Electric Machines I
DC Machines - DC Generators
Dr. Firas Obeidat
1
Working of Simple Loop Generator
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.
6
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
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 commutativelycompound. In series field oppose the shunt field, the generator is said to
be differentially compounded.
7
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Types of DC Generators
Separately Excited Generators
Short Shunt Generators
Shunt Wound Generators
Series Wound Generators
Long Shunt Generators
8
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Types of DC Generators
9
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
The Terminal Characteristic of a Separately Excited DC Generator
For Separately Excited DC Generator
𝐼𝐴 = 𝐼𝐿
IF
+
𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 𝑅𝐴
VF
𝑉𝐹 = 𝐼𝐹 𝑅𝐹
-
𝐸𝐴 = 𝑘ϕ𝜔𝑚
The terminal voltage can be controlled by:
1. Change the speed of rotation: If 𝜔
increases, then 𝐸𝐴=𝑘 𝜔𝑚 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=𝑘 𝜔𝑚 must rise too, so
𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 𝑅𝐴 increases.
IA
RF
LF
IL
+
+ RA
VT
EA
-
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
: is the flux
𝜔m: is the rotor angular speed
10
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
The Terminal Characteristic of a Self Excited Shunt DC Generator
For Self Excited Shunt DC Generator
𝐼𝐴 = 𝐼𝐹 + 𝐼𝐿
IA
+ RA RF
IF
EA
LF
-
𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 𝑅𝐴
𝑉𝑇 = 𝐼𝐹 𝑅𝐹
IL
+
VT
-
𝐸𝐴 = 𝑘ϕ𝜔𝑚
The terminal voltage can be controlled by:
1. Change the speed of rotation: If 𝜔
increases, then 𝐸𝐴=𝑘 𝜔𝑚 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=𝑘 𝜔𝑚 must rise too, so
𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 𝑅𝐴 increases.
11
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
The Terminal Characteristic of a Self Excited Series DC Generator
For Self Excited Series DC Generator
𝐼𝐴 = 𝐼𝑠 = 𝐼𝐿
IA
+ RA
EA
-
𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 (𝑅𝐴 +𝑅𝑠 )
𝐸𝐴 = 𝑘ϕ𝜔𝑚
Is
Rs
IL
Ls +
VT
-
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.
12
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
The Terminal Characteristic of Cumulatively Compound DC Generator
For Long Shunt Cumulatively Compound DC Generator
IA
𝐼𝐴 = 𝐼𝐹 + 𝐼𝐿
+ RA
EA
-
𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 (𝑅𝐴 +𝑅𝑠 )
𝑉𝑇 = 𝐼𝐹 𝑅𝐹
𝐸𝐴 = 𝑘ϕ𝜔𝑚
IL
Rs
Ls
+
RF
IF
LF
VT
-
For Short Shunt Cumulatively Compound DC Generator
𝐼𝐴 = 𝐼𝐹 + 𝐼𝐿
IA
𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 𝑅𝐴 −𝐼𝐿 𝑅𝑠
+ RA RF
IF
EA
LF
-
𝐸𝐴 = 𝑘ϕ𝜔𝑚
IL
Rs
Ls
+
VT
-
The terminal voltage Cumulatively Compound DC Generator can be controlled by:
1. Change the speed of rotation: If 𝜔 increases, then 𝐸𝐴=𝑘 𝜔𝑚 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=𝑘
𝜔𝑚 must rise too, so 𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 𝑅𝐴 increases.
13
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Examples
Example: A shunt DC generator delivers 450A at 230V and the resistance of
the shunt field and armature are 50 and 0.3 respectively. Calculate emf.
IA
𝐼𝐴 = 𝐼𝐹 + 𝐼𝐿 = 4.6 + 450 = 454.6𝐴
𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴 𝑅𝐴 = 230 + 454.6 × 0.3 = 243.6V
+ RA RF
IF
EA
LF
-
IL=450A
+
-
VT=230V
230
𝐼𝑓 =
= 4.6𝐴
50
VT=500V
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
0.05 , 0.03 and 250 respectively. Calculate the generated voltage and the
armature current. Allow 1V per brush for contact drop.
500
𝐼𝐹 =
= 2𝐴
250
IA
IL=50A
+
𝐼𝐴 = 𝐼𝐹 + 𝐼𝐿 = 2 + 50 = 52𝐴
+ RA Rs Ls RF
IF
Voltage drop across series winding=𝐼A𝑅s=52×0.03=1.56V EA
LF
Armature voltage drop=𝐼A𝑅A=52×0.05=2.6V
Drop at brushes=2 1=2V
𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴 𝑅𝐴 + 𝑠𝑒𝑟𝑖𝑒𝑠 𝑑𝑟𝑜𝑝 + 𝑏𝑟𝑢𝑠ℎ𝑒𝑠 𝑑𝑟𝑜𝑝 = 500 + 2.6 + 1.56 + 2 = 506.16V
14
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Examples
Example: A short shunt compound DC generator delivers a load current of
30A at 220V and has armature, series field and shunt field resistances of
0.05 , 0.3 and 200 respectively. Calculate the induced emf and the
armature current. Allow 1V per brush for contact drop.
Voltage drop across series winding=𝐼L𝑅s
=30 ×0.3=9V
Voltage across shunt winding=220 + 9=229V
IA
𝐼𝐴 𝑅𝐴
= 31.145 × 0.05 = 1.56V
Drop at brushes=2 1=2V
+ RA RF
IF
EA
LF
-
Rs
IL=30A
+
Ls
-
VT=220V
229
𝐼𝐹 =
= 1.145𝐴
200
𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴 𝑅𝐴 + 𝑠𝑒𝑟𝑖𝑒𝑠 𝑑𝑟𝑜𝑝 + 𝑏𝑟𝑢𝑠ℎ𝑒𝑠 𝑑𝑟𝑜𝑝 = 220 + 9 + 1.56 + 2 = 232.56V
15
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Examples
VT=230V
Example: A long shunt compound DC generator delivers a load current of
150A at 230V and has armature, series field and shunt field resistances of
0.032 , 0.015 and 92 respectively. Calculate (i) induced emf (ii) total
power generated and (iii) distribution of this power.
IA
IL=150A
230
(i ) 𝐼 =
+
= 2.5𝐴
𝐹
L
R
R
R
s
A
s
F
92
+
IF
EA
𝐼𝐴 = 𝐼𝐹 + 𝐼𝐿 = 2.5 + 150 = 152.5𝐴
LF
Voltage drop across series winding=
𝐼A𝑅s=152.5 ×0.015=2.2875V
Armature voltage drop=𝐼A𝑅A=152.5 ×0.032=4.88V
𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴 𝑅𝐴 + 𝐼 𝑅 = 230 + 2.2875 + 4.88 = 237.1675V
(ii) Total power generated by the armature=𝐸𝐴 𝐼𝐴 =237.1675 152.5=36168.04375W
(iii) Power lost in armature=𝐼𝐴 2 𝑅𝐴 =152.52 0.032=744.2W
Power dissipated in shunt winding=𝑉𝑇 𝐼𝐹 =230 2.5=575W
Power dissipated in series winding=𝐼𝐴 2 𝑅𝑠 =152.52 0.015=348.84375W
Power delivered to the load=𝑉𝑇 𝐼𝐿 =230 150=34500W
Total power generated by the armature=744.2 + 575 + 348.843 + 34500=36168.04375W
16
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
E.M.F. Equation of DC Generator
Let
: 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=d /dt volt
Flux cut/conductor in one revolution d = P Wb
Number of revolutions /second=N/60
Time for one revolution dt=60/N second
E.M.F. generated/conductor= d /dt= PN/60 volt
17
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
E.M.F. Equation of DC Generator
For simplex wave-wound generator
Number of parallel paths=2
Number of conductors (in series) in one path=Z/2
𝜙𝑃𝑁 𝑍 𝜙𝑃𝑍𝑁
× =
𝑣𝑜𝑙𝑡
60
2
120
For simplex lap-wound generator
𝐸. 𝑀. 𝐹. 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑/𝑝𝑎𝑡ℎ(𝐸𝐴 ) =
Number of parallel paths=P
Number of conductors (in series) in one path=Z/P
𝜙𝑃𝑁 𝑍 𝜙𝑍𝑁
𝐸. 𝑀. 𝐹. 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑/𝑝𝑎𝑡ℎ(𝐸𝐴 ) =
× =
𝑣𝑜𝑙𝑡
60
𝑃
60
In general
where
𝜙𝑍𝑁 𝑃
A=2 for simplex wave-winding
𝐸𝐴 =
× 𝑣𝑜𝑙𝑡
60
𝐴
A=P for simplex lap-winding
1 2π𝑁
𝑃
𝑍𝑃
2π𝑁
Where 𝜔𝑚 = 60
𝐸𝐴 =
×
× 𝜙𝑍 × =
𝜙𝜔𝑚 𝑣𝑜𝑙𝑡
2π
60
𝐴 2π𝐴
For a given DC machine Z,P and A are constant
𝑍𝑃
Where 𝑘 = 2π𝐴
𝐸𝐴 = 𝑘𝜙𝜔𝑚 𝑣𝑜𝑙𝑡
18
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
E.M.F. Equation of DC Generator
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?
𝜙𝑍𝑁 𝑃 7 × 10−3 × 51 × 20 × 1500 4
𝐸𝐴 =
× =
× = 357𝑣𝑜𝑙𝑡
60
𝐴
60
2
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?
With lap-wound, P=A=8
𝜙𝑍𝑁 𝑃 0.05 × 500 × 1200 8
𝐸𝐴 =
× =
× = 500𝑣𝑜𝑙𝑡
60
𝐴
60
8
With wave-wound, P=8, A=2
𝜙𝑍𝑁 𝑃 0.05 × 500 × 𝑁 8
𝐸𝐴 =
× =
× = 500
60
𝐴
60
2
→
𝑁 = 300𝑟𝑝𝑚
19
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
E.M.F. Equation of DC Generator
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 0.2 and a field resistance of 250 .
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
(1)
IA
With lap-wound, P=A=4
RF=250Ω
IF
LF
-
5kW Geyser
2.5kW lighting
+
EA
-
IL
+
VT=250V
5000 + 2500
= 30𝐴
250
250
𝐼𝐹 =
= 1𝐴
250
𝐼𝐿 =
RA=0.2Ω
𝐼𝐴 = 𝐼𝐿 + 𝐼𝐹 = 30 + 1 = 31𝐴
𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴 𝑅𝐴 + 𝑏𝑟𝑢𝑠ℎ𝑒𝑠 𝑑𝑟𝑜𝑝 = 250 + 31 × 0.2 + 2 × 1 = 258.2V
𝐸𝐴 =
𝜙𝑍𝑁 𝑃 𝜙 × 120 × 1000 4
× =
× = 258.2𝑣𝑜𝑙𝑡
60
𝐴
60
4
→
𝜙 = 129.1𝑚𝑊𝑏
(2) Armature current per parallel path=31/4=7.75A
20
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
E.M.F. Equation of DC Generator
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 0.04 and
brush drop 2V.
𝐸𝐴1 = 𝑉𝑇1 + 𝐼𝐴 𝑅𝐴 + 𝑏𝑟𝑢𝑠ℎ𝑒𝑠 𝑑𝑟𝑜𝑝 = 125 + 200 × 0.04 + 2 = 135V
𝑁𝐴1 = 1000𝑟𝑝𝑚
𝑁𝐴2
800
= 𝐸𝐴1
= 135
= 108𝑉
𝑁𝐴1
1000
𝑅𝑙𝑜𝑎𝑑
125
=
= 0.625Ω
200
+
RF
VF
LF
+
EA1
-
RA
-
𝐸𝐴2 = 𝑉𝑇2 + 𝐼𝐴2 𝑅𝐴 + 𝑏𝑟𝑢𝑠ℎ𝑒𝑠 𝑑𝑟𝑜𝑝
108 = 𝐼𝐴2 × 0.625 + 𝐼𝐴2 × 0.04 + 2
108 − 2
𝐼𝐴2 =
= 159.4A
0.625 + 0.04
𝑉𝑇2 = 𝐼𝐴2 𝑅𝑙𝑜𝑎𝑑 = 159.4 × 0.625 = 99.6𝑉
IL=159.4A
VF
-
RF
LF
+
EA2
-
RA
+
VT2=99.6V
𝑉𝑇2 = 𝐼𝐴2 𝑅𝑙𝑜𝑎𝑑
IF
+
+
VT1=125V
𝐸𝐴2
IL=200A
IF
-
21
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Total Loss in a DC Generator
(A) Copper Losses
(i) Armature copper losses=Ia2Ra
This loss is about 30-40% of full load losses.
(ii) Field copper loss:
In case of shunt generator, field copper losses=IF2RF
In case of shunt generator, field copper losses=IL2Rs
This loss is about 20-30% of full load losses.
(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.
This loss is about 20-30% of full load losses.
(C) Mechanical Losses
(i) Friction Loss at bearing and commutator.
(ii) Air Friction or Windage Loss of rotating armature
This loss is about 10-20% of full load losses.
22
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Total Loss in a DC Generator
Armature Cu Loss
Total Losses
Copper Losses
Shunt Cu Loss
Series Cu Loss
Hysteresis Loss
Iron Losses
Eddy Current Loss
Friction Loss
Mechanical Losses
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).
23
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Power Stages and Efficiency
Mechanical Efficiency
𝜂𝑚 =
𝑇𝑜𝑡𝑎𝑙 𝑤𝑎𝑡𝑡𝑠 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑖𝑛 𝑎𝑟𝑚𝑎𝑡𝑢𝑟𝑒
𝐸𝐴 𝐼𝐴
× 100% =
× 100%
𝑀𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑝𝑜𝑤𝑒𝑟 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑
𝑂𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑑𝑟𝑖𝑣𝑖𝑛𝑔 𝑒𝑛𝑔𝑖𝑛𝑒
Electrical Efficiency
𝜂𝑒 =
𝑊𝑎𝑡𝑡𝑠 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑖𝑛 𝑙𝑜𝑎𝑑 𝑐𝑖𝑟𝑢𝑖𝑡
𝑉𝐼𝐿
× 100% =
× 100%
𝑇𝑜𝑡𝑎𝑙 𝑤𝑎𝑡𝑡𝑠 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑖𝑛 𝑎𝑟𝑚𝑎𝑡𝑢𝑟𝑒
𝐸𝐴 𝐼𝐴
Overall or Commercial Efficiency
𝜂𝑐 = 𝜂 𝑚 × 𝜂𝑒 =
𝑊𝑎𝑡𝑡𝑠 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑖𝑛 𝑙𝑜𝑎𝑑 𝑐𝑖𝑟𝑢𝑖𝑡
𝑉𝐼𝐿
× 100% =
× 100%
𝑀𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑝𝑜𝑤𝑒𝑟 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑
𝑂𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑑𝑟𝑖𝑣𝑖𝑛𝑔 𝑒𝑛𝑔𝑖𝑛𝑒
24
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Power Stages and Efficiency
Example: A shunt generator delivers 195A at terminal voltage of 250V. The
armature resistance and shunt field resistance are 0.02 and 50 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)
250
𝐼𝑓 =
= 5𝐴
50
𝐼𝐴 = 𝐼𝐹 + 𝐼𝐿 = 5 + 195 = 200𝐴
𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴 𝑅𝐴 = 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
25
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Power Stages and Efficiency
𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑖𝑛𝑝𝑢𝑡 = 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑜𝑢𝑡𝑝𝑢𝑡 + 𝑡𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠𝑒𝑠 = 48750 + 3000 = 51750𝑊
𝑂𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑖𝑚𝑒 𝑚𝑜𝑡𝑜𝑟 = 51750𝑊
(c) 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑝𝑜𝑤𝑒𝑟(𝐸𝐴 𝐼𝐴 ) = 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑖𝑛𝑝𝑢𝑡 − 𝑠𝑡𝑟𝑎𝑦 𝑙𝑜𝑠𝑠
𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑝𝑜𝑤𝑒𝑟(𝐸𝐴 𝐼𝐴 ) = 51750 − 950 = 50800𝑊
𝐸𝐴 𝐼𝐴
50800
× 100% =
× 100% = 98.2%
𝑂𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑑𝑟𝑖𝑣𝑖𝑛𝑔 𝑒𝑛𝑔𝑖𝑛𝑒
51750
𝜂𝑚 =
𝑉𝐼𝐿
48750
𝜂𝑒 =
=
× 100% = 95.9%
𝐸𝐴 𝐼𝐴 50800
𝜂𝑐 =
𝑉𝐼𝐿
48750
× 100% =
× 100% = 94.2%
𝑂𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑑𝑟𝑖𝑣𝑖𝑛𝑔 𝑒𝑛𝑔𝑖𝑛𝑒
51750
26
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Power Stages and Efficiency
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 55 . If it has full load
efficiency of 88%, find the armature resistance.
𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑜𝑢𝑡𝑝𝑢𝑡 = 𝑉𝐼𝐿 = 220 × 196 = 43120𝑊
𝑉𝐼𝐿
𝜂𝑒 =
× 100% = 88%
→ 𝐸𝐴 𝐼𝐴 = 43120 ÷ 0.88 = 49000𝑊
𝐸𝐴 𝐼𝐴
𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠𝑒𝑠 = 49000 − 43120 = 5880𝑊
𝐼𝑓 = 220 ÷ 55 = 4𝐴
𝑆ℎ𝑢𝑛𝑡 𝐶𝑢 𝑙𝑜𝑠𝑠 = 𝐼𝑓 𝑉 = 4 × 220 = 880𝑊
Constant losses=Shunt Cu losses+stray losses=880+720=1600W
Total losses=Armature losses + Constant losses=𝐼𝐴 2 𝑅𝐴 +1600=5880
𝐼𝐴 2 𝑅𝐴 = 5880 − 1600 = 4280𝑊
𝐼𝐴 = 𝐼𝐿 + 𝐼𝑓 = 195 + 4 = 199𝐴
𝑅𝐴 = 4280 ÷ 1992 = 0.108Ω
27
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Voltage Regulation
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
𝐸𝐴 − 𝑉𝐹𝐿
× 100% =
× 100%
𝑉𝐹𝐿
𝑉𝐹𝐿
Example: A 4-pole shunt DC generator is delivering 20A to a load of 10 . If
the armature resistance is 0.5
and the shunt field resistance is 50 ,
calculate the induced emf and the efficiency of the machine. Allow a drop of
1V per brush.
𝑇𝑒𝑟𝑚𝑖𝑛𝑎𝑙 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 = 𝐼𝐿 𝑅 = 20 × 10 = 200𝑉
𝐼𝑓 = 200 ÷ 50 = 4𝐴
𝐼𝐴 = 𝐼𝐿 + 𝐼𝑓 = 20 + 4 = 24𝐴
𝐼𝐴 𝑅𝐴 = 24 × 0.5 = 12𝑉
𝐸𝐴 = 𝐼𝐴 𝑅𝐴 + 𝑉 + 𝑏𝑟𝑢𝑠ℎ 𝑑𝑟𝑜𝑝 = 12 + 200 + 2 = 214𝑉
𝑉𝐼𝐿
200 × 20
× 100% =
× 100% = 77.9%
𝐸𝐴 𝐼𝐴
214 × 24
𝐸𝐴 − 𝑉𝐹𝐿
214 − 200
𝑉𝑅 =
× 100% =
× 100% = 7%
𝑉𝐹𝐿
200
𝜂𝑒 =
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
28
Uses of DC Generators
Shunt 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.
Series Generators
Series generators are used as boosters in a certain types of
distribution systems particularly in railway service.
Compound 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.
29
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
30