30a, the pole flux is shown by itself. In Figure 8- 30b, the rotor fl ux and the compensating winding fl ux are shown. Figure 8- 3Oc represents the sum of these three fluxes, which isjust equal to the original pole flux by itself. Figure 8- 3 shows a more careful development of the effect of compensat- ing windings on a dc machine. Notice that the magnetomotive force due to the
- 537. IX: MACHINERY FUNDAMENTALS 513 Stator ~~~i"" - L---~s------'>.~ L---~N------'>.~ j ! ! Rotor ~~~ClIT0J:]0~·~0c·IT0J:]6 ~.~" []"'iJ,,~®~~,,~,, ~~ - Motionof ::f,A· turns Pole magnetomotive force Compensating ,,(!"inding /~ / r-~ Rotor genel1ltor - Motion of mmm magnetomotiveforce '-_____-':1..., =::fpole +::fR +::f"" ::f. A • turns FIGURE 8-3 1 I Neutral plane 00' shifted :1..., =::fpole The flux and magnetomotive forces in a de machine with compensating windings. compensating windings is equal and opposite to the magnetomotive force due to the rotor at every point under the pole faces. The resulting net magnetomotive force is just the magnetomotive force due to the poles, so the flux in the machine is unchanged regardless of the load on the machine. The stator of a large dc ma- chine with compensating windings is shown in Figure 8- 32. The major disadvantage of compensating windings is that they are expen- sive, since they must be machined into the faces of the poles. Any motor that uses them must also have interpoles, since compensating windings do not cancel L dildt effects. The interpoles do not have to be as strong, though, since they are canceling only L dildt voltages in the windings, and not the voltages due to neutral-plane shifting. Because of the expense of having both compensating wind- ings and interpoles on such a machine, these windings are used only where the ex- tremely severe nature of a motor's duty demands them.
- 538. 514 ELECTRIC MACHINERY RJNDAMENTALS ""GURE 8-32 The stator of a six-pole dc machine with imerpoIes and compensating windings. (Courtesy of Westinghouse Electric Company.) 8.5 THE INTERNAL GENERATED VOLTAGE AND INDUCED TORQUE EQUATIONS OF REAL DC MACHINES How much voltage is produced by a real dc machine? The induced voltage in any given machine depends on three factors: I. The flux <p in the machine 2. The speed W of the machine's rotor 3. A constant depending on the construction of the machine How can the voltage in the rotor windings of a real machine be detennined? The voltage out of the annature of a real machine is equal to the number of conductors per current path times the voltage on each conductor. The voltage in any single conductor under the polefaces was previously shown to be eind = e = vBI TIle voltage out of the annature of a real machine is thus E = ZvBI A a (8- 3 1) (8- 32)
- 539. IX: MACHINERY FUNDAMENTALS 515 where Z is the total number of conduct.ors and a is the number of current paths. The velocity of each conductor in the rotor can be expressed as v = rw, where r is the radius of the rotor, so E = ZrwBI A a (8- 33) nlis voltage can be reexpressed in a more convenient form by noting that the nux of a pole is equal to the nux density under the pole times the pole's area: 4> = BAp The rotor of the machine is shaped like a cylinder, so its area is A = 27frl (8- 34) If there are P poles on the machine, the n the portion of the area associated with each pole is the total area A divided by the number of poles P: A = A = 27frl p P P (8- 35) The total flux per pole in the machine is thus 4> = BAp = B(27frl) = 27fr1B p p (8- 36) lllerefore, the internal generated voltage in the machine can be expressed as Finally, where E _ ZrwBI A- a (8- 33) (8- 37) (8- 38) (8- 39) In modern industrial practice, it is common to express the speed of a ma- chine in revolutions per minute instead of radians per second. The conversion from revolutions per minute to radians per second is (8-40) so the voltage equation with speed expressed in tenns of revolutions per minute is
- 540. 516 ELECTRIC MACHINERY RJNDAMENTALS I E, ~ K'<I>" I (8-4 1) where I K'= ~ I (8-42) How much torque is induced in the annature of a real dc machine? The torque in any dc machine depends on three factors: I. The flux <p in the machine 2. The armature (or rotor) current I... in the machine 3. A constant depending on the construction of the machine How can the torque on the rotor of a real machine be determined? The torque on the armature of a real machine is equal to the number of conductors Z times the torque on each conductor. TIle torque in any single conductor under the polefaces was previously shown to be (8-43) If there are a current paths in the machine, then the total annature current I... is split among the a current paths, so the current in a single conductor is given by I, leo"" = a and the torque in a single conductor on the motor may be expressed as dAIB Tcood = - a - (8-44) (8-45) Since there are Z conductors, the total induced torque in a dc machine rotor is (8-46) The flux per pole in this machine can be expressed as J.. =BA =B(2 7rrD =27rr1B '+' p P P (8-47) so the total induced torque can be reexpressed as (8-48) Finally, I Tind = K<pIA I (8-49) where I K= ~ I (8- 39)
- 541. IX: MACHINERY FUNDAMENTALS 517 Both the internal generated voltage and the induced torque equations just given are only approximations, because not alJ the conductors in the machine are under the pole faces at any given time and also because the surfaces of each pole do not cover an entire liP of the rotor's surface. To achieve greater accuracy, the number ofconductors under the pole faces could be used instead of the total num- ber of conductors on the rotor. Example &-3. A duplex lap-wound annature is used in a six-pole dc machine with six brush sets. each spanning two conunutator segments. There are 72 coils on the arma- ture, each containing 12 turns. The flux per pole in the machine is 0.039 Wb, and the ma- chine spins at 400 r/min. (a) How many current paths are there in this machine? (b) What is its induced voltage EA? Solutioll (a) The number of current paths in this machine is a = mP = 2(6) = 12 current paths (b) The induced voltage in the machine is EA = K'qm K' = ZP 60a The number of conductors in this machine is Z = 2CNc = 2(72)(12) = 1728 conductors Therefore, the constant K' is K' = ZP = (1728X6) = 144 60a (60XI2) . and the voltage EA is EA = K'cpn = (14.4)(0.039 Wb)(400 r/min) = 224.6 V (8--26) (8-41) (8-42) (8-22) EXllmple 8-4. A 12-pole dc generator has a simplex wave-wound annature con- taining 144 coils of 10 turns each. The resistance of each turn is 0.011 n. Its flux per pole is 0.05 Wb, and it is turning at a speed of 200 r/min. (a) How many current paths are there in this machine? (b) What is the induced annature voltage of this machine? (c) What is the effective annature resistance of this machine? (d) If a I-ill resistor is connected to the tenninals of this generator, what is the re- sulting induced countertorque on the shaft of the machine? (Ignore the internal annature resistance of the machine.)
- 542. 518 ELECTRIC MACHINERY RJNDAMENTALS Solutio" (a) There are a = 2m = 2 current paths in this winding. (b) There are Z = 2CNc = 2(144X 10) = 2880 conductors on this generator's rotor. Therefore, K' = ZP = (2880XI2) = 288 60a (60)(2) Therefore, the induced voltage is E}, = K'q>n = (288XO.OS Wb)(200 r/min) = 2880 V (e) There are two parallel paths through the rotor of this machine, each one consist- ing of 712 = 1440 conductors, or 720 turns. Therefore, the resistance in each current path is Resistancelpath = (720 turnsXO.OII !lIturn) = 7.92 n Since there are two parallel paths, the effective armature resistance is R}, = 7.9; n = 3.96 n (d) If a lOOO~ load is connected to the tenninals of the generator, and if R}, is ig- nored, then a current of 1 = 2880 V/HXX) n = 2.88 A flows. The constant K is given by K = ZP = (2880)(12) = 2750:2 27TO" (27T)(2) . Therefore, the countertorque on the shaft of the generator is "Tind = Kt$/}, = (27S0.2XO.05 Wb)(2.88 A) = 396 Nom 8.6 THE CONSTRUCTION OF DC MACHINES A simplified sketch of a dc machine is shown in Figure 8- 33, and a more detailed cutaway diagram of a dc machine is shown in Figure 8- 34. TIle physical structure of the machine consists oftwo parts: the stator or sta- tionary part and the rotor or rotating part. TIle stationary part of the machine con- sists of the frame, which provides physical support, and the pole pieces, which project inward and provide a path for the magnetic flux in the machine. TIle ends of the pole pieces that are near the rotor spread out over the rotor surface to dis- tribute its flux evenly over the rotor surface. TIlese ends are called the pole shoes. 1lle exposed surface of a pole shoe is called a poleface, and the distance between the pole face and the rotor is calJed the air gap.
- 543. IX: MACHINERY FUNDAMENTALS 519 Field pole and r.J"Oid Nameplate -+--,'1--'<:. Yoke ----j'!- Frame FIGURE 8-33 A simplified diagram of a de machine. (a) FIGURE 8-34 (a) A cutaway view of a 4O(X)..hp, 700,V. 18-pole de machine showing compensating windings. interpoles. equalizer. and commutator. (Courtesy ofGeneml Electric Company.) (b) A cutaway view of a smaller four'pole de motor including interpoles but without compensating windings. (Courtesy ofMagneTek lncorpomted.)
- 544. 520 ELECTRIC MACHINERY RJNDAMENTALS TIlere are two principal windings on a dc machine: the annature windings and the field windings. The armature windings are defined as the windings in which a voltage is induced, and thefield windings are defined as the windings that produce the main magnetic flux in the machine. In a nonnal dc machine, the annature windings are located on the rotor, and the field windings are located on the stator. Because the annature windings are located on the rotor, a dc machine's rotor itself is sometimes called an armature. Some major features of typical dc motor construction are described below. Pole and Frame Construction TIle main poles of older dc machines were often made of a single cast piece of metal, with the field windings wrapped around it. TIley often had bolted-on lami- nated tips to reduce core losses in the pole faces. Since solid-state drive packages have become common, the main poles of newer machines are made entirely of laminated material (see Figure 8- 35). This is true because there is a much higher ac content in the power supplied to dc motors driven by solid-state drive pack- ages, resulting in much higher eddy current losses in the stators of the machines. TIle pole faces are typically either chamfered or eccentric in construction, mean- ing that the outer tips of a pole face are spaced slightly further from the rotor's surface than the center of the pole face is (see Figure 8- 36). This action increases the reluctance at the tips of a pole face and therefore reduces the flux-bunching ef- fect of annature reaction on the machine. ""GURE 8-35 Main field pole assembly for a de motor. Note the pole laminations and compensating windings. (Courtesy of General Electric Company.)
- 545. IX: MACHINERY FUNDAMENTALS 521 The poles on dc machines are called salient poles, because they stick out from the surface of the stator. The interpoles in dc machines are located between the main poles. TIley are more and more commonly of laminated construction, because of the same loss problems that occur in the main poles. Some manufacturers are even constructing the portion of the frame that serves as the magnetic flux 's return path (the yoke) with laminations, to further re- duce core losses in electronically driven motors. Rotor or Armature Construction The rotor or armature of a dc machine consists of a shaft machined from a steel bar with a core built up over it. The core is composed of many laminations stamped from a stccl plate, with notches along its outer surface to hold the arrna- ture windings. TIle commutator is built onto the shaft of the rotor at one end of the core. TIle annature coils are laid into the slots on the core, as described in Section 8.4, and their ends are connected to the commutator segments. A large dc machine rotor is shown in Figure 8- 37. Commutator and Brushes The commutator in a dc machine (Figure 8- 38) is typically made of copper bars insulated by a mica-type material. TIle copper bars are made sufficiently thick to pennit normal wear over the lifetime of the motor. The mica insulation between commutator segments is harder than the commutator material itself, so as a ma- chine ages, it is often necessary to undercut the commutator insulation to ensure that it does not stick up above the level of the copper bars. The brushes ofthe machine are made of carbon, graphite, metal graphite, or a mixture of carbon and graphite. They have a high conductivity to reduce elec- tricallosses and a low coefficient of friction to reduce excessive wear. They are N s ,., ,b, FIGURE 8-36 Poles with extra air-gap width at the tips to reduce armature reaction. (a) Chamfered poles; (b) eccentric or uniformly graded poles. s
- 546. 522 ELECTRIC MACHINERY RJNDA MENTALS ""GURE 8-37 Photograph of a dc machine with the upper stator half removed shows the construction of its rotor. (Courtesy of General Electric Company.) ""GURE 8-38 Close-up view of commutator and brushes in a large dc machine. (Courtesy of General Electric Company.)
- 547. IX: MACHINERY FUNDAMENTALS 523 deliberately made of much softer material than that of the commutator segments, so that the commutator surface will experience very little wear. The choice of brush hardness is a compromise: If the brushes are too soft, they will have to be replaced too often ; but if they are too hard, the commutator surface will wear ex- cessively over the life of the machine. All the wear that occurs on the commutator surface is a direct resu It of the fact that the brushes must rub over them to convert the ac voltage in the rotor wires to dc voltage at the machine's terminals. If the pressure of the brushes is too great, both the brushes and commutator bars wear excessively. However, if the brush pressure is too small, the brushes tend to jump slightly and a great deal of sparking occurs at the brush-commutator segment interface. This sparking is equally bad for the brushes and the commutator surface. TIlerefore, the brush pressure on the commutator surface must be carefully adjusted for maximum life. Another factor which affects the wear on the brushes and segments in a dc machine commutator is the amount of current fl owing in the machine. llle brushes normally ride over the commutator surface on a thin oxide layer, which lubricates the motion of the brush over the segments. However, if the current is very small, that layer breaks down, and the friction between the brushes and the commutator is greatly increased. lllis increased friction contributes to rapid wear. For maximum brush life, a machine should be at least partially loaded all the time. Winding Insulation Other than the commutator, the most critical part of a dc motor's design is the in- sulation of its windings. If the insulation of the motor windings breaks down, the motor shorts out. The repair of a machine with shorted insulation is quite expen- sive, if it is even possible. To prevent the insulation of the machine windings from breaking down as a result of overheating, it is necessary to limit the temperature of the windings. This can be partially done by providing a cooling air circulation over them, but ultimately the maximum winding temperature limits the maximum power that can be supplied continuously by the machine. Insulation rarely fails from immediate breakdown at some critical tempera- ture. Instead, the increase in temperature produces a gradual degradation of the in- sulation, making it subject to failure due to another cause such as shock, vibration, or electrical stress. There is an old rule of thumb which says that the life ex- pectancy of a motor with a given insulation is halved for each JO percent rise in winding temperature. This rule still applies to some extent today. To standardize the temperature limits of machine insulation, the National Electrical Manufacturers Association (NEMA) in the United States has defined a series of insulation system classes. Each insulation system class specifies the max- imum temperature rise permissible for each type of insulation. lllere are four stan- dard NEMA insulation classes for integral-horsepower dc motors: A, 8, F, and H. Each class represents a higher pennissible winding temperature than the one before it. For example, if the annature winding temperature rise above ambient tempera- ture in one type of continuously operating dc motor is measured by thermometer,
- 548. 524 ELECTRIC MACHINERY RJNDAMENTALS it must be limited to 70°C for class A, WO°C for class B, 30°C for class F, and ISSoC for class H insulation. TIlese temperature specifications arc set out in great detail in NEMA Stan- dard MG I-1993, Motors and Generators. Similar standards have been defined by the International Electrotechnical Commission (lEC) and by various national stan- dards organizations in other countries. 8.7 POWER FLOW AND LOSSES IN DC MACHINES DC generators take in mechanical power and produce electric power, while dc motors take in electric power and pnxluce mechanical power. In either case, not all the power input to the machine appears in useful fonn at the other end-there is always some loss associated with the process. TIle efficiency of a dc machine is defined by the equation (8- 50) TIle difference between the input power and the output power of a machine is the losses that occur inside it. Therefore, Poot - ~oss .,, = x 100% P;.o (8- 51) The Losses in DC Machines The losses that occur in dc machines can be divided into fi ve basic categories: I. Electrical or copper losses (l lR losses) 2. Brush losses 3. Core losses 4. Mechanical losses 5. Stray load losses ELECTRICAL OR COPPER LOSSES. Copper losses are the losses that occur in the armature and field windings of the machine. TIle copper losses for the anna- ture and field windings are given by Armature loss: Field loss: where P), - annature loss PF - field circuit loss PA = li RA PF = I} RF (8- 52) (8- 53)
- 549. IX: MACHINERY FUNDAMENTALS 525 I, - annature current I, - field current R, - annature resistance R, - field resistance The resistance used in these calculations is usually the winding resistance at nor- mal operating temperature. BRUSH LOSSES. 1lle brush drop loss is the power lost across the contact poten- tial at the brushes of the machine. It is given by the equation IpsD - VsDIA I where PBD = brush drop loss VBD = brush voltage drop IA = annature current (8- 54) The reason that the brush losses are calculated in this manner is that the voltage drop across a set of brushes is approximately constant over a large range ofarrna- ture currents. Unless otherwise specified. the brush voltage drop is usually as- sumed to be about 2 V. CORE LOSSES. The core losses are the hysteresis losses and eddy current losses occurring in the metal of the motor. These losses are described in Chapter 1. lllese losses vary as the square of the flux density (B2) and. for the rotor, as the 1.5th power of the speed of rotation (nI.5) . 1" IECHANICAL LOSSES. 1lle mechanical losses in a dc machine are the losses associated with mechanical effects. There are two basic types of mechanical losses:friction and windage. Friction losses are losses caused by the friction of the bearings in the machine, while windage losses are caused by the friction between the moving parts of the machine and the air inside the motor's casing. These losses vary as the cube of the speed of rotation of the machine. STRAY LOSSES (OR MISCELLANEOUS LOSSES). Stray losses are losses that cannot be placed in one of the previous categories. No matter how carefully losses are accounted for, some always escape incl usion in one of the above categories. All such losses are lumped into stray losses. For most machines, stray losses are taken by convention to be I percent of full load. The Power-Flow Diagram One of the most convenient techniques for accounting for power losses in a ma- chine is the power-flow diagram. A power-flow diagram for a dc generator is shown in Figure 8- 39a. In this figure, mechanical power is input into the machine,
- 550. 526 ELECTRIC MACHINERY RJNDAMENTALS Stray losses , , , , , , Mechanical losses EA IA '" iDd Ul m I'R losses ""GURE 8-39 Core losses p~ , Core losses ,,' Mechanical losses ,b, I'R losses Stray losses Power-flow diagrams for de machine: (a) generator: (b) motor. and then the stray losses, mechanical losses, and core losses are subtracted. After they have been subtracted, the remaining power is ideally converted from me- chanical to electrical fonn at the point labeled P"ODV.TIle mechanical power that is converted is given by I P"onv - Tindw m I (8- 55) and the resulting electric power produced is given by (8- 56) However, this is not the power that appears at the machine's tenninals. Be- fore the terminals are reached, the electricalllR losses and the brush losses must be subtracted. In the case of dc motors, this power-now diagram is simply reversed. The power-now diagram for a motor is shown in Figure 8- 39b.
- 551. IX: MACHINERY FUNDAMENTALS 527 Example problems involving the calculation of motor and generator effi- ciencies will be given in the next two chapters. S.S SUMMARY DC machines convert mechanical power to dc electric power, and vice versa. In this chapter, the basic principles of dc machine operation were explained first by looking at a simple linear machine and then by looking at a machine consisting of a single rotating loop. The concept of commutation as a technique for converting the ac voltage in rotor conductors to a dc output was introduced, and its problems were explored. The possible winding arrangements of conductors in a dc rotor (lap and wave windings) were also examined. Equations were then derived for the induced voltage and torque in a dc ma- chine, and the physical construction of the machines was described. Finally, the types of losses in the dc machine were described and related to its overall operat- ing efficiency. QUESTIONS 8-1, What is conunutation? How can a commutator convert ac voltages on a machine's armature to dc voltages at its terminals? 8-2, Why does curving the pole faces in a dc machine contribute to a smoother dc output voltage from it? 8-3, What is the pitch factor of a coil? 8-4, Explain the concept of electrical degrees. How is the electrical angle of the voltage in a rotor conductor related to the mechanical angle of the machine's shaft? 8-5, What is conunutator pitch? 8-6, What is the plex of an armature winding? 8-7, How do lap windings differ from wave windings? 8-8, What are equalizers? Why are they needed on a lap-wound machine but not on a wave-wolUld machine? 8-9, What is annature reaction? How does it affect the operation of a dc machine? 8-10, Explain the L dildt voltage problem in conductors lUldergoing commutation. 8-11, How does brush shifting affect the sparking problem in dc machines? 8-12, What are conunutating poles? How are they used? 8-13, What are compensating windings? What is their most serious disadvantage? 8-14, Why are laminated poles used in modem dc machine construction? 8-15, What is an insulation class? 8-16, What types of losses are present in a dc machine? PROBLEMS 8-1, The following infonnation is given about the simple rotating loop shown in fig- ure 8--6:
- 552. 528 ELECTRIC MACHINERY RJNDA MENTALS B = O.S T l = O.5 m r = O.I25 m VB = 24 V R = 0.4 0 w = 250 radls (a) Is this machine operating as a motor or a generator? Explain. (b) What is the current i flowing into or out of the machine? What is the power flowing into or out of the machine? (c) If the speed of the rotor were changed to 275 rad/s, what would happen to the current flow into or out of the machine? (d) If the speed of the rotor were changed to 225 rad/s, what would happen to the current flow into or out of the machine? &-2. Refer to the simple two-pole eight-coil machine shown in Figure PS- I. The follow- ing information is given about this machine: ~ad I I_ _ wne __' , , , w , I .. I ---------:::/~_~'',. , - 5' -i/_-"'=:::--------- N , , , I I , , I I , , , , , I I , , , I I , , , I I ' , ,, ,, 7""~-~-~-:=-~-~~8~. ~,~ O~6~.~5~~-~-~------,£~7 3 3' - -- -- I 4 ---- - , , I I , I , I , , , ,,,,, / 1' : 5 , , , r----'-----"' I 20" I 20° I , , , ,, , , , , , , , s Given: II '" 1.0 T in the air gap I '" 0.3 m (length of sides) r '" 0.08 m (radius of coils) n '" 1700 r!min - - - - Lines on this side of rotor - - - - Lines on other side of rotor Jo'IGURE 1'8-1 The machine in Problem 8- 2.
- 553. IX: MACHINERY FUNDAMENTALS 529 B = 1.0T in air gap 1= 0.3 m (length of coil sides) r = 0.08 m (radius of coils) n = 1700 rlmin ccw The resistance of each rotor coil is 0.04 n. (a) Is the armature winding shown a progressive or retrogressive winding? (b) How many current paths are there through the armature of this machine? (c) What are the magnitude and the polarity of the voltage at the brushes in this machine? (d) What is the annature resistance RA of this machine? (e) If a 1O~ resistor is connected to the tenninals of this machine, how much cur- rent flows in the machine? Consider the internal resistance of the machine in de- tennining the current flow. (f) What are the magnitude and the direction of the resulting induced torque? (g) Assuming that the speed of rotation and magnetic flux density are constant, plot the terminal voltage of this machine as a flUlction of the current drawn from it. 8-3. Prove that the equation for the induced voltage of a single simple rotating loop 2 , . , = - "'w In 7T '+' (8-6) is just a special case of the general equation for induced voltage in a dc machine (8--38) 8-4. A dc machine has eight poles and a rated current of 100 A. How much current will flow in each path at rated conditions if the armature is (a) simplex lap-wound, (b) duplex lap-wound, (c) simplex wave-wound? 8-5. How many parallel current paths will there be in the armature of a 12-pole machine if the armature is (a) simplex lap-wolUld, (b) duplex wave-wound, (c) triplex lap- wound, (d) quadruplex wave-wolUld? 8-6. The power converted from one fonn to another within a dc motor was given by Use the equations for EA and "Tiod [Equations (8--38) and (8-49)] to prove that EAtA = "Tiodw..; that is. prove that the electric power disappearing at the point of power conversion is exactly equal to the mechanical power appearing at that point. 8-7. An eight-pole, 25-kW, 120-V dc generator has a duplex lap-wound armature which has 64 coils with 16 turns per coil. Its rated speed is 2400 r/min. (a) How much flux per pole is required to produce the rated voltage in this genera- tor at no-load conditions? (b) What is the current per path in the annature of this generator at the rated load? (c) What is the induced torque in this machine at the rated load? (d) How many brushes must this motor have? How wide must each one be? (e) If the resistance of this winding is 0.011 0: per turn, what is the armature resis- tance RA of this machine? 8-8. Figure PS- 2 shows a small two-pole dc motor with eight rotor coils and four turns per coil. The flux per pole in this machine is 0.0125 Wh.
- 554. 530 ELECTRIC MACHINERY RJNDAMENTALS 2 3 N s ""GURE 1'8-2 The machine in Problem 8--8. (a) If this motor is cOIUlected to a 12-V dc car battery, whal will the no-load speed of the motor be? (b) If the positive terminal of the battery is connected to the rightmost brush on the motor, which way will it rotate? (c) If this motor is loaded down so that it consrunes 50 W from the battery, what will the induced torque of the motor be? (Ignore any internal resistance in the motor.) &-9. Refer to the machine winding shown in Figure P8-3. (a) How many parallel current paths are there through this annature winding? (b) Where should the brushes be located on this machine for proper commutation? How wide should they be? (c) What is the plex of this machine? (d) If the voltage on any single conductor lUlder the pole faces in this machine is e, what is the voltage at the lenninals of this machine? 8-10. Describe in detail the winding of the machine shown in Figure PS-4. If a positive voltage is applied to the brush under the north pole face, which way will this motor rotate? REFERENCES I. Del Toro. V. Electric Machines and Pov.·er Systems. Englewood Cliffs. N.J.: Prentice-Ha.lt. 1985. 2. Fitzgerald. A. E., C. Kingsley. Jr.. and S. D. Umans. Electric Machinery. 5th ed. New York: McGraw-Hilt. 1990. 3. Hubert. Charles I. Preventative Maintenance ofElectrical Equipment. 2nd ed. New York: McGraw-Hilt. 19ff}. 4. Kosow. Irving L. Electric Machinery and Transfanners. Englewood Cliffs. N.J.: Premice-HatJ. 1972. 5. Na.tional Electrical Manufacturers Association. Motors and Generators, Publication MG1-1993. Washington, D.C.. 1993. 6. Siskind. Charles. Direct Current Machinery. New York: McGraw-Hilt. 1952. 7. Werninck. E. H. (ed.). Electric Motor Handbook.. London: McGraw-Hilt. 1978.
- 555. IX: MACHINERY FUNDAMENTALS 531 9 10 8 11 / 16' I 3' 7 I , I / 12 , I / - 15' I , , I / - -- 6 I I g h 13 j , N , I , , , , 6' 5 " m , 14 , P " " I - / I , I -- - / , - / , 8' I 11' / 0'/ 3 16 2 ('1 :x ~~ :XXX -- --- -- , , , , , " , N , , , , , , , , , , , , --- --- ~ m " " p " b , d , f g h ; j , , m " (hI FIGURE P8-J (a) The machine in Problem 8--9. (b) The annature winding diagram of this machine.
- 556. 532 ELECTRIC MACHINERY RJNDAMENTALS ' __ 10 N s 14 2 ""GURE 1'8-4 The machine in Problem 8- 10.
- 557. CHAPTER 9 DC MOTORS AND GENERATORS Dc motors are de machines used as motors, and de generators are de machines used as generators. As noted in Chapter 8, the same physical machine can op- erate as either a motor or a generator-it is simply a question of the direction of the power now through it. This chapter will examine the different types of de mo- tors that can be made and explain the advantages and disadvantages of each. It will include a discussion of de motor starting and solid-state controls. Finally, the chapter will conclude with a discussion of de generators. 9.1 INTRODUCTION TO DC MOTORS The earliest power systems in the United States were de systems, but by the 1890s ac power systems were clearly winning out over de systems. Despite this fact, de motors continued to be a significant fraction of the machinery purchased each year through the 1960s (that fraction has declined in the last 40 years). Why were dc motors so common, when dc power systems themselves were fairly rare? There were several reasons for the continued popularity of dc motors. One was that dc power systems are still common in cars, trucks, and aircraft. When a vehicle has a dc power system, it makes sense to consider using dc motors. An- other application for dc motors was a situation in which wide variations in speed are needed. Before the widespread use of power electronic rectifier-inverters, dc motors were unexcelled in speed control applications. Even ifno dc power source were available, solid-state rectifier and chopper circuits were used to create the necessary dc power, and dc motors were used to provide the desired speed control. 533
- 558. 534 ELECTRIC MACHINERY RJNDA MENTALS ,., ,b , ""GURE 9- 1 Early de motors. (a) A very early de motor built by Elihu Thompson in 1886. It was rated at about oj hp. (Courtesy ofGeneml Electric Company.) (b) A larger four-pole de motor from about the turn of the century. Notice the h.andle for shifting the brush.es to the neutral plane. (Courtesy ofGeneml Electric Company. ) (Today, induction motors with solid-state drive packages are the preferred choice over dc motors for most speed control applications. However, there are still some applications where dc motors are preferred.) DC motors are often compared by their speed regulations. TIle speed regu- lation (SR) of a motor is defined by ' I S -R -~ --: W~"- ' Wn =~W~ "-X -l- ()() - %'1 W-l )
- 559. rx: MmDRS AND GENERATORS 535 I SR = nul ~ nfl x 100% I (9-2) It is a rough measure of the shape of a motor's torque- speed characteristic-a positive speed regulation means that a motor's speed drops with increasing load, and a negative speed regulation means a motor's speed increases with increasing load. The magnitude of the speed regulation tells approximately how steep the slope of the torque- speed curve is. DC motors are, of course, driven from a dc power supply. Unless otherwise specified, the input voltage to a de motor is assumed to be constant, because that assumption simplifies the analysis of motors and the comparison between differ- ent types of motors. There are five major types of dc motors in general use: I. 1lle separately excited dc motor 2. 1lle shunt dc motor 3. 1lle pennanent-magnet dc motor 4. 1lle series dc motor 5. 1lle compounded dc motor Each of these types will be examined in turn. 9.2 THE EQUIVALENT CIRCUIT OFADC MOTOR The equivalent circuit of a dc motor is shown in Figure 9-2. In this figure, the ar- mature circuit is represented by an ideal voltage source E), and a resistor R),. This representation is really the Thevenin equivalent of the entire rotor structure, in- cluding rotor coils, interpoles, and compensating windings, if present. The brush voltage drop is represented by a small battery V bru<h opposing the direction of cur- rent flow in the machine. 1lle field coils, which produce the magnetic flux in the generator, are represented by inductor LF and resistor RF. The separate resistor R odj represents an external variable resistor used to control the amount of current in the field circuit. There are a few variations and simplifications of this basic equivalent cir- cuit. 1lle brush drop voltage is often only a very tiny fraction of the generated voltage in a machine. Therefore, in cases where it is not too critical, the brush drop voltage may be left out or approximately included in the value of R),. Also, the internal resistance of the field coils is sometimes lumped together with the variable resistor, and the total is called RF (see Figure 9-2b). A third variation is that some generators have more than one field coil, all ofwhich will appear on the equivalent circuit. The internal generated voltage in this machine is given by the equation (8- 38)
- 560. 536 ELECTRIC MACHINERY RJNDAMENTALS --iJ:- V~ I, I 'VI/v , R, E, L, (a) R, I, _ A, R, + E, L, A, ,b, ""GURE 9-2 (a) The equivalent circuit of a dc motor. (b) A simplified equivalent circuit eliminating the brush voltage drop and combining R..., with the field resistance. and the induced torque developed by the machine is given by (8-49) TIlese two equations, the Kirchhoff's voltage law equation of the annature circuit and the machine's magnetization curve, are all the tools necessary to analyze the behavior and performance of a dc motor. 9.3 THE MAGNETIZATION CURVE OF A DC MACHINE The internal generated voltage Ell of a dc motor or generator is given by Equation (8- 38), (8- 38) Therefore, Ell is directly proportional to the nux in the machine and the speed of rotation of the machine. How is the internal generated voltage related to the field current in the machine?
- 561. rx: MmDRS AND GENERATORS 537 ~.Wb '--- - - - - - - - - - - - - - 3'. A· turns FIGURE 9-3 The magnetization curve of a ferromagnetic material (4) versus 3'). "'="'0 n ="0 (constant) ' - - - - - - - - - - - - - - - - IF [=~; ] FIGURE 9-4 The magnetization curve of a dc machine expressed as a plot of E,t versus IF. for a fixed speed .".. The field current in a dc machine produces a field magnetomotive force given by '?} = NFIF. nlis magnetomotive force produces a flux in the machine in accordance with its magnetization curve (Figure 9- 3). Since the field current is di- rectly proportional to the magnetomoti ve force and since EA is directly propor- tional to the flux, it is customary to present the magnetization curve as a plot of EA versus field current for a given speed Wo (Figure 9--4). It is worth noting here that, to get the maximum possible power per pound of weight out of a machine, most motors and generators are designed to operate near the saturation point on the magnetization curve (at the knee of the curve). This implies that a fairly large increase in field current is often necessary to get a small increase in EAwhen operation is near full load. The dc machine magnetization curves used in this book are also available in electronic form to simplify the solution of problems by MATLAB. Each magneti- zation curve is stored in a separate MAT file. Each MAT file contains three
- 562. 538 ELECTRIC MACHINERY RJNDAMENTALS variables: if_val ues, containing the values of the field current; ea_val ues, containing the corresponding val ues of E).; and n_O, containing the speed at which the magnetization curve was measured in units of revolutions per minute. 9.4 SEPARATELY EXCITED AND SHUNT DC MOTORS TIle equivalent circuit of a separately excited dc motor is shown in Figure 9- 5a, and the equivalent circuit of a shunt dc motor is shown in Figure 9- 5b. A sepa- rately excited dc motor is a motor whose field circuit is supplied from a separate I, - + R., R, V, L, + E, ""GURE 9-5 R, Sometimes lumped together and called RF C)E, V, IF = - R, Vr = E). + lARA lL=IA R, Lumped together and called RF V, IF= RF [ (a) I, - l,j Vr = EA + lARA 'bJ I, - R... R, L, I, I, - - + V, + V, (a) The equivalent circuit of a separately excited dc lootor. (b) The equivalent circuit of a shunt dc motor.
- 563. rx: MmDRS AND GENERATORS 539 constant-voltage power supply, while a shunt dc motor is a motor whose field circuit gets its power directly across the armature terminals of the motor. When the supply voltage to a motor is assumed constant, there is no practical difference in behavior between these two machines. Unless otherwise specified, whenever the behavior of a shunt motor is described, the separately excited motor is included, too. The Kirchhoff's voltage law (KVL) equation for the armature circuit of these motors is (9- 3) The Terminal Characteristic of a Shunt DC Motor A tenninal characteristic of a machine is a plot of the machine's output quantities versus each other. For a motor, the output quantities are shaft torque and speed, so the terminal characteristic of a motor is a plot of its output torque versus speed. How does a shunt dc motor respond to a load? Suppose that the load on the shaft of a shunt motor is increased. Then the load torque "Tlood will exceed the in- duced torque "TiOO in the machine, and the motor wil I start to slow down. When the motor slows down, its internal generated voltage drops (EA = K4>wJ.), so the ar- mature current in the motor IA = (VT - EAJ.)/RA increases. As the annature current rises, the induced torque in the motor increases (1]00 = K4> IAi ), and finally the in- duced torque will equal the load torque at a lower mechanical speed of rotation w. TIle output characteristic of a shunt dc motor can be derived from the in- duced voltage and torque equations of the motor plus Kirchhoff's voltage law. (KVL) TIle KVL equation for a shunt motor is VT = EA + lARA The induced voltage EA= K4>w, so VT = K¢w + lARA Since "Tind = K4>IA, current IA can be expressed as "Tind IA = K4> Combining Equations (9-4) and (9-5) prOOuces Finally, solving for the motor's speed yields (9- 3) (9-4) (9- 5) (9-6) (9- 7) This equation is just a straight line with a negative slope. TIle resulting torque- speed characteristic of a shunt dc motor is shown in Figure 9--6a.
- 564. 540 ELECTRIC MACHINERY RJNDAMENTALS "---------------------------- '00 (a) ------ WithAR ---NoAR "---------------------------- '00 ,b, ""GURE 9--6 (a) Torque-speed characteristic of a shunt or separately excited dc motor with compensating windings to eliminate armature reaction. (b) Torque-speed characteristic of the motor with annature reaction present. II is important to realize that, in order for the speed of the motor to vary lin- early with torque, the other terms in this expression must be constant as the load changes. TIle tenninal voltage supplied by the dc power source is assumed to be constant- if it is not constant, then the voltage variations will affect the shape of the torque- speed curve. Another effect internal to the motor that can also affect the shape of the torque-speed curve is armature reaction. If a motor has annature reaction, then as its load increases, the flux-weakening effects reduce its flux. As Equation (9-7) shows, the effect ofa reduction in flux is to increase the motor's speed at any given load over the speed it would run at without armature reaction. The torque-speed characteristic of a shunt motor with annature reaction is shown in Figure 9--6b. If a motor has compensating windings, of course there will be no flux-weakening problems in the machine, and the flux in the machine will be constant.
- 565. rx: MmDRS AND GENERATORS 541 " R, " - - + 0.060 R;).~ I', son + R, < ~ E, VT",250V L, NF '" 1200tuTns FIGURE 9- 7 The shunt motor in Example 9--1. Ifa shunt dc motor has compensating windings so that its flux is constant regardless ofload, and the motor's speed and armature current are known at any one value of load, then it is possible to calculate its speed at any other value of load, as long as the armature current at that load is known or can be detennined. Example 9- 1 illustrates this calculation. Example 9-1. A 50-hp, 250-V, 1200 r/min dc shunt motor with compensating windings has an armature resistance (including the brushes, compensating windings, and interpoles) of 0.06 n. Its field circuit has a total resistance Rodj + RF of 50 fl, which pro- duces a no-load speed of 1200 r/min. There are 1200 tlU1lSper pole on the shunt field wind- ing (see Figure 9-7). (a) Find the speed of this motor when its input current is 100 A. (b) Find the speed of this motor when its input current is 200 A. (c) Find the speed of this motor when its input current is 300 A. (d) Plot the torque-speed characteristic of this motor. Solutioll The internal generated voltage of a dc machine with its speed expressed in revolutions per minute is given by (8-41) Since the field current in the machine is constant (because Vr and the field resistance are both constant), and since there are no annature reaction effects, the flux in this motor is constant.The relationship between the speeds and internal generated voltages of the motor at two different load conditions is thus (9-8) The constant K ' cancels, since it is a constant for any given machine, and the flux <p can- cels as described above. Therefore, (9-9)
- 566. 542 ELECTRIC MACHINERY RJNDAMENTALS At no load, the armature current is zero, so E"I = Vr = 250 V, while the speed nl = 12DO r/min. If we can calculate the internal generated voltage at any other load, it will be possible to detennine the motor speed at that load from Equation (9--9). (a) If h = lDO A, then the armature current in the motor is V, I" = It. - IF = It. - RF - lOOA _ 250 V - 95A - son - Therefore, E" at this load will be E" = Vr - I"R" = 250 V - (95 A)(O.06 ll) = 244.3 V The resulting speed of the motor is E"2 244.3 V 1200 / . n2 = E"I nl = 250V rmm = 1173 r/min (b) If h = 2DO A, then the armature current in the motor is I _ 2DOA _ 250V - 195A ,, - son - Therefore, E" at this load will be E" = Vr - I"R" = 250 V - (195 A)(O.()5 ll) = 238.3 V The resulting speed of the motor is E"2 238.3 V 1200 / . 1144 / . n2= Enl= 250V rmm = rmm " (e) If h = 3DO A, then the armature current in the motor is V, I" = It. - IF = It. - RF - 300 A _ 250 V - 295 A - son - Therefore, E" at this load will be E" = Vr - I"R" = 250 V - (295 A)(O.()5 ll) = 232.3 V The resulting speed of the motor is E"2 232.3 V 1200 / . n2 = E"I nl = 250V rmm = 1115r/min (d) To plot the output characteristic of this motor, it is necessary to find the torque corresponding to each value of speed. At no load, the induced torque "Tind is clearly zero. The induced torque for any other load can be fOlUld from the fact that power converted in a dc motor is
- 567. rx: MmDRS AND GENERATORS 543 I PCQDV Ell!'" 7indW I From this equation, the induced torque in a motor is E",!", 7iod = -- w Therefore, the induced torque when !L = 100 A is _ (244.3 V)(95 A) _ 7 ;00 - (1173 r/minXI min/60sX27T rad!r) - The induced torque when h = 200 A is (238.3 V)(95 A) = 388 N • m 7 ;00 = (1144 r/minXI min/60sX27T rad!r) The induced torque when !L = 300 A is (232.3 VX295 A) = 587 N • m 7iOO = (1115 r/minXI min/60sX27T rad!r) (8- 55,8--56) (9--10) The resulting torque-speed characteristic for this motor is plotted in Figure 9--8. Nonlinear Analysis of a Shunt DC Motor T he nux $ and hence the internal generated vo ltage E", of a dc machine is a non- linear function of its magnetomotive force. T herefore, anything that changes the 1200 11110 0 11X1O ~ • 0 900 800 7110 T o 200 400 600 800 FIGURE 9- 8 The torque-speed characteristic of the motor in Example 9--1. f iOO' N·m
- 568. 544 ELECTRIC MACHINERY RJNDAMENTALS magnetomotive force in a machine will have a nonlinear effect on the internal generated voltage of the machine. Since the change in Ell cannot be calculated an- alytically, the magnetization curve of the machine must be used to accurately de- tennine its Ell for a given magnetomotive force. The two principal contributors to the magnetomotive force in the machine are its field current and its annature re- action, if present. Since the magnetization curve is a direct plot of Ell versus IF for a given speed wo , the effect of changing a machine's field current can be detennined di- rectly from its magnetization curve. If a machine has annature reaction, its flux will be reduced with each increase in load. The total magnetomotive force in a shunt dc motor is the field circuit magnetomotive force less the magnetomotive force due to annature re- action (AR): (9-11 ) Since magnetization curves are expressed as plots of Ell versus field current, it is customary to define an equivalentfield current that wou ld produce the same out- put voltage as the combination of all the magnetomotive forces in the machine. 1lle resulting voltage Ell can then be detennined by locating that equivalent field current on the magnetization curve.1lle equivalent field current of a shunt dc mo- tor is given by (9- 12) One other effect must be considered when nonlinear analysis is used to de- termine the internal generated voltage ofa dc motor. The magnetization curves for a machine are drawn for a particular speed, usually the rated speed of the ma- chine. How can the effects of a given field current be determined if the motor is turning at other than rated speed? 1lle equation for the induced voltage in a dc machine when speed is ex- pressed in revolutions per minute is Ell = K'cp n (8-4 1) For a given effective field current, the flu x in a machine is fixed, so the internal generated voltage is related to speed by (9- 13) where Ello and 110 represent the reference values of voltage and speed, respectively. If the reference conditions are known from the magnetization curve and the actual Ell is known from Kirchhoff's voltage law, then it is possible to determine the ac- tual speed n from Equation (9-1 3).1lle use of the magnetization curve and Equa- tions (9-1 2) and (9-1 3) is illustrated in the following example, which analyzes a dc motor with armature reaction.
- 569. > t " ., § ." " .~ , !. a 300 250 233 200 150 100 50 I I o 0.0 I I / I I 1.0 2.0 rx: MmDRS AND GENERATORS 545 ,/ V / / V / 3.0 4.0 4.3 5.0 6.0 7.0 8.0 9.0 10.0 Field current, A FIGURE 9-9 The magnetization curve of a typical 25()'Vdc motor. taken at a speed of 1200 r/min. EXllmple 9-2. A 5O-hp, 250-V, 1200 r/min dc shlUlt motor without compensating windings has an armature resistance (including the brushes and interpoles) of 0.06 n. Its field circuit has a total resistance RF + Radj of 50 n, which produces a no-load speed of 1200 r/min. There are 1200 turns per pole on the shunt field winding, and the armature re- action produces a demagnetizing magnetomotive force of 840 A • turns at a load current of 200 A. The magnetization curve of this machine is shown in Figure 9-9. (a) Find the speed of this motor when its input current is 200 A. (b) This motor is essentially identical to the one in Example 9- 1 except for the ab- sence of compensating windings. How does its speed compare to that of the pre- vious motor at a load current of 200 A? (c) Calculate and plot the torque-speed characteristic for this motor. Solutioll (a) If IL = 200 A, then the armature current of the motor is V, lit = IL - IF = h - R, - 2ooA - 250 V - 195A - 50n -
- 570. 546 ELECTRIC MACHINERY RJNDA MENTALS Therefore, the internal generated voltage of the machine is EA = Vr - lARA = 250 V - (195 A)(O.()5 fi) = 238.3 V At h = 200 A, the demagnetizing magnetomoti ve force due to armature reac- tion is 840 A · turns, so the effective shunt field current of the motor is "'" r,,= IF - N, (9-12) From the magnetization curve, this effective field current would produce an in- ternal generated voltage EAOof 233 Vat a speed 110 of 1200 r/min. We know that the internal generated voltage EAO would be 233 V at a speed of 1200 r/min. Since the actual internal generated voltage EA is 238.3 V, the ac- tual operating speed of the motor must be (9-13) EA 238.3 V (1200 I . ) n= £ no = 233V rmm = " 1227 r/min (b) At 200 A of load in Example 9--1, the motor's speed was n = 1144 r/min. In this example, the motor's speed is 1227 r/min. Notice that the speed of the motor with armature reaction is higher than the speed ofthe motor with no armature reaction .This relative increase in speed is due to the flux weakening in the ma- chine with armature reaction. (c) To derive the torque- speed characteristic of this motor, we must calculate the torque and speed for many different conditions of load. Unfortunately, the de- magnetizing armature reaction magnetomotive force is only given for one con- dition of load (200 A). Since no additional information is available, we will as- srune that the strength of '.JAR varies linearly with load current. A MATLAB M-file which automates this calculation and plots the resulting torque-speed characteristic is shown below. It peIfonns the same steps as part a to detennine the speed for each load current, and then calculates the induced torque at that speed. Note that it reads the magnetization curve from a file called f i g9_ 9 .ma t. This file and the other magnetization curves in this chapter are available for download from the book's World Wide Web site (see Preface for details). ~ M-file: s hunt_ t s_curve. m ~ M-file c r eat e a p l ot o f the t o r qu e - speed cu rve o f the ~ the s hunt dc mot or with a r ma ture r eaction in ~ Exampl e 9- 2. ~ Get the magne ti zation c u rve. Thi s fil e conta ins the ~ three variabl es if_va lu e, ea_va lue, a nd n_O. l oad fi g9_9. ma t ~ Firs t , initia li ze the va lues n eeded in thi s p r ogr am. v_t = 250; % Te rmina l vo lt age (V)
- 571. rx: MmDRS AND GENERATORS 547 c- f 0 50 ; •Fi e l d r es i s t a nce (ohms) c-" 0 0 . 06 ; •Arma tur e r es i s t a nce (ohms) i 1 0 1 0, 1 0,300 ; •Li ne curre nt s (A ) 0 - f 0 1200 ; •Number of turns 0 0 fi e l d f -" c O 0 84 0 ; •Arma tur e r eact i on , 200 A % Ca l c u l a t e the a rma ture c urre nt f or each l oad . i _a = i _ l - v_t I r _ f ; % Now ca l c u l a t e the i nt e rna l gene r a t ed vo l t age f or % each a rma ture c urre nt. e_a = v_t - i _a * r _a; (A- t i m) % Ca l c u l a t e the a rma ture r eact i on MMF f or each a rma ture % c urre nt. f _ar = ( i _a I 200) * f _ar O: % Ca l c u l a t e the e ff ect i ve fi e l d c urre nt. i f = v_t I r _ f - f _ar I n_ f : % Ca l c u l a t e the r esu l t i ng i nt e rna l gene r a t ed vo l t age a t % 1200 r / mi n by i nt e r po l a t i ng the mot or 's magne t i za t i on % c urve . e_aO = i nt e r p l ( if_va l ues, ea_va l ue s, i _ f , ' sp line ' ) ; % Ca l c u l a t e the r esult i ng speed f r om Equa t i on (9 - 13 ) . n = ( e_a . 1 e_aO ) .. n_O: % Ca l c u l a t e the i nduced t or que co rrespond i ng t o each % speed f r om Equa t i on s ( 8- 55 ) a nd (8- 56 ) . t _ i nd = e_a . * i _a .1 (n .. 2 .. p i I 60): % Pl ot the t or que - speed c urve p l ot (t _ i nd, n , 'Col or' , 'k' , 'LineWi dth' ,2 . 0 ) ; h o l d on ; x l abe l ( ' t au_( i nd} (N- m) ' , ' Fontwei ght' , 'Bo l d ' ) ; y l abe l ( ' i tn_( m} rm b f (r / mi n ) ' , 'Fontwei ght' , 'Bo l d ' ) : ( ' b f Shunt OC mot or t or que - speed c h a r act e r i s t i c ' ) ax i s( [ 0 600 1100 1300 ] ) ; g r i d on ; h o l d o ff ; The resulting torque-speed characteristic is shown in Figure 9-10. Note that for any given load. the speed of the motor with armature reaction is higher than the speed of the motor without armature reaction. Speed Control of Shunt DC Motors How can the speed of a shunt dc motor be controlled? There are two common methods and one less common method in use.1lle common methods have already been seen in the simple linear machine in Chapter 1 and the simple rotating loop in Chapter 8. The two common ways in which the speed of a shunt dc machine can be controlled are by
- 572. 548 ELECTRIC MACHINERY RJNDAMENTALS 1300 ,------,--,----,------,--,------, 1280 1260 1240 1220 t--~---'---: ~ 1200 ./"80 ]]60 ]]40 ]]20 ]]00 '-----c'cC_-~-__C'c---c'cC_-~-~ o 100 200 300 400 500 600 'tiD<! N·m ""GURE 9- 10 The torque--speed characteristic of the motor with armature reaction in Example 9--2. I. Adjusting the field resistance RF (and thus the field nux) 2. Adjusting the tenninal voltage applied to the annature. The less common method of speed control is by 3. Inserting a resistor in series with the armature circuit. E:1.ch of these methods is described in detail below. CHANGING THE FIELD RESISTANCE. To understand what happens when the field resistor of a dc motor is changed. assume that the field resistor increases and observe the response. If the field resistance increases, then the field current de- creases (IF = Vr lRF i ), and as the field current decreases, the nux <P decreases with it. A decrease in nux causes an instantaneous decrease in the internal gener- ated voltage EA(= K<p-tw), which causes a large increase in the machine's anna- ture current, since 1lle induced torque in a motor is given by "TiDd = K<pfA' Since the nux <p in this machine decreases while the current fA increases, which way does the induced torque change? 1lle easiest way to answer this question is to look at an example. Figure 9- 1I shows a shunt dc motor with an internal resistance of 0.25 O. lt is currently operating with a tenninal voltage of 250 V and an internal generated voltage of 245 V. 1llerefore, the annature current now is fA = (250 V -
- 573. rx: MmDRS AND GENERATORS 549 RA '" 0.250 + + EA",245 V'" Kfw VT", 250 V FIGURE 9- 11 A 250-V shunt de motor with typical values of EA and RA. 245 V)/0.25 n = 20 A. What happens in this motor ifthere is a I percent decrease influx? If the flux decreases by I percent, then EA must decrease by I percent too, because EA = K4>w. Therefore, EAwill drop to EA2 = 0.99 EAt = 0.99(245 V) = 242.55 V The annature current must then rise to f = 250 V - 242.55 V = 298 A A 0.25 n . Thus a I percent decrease in flux produced a 49 percent increase in armature current. So to get back to the original discussion, the increase in current predomi- nates over the decrease in flux, and the induced torque rises: U Tjnd = K4> fA Since Tind > Tto"," the motor speeds up. However, as the motor speeds up, the internal generated voltage EA rises, causing fAto fall. As fAfalls, the induced torque Tind falls too, and fmally T;Dd again equals Ttood at a higher steady-state speed than originally. To summarize the cause-and-effcct behavior involved in this method of speed control: I. Increasing RF causes fF(= VT IRFi ) to decrease. 2. Decreasing IF decreases 4>. 3. Decreasing 4> lowers EA(= K4>J..w). 4. Decreasing EAincreases fA(= VT -EAJ..)IRA- 5. Increasing fA increases T;od(= K4>UAfI), with the change in fA dominant over the change in flux). 6. Increasing Tind makes T;od > Ttood, and the speed w increases. 7. Increasing to increases EA = Kcf>wi again.
- 574. 550 ELECTRIC MACHINERY RJNDAMENTALS (a) ,b, 8. Increasing Elt decreases lit- FlGURE 9-12 The effect of field resistance speed control on a shunt motor's torque-speed characteristic: (a) over the motor's normal operating range: (b) over the entire range from no-load to stall conditions. 9. Decreasing lit decreases "Tind until "Tind = "TJoad at a higher speed w. The effect of increasing the field resistance on the output characteristic of a shu nt motor is shown in Figure 9-1 2a. Notice that as the flux in the machine decreases, the no-load speed of the motor increases, while the slope of the torque-speed curve becomes steeper. Naturally, decreasing RF would reverse the whole process, and the speed of the motor would drop. A WARNING ABOUT FIELD RESISTANCE SPEED CONTROL. TIle effect of in- creasing the field resistance on the output characteristic of a shunt dc motor is shown in Figure 9-1 2. Notice that as the flux in the machine decreases, the no- load speed of the motor increases, while the slope of the torque-speed curve be- comes steeper. This shape is a consequence of Equation (9- 7), which describes the tenninal characteristic of the motor. In Equation (9- 7), the no-load speed is
- 575. R, I, Variable - + voltage controller + E, V, - - FIGURE 9-13 /- rx: MmDRS AND GENERATORS 551 I, - y,,!I, ~ R, L, VTis oonstant VA is variable + V, Armature voltage control of a shunt (or separately excited) dc motor. proportional to the reciprocal of the nUX: in the motor, while the slope of the curve is proportional to the reciprocal of the flux squared. Therefore, a decrease in flux causes the slope of the torque- speed curve to become steeper. Figure 9- I2a shows the tenninal characteristic of the motor over the range from no-load to full-load conditions. Over this range, an increase in field resis- tance increases the motor's speed, as described above in this section. For motors operating between no-load and full-l oad conditions, an increase in RF may reliably be expected to increase operating speed. Now examine Figure 9- 12h. This figure shows the tenninal characteristic of the motor over the full range from no-load to stall conditions. It is apparent from the figure that at very slow speeds an increase in field resistance will actually de- crease the speed of the motor. TIlis effect occurs because, at very low speeds, the increase in annature current caused by the decrease in Ell. is no longer large enough to compensate for the decrease in flux in the induced torque equation. With the flux decrease actually larger than the armature current increase, the in- duced torque decreases, and the motor slows down. Some small dc motors used for control purposes actually operate at speeds close to stall conditions. For these motors, an increase in field resistance might have no effect, or it might even decrease the speed of the motor. Since the results are not predictable, field resistance speed control should not be used in these types of dc motors. Instead, the annature voltage method of speed control shou ld be employed. CHANGING THE ARMATURE VOLTAGE. TIle second form of speed control in- volves changing the voltage applied to the armature of the motor without chang- ing the voltage applied to the field. A connection similar to that in Figure 9- 3 is necessary for this type of control. In effect, the motor must be separately excited to use armature voltage control. If the voltage VA is increased, then the annature current in the motor must rise [Ill. = (VAi - EA)IRAl As /11. increases, the induced torque "Tind = K4>IAi in- creases, making "Tind > "TJoad, and the speed w of the motor increases.
- 576. 552 ELECTRIC MACHINERY RJNDAMENTALS v" ' - - - - - - - - - - - - - - - - f ind ""GURE 9-14 The effect of armature voltage speed control on a shunt motor's torque-speed characteristic. Bul as the speed w increases, the internal generaled voltage EA(= K4>wi) increases, causing the armature current to decrease. This decrease in JA decreases the induced torque, causing Tind to equal Ttoad at a higher rotational speed w. To summarize the cause-and-effect behavior in this method of speed control: I . An increase in VAincreases JA [= (VA i - EA)/RA]. 2. Increasing JA increases Tind (= K4>JAi ). 3. Increasing Tind makes TiDd >TJoad increasing w. 4. Increasing w increases EA(= K4>wi). 5. Increasing EA decreases JA [= (VA i - EA)/RAl 6. Decreasing JA decreases Tind until Tind = TJoad at a higher w. TIle effect of an increase in VA on the torque-speed characteristic of a sepa- rately excited motor is shown in Figure 9- 14. Notice that the no-load speed of the motor is shifted by this method of speed control, but the slope of the curve re- mains constant. INSERTING A RESISTOR IN SERIES WITH THE ARll ATURE c m.CUIT. If a resistor is inserted in series with the annature circuit, the effect is to drastically in- crease the slope of the motor's torque-speed characteristic, making it operate more slowly if loaded (Figure 9- 15). lllis fact can easily be seen from Equation (9- 7). The insertion of a resistor is a very wasteful method of speed control, since the losses in the inserted resistor are very large. For this reason, it is rarely used. It will be found only in applications in which the motor spends almost all its time operating at fuJI speed or in applications too inexpensive to justify a better form of speed control. TIle two most common methods of shunt motor speed control- field resis- tance variation and armature voltage variation- have different safe ranges of operation.
- 577. rx: MmDRS AND GENERATORS 553 '------------------ '00 FIGURE 9-15 The effect of armature resistance speed control on a shunt motor's torque-speed characteristic. In field resistance control, the lower the field current in a shunt (or sepa- rately excited) dc motor, the faster it turns: and the higher the field current, the slower it turns. Since an increase in field current causes a decrease in speed, there is always a minimum achievable speed by field circuit control. This minimum speed occurs when the motor's field circuit has the maximum permissible current flowing through it. If a motor is operating at its rated terminal voltage, power, and field current, then it will be running at rated speed, also known as base speed. Field resistance control can control the speed of the motor for speeds above base speed but not for speeds below base speed. To achieve a speed slower than base speed by field cir- cuit control would require excessive field current, possibly burning up the field windings. In armature voltage control, the lower the armature voltage on a separately excited dc motor, the slower it turns; and the higher the armature voltage, the faster it turns. Since an increase in annature voltage causes an increase in speed, there is always a maximum achievable speed by armature voltage control. This maximum speed occurs when the motor's armature voltage reaches its maximum permissible level. If the motor is operating at its rated voltage, field current, and power, it will be turning at base speed. Annature voltage control can control the speed of the motor for speeds below base speed but not for speeds above base speed. To achieve a speed faster than base speed by armature voltage control would require excessive annature voltage, possibly damaging the annature circuit. These two techniques of speed control are obviously complementary. Ar- mature voltage control works well for speeds below base speed, and field resis- tance or field current control works well for speeds above base speed. By com- bining the two speed-control techniques in the same motor, it is possible to get a range of speed variations of up to 40 to I or more. Shunt and separately excited dc motors have excellent speed control characteristics.
- 578. 554 ELECTRIC MACHINERY RJNDAMENTALS Maximum torque fmax Maximum powerPmu fmax constant Pmu constant VA control RFcontrol "------cC------ ,- ,~ ""GURE 9-16 fmu constant so Prmx constant PIOU '" f mn w..,-______ V" control VA control <-----~------- ,. ,~ Power and torque limits as a function of speed for a shunt motor under annature volt and field resistance control. TIlere is a significant difference in the torque and power limits on the ma- chine under these two types of speed control. The limiting factor in either case is the heating of the annature conductors, which places an upper limit on the mag- nitude of the annature current Ill.. For annature voltage control, the flux in the motor is constant, so the maxi- mum torque in the motor is (9-1 4) This maximum torque is constant regardless ofthe speed ofthe rotation of the mo- tor. Since the power out of the motor is given by P = "TW, the maximum power of the motor at any speed under annature voltage control is (9- 15) TIlUS the maximum power out ofthe motor is directly proportional to its operat- ing speed under armature voltage control. On the other hand, when field resistance control is used, the flux does change. In this form of control, a speed increase is caused by a decrease in the ma- chine's flux. In order for the annature current limit not to be exceeded, the in- duced torque limit must decrease as the speed of the motor increases. Since the power out of the motor is given by P = "TW, and the torque limit decreases as the speed of the motor increases, the maximum power out ofa dc motor underfield current control is constant, while the maximum torque varies as the reciprocal of the motor's speed. TIlese shunt dc motor power and torque limitations for safe operation as a function of speed are shown in Figure 9-16. TIle following examples illustrate how to fmd the new speed of a dc motor if it is varied by field resistance or annature voltage control methods.
- 579. rx: MaJ"ORS AND GENERATORS 555 RA '" 0.03 n - - ", +~-------, ,-----v./v--~ + + ,b, FIGURE 9-17 (3) The shunt motor in Example 9--3. (b) The separately excited de motor in Example 9--4. Example 9-3. Figure 9--17a shows a 1000hp. 250-V, 1200 rhnin shlUlt dc motor with an armature resistance of 0.03 n and a field resistance of 41.67 O. The motor has compen- sating windings. so armature reaction can be ignored. Mechanical and core losses may be as- sumed to be negligible for the purposes of this problem. The motor is assmned to be driving a load with a line current of 126 A and an initial speed of 1103 r/min. To simplify the prob- lem. assmne that the amolUlt of armature current drawn by the motor remains constant. (a) If the machine's magnetization curve is shown in Figure 9-9. what is the mo- tor's speed if the field resistance is raised to 50 n1 (b) Calculate and plot the speed of this motor as a ftmction of the field resistance RI' assuming a constant-current load. Solutioll (a) The motor has an initial line current of 126 A, so the initial armature current is 150 V IA I = lu - 1Ft = 126A - 41.67 n = 120A Therefore, the internal generated voltage is EAI = VT - IA tRA = 250 V - (120 A)(0.03fi) = 246.4 V After the field resistance is increased to 50 n, the field current will become
- 580. 556 ELECTRIC MACHINERY RJNDAMENTALS I - VT _2S0V -SA F2 - RF - 50 n - The ratio of the internal generated voltage at one speed to the internal generated voltage at another speed is given by the ratio of Equation (&--41) at the two speeds: Elll _ K'q,l n2 Elll - K'q,[n[ (9- 16) Because the armature current is assumed constant, EIlI = Elll, and this equation reduces to ~, n2=q,lnl (9- 17) A magnetization curve is a plot of Ell versus IF for a given speed. Since the val- ues of Ell on the curve are directly proportional to the flux, the ratio of the inter- nal generated voltages read off the curve is equal to the ratio of the fluxes within the machine. At IF =5 A, Ello =250 V, while at IF =6 A, Ello =268 V. There- fore, the ratio of fluxes is given by q,[ 268 V = = 1.076 q,l 250 V and the new s~ed of the motor is n2 =!:nl =(1.076XII03r/min) = 1187r/min (b) A MATLAB M-file that calculates the s~ed of the motor as a ftmction of RF is shown below. ~ M -file, r f _speed_contro l .m ~ M -file c reate a p l ot o f the speed of a s hunt dc ~ mot or as a f unc t i on o f f i e l d r es i s tance, assumi ng ~ a con s tant armature c urrent (Exampl e 9- 3). ~ Get the magnet i zat i on c urve. Thi s f il e conta i n s the ~ three var i abl es if_va l u e, ea_va l u e, and n_O. l oad fi g9_9.mat ~ Fir s t , i nit i a li ze v_t = 250; c- f 0 40,1:70; c-" 0 0.03; i -" 0 12 0; the va l u es n eeded i n thi s program. ~ Termi na l vo l tage (V) ~ Fie l d re s i s tance (ohms) ~ Armature re s i s tance (ohms) ~ Armature c urrent s (A) ~ The approach here i s t o ca l c u l ate the e_aO at the ~ re f erence fi e l d c urrent , and then to ca l cu l ate the ~ e_aO f or every fi e l d c urrent. The r e f e rence speed i s ~ 1103 r / mi n , so by knowi ng the e_aO and r e f e rence ~ speed, we will be abl e to ca l cu l ate the speed at the ~ other fi e l d c urrent.
- 581. rx: M mDRS AND GENERATORS 557 % Ca l c ula t e the int e rna l gen e r a t ed vo lt age a t 1200 r / min % f or the r e f e r e nce fi e l d c urre nt (5 A) by int e r po l a ting % the mot or 's magne tiz a tion c urve. The r e f e r e nce speed % correspond ing t o thi s fi e l d c urre nt i s 1103 r / min . e_aO_r e f = int e r p l ( if_va lues, ea_va lues, 5, ' sp line ' ) ; n_ r e f = 1103; % Ca l c ula t e the fi e l d c urre nt f or each va lue o f fi e l d % r es i s t a nce. i _ f = v_t . / r _ f ; % Ca l c ula t e the E_aO f or eac h fi e l d c urre nt by % int e r po l a ting the mot or 's magne ti zation c urve. e_aO = int e r p l ( if_va lues, ea_va lues, i _ f , ' sp line ' ) ; % Ca l c ula t e the r esulting speed from Equa tion (9 -1 7): % n2 = (phil / phi2 ) .. nl = (e_aO_l / e_aO_2 ) .. nl n2 = ( e_aO_r e f . / e_aO ) .. n_ r e f ; % Plot the speed ver s u s r _ f c urve. p l ot (r _ f , n2, 'Col or' , 'k' , 'LineWi dth' ,2.0 ) ; h o l d on ; x l abe l ( 'Fi e l d r es i s t a nce, Omega ' , 'Fontwei ght' , 'Bo l d ' ) ; y l abe l ( ' itn_( m} rm b f (r / min ) ' , 'Fontwei ght' , 'Bo l d ' ) ; title ( ' Speed vs itR_( F ) rm b f f or a Shunt IX: M ot or' , 'Fontwei ght' , 'Bo l d ' ) ; ax i s( [40 70 0 1400 ] ) ; g rid on ; h o l d o ff ; The resulting plot is shown in Figure 9-1 8. 1400 1200 , , IlXXl .§ 800 ~ •600 0 400 200 0 40 45 60 65 70 Field resistance. n FIGURE 9- 18 Plot of speed versus field resistance for the shunt dc motor of Example 9-3.
- 582. 558 ELECTRIC MACHINERY RJNDAMENTALS Note that the assumption of a constant annature current as RF changes is not a very go<Xl one for real loads. The current in the annature will vary with speed in a fashion dependent on the torque required by the type of load attached to the mo- tor. These differences will cause a motor 's speed-versus-RF curve to be slightly different than the one shown in Figure 9-1 8, but it will have a similar shape. Example 9-4. The motor in Example 9- 3 is now cotUlected separately excited, as shown in Figure 9-17b. The motor is initially flmning with VA = 250 V, IA = 120 A, and n = 1103 rlmin, while supplying a constant-torque load. What will the speed of this motor be if VA is reduced to 200 V? Solutio" The motor has an initial line current of 120 A and an armature voltage VA of 250 V, so the internal generated voltage EA is EA = Vr - lARA = 250 V - (l20AXO.03!l) = 246.4 V By applying Equation (9-16) and realizing that the flux ~ is constant, the motor's speed can be expressed as To find EJa use Kirchhoff's voltage law: ", = EJa = Vr - lJaRA (9-16) Since the torque is constant and the flux is constant, IA is constant. This yields a voltage of EJa = 200 V - (120 AXO.03 !l) = 196.4 V The final speed of the motor is thus E A2 196.4V,'03 /· 879/· ~ = EAt nt = 246.4 V r mill = r min The Effect of an Open Field Circuit TIle previous section of this chapter contained a discussion of speed control by varying the field resistance of a shunt motor. As the field resistance increased, the speed of the motor increased with it. What would happen if this effect were taken to the extreme, if the field resistor really increased? What would happen if the field circuit actually opened while the motor was running? From the previous dis- cussion, the flux in the machine would drop drastically, all the way down to ~res, and EA(= K~w) would drop with it. This would cause a really enonnous increase in the armature current, and the resulting induced torque would be quite a bit higher than the load torque on the motor. TIlerefore, the motor's speed starts to rise and just keeps going up.
- 583. rx: MmDRS AND GENERATORS 559 The results of an open field circuit can be quite spectacular. When the au- thor was an undergraduate, his laboratory group once made a mistake of this sort. The group was working with a small motor-generator set being driven by a 3-hp shunt dc motor. The motor was connected and ready to go, but there was just one little mistake- when the field circuit was connected, it was fused with a O.3-A fuse instead of the 3-A fu se that was supposed to be used. When the motor was started, it ran nonnally for about 3 s, and then sud- denly there was a flash from the fuse. ]mmediately, the motor's speed skyrock- eted. Someone turned the main circuit breaker off within a few seconds, but by that time the tachometer attached to the motor had pegged at 4000 r/min. TIle mo- tor itself was only rated for 800 rimin. Needless to say, that experience scared everyone present very badly and taught them to be most careful about fie ld circuit protection. In dc motor starting and protection circuits, afield loss relay is nonnally included to disconnect the motor from the line in the event of a loss of field current. A similar effect can occur in ordinary shunt dc motors operating with light fields if their annature reaction effects are severe enough. If the annature reaction on a dc motor is severe, an increase in load can weaken its flux enough to actually cause the motor's speed to rise. However, most loads have torque-speed curves whose torque increases with speed, so the increased speed of the motor increases its load, which increases its annature reaction, weakening its flux again. TIle weaker flux causes a further increase in speed, further increasing load, etc., until the motor overspeeds. nlis condition is known as runaway. In motors operating with very severe load changes and duty cycles, this flux- weakening problem can be solved by installing compensating windings. Unfortunately, compensating windings are too expensive for use on ordinary run-of- the-mill motors. TIle solution to the runaway problem employed for less-expensive, less-severe duty motors is to provide a turn or two of cumulative compounding to the motor's poles. As the load increases, the magnetomotive force from the series turns increases, which counteracts the demagnetizing magnetomotive force of the annature reaction. A shunt motor equipped with just a few series turns like this is calJed a stabilized shunt motor. 9.5 THE PERMANENT-MAGNET DC MOTOR A permanent-magnet de (PMDC) motor is a dc motor whose poles are made of pennanent magnets. Permanent-magnet dc motors offer a number of benefits compared with shunt dc motors in some applications. Since these motors do not require an external field circuit, they do not have the field circuit copper losses as- sociated with shunt dc motors. Because no field windings are required, they can be smaller than corresponding shunt dc motors. PMDC motors are especially common in smaller fractional- and subfractional-horsepower sizes, where the ex- pense and space of a separate field circuit cannot be justified. However, PMDC motors also have disadvantages. Pennanent magnets can- not produce as high a flux density as an externally supplied shunt field, so a
- 584. 560 ELECTRIC MACHINERY RJNDAMENTALS Residual flux density n... ---- --------------.f~t-----------------H ~r~) ""GURE 9-19 Coercive Magnetizing intensity He ----- (.j (a) The magnetization curve of a typical ferromagnetic material. Note the hysteresis loop. After a large ntagnetizing intensity H is applied to the core and then removed. a residual flux density n... rentains behind in the core. This flux can be brought to zero if a coercive magnetizing intensity He is applied to the core with the opposite polarity. In this case. a relatively sntall value of it will demagnetize the core. PMDC motor will have a lower induced torque "rind per ampere of annature cur- rent lit than a shunt motor of the same size and construction. In addition, PMDC motors run the risk of demagnetization. As mentioned in Chapler 8, the annature current lit in a dc machine produces an annature magnetic field of its own. The ar- mature mlllf subtracts from the mmf of the poles under some portions of the pole faces and adds to the rnrnfofthe poles under other portions of the pole faces (see Figures 8- 23 and 8- 25), reducing the overall net nux in the machine. This is the armature reaction effect. In a PMDC machine, the pole nux is just the residual flux in the pennanent magnets. If the armature current becomes very large, there is some risk that the armature mmf may demagnetize the poles, pennanenlly re- ducing and reorienting the residual flux in them. Demagnetizalion may also be caused by the excessive heating which can occur during prolonged periods of overload. Figure 9- 19a shows a magnetization curve for a typical ferromagnetic ma- terial. II is a plot of flux density B versus magnelizing intensity H (or equivalently, a plot of flux <p versus mlllf ?]i). When a strong external magnetomotive force is applied 10 this material and then removed, a residual flux B.... will remain in the material. To force the residual flux to zero, it is necessary to apply a coercive mag- netizing inlensity H e with a polarity opposite 10 the polarity of the magnetizing in- lensity H that originally established the magnetic field. For normal machine
- 585. rx: M mDRS AND GENERATORS 561 II (or¢) HcC3'C)~ -----'f----+--+-------H (or 3') ,b , Alnico 5 Samarium cobalt Ceramic 7 _~, B. T I., 1.4 1.3 1.2 U 1.0 0.9 0.8 0.7 0.6 0.' 0.4 0.3 0.2 0.1 - 1000 - 900 - 800 - 700 - 600 - 500 - 400 - 300 - 200 - 100 0 H.kAlm "I FIGURE 9- 19 (roncludtd) (b) The magnetization curve of a ferromagnetic material suitable for use in permanent magnets. Note the high residual flux density II... and the relatively large coercive magnetizing intensity He. (c) The second quadrant of the magnetization curves of some typical magnetic materials. Note that the rare- earth magnets combine both a high residual flux and a high coercive magnetizing intensity.
- 586. 562 ELECTRIC MACHINERY RJNDAMENTALS applications such as rotors and stators, a ferromagnetic material should be picked which has as small a Br.. and Hc as possible, since such a material will have low hysteresis losses. On the other hand, a good material for the poles of a PMDC motor should have as large a residualflux density Bros as possible, while simultaneously having as large a coercive magnetizing intensity Hcas possible. The magnetization curve of such a material is shown in Figure 9- I9b. TIle large Br• s produces a large fl ux in the machine, while the large Hc means that a very large current would be re- quired to demagnetize the poles. In the last 40 years, a number of new magnetic materials have been devel- oped which have desirable characteristics for making pennanent magnets. The major types of materials are the ceramic (ferrite) magnetic materials and the rare- earth magnetic materials. Figure 9- I9c shows the second quadrant of the magne- tization curves of some typical ceramic and rare-earth magnets, compared to the magnetization curve of a conventional ferromagnetic alloy (Alnico 5). It is obvi- ous from the comparison that the best rare-earth magnets can produce the same residual flux as the best conventional ferromagnetic alloys, while simultaneously being largely immune to demagnetization problems due to annature reaction. A pennanent-magnet dc motor is basically the same machine as a shunt dc motor, except that the flux ofa PMDC motor isfixed. TIlerefore, it is not possible to control the speed of a PMDC motor by varying the field current or flux. The only methods of speed control available for a PMDC motor are armature voltage control and annature resistance control. For more information about PMDC motors, see References 4 and 10. 9.6 THE SERIES DC MOTOR A series dc motor is a dc motor whose field windings consist of a relatively few turns connected in series with the annature circuit. TIle equivalent circuit of a se- ries dc motor is shown in Figure 9- 20. In a series motor, the annature current, field current, and line current are all the same. TIle Kirchhoff's voltage law equa- tion for this motor is (9- 18) Induced Torque in a Series DC Motor TIle tenninal characteristic of a series dc motor is very different from that of the shunt motor previously studied. The basic behavior of a series dc motor is due to the fact that the flux is directly proP011iolUll to the armature current, at least until saturation is reached. As the load on the motor increases, its flux increases too. As seen earlier, an increase in flux in the motor causes a decrease in its speed. TIle re- sult is that a series motor has a sharply drooping torque-speed characteristic. TIle induced torque in this machine is given by Equation (8-49): (8-49)
- 587. rx: MmDRS AND GENERATORS 563 + v, llt = ls= IL VT= EIt + lit (RIt + Rs) FIGURE 9- 10 The equiva.lent circuit of a series dc motor. The nux in this machine is directly proportional to its armature current (at least until the metal saturates).lllerefore, the nux in the machine can be given by (9-1 9) where c is a constant of proportionality. TIle induced torque in this machine is thus given by (9- 20) In other words, the torque in the motor is proportional to the square of its anna- ture current. As a result of this relationship, it is easy to see that a series motor gives more torque per ampere than any other dc motor. It is therefore used in ap- plications requiring very high torques. Examples of such applications are the starter motors in cars, elevator motors, and tractor motors in locomotives. The Terminal Characteristic of a Series DC Motor To detennine the tenninal characteristic of a series dc motor, an analysis will be based on the assumption of a linear magnetization curve, and then the effects of saturation will be considered in a graphical analysis. The assumption of a linear magnetization curve implies that the nux in the motor will be given by Equation (9-1 9): (9-1 9) This equation will be used to derive the torque-speed characteristic curve for the series motor. The derivation of a series motor's torque-speed characteristic starts with Kirchhoff's voltage law: VT = Elt + IIt(RIt + Rs) (9-1 8) From Equation (9- 20), the armature current can be expressed as
- 588. 564 ELECTRIC MACHINERY RJNDAMENTALS Also, Ell = K~w. Substituting these expressions in Equation (9- 18) yields (T:::; VT = K~w + Vic(RA + Rs) (9- 21) If the nux can be eliminated from this expression, it will directly relate the torque of a motor to its speed. To eliminate the nux from the expression, notice that and the induced torque equation can be rewritten as K Tind = cq? TIlerefore, the nux in the motor can be rewritten as ~ = H'Jrind (9- 22) Substituting Equation (9- 22) into Equation (9- 21) and solving for speed yields " {T:::; VT = Kv; ('Jrindw + VKc (RA + Rs) RA + Rs 'I/KC VTindW = VT - ',IKe VTind VT RA + Rs w = VKc VTind - Ke TIle resulting torque- speed relationship is VT I RA + Rs w - ----- - - VKc VTind Ke (9- 23) Notice that for an unsaturated series motor the speed of the motor varies as the reciprocal of the square root of the torque. TImt is quite an unusual relationship! TIlis ideal torque-speed characteristic is plotted in Figure 9- 21. One disadvantage of series motors can be seen immediately from this equa- tion. When the torque on this motor goes to zero, its speed goes to infinity. In practice, the torque can never go entirely to zero because of the mechanical, core, and stray losses that must be overcome. However, if no other load is connected to the motor, it can turn fast enough to seriously damage itself. Never completely un- load a series motor, and never connect one to a load by a belt or other mechanism that could break. If that were to happen and the motor were to become unloaded while running, the results could be serious. TIle nonlinear analysis of a series dc motor with magnetic saturation effects, but ignoring armature reaction, is illustrated in Example 9- 5.
- 589. rx: MmDRS AND GENERATORS 565 'n FIGURE 9-21 The torque-speed characteristic of a series dc motor. Example 9-5. Figure 9--20 shows a 250-V series dc motor with compensating windings. and a total series resistance RA + Rs of 0.08 ll. The series field consists of 25 turns per pole. with the magnetization curve shown in Figure 9- 22. (a) Find the speed and induced torque of this motor for when its armature current is 50A. (b) Calculate and plot the torque-speed characteristic for this motor. Solutioll (a) To analyze the behavior of a series motor with saturation. pick points along the operating curve and find the torque and speed for each point. Notice that the magnetization curve is given in lUlits of magnetomotive force (ampere-turns) versus EA for a speed of 1200 r/min. so calculated EA values must be compared to the equivalent values at 1200 r/min to detennine the actual motor speed. For IA = 50 A. EA = Vr - IA(RA + Rs) = 250 V - (50AXO.080) = 246 V Since IA = I" = 50 A. the magnetomotive force is ?:f = NI = (25turnsX50A) = 1250A o tums From the magnetization curve at?:f = 1250 A ° turns. EAO = 80 V. To get the correct speed of the motor. remember that. from Equation (9- 13). E, " =- "" E" = 286; 120 r/min = 3690 r/min To find the induced torque supplied by the motor at that speed. ra;all that p00IfV = EAIA = 7;....W. Therefore.
- 590. 566 ELECTRIC MACHINERY RJNDAMENTALS > J • 00 • ~ " • , , • 00 • 0 ~ = (246 VX50 A) _ • (3690r/minXlmin/60sX27Tradlr) - 31.8N m (b) To calculate the complete torque-speed characteristic, we must repeat the steps in a for many values of armature current. A MATLAB M-file that calculates the torque-speed characteristics of the series dc motor is shown below. Note that the magnetization curve used by this program works in terms of field magnetomo- tive force instead of effective field current. % M-fi l e : seri es_t s_curve .m % M-fi l e c r eat e a p l ot of the t orque- speed c urve o f the % the seri es dc mot or with armat ure reac tion in % Exampl e 9- 5. % Get the magne tizati on c urve . Thi s fil e cont a in s the % three variabl es mmf_values, ea_va lues, a nd n_O. l oad fig 9_22 .mat 300 2'" ./" .....- / II.. '" 1200 rhnin / / 200 / / I'" / II 100 I I1000 2000 3000 4(xx) 5000 6(XXl 7000 g(XX) 9000 1O.(xx) Field magnetomotive force 3'. A . turns HGURE 9-12 The magnetization curve of the motor in Example 9-5. This curve was taken at speed II,. = 1200 r/min.
- 591. rx: MmDRS AND GENERATORS 567 % Firs t , i n itia liz e v_t = 250; the values needed in thi s program. % Te rminal vo lt age (V) r _0 i 0 - n_ o 0 0.08; 0 1 0, 1 0,300; 0 25; % Armature + fi e l d r es i s tance (ohms) % Armature (line) c urre nt s (A) % Numbe r of series turn s on fi e l d % Ca l culat e the MMF f o r each l oad f=n_s *i_a; % Ca l culat e the int e rnal generated volt age e_a . e_a = v_t - i _a * r _a; % Ca l culat e the r esulting interna l generated volt age at % 1 200 r / min by int e rpolating the mot or 's mag ne tization % curve. e_aO = int e rpl (mmf_va lues,ea_values, f ,'spline'); % Ca l culat e the motor's speed fr om Equat i on (9 -1 3) . n = (e_a . 1 e_aO) * n_O; % Ca l culat e the induced t orque corresponding t o each % speed from Equations (8 - 55 ) and (8 - 56 ) . t _ ind = e_a .* i _a . 1 (n * 2 * p i I 60); % Plo t the t o rque - speed curve p l o t (t _ ind, n , ' Co l or' , 'k' , 'LineWi dth', 2 . 0) ; ho l d on; x l abel ( ' tau_ {ind) (N-m) ' , ' Fontwe i ght ' , 'Sol d' ) ; y l abel ( ' itn_ {m) nnbf (r I min ) , , 'Fontwe i ght ' , 'Sol d' ) ; titl e ( ' Serie s IX: M ot or To rque - Speed Chara c t e ri s ti c ' , 'Fontwe i ght , , 'So l d' ) ; axi s( [ 0 700 0 5000 ] ) ; gri d on; ho l d o ff; The resulting motor torque-speed characteristic is shown in Figure 9- 23. Notice the severe overspeeding at very small torques. Speed Control of Series DC Motors Unlike with the shunt dc motor, there is only one efficient way to change the speed of a series dc motor. That method is to change the terminal voltage of the motor. If the terminal voltage is increased, the first term in Equation (9-23) is in- creased, resulting in a higher speedfor any given torque. The speed of series dc motors can also be controlled by the insertion of a se- ries resistor into the motor circuit, but this technique is very wasteful of power and is used only for intennittent periods during the start-up of some motors. Until the last 40 years or so, there was no convenient way to change VT, so the only method of speed control available was the wasteful series resistance method. That has all changed today with the introduction of solid-state control circuits. Techniques of obtaining variable terminal voltages were discussed in Chapter 3 and will be considered further later in this chapter.
- 592. 568 ELECTRIC MACHINERY RJNDAMENTALS 5(XX) 4500 4(xx) 3500 " 3(xx) ~ 2500 • " 2(xx) 1500 I(xx) 500 00 100 2lXl 3Ol 400 '00 6lll 700 fiDd· N · m ""GURE 9-23 The torque-speed characteristic of the series dc motor in Example 9--5. 9.7 THE COMPOUNDED DC MOTOR A compounded dc motor is a motor with both a shunt and a seriesfield. Such a mo- tor is shown in Figure 9- 24. The dots that appear on the two field coils have the same meaning as the dots on a transfonner: Cu"ent flowing into a dot produces a positive magnetonwtive force. If current fl ows into the dots on both field coils, the resulting magnetomotive forces add to produce a larger total magnetomotive force. This situation is known as cumulative compounding. If current flows into the dot on one field coil and out of the dot on the other field coil, the resulting magnetomotive forces subtract. In Figure 9-24 the round dots correspond to cumulative compound- ing of the motor, and the squares correspond to differential compounding. 1lle Kirchhoff's voltage law equation for a compounded dc motor is Vr = E), + f),(R), + Rs) 1lle currents in the compounded motor are related by fA= IL - 1F VT fF = - RF (9- 24) (9- 25) (9- 26) 1lle net magnetomotive force and the effective shunt field current in the com- pounded motor are given by ,nd ~-~~-~ I 9i'oet - 9i'F ~ 9i'SE 9i'AR I (9- 27) (9- 28)
- 593. rx: MmDRS AND GENERATORS 569 (b' FIGURE 9-24 The equiva.lent circuit of contpounded dc motors: (a.) Ions-shunt connection: (b) shon-shunt connection. where the positive sign in the equations is associated with a cumulatively com- pounded motor and the negative sign is associated with a differentially com- pounded motor. The Torque-Speed Characteristic of a Cumulatively Compounded DC Motor In the cumulatively compounded dc motor, there is a component of flux which is constant and another component which is proportional to its annature current (and thus to its load). TIlerefore, the cumulatively compounded motor has a higher starting torque than a shunt motor (whose flux is constant) but a lower starting torque than a series motor (whose entire flux is proportional to armature current). In a sense, the cumulatively compounded dc motor combines the best fea- tures of both the shunt and the series motors. Like a series motor, it has extra torque for starting; like a shunt motor, it does not overspeed at no load. At light loads, the series field has a very small effect, so the motor behaves approximately as a shunt dc motor. As the load gets very large, the series flux
- 594. 570 ELECTRIC MACHINERY RJNDAMENTALS "m' rhnin Shunt Cumulatively compounded ,~- Series L _______~________ fjnd ,,' ,~_ Shunt Cumulatively compounded ~----=:::::::=L._ ,~ ,b, ""GURE 9-25 (a) The torque-speed characteristic of a cumulatively compounded dc motor compared to series and shunt motors with the same full-load rating. (b) The torque-speed characteristic of a cumulatively compounded dc motor compared to a shunt motor with the same no-lood speed. becomes quite important and the torque-speed curve begins to look like a series motor's characteristic. A comparison of the torque-speed characteristics of each of these types of machines is shown in Figure 9- 25. To detennine the characteristic curve ofa cumulatively compounded dc mo- tor by nonlinear analysis, the approach is similar to that for the shunt and series motors seen before. Such an analysis will be illustrated in a later example. The Torque-Speed Characteristic of a Differentially Compounded DC Motor In a differentially compounded dc motor, the shunt magnetomotive force and se- ries magnetomotiveforce subtractfrom each other. This means that as the load on the motor increases, lit increases and the flux in the motor decreases. But as the flux decreases, the speed of the motor increases. This speed increase causes an- other increase in load, which further increases lit, further decreasing the flux, and increasing the speed again. The result is that a differentially compounded motor is
- 595. rx: MmDRS AND GENERATORS 571 H GURE 9-26 The torque-speed characteristic ofa L _____________ fjmd differentially compounded dc motor. unstable and tends to run away. This instability is much worse than that of a shunt motor with armature reaction. It is so bad that a differentially compounded motor is unsuitable for any application. To make matters worse, it is impossible to start such a motor. At starting con- ditions the annature current and the series field current are very high. Since the se- ries flux subtracts from the shunt flux, the series field can actually reverse the mag- netic polarity of the machine's poles. The motor will typically remain still or turn slowly in the wrong direction while burning up, because of tile excessive annature current. When this type of motor is to be started, its series field must be short- circuited, so that it behaves as an ordinary shunt motor during the starting perioo. Because of the stability problems of the differentially compounded de motor, it is almost never intentionally used. However, a differentially compounded motor can result if the direction of power flow reverses in a cumulatively compounded generator. For that reason, if cumulative ly compounded dc generators are used to supply power to a system, they will have a reverse-power trip circuit to disconnect them from the line if the power flow reverses. No motor- generator set in which power is expected to fl ow in both directions can use a differentially compounded motor, and therefore it cannot use a cumulatively compounded generator. A typical terminal characteristic for a differentially compounded dc motor is shown in Figure 9- 26. The Nonlinear Analysis of Compounded DC Motors The determination of the torque and speed of a compounded dc motor is ill us- trated in Example 9--6. EXllmple 9-6. A lOO-hp, 250-V compounded dc motor with compensating wind- ings has an internal resistance, including the series winding, of 0.04 O. There are J(X)O turns per pole on the shunt field and 31lU1lS per pole on the series winding. The machine is shown in Figure 9-27, and its magnetization curve is shown in Figure 9-9. At no load, the field resistor has been adjusted to make the motor run at 1200 r/min. The core, mechanical, and stray losses may be neglected.
- 596. 572 ELECTRIC MACHINERY RJNDAMENTALS + O.04n < L, NF = ](XXl turns per pole •• • Cumulatively compounded • Differentially compounded Vr= 250 V ""GURE 9-27 The compounded dc motor in Example 9--6. (a) What is the shunt field current in this machine al no load? (b) If the motor is cumulatively compOlUlded, find its speed when I}, = 200 A. (c) If the motor is differentially compounded, find its speed when I}, = 200 A. Solutioll (a) Al no load, the armature ClUTent is zero, so the internal generated vollage of the motor must equal Vr, which means that it must be 250 V. From the magnetiza- tion curve, a field current of 5 A will produce a vollage E}, of 250 V at 1200 r/min. Therefore, the shlUll field ClUTent must be 5 A. (b) When an armature ClUTent of 200 A flows in the motor, the machine's internal generated vollage is E}, = Vr - I},(R}, + Rs) = 250 V - (200 A)(0.04fi) = 242 V The effective field clUTenl of this cumulatively compolUlded motor is • A;E ~AR IF = IF + HI}, - N (9- 28) F , 3 = 5A + I()(X) 200 A = 5.6A From the magnetization curve, E},o = 262 V at speed no = 1200 r/min. There- fore, the motor's speed will be E, n = - n" E" 242 V . = 262 V 1200 rlnnn = 1108 rlmin (c) If the machine is differentially compounded, the effective field current is • NSE '3'AR IF = IF - NF I}, - NF (9- 28) 3 = 5A - 1000 200 A = 4.4 A
- 597. rx: MmDRS AND GENERATORS 573 From the magnetization curve, EAO = 236 V at speed fit! = 1200 r/min. There- fore, the motor's speed will be E, n=rno " = ~~ ~ 1200 r/min = 1230 r/min Notice that the speed of the cumulatively compounded motor decreases with load, while the speed of the differentially compounded motor increases with load. Speed Control in the Cumulatively Compounded DC Motor The techniques available for the control of speed in a cumulatively compounded dc motor are the same as those available for a shunt motor: I. Change the field resistance RF . 2_ Change the armature voltage VA' 3. Change the armature resistance RA. The arguments describing the effects of changing RF or VA are very similar to the arguments given earlier for the shunt motor. Theoretically, the differentially compounded dc motor could be controlled in a similar manner. Since the differentially compounded motor is almost never used, that fact hardly matters. 9.8 DC MOTOR STARTERS In order for a dc motor to function properly on the job, it must have some special control and protection equipment associated with it. The purposes of this equip- ment are I. To protect the motor against damage due to short circuits in the equipment 2_ To protect the motor against damage from long-tenn overloads 3. To protect the motor against damage from excessive starting currents 4. To provide a convenient manner in which to control the operating speed of the motor The first three functions will be discussed in this section, and the fourth function will be considered in Section 9.9. DC Motor Problems on Starting In order for a dc motor to function properly, it must be protected from physical damage during the starting period. At starting conditions, the motor is not turning, and so EA = 0 V. Since the internal resistance of a normal dc motor is very low
- 598. 574 ELECTRIC MACHINERY RJNDAMENTALS o.osn R.~ I, I, + R, IA 2A 3A R., + 1'1 R, V, E, L, ""GURE 9-28 A shunt motor with a starting resistor in series with its annature. Contacts lA. 2A. and 3A short· circuit portions of the starting resistor when they close. compared to its size (3 to 6 percent per unit for medium·size motors), a very high current fl ows. Consider, for example, the 50·hp, 250·Y motor in Example 9-1. This motor has an armature resistance Rio. of 0.06 n, and a full·load current less than 200 A, but the current on starting is VT - EA RA _ 250 Y -O Y =4 167A 0.060 lllis current is over 20 times the motor's rated full·load current. It is possible for a motor to be severely damaged by such currents, even if they last for only a moment. A solution to the problem of excess current during starting is to insert a starting resistor in series with the annature to limit the current flow until EIo. can build up to do the limiting. This resistor must not be in the circuit pennanently, be· cause it would result in excessive losses and would cause the motor's torque-speed characteristic to drop off excessively with an increase in load. 1l1erefore, a resistor must be inserted into the annature circuit to limit cur· rent flow at starting, and it must be removed again as the speed of the motor builds up. In mooem practice, a starting resistor is made up of a series of pieces, each of which is removed from the motor circuit in succession as the motor speeds up, in order to limit the current in the motor to a safe value while never reducing it to too Iowa value for rapid acceleration. Figure 9-28 shows a shunt motor with an extra starting resistor that can be cut out of the circuit in segments by the closing of the lA, 2A, and 3A contacts. Two actions are necessary in order to make a working motor starter. The first is to pick the size and number of resistor segments necessary in order to limit the starting current to its desired bounds. The second is to design a control circuit that
- 599. rx: MmDRS AND GENERATORS 575 3 Off R." f----~+ v, FIGURE 9-29 A manual de motor starter. shuts the resistor bypass contacts at the proper time to remove those parts of the resistor from the circuit. Some older de motor starters used a continuous starting resistor which was gradually cut out of the circuit by a person moving its handle (Figure 9- 29). nlis type of starter had problems, as it largely depended on the person starting the mo- lar not to move its handle too quickly or too slowly. Irthe resistance were cut out too quickly (before the motor could speed up enough), the resulting current flow would be too large. On the other hand, ifthe resistance were cut out too slowly, the starting resistor could burn up. Since they depended on a person for their correct op- eration, these motor starters were subject to the problem of human error. 1lley have almost entirely been displaced in new installations by automatic starter circuits. Example 9- 7 illustrates the selection of the size and number of resistor seg- ments needed by an automatic starter circuit. The question of the timing required to cut the resistor segments out of the annature circuit will be examined later. EXllmple 9-7. Figure 9- 28 shows a lOO-hp. 250-V. 350-A shunt dc motor with an armature resistance of 0.05 O. It is desired to design a starter circuit for this motor which will limit the maximum starting current to twice its rated value and which will switch out sections of resistance as the armature current falls to its rated value. (a) How many stages of starting resistance will be required to limit the current to the range specified? (b) What must the value ofeach segment of the resistor be? At what voltage should each stage of the starting resistance be cut out? Solutioll (a) The starting resistor must be selected so that the current flow equals twice the rated current of the motor when it is first connected to the line. As the motor starts to speed up. an internal generated voltage EA will be produced in the
- 600. 576 ELECTRIC MACHINERY RJNDAMENTALS motor. Since this voltage opposes the terminal voltage of the motor, the increas- ing internal generated voltage decreases the current flow in the motor. When the current flowing in the motor falls to rated current, a section of the starting resis- tor must be taken out to increase the starting ClUTent back up to 200 percent of rated current. As the motor continues to speed up, EA continues to rise and the annature current continues to fall. When the current flowing in the motor falls to rated current again, another section of the starting resistor must be taken out. This process repeats lUltil the starting resistance to be removed at a given stage is less than the resistance of the motor's annature circuit. At that point, the mo- tor's annature resistance will limit the current to a safe value all by itself. How many steps are required to accomplish the current limiting? To find out, define R'rA as the original resistance in the starting circuit. So R'rA is the sum of the resistance of each stage of the starting resistor together with the resistance of the armature circuit of the motor: + R, (9- 29) Now define R'rA.i as the total resistance left in the starting circuit after stages I to i have been shorted out. The resistance left in the circuit after removing stages I through i is + R, Note also that the initial starting resistance must be V, R,<J, = - 1- m.. (9- 30) In the first stage of the starter circuit, resistance RI must be switched out of the circuit when the current I.It falls to VT - EA fA = R = fmin ,. After switching that part of the resistance out, the annature current must jump to VT - EA R = fm:lx ,<J,.l fA = Since E,It (= Kef-) is directly proportional to the speed of the motor, which can- not change instantaneously, the quantity VT- E,It must be constant at the instant the resistance is switched out. Therefore, frrm,R'rA = VT - E,It = fmnR,ot.l or the resistance left in the circuit after the first stage is switched out is fmin = - I - R~ - (9- 31) By direct extension, the resistance left in the circuit after the nth stage is switched out is (9- 32)
- 601. rx: MmDRS AND GENERATORS 577 The starting process is completed when R'<AJI for stage n is less than or equal to the internal annature resistance RA of the motor. At that point, RA can limit the ClUTent to the desired value all by itself. At the boundary where RA = R,OC,II (9--33) R, = ('m'")" RkJ! lmIlJ. (9--34) Solving for n yields (9--35) where n must be rounded up to the next integer value, since it is not possible to have a fractional number of starting stages. If n has a fractional part, then when the final stage of starting resistance is removed, the armature current of the mo- tor will jwnp up to a value smaller than lmu. In this particular problem, the ratio lminllr=. = 0.5, and RIOt is VT 250 V R~ - - , - = 700 A = 0.357 n m n log (RAiR,,J log (0.05 fIA).357 fi) n= = =2 84 log (lmin1Imax) log (350 MOO A) . The number of stages required will be three. (b) The annature circuit will contain the annature resistor RA and three starting re- sistors RI, R2, and RJ. This arrangement is shown in Figure 9--28. At first, EA = 0 V and IA = 700 A, so _ Vr _ IA - R + R + R + R - 700 A A I 2 J Therefore, the total resistance must be (9--36) This total resistance will be placed in the circuit WItii the ClUTent falls to 350 A. This occurs when EA = Vr - IAR,,,, = 250 V - (350 AX0.357 !l) = 125 V When EA = 125 V,IAhas fallen to 350 A and it is time to cut out the first starting resistor RI . When it is cut out, the ClUTent should jump back to 7ooA. Therefore, R + R + R = Vr - EA = 250 V - 125 V = 0 1786 n A 2 J I 700 A . mn (9--37) This total resistance will be in the circuit untillA again falls to 350A. This occurs when EA reaches EA = Vr - IAR,,,, = 250 V - (350A)(0.1786!l) = 187.5 V
- 602. 578 ELECTRIC MACHINERY RJNDAMENTALS When EA = 187.5 V, fA has fallen to 350 A and it is time to cut out the second starting resistor R2. When it is cut out, the current should jump back to 700 A. Therefore, R +R = VT -EA =250V-187.5V =008930 A 1 I 700 A . ~ (9- 38) This total resistance will be in the circuit until fA again falls to 350 A. This occurs when EA reaches EA = VT - fAR,o< = 250V - (350A)(0.08930) = 218.75V When EA = 218.75 V, fA has fallen to 350 A and it is time to cut out the third starting resistor R1. When it is cut out, only the internal resistance of the motor is left. By now, though, RA alone can limit the motor's current to VT - EA 250 V - 218.75 V lAo = R = 0.050 , = 625 A (less than allowed maximrun) From this point on, the motor can speed up by itself. From Equations (9- 34) to (9- 36), the required resistor values can be calculated: RJ = R'OI.3 - RA = 0.08930 - 0.05 0 = 0.03930 R2 = R'0I2 - RJ - RA = 0.1786 0 - 0.0393 0 - 0.05 0 = 0.0893 0 R] = R"".l - R2- R3 - RA = 0.357 0 - 0.1786 0 - 0.0393 0 - 0.05 0 = 0.1786 0 And RJ, Rl., and RJ are cut out when EA reaches 125, 187.5, and 218.75 V, respectively. DC Motor Starting Circuits Once the starting resistances have been selected, how can their shorting contacts be controlled to ensure that they shut at exactly the correct moment? Several dif- ferent schemes are used to accomplish this switching, and two of the most com- mon approaches will be examined in this section. Before that is done, though, it is necessary to intrrxluce some of the components used in motor-starting circuits. Figure 9- 30 illustrates some of the devices commonly used in motor- control circuits. 1lle devices illustrated are fuses, push button switches, relays, time delay relays, and overloads. Figure 9- 30a shows a symbol for a fuse. The fuses in a motor-control cir- cuit serve to protect the motor against the danger of short circuits. They are placed in the power supply lines leading to motors. If a motor develops a short circuit, the fuses in the line leading to it will burn out, opening the circuit before any damage has been done to the motor itself. Figure 9-30b shows spring-type push button switches. There are two basic types of such switches-normally open and nonnally shut. Nonnally open con- tacts are open when the button is resting and closed when the button has been
- 603. rx: MmDRS AND GENERATORS 579 -~o O~-- 0 1 0 Normally open Normally closed ,,' FIGURE 9-30 1 T ,b, Normally Nonnally open closed ,d, OL Heater ,., (a) A fuse. (b) Normally open and normally closed push button switches. (c) A relay coil and comacts. (d) A time delay relay and contacts. (e) An overload and its normally closed contacts. pushed, while normally closed contacts are closed when the button is resting and open when the button has been pushed. A relay is shown in Figure 9-30c. It consists of a main coil and a number of contacts.1lle main coil is symbolized by a circle, and the contacts are shown as par- allel1ines. TIle contacts are of two types- nonnally open and nonnally closed. A normally open contact is one which is open when the relay is deenergized, and a normally closed contact is one which is closed when the relay is deenergized. When electric power is applied to the relay (the relay is energized), its contacts change state:1lle nonnally open contacts close, and the nonnally closed contacts open. A time delay relay is shown in Figure 9- 3Od. It behaves exactly like an or- dinary relay except that when it is energized there is an adjustable time delay be- fore its contacts change state. An overload is shown in Figure 9- 30e. It consists ofa heater coil and some nonnally shut contacts. The current flowing to a motor passes through the heater coils. If the load on a motor becomes too large, then the current flowing to the mo- tor wi 11 heat up the heater coils, which will cause the normally shut contacts of the overload to open. TIlese contacts can in turn activate some types of motor protec- tion circuitry. One common motor-starting circuit using these components is shown in Fig- ure 9- 31. In this circuit, a series of time delay relays shut contacts which remove each section of the starting resistor at approximately the correct time after power is
- 604. 580 ELECTRIC MACHINERY RJNDAMENTALS + 1 1 F, F, R.. R, L, FL E, M R"., M + - OL F, lID 2TD 3TO F, SO", ~ Srop I 0 J , FL OL lID M 2ID ITO 3ID 2TO ""GURE 9-31 A dc motor starting circuit rising time delay relays to cut out the starting resistor. applied to the motor. When the start button is pushed in this circuit, the motor's ar- mature circuit is connected to its power supply, and the machine starts with all re- sistance in the circuit. However, relay ITO energizes at the same time as the motor starts, so after some delay the ITO contacts will shut and remove part of the start- ing resistance from the circuit. Simultaneously, relay 2m is energized, so after an- other time delay the 2TD contacts wil Ishut and remove the second part of the tim- ing resistor. When the 2TD contacts shut, the 3TD relay is energized, so the process repeats again, and finally the motor runs at full speed with no starting resistance present in its circuit. Ifthe time delays are picked properly, the starting resistors can be cut out at just the right times to limit the motor's current to its design values.
- 605. rx: MmDRS AND GENERATORS 581 + I I Radj OL S"rt FL OL ~ Stop ~ J~IO r------{0 M IA }---1 IAR ~-j~---------{ 2A }---1 2AR 3AR ~-Ie---------~ 3A }--~ FIGURE 9-32 (a) A de motor starting circuit using oountervoltage-sensing relays to eut out the starting resistor. Another type of motor starter is shown in Figure 9- 32. Here, a series of re- lays sense the value of Ell in the motor and cut out the starting resistance as Ell rises to preset levels. This type of starter is better than the previous one, since if the motor is loaded heavily and starts more slowly than normal, its armature re- sistance is still cut out when its current falls to the proper value. Notice that both starter circuits have a relay in the field circuit labeled FL. This is afield loss relay. If the field current is lost for any reason, the field loss
- 606. 582 ELECTRIC MACHINERY RJNDAMENTALS I, I A 2A 3 A 700 A k----'T'----~T----"-''------ " I, I, ,b , ""GURE 9- 32 (conclud...>d) (b) The armature current in a dc motor during starting. relay is deenergized, which turns off power to the M relay. When the M relay deenergizes, its nonnally open contacts open and disconnect the motor from the power supply. This relay prevents the motor from running away if its field current is lost. Notice also that there is an overload in each motor-starter circuit. If the power drawn from the motor becomes excessive, these overloads will heat up and open the OL nonnally shut contacts, thus turning off the M relay. When the M re- lay deenergizes, its nonnally open contacts open and disconnect the motor from the power supply, so the motor is protected against damage from prolonged ex- cessive loads. 9.9 THE WARD·LEONARD SYSTEM AND SOLID·STATE SPEED CONTROLLERS TIle speed of a separately excited, shunt, or compounded dc motor can be varied in one of three ways: by changing the field resistance, changing the annature volt- age, or changing the armature resistance. Of these methods, perhaps the most use- ful is annature voltage control, since it permits wide speed variations without af- fecting the motor's maximum torque. A number of motor-control systems have been developed over the years to take advantage of the high torques and variable speeds available from the anna- ture voltage control ofdc motors. In the days before solid-state electronic compo- nents became available, it was difficult to produce a varying dc voltage. In fact, the nonnal way to vary the armature voltage of a dc motor was to provide it with its own separate dc generator. An armature voltage control system of this sort is shown in Figure 9- 33. TIlis figure shows an ac motor serving as a prime mover for a dc generator, which
- 607. rx: M mDRS AND GENERATORS 583 IX generator rx: motor R" I" I" R" + + wm E" v" Vn E" Three-phase motor (induction or synchronous) Rn Ln Rn Ln In I In I Three-phase rectifier Three-phase rectifier and control circuit and control circuit (a) + - Switch to , , ' IX"" reverse power >' connections ' ' , ' weroutput + - D, D, D, D, D, D. Three-phase input power ,b , FIGURE 9-33 (a) A Ward-Leonard system for dc motor speed control. (b) The circuit for producing field current in the dc generator and dc motor. in turn is used to supply a dc voltage to a dc motor. Such a system of machines is called a Ward-Leonard system, and it is extremely versatile. In such a motor-control system, the annature voltage of the motor can be controlled by varying the field current of the dc generator. This annature voltage
- 608. 584 ELECTRIC MACHINERY RJNDAMENTALS Generator operation (r reversed and ro normal) - rind Motor operation (both rand ro reversed) ""GURE 9-34 Motor operation (normal r andro) Torque-speed curves Generator operation (r normal and ro reversed) The operating range of a Ward-Leonan:l motor-control system. The motor can operate as a motor in either the forward (quadram 1) or reverse (quadrant 3) direction and it can also regenerate in quadrams 2 and 4. allows the motor's speed to be smoothly varied between a very small value and the base speed. The speed of the motor can be adjusted above the base speed by reducing the motor's field current. With such a flexible arrangement, total motor speed control is possible. Furthermore, if the field current of the generator is reversed, then the polar- ity of the generator's annature voltage will be reversed, too. This will reverse the motor's direction of rotation. Therefore, it is possible to get a very wide range of speed variations in either direction ofrotation out of a Ward-Leonard dc motor- control system. Another advantage of the Ward-Leonard system is that it can "regenerate," or return the machine's energy of motion to the supply lines. If a heavy load is first raised and then lowered by the dc motor ofa Ward-Leonard system, when the load is falling, the dc motor acts as a generator, supplying power back to the power system. In this fashion, much of the energy required to lift the load in the first place can be recovered, reducing the machine's overall operating costs. TIle possible modes of operation of the dc machine are shown in the torque- speed diagram in Figure 9- 34. When this motor is rotating in its nonnal direction and supplying a torque in the direction of rotation, it is operating in the first quadrant of this figure. If the generator's field current is reversed, that will re- verse the tenninal voltage of the generator, in turn reversing the motor's annature voltage. When the armature voltage reverses with the motor field current remain- ing unchanged, both the torque and the speed of the motor are reversed, and the machine is operating as a motor in the third quadrant of the diagram. If the torque or the speed alone of the motor reverses while the other quantity does not, then the machine serves as a generator, returning power to the dc power system. Because
- 609. rSCR] Th~ r ph= SCR., input FIGURE 9-35 lrSCR2 l,- SCR~ II I, Operation "0' possible lrSCRJ V-SC~ (., (b, rx: MmDRS AND GENERATORS 585 ,Free- wheeling diode) D, v, K. Operation "oj possible + I v, E,C) - (a) A two-quadrant solid-state dc motor controller. Since current cannot flow out of the positive terminals of the armature. this motor cannot act as a generator. returning power to the system. (b) The possible operating quadrants of this motor controller. a Ward-Leonard system pennits rotation and regeneration in either direction, it is called afour-quadrant control system. The disadvantages of a Ward-Leonard system should be obvious. One is that the user is forced to buy three full machines of essentially equal ratings, which is quite expensive. Another is that three machines will be much less efficient than one. Because of its expense and relatively low efficiency, the Ward-Leonard sys- tem has been replaced in new applications by SCR-based controller circuits. A simple dc armature voltage controller circuit is shown in Figure 9- 35. The average voltage applied to the armature of the motor, and therefore the aver- age speed of the motor, depends on the fraction of the time the supply voltage is applied to the armature. This in turn depends on the relative phase at which the
- 610. 586 ELECTRIC MACHINERY RJNDAMENTALS -. Th ph inp '" m tI " - Generator (regeneration) -... ""GURE 9-36 (a) ,b, V + V, E, _I Motor Generator __ (regeneration) - I' + - (a) A four-quadrant solid-state dc motor controller. (b) The possible operating quadrants of this motor controller. SCRs in the rectifier circuit are triggered. This particular circuit is only capable of supplying an annature voltage with one polarity, so the motor can only be re- versed by switching the polarity of its field connection. Notice that it is not possi- ble for an annature current to now out the positive terminal of this motor, since current cannot now backward through an SCR. nlerefore, this motor cannot re- generate, and any energy supplied to the motor cannot be recovered. This type of control circuit is a two-quadrant controller, as shown in Figure 9- 35b. A more advanced circuit capable of supplying an annature voltage with ei- ther polarity is shown in Figure 9- 36. n lis annature voltage control circuit can
- 611. rx: MmDRSAND GENERATORS 587 (a) FIGURE 9-37 (a) A typical solid-state shunt dc motor drive. (Courtesy ofMagneTek, Inc. ) (b) A close-up view of the low-power electronics circuit board. showing the adjustmems for current limits. acceleration rate. deceleration rate. minimum speed. and maximum speed. (Courtesy ofMagneTel;. Inc.) pennit a current fl ow out of the positive terminals of the generator, so a motor with this type of controller can regenerate. If the polarity of the motor field circuit can be switched as well, then the solid-state circuit is a full four-quadrant con- troller like the Ward-Leonard system. A two-quadrant or a full four-quadrant controller built with SCRs is cheaper than the two extra complete machines needed for the Ward-Leonard system, so solid-state speed-control systems have largely displaced Ward-Leonard systems in new applications. A typical two-quadrant shunt dc motor drive with armature voltage speed control is shown in Figure 9- 37, and a simplified block diagram of the drive is shown in Figure 9- 38. nlis drive has a constant field voltage supplied by a three- phase full -wave rectifier, and a variable annature terminal voltage supplied by six SCRs arranged as a three-phase full-wave rectifier. The voltage supplied to the ar- mature of the motor is controlled by adjusting the firing angle of the SCRs in the bridge. Since this motor controller has a fixed field voltage and a variable anna- ture voltage, it is only able to control the speed of the motor at speeds less than or equal to the base speed (see "Changing the Armature Voltage" in Section 9.4). The controller circuit is identical with that shown in Figure 9- 35, except that all of the control electronics and feedback circuits are shown.
- 612. ~ ~ : Pr~-;ct~~ci~i~ -----;l~~---: t- : (Protective devices Current- 1 : tied to fault relay) In;~~ng ! : Sample foc 1 1 trips J L__________________ _ 341 P()',o"er 3-phase , I Full-wave bridge ~I- - , (diodes) Current feedback Voc Sync voltage T341 P()',o"er 1------- Speed<: i Low-pow~;I.;;:;~~;T-~ adj ~ Accelerutionl I I : deceleration r-- s",,,, regulator Current ---,I I Firing I ! • circuit r-r-: I--- I OC~_~ rx: motor 1 Three-phase 1 full wave SCR-bridge Main H V limiter I L --,- FlGURE 9-38 I I ____I T achometer speed feedback contacts -----------------4---, r-t~--r-~-- Start stop circuit Fault relay Run -.l .:L I R" I -l---colf-t-T ~ re~a! I , 1_ _ _ _ _ _ _ _ _ _ _ _ _ -.l L ______________________________ _ I Tachometer Shunt field A simplified block diagram of the t)pical solid-state shunt dc motoc drive shown in Figure 9--37. (Simplified/rom a block diagmm provided l7y MagneTek, Inc.)
- 613. rx: MmDRS AND GENERATORS 589 The major sections of this dc motor drive include: I. A protection circuit section to protect the motor from excessive armature cur- rents. low tenninal voltage. and loss of field current. 2. A start/stop circuit to connect and disconnect the motor from the line. 3. A high-power electronics section to convert three-phase ac power to dc power for the motor's annature and field circuits. 4. A low-power electronics section to provide firing pulses to the SCRs which supply the annature voltage to the motor. This section contains several major subsections, which will be described below. Protection Circuit Section The protection circuit section combines several different devices which together ensure the safe operation of the motor. Some typical safety devices included in this type of drive are I. Current-limiting fuses, to disconnect the motor quickly and safely from the power line in the event of a short circuit within the motor. Current-limiting fuses can interrupt currents of up to several hundred thousand amperes. 2. An instantaneous static trip, which shuts down the motor if the annature cur- rent exceeds 300 percent of its rated value. If the armature current exceeds the maximum allowed value, the trip circuit activates the fault relay, which deenergizes the run relay, opening the main contactors and disconnecting the motor from the line. 3. An inverse-time overload trip, which guards against sustained overcurrent conditions not great enough to trigger the instantaneous static trip but large enough to damage the motor if allowed to continue indefinitely. TIle term in- verse time implies that the higher the overcurrent flowing in the motor, the faster the overload acts (Figure 9- 39). For example, an inverse-time trip might take a full minute to trip if the current flow were 150 percent of the rated current of the motor, but take 10 seconds to trip if the current now were 200 percent of the rated current of the motor. 4. An undefi!oltage trip, which shuts down the motor if the line voltage supply- ing the motor drops by more than 20 percent. 5. Afield loss trip, which shuts down the motor if the field circuit is lost. 6. An overtemperature trip, which shuts down the motor if it is in danger of overheating. StartlStop Circuit Section The start/stop circuit section contains the controls needed to start and stop the mo- tor by opening or closing the main contacts connecting the motor to the line. The motor is started by pushing the run button, and it is stopped either by pushing the
- 614. 590 ELECTRIC MACHINERY RJNDAMENTALS I 3/",oed I",oed ---------------------------- "--':10'--- '2"0- -30 "'- --'40"- ': 50 '--- '60"- - Trip time. s ""GURE 9-39 An inverse-time trip characteristic. stop button or by energizing the fault relay. In either case, the run relay is deener- gized, and the main contacts connecting the motor to the line are opened. High-Power Electronics Section The high-power electronics section contains a three-phase full-wave diode recti- fier to provide a constant voltage to the fie ld circuit of the motor and a three-phase full-wave SCR rectifier to provide a variable voltage to the annature circuit of the motor. Low-Power Electronics Section TIle low-power electronics section provides firing pulses to the SCRs which sup- ply the annature voltage to the motor. By adjusting the firing time of the SCRs, the low-power electronics section adjusts the motor's average armature voltage. TIle low-power electronics section contains the following subsystems: I. Speed regulation circuit. TIlis circuit measures the speed of the motor with a tachometer, compares that speed with the desired speed (a reference voltage level), and increases or decreases the annature voltage as necessary to keep the speed constant at the desired value. For exrunple, suppose that the load on the shaft of the motor is increased. If the load is increased, then the motor will slow down. TIle decrease in speed wi II reduce the voltage generated by the tachometer, which is fed into the speed regulation circuit. Because the voltage level corresponding to the speed of the motor has fallen below the reference voltage, the speed regulator circuit will advance the firing time of the SCRs, producing a higher annature voltage. The higher armature voltage will tend to increase the speed of the motor back to the desired level (see Figure 9-40).
- 615. rx: MmDRS AND GENERATORS 591 Speed adjustment potentiometer +~,J 1--------1 -----' C'- _0 + + I I I Voltage I V..r I I I regulator I t--'~ c - '-~O - , , l ______ -.l (V,oo:b ex speed) Tachometer rx: motor ,., 2 " , ,b, FIGURE 9-40 (a) The speed regulator cin:uit produces an output voltage which is proportional to the difference between the desired speed of the motor (set by V<d') and the actual speed of the nx>tor (measured by Vtoob). This output voltage is applied to the firing cin:uit in such a way that the larger the output voltage becomes, the earlier the SCRs in the drive turn on and the higher the average terminal voltage becomes. (b) The effect of increasing load on a shunt dc motor with a speed regulator. The load in the nx>tor is increased. If no regulator were present. the motor would slow down and operate at point 2. When the speed regulator is present. it detects the decrease in speed and boosts the armature voltage of the motor to compensate. This raises the whole torque-speed characteristic curve of the motor. resulting in operation at point 2'. With proper design, a circuit of this type can provide speed regulations of 0. 1 percent between no-load and full-load conditions. 1lle desired operating speed of the motor is controlled by changing the ref- erence voltage level. The reference voltage level can be adjusted with a small potentiometer, as shown in Figure 9-40.
- 616. 592 ELECTRIC MACHINERY RJNDAMENTALS 2. Current-limiting circuit. This circuit measures the steady-state current flow- ing to the motor, compares that current with the desired maximum current (set by a reference voltage level), and decreases the annature voltage as nec- essary to keep the current from exceeding the desired maximum value. The desired maximum current can be adjusted over a wide range, say from 0 to 200 percent or more of the motor's rated current. This current limit should typically be set at greater than rated current, so that the motor can accelerate under full-l oad conditions. 3. Acceleration/deceleration circuit. TIlis circuit limits the acceleration and de- celeration of the motor to a safe valUC. Whenever a dramatic speed change is commanded, this circuit intervenes to ensure that the transition from the orig- inal speed to the new speed is smooth and does not cause an excessive anna- ture current transient in the motor. TIle acceleration/deceleration circuit completely eliminates the need for a starting resistor, since starting the motor is just another kind of large speed change, and the acceleration/deceleration circuit acts to cause a smooth increase in speed over time. TIlis gradual smooth increase in speed limits the current flow- ing in the machine's annature to a safe value. 9.10 DC MOTOR EFFICIENCY CALCULATIONS To calculate the efficiency of a dc motor, the following losses must be detennined: I. Copper losses 2. Brush drop losses 3. Mechanical losses 4. Core losses 5. Stray losses TIle copper losses in the motor are the { 2R losses in the armature and field circuits of the motor. These losses can be found from a knowledge of the currents in the machine and the two resistances. To determine the resistance of the anna- ture circuit in a machine, block its rotor so that it cannot turn and apply a small dc voltage to the armature tenninals. Adjust that voltage until the current flowing in the annature is equal to the rated annature current of the machine. TIle ratio of the applied voltage to the resulting armature current fl ow is Rio.' The reason that the current should be about equal to full-load value when this test is done is that Rio. varies with temperature, and at the full-l oad value of the current, the annature windings will be near their normal operating temperature. TIle resulting resistance will not be entirely accurate, because I. The cooling that normally occurs when the motor is spinning will not be present.
- 617. rx: MmDRS AND GENERATORS 593 2. Since there is an ac voltage in the rotor conductors during nonnal operation, they suffer from some amount of skin effect, which further raises armature resistance. IEEE Standard 11 3 (Reference 5) deals with test procedures for dc machines. It gives a more accurate procedure for determining RA , which can be used if needed. The field resistance is detennined by supplying the full-rated field voltage to the field circuit and measuring the resulting field current. TIle field resistance RF is just the ratio of the field voltage to the field current. Brush drop losses are often approximately lumped together with copper losses. If they are treated separately, they can be detennined from a plot of contact potential versus current for the particular type of brush being used. The brush drop losses are just the product ofthe brush voltage drop VBO and the annature current IA. The core and mechanical losses are usually detennined together. If a motor is allowed to tum freely at no load and at rated speed, then there is no output power from the machine. Since the motor is at no load, IAis very small and the annature copper losses are negligible. Therefore , if the field copper losses are subtracted from the input power of the motor, the remaining input power must consist of the mechanical and core losses of the machine at that speed.TIlese losses are called the no-load rotational losses ofthe motor. As long as the motor's speed remains nearly the same as it was when the losses were measured, the no-load rotational losses are a gD<Xl estimate of mechanical and core losses under load in the machine. An example of the detennination of a motor's efficiency is given below. Example 9-8. A SO-hp. 2S0-V. 1200 rlmin shunt dc motor has a rated armature current of 170 A and a rated field current of 5 A. When its rotor is blocked. an armature voltage of 10.2 V (exclusive of brushes) produces 170 A of current flow. and a field volt- age of2S0 V produces a field current flow of S A. The brush voltage drop is assumed to be 2 V. At no load with the terminal voltage equal to 240 V. the annature current is equal to 13.2 A. the field current is 4.8 A. and the motor's speed is IISO r/min. (a) How much power is output from this motor at rated conditions? (b) What is the motor's efficiency? Solutioll The armature resistance of this machine is approximately RA = 10.2 V =006 0 170A . and the field resistance is R ~=2S0 V =50 0 , 5 A Therefore, at full load the armature [2R losses are P A = (l70A)2(0.06!l) = 1734 W and the field circuit [ 2R losses are
- 618. 594 ELECTRIC MACHINERY RJNDAMENTALS P F = (SA:Y(500) = 12S0W The brush losses at full load are given by Pbtuob = VBofA = (2 V)(170A) = 340W The rotational losses at full load are essentially equivalent to the rotational losses at no load, since the no-load and full-load speeds of the motor do not differ too greatly. These losses may be ascertained by detennining the input power to the armature circuit at no load and assuming that the annature copper and brush drop losses are negligible, meaning that the no-load armature input power is equal to the rotationa1losses: P,ot. = POOle + P_ = (240 VX13.2 A) = 3168W (a) The input power of this motor at the rated load is given by P~ = VrlL = (250VXI7SA) = 43,7S0W Its output power is given by P00' = Pm - Pb<wb - P<:U - P<Ote - P1DOC.b - P !MaY = 43,7S0W - 340 W - 1734 W - 1250W - 3168W - (O.OIX43,750W) = 36,82oW where the stray losses are taken to be I percent of the input power. (b) The efficiency of this motor at full10ad is T] = ~ou, x 100% ~ _ 36,820W - 43,7SoW x 100% = 84.2% 9.11 INTRODUCTION TO DC GENERATORS DC generators are dc machines used as generators. As previously pointed out, there is no real difference between a generator and a motor except for the direc- tion of power fl ow. There are five major types of dc generators, classified accord- ing to the manner in which their field flux is produced: I. Separately excited generator. In a separately excited generator, the field flux is derived from a separate power source independent of the generator itself. 2. Shunt generator. In a shunt generator, the field flux is derived by connecting the field circuit directly across the tenninals of the generator. 3. Series generator. In a series generator, the field flux is produced by connect- ing the field circuit in series with the armature of the generator. 4. Cumulatively compounded generator. In a cumulatively compounded gener- ator, both a shunt and a series field are present, and their effects are additive. S. Differentially compounded generator. In a differentially compounded genera- tor, both a shunt and a series field are present, but their effects are subtracti ve. TIlese various types of dc generators differ in their terminal (voJtage--current) characteristics, and therefore in the applications to which they are suited.
- 619. rx: MmDRS AND GENERATORS 595 FI GURE 9-41 The first practical dc generator. This is an exact duplicate of the "long- legged Mary Ann." Thomas Edison's first commerdal dc generator. which was built in 1879. It was rated at 5 kW. 100 V. and 1200 r/min. (Courtesy ofGeneml Electric Company.) DC generators are compared by their voltages, power ratings, efficiencies, and voltage regulations. Voltage regulation (VR) is defined by the equation I VR = Vol ~ Va x 100% I (9- 39) where V.. is the no-load tenninal voltage of the generator and V fI is the full-load ter- minal voltage ofthe generator. It is a rough measure of the shape ofthe generator's voltage-current characteristic-a positive voltage regulation means a drooping characteristic, and a negative voltage regulation means a rising characteristic. All generators are driven by a source of mechanical power, which is usually called the prime nwver of the generator. A prime mover for a dc generator may be a steam turbine, a diesel engine, or even an electric motor. Since the speed of the prime mover affects the output voltage of a generator, and since prime movers can vary widely in their speed characteristics, it is customary to compare the voltage regulation and output characteristics of different generators, assuming constant- speed prime movers. Throughout this chapter, a generator's speed will be assumed to be constant unless a specific statement is made to the contrary. DC generators are quite rare in modern power systems. Even dc power sys- tems such as those in automobiles now use ac generators plus rectifiers to produce dc power. The equivalent circuit ofa dc generator is shown in Figure 9-42, and a sim- plified version of the equivalent circuit is shown in Figure 9-43. They look simi- lar to the equivalent circuits ofa dc motor, except that the direction ofcurrent flow and the brush loss are reversed.
- 620. 596 ELECTRIC MACHINERY RJNDAMENTALS + ""GURE 9-42 The equivalent ein:uit of a de generator. v, ""GURE 9-43 I, - ~----AN .. V---A-;> , + R, v, A simplified equivalent ein:uit of a de generator. with RF eombining the resistances of the field coils and the variable control resistor. 9.12 THE SEPARATELY EXCITED GENERATOR A separately excited dc generator is a generator whose field current is supplied by a separate external dc voltage source. The equivalent circuit of such a machine is shown in Figure 9-44. In this circuit, the voltage VT represents the actual voltage measured at the tenninals of the generator, and the current II. represents the cur- rent flowing in the lines connected to the terminals. The internal generated volt- age is Ell, and the armature current is lit- It is clear that the annature current is equal to the line current in a separately excited generator: The Terminal Characteristic of a Separately Excited DC Generator (9-40) TIle terminnl characteristic of a device is a plot of the output quantities of the de- vice versus each other. For a dc generator, the output quantities are its tenninal volt- age and line current. llle tenninal characteristic of a separately excited generator is
- 621. rx: MmDRS AND GENERATORS 597 I, - ,-----".fII'V--~+ + v, FIGURE 9-44 A separately excited de generator. thus a plot of VT versus IL for a constant speed w. By Kirchhoff's voltage law, the terminal voltage is (9-41) Since the internal generated voltage is independent of lA, the tenninal characteris- tic of the separately excited generator is a straight line, as shown in Figure 9-45a. What happens in a generator of this sort when the load is increased? When the load supplied by the generator is increased, IL (and therefore IA) increases. As the annature current increases, the lARA drop increases, so the terminal voltage of the generator falls. This tenninal characteristic is not always entirely accurate. In generators without compensating windings, an increase in IAcauses an increase in armature reaction, and armature reaction causes flux weakening. This flux weakening causes a decrease in EA = K<p J..w which further decreases the tenninal voltage of the generator. TIle resulting tenninal characteristic is shown in Figure 9-45b. In all future plots, the generators will be assumed to have compensating windings unless stated otherwise. However, it is important to realize that annature reaction can modify the characteristics if compensating windings are not present. Control of Terminal Voltage The tenninal voltage of a separately excited dc generator can be controlled by changing the internal generated voltage EA of the machine. By Kirchhoff's volt- age law VT = EA - lARA, so if EAincreases, VTwill increase, and if EA decreases, VT will decrease. Since the internal generated voltage EAis given by the equation EA = K<pw, there are two possible ways to control the voltage of this generator: I. Change the speed of rotation. If w increases, then EA = K<pwi increases, so VT = EAi - lARA increases too.
- 622. 598 ELECTRIC MACHINERY RJNDAMENTALS V, E, 1:::::::::::::::::=====I"R,drop v, "--------------------- ~ ,,' ---------- l"R'drop ----- ARdrop "----------cc--------- ~ ,b, FlGURE 9-45 The terminal characteristic of a separately excited dc generator (a) with and (b) without compensating windings. 2. Change the field current. If RF is decreased. then the field current increases (IF = VF IR~) TIlerefore, the nux <f.> in the machine increases. As the nux rises, EA = K<f.>iw must rise too, so VT = EAi -lARAincreases. In many applications, the speed range of the prime mover is quite limited, so the tenninal voltage is most commonly controlled by changing the field cur- rent. A separately excited generator driving a resistive load is shown in Figure 9-46a. Figure 9-46b shows the effect of a decrease in field resistance on the ter- minal voltage of the generator when it is operating under a load. Nonlinear Analysis of a Separately Excited DC Generator Because the internal generated voltage of a generator is a nonlinear function of its magnetomotive force, it is not possible to calculate simply the value of EA to be expected from a given field current. TIle magnetization curve of the generator must be used to accurately calculate its output voltage for a given input voltage. In addition, if a machine has armature reaction, its nux will be reduced with each increase in load, causing EA to decrease. TIle only way to accurately deter- mine the output voltage in a machine with annature reaction is to use graphical analysis. TIle total magnetomotive force in a separately excited generator is the field circuit magnetomotive force less the magnetomotive force due to annature reac- tion (AR):
- 623. + v, FIGURE 9-46 I, v, v' , R, L, ( ?- 1 (a) VT -------- "~"- 'bJ rx: MmDRS AND GENERATORS 599 R, I, + E, v, R~ (a) A separately excited de generator with a resistive load. (b) The effect of a decrease in field resistance on the output voltage of the generator. (9-42) As with de motors, it is customary to define an equivalentfield current that would produce the same output voltage as the combination of all the magnetomotive forces in the machine. The resulting voltage EAO can then be detennined by locat- ing that equivalent field current on the magnetization curve. The equivalent field current of a separately excited de generator is given by • "'AR IF=IF-N F (9-43) Also, the difference between the speed of the magnetization curve and the real speed of the generator must be taken into account using Equation (9- 13): EA _!!.. EAO no (9- 13) The following example illustrates the analysis of a separately excited de generator.
- 624. 600 ELECTRIC MACHINERY RJNDA MENTALS + v".. I, 0.05.n I, 0-300.1~ + 20fi R, R, VF", 430 V (+) E, V, NF",!()(Xlturns L, ""GURE 9-47 The separately excited dc generator in Example 9--9. Example 9-9. A separately excited dc generator is rated at 172 kW. 430 V. 400 A. and 1800 r/min.1t is shown in Figure 9-47. and its magnetization curve is shown in Fig- ure 9-48.This machine has the following characteristics: RA = 0.05 n RF = 20 n Rodj = Ot03oo n VF = 430V NF = J()(X) turns per pole (a) Ifthe variable resistor Rodj in this generator's field circuit is adjusted to 63 n and the generator's prime mover is driving it at 1600 rlmin, what is this generator's no-load tenninal voltage? (b) What would its voltage be if a 360-A load were connected to its tenninals? As- swne that the generator has compensating windings. (c) What would its voltage be if a 360-A load were connected to its terminals but the generator does not have compensating windings? Asswne that its annature reaction at this load is 450 A · turns. (d) What adjustment could be made to the generator to restore its tenninal voltage to the value found in part a? (e) How much field current would be needed to restore the terminal voltage to its no-load value? (Assume that the machine has compensating windings.) What is the required value for the resistor R ..Jj to accomplish this? Solutio" (a) If the generator's total field circuit resistance is RF + R ..Jj = 83 n then the field current in the machine is VF 430 V IF = RF = 83 n = 5.2 A From the machine's magnetization curve, this much current would produce a voltage EAO = 430 V at a speed of 1800 r/min. Since this generator is actually turning at n.. = 1600 rlmin, its internal generated voltage EA will be E, " = (9- 13)
- 625. 500 450 430 410 400 300 200 100 o I I rx: M mDRS AND GENERATORS 601 /' / V 0.0 1.0 2.0 3.0 4.0 /5.0 6.0 7.0 8.0 9.0 10.0 4.75 5.2 6.15 Field current. A Note: When the field current is zero, E" is about 3 V. FIGURE 9-48 The magnetization curve for the generator in Example 9--9. E = 1600 r/m~n 430 V = 382 V " 1800 r/mm Since Vr = E" at no-load conditions, the output voltage of the generator is Vr = 382 V. (b) If a 360-A load were cotUlected to this generator's terminals, the tenninal volt- age of the generator would be Vr = E" - I"R" = 382 V - (360 AXO.05 0 ) = 364 V (c) If a 360-A load were connected to this generator's terminals and the generator had 450 A ° turns of armature reaction, the effective field current would be 1* = I _ ~AR = S.2 A _ 450A o turns = 4.75A F F NF J(X)() turns
- 626. 602 ELECTRIC MACHINERY RJNDAMENTALS From the magnetization curve, EAO = 410 V, so the internal generated voltage at 1600 rlmin would be = EAO no E = 1600 r/m~n 410 V = 364 V A 1800 r/mlll Therefore, the tenninal voltage of the generator would be Vr = EA - lARA = 364 V - (360 AXO.05 0) = 346 V It is lower than before due to the annature reaction. (9-13) (d) The voltage at the tenninals of the generator has fallen, so to restore it to its original value, the voltage of the generator must be increased. This requires an increase in EA, which implies that R..tj must be decreased to increase the field current of the generator. (e) For the terminal voltage to go back up to 382 V, the required value of EA is EA = Vr + lARA = 382 V + (360 AXO.05 0) = 400 V To get a voltage EA of 400 V at n.. = 1800 rlmin would be 1600 rlmin, the equivalent voltage at EA _ .!!. (9-13) E = 18oor/m~n 400 V = 450 V AO 1600 r/mlll From the magnetization curve, this voltage would require a field current of IF = 6.15 A. The field circuit resistance would have to be V, RF + Radj = T, 430V 200 + Radj = 6.15A = (:f).90 Radj = 49.90 - 500 Notice that, for the same field current and load current, the generator with annature reaction had a lower output voltage than the generator without annature reaction. The annature reaction in this generator is exaggerated to illustrate its ef- fects- it is a good deal smaller in well-designed modem machines. 9.13 THE SHUNT DC GENERATOR A shunt dc generator is a de generator that supplies its own field current by hav- ing its field connected directly across the terminals of the machine. The equi va- lent circuit of a shunt dc generator is shown in Figure 9-49. In this circuit, the ar- mature current of the machine supplies both the field circuit and the load attached to the machine: (9-44)
- 627. + E, I, " - R, "1 /. IA = IF+h Vr = EA - lARA V, IF = - R, I, - v., L, rx: MmDRS AND GENERATORS 603 + v, HGURE 9-49 The equivalent circuit of a shunt de generator. The Kirchhoff's voltage law equation for the armature circuit of this machine is (9-45) This type of generator has a distinct advantage over the separately excited dc generator in that no external power supply is required for the field circuit. But that leaves an important question unanswered: If the generator supplies its own field current, how does it get the initial field nux to start when it is first turned on? Voltage Buildup in a Shunt Gener ator Assume that the generator in Figure 9-49 has no load connected to it and that the prime mover starts to turn the shaft of the generator. How does an initial voltage appear at the terminals of the machine? The voltage buildup in a dc generator depends on the presence ofa residual flux in the poles of the generator. When a generator first starts to turn, an internal voltage will be generated which is given by EA = K<Pre.w This voltage appears at the tenninals of the generator (it may only be a volt or two). But when that voltage appears at the tenninals, it causes a current to fl ow in the generator's field coil (IF = Vr i /RF). This field current produces a magneto- motive force in the poles, which increases the nux in them. 1lle increase in nux causes an increase in EA = K<p iw, which increases the terminal voltage Vr. When Vr rises, IF increases further, increasing the fl ux <p more, which increases EA, etc. This voltage buildup behavior is shown in Figure 9- 50. Notice that it is the effect of magnetic saturation in the pole faces which eventually limits the termi- nal voltage of the generator.
- 628. 604 ELECTRIC MACHINERY RJNDAMENTALS EA (and VT), V v,. VTversus IF EA versus IF " -------------- ~ --1"" --?f__ Magnetization curve EA, res "------------~------ IF' A IF " ""GURE 9-50 Voltage buildup on starting in a shunt dc generator, Figure 9-50 shows the voltage buildup as though it occurred in discrete steps, These steps are drawn in to make obvious the positive feedback between the generator's internal voltage and its field current. In a real generator, the voltage does not build up in discrete steps: Instead both E A and IF increase simultaneously until steady-state conditions are reached, What if a shunt generator is started and no voltage builds up? What could be wrong? nlere are several possible causes for the voltage to fail to build up during starting, Among them are L There may be no residual magnetic flux in the generator to start the process going, If the residual flux ~re. = 0, then E A = 0, and the voltage never builds up, If this problem occurs, disconnect the field from the armature circuit and connect it directly to an external dc source such as a battery, 1lle current flow from this external dc source will leave a residual flux in the poles, which will then allow normal starting, nlis procedure is known as "flashing the field," 2. The direction ofrotation ofthe generator may have been reversed, or the con- nections of the field may have been reversed, In either case, the residual flux produces an internal generated voltage EA, The voltage EA produces a field current which produces a flux opposing the residual flux, instead of adding to it. Under these circumstances, the flux actually decreases below ~r•• and no voltage can ever build up, If this problem occurs, it can be fixed by reversing the direction of rota- tion, by reversing the field connections, or by flashing the field with the op- posite magnetic polarity,
- 629. v" v, v, R, rx: MmDRS AND GENERATORS 605 ~ '------------------------------ /F.A FIGURE 9-51 The elTect of shunt fietd resistance on no-load tenninal voltage in a dc generator. If Rr > Rl (the critical resistance). then the generator's voltage will never build up. 3. The field resistance may be adjusted to a value greater than the critical re- sistance. To understand this problem, refer to Figure 9- 51. Nonnally, the shunt generator will build up to the point where the magnetization curve in- tersects the field resistance line. If the field resistance has the value shown at Rl in the figure, its line is nearly parallel to the magnetization curve. At that point, the voltage of the generator can fluctuate very widely with only tiny changes in RF or lit. nlis value of the resistance is called the critical resis- tance. If RF exceeds the critical resistance (as at R3in the figure), then the steady-state operating voltage is essentially at the residual level, and it never builds up. The solution to this problem is to reduce Rr. Since the voltage of the magnetization curve varies as a function of shaft speed, the critical resistance also varies with speed. In general, the lower the shaft speed, the lower the critical resistance. The Terminal Characteristic of a Shunt DC Generator The terminal characteristic of a shunt dc generator differs from that of a separately excited dc generator, because the amount of field current in the machine depends on its terminal voltage. To understand the terminal characteristic of a shunt gen- erator, start with the machine unloaded and add loads, observing what happens. As the load on the generator is increased, II- increases and so lit = IF + IL i also increases. An increase in lit increases the annature resistance voltage drop IItRIt, causing Vr = Elt - lit i RIt to decrease. This is precisely the same behavior observed in a separately excited generat.or. However, when Vr decreases, the field current in the machine decreases with it. This causes the flux in the machine to
- 630. 606 ELECTRIC MACHINERY RJNDAMENTALS v, ---=::---===--------------}- :~A --- I -;i:Id weakening effect L-____________________________ ~ ""GURE 9-52 The terminal characteristic of a shunt dc generator. decrease, decreasing EA- Decreasing EAcauses a further decrease in the terminal voltage Vr = EAJ.. - lARA' TIle resulting terminal characteristic is shown in Figure 9-52. Notice that the voltage drop-off is steeper than just the lARA drop in a sepa- rately excited generator. In other words, the voltage regulation of this generator is worse than the voltage regulation of the same piece of equipment connected sep- aratelyexcited. Voltage Control for a Shunt DC Generator As with the separately excited generator, there are two ways to control the voltage of a shunt generator: I. Change the shaft speed W m of the generator. 2. Change the field resistor of the generator, thus changing the field current. Changing the field resistor is the principal method used to control tenninal voltage in real shunt generators. If the field resistor RF is decreased, then the field current IF = VrIRFJ.. increases. When IF increases, the machine's flux <P increases, causing the internal generated voltage EA to increase. The increase in EA causes the tenninal voltage of the generator to increase as well. The Analysis of Shunt DC Generators TIle analysis of a shunt dc generator is somewhat more complicated than the analysis of a separately excited generator, because the field current in the machine depends directly on the machine's own output voltage. First the analysis of shunt generators is studied for machines with no armature reaction, and afterward the effects are annature reaction are included.
- 631. rx: MmDRS AND GENERATORS 607 v, V,. V,~ VT versus IF EA reduction --------------->6/ -r~ , , EA versus IF '-________________-.L-________ ~ 11'01 FIGURE 9-53 Graphical analysis of a shunt dc generator with contpensating windings. Figure 9- 53 shows a magnetization curve for a shunt dc generator drawn at the actual operating speed of the machine. The field resistance RF> which is just equal to VTIIF> is shown by a straight line laid over the magnetization curve. At no load, VT = E A and the generator operates at the voltage where the magnetization curve intersects the field resistance line. The key to understanding the graphical analysis of shunt generators is to re- member Kirchhoff's voltage law (KVL): (9-45) (9-46) The difference between the internal generated voltage and the tenninal voltage is just the lARA drop in the machine. The line of a11 possible values of EA is the mag- netization curve, and the line of all possible tenninal voltages is the resistor line (IF = VT IRF)· Therefore, to find the tenninal voltage for a given load, just deter- mine the lARA drop and locate the place on the graph where that drop fits exactly between the E A line and the V T line. There are at most two places on the curve where the lARAdrop will fit exactly. If there are two possible positions, the one nearer the no-load voltage will represent a normal operating point. A detailed plot showing several di fferent points on a shunt generator's char- acteristic is shown in Figure 9- 54. Note the dashed line in Figure 9- 54b. lllis line is the terminal characteristic when the load is being reduced. llle reason that it does not coincide with the line of increasing load is the hysteresis in the stator poles of the generator.
- 632. 608 ELECTRIC MACHINERY RJNDAMENTALS , L V, 1. lARA drop --- I V VIT> -~ , :; , ,". , , , I , 1 ".// VV / /,/ /// , , ~ ,/ , , I, ,,' ,b, ""GURE 9-54 Graphical derivation of the terminal characteristic of a shunt dc generator. If annature reaction is present in a shunt generator, this process becomes a little more complicated. The armature reaction produces a demagnetizing magne- tomotive force in the generator at the same time that the lARA drop occurs in the machine. To analyze a generator with annature reaction present, assume that its ar- mature current is known. Then the resistive voltage drop lARA is known, and the demagnetizing magnetomotive force of the annature current is known.1lle tenni- nal voltage of this generator must be large enough to supply the generator's nux after the demagnetizing effects of armature reaction have been subtracted. To meet this requirement both the annature reaction magnetomotive force and the lARA drop must fit between the Ell. line and the VT line. To detennine the output voltage for a given magnetomotive force, simply locate the place under the mag- netization curve where the triangle formed by the armature reaction and lARA ef- fects exactlyfits between the line of possible VT values and the line of possible Ell. values (see Figure 9-55). 9.14 THE SERIES DC GENERATOR A series dc generator is a generator whose field is connected in series with its ar- mature. Since the annature has a much higher current than a shunt field, the series field in a generator of this sort will have only a very few turns of wire, and the wire used will be much thicker than the wire in a shunt field. Because magneto- motive force is given by the equation'?} = NI, exactly the same magnetomotive force can be produced from a few turns with high current as can be produced from many turns with low current. Since the fu ll-load current flows through it, a series field is designed to have the lowest possible resistance. 1lle equivalent circuit of a series dc generator is shown in Figure 9-56. Here, the annature current, field
- 633. rx: MmDRS AND GENERATORS 609 Ell. '" Vr at no load r ---------:;::'J.........C lARA drop Ell. with load r---"-''----'-''<;::/lr( Ell. versus IF Vr with load f------y"-"'k;Y Demagnetizing mmf (converted to an equivalent field current) '-------------------------- ~ FIGURE 9-55 Graphical analysis of a shunt dc generator with annature reaction. (NSF. turns) + v, HGURE 9-S6 111. "' ls"'IL Vr",EA -IA(RA+Rs) The equivalent circuit of a series dc generator. currenl, and line currenl all have the same value. The Kirchhoff's voltage law equation for this machine is (9-47) The Terminal Characteristic of a Series Generator The magnetization curve of a series dc generator looks very much like the magne- tization curve of any other generator. At no load, however, there is no field current, so Vr is reduced to a small level given by the residual nux in the machine. As the load increases, the field current rises, so Ell. rises rapidly. The iA(RA + Rs) drop goes up too, but at first the increase in Ell. goes up more rapidly than the iA(RA + Rs) drop rises, so Vr increases. After a while, the machine approaches saturation, and Ell.
- 634. 610 ELECTRIC MACHINERY RJNDAMENTALS / / b / / / ~~---T)- /","'" I 111. (RA+ Rs) drop / / V, '--------------- h (= Is= 111.) ""GURE 9-57 Derivation of the terminal characteristic for a series dc generator. V, Armature reaction '--____________-'L-~ ""GURE 9-58 A series generator tenninal characteristic with large armature reaction effects. suitable for electric welders. becomes almost constant. At that point, the resistive drop is the predominant effect, and VT starts to fall. lllis type of characteristic is shown in Figure 9- 57. It is obvious that this machine would make a bad constant-voltage source. In fact, its voltage regulation is a large negative number. Series generators are used only in a few specialized applications, where the steep voltage characteristic of the device can be exploited. One such application is arc welding. Series generators used in arc welding are deliberately designed to have a large annature reaction, which gives them a terminal characteristic like the one shown in Figure 9- 58. Notice that when the welding electrodes make contact with each other before welding commences, a very large current flows. As the op- erator separates the welding electrodes, there is a very steep rise in the generator's voltage, while the current remains high. This voltage ensures that a welding arc is maintained through the air between the electrodes.
- 635. OCMmDRSANDGENERATORS 611 + FIGURE 9-59 IIt "'h+IF vT '" Elt -11t(RIt + Rsl v, IF"'1fF :¥.... '" NF IF+ NSFJIt - ::fAR • v, The equivalent cirwit of a cumulatively compounded dc generator with a long-shunt connection. 9.15 THE CUMULATIVELY COMPOUNDED DC GENERATOR A cumulatively compounded dc generator is a dc generator with both series and shuntfields, connected so that the magnetomotive forces from the two fields are additive. Figure 9- 59 shows the equivalent circuit ofa cumulatively compounded dc generator in the "long-shunt" connection. TIle dots that appear on the two field coils have the same meaning as the dots on a transfonner: Current flowing into a dot produces a positive magnetomotive force. Notice that the annature current flows into the dotted end of the series field coil and that the shunt current IF flows into the dotted end of the shunt field coil. Therefore, the total magnetomotive force on this machine is given by (9-48) where 'ifF is the shunt field magnetomotive force, 'ifSE is the series field magneto- motive force, and 2FAR is the armature reaction magnetomotive force. The equiva- lent effective shunt field current for this machine is given by NFl; = NFIF + NsEJA - 2FAR The other voltage and current relationships for this generator are IIA - iF + I, I (9-49) (9- 50) (9- 51)
- 636. 612 ELECTRIC MACHINERY RJNDAMENTALS • v, ""GURE 9-60 The equivalent cin;uit of a cumulatively compounded dc generator with a short-shunt connection. (9- 52) TIlere is another way to hook up a cumulatively compounded generator. It is the "short-shunt" connection, where the series field is outside the shunt field circuit and has current IL flowing through it instead of I.... A short-shunt cumula- tively compounded dc generator is shown in Figure 9-60. The Terminal Characteristic of a Cumulatively Compounded DC Generator To understand the terminal characteristic of a cumulatively compounded dc gen- erator, it is necessary to understand the competing effects that occur within the machine. Suppose that the load on the generator is increased. Then as the load in- creases, the load current IL increases. Since IA = IF + ILi , the annature current IA increases too. At this point two effects occur in the generator: I. As IAincreases, the IA(RA + Rs) voltage drop increases as well.1l1is tends to cause a decrease in the terminal voltage VT = EA - IA i (RA + Rs). 2. As IA increases, the series field magnetomotive force 91'SE = NSEIA increases too. This increases the total magnetomotive force 91"0' = NFIF + NSEIAi which increases the flux in the generator.1l1e increased flux in the generator increases EA, which in turn tends to make VT = EA i - IA(RA + Rs) rise. TIlese two effects oppose each other, with one tending to increase VT and the other tending to decrease VT. Which effect predominates in a given machine? It all depends on just how many series turns were placed on the poles of the ma- chine. The question can be answered by taking several individual cases: I. Few series turns (NSE smnll). If there are only a few series turns, the resistive voltage drop effect wins hands down. The voltage falls off just as in a shunt
- 637. OCMmDRSANDGENERATORS 613 v, Undercompounded Shum L-___________________/~-------~ " FIGURE 9-61 Terminal characteristics of cumulatively compounded dc generators. generator, but not quite as steeply (Figure 9--61). This type of construction, where the full-load tenninal voltage is less than the no-load tenninal voltage, is called undercompounded. 2. More series turns (NSE larger). If there are a few more series turns of wire on the poles, then at first the flux-strengthening effect wins, and the terminal voltage rises with the load. However, as the load continues to increase, mag- netic saturation sets in, and the resistive drop becomes stronger than the flux increase effect. In such a machine, the terminal voltage first rises and then falls as the load increases. If VTat no load is equal to VTat full load, the gen- erator is called flat-compounded. 3. Even more series turns are added (NSE large). If even more series turns are added to the generator, the flux-strengthening effect predominates for a longer time before the resistive drop takes over. The result is a characteristic with the fu ll-load tenninal voltage actually higher than the no-load tenninal voltage. If VTat a full load exceeds VTat no load, the generator is called over- compounded. All these possibilities are illustrated in Figure 9--61. lt is also possible to realize all these voltage characteristics in a single gen- erator if a diverter resistor is used. Figure 9--62 shows a cumulatively com- pounded dc generator with a relatively large number of series turns NSE. A diverter resistor is connected around the series field. If the resistor Rdh is adjusted to a large value, most of the annature current flows through the series field coil, and the generator is overcompounded. On the other hand, ifthe resistor RcJjy is adjusted to a small value, most of the current flows around the series field through RcJjy, and the generator is undercompounded. lt can be smoothly adjusted with the resistor to have any desired amount of compounding.
- 638. 614 ELECTRIC MACHINERY RJNDAMENTALS "/. Y I, R, R, L, I, - • - • • I, I y.: (+)£, /- R, • L, ""GURE 9-62 A cumulatively compounded dc generator with a series diverter resistor. Voltage Control of Cumulatively Compounded DC Generators + v, - TIle techniques available for controlling the tenninal voltage of a cumulatively compounded dc generator are exactly the same as the techniques for controlling the voltage of a shunt dc generator: I. Change the speed of rotation. An increase in w causes Ell = KtPwi to in- crease, increasing the terminal voltage VT = Ell i - IIl(RIl + Rs). 2. Change the field current. A decrease in RF causes IF = VT I RFJ.. to increase, which increases the total magnetomotive force in the generator. As ?ft", increases, the nux <p in the machine increases, and Ell = KtPiw increases. Finally, an increase in Ell raises VT. Analysis of Cumulatively Compounded DC Generators Equations (9- 53) and (9- 54) are the key to describing the tenninal characteristics of a cumulatively compounded dc generator. The equivalent shunt field current leq due to the effects of the series field and armature reaction is given by I NSE ?fAR leq = NF IA - NF Therefore, the total effective shunt field current in the machine is 1 ;= IF + leq (9- 53) (9- 54) TIlis equivalent current leq represents a horizontal distance to the left or the right of the field resistance line (RF = VTI RF) along the axes of the magnetization curve.
- 639. OCMmDRS ANDGENERATORS 615 E" and Vr Magnetization curve (E" versus IF) ~__~E~,".~IO~ "~ ~ d~__~~",__------ E" and Vr. no load 1-::=====7::fi~.=~} IR drop Vr· loaded I '-------------------------- ~ FIGURE 9-63 Graphical analysis of a cumulatively compounded dc generator. The resistive drop in the generator is given by I"(R,, + Rs), which is a length along the vertical axis on the magnetization curve. Both the equivalent current le<j and the resistive voltage drop I"(R,, + Rs) depend on the strength of the armature current I". 1llerefore, they form the two sides of a triangle whose magnitude is a function of I", To find the output voltage for a given load, detennine the size of the triangle and fmd the one point where it exactly fits between the field current line and the magnetization curve. This idea is illustrated in Figure 9--63.1lle tenninal voltage at no-load con- ditions will be the point at which the resistor line and the magnetization curve in- tersect, as before. As load is added to the generator, the series field magnetomotive force increases, increasing the equivalent shunt field current leq and the resistive voltage drop I"(R,, + Rs) in the machine. To find the new output voltage in this generator, slide the leftmost edge of the resulting triangle along the shunt field cur- rent line until the upper tip of the triangle touches the magnetization curve. The up- per tip of the triangle then represents the internal generated voltage in the machine, while the lower line represents the tenninal voltage of the machine. Figure 9--64 shows this process repeated several times to construct a com- plete terminal characteristic for the generator. 9.16 THE DIFFERENTIALLY COMPOUNDED DC GENERATOR A differentially compounded dc generator is a generator with both shunt and se- ries fields, but this time their magnetomotiveforces subtractfrom each other.The
- 640. 616 ELECTRIC MACHINERY RJNDAMENTALS v, L-_ _ _ _ _ _ _ _ _ ~ '------------ 1, ""GURE 9-64 Graphical derivation of the terminal characteristic of a cumulatively compounded dc generator. I, I, " - • + R, R, L, I, I v., /- I" = IL+I,, + V, E, • V, 1,,=][; , L, Vr = E" - I" (R" + Rsl ""GURE 9-65 The equivalent cin:uit of a differentially compounded dc generator with a long-shunt connection. equivalenl circuit of a differentially compounded dc generator is shown in Figure 9--65. Notice that the armature current is now fl owing out of a dotted coil end, while the shunt field current is fl owing into a dotted coil end. In this machine, the net magnetornotive force is (9- 55) ~AR I (9- 56)
- 641. OCMmDRSANDGENERATORS 617 and the equivalent shunt field current due to the series field and annature reaction is given by NSE :fAR leq = - NF IA - -N - F - (9- 57) The total effective shunt field current in this machine is (9- 58a) (9- 58b) Like the cumulatively compounded generator, the differentially com- pounded generator can be connected in either long-shunt or short-shunt fashion. The Terminal Characteristic of a Differentially Compounded DC Generator In the differentially compounded dc generator, the same two effects occur that were present in the cumulatively compounded dc generator.lltis time, though, the effects both act in the same direction. They are I. As Iii increases, the lli(RIi + Rs) voltage drop increases as well. This increase tends to cause the terminal voltage to decrease VT = Eli -Iiii (Rli + Rs). 2. As Iii increases, the series field magnetomotive force '3'SE = NSEIIi increases too. lltis increase in series field magnetomotive force reduces the net mag- netomotive force on the generator (2F,OI = NFIF - NSE IIii ), which in turn re- duces the net flux in the generator. A decrease in flux decreases Eli, which in turn decreases VT. Since both these effects tend to decrease Vn the voltage drops drastically as the load is increased on the generator. A typical tenninal characteristic for a dif- ferentially compounded dc generator is shown in Figure 9-66. Voltage Control of Differentially Compounded DC Generators Even though the voltage drop characteristics of a differentially compounded dc generator are quite bad, it is still possible to adjust the terminal voltage at any given load setting. The techniques available for adjusting tenninal voltage are ex- actly the same as those for shunt and cumulatively compounded dc generators: I. Change the speed of rotation W m. 2. Change the field current IF.
- 642. 618 ELECTRIC MACHINERY RJNDAMENTALS v, Differentially contpounded Shunt L - - - - - - - - - - - - - - _ I, ""GURE 9-66 The terminal characteristic of a differentially contpounded dc generator. EAnJ and Vrnl f------CCCO---7r' IRdrop EA' loaded f::====::;;2ti¥ Vr. loaded r- I., '--------------1, ""GURE 9-67 Graphical analysis of a differentially contpounded dc generator. Craphical Analysis of a Differentially Compounded DC Generator TIle voltage characteristic of a differentially compounded dc generator is graphi- cally detennined in precisely the same manner as that used for the cumulatively compounded dc generator. To find the terminal characteristic of the machine, re- fer to Figure 9--67.
- 643. OCMmDRSANDGENERATORS 619 v, 1 ~7"1'----------t----4;- DifferentiallY leq compounded '---------- l, '------------ l, FIGURE 9-68 Graphical derivation of the terminal characteristic of a differentially contpounded dc generator. The portion of the effecti ve shunt field current due to the actual shunt field is always equal to VT IRF , since that much current is present in the shunt field. The remainder of the effective field current is given by Ieq and is the sum of the series field and annature reaction effects. This equivalent current 1O<j represents a nega- tive horizontal distance along the axes of the magnetization curve, since both the series field and the armature reaction are subtractive. The resistive drop in the generator is given by IA(RA + Rs), which is a length along the vertical axis on the magnetization curve. To fmd the output voltage for a given load, detennine the size of the triangle formed by the resistive voltage drop and 1O<j' and find the one point where it exactly fits between the field current line and the magnetization curve. Figure 9--68 shows this process repeated several times to construct a com- plete terminal characteristic for the generator. 9.17 SUMMARY There are several types of dc motors, differing in the manner in which their field fluxes are deri ved. These types of motors are separately excited, shunt, permanent-magnet, series, and compounded. TIle manner in which the flux is de- rived affects the way it varies with the load, which in turn affects the motor's overall torque-speed characteristic. A shunt or separately excited dc motor has a torque- speed characteristic whose speed drops linearly with increasing torque. Its speed can be controlled by changing its field current, its annature voltage, or its annature resistance. A pennanent-magnet dc motor is the same basic machine except that its flux is derived from pennanent magnets. Its speed can be controlled by any of the above methods except varying the field current.
- 644. 620 ELECTRIC MACHINERY RJNDAMENTALS A series motor has the highest starting torque of any dc motor but tends to overspeed at no load. It is used for very high-torque applications where speed reg- ulation is not important, such as a car starter. A cumulatively compounded dc motor is a compromise between the series and the shunt motor, having some of the best characteristics of each. On the other hand, a differentially compounded dc motor is a complete disaster. It is unstable and tends to overspeed as load is added to it. DC generators are dc machines used as generators. TIlere are several differ- ent types of dc generators, differing in the manner in which their field fluxes are derived. These methods affect the output characteristics of the different types of generators. The common dc generator types are separately excited, shunt, series, cumulatively compounded, and differentially compounded. TIle shunt and compounded dc generators depend on the nonlinearity of their magnetization curves for stable output voltages. If the magnetization curve of a dc machine were a straight line, then the magnetization curve and the tenni- nal voltage line of the generator would never intersect. TIlere would thus be no stable no-load voltage for the generator. Since nonlinear effects are at the heart of the generator's operation, the output voltages of dc generators can only be deter- mined graphically or numerically by using a computer. Today, dc generators have been replaced in many applications by ac power sources and solid-state electronic compone nts. TIlis is true even in the automobile, which is one of the most common users of dc power. QUESTIONS 9-1. What is the speed regulation of a dc motor? 9-2. How can the speed of a shunt dc motor be controlled? Explain in detail. 9-3. What is the practical difference between a separately excited and a shunt dc motor? 9-4. What effect does annature reaction have on the torque-speed characteristic of a shunt dc motor? Can the effects of annature reaction be serious? What can be done to remedy this problem? 9-5. What are the desirable characteristics of the permanent magnets in PMDC machines? 9-6. What are the principal characteristics of a series dc motor? What are its uses? 9-7. What are the characteristics of a cumulatively compOlmded dc motor? 9-8. What are the problems associated with a differentially compounded dc motor? 9-9. What happens in a shlUlt dc motor if its field circuit opens while it is flmning? 9-10. Why is a starting resistor used in dc motor circuits? 9-11. How can a dc starting resistor be cut out of a motor's armature circuit at just the right time during starting? 9-12. What is the Ward-Leonard motor control system? What are its advantages and disadvantages? 9-13. What is regeneration? 9-14. What are the advantages and disadvantages of solid-state motor drives compared to the Ward-Leonard system?
- 645. rx: MmDRS AND GENERATORS 621 9-15. What is the pwpose of a field loss relay? 9-16. What types of protective features are included in typical solid-state dc motor drives? How do they work? 9-17. How can the direction of rotation of a separately excited dc motor be reversed? 9-18. How can the direction of rotation of a shunt dc motor be reversed? 9-19. How can the direction of rotation of a series dc motor be reversed? 9-20. Name and describe the features of the fi ve types of generators covered in this chapter. 9-21. How does the voltage buildup occur in a shunt dc generator during starting? 9-22. What could cause voltage buildup on starting to fail to occur? How can this problem be remedied? 9-23. How does armature reaction affect the output voltage in a separately excited dc generator? 9-24. What causes the extraordinarily fast voltage drop with increasing load in a differen- tially compounded dc generator? PROBLEMS Problems 9-1 to 9-12 refer to the following dc motor: Prned = 15 hp Vr=240 V nrned = 1200 r/min RA = 0.40 n Rs = 0.04 n h..,,'od = 55 A NF = 2700 turns per pole NSE = 27 turns per pole RF = 100 0 Rodj = 100 to 400 0 Rotational losses are 1800 W at full load. Magnetization curve is as shown in Figure P9--I. In Problems 9-1 through 9- 7. assrune that the motor described above can be con- nected in shlUlt. The equivalent circuit of the shunt motor is shown in Figure P9-2. 9-1. If the resistor Rodj is adjusted to 175 n what is the rotational speed of the motor at no-load conditions? 9-2. Assuming no annature reaction. what is the speed of the motor at full load? What is the speed regulation of the motor? 9-3. If the motor is operating at full load and if its variable resistance Rodj is increased to 250 n.what is the new speed of the motor? Compare the full-load speed of the mo- tor with Rodj = 175 n to the full-load speed with Rodj = 250 O. (Assume no anna- ture reaction. as in the previous problem.) 9-4. Assume that the motor is operating at full load and that the variable resistor Rodj is again 175 n. Ifthe annature reaction is 1200 A· turns at full load. what is the speed of the motor? How does it compare to the result for Problem 9--2? 9-5. If Rodj can be adjusted from 100 to 400 n.what are the maximum and minimrun no- load speeds possible with this motor? 9-6. What is the starting ClUTent of this machine if it is started by connecting it directly to the power supply Vr? How does this starting current compare to the full-load cur- rent of the motor?
- 646. 622 ELECTRIC MACHINERY RJNDAMENTALS > -' f ~ ~ !, " E • • 320 300 Speed 1200r/min 280 260 V /' 240 220 V 200 180 / 160 140 / 120 100 / 80 60 / 40 20 / ,II o 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 l.l 1.2 1.3 1.4 Shum field currem. A ""GURE 1 '9- 1 The masnetization curve for the dc motor in Problems 9--1 to 9- 12. This curve was made at a constam speed of 1200 r/min. 9-7. Plot the torque-speed characteristic of this motor assuming no armature reaction. and again assuming a full-load armature reaction of 1200 A ollU1lS. For Problems 9--8 and 9-9. the shunt dc motor is reconnected separately excited. as shown in Figure P9- 3. It has a fixed field voltage VI' of 240 V and an annature voltage VA that can be varied from 120 to 240 V. 9-8. What is the no-load speed of this separately excited motor when R>dj = 175 nand (a) VA = 120 V. (b) VA = 180 V. (c) VA = 240 V? 9-9. For the separately excited motor of Problem 9--8: (a) What is the maximwn no-load speed attainable by varying both VA and R>dj? (b) What is the minimwn no-load speed attainable by varying both VA and R>dj?
- 647. rx: MmDRS AND GENERATORS 623 - - 0.4On ~ IF I ~ Rodj + lOon Vr = 240 V FIGURE P9-2 The equiva.lent circuit of the shunt ntotor in Problems 9- 1 to 9- 7. I, I, R, I, - - " - + + 0.40 n R., + VF= 240 V RF = 100 n E, V A =120to240V FIGURE P9-J The equiva.lent circuit of the separately excited motor in Problems 9--8 and 9--9. 9-10. If the motor is connected cumulatively compOlUlded as shown in Figure P9-4 and if R adj = 175 O. what is its no-load speed? What is its full-load speed? What is its speed regulation? Calculate and plot the torque-speed characteristic for this motor. (Neglect annature effects in this problem.) 9- 11. The motor is connected cumulatively compounded and is operating at full load. What will the new speed of the motor be if Radj is increased to 250 O ? How does the new speed compare to the full-load speed calculated in Problem 9--1O? 9-12. The motor is now connected differentially compounded. (a) If Radj = 175 O. what is the no-load speed of the motor? (b) What is the motor's speed when the annature current reaches 20A? 40 A? 60A? (c) Calculate and plot the torque-speed characteristic curve of this motor. 9-13. A 7.5-hp. l20-V series dc motor has an armature resistance of 0.2 0 and a series field resistance of 0.16 O. At full load. the current input is 58 A. and the rated speed is
- 648. 624 ELECTRIC MACHINERY RJNDAMENTALS O.4411",RA +RS . '" Cumulatively compounded • '" Differentially compounded + 100 n VT", 240 V •• ""GURE 1'9-4 The equivalent cin:uit of the compounded motor in Problems 9- 10 to 9--12. 1050 r/min. Its magnetization curve is shown in Figure P9-5. The core losses are 200 W, and the mechanical losses are 240 W al full load. Assume that the mechanical losses vary as the cube of the speed of the motor and that the core losses are constant. (a) What is the efficiency of the motor at fullload1 (b) What are the speed and efficiency of the motor ifit is operating at an annature current of 35 A1 (c) Plot the torque-speed characteristic for this motor. 9-14. A 20-hp, 240-V, 76-A, 900 r/min series motor has a field winding of 33 turns per pole. Its annature resistance is 0.09 n, and its field resistance is 0.06 n. The mag- netization curve expressed in terms of magnetomotive force versus EA at 900 r/min is given by the following table: 95 188 212 229 243 :Ji. A• turns '00 1500 2000 2500 3000 Annature reaction is negligible in this machine. (a) CompUle the motor's torque, speed, and output power a133, 67,100, and 133 percent of full-load armalure ClUTent. (Neglect rotational losses.) (b) Plot the torque-speed characteristic of this machine. 9-15. A300-hp, 44O-V, 560-A, 863 r/min shunt dc motor has been tested, and the follow- ing data were taken: Blocked-rotor test: VA = 16.3 V exclusive of brushes 110. = 500 A No-load operation: VA = 16.3 V including brushes 110. = 23.1 A VF = 440 V IF = 8.86A IF = 8.76A n = 863 r/min
- 649. > J • ~ • " ~ • • , , a 1! ~ 160 150 140 130 120 110 100 90 80 70 60 50 / 40 30 20 / / / 10 / o o 10 / / rx: MmDRS AND GENERATORS 625 5,.., 1200r~ /" / / / / / 20 30 40 60 70 Series field current. A FIGURE 1'9-5 The magnetization curve for the series moior in Problem 9--13. This curve was taken al a constant speed of 1200 r/min. What is this motor's efficiency at the rated conditions? [Note:Assrune that ( 1) the brush voltage drop is 2 V, (2) the core loss is to be determined at an armature volt- age equal to the armature voltage lUlder full load, and (3) stray load losses are 1 per- cent of full load.] Problems 9- 16 to 9--19 refer to a 240-V, IOO-A de motor which has both shunt and series windings. Its characteristics are RA =O.14f.! Rs = 0.04 n R" = 200 n Radj = 0 to 300 n,currently set to 120 n N" = 1500 turns Nsf', = 12 turns nO. = 1200 r/min
- 650. 626 ELECTRIC MACHINERY RJNDAMENTALS 300 250 > 200 j "' ~ / ~ 150 , ~ • E • • 100 / / o 0.0 ""GURE 1'9-6 0.25 S"",d 1200 r/min / / / 0.50 0.75 Field current. A 1.00 The masnetization curve for the dc motor in Problems 9--16 to 9--19. 1.25 1.50 This motor has compensating windings and interpoles. The magnetization curve for this motor at 1200 rfmin is shown in Figure P9--6. 9-16. The motor described above is connected in shunt. (a) What is the no-load speed of this motor when R odj = 120 0 ? (b) What is its full-load speed? (c) Under no-load conditions. what range of possible speeds can be achieved by adjusting Rodj? 9-17. This machine is now connected as a cumulatively compounded dc motor with Rodj = 120 n. (a) What is the full-load speed of this motor? (b) Plot the torque-speed characteristic for this motor. (c) What is its speed regulation? 9-18. The motor is recotulected differentially compolUlded with R adj = 120 O. Derive the shape of its torque-speed characteristic.
- 651. rx: MmDRS AND GENERATORS 627 9-19. A series motor is now constructed from this machine by leaving the shlUlt field out entirely. Derive the torque-speed characteristic of the resulting motor. 9-20. An automatic starter circuit is to be designed for a shlUlt motor rated at 15 hp. 240 V. and 60 A. The annature resistance of the motor is 0.15 n. and the shunt field re- sistance is 40 n. The motor is to start with no more than 250 percent of its rated ar- mature current. and as soon as the current falls to rated value. a starting resistor stage is to be cut out. How many stages of starting resistance are needed. and how big should each one be? 9-2 1. A 15-hp. 230-V. 1800 rlmin shunt dc motor has a full-load armature current of 60 A when operating at rated conditions. The annature resistance of the motor is RA = 0.15 n. and the field resistance R" is 80 n.The adjustable resistance in the field cir- cuit R>dj may be varied over the range from 0 to 200 n and is currently set to 90 n. Annature reaction may be ignored in this machine. The magnetization curve for this motor. taken at a speed of 1800 r/min. is given in tabular fonn below: 0.80 1.00 1.28 242 8.' I 150 I 180 0.00 2.88 (a) What is the speed of this motor when it is ruIllling at the rated conditions spec- ified above? (b) The output power from the motor is 7.5 hp at rated conditions. What is the out- put torque of the motor? (c) What are the copper losses and rotational losses in the motor at full load (ignore stray losses)? (d) What is the efficiency of the motor at full load? (e) If the motor is now unloaded with no changes in tenninal voltage or R>dj' what is the no-load speed of the motor? (f) Suppose that the motor is running at the no-load conditions described in part e. What would happen to the motor if its field circuit were to open? Ignoring ar- mature reaction. what would the final steady-state speed of the motor be under those conditions? (g) What range of no-load speeds is possible in this motor. given the range offield resistance adjustments available with Radj? 9-22. The magnetization curve for a separately excited dc generator is shown in Figure P9- 7. The generator is rated at 6 kW, 120 V. 50 A. and ISOO rlmin and is shown in Figure P9-8. Its field circuit is rated at SA. The following data are known about the machine: RA = O.ISO ~j = Ot030n N" = 1000 turns per pole V,, =120V R,, =24n Answer the following questions about this generator. assruning no armature reaction. (a) If this generator is operating at no load. what is the range of voltage adjustments that can be achieved by changing Radj? (b) If the field rheostat is allowed to vary from 0 to 30 n and the generator's speed is allowed to vary from 1500 to 2()(x) r/min. what are the maximwn and mini- mum no-load voltages in the generator?
- 652. 628 ELECTRIC MACHINERY RJNDAMENTALS 160 150 140 a --- V- a a /" 13 12 II / / I I a a I / / II 40 / a / a / ,II 30 2 a 2 3 4 , 6 7 Shunt field current. A a 1000 2000 3000 4000 5000 6000 7000 Field mmf. A· turns ""GURE 1'9-7 The magnetization curve for Problems 9--22 to 9--28. This curve was taken at a speed of 1800 r/min. 9-23. If the armature current of the generator in Problem 9--22 is 50 A. the speed of the generator is 1700 r/min. and the tenninal voltage is 106 V, how much field current must be flowing in the generator? 9-24. Assuming that the generator in Problem 9- 22 has an annature reaction at full load equivalent to 400 A • turns of magnetomotive force. what will the terminal voltage of the generator be when I" = 5 A. n.. = 1700 r/min. and IA = 50 A? 9-25. The machine in Problem 9--22 is reconnected as a shunt generator and is shown in Figure P9-9. The shunt field resistor R>dj is adjusted to 10 n. and the generator's speed is 1800 r/min.
- 653. rx: M mDRS AND GENERATORS 629 I, R, I, I, - - - + O.~8n + :/ R~ 120 V V, RF=24f1 ~ c,JE ' V, L, FIGURE 1'9-8 The separately excited de generator in Problems 9--22 to 9--24. R, ,-~-----"V'VV'--------~e--C---- ~ + 0.18 n i'" JIF R-». + 24 n ~ RF V, FIGURE 1'9-9 The shunt de generator in Problems 9- 25 and 9--26. (a) What is the no-load lennina! voltage of the generator? (b) Assruning no armature reaction, what is the terminal voltage of the generator with an armature current of 20 A? 40 A? (c) Assruning an annature reaction equal to 300 A • turns at [unload, what is the lennina! voltage of the generator with an armature current of 20 A? 40 A? (d) Calculate and plot the terminal characteristics of this generator with and with- out armature reaction. 9-26. If the machine in Problem 9--25 is running at 1800 r/min with a field resistance Rodj = 10 n and an annature current of 25 A, what will the resulting terminal voltage be? If the field resistor decreases to 5 n while the armature current remains 25 A, what will the new terminal voltage be? (Assrune no armature reaction.) 9-27. A 120-V, 50-A cumulatively compounded dc generator has the following characteristics: RA + Rs = 0.2 1 n RF = 20 n R>dj = 0 to 30 n,set to 10 n NF = J()(X) turns NSE = 20 turns n.. = 1800 r/min
- 654. 630 ELECTRIC MACHINERY RJNDA MENTALS 0.21 n + 200 v, • L" N,,= 1000 turns ""GURE 1'9- 10 The compounded dc generator in Problems 9--27 and 9- 28. The machine has the magnetization curve shown in Figure P9- 7. Its equivalent cir- cuit is shown in Figure P9--1O. Answer the following questions about this machine, assuming no annature reaction. (a) If the generator is operating at no load, what is its terminal voltage? (b) If the generator has an armature current of 20 A, what is its tenninal voltage? (c) If the generator has an armature current of 40 A, what is its tenninal voltage? (d) Calculate and plot the tenninal characteristic of this machine. 9-28. If the machine described in Problem 9- 27 is reconnected as a differentially com- pounded dc generator, what will its teoninal characteristic look like? Derive it in the same fashion as in Problem 9-27. 9-29. A cumulatively compounded dc generator is operating properly as a flat- compounded dc generator. The machine is then shut down, and its shunt field con- nections are reversed. (a) Ifthis generator is turned in the same direction as before, will an output voltage be built up at its tenninals? Why or why not? (b) Will the voltage build up for rotation in the opposite direction? Why or why not? (c) For the direction of rotation in which a voltage builds up, will the generator be cumulatively or differentially compolUlded? 9-30. A three-phase synchronous machine is mechanically connected to a shlUlt dc ma- chine, fonning a motor-generator set, as shown in Figure P9--ll. The dc machine is connected to a dc power system supplying a constant 240 V, and the ac machine is connected to a 480-V, 60-Hz infinite bus. The dc machine has four poles and is rated at 50 kW and 240 V. It has a per-unit annature resistance of 0.04. The ac machine has four poles and is V-connected. It is rated at 50 kVA, 480 V, and 0.8 PF, and its saturated synchronous reactance is 2.0 n per phase. All losses except the dc machine's annature resistance may be neglected in this problem. Assume that the magnetization curves of both machines are linear. (a) Initially, the ac machine is supplying 50 kVA at 0.8 PF lagging to the ac power system.
- 655. rx: MmDRS AND GENERATORS 631 MGset rx: machine AC machine 1'1 R, 6 R, ~ AC power , V, E, system , (infinite bus) & L, ~ R, L, + V, FIGURE 1'9-11 The motor-generator set in Problem 9- 30. I. How much power is being supplied to the dc motor from the dc power system? 2. How large is the internal generated voltage EA ofthe dc machine? 3. How large is the internal generated voltage EA of the ac machine? (b) The field current in the ac machine is now increased by 5 percent. What effect does this change have on the real power supplied by the motor- generator set? On the reactive power supplied by the motor- generator set? Calculate the real and reactive power supplied or consumed by the ac machine under these condi- tions. Sketch the ac machine's phasor diagram before and after the change in field current. (c) Starting from the conditions in part b. the field current in the dc machine is now decreased by I percent. What effect does this change have on the real power supplied by the motor-generator set? On the reactive power supplied by the motor- generator set? Calculate the real and reactive power supplied or con- swned by the ac machine lUlder these conditions. Sketch the ac machine's pha- sor diagram before and after the change in the dc machine's field current. (d) From the above results. answer the following questions: I. How can the real power flow through an ac-dc motor- generator set be controlled? 2. How can the reactive power supplied or consumed by the ac machine be controlled without affecting the real power flow? REFERENCES 1. Chaston. A. N. Electric Machinery. Reston. Va.: Reston Publications. 1986. 2. Fitzgerald. A. E.. and C. Kingsley. Jr. Electric Machinery. New YorK: McGraw-Hill. 1952. 3. Fitzgerald. A. E.. C. Kingsley. Jr.• and S. D. Umans. Electric Machinery. 5th ed. New York: McGraw-Hill. 1990. 4. Heck. C. Magnetic Materials and Their AppliCllTions. London: Butterwonh & Co.. 1974.
- 656. 632 ELECTRIC MACHINERY RJNDAMENTALS 5. IEEE Standard 113-1985. Guide on Test Procedures for DC Machines. Piscataway. N.J.: IEEE. 1985. (Note that this standard has been officially withdrawn but is still available.) 6. Kloeftler. S. M.• R. M. Ken;hner. and J. L. Brenneman. Direct Current Machinery. Rev. ed. New York: Macmillan. 1948. 7. Kosow. Irving L. Electric Machinery atuf Tmnsjormers. Englewood ClilTs. N.J.: Prentice-Hall. 1972. 8. McPherson. George. An Introduction 10 Electrical Machines and Tmnsfonners. New York: Wiley. 1981. 9. Siskind. Charles S. Direct-Current Machinery. New York: McGraw-Hill. 1952. 10. Siemon. G. R.• and A. Straughen. Electric Machines. Reading. Mass.: Addison-Wesley. 1980. II. Werninck. E. H. (ed.). Electric MOlOr Handbook. London: McGraw-Hili. 1978.
- 657. CHAPTER 10 SINGLE-PHASE AND SPECIAL-PURPOSE MOTORS Chapters 4 through 7 were devoted to the operation orthe two major classes of ac machines (synchronous and induction) on three-phase power systems. Motors and generators of these types are by far the most common ones in larger commercial and industrial settings. However, most homes and small businesses do not have three-phase power available. For such locations, all motors must run from single-phase power sources. nlis chapter deals with the theory and operation of two major types of single-phase motors: the universal motor and the single- phase induction motor. The universal motor, which is a straightforward extension of the series de motor, is descri bed in Section 10.1. The single-phase induction motor is described in Sections 10.2 to 10.5. The major problem associated with the design of single-phase induction motors is that, unlike three-phase power sources, a single-phase source does not produce a rotat- ing magnetic field. Instead, the magnetic field produced by a single-phase source remains stationary in position and pulses with time. Since there is no net rotating magnetic field, conventional induction motors cannot function, and special de- signs are necessary. In addition, there are a number of special-purpose motors which have not been previously covered. These include reluctance motors, hysteresis motors, stepper motors, and brushless dc motors.1lley are included in Section 10.6. 633
- 658. 634 ELECTRIC MACHINERY RJNDAMENTALS v, ""CURE 10-1 Equivalent cin;uit of a univeTS3.1 motor. 10.1 THE UNIVERSAL MOTOR Perhaps the simplest approach to the design of a motor that will operate on a single-phase ac power source is to take a dc machine and run it from an ac supply. Recall from Chapter 8 that the induced torque of a dc motor is given by (8-49) If the polarity ofthe voltage applied to a shunt or series dc motor is reversed, both the direction of the field flux and the direction ofthe armature current reverse, and the resulting induced torque continues in the same direction as before. Therefore, it should be possible to achieve a pulsating but unidirectional torque from a dc motor connected to an ac power supply. Such a design is practical only for the series dc motor (see Figure 10- 1), since the annature current and the field current in the machine must reverse at ex- actly the same time. For shunt dc motors, the very high field inductance tends to delay the reversal of the field current and thus to unacceptably reduce the average induced torque of the motor. In order for a series dc motor to function effectively on ac, its field poles and stator frame must be completely laminated. If they were not completely lam- inated, their core losses would be enonnous. When the poles and stator are lami- nated, this motor is often called a universal motor, since it can run from either an ac or a dc source. When the motor is running from an ac source, the commutation will be much poorer than it would be with a dc source. The extra sparking at the brushes is caused by transfonner action inducing voltages in the coils undergoing com- mutation. These sparks significantly shorten brush life and can be a source of radio-frequency interference in certain environments. A typical torque-speed characteristic of a universal motor is shown in Fig- ure 10- 2. It differs from the torque-speed characteristic of the same machine op- erating from a dc voltage source for two reasons: I. The armature and field windings have quite a large reacta