COMPOUND DC GENERATOR
Lecture # 5
C
OMPOUND
DC
G
ENERATOR
32
C
OMPOUND
DC
G
ENERATOR
o
When the series field is used along with the shunt field, we have
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C
OMPOUND
DC
G
ENERATOR
33
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:
1.
D
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y
.
2.
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)
.
1.
D
i
f
f
e
r
e
n
t
i
a
l
C
o
m
p
o
u
n
d
D
C
G
e
n
e
r
a
t
o
r
:
o
As
the
current
I
passes through the series field,
the total
flux will go down
rapidly since the polarities of the series and shunt fields
oppose each
other
.
Therefore,
E
g
will
decrease.
o
As
load increases, the
voltage
will drop sharply because of the
flux
decrease.
o
It is good for
constant-current
applications.
C
OMPOUND
DC
G
ENERATOR
34
2.
C
u
m
u
l
a
t
i
v
e
C
o
m
p
o
u
n
d
D
C
G
e
n
e
r
a
t
o
r
:
o
Both the
shunt field and series field have the
same polarity
and the
series
field aids the shunt field (add
flux
together).
o
The flux
will not
decrease as sharp as it did with
differential
compound
generator.
o
T
y
p
e
s
o
f
C
u
m
u
l
a
t
i
v
e
C
o
m
p
o
u
n
d
D
C
G
e
n
e
r
a
t
o
r
:
•
O
v
e
r
c
o
m
p
o
u
n
d
i
n
g
:
t
h
e
f
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l
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-
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o
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h
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n
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r
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s
f
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d
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s
s
t
r
o
n
g
.
•
F
l
a
t
c
o
m
p
o
u
n
d
i
n
g
:
t
h
e
f
u
l
l
-
l
o
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d
v
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t
a
g
e
i
s
r
e
l
a
t
i
v
e
l
y
e
q
u
a
l
t
o
no-load
voltage.
•
U
n
d
e
r
c
o
m
p
o
u
n
d
i
n
g
:
t
h
e
f
u
l
l
-
l
o
a
d
v
o
l
t
a
g
e
i
s
l
e
s
s
t
h
a
n
t
h
e
no-load
voltage.
C
OMPOUND
DC
G
ENERATOR
36
a)
The load
current:
P
o
=
V
t
.
I
L
,
then
I
L
=
P
o
/V
t
= 5000/125 =
40A
b)
The field
current:
I
f
= V
f
/R
f
= V
t
/R
f
= 125/125 =
1A
c)
The
armature current:
[KCL]
I
a
=
I
L
+
I
f
=
41A
S
o
l
.
E
x
a
m
p
l
e
6
:
A long shunt compound generator rated
5-KW,
125-V
has an
efficiency
of
80%
when supplying rated load. If R
f
=
125
,
R
a
=
0.2
,
and R
s
=
0.05
,
find at
full-
load:
+
V
f
_
C
OMPOUND
DC
G
ENERATOR
37
E
x
a
m
p
l
e
6
:
A long shunt compound generator rated
5-KW,
125-V
has an
efficiency
of
80%
when supplying rated load. If R
f
=
125
,
R
a
=
0.2
,
and R
s
=
0.05
,
find at
full-
load:
S
o
l
.
d)
The copper
loss:
P
cu
R
f
I
2
R
R
I
2
f
a
s
a
P
cu
125x1
2
(0.2
0.05)x41
2
545.25
W
e)
The stray power
loss:
P
o
P
st
P
cu
P
conv
%
P
o
100
P
i
P
o
100
5000
100
6250
W
P
i
80
P
i
=
P
o
+
P
cu
+
P
st
P
st
=
P
i
–
P
o
–
P
cu
P
i
P
st
=
6250
–
5000
–
545.25 = 704.75
W
C
OMPOUND
DC
G
ENERATOR
38
a)
The load
current:
P
o
=
V
t
.
I
L
,
then
I
L
=
P
o
/
V
t
= (3
10
3
)/200 = 15
A
b)
The shunt
field current:
Solve for
V
f
:
[KVL]
Loop
1
–
V
f
+
R
s
I
L
+
V
t
=
0
,
then
V
f
=
R
s
I
L
+
V
t
f
V = (0.2
15) + 200=203
V
I
f
=
V
f
/ R
f
= 203/100 = 2.03
A
c)
The
armature current:
[KCL]
I
a
=
I
L
+
I
f
= 15+2.03=17.03
A
S
o
l
.
+
V
f
E
x
a
m
p
l
e
7
:
A short shunt compound generator rated
3-KW,
200-V
has
stray
losses of
120-W
at full load. If R
f
=
100
, R
a
=
0.9
, and
R
s
=
0.2
, find
at full-
load:
L
o
o
p
2
L
o
o
p
1
C
OMPOUND
DC
G
ENERATOR
P
cu
R
f
I
2
R
I
2
R
I
2
f
a
a
s
L
P
cu
412
261
45
718
W
g)
The
efficiency:
P
i
=
P
conv
+
P
st
P
i
=
3718
+
120
=
3838
W
%
P
o
100
3000
100
78.2
%
P
i
3838
or
P
i
P
o
P
st
P
cu
P
conv
E
x
a
m
p
l
e
7
:
A short shunt compound generator rated
3-KW,
200-V
has
stray
losses of
120-W
at full load. If R
f
=
100
, R
a
=
0.9
, and
R
s
=
0.2
, find
at full-
load:
S
o
l
.
39
E
g
I
a
R
a
V
f
E
g
17.03
0.9
203
218.33
V
e)
The converted
power:
P
conv
=
E
g
I
a
=218.33
17.03 =
3718W
f)
The total copper
loss
P
cu
P
conv
P
o
P
cu
3
718
3.0
10
3
718W
d)
The generated :
[KVL]
Loop
2
E
g
I
a
R
a
V
f
0
,
then
S
OLVED
P
ROBLEMS
a)
The
armature
current:
I
a
=
I
L
+
I
f
=
V
t
/I
L
+
V
f
/I
f
;
V
f
=
V
t
= (250/12.5) + (250/250) =
21A
b)
The induced emf:
(
Loop
1
)
E
g
=
R
a
I
a
+
V
t
E
g
= (0.24
21) +
250= 255
V
c)
The flux per
pole:
2
2
255
9.83
mWb
778
8
52.36
N
syn
2
500
2
52.36
rad
/
s
60
60
S
o
l
.
+
V
f
E
x
a
m
p
l
e
8
:
An 8-pole dc shunt generator with
778
wave-connected
armature
conductors and
running at 500rpm supplies a load of 12.5 Ω
resistance
at
terminal
voltage of
250
V.
The
armature
resistance is
0.24Ω
and the
field
resistance is
250
Ω. Find
a) the
armature
current,
b)
the
induced
emf
and c) the flux per
pole.
2
a
E
g
Z
P
E
g
2
a
Z
P
L
o
o
p
1
49
S
OLVED
P
ROBLEMS
41
turns per coil are 8. Knowing that the
armature
is
simplex
wave-wound.
a) Since the induced voltage
E
g
is
more
than
the terminal
voltage
V
t
;
[E
g
>
V
t
]
,
then
the dc
machine
is working as a
generator.
g
E
N
P
N
a
E
g
a
P
2
240
N
400
4
10
10
3
94.25
a
b
)
I
R
a
E
g
V
t
240
200
100
A
0.2
Since there are 8 turns in a coil, it
means
there are 16
active conductors/coil.
Hence,
the
Number of Coils
= 400/8 = 50
coils.
S
o
l
.
c)
N
syn
2
900
2
94.25
rad
/
s
60
60
a
2
Plex
2
1
2
pathes
E
x
a
m
p
l
e
9
:
A 4-pole,
900
rpm dc
machine
has a
terminal
voltage of
220
V and an induced
voltage of
240
V at rated speed. The
armature
circuit resistance is 0.2
Ω.
a) Is the
machine
operating as a generator or a
motor
? b) Compute the
armature
current
and
c) the number
of
armature
coils if the
air-gap
flux/pole is 10
mWb
and the
armature
1)
A generator
turning
at
200 rad/s
develops
252V. As
a result
of
a
short
in the
field
winding,
the
flux
drops
by
30%. This
causes
a drop in
voltage. How fast should it
turn to
restore
the
voltage to
its
original
value?
2)
A
220
V,
10
kW
generator
has
255
V
measured
at
its
terminal
under
no-load
condition.
What
is the percent voltage
regulation?
3)
A
1
kW,
120
V
generator
has
a
full-load regulation
of 18%. What
voltage
should be
measured
at
its
terminals at
no-load?
4)
A
5
kW,
230
V
generator
requires
9
hp
when
supplying
rated
load.
What
is
the
full
load
efficiency?
5)
A
shunt
generator whose
no-load
voltage is
480
V is
connected
to
a 440
V bus.
If
R
a
= 0.6
and
R
f
= 165
,
find the power it
will
supply to the
bus?
6)
A
100 kW, 600
V
short shunt
compound
generator
has R
f
= 300
,
R
a
= 0.1
, R
s
=
0.05
,
and constant stray power
losses
of 2.4
kW.
Find the efficiency at
full-load.
Extra
Problems
42
7)
For
the
generator characteristics shown
in
the figure below, calculate the percent
voltage
regulation
for
each generator
at
full-load current of
12A.
Extra
Problems
43
150
100
50
250
200
400
350
300
0
3
6
9
12
15
V
t
[V]
I
L
[A]
Gen.
1
Gen.
2
Gen.
3
Gen.
4
Exploring the operation and characteristics of a compound DC generator. This lecture delves into the design, working principles, and applications of this type of generator in various industries. Gain insights into the unique features that distinguish a compound DC generator from other types, including its efficiency and performance. Understand how it generates electrical power and the benefits it offers for different power generation needs. Dive into the intricacies of its components and the role they play in producing electricity consistently and reliably.
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Presentation Transcript
Lecture # 5 COMPOUND DC GENERATOR
COMPOUND DC GENERATOR o When the series field is used along with the shunt field, we havea compound dc generator. o The purpose of the compound generator is to improve theload characteristics of the shunt generator as loadincreases. (a) Long Shunt (b) Short Shunt Types of compound dc generators 32
COMPOUND DC GENERATOR Two types of compound dc generators: 1.Differential compound generator: the series field polarity opposes the shunt field polarity. 2.Cumulative compound generator: both the shunt field and series field have the same polarity and the series field aids the shunt field (add flux together). 1.Differential Compound DC Generator: o As the current I passes through the series field, the total flux will go down rapidly since the polarities of the series and shunt fields oppose eachother. Therefore, Eg will decrease. o As load increases, the voltage will drop sharply because of the flux decrease. o It is good for constant-currentapplications. 33
COMPOUND DC GENERATOR 2. Cumulative Compound DC Generator: o Both the shunt field and series field have the same polarity and theseries field aids the shunt field (add flux together). o The flux will not decrease as sharp as it did with differentialcompound generator. o Types of Cumulative Compound DC Generator: Over compounding: the full-load voltage is larger than no-loadvoltage when the series field is strong. Flat compounding: the full-load voltage is relatively equal to no-load voltage. Under compounding: the full-load voltage is less than the no-load voltage. 34
COMPOUND DC GENERATOR Example 6: A long shunt compound generator rated 5-KW, 125-V has an efficiency of 80% when supplying rated load. If Rf = 125 , Ra = 0.2 , and Rs = 0.05 , find at full- load: Sol. a) The load current: Po = Vt. IL IL = Po/Vt = 5000/125 =40A + , then Vf b) The field current: If = Vf/Rf = Vt/Rf = 125/125 =1A _ c) The armature current:[KCL] Ia= IL + If=41A 36
COMPOUND DC GENERATOR Example 6: A long shunt compound generator rated 5-KW, 125-V has an efficiency of 80% when supplying rated load. If Rf = 125 , Ra = 0.2 , and Rs = 0.05 , find at full- load: Sol. d)The copper loss: Pcu = Rf I2+ (R + R )I2 f a s a Pcu=125x12+ (0.2 + 0.05)x412= 545.25W e) The stray powerloss: % =Po 100 Pi =Po 100 =5000 100 = 6250 W Pi Pi= Po+ Pcu+ Pst Pst= Pi Po Pcu 80 Po Pi Pconv Pst = 6250 5000 545.25 = 704.75W Pst Pcu 37
COMPOUND DC GENERATOR Example 7: A short shunt compound generator rated 3-KW, 200-V has stray losses of120-W at full load. If Rf = 100 , Ra = 0.9 , and Rs = 0.2 , find at full- load: Sol. a) The load current: + Po = Vt. IL , then Loop 2 Vf Loop 1 IL = Po / Vt = (3 103)/200 = 15A b) The shunt field current: Solve for Vf : [KVL] Loop1 If = Vf / Rf = 203/100 = 2.03A Vf + Rs IL +Vt = 0 , then c) The armature current:[KCL] Vf = Rs IL +Vt V = (0.2 15) + 200=203V Ia = IL + If = 15+2.03=17.03 A f 38
COMPOUND DC GENERATOR Example 7: A short shunt compound generator rated 3-KW, 200-V has stray losses of120-W at full load. If Rf = 100 , Ra = 0.9 , and Rs = 0.2 , find at full- load: Sol. d) The generated : [KVL] Loop 2 Eg + Ia Ra+ Vf= 0 , then Pi Po Pconv Pst Pcu Eg = Ia Ra + Vf Eg= (17.03 0.9) + 203 = 218.33 V e) The converted power: P conv = EgIa =218.33 17.03 = 3718W f) The total copperloss Pcu = Pconv Po Pcu= 3718 3.0 103= 718W or Pcu= RfI2+ R I2+ R I2 a f a s L Pcu= 412 + 261+ 45 = 718W g) The efficiency: Pi = Pconv +Pst Pi = 3718 + 120 = 3838W % =Po 100 =3000 100 = 78.2% Pi 3838 39
SOLVED PROBLEMS Example 8: An 8-pole dc shunt generator with 778 wave-connected armature conductors and running at 500rpm supplies a load of 12.5 resistance at terminal voltage of 250V. The armature resistance is 0.24 and the field resistance is 250 . Find a) the armature current, b) the induced emf and c) the flux perpole. Sol. a) The armature current: Ia = IL + If = Vt/IL+ Vf/If ; Vf =Vt = (250/12.5) + (250/250) =21A b)The induced emf: (Loop 1) Eg = Ra Ia + Vt Eg = (0.24 21) + 250= 255V c) The flux perpole: = Nsyn2 = 500 2 = 52.36 rad/s 60 + Loop 1 Vf 2 a Eg Z P Z P 2 a Eg= = 2 2 255= 9.83mWb 778 8 52.36 = 60 49
SOLVED PROBLEMS Example 9: A 4-pole, 900 rpm dc machine has a terminal voltage of 220 V and an induced voltage of 240 V at rated speed. The armature circuit resistance is 0.2 . a) Is the machine operating as a generator or a motor ? b) Compute the armature currentand c) the number of armature coils if the air-gap flux/pole is 10 mWb and thearmature turns per coil are 8. Knowing that the armature is simplexwave-wound. Sol. a) Since the induced voltage Eg is more than the terminal voltage Vt ; [Eg > Vt], then the dc machine is working as a generator. N = aEg E =N P a Eg Vt g =240 200=100A 0.2 P = b) I a Ra 2 240 4 10 10 3 94.25 Since there are 8 turns in a coil, it means there are 16 active conductors/coil.Hence, the Number of Coils = 400/8 = 50coils. N = = 400 c) = Nsyn 2 = 900 2 = 94.25 rad /s 60 a = 2 Plex = 2 1= 2 pathes 60 41
Extra Problems 1) A generator turning at 200 rad/s develops 252V. As a result of a short in the field winding, the flux drops by 30%. This causes a drop in voltage. How fast should it turn to restore the voltage to its original value? 2) A 220 V, 10 kW generator has 255 V measured at its terminal under no-load condition. What is the percent voltage regulation? A 1 kW, 120 V generator has a full-load regulation of 18%. What voltage should be measured at its terminals at no-load? 3) 4) A 5 kW, 230 V generator requires 9 hp when supplying rated load. What is the full load efficiency? 5) A shunt generator whose no-load voltage is 480 V is connected to a 440 V bus. If Ra = 0.6 and Rf = 165 , find the power it will supply to thebus? 6) A 100 kW, 600 V short shunt compound generator has Rf= 300 , Ra= 0.1 , Rs= 0.05 , and constant stray power losses of 2.4 kW. Find the efficiency at full-load. 42
Extra Problems 7) For the generator characteristics shown in the figure below, calculate the percent voltage regulation for each generator at full-load current of 12A. 400 Gen.1 350 Gen.2 300 250 Gen.3 Vt[V] 200 150 Gen.4 100 50 0 3 6 9 12 15 IL[A] 43
