1337683699.3104Lecture 07 - Synchronous machines.ppt

ELEN 3441 Fundamentals of Power Engineering Spring 2008
1
Instructor:
Dr. Gleb V. Tcheslavski
Contact:
gleb@ee.lamar.edu
Office Hours:
TBD; Room 2030
Class web site:
http://ee.lamar.edu/gleb/
Index.htm
Lecture 7: Synchronous machines

2
Construction of synchronous
machines
Synchronous machines are AC machines that have a field
circuit supplied by an external DC source.
In a synchronous generator, a DC current is applied to the rotor winding
producing a rotor magnetic field. The rotor is then turned by external means
producing a rotating magnetic field, which induces a 3-phase voltage within
the stator winding.
In a synchronous motor, a 3-phase set of stator currents produces a
rotating magnetic field causing the rotor magnetic field to align with it. The
rotor magnetic field is produced by a DC current applied to the rotor
winding.
Field windings are the windings producing the main magnetic field (rotor
windings for synchronous machines); armature windings are the windings
where the main voltage is induced (stator windings for synchronous
machines).

3
machines
The rotor of a synchronous machine is a large electromagnet. The magnetic poles
can be either salient (sticking out of rotor surface) or non-salient construction.
Non-salient-pole rotor: usually two- and four-pole rotors. Salient-pole rotor: four
and more poles.
Rotors are made laminated to reduce eddy current losses.

4
machines
Salient pole with field
windings
Salient pole without
field windings –
observe laminations
A synchronous rotor with 8 salient poles

5
machines
Two common approaches are used to supply a DC current to the field circuits on
the rotating rotor:
1. 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
DC power source mounted directly on
the shaft of the machine.
Slip rings are metal rings completely encircling the shaft of a machine but insulated
from it. One end of a DC rotor winding is connected to each of the two slip rings on
the machine’s shaft. Graphite-like carbon brushes connected to DC terminals ride on
each slip ring supplying DC voltage to field windings regardless the position or speed
of the rotor.

6
machines
Slip rings
Brush

7
machines
Slip rings and brushes have certain disadvantages: increased friction and wear
(therefore, needed maintenance), brush voltage drop can introduce significant
power losses. Still this approach is used in most small synchronous machines.
On large generators and motors, brushless exciters are used.
A brushless exciter is a small AC generator whose field circuits are
mounted on the stator and armature circuits are mounted on the rotor
shaft. The exciter generator’s 3-phase output is rectified to DC by a 3-
phase rectifier (mounted on the shaft) and fed into the main DC field
circuit. It is possible to adjust the field current on the main machine by
controlling the small DC field current of the exciter generator (located on
the stator).
Since no mechanical contact occurs between the rotor and the stator, exciters of
this type require much less maintenance.

8
machines
A brushless exciter: a
low 3-phase current is
rectified and used to
supply the field circuit
of the exciter (located
on the stator). The
output of the exciter’s
armature circuit (on the
rotor) is rectified and
used as the field
current of the main
machine.

9
machines
To make the
excitation of a
generator completely
independent of any
external power
source, a small pilot
exciter is often added
to the circuit. The pilot
exciter is an AC
generator with a
permanent magnet
mounted on the rotor
shaft and a 3-phase
winding on the stator
producing the power
for the field circuit of
the exciter.

10
machines
A rotor of large
synchronous machine
with a brushless exciter
mounted on the same
shaft.
Many synchronous
generators having
brushless exciters also
include slip rings and
brushes to provide
emergency source of
the field DC current.

11
machines
A large
synchronous
machine with
the exciter
and salient
poles.

12
Rotation speed of synchronous
generator
By the definition, synchronous generators produce electricity whose
frequency is synchronized with the mechanical rotational speed.
120
m
e
n P
f  (7.11.1)
Where fe is the electrical frequency, Hz;
nm is mechanical speed of magnetic field (rotor speed for synchronous
machine), rpm;
P is the number of poles.
Steam turbines are most efficient when rotating at high speed; therefore,
to generate 60 Hz, they are usually rotating at 3600 rpm and turn 2-pole
generators.
Water turbines are most efficient when rotating at low speeds (200-300
rpm); therefore, they usually turn generators with many poles.

13
Internal generated voltage of a
synchronous generator
The magnitude of internal generated voltage induced in a given stator is
2
A C
E N f K
  
 
where K is a constant representing the construction of the machine,  is flux in it
and  is its rotation speed.
Since flux in the
machine depends
on the field current
through it, the
internal generated
voltage is a
function of the
rotor field current.
Magnetization curve (open-circuit characteristic) of a
synchronous machine

14
Equivalent circuit of a synchronous
generator
The internally generated voltage in a single phase of a
synchronous machine EA is not usually the voltage appearing
at its terminals. It equals to the output voltage V only when
there is no armature current in the machine. The reasons
that the armature voltage EA is not equal to the output
voltage V are:
1. Distortion of the air-gap magnetic field caused by the
current flowing in the stator (armature reaction);
2. Self-inductance of the armature coils;
3. Resistance of the armature coils;
4. Effect of salient-pole rotor shapes.

15
generator
Armature reaction (the
largest effect):
When the rotor of a synchronous
generator is spinning, a voltage
EA is induced in its stator. When
a load is connected, a current
starts flowing creating a
magnetic field in machine’s
stator. This stator magnetic field
BS adds to the rotor (main)
magnetic field BR affecting the
total magnetic field and,
therefore, the phase voltage.
Lagging
load

16
generator
Assuming that the generator is connected to a lagging load, the load current IA will
create a stator magnetic field BS, which will produce the armature reaction voltage
Estat. Therefore, the phase voltage will be
A stat
V E E
   (7.16.1)
The net magnetic flux will be
net R S
B B B
  (7.16.2)
Rotor field Stator field
Note that the directions of the net magnetic flux and the phase voltage are the
same.

17
generator
Assuming that the load reactance is X, the armature reaction voltage is
stat A
E jXI
  (7.17.1)
The phase voltage is then A A
V E jXI
   (7.17.2)
Armature reactance can be modeled by the following
circuit…
However, in addition to armature reactance effect,
the stator coil has a self-inductance LA (XA is the
corresponding reactance) and the stator has
resistance RA. The phase voltage is thus
A A A A A
V E jXI jX I RI
     (7.17.3)

18
generator
Often, armature reactance and self-inductance are combined into the synchronous
reactance of the machine:
S A
X X X
  (7.18.1)
A S A A
V E jX I RI
   
Therefore, the phase voltage is
(7.18.2)
The equivalent circuit of a 3-phase
synchronous generator is shown.
The adjustable resistor Radj controls the
field current and, therefore, the rotor
magnetic field.

19
generator
A synchronous generator can be Y- or -connected:
The terminal voltage will be
3
T T
V V for Y V V for
 
      
(7.19.1) (7.19.2)

20
generator
Note: the discussion above assumed a balanced load on the generator!
Since – for balanced loads – the three phases of a synchronous generator are
identical except for phase angles, per-phase equivalent circuits are often used.

21
Phasor diagram of a synchronous
generator
Since the voltages in a synchronous generator are AC voltages, they are usually
expressed as phasors. A vector plot of voltages and currents within one phase is
called a phasor diagram.
A phasor diagram of a synchronous generator
with a unity power factor (resistive load)
Lagging power factor (inductive load): a larger
than for leading PF internal generated voltage
EA is needed to form the same phase voltage.
Leading power factor (capacitive load).
For a given field current and magnitude of
load current, the terminal voltage is lower for
lagging loads and higher for leading loads.

22
Power and torque in synchronous
generators
A synchronous generator needs to be connected to a prime mover whose speed is
reasonably constant (to ensure constant frequency of the generated voltage) for
various loads.
The applied mechanical power
in app m
P  

is partially converted to electricity
3 cos
conv ind m A A
P E I
  
 
(7.22.1)
(7.22.2)
Where  is the angle between
EA and IA.
The power-flow diagram of a
synchronous generator.

23
generators
The real output power of the synchronous generator is
3 cos 3 cos
out T L A
P V I V I

 
 
The reactive output power of the synchronous generator is
3 sin 3 sin
out T L A
Q V I V I

 
 
(7.23.1)
(7.23.2)
Recall that the power factor angle  is the angle between V and IA and not the
angle between VT and IL.
In real synchronous machines of any size, the
armature resistance RA << XS and, therefore,
the armature resistance can be ignored. Thus,
a simplified phasor diagram indicates that
sin
cos A
A
S
E
I
X

  (7.23.3)

24
generators
Then the real output power of the synchronous generator can be approximated as
3 sin
A
out
S
V E
P
X
 
 (7.24.1)
We observe that electrical losses are assumed to be zero since the resistance is
neglected. Therefore:
conv out
P P
 (7.24.2)
Here  is the torque angle of the machine – the angle between V and EA.
The maximum power can be supplied by the generator when  = 900:
max
3 A
S
V E
P
X

 (7.24.3)

25
generators
The maximum power specified by (7.24.3) is called the static stability limit
of the generator. Normally, real generators do not approach this limit: full-
load torque angles are usually between 150 and 200.
The induced torque is
sin
ind R S R net R net
kB B kB B kB B
 
     (7.25.1)
Notice that the torque angle  is also the angle between the rotor magnetic field
BR and the net magnetic field Bnet.
Alternatively, the induced torque is
3 sin
A
ind
m S
V E
X
 


 (7.25.2)

26
Measuring parameters of
synchronous generator model
The three quantities must be determined in order to describe the generator model:
1. The relationship between field current and flux (and therefore between the field
current IF and the internal generated voltage EA);
2. The synchronous reactance;
3. The armature resistance.
We conduct first the open-circuit test on the synchronous generator: the generator
is rotated at the rated speed, all the terminals are disconnected from loads, the
field current is set to zero first. Next, the field current is increased in steps and the
phase voltage (whish is equal to the internal generated voltage EA since the
armature current is zero) is measured.
Therefore, it is possible to plot the dependence of the internal generated voltage
on the field current – the open-circuit characteristic (OCC) of the generator.

27
Since the unsaturated core of the machine has a
reluctance thousands times lower than the
reluctance of the air-gap, the resulting flux
increases linearly first. When the saturation is
reached, the core reluctance greatly increases
causing the flux to increase much slower with
the increase of the mmf.
We conduct next the short-circuit test on the synchronous generator: the generator
is rotated at the rated speed, all the terminals are short-circuited through
ammeters, the field current is set to zero first. Next, the field current is increased in
steps and the armature current IA is measured as the field current is increased.
The plot of armature current (or line current) vs. the field current is the short-circuit
characteristic (SCC) of the generator.

28
The SCC is a straight line since, for the
short-circuited terminals, the magnitude of
the armature current is
2 2
A
A
A S
E
I
R X


(7.28.1)
The equivalent generator’s circuit during SC
The resulting
phasor diagram
The magnetic
fields during
short-circuit test
Since BS almost cancels BR, the
net field Bnet is very small.

29
An approximate method to determine the synchronous reactance XS at a given
field current:
1. Get the internal generated voltage EA from the OCC at that field current.
2. Get the short-circuit current IA,SC at that field current from the SCC.
3. Find XS from
,
A
S
A SC
E
X
I

Since the internal machine impedance is
 
2 2
,
since
A
S A S S S A
A SC
E
Z R X X X R
I
     
(7.29.1)
(7.29.2)

30
A drawback of this method is that the internal generated voltage EA is measured
during the OCC, where the machine can be saturated for large field currents, while
the armature current is measured in SCC, where the core is unsaturated.
Therefore, this approach is accurate for unsaturated cores only.
The approximate value of synchronous
reactance varies with the degree of
saturation of the OCC.
Therefore, the value of the synchronous
reactance for a given problem should be
estimated at the approximate load of the
machine.
The winding’s resistance can be
approximated by applying a DC voltage
to a stationary machine’s winding and
measuring the current. However, AC
resistance is slightly larger than DC
resistance (skin effect).

31
synchronous generator model: Ex
Example 7.1: A 200 kVA, 480 V, 50 Hz, Y-connected synchronous generator with a
rated field current of 5 A was tested and the following data were obtained:
1. VT,OC = 540 V at the rated IF.
2. IL,SC = 300 A at the rated IF.
3. When a DC voltage of 10 V was applied to two of the terminals, a current of 25 A
was measured.
Find the generator’s model at the rated conditions (i.e., the armature resistance and
the approximate synchronous reactance).
Since the generator is Y-connected, a DC
voltage was applied between its two
phases. Therefore:
2
10
0.2
2 2 25
DC
A
DC
DC
A
DC
V
R
I
V
R
I

   


32
synchronous generator model: Ex
The internal generated voltage at the rated field current is
,
540
311.8
3 3
T
A OC
V
E V V

    
The synchronous reactance at the rated field current is precisely
2 2
2 2 2 2
2 2
,
311.8
0.2 1.02
300
A
S S A A
A SC
E
X Z R R
I
       
We observe that if XS was estimated via the approximate formula, the result would
be:
,
311.8
1.04
300
A
S
A SC
E
X
I
   
Which is close to the previous result.
The error ignoring RA is much smaller
than the error due to core saturation.
The equivalent circuit

33
The Synchronous generator
operating alone
The behavior of a synchronous generator varies greatly under
load depending on the power factor of the load and on
whether the generator is working alone or in parallel with other
synchronous generators.
Although most of the synchronous generators in the world
operate as parts of large power systems, we start our
discussion assuming that the synchronous generator works
alone.
Unless otherwise stated, the speed of the generator is
assumed constant.

34
operating alone
Effects of load changes
A increase in the load is an
increase in the real and/or
reactive power drawn from the
generator.
Since the field resistor is unaffected, the field current is constant and, therefore, the
flux  is constant too. Since the speed is assumed as constant, the magnitude of
the internal generated voltage is constant also.
Assuming the same power factor of the load, change in load will change the
magnitude of the armature current IA. However, the angle will be the same (for a
constant PF). Thus, the armature reaction voltage jXSIA will be larger for the
increased load. Since the magnitude of the internal generated voltage is constant
A S A
E V jX I

  (7.34.1)
Armature reaction voltage vector will “move parallel” to its initial position.

35
operating alone
Increase load effect on generators with
Lagging PF
Leading PF
Unity PF

36
operating alone
1. For lagging (inductive) loads, the phase (and terminal) voltage
decreases significantly.
2. For unity power factor (purely resistive) loads, the phase (and
terminal) voltage decreases slightly.
3. For leading (capacitive) loads, the phase (and terminal) voltage rises.
Generally, when a load on a synchronous generator is added, the following
changes can be observed:
Effects of adding loads can be described by the voltage regulation:
100%
nl fl
fl
V V
VR
V

 (7.36.1)
Where Vnl is the no-load voltage of the generator and Vfl is its full-load voltage.

37
operating alone
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. A synchronous generator operating
at a leading power factor often has a negative voltage regulation.
Normally, a constant terminal voltage supplied by a generator is desired. Since the
armature reactance cannot be controlled, an obvious approach to adjust the
terminal voltage is by controlling the internal generated voltage EA = K. This
may be done by changing flux in the machine while varying the value of the field
resistance RF, which is summarized:
1. Decreasing the field resistance increases the field current in the generator.
2. An increase in the field current increases the flux in the machine.
3. An increased flux leads to the increase in the internal generated voltage.
4. An increase in the internal generated voltage increases the terminal voltage of
the generator.
Therefore, the terminal voltage of the generator can be changed by adjusting the
field resistance.

38
operating alone: Example
Example 7.2: A 480 V, 60 Hz, Y-connected six-pole synchronous generator has a
per-phase synchronous reactance of 1.0 . Its full-load armature current is 60 A at
0.8 PF lagging. Its friction and windage losses are 1.5 kW and core losses are 1.0
kW at 60 Hz at full load. Assume that the armature resistance (and, therefore, the
I2R losses) can be ignored. The field current has been adjusted such that the no-load
terminal voltage is 480 V.
a. What is the speed of rotation of this generator?
b. What is the terminal voltage of the generator if
1. 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 of this generator (ignoring the unknown 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 the full load?
how large is the induced countertorque?
e. What is the voltage regulation of this generator at 0.8 PF lagging? at 1.0 PF? at
0.8 PF leading?

39
Since the generator is Y-connected, its phase voltage is
3 277
T
V V V
   
At no load, the armature current IA = 0 and the internal generated voltage is EA =
277 V and it is constant since the field current was initially adjusted that way.
a. The speed of rotation of a synchronous generator is
120 120
60 1200
6
m e
n f rpm
P
   
which is
1200
2 125.7
60
m rad s
 
  
b.1. For the generator at the rated current and the 0.8
PF lagging, the phasor diagram is shown. The phase
voltage is at 00, the magnitude of EA is 277 V,

40
1 60 36.87 60 53.13
S A
jX I j
       
and that
Two unknown quantities are the magnitude of V and the angle  of EA. From the
phasor diagram:
   
2 2
2
sin cos
A S A S A
E V X I X I
  
  
 
2
2
cos sin 236.8
A S A S A
V E X I X I V
  
    
Then:
Since the generator is Y-connected,
3 410
T
V V V

  

41
b.2. For the generator at the rated current and
the 1.0 PF, the phasor diagram is shown.
Then:
 
2
2
cos sin 270.4
A S A S A
V E X I X I V
  
    
3 468.4
T
V V V

  
and
b.3. For the generator at the rated current and the
0.8 PF leading, the phasor diagram is shown.
Then:
 
2
2
cos sin 308.8
A S A S A
V E X I X I V
  
    
3 535
T
V V V

  
and

42
c. The output power of the generator at 60 A and 0.8 PF lagging is
3 cos 3 236.8 60 0.8 34.1
out A
P V I kW
 
      
The mechanical input power is given by
34.1 0 1.0 1.5 36.6
in out elec loss coreloss mech loss
P P P P P kW
  
         
The efficiency is
34.1
100 % 100% 93.2%
36.6
out
in
P
P
      
d. The input torque of the generator is
36.6
291.2
125.7
in
app
m
P
N m


    -

43
The induced countertorque of the generator is
e. The voltage regulation of the generator is
34.1
271.3
125.7
conv
app
m
P
N m


    -
Lagging PF:
480 410
100% 17.1%
410
VR

  
Unity PF:
Lagging PF:
480 468
100% 2.6%
468
VR

  
480 535
100% 10.3%
535
VR

   

44
Terminal characteristics of
synchronous generators
All generators are driven by a prime mover, such as a steam, gas, water, wind
turbines, diesel engines, etc. Regardless the power source, most of prime movers
tend to slow down with increasing the load. This decrease in speed is usually
nonlinear but governor mechanisms of some type may be included to linearize this
dependence.
The speed drop (SD) of a prime mover is defined as:
100%
nl fl
fl
n n
SD
n

 
Most prime movers have a speed drop from 2% to 4%. Most governors have a
mechanism to adjust the turbine’s no-load speed (set-point adjustment).
(7.44.1)

45
A typical speed
vs. power plot
Since the shaft speed is linked to the electrical frequency as
120
m
e
n P
f  (7.45.1)
the power output from the generator is related to its frequency:
A typical
frequency vs.
power plot
 
p nl sys
P s f f
  (7.45.2)
Operating frequency of the system
Slope of curve, W/Hz

46
A similar relationship can be derived for the reactive power Q and terminal voltage
VT. When adding a lagging load to a synchronous generator, its terminal voltage
decreases. When adding a leading load to a synchronous generator, its terminal
voltage increases.
The plot of terminal voltage vs.
reactive power is not necessarily
linear.
Both the frequency-power and
terminal voltage vs. reactive power
characteristics are important for
parallel operations of generators.
When a generator is operating alone supplying the load:
1. The real and reactive powers are the amounts demanded by the load.
2. The governor of the prime mover controls the operating frequency of the system.
3. The field current controls the terminal voltage of the power system.

47
synchronous generators: Example
Example 7.3: A generator with no-load frequency
of 61.0 Hz and a slope sp of 1 MW/Hz is
connected to Load 1 consuming 1 MW of real
power at 0.8 PF lagging. Load 2 (that is to be
connected to the generator) consumes a real
power of 0.8 MW at 0.707 PF lagging.
a. Find the operating frequency of the system before the switch is closed.
b. Find the operating frequency of the system after the switch is closed.
c. What action could an operator take to restore the system frequency to 60 Hz
after both loads are connected to the generator?
The power produced by the generator is
 
p nl sys
P s f f
 
Therefore: sys nl
p
P
f f
s
 

48
synchronous generators: Example
a. The frequency of the system with one load is
1
61 60
1
sys nl
p
P
f f Hz
s
     
b. The frequency of the system with two loads is
1.8
61 59.2
1
sys nl
p
P
f f Hz
s
     
c. To restore the system to the proper operating frequency, the operator should
increase the governor no-load set point by 0.8 Hz, to 61.8 Hz. This will restore
the system frequency of 60 Hz.

49
Parallel operation of synchronous
generators
Most of synchronous generators are operating in parallel with other
synchronous generators to supply power to the same power system.
Obvious advantages of this arrangement are:
1. Several generators can supply a bigger load;
2. A failure of a single generator does not result in a total power loss to the load
increasing reliability of the power system;
3. Individual generators may be removed from the power system for maintenance
without shutting down the load;
4. A single generator not operating at near full load might be quite inefficient.
While having several generators in parallel, it is possible to turn off some of
them when operating the rest at near full-load condition.

50
Conditions required for paralleling
A diagram shows that Generator 2
(oncoming generator) will be connected
in parallel when the switch S1 is closed.
However, closing the switch at an
arbitrary moment can severely
damage both generators!
If voltages are not exactly the same in both lines (i.e. in a and a’, b and b’ etc.), a
very large current will flow when the switch is closed. Therefore, to avoid this,
voltages coming from both generators must be exactly the same. Therefore, the
following conditions must be met:
1. The rms line voltages of the two generators must be equal.
2. The two generators must have the same phase sequence.
3. The phase angles of two a phases must be equal.
4. The frequency of the oncoming generator must be slightly higher than the
frequency of the running system.

51
Conditions required for paralleling
If the phase sequences are different,
then even if one pair of voltages
(phases a) are in phase, the other two
pairs will be 1200 out of phase creating
huge currents in these phases.
If the frequencies of the generators are different, a large power transient may occur
until the generators stabilize at a common frequency. The frequencies of two
machines must be very close to each other but not exactly equal. If frequencies
differ by a small amount, the phase angles of the oncoming generator will change
slowly with respect to the phase angles of the running system.
If the angles between the voltages can be observed, it is possible to close the
switch S1 when the machines are in phase.

52
General procedure for paralleling
generators
When connecting the generator G2 to the running system, the following steps
should be taken:
1. Adjust the field current of the oncoming generator to make its terminal voltage
equal to the line voltage of the system (use a voltmeter).
2. Compare the phase sequences of the oncoming generator and the running
system. This can be done by different ways:
1) Connect a small induction motor to the terminals of the oncoming generator
and then to the terminals of the running system. If the motor rotates in the
same direction, the phase sequence is the same;
2) Connect three light bulbs across the
open terminals of the switch. As the phase
changes between the two generators, light
bulbs get brighter (large phase difference)
or dimmer (small phase difference). If all
three bulbs get bright and dark together,
both generators have the same phase
sequences.

53
General procedure for paralleling
generators
If phase sequences are different, two of the conductors on the oncoming
generator must be reversed.
3. The frequency of the oncoming generator is adjusted to be slightly higher than
the system’s frequency.
4. Turn on the switch connecting G2 to the system when phase angles are equal.
The simplest way to determine the moment when two generators are in phase is by
observing the same three light bulbs. When all three lights go out, the voltage
across them is zero and, therefore, machines are in phase.
A more accurate way is to use a synchroscope – a meter
measuring the difference in phase angles between two a
phases. However, a synchroscope does not check the
phase sequence since it only measures the phase
difference in one phase.
The whole process is usually automated…

54
Operation of generators in parallel
with large power systems
Often, when a synchronous generator is added to a power system, that system is
so large that one additional generator does not cause observable changes to the
system. A concept of an infinite bus is used to characterize such power systems.
An infinite bus is a power system that is 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 and reactive power-voltage characteristics are:

55
Consider adding a generator to an
infinite bus supplying a load.
The frequency and terminal voltage of all
machines must be the same. Therefore,
their power-frequency and reactive
power-voltage characteristics can be
plotted with a common vertical axis.
Such plots are called sometimes as house
diagrams.

56
If the no-load frequency of the oncoming
generator is slightly higher than the
system’s frequency, the generator will be
“floating” on the line supplying a small
amount of real power and little or no
reactive power.
If the no-load frequency of the oncoming
generator is slightly lower than the
system’s frequency, the generator will
supply a negative power to the system:
the generator actually consumes energy
acting as a motor!
Many generators have circuitry
automatically disconnecting them from the
line when they start consuming energy.

57
If the frequency of the generator is
increased after it is connected to the
infinite bus, the system frequency
cannot change and the power supplied
by the generator increases.
Notice that when EA stays constant (field
current and speed are the same), EAsin
(which is proportional to the output power
if VT is constant) increases.
If the frequency of the generator is further increased, power output from the
generator will be increased and at some point it may exceed the power consumed by
the load. This extra power will be consumed by the load.

58
After the real power of the generator is adjusted to the desired value, the generator
will be operating at a slightly leading PF acting as a capacitor that consumes
reactive power. Adjusting the field current of the machine, it is possible to make it to
supply reactive power Q to the system.
Summarizing, when the generator is operating in parallel to an infinite bus:
1. The frequency and terminal voltage of the generator are controlled by the
system 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 generator’s field current controls the reactive power supplied by the
generator to the system.

59
Generators in parallel with other
generators of the same size
When a generator is working alone, its real and reactive power are fixed and
determined by the load.
When a generator is connected to an infinite bus, its frequency and the terminal
voltage are constant and determined by a bus.
When two generators of the same size
are connected to the same load, the
sum of the real and reactive powers
supplied by the two generators must
equal the real and reactive powers
demanded by the load:
1 2
1 2
tot load G G
tot load G G
P P P P
Q Q Q Q
  
  
(7.59.1)
(7.59.2)

60
Since the frequency of G2 must be slightly
higher than the system’s frequency, the
power-frequency diagram right after G2 is
connected to the system is shown.
If the frequency of G2 is next
increased, its power-frequency
diagram shifts upwards. Since the total
power supplied to the load is constant,
G2 starts supplying more power and
G1 starts supplying less power and the
system’s frequency increases.

61
Therefore, when two generators are operating together, an increase in
frequency (governor set point) on one of them:
1. Increases the system frequency.
2. Increases the real power supplied by that generator, while reducing the
real power supplied by the other one.
When two generators are operating
together, an increase in the field current
on one of them:
1. Increases the system terminal
voltage.
2. Increases the reactive power supplied
by that generator, while reducing the
reactive power supplied by the other.
If the frequency-power curves of both generators are known, the powers supplied by
each generator and the resulting system frequency can be determined.

62
generators of the same size: Ex
Example 7.4: Two generators are set to supply
the same load. Generator 1 has a no-load
frequency of 61.5 Hz and a slope sp1 of 1
MW/Hz. Generator 2 has a no-load frequency
of 61.0 Hz and a slope sp2 of 1 MW/Hz. The two
generators are supplying a real load of 2.5 MW
at 0.8 PF lagging.
a. Find the system frequency and power supplied by each generator.
b. Assuming that an additional 1 MW load is attached to the power system, find the
new system frequency and powers supplied by each generator.
c. With the additional load attached (total load of 3.5 MW), find the system
frequency and the generator powers, if the no-load frequency of G2 is increased
by 0.5 Hz.
The power produced by a synchronous generator with a given slope and a no-load
frequency is
 
p nl sys
P s f f
 

63
The total power supplied by the generators equals to the power consumed by the
load:
1 2
load
P P P
 
a. The system frequency can be found from:
   
1 2 1 ,1 2 ,2
load p nl sys p nl sys
P P P s f f s f f
     
as
1 ,1 2 ,2
1 2
1 61.5 1 61.0 2.5
60.0
1 1
p nl p nl load
sys
p p
s f s f P
f Hz
s s
     
   
 
   
   
1 1 ,1
2 2 ,2
1 61.5 60 1.5
1 61.0 60 1
p nl sys
p nl sys
P s f f MW
P s f f MW
      
      
The powers supplied by each generator are:

64
b. For the new load of 3.5 MW, the system frequency is
1 ,1 2 ,2
1 2
1 61.5 1 61.0 3.5
59.5
1 1
p nl p nl load
sys
p p
s f s f P
f Hz
s s
     
   
 
   
   
1 1 ,1
2 2 ,2
1 61.5 59.5 2.0
1 61.0 59.5 1.5
p nl sys
p nl sys
P s f f MW
P s f f MW
      
      
The powers are:
c. If the no-load frequency of G2 increases, the system frequency is
1 ,1 2 ,2
1 2
1 61.5 1 61.5 3.5
59.75
1 1
p nl p nl load
sys
p p
s f s f P
f Hz
s s
     
   
 
   
1 2 1 ,1 1 61.5 59.75 1.75
p nl sys
P P s f f MW
       
The powers are:

65
When two generators of the same size are working in parallel, a change in
frequency (governor set points) of one of them changes both the system frequency
and power supplied by each generator.
To adjust power sharing without changing
the system frequency, we need to increase
the frequency (governor set points) of one
generator and simultaneously decrease the
frequency of the other generator.
To adjust the system frequency without
changing power sharing, we need to
simultaneously increase or decrease the
frequency (governor set points) of both
generators.

66
Similarly, to adjust the reactive power
sharing without changing the terminal
voltage, we need to increase
simultaneously the field current of one
generator and decrease the field current of
the other generator.
To adjust the terminal voltage without
changing the reactive power sharing, we
need to simultaneously increase or
decrease the field currents of both
generators.

67
It is important that both generators being paralleled have dropping
frequency-power characteristics.
If two generators have flat or almost
flat frequency-power characteristics,
the power sharing between them can
vary widely with only finest changes in
no-load speed. For good of power
sharing between generators, they
should have speed drops of 2% to 5%.

68
Synchronous motors
The field current IF of the motor
produces a steady-state rotor
magnetic field BR. A 3-phase set of
voltages applied to the stator produces
a 3-phase current flow in the windings.
A 3-phase set of currents in an
armature winding produces a uniform
rotating magnetic field Bs.
Two magnetic fields are present in the machine, and the rotor field tends to align
with the stator magnetic field. Since the stator magnetic field is rotating, the rotor
magnetic field will try to catch up pulling the rotor.
The larger the angle between two magnetic fields (up to a certain maximum), the
greater the torque on the rotor of the machine.

69
Synchronous motor equivalent
circuit
A synchronous motor has the same
equivalent circuit as synchronous
generator, except that the direction of
power flow (and the direction of IA) is
reversed. Per-phase circuit is shown:
A change in direction of IA changes the Kirchhoff’s voltage law equation:
A S A A A
V E jX I R I
   
A S A A A
E V jX I R I

  
Therefore, the internal generated voltage is
We observe that this is exactly the same equation as the equation for the generator,
except that the sign on the current terms is reversed.
(7.69.1)
(7.69.2)

70
Synchronous motor vs.
Let us suppose that a phasor
diagram of synchronous
generator is shown. BR produces
EA, Bnet produces V, and BS
produces Estat = -jXSIA. The
rotation on both diagrams is
counterclockwise and the
induced torque is
ind R net
kB B
   (7.70.1)
clockwise, opposing the direction of rotation. In other words, the induced torque in
generators is a counter-torque that opposes the rotation caused by external torque.
If the prime mover loses power, the rotor will slow down and the rotor field BR will
fall behind the magnetic field in the machine Bnet. Therefore, the operation of the
machine changes…

71
Synchronous motor vs.
The induced torque becomes
counter-clockwise, being now in the
direction of rotation. The machine
starts acting as a motor.
The increasing torque angle 
results in an increasing torque in
the direction of rotation until it
equals to the load torque.
At this point, the machine operates at steady state and synchronous speed but as a
motor.
Notice that, since the direction of IA is changed between the generator and motor
actions, the polarity of stator voltage (-jXSIA) also changes.
In a summary: in a generator, EA lies ahead of V, while in a motor, EA lies
behind V.

72
Steady-state operation of motor:
Torque-speed curve
Usually, synchronous motors are connected to large power systems (infinite bus);
therefore, their terminal voltage and system frequency are constant regardless
the motor load. Since the motor speed is locked to the electrical frequency, the
speed should be constant regardless the load.
The steady-state speed of the motor is
constant from no-load to the maximum torque
that motor can supply (pullout torque).
Therefore, the speed regulation of
synchronous motor is 0%.
The induced torque is
sin
ind R net
kB B
 

or 3
sin
A
ind
m S
V E
X

 


(7.72.1)
(7.72.2)

73
Torque-speed curve
The maximum pullout torque occurs when  = 900:
Normal full-load torques are much less than that (usually, about 3 times smaller).
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. It starts
to slip behind them. As the motor slows down, the stator magnetic field “laps” it
repeatedly, and the direction of the induced torque in the rotor reverses with each
pass. As a result, huge torque surges of alternating direction cause the motor
vibrate severely. The loss of synchronization after the pullout torque is exceeded is
known as slipping poles.
max
3
R net
A
m S
V E
X
kB B 


  (7.73.1)

74
Effect of torque changes
Assuming that a synchronous motor operates
initially with a leading PF.
If the load on the motor increases, the rotor
initially slows down increasing the torque angle
. As a result, the induced torque increases
speeding up the rotor up to the synchronous
speed with a larger torque angle .
Since the terminal voltage and frequency
supplied to the motor are constant, the
magnitude of internal generated voltage
must be constant at the load changes
(EA = K and field current is constant).

75
Effect of torque changes
Assuming that the armature resistance is negligible, the power converted from
electrical to mechanical form in the motor will be the same as its input power:
3
3 cos sin
A
A
S
V E
P V I
X

  
  (7.73.1)
Since the phase voltage is constant, the quantities IAcos and EAsin are directly
proportional to the power supplied by (and to) the motor. When the power
supplied by the motor increases, the distance proportional to power increases.
Since the internal generated voltage is
constant, its phasor “swings down” as
load increases. The quantity jXSIA has to
increase; therefore, the armature current
IA increases too.
Also, the PF angle changes too moving
from leading to lagging.

76
Effect of field current changes
Assuming that a synchronous motor operates
initially with a lagging PF.
If, for the constant load, the field current on the
motor increases, the magnitude of the internal
generated voltage EA increases.
Since changes in IA do not affect the shaft
speed and the motor load is constant, the
real power supplied by the motor is
unchanged. Therefore, the distances
proportional to power on the phasor
diagram (EAsin and IAcos) must be
constant.
Notice that as EA increases, the magnitude of the armature current IA first
decreases and then increases again. At low EA, the armature current is lagging and
the motor is an inductive load that consumes reactive power Q. As the field current
increases , IA eventually lines up with V, and the motor is purely resistive. As the
field current further increases, IA becomes leading and the motor is a capacitive
load that supplies reactive power Q to the system (consumes –Q).

77
A plot of armature current vs. field current is
called a synchronous motor V curve. V
curves for different levels of real power
have their minimum at unity PF, when only
real power is supplied to the motor. For field
currents less than the one giving the
minimum IA, the armature current is lagging
and the motor consumes reactive power.
For field currents greater than the one
giving the minimum IA, the armature current
is leading and the motor supplies reactive
power to the system.
Therefore, by controlling the field current of a synchronous
motor, the reactive power consumed or supplied to the power
system can be controlled.

78
When the projection of the phasor EA onto
V (EAcos) is shorter than V, a
synchronous motor has a lagging current
and consumes Q. Since the field current is
small in this situation, the motor is sais to
be under-excited.
When the projection of the phasor EA
onto V (EAcos) is longer than V, a
synchronous motor has a leading
current and supplies Q to the system.
Since the field current is large in this
situation, the motor is sais to be over-
excited.

79
power factor correction
Assuming that a load contains a
synchronous motor (whose PF
can be adjusted) in addition to
motors of other types. What
does the ability to set the PF of
one of the loads do for the
power system?
Let us consider a large power system operating at 480 V. Load 1 is an induction
motor consuming 100 kW at 0.78 PF lagging, and load 2 is an induction motor
consuming 200 kW at 0.8 PF lagging. Load 3 is a synchronous motor whose real
power consumption is 150 kW.
a. If the synchronous motor is adjusted to 0.85 PF lagging, what is the line current?
b. If the synchronous motor is adjusted to 0.85 PF leading, what is the line current?
c. Assuming that the line losses are PLL = 3IL
2RL, how du these losses compare in
the two cases?

80
a. The real power of load 1 is 100 kW, and the reactive power of load 1 is
 
1
1 1 tan 100tan cos 0.78 80.2
Q P kVAR
 
   
The real power of load 2 is 200 kW, and the reactive power of load 2 is
 
1
2 2 tan 200tan cos 0.8 150
Q P kVAR
 
   
The real power of load 3 is 150 kW, and the reactive power of load 3 is
 
1
3 3 tan 150tan cos 0.85 93
Q P kVAR
 
   
The total real load is 1 2 3 100 200 150 450
tot
P P P P kW
       
The total reactive load is 1 2 3 80.2 150 93
tot
Q Q Q Q kVAR
      
The equivalent system PF is
1 1 323.2
cos cos tan cos tan 0.812
450
Q
PF lagging
P
  
   
    
   
   
The line current is
450 000
3 cos 3 480 0.812
667
tot
L
L
I
V
A
P


 
 



81
b. The real and reactive powers of loads 1 and 2 are the same. The reactive power
of load 3 is
 
1
3 3 tan 150tan cos 0.85 93
Q P kVAR
 
     
The total real load is 1 2 3 100 200 150 450
tot
P P P P kW
       
The total reactive load is 1 2 3 80.2 150 93
tot
Q Q Q Q kVAR
      
The equivalent system PF is
1 1 137.2
cos cos tan cos tan 0.957
450
Q
PF lagging
P
  
   
    
   
   
The line current is
450 000
3 cos 3 480 0.957
566
tot
L
L
I
V
A
P


 
 



82
c. The transmission line losses in the first case are
2
1 7 0
3 344 0
LL L L L
P I R R
 
  
The transmission line losses in the second case are
2
6 70
3 9 1
LL L L L
P I R R
  

We notice that the transmission power losses are 28% less in the second
case, while the real power supplied to the loads is the same.

83
The ability to adjust the power factor of one or more loads in a power system can
significantly affect the efficiency of the power system: the lower the PF, the greater
the losses in the power lines. Since most loads in a typical power system are
induction motors, having one or more over-excided synchronous motors (leading
loads) in the system is useful for the following reasons:
1. A leading load supplies some reactive power to lagging loads in the system.
Since this reactive power does not travel along the transmission line,
transmission line current is reduced reducing power losses.
2. Since the transmission line carries less current, the line can be smaller for a
given power flow reducing system cost.
3. The over-excited mode of synchronous motor increases the motor’s maximum
torque.
Usage of synchronous motors or other equipment increasing the overall system’s
PF is called power-factor correction. Since a synchronous motor can provide PF
correction, many loads that can accept constant speed are driven by over-excited
synchronous motors.

84
Starting synchronous motors
Consider a 60 Hz synchronous motor.
When the power is applied to the stator windings, the rotor (and,
therefore its magnetic field BR) is stationary. The stator magnetic field
BS starts sweeping around the motor at synchronous speed.
Note that the induced torque on the shaft
ind R S
kB B
   (7.84.1)
is zero at t = 0 since both magnetic fields are aligned.
At t = 1/240 s the rotor has barely moved but the stator
magnetic field BS has rotated by 900. Therefore, the torque
on the shaft is non-zero and counter-clockwise.

85
At t = 1/120 s the rotor and stator magnetic fields point in opposite
directions, and the induced torque on the shaft is zero again.
At t = 3/240 s the stator magnetic fields point to the
right, and the induced torque on the shaft is non-
zero but clockwise.
Finally, at t = 1/60 s the rotor and stator magnetic fields are aligned
again, and the induced torque on the shaft is zero.
During one electrical cycle, the torque was counter-clockwise
and then clockwise, and the average torque is zero. The
motor will vibrate heavily and finally overheats!

86
Three basic approaches can be used to safely start a synchronous motor:
1. Reduce the speed of the stator magnetic field to a low enough value
that the rotor can accelerate and two magnetic fields lock in during one
half-cycle of field rotation. This can be achieved by reducing the
frequency of the applied electric power (which used to be difficult but
can be done now).
2. Use an external prime mover to accelerate the synchronous motor up
to synchronous speed, go through the paralleling procedure, and bring
the machine on the line as a generator. Next, turning off the prime
mover will make the synchronous machine a motor.
3. Use damper windings or amortisseur windings – the most popular.

87
Motor starting by amortisseur or
damper windings
Amortisseur (damper) windings are special bars
laid into notches carved in the rotor face and then
shorted out on each end by a large shorting ring.

88
damper windings
A diagram of a salient 2-pole rotor with an
amortisseur winding, with the shorting bars on the
ends of the two rotor pole faces connected by wires
(not quite the design of actual machines).
We assume initially that the
rotor windings are
disconnected and only a 3-
phase set of voltages are
applied to the stator.
As BS sweeps along in s counter-clockwise direction, it
induces a voltage in bars of the amortisseur winding:
At t = 0, assume that BS
(stator field) is vertical.
 
ind
e v B l
   (7.88.1)

89
damper windings
Here v – the velocity of the bar relative to the
magnetic field;
B – magnetic flux density vector;
l – length of conductor in the magnetic field.
The bars at the top of the rotor are moving to the right
relative to the magnetic field: a voltage, with direction
out of page, will be induced. Similarly, the induced
voltage is into the page in the bottom bars. These
voltages produce a current flow out of the top bars and
into the bottom bars generating a winding magnetic
field Bw to the right. Two magnetic fields will create a
torque
ind W S
kB B
   (7.89.1)
The resulting induced torque will be counter-clockwise.

90
damper windings
At t = 1/240 s, BS has rotated 900 while the rotor has barely
moved. Since v is parallel to BS, the voltage induced in the
amortisseur windings is zero, therefore, no current in wires
create a zero-torque.
At t = 1/120 s, BS has rotated
another 900 and the rotor is still.
The voltages induced in the bars
create a current inducing a
magnetic field pointing to the left.
The torque is counter-clockwise.
Finally, at t = 3/240 s, no voltage is
induced in the amortisseur windings
and, therefore, the torque will be zero.

91
damper windings
We observe that the torque is either counter-clockwise or zero, but it is always
unidirectional. Since the net torque is nonzero, the motor will speed up.
However, the rotor will never reach the synchronous speed! If a rotor was running
at the synchronous speed, the speed of stator magnetic field BS would be the same
as the speed of the rotor and, therefore, no relative motion between the rotor and
the stator magnetic field. If there is no relative motion, no voltage is induced and,
therefore, the torque will be zero.
Instead, when the rotor’s speed is close to synchronous, the regular field current
can be turned on and the motor will operate normally. In real machines, field circuit
are shorted during starting. Therefore, if a machine has damper winding:
1. Disconnect the field windings from their DC power source and short them out;
2. Apply a 3-phase voltage to the stator and let the rotor to accelerate up to near-
synchronous speed. The motor should have no load on its shaft to enable motor
speed to approach the synchronous speed as closely as possible;
3. Connect the DC field circuit to its power source: the motor will lock at
synchronous speed and loads may be added to the shaft.

92
Relationship between synchronous
generators and motors
Synchronous generator and synchronous
motor are physically the same machines!
A synchronous machine can supply real
power to (generator) or consume real
power (motor) from a power system. It
can also either consume or supply
reactive power to the system.
1. The distinguishing characteristic of a
synchronous generator (supplying P)
is that EA lies ahead of V while for a
motor EA lies behind V.
2. The distinguishing characteristic of a
machine supplying reactive power Q
is that Eacos > V (regardless
whether it is a motor or generator).
The machine consuming reactive
power Q has Eacos < V .

93
Synchronous machine ratings
The speed and power that can be obtained from a synchronous motor or generator
are limited. These limited values are called ratings of the machine. The purpose of
ratings is to protect the machine from damage. Typical ratings of synchronous
machines are voltage, speed, apparent power (kVA), power factor, field current and
service factor.
1. Voltage, Speed, and Frequency
The rated frequency of a synchronous machine depends on the power system to
which it is connected. The commonly used frequencies are 50 Hz (Europe, Asia),
60 Hz (Americas), and 400 Hz (special applications: aircraft, spacecraft, etc.).
Once the operation frequency is determined, only one rotational speed in possible
for the given number of poles:
120 e
m
f
n
P
 (7.93.1)

94
A generator’s voltage depends on the flux, the rotational speed, and the
mechanical construction of the machine. For a given design and speed, the higher
the desired voltage, the higher the flux should be. However, the flux is limited by
the field current.
The rated voltage is also limited by the windings insulation breakdown limit, which
should not be approached closely.
Is it possible to operate a synchronous machine at a frequency other than the
machine is rated for? For instance, can a 60 Hz generator operate at 50 Hz?
The change in frequency would change the speed. Since EA = K, the
maximum allowed armature voltage changes when frequency changes.
Specifically, if a 60 Hz generator will be operating at 50 Hz, its operating
voltage must be derated to 50/60 or 83.3 %.

95
2. Apparent power and Power factor
Two factors limiting the power of electric machines are
1) Mechanical torque on its shaft (usually, shaft can handle much more torque)
2) Heating of the machine’s winding
The practical steady-state limits are set by heating in the windings.
The maximum acceptable armature current sets the apparent power rating for a
generator:
3 A
S V I


If the rated voltage is known, the maximum accepted armature current determines
the apparent power rating of the generator:
, ,max , ,max
3 3
rated A L rated L
S V I V I

 
(7.95.1)
(7.95.2)
The power factor of the armature current is irrelevant for heating the
armature windings.

96
The stator cupper losses also do not depend on the current angle:
2
3
SCL A A
P I R
 (7.96.1)
The rotor (field winding) cupper losses are:
Since the current angle is irrelevant to the armature heating, synchronous
generators are rated in kVA rather than in KW.
2
RCL F F
P I R
 (7.96.2)
Allowable heating sets the maximum field current,
which determines the maximum acceptable
armature voltage EA. These translate to
restrictions on the lowest acceptable power factor:
The current IA can have different angles (that
depends on PF). EA is a sum of V and jXSIA. We
see that, (for a constant V) for some angles the
required EA exceeds its maximum value.

97
If the armature voltage exceeds its maximum allowed value, the windings could be
damaged. The angle of IA that requires maximum possible EA specifies the rated
power factor of the generator. It is possible to operate the generator at a lower
(more lagging) PF than the rated value, but only by decreasing the apparent power
supplied by the generator.
Synchronous motors are usually rated in terms of real output power and the lowest
PF at full-load conditions.
3. Short-time operation and service factor
A typical synchronous machine is often able to supply up to 300% of its rated
power for a while (until its windings burn up). This ability to supply power above the
rated values is used to supply momentary power surges during motor starts.
It is also possible to use synchronous machine at powers exceeding the rated
values for longer periods of time, as long as windings do not have time to hit up too
much before the excess load is removed. For instance, a generator that could
supply 1 MW indefinitely, would be able to supply 1.5 MW for 1 minute without
serious harm and for longer periods at lower power levels.

98
The maximum temperature rise that a machine can stand depends on the
insulation class of its windings. The four standard insulation classes with they
temperature ratings are:
A – 600C above the ambient temperature
B – 800C above the ambient temperature
F – 1050C above the ambient temperature
H – 1250C above the ambient temperature
The higher the insulation class of a given machine, the greater the power that can
be drawn out of it without overheating its windings.
The overheating is a serious problem and synchronous machines should not be
overheated unless absolutely necessary. However, power requirements of the
machine not always known exactly prior its installation. Because of this, general-
purpose machines usually have their service factor defined as the ratio of the
actual maximum power of the machine to the rating on its plate.
For instance, a machine with a service factor of 1.15 can actually be operated at
115% of the rated load indefinitely without harm.

1337683699.3104Lecture 07 - Synchronous machines.ppt

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