6 DC machimes- DC generators.pdf

Electric Machines I
DC Machines - DC Generators
1
Dr. Firas Obeidat

2
Table of contents
1 • Construction of Simple Loop Generator
2 • Working of Simple Loop Generator
3 • Types of DC Generators
4 • The Terminal Characteristic of a Separately Excited DC Generator
5 • The Terminal Characteristic of a Self Excited Shunt DC Generator
6 • The Terminal Characteristic of a Self Excited Series DC Generator
7 • The Terminal Characteristic of Cumulatively Compound DC Generator
8 • E.M.F. Equation of DC Generator
9 • Total Loss in a DC Generator
10 • Power Stages and Efficiency
11 • Voltage Regulation
12 • Uses of DC Generators
Dr. Firas Obeidat Faculty of Engineering Philadelphia University

3
Construction of Simple Loop Generator
 A single turn rectangular copper ABCD rotating about its own axis in a
magnetic field provided by either permanent magnet or electromagnet.
The two ends of the coil are joined to slip ring ‘a’ and ‘b’ which are
insulated from each other and from the central shaft. Two collecting
brushes press against the slip rings; their function is to collect the
current induced in the coil and to convey it to external load resistance.
 The rotating coil is called the ‘armature’.

4
Working of Simple Loop Generator
 Imagine the coil to be rotating in clockwise
direction. As the coil assumes successive
positions in the field, the flux linked with it
changes. An emf is induced in it which is
proportional to the rate of change of flux
linkages (e=Ndϕ/dt).
 When the plane of coil is in position 1, then
flux linked with the coil is maximum but rate
of change of flux linkage is minimum. Hence,
there is no induced emf in the coil.
 As the coil continues rotating further, the rate
of change of flux linkages (and hence induced
emf in it) increases, till position 3 is reached
where θ=90o. The coil plane is horizontal
(parallel to the lines of flux). The flux linked
with the coil is minimum but rate of change of
flux linkage is maximum. Hence, maximum
emf is induced in the coil when in this position.
In the second half revolution, the direction of
the current flow is DCMLBA. Which is just
the reverse of the previous direction of flow.

5
 In the next quarter revolution (from 90o
to 180o), the flux linked with the coil
gradually increases but the rate of change
of flux linkages decreases. Hence the
induced emf decreases gradually till in
position 5 of the coil, it reduces to zero
value.
 In the first half revolution of the coil, no
emf is induced in it when in position 1,
maximum when in position 3 and no emf
when in position 5. In this half revolution,
the direction of the current flow is
ABMLCD. The current through the load
resistor R flows from M to L during the
first half revolution of the coil.
 In the next half revolution (from 180o to
360o), the variations in the magnitude of
emf are similar to those in the first half
revolution. Its value is maximum when
the coil is in position 7 and minimum
when it in position 1.

6
 The current which is obtained from
such a simple generator reverses its
direction after every half
revolution, this current is known as
alternating current. To make the
flow of current unidirectional in
the external circuit, the slip rings
are replaced by split rings.
 In the first half revolution segment
‘a’ is connected to brush 1 and
segment ‘b’ is connected to brush 2,
while in the second half revolution
segment ‘b’ is connected to brush 1
and segment ‘a’ is connected to
brush 2. In this case the current
will flow in the resistor from M to
L in the two halves of revolution.
 The resulting current is
unidirectional but not continuous
like pure direct current.

7
Types of DC Generators
Generators are usually classified according to the way in which their fields
are excited
A. Separately Excited Generators: are those whose field magnets are
energized from an independent external source of DC current.
B. Self Excited Generators: are those whose field magnets are energized by
current produced by the generators themselves. There are three types of
self excited generators named according to the manner in which their
field coils are connected to the armature.
i. Shunt Wound: the field windings are connected across or in parallel
with the armature conductors and have the full voltage of the generator
applied across them.
ii. Series Wound: the field windings are joined in series with the armature
conductors
iii.Compound Wound: it is a combination of a few series and a few shunt
windings and can be either short-shunt or long-shunt. In compound
generator, the shunt field is stronger than the series field. When series
field aids the shunt field, generator is said to be commutatively-
compound. In series field oppose the shunt field, the generator is said to
be differentially compounded.

8
Separately Excited Generators Shunt Wound Generators Series Wound Generators
Short Shunt Generators Long Shunt Generators

9

10
The Terminal Characteristic of a Separately Excited DC Generator
𝐼𝐴 = 𝐼𝐿
𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 𝑅𝐴
𝑉𝐹 = 𝐼𝐹 𝑅𝐹
For Separately Excited DC Generator
Where
IA: is the armature current
IL: is the load current
EA: is the internal generated voltage
VT: is the terminal voltage
IF: is the field current
VF: is the field voltage
RA: is the armature winding resistance
RF: is the field winding resistance
ϕ: is the flux
𝜔m: is the rotor angular speed
𝐸𝐴 = 𝑘ϕ𝜔𝑚
The terminal voltage can be controlled by:
1. Change the speed of rotation: If 𝜔
increases, then 𝐸𝐴=𝑘ϕ𝜔𝑚 increases, so
𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 𝑅𝐴 increases as well.
2. Change the field current. If RF is
decreased. then the field current
increases (𝑉𝐹 = 𝐼𝐹 𝑅𝐹 ). Therefore, the
flux in the machine increases. As the
flux rises, 𝐸𝐴=𝑘ϕ𝜔𝑚 must rise too, so
𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 𝑅𝐴 increases.
EA
-
+ RA
IA IL
-
+
VT
RF
LF
-
+
VF
IF

11
The Terminal Characteristic of a Self Excited Shunt DC Generator
𝐼𝐴 = 𝐼𝐹 + 𝐼𝐿
𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 𝑅𝐴
𝑉𝑇 = 𝐼𝐹 𝑅𝐹
For Self Excited Shunt DC Generator
The terminal voltage can be controlled by:
1. Change the speed of rotation: If 𝜔
increases, then 𝐸𝐴=𝑘ϕ𝜔𝑚 increases, so
2. Change the field current. If RF is
decreased. then the field current
increases (𝑉𝐹 = 𝐼𝐹 𝑅𝐹 ). Therefore, the
flux in the machine increases. As the
flux rises, 𝐸𝐴=𝑘ϕ𝜔𝑚 must rise too, so
𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 𝑅𝐴 increases.
EA
-
+ RA
IA IL
-
+
VT
RF
LF
IF

12
The Terminal Characteristic of a Self Excited Series DC Generator
𝐼𝐴 = 𝐼𝑠 = 𝐼𝐿
𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴(𝑅𝐴+𝑅𝑠)
For Self Excited Series DC Generator
At no load, there is no field current, so VT is
reduced to a small level given by the
residual flux in the machine. As the load
increases, the field current rises, so EA rises
rapidly The IA(RA+ Rs) drop goes up too,
but at first the increase in EA goes up more
rapidly than the IA(RA+ Rs) drop rises, so
VT increases. After a while, the machine
approaches saturation, and EA becomes
almost constant. At that point, the resistive
drop is the predominant effect, and VT
starts to fall.
EA
-
+ RA
IA IL
-
+
VT
Rs Ls
Is

13
The Terminal Characteristic of Cumulatively Compound DC Generator
𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴(𝑅𝐴+𝑅𝑠)
For Long Shunt Cumulatively Compound DC Generator
EA
-
+ RA
IA IL
-
+
VT
Rs Ls RF
LF
IF
For Short Shunt Cumulatively Compound DC Generator
𝑉𝑇 = 𝐼𝐹 𝑅𝐹
𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴𝑅𝐴 −𝐼𝐿𝑅𝑠
EA
-
+ RA
IA IL
-
+
VT
Rs Ls
RF
LF
IF
The terminal voltage Cumulatively Compound DC Generator can be controlled by:
1. Change the speed of rotation: If 𝜔 increases, then 𝐸𝐴=𝑘ϕ𝜔𝑚 increases, so
2. Change the field current. If RF is decreased. then the field current increases
(𝑉𝐹 = 𝐼𝐹 𝑅𝐹). Therefore, the flux in the machine increases. As the flux rises, 𝐸𝐴=𝑘ϕ
𝜔𝑚 must rise too, so 𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 𝑅𝐴 increases.

14
Examples
Example: A shunt DC generator delivers 450A at 230V and the resistance of
the shunt field and armature are 50Ω and 0.3 Ω respectively. Calculate emf.
𝐼𝑓 =
230
50
= 4.6𝐴
𝐼𝐴 = 𝐼𝐹 + 𝐼𝐿 = 4.6 + 450 = 454.6𝐴
𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴𝑅𝐴 = 230 + 454.6 × 0.3 = 243.6V
Example: A long shunt compound DC generator delivers a load current of
50A at 500V and has armature, series field and shunt field resistances of
0.05Ω, 0.03Ω and 250Ω respectively. Calculate the generated voltage and the
armature current. Allow 1V per brush for contact drop.
𝐼𝐹 =
500
250
= 2𝐴
𝐼𝐴 = 𝐼𝐹 + 𝐼𝐿 = 2 + 50 = 52𝐴
Voltage drop across series winding=𝐼A𝑅s=52×0.03=1.56V
Armature voltage drop=𝐼A𝑅A=52×0.05=2.6V
Drop at brushes=2×1=2V
𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴𝑅𝐴 + 𝑠𝑒𝑟𝑖𝑒𝑠 𝑑𝑟𝑜𝑝 + 𝑏𝑟𝑢𝑠ℎ𝑒𝑠 𝑑𝑟𝑜𝑝 = 500 + 2.6 + 1.56 + 2 = 506.16V
EA
-
+ RA
IA IL=450A
-
+
V
T
=230V
RF
LF
IF
EA
-
+ RA
IA
-
+
Rs Ls RF
LF
IF
V
T
=500V
IL=50A

15
Examples
Example: A short shunt compound DC generator delivers a load current of
0.05Ω, 0.3Ω and 200Ω respectively. Calculate the induced emf and the
armature current. Allow 1V per brush for contact drop.
𝐼𝐹 =
229
200
= 1.145𝐴
Armature voltage drop=𝐼𝐴𝑅𝐴
= 31.145 × 0.05 = 1.56V
Voltage drop across series winding=𝐼L𝑅s
=30 ×0.3=9V
Drop at brushes=2×1=2V
𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴𝑅𝐴 + 𝑠𝑒𝑟𝑖𝑒𝑠 𝑑𝑟𝑜𝑝 + 𝑏𝑟𝑢𝑠ℎ𝑒𝑠 𝑑𝑟𝑜𝑝 = 220 + 9 + 1.56 + 2 = 232.56V
Voltage across shunt winding=220 + 9=229V
EA
-
+ RA
IA
-
+
Rs Ls
RF
LF
IF
V
T
=220V
IL=30A

16
Examples
Example: A long shunt compound DC generator delivers a load current of
0.032Ω, 0.015Ω and 92Ω respectively. Calculate (i) induced emf (ii) total
power generated and (iii) distribution of this power.
EA
-
+ RA
IA
-
+
Rs Ls RF
LF
IF
V
T
=230V
IL=150A
𝐼𝐹 =
230
92
= 2.5𝐴
𝐼𝐴 = 𝐼𝐹 + 𝐼𝐿 = 2.5 + 150 = 152.5𝐴
Voltage drop across series winding=
𝐼A𝑅s=152.5 ×0.015=2.2875V
Armature voltage drop=𝐼A𝑅A=152.5 ×0.032=4.88V
Total power generated by the armature=𝐸𝐴𝐼𝐴=237.1675×152.5=36168.04375W
𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴𝑅𝐴 + 𝐼A𝑅s = 230 + 2.2875 + 4.88 = 237.1675V
(i)
(ii)
(iii) Power lost in armature=𝐼𝐴
2
𝑅𝐴=152.52
×0.032=744.2W
Power dissipated in shunt winding=𝑉𝑇𝐼𝐹=230×2.5=575W
Power dissipated in series winding=𝐼𝐴
2
𝑅𝑠=152.52×0.015=348.84375W
Power delivered to the load=𝑉𝑇𝐼𝐿=230×150=34500W
Total power generated by the armature=744.2 + 575 + 348.843 + 34500=36168.04375W

17
E.M.F. Equation of DC Generator
Let
ϕ: flux/pole in weber.
Z: total number of armature conductors
Z=number of slots × number of conductors/slot
A: number of parallel paths in armature
N: armature rotation in rpm
E: emf induced in any parallel path in armature
Generated emf EA=emf generated in any one of the parallel paths
Average emf generated/conductor=dϕ/dt volt
Flux cut/conductor in one revolution dϕ=ϕP Wb
Number of revolutions /second=N/60
Time for one revolution dt=60/N second
E.M.F. generated/conductor= dϕ/dt= ϕPN/60 volt

18
For simplex wave-wound generator
Number of parallel paths=2
Number of conductors (in series) in one path=Z/2
𝐸. 𝑀. 𝐹. 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑/𝑝𝑎𝑡ℎ(𝐸𝐴) =
𝜙𝑃𝑁
60
×
𝑍
2
=
𝜙𝑃𝑍𝑁
120
𝑣𝑜𝑙𝑡
For simplex lap-wound generator
Number of parallel paths=P
Number of conductors (in series) in one path=Z/P
𝐸. 𝑀. 𝐹. 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑/𝑝𝑎𝑡ℎ(𝐸𝐴) =
𝜙𝑃𝑁
60
×
𝑍
𝑃
=
𝜙𝑍𝑁
60
𝑣𝑜𝑙𝑡
In general
𝐸𝐴 =
𝜙𝑍𝑁
60
×
𝑃
𝐴
𝑣𝑜𝑙𝑡
where
A=2 for simplex wave-winding
A=P for simplex lap-winding
𝐸𝐴 =
1
2π
×
2π𝑁
60
× 𝜙𝑍 ×
𝑃
𝐴
=
𝑍𝑃
2π𝐴
𝜙𝜔𝑚 𝑣𝑜𝑙𝑡 Where 𝜔𝑚 =
2π𝑁
60
For a given DC machine Z,P and A are constant
𝐸𝐴 = 𝑘𝜙𝜔𝑚 𝑣𝑜𝑙𝑡 Where 𝑘 =
𝑍𝑃
2π𝐴

19
Example: A four pole generator, having wave wound armature winding has 51
slots, each slot containing 20 conductor. What will be the voltage generated in
the machine when driven at 1500 rpm assuming the flux per pole to be
7mWb?
𝐸𝐴 =
𝜙𝑍𝑁
60
×
𝑃
𝐴
=
7 × 10−3 × 51 × 20 × 1500
60
×
4
2
= 357𝑣𝑜𝑙𝑡
Example: An 8 pole Dc generator has 500 armature conductors, and a useful
flux of 0.05Wb per pole. what will be the emf generated if it is lap-connected
and runs at 1200 rpm? What must be the speed at which it is to be driven
produce the same emf if it is wave-wound?
𝐸𝐴 =
𝜙𝑍𝑁
60
×
𝑃
𝐴
=
0.05 × 500 × 1200
60
×
8
8
= 500𝑣𝑜𝑙𝑡
With lap-wound, P=A=8
With wave-wound, P=8, A=2
𝐸𝐴 =
𝜙𝑍𝑁
60
×
𝑃
𝐴
=
0.05 × 500 × 𝑁
60
×
8
2
= 500 → 𝑁 = 300𝑟𝑝𝑚

20
Example: A four pole lap-connected armature of a DC shunt generator is
required to supply the loads connected in parallel:
(a) 5kW Geyser at 250 V and (b) 2.5kW lighting load also at 250V.
The generator has an armature resistance 0.2Ω and a field resistance of 250Ω.
The armature has 120 conductors in the slots and runs at 1000 rpm. Allowing
1V per brush for contact drops, find
(1) Flux per pole, (2) armature current per parallel path
𝐸𝐴 =
𝜙𝑍𝑁
60
×
𝑃
𝐴
=
𝜙 × 120 × 1000
60
×
4
4
= 258.2𝑣𝑜𝑙𝑡 → 𝜙 = 129.1𝑚𝑊𝑏
With lap-wound, P=A=4
𝐼𝐿 =
5000 + 2500
250
= 30𝐴
𝐼𝐹 =
250
250
= 1𝐴
𝐼𝐴 = 𝐼𝐿 + 𝐼𝐹 = 30 + 1 = 31𝐴
𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴𝑅𝐴 + 𝑏𝑟𝑢𝑠ℎ𝑒𝑠 𝑑𝑟𝑜𝑝 = 250 + 31 × 0.2 + 2 × 1 = 258.2V
EA
-
+
RA=0.2Ω
IA IL
-
+
LF
IF
V
T
=250V
RF=250Ω
5kW
Geyser
2.5kW
lighting
(2)
(1)
Armature current per parallel path=31/4=7.75A

21
Example: A separately excited DC generator, when running at 1000 rpm
supplied 200A at 125V. What will be the load current when the speed drops to
800 rpm if IF is unchanged? Given that the armature resistance 0.04Ω and
brush drop 2V.
𝐸𝐴2 = 𝑉𝑇2 + 𝐼𝐴2𝑅𝐴 + 𝑏𝑟𝑢𝑠ℎ𝑒𝑠 𝑑𝑟𝑜𝑝
𝐸𝐴1 = 𝑉𝑇1 + 𝐼𝐴𝑅𝐴 + 𝑏𝑟𝑢𝑠ℎ𝑒𝑠 𝑑𝑟𝑜𝑝 = 125 + 200 × 0.04 + 2 = 135V
𝑁𝐴1 = 1000𝑟𝑝𝑚
𝐸𝐴2 = 𝐸𝐴1
𝑁𝐴2
𝑁𝐴1
= 135
800
1000
= 108𝑉
𝑅𝑙𝑜𝑎𝑑 =
125
200
= 0.625Ω
108 = 𝐼𝐴2 × 0.625 + 𝐼𝐴2 × 0.04 + 2
𝑉𝑇2 = 𝐼𝐴2𝑅𝑙𝑜𝑎𝑑
𝐼𝐴2 =
108 − 2
0.625 + 0.04
= 159.4A
𝑉𝑇2 = 𝐼𝐴2𝑅𝑙𝑜𝑎𝑑 = 159.4 × 0.625 = 99.6𝑉
EA2
-
+ RA
IL=159.4A
-
+
RF
LF
-
+
VF
IF
V
T2
=99.6V
EA1
-
+ RA
IL=200A
-
+
RF
LF
-
+
VF
IF
V
T1
=125V

22
Total Loss in a DC Generator
(A) Copper Losses
(i) Armature copper losses=Ia
2Ra
This loss is about 30-40% of full load losses.
(ii) Field copper loss:
In case of shunt generator, field copper losses=IF
2RF
In case of shunt generator, field copper losses=IL
2Rs
(iii) The loss due to brush contact resistance.
(B) Magnetic (Iron or Core) Losses
(i) Hysteresis Loss, 𝑾𝒉 ∝ 𝑩𝒎𝒂𝒙
𝟏.𝟔
𝒇
(ii) Eddy Current Loss, 𝑾𝒆 ∝ 𝑩𝒎𝒂𝒙
𝟐
𝒇𝟐
These losses are practically constant for shunt and compound wound
generators, because in their case, field current is approximately constant.
(C) Mechanical Losses
(i) Friction Loss at bearing and commutator.
(ii) Air Friction or Windage Loss of rotating armature

23
Total Loss in a DC Generator
Total
Losses
Copper Losses
Armature Cu Loss
Shunt Cu Loss
Series Cu Loss
Iron Losses
Hysteresis Loss
Eddy Current Loss
Mechanical Losses
Friction Loss
Air Friction or Windage Loss
Stray Losses
Iron and mechanical losses are collectively known as Stray (Rotational) losses.
Constant or Standing Losses
Field Cu losses is constant for shunt and compound generators. Stray losses
and shunt Cu loss are constant in their case. These losses are together known
as Constant or Standing Losses (Wc).

24
Power Stages and Efficiency
Mechanical Efficiency
𝜂𝑚 =
𝑇𝑜𝑡𝑎𝑙 𝑤𝑎𝑡𝑡𝑠 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑖𝑛 𝑎𝑟𝑚𝑎𝑡𝑢𝑟𝑒
𝑀𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑝𝑜𝑤𝑒𝑟 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑
× 100% =
𝐸𝐴𝐼𝐴
𝑂𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑑𝑟𝑖𝑣𝑖𝑛𝑔 𝑒𝑛𝑔𝑖𝑛𝑒
× 100%
Electrical Efficiency
𝜂𝑒 =
𝑊𝑎𝑡𝑡𝑠 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑖𝑛 𝑙𝑜𝑎𝑑 𝑐𝑖𝑟𝑢𝑖𝑡
𝑇𝑜𝑡𝑎𝑙 𝑤𝑎𝑡𝑡𝑠 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑖𝑛 𝑎𝑟𝑚𝑎𝑡𝑢𝑟𝑒
× 100% =
𝑉𝐼𝐿
𝐸𝐴𝐼𝐴
× 100%
Overall or Commercial Efficiency
𝜂𝑐 = 𝜂𝑚 × 𝜂𝑒 =
𝑊𝑎𝑡𝑡𝑠 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑖𝑛 𝑙𝑜𝑎𝑑 𝑐𝑖𝑟𝑢𝑖𝑡
𝑀𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑝𝑜𝑤𝑒𝑟 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑
× 100% =
𝑉𝐼𝐿
× 100%

25
Example: A shunt generator delivers 195A at terminal voltage of 250V. The
armature resistance and shunt field resistance are 0.02Ω and 50Ω respectively.
The iron and friction losses equal 950W. Find
(a) emf generated (b) Cu losses (c) output of the prime motor
(d) commercial, mechanical and electrical efficiencies.
(a) 𝐼𝑓 =
250
50
= 5𝐴
𝐼𝐴 = 𝐼𝐹 + 𝐼𝐿 = 5 + 195 = 200𝐴
𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴𝑅𝐴 = 250 + 200 × 0.02 = 254V
(b) 𝐴𝑟𝑚𝑎𝑡𝑢𝑟𝑒 𝐶𝑢 𝑙𝑜𝑠𝑠 = 𝐼𝐴
2
𝑅𝐴 = 2002 × 0.02 = 800𝑊
𝑆ℎ𝑢𝑛𝑡 𝐶𝑢 𝑙𝑜𝑠𝑠 = 𝐼𝑓
2
𝑅𝑓 = 52 × 50 = 1250𝑊
𝑇𝑜𝑡𝑎𝑙 𝐶𝑢 𝑙𝑜𝑠𝑠 = 800 + 1250 = 2050𝑊
(c) Stray losses=950W
Total losses=950+2050=3000W
𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑜𝑢𝑡𝑝𝑢𝑡 = 𝑉𝐼𝐿 = 250 × 195 = 48750𝑊
𝑂𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑖𝑚𝑒 𝑚𝑜𝑡𝑜𝑟 = 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑖𝑛𝑝𝑢𝑡

26
𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑖𝑛𝑝𝑢𝑡 = 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑜𝑢𝑡𝑝𝑢𝑡 + 𝑡𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠𝑒𝑠 = 48750 + 3000 = 51750𝑊
𝑂𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑖𝑚𝑒 𝑚𝑜𝑡𝑜𝑟 = 51750𝑊
𝜂𝑚 =
𝐸𝐴𝐼𝐴
× 100% =
50800
51750
× 100% = 98.2%
𝜂𝑒 =
𝑉𝐼𝐿
𝐸𝐴𝐼𝐴
=
48750
50800
× 100% = 95.9%
𝜂𝑐 =
𝑉𝐼𝐿
× 100% =
48750
51750
× 100% = 94.2%
(c) 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑝𝑜𝑤𝑒𝑟(𝐸𝐴𝐼𝐴) = 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑖𝑛𝑝𝑢𝑡 − 𝑠𝑡𝑟𝑎𝑦 𝑙𝑜𝑠𝑠
𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑝𝑜𝑤𝑒𝑟(𝐸𝐴𝐼𝐴) = 51750 − 950 = 50800𝑊

27
Example: A shunt generator has a full load current of 196 A at 220V. The stray
lassos are 720W and the shunt field coil resistance is 55Ω. If it has full load
efficiency of 88%, find the armature resistance.
𝑆ℎ𝑢𝑛𝑡 𝐶𝑢 𝑙𝑜𝑠𝑠 = 𝐼𝑓𝑉 = 4 × 220 = 880𝑊
Constant losses=Shunt Cu losses+stray losses=880+720=1600W
𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑜𝑢𝑡𝑝𝑢𝑡 = 𝑉𝐼𝐿 = 220 × 196 = 43120𝑊
𝜂𝑒 =
𝑉𝐼𝐿
𝐸𝐴𝐼𝐴
× 100% = 88% → 𝐸𝐴𝐼𝐴 = 43120 ÷ 0.88 = 49000𝑊
𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠𝑒𝑠 = 49000 − 43120 = 5880𝑊
𝐼𝑓 = 220 ÷ 55 = 4𝐴
𝐼𝐴 = 𝐼𝐿 + 𝐼𝑓 = 195 + 4 = 199𝐴
Total losses=Armature losses + Constant losses=𝐼𝐴
2
𝑅𝐴+1600=5880
𝐼𝐴
2
𝑅𝐴 = 5880 − 1600 = 4280𝑊
𝑅𝐴 = 4280 ÷ 1992 = 0.108Ω

28
Voltage Regulation
Example: A 4-pole shunt DC generator is delivering 20A to a load of 10Ω. If
the armature resistance is 0.5 Ω and the shunt field resistance is 50 Ω,
calculate the induced emf and the efficiency of the machine. Allow a drop of
1V per brush.
𝑇𝑒𝑟𝑚𝑖𝑛𝑎𝑙 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 = 𝐼𝐿𝑅 = 20 × 10 = 200𝑉
𝜂𝑒 =
𝑉𝐼𝐿
𝐸𝐴𝐼𝐴
× 100% =
200 × 20
214 × 24
× 100% = 77.9%
𝐼𝑓 = 200 ÷ 50 = 4𝐴 𝐼𝐴 = 𝐼𝐿 + 𝐼𝑓 = 20 + 4 = 24𝐴
𝐼𝐴𝑅𝐴 = 24 × 0.5 = 12𝑉
𝐸𝐴 = 𝐼𝐴𝑅𝐴 + 𝑉 + 𝑏𝑟𝑢𝑠ℎ 𝑑𝑟𝑜𝑝 = 12 + 200 + 2 = 214𝑉
The voltage regulation (VR) is defined as the difference between the no-load
terminal voltage (VNL) to full load terminal voltage (VFL) and is expressed as
a percentage of full load terminal voltage. It is therefore can be expressed as,
𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑅𝑒𝑔𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑉𝑅 =
𝑉𝑁𝐿 − 𝑉𝐹𝐿
𝑉𝐹𝐿
× 100% =
𝐸𝐴 − 𝑉𝐹𝐿
𝑉𝐹𝐿
× 100%
𝑉𝑅 =
𝐸𝐴 − 𝑉𝐹𝐿
𝑉𝐹𝐿
× 100% =
214 − 200
200
× 100% = 7%

29
Uses of DC Generators
• Shunt generators with field regulators are used for ordinary
lighting and power supply purposes. They are also used for
charging batteries because their terminal voltages are almost
constant.
Shunt Generators
• Series generators are used as boosters in a certain types of
distribution systems particularly in railway service.
Series Generators
• The cumulatively compound generator is the most used DC
generator because its external characteristics can be adjusted
for compensating the voltage drop in the line resistance.
Cumulatively compound generators are used for motor driving
which require DC supply at constant voltage, for lamp loads
and for heavy power service such as electric railways.
• The differential compound DC generator has an external
characteristic similar to that of shunt generator but with large
demagnetization armature reaction. Differential compound DC
generators re widely used in arc welding where larger voltage
drop is desirable with increase in current.
Compound Generators

6 DC machimes- DC generators.pdf

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