SOLVED PROBLEMS
UNIT-2
EXAMPLE 9.1 PAGE NO: 9.28



A 5KW, 250V, 4 pole, 1500 rpm shunt generator is designed to have
a square pole face. The loadings are: Average flux density in the
gap=0.42Wb/m2 and ampere conductors per metre=15000. Find
the main dimensions of the machine. Assume full load
efficiency=0.87 and ratio of pole arc to pole pitch=0.66
Given datas: Output power P=5KW, Terminal voltage V=250V
No of poles p=4, Speed N=1500rpm, Bav=0.42 Tesla,
ac=30000, Full load efficiency η=0.87, ratio of pole arc to pole pitch
ψ=0.66, DC Shunt Generator
To find Main dimensions D and L for square pole face.
EXAMPLE 9.1 PAGE NO: 9.28

Solution:
P
5 KW
Power developed in the armature Pa  
 5.75KW
 0.87
N 1500
Speed n 

 25 rps
60
60
Output co  efficient Co   2 Bav ac  103   2  0.42  15000  103  62.1
Pole arc
b L

   0.66 ( for square pole face b  L)
Pole pitch  
0.66  
 0.66  L  0.66  L  0.66

D  0.518 D

p
4
L  0.518 D
L
D
Output equation Pa  C0 D 2 Ln
5.75  62.1 D 2  0.518 D  25  D  0.193m and L  0.1m
Example 9.2, Page 9.28


A design is required for a 50KW, 4 pole, 600rpm, DC shunt generator, the
full load terminal voltage being 220V. If the maximum gap density is 0.83
Wb/m2 and the armature ampere conductors per metre are 30000,
calculate suitable dimensions of armature core to give a square pole face.
Assume that the full load armature voltage drop is 3% of the rated
terminal voltage, and that the field current is 1% of rated full load current.
Ratio of pole arc to pole pitch is 0.67.
Given Datas: Output power=50KW, Pole p=4, Speed N=600 rpm,
Full load terminal voltage V=220V, Maximum air gap flux density Bg=0.83T,
ac=30000, Square pole face, DC shunt generator, full load armature
voltage drop IaRa=3% of V, Field current If=1% of IL
To find: Main dimensions D and L
Example 9.2, Page 9.28

Solution:

Bav
 Bav   Bg  0.67  0.83  0.5561T  Bav
Bg
Output Co  efficient C0   2 Bav ac 103
C0   2  0.5561 30000 10 3  167  C0
Speed n 
N 600

 10 rps  n
60 60
3
 220  6.6V (3% of ter min al voltage)
100
E  V  I a Ra  220  6.6  226.6V  E
I a Ra 
Full load current I L 
P 50 KW

 227 A
V
220
1
 227  2.27 A
100
I a  I L  I f ( for DC shunt generator )
Field current I f 
I a  227  2.27  229.27 A  I a
Power developed in the armature Pa  EI a 103
Pa  226.6V  229.27 A 10 3  Pa  51.8 KW
Example 9.2, Page 9.28
Pole arc
b L

   0.67
Pole pitch  
(b  L for square pole face)
D
0.67
L  0.67  L  0.67

D  L  0.526 D
p
4
Output Equation Pa  C0 D 2 Ln
51.8KW  167  D 2  0.526 D 10
D 3  0.0591 m3  D  0.389m and L  0.526  0.389
L  0.21m
Example 9.3, Page: 9.29

Determine the main dimensions, number of poles and the length of
air gap of a 600KW, 500V, 900rpm generator. Assume average gap
density as 0.6 Wb/m2 and ampere conductors per metre as 35000.
The ratio of pole arc to pole pitch is 0.75 and the efficiency is 91%.
The following are the design constraints: Peripheral speed should
not be greater than 40 m/s, Frequency of flux reversals should not
be greater than 50Hz, Current per brush arm should not be greater
than 400A and armature mmf per pole should not be greater than
7500A. The mmf required for air gap is 50% of armature mmf and
gap contraction factor is 1.15.
Example 9.3, Page: 9.29

Given Datas: Output Power P=600KW,Voltage V=500V, Speed N=900 rpm, generator,
Bav=0.6T, ac=35000, ψ=0.75,
Efficiency η=91%, Atg=50% of Ata, Kg=1.15
Output co  efficient C0   2 Bav ac  103
Solution:
C0   2  0.6  35000 103  C0  207
Power developed by armature Pa 
P


600
 Pa  660 KW
0.91
N 900

 n  15 rps
60 60
P
660
D2 L  a 
 0.2126m3
C0 n 207 15
Speed n 
No. of poles (i ) Frequency f 
pn
Hz
2
4 15
 30 Hz
2
6  15
p 6 f 
 45Hz  Choose p  6
2
8 15
p 8 f 
 60 Hz 
2
I
P 600 KW
(ii ) Current per brush arm I b  2 I z  2 a  I a  
 1200 A
p
V
500
p 4 f 
Example 9.3, Page: 9.29
Ia
1200
Current / brush arm I b  2  2
 600 A ( for p  4) 
p
4
Ia
1200
Current / brush arm I b  2  2
 400 A ( for p  6)
p
6
So choose no. of poles p  6
Main dim ensions : 
b


L

 0.75( for square pole face b  L )
0.75 D
 0.75  L  0.75  L 
 L  0.393D

( p  6)
L
D 2 L  0.2126m3
D 2  0.393D  0.2126m3  D  0.8m and L  0.33m
Example 9.3, Page: 9.29

Checks: (i) Peripheral speed :Va   Dn    0.8 15
Va  37.7 m / s (Within specified lim it )
Pole pitch  
D
p

  0.8
6
 0.42
ac  35000  0.42

2
2
ATa  7330 (Within specified lim it )
Armature MMF per pole ATa 
ATg  50% of ATa  ATg 
50
 7330  ATg  3665 A
100
Bav
Bav
0.6

 Bg 

 0.8  Bg
Bg

0.75
MMF required for air gap ATg  800000 K g Bg lg
lg 
ATg
800000 K g Bg
lg  5mm

3665
 lg  0.05m
800000 1.15  0.8
Example 9.4, page: 9.30

Calculate the diameter and length of armature for a 7.5KW, 4 pole,
1000 rpm, 220V shunt motor. Given full load efficiency=0.83;
maximum gap flux density=0.9Wb/m2; specific electric
loading=30000 ampere conductor per metre; field form factor=0.7.
Assume that the maximum efficiency occures at full load and the
field current is 2.5% of rated current.The pole face is square.

Given:Power output P=7.5KW; Pole p=4, Speed N=1000rpm, Voltage
V=220V, Efficiency η=0.83, Max Flux density Bg=0.9 T; ac=30000, Field
form factor Kf=ψ=0.7, Field current If=2.5% of IL, Square pole face, DC

Shunt motor.
Solution:
Example 9.4, page: 9.30
7.5 KW
 9036W

0.83
Total losses at full load  9036W  7500W  1536W
Power input 
P

At full load
Total loss 1536
Copper loss  Iron loss 

 768W
2
2
Power input 9036
Motor current at full load I L 

 41.1A
Voltage
220
2.5
Field current I f  2.5% of I L  I f 
 41.1  I f  1.0275 A
100
Field copper loss  I 2f R  V  I f  220  1.0275  226.05W
Friction and windage losses  768  226  542W
Power developed in armature Pa  P  Friction and windage losses
Pa  7500  542  8.042 KW
Example 9.4, page: 9.30
Average flux density Bav   Bg  0.7  0.9  0.63T
Output co  efficient C0   2 Bav ac  103
C0   2  0.63  30000  103  C0  186.5355
N 1000
Speed n 

 16.6667 rps
60
60
Pole arc
b L
  Kf 
   0.7 ( Square pole face b  L)
Pole pitch  
L

 0.7  L  0.7  L  0.7
D
4
L  0.5498D
Output equation Pa  C0 D 2 Ln  186.5355  D 2  0.5498D  16.6667
8.042  1709.2904 D 3  D 3  0.0047 m3  D  0.1676 & L  0.0921m
Tutorial Problem

Find the main dimensions and the number of poles of a
37KW, 230V, 1400rpm shunt motor, so that a square pole
face is obtained. The average gap density is 0.5 T and the
ampere conductors per metre are 22000. The ratio of
pole arc to pole pitch is 0.7 and the full load efficiency is
90%.Assume field current is 2.5% of full load current.

Ans: C0=108.5656, Pi=41.1111KW, n=23.3333rps
Total loss=4111.1111KW, Iron loss=2055.5556W,
IL=178.744A, If=4.4686A, Field cu loss=1027.7778W, FW
loss=1027.7778W, Pa=38KW, L=0.5498D, D=0.3010m,
L=0.1655m, p=4

Tutorial Problem


For a preliminary design of a 50HP, 230V, 1400 rpm, DC
shunt motor. Calculate the armature diameter and core
length, the number of poles and peripheral speed. Take
Bav=0.5T and ac=25000, efficiency=0.9.
(Assume Field current is 3%of full load current and
L/τ=0.7)
Example 9.5, Page: 9.31




A 150KW, 230V, 500 rpm, DC shunt motor has a square field
coil. Find the number of poles and the main dimensions and
air gap length. Assume average flux density over the pole arc as
0.85Wb/m2, and the ampere conductors per metre as 29000.
The ration of width of pole body to pole pitch is 0.55 and the
ration of pole arc to pole pitch is 0.7. The efficiency is 0.91.
Assume that the MMF required for air gap is 55% of armature
mmf and the gap contraction factor is 1.15.
Given Data: P=150KW, Voltage V=230V, N=500rpm, DC shunt
motor, square pole, Bg=0.85T(over one pole arc, not entire
pole pitch), ac=29000, L/τ=0.55, Ψ=b/τ=0.7, Efficiency
η=91%, Gap contraction factor Kg=1.15,
Atg=55% of Ata
To find: No. of poles p, Main dimensions D,L and lg.
Example 9.5, Page: 9.31

Solution:
Power developed by the armature Pa  P ( L arg e Motor )
Pa  150 KW
N 500
Speed n 

 8.3333 rps
60 60
Bav

 Bav   Bg  0.7  0.85  0.595Wb / m 2  Bav
Bg
C0   2 Bav ac 103   2  0.595  29000 10 3  C0  170.3
f 
f 
f 
f 
f 
pn

2
pn

2
pn

2
pn

2
pn

2
4  8.3333
p4 f 
 f  16.6667 Hz 
2
6  8.3333
p6 f 
 f  25Hz
2
8  8.3333
p 8 f 
 f  33.3332 Hz
2
10  8.3333
p  10  f 
 f  41.6665 Hz
2
12  8.3333
p  12  f 
 f  50 Hz  Taken
2
Example 9.5, Page: 9.31
 Length of the machine L  Width of Pole body  0.55
D
D
L  0.55  L  0.55
 0.55
 L  0.144 D
p
12
Main Dimensions D & L
Pa  C0 D 2 Ln  150 KW  170.3  D 2  0.144 D  8.3333
 D 3  0.7340 m3  D  0.9021m & L  0.1299m
Pole pitch  
D
p

  0.9021
12
Armature MMF per pole ATa 
 0.2362m
ac  29000  0.2362

2
2
ATa  3424.4538 A
Air gap MMF ATg  55% of ATa  0.55  3424.4538  1883.4496 A
ATa  1883.4496 A
ATg  800000 K g Bg lg  1883.4496  800000 1.15  0.85  lg
lg  2.4mm
Example 9.13, Page: 9.49



A 250KW, 500V, 600rpm DC generator is built with an
armature diameter of 0.75m and a core length of 0.3m. The lap
connected armature has 720 conductors. Using the data
obtained from this machine, determine the armature diameter,
core length, number of armature slots, armature conductors
and commutator segments for a 350KW, 440V, 720 rpm, 6
pole DC generator.
Assume a square pole face with ratio f pole arc to pole pitch
equal to 0.66. The full load efficiency is 0.91 and the internal
voltage drop is 4% of rated voltage.
The voltage between adjacent segments should not exceed
15V at no load.
Example 9.13, Page: 9.49

Solution:
250 KW DC Shunt Generator
Power developed by Armature Pa 
P 250

 Pa  274.7253KW
 0.91
N 600

 10rps  n
60 60
Pa
274.7253
Output Co  efficient Co  2 
 162.8  Co
2
D Ln 0.75  0.3 10
Generated voltage Eg  V  I a Ra  Eg  V  4% of V
Speed n 
Eg  500  0.04  500  520V  Eg
Generated Voltage Eg 
 Bav 
Eg  ( A  p )
 DLnZ

( B  DL)nZ
p nZ
p
 av
 Bav 
 p  Bav DL
A
A
 DL
520  6
 0.613Tesla  Bav
  0.75  0.3 10  720
Example 9.13, Page: 9.49
350 KW DC Shunt Generator
Co  162.8
Power developed by Armature Pa 
P 350

 Pa  384.6154 KW
 0.91
Bav  0.613Tesla
N 720

 12rps  n
60 60
P
384.6154
D2 L  a 
 0.1969  D 2 L
Co n 162.8 12
Speed n 
Pole arc (b  L)
D
 0.66  L  0.66  L  0.66
 L  0.3456 D
Pole pitch 
( p  6)
D 2 L  0.1969  D 2 (0.3456 D)  0.1969  D 3  0.5698  D  0.829m & L  0.2865m
Generated voltage Eg  V  I a Ra  Eg  V  4% of V
Eg  440  0.04  440  457.6V  Eg
B  DL 0.613    0.829  0.2865
p
   av

   76.2mWb
 DL
p
6
Eg A Eg
p nZ
Generated Voltage Eg 
Z 

Lap winding A  p
A
p n  n
Eg
457.6
Z

 Z  500
 n 76.2 103 12
Bav 
Example 9.13, Page: 9.49
NUMBER OF SLOTS
No of slots S 
D
Ys
Ys  35mm  S 
whereYs  Slot pitch var ies from 35mm to 25mm
  0.829
3
 74.4109
74 Slots
35 10
  0.829
Ys  25mm  S 
 104.1752 104 Slots
3
25  10
S  74 to 104 slots  equation A
The slots per pole ( S / p ) should lie between 9 to 16 for proper commutation
S
 (9 to 16)  S  p (9 to 16) where p  6 poles
p
S  6(9 to 16)  S  54 to 96 slots  equation B
Combining equation A & B
No of slots S  (74 to 96)
NUMBER OF SLOTS
Example
9.13, Page: 9.49
In a lap winding the no. of slots should be multiple of pole pairs.
Pole pairs
p 6
  3  S  75, 78,81,84,87, 90, 93, 96
2 2
It is desirable that the no. of slots / pole( S / p )  Integer 
1
2
75
1
 12  
6
2

81
1
S  81 
 13 
6
2
 All values are desirable, so S  75,81,87, 93
87
1
S  87 
 14  
6
2
93
1
S  93 
 15  
6
2
S  75 
To reduce flux pulsation loss ,   (no. of slots / pole ) should be integer 
S
1
( S  75,81,87, 93)
1
 Integer   0.66 
 Integer 
p
2
6
2
(87)
1
for S  87  0.66 
 9.57 9  (nearest value), so choose S  87
6
2

1
2
Example 9.13, Page: 9.49
WINDING
Minimum no. of coils required C 
E p 457.6  6

 183.04
15
15
C  183
Z 500

 5.7471 6  Z s
S 87
Total no. of conductor Z  Z s  S  6  87  Z  522
Conductors / Slot Z s 
NUMBER OF COIL SIDES PER SLOT (U ) ( LapWinding )
 p
Zs / U should be Integer (ie divisible byU ) and should be multiples of pole pair    3
2
1
No. of coil C  US
( S  87)
2
1
U  2  C   2  87  87  (as min . no. of coils req C  183)
2
1
U  4  C   4  87  174  (as min . no. of coils req C  183)
2
1
U  6  C   6  87  261  Z s / 6  6 / 6  1(divisableby 'U ' & multiples of 3)
2
1
U  8  C   8  87  348   Z s / 6  8 / 6  1.3333( not divisable)
Example 9.13, Page: 9.49
1
1
C  US   6  87  C  261
2
2
No. of Commutator segments  No. of Armature coils
No. of Commutator segments  261
Diameter of the Commutator Dc  0.7  Armature diameter
Dc  0.7  0.829  0.58m
Commutator pole pitch  c 
 Dc
C

  0.58
261
 0.007m
 c  0.007 m  7 mm  4mm
Z
522
Turns per coil Tc 

 1 ( Single turn coil used )
2C 2  261
Turns per coil , Tc should be Integer.
Example 9.14, Page: 9.51
A 100KW, 500V, 6 pole, 450rpm, DC shunt motor has the
following
data:
Armature
diameter=0.54m,
Length=0.245m, Average flux density 0.55 T, Number of
ducts=2, width of each duct=10mm, stacking factor=0.92.
The commutator diameter is 0.65 of Armature Diameter.
The ratio of pole arc to pole pitch is 0.66, efficiency=0.89.
 Find the no. of armature slots and work out the details of
a suitable armature winding. Assume an armature
voltage drop of 5% of rated voltage and field current 1%
of line current. Check for the following:
1. The slot loading should not exceed 1500A
2. Commutator pitch βc should not be less than 4mm
3. Voltage between adjacent segments not to exceed 15V.

Example 9.14, Page: 9.51

Solution:
TYPE OF WINDING ( DC Shunt motor )
Back EMF Eb  V  I a Rm  V  5% of V ( given)
Back EMF Eb  500  0.05  500  475V
P
100 KW
Input power Pi  
 112.3596 KW

0.89
P 112.3596 KW
Line current I L  i 
225 A
V
500V
Field current I f  1% of Line current I L
Field current I f  0.01 225  2.25 A
Armature current I a  I L  I f  225  2.25  222.75 A
I a 222.75
Current per Parallel path I z  
 111.375 A  200 A
A
2
So wave winding can be used ( A  2) for simple and economical reason
Current per brush arm I b  2 I z  2 111.375 A  222.75A
Example 9.14, Page: 9.51
No. of Conductors :
p
 p  Bav   DL  0.55    0.54  0.245
 DL
450
p  0.2286 ; n 
 7.5 rps
60
p nZ
Back EMF Eb 
( wave winding )  Z
( A  2)
E A
475  2
Z b

554
( p ) n 0.2286  7.5
Bav 
No. of Slots :
The slot pitchYs var ies from 35mm to 25mm & Ys 
No. of Slots S 
D
Ys  35mm  S 
Ys
D
S
, for Ys  35 mm and Ys  25mm
  0.54
35 103
  0.54
Ys  25mm  S 
25 103

48 

 No. of slots S  48 to 68
68

Example 9.14, Page: 9.51
The no. of slots per pole( S / p )  9 to 16
 S  p (9 to 16) p  6
S  6  (9 to 16)  S  54 to 96
Comparing with the previous slot range
S  54 to 68
In wave winding No. of slots should not be the multiples of pole pair
Pole pair p / 2  6 / 2  3
 S  55,56,58,59, 61, 62, 64, 65, 67, 68
S
1
 Integer 
p
2
1
( S  55,56,58,59, 61, 62, 64, 65, 67, 68)
 Integer   0.66
2
6
1
( S  59, 68)
1
 Integer   0.66
 Integer  (nearly )
2
6
2
In order to avoid flux pulsation loss 
S

p
S

p
Example 9.14, Page: 9.51
Winding :Voltage b / w adjacent segment should not exceed 15V
E p 475  6

 190
15
15
No. of coils sides in wave winding should not be multiples of pole pair (3).
Minimum no. of coils required C 
With S  59, Conductor per Slot Z s  Z / S  554 / 59  9.39 10
Now Z  S  Z s  59  10  Z  590
1
No. of coils C  US , where U  no. of coil sides (U  2, 4, 6,8,10,12,...)
2
1
1
U  2  C  2  59  59 :  U  4  C  4  59  118 
2
2
1
1
U  6  C  6  59  177:  U  8  C  8  59  236
2
2
1
U  10  C  10  59  295
2
Minimum no. of coils U  2, 4, 6 ruled out , thus we left withU  8,10
Z
590

 1.25(not Integer ) 
2C 2  236
Z
590
forU  10, C  295, Turns per coil Tc 

 1( Integer ) Single turn coil
2C 2  295
Zs / U  10 / 10  1 ( divisible)
forU  8, C  236, Turns per coil Tc 
Example 9.14, Page: 9.51
With S  68, Conductor per Slot Z s  Z / S  554 / 68  8.1471 8
Now Z  S  Z s  68  8  Z  544
1
No. of coils C  US , where U  no. of coil sides (U  2, 4, 6,8,10,12,...)
2
1
1
U  2  C  2  68  68 :  U  4  C  4  68  136 
2
2
1
U  6  C  6  68  204:  multiples of pole pair (3)
2
1
1
U  8  C  8  68  272 : U  10  C  10  68  340
2
2
Minimum no. of coils U  2, 4, 6 ruled out , thus we left withU  8,10
Z
544

 1( Integer ) Single turn coil
2C 2  272
Z
544
forU  10, C  295, Turns per coil Tc 

 0.8( Not Integer ) 
2C 2  340
Z s / U  8 / 8  1( divisible), C  272, Tc  1, S  68,U  8, Z  544
forU  8, C  272, Turns per coil Tc 
Example 9.14, Page: 9.51
(i ) Slot loading : I z Z s
for Zs  10, Iz  111.375 A  Slot loading  111.375 10
Slot loading  1113.75 A  1500 A
for Zs  8, Iz  111.375 A  Slot loading  111.375  8
Slot loading  891A  1500 A
 Dc
Commutator Pitch  c 
where Dc  0.65  0.54  0.351m
C
  0.351
for C  295   c 
 3.738mm  4mm 
295
  0.351
for C  236   c 
 4.6725mm  4mm so choose S  68
236
Exercise 35: Page: 9.112


Determine the diameter and length of armature core for a
55KW, 110V,1000rpm,4-pole Dc shunt generator, assuming the
specific electric and specific magnetic loadings as 26000 and
0.5tesla respectively. The pole arc should be 70% of pole pitch
and length of the core about 1.1 times the pole arc. Allow 10A
for the field current and assume a voltage drop of 4V for the
armature circuit. Specify the winding used and also determine
suitable values for the number of armature conductors and
the number of slots.
Given Data:
P  55KW , V  110V , N  1000rpm, p  4, ac  26000,
Bav  0.5Wb / m2 , b  0.7 , L  1.1b, I f  10 A, I a Ra  4V
Exercise 35: Page: 9.112

Solution:
Type of Winding
P 55000

 500 A
V
110
Armature current I a  I L  I f  500  10  510 A
Line Current I L 
Ia
A
510
A  2 forWave winding  I z 
 255 A  200 A , so 
2
510
A  P  4 for Lap winding  I z 
 127.5 A  200 A ,
4
Thus we can opt Lap winding for these data
Current per parallel path I z 
Generated voltage Eg  V  I a Ra  110  4  114V
Power developed in Armature Pa  Eg  I a  103
Pa  114  510  103  Pa  58.14 KW
Exercise 35: Page: 9.112
N 1000
Speed , n 

 16.6667 rps
60
60
Output co  efficient Co   2 Bav ac  103
Co   2  0.5  26000  103  128.3049
L  1.1 (0.7 )  0.77  0.77
D
p4
L  0.6048 D
Main Dimensions D & L
Pa
58.14
Pa  C o D Ln  D L 

Co n 128.3049 16.6667
2
2
D 2 L  0.0272
D 2 (0.6048D)  0.0272  D 3  0.045 m3
D  0.3556m and L  0.215m
Exercise 35: Page: 9.112
No. of conductors ( Z )
Specific Electric Loading ac 
IzZ
ac  D
Z 
D
Iz
26000    0.3556
 228  Z
127.5
D
No. of Slots ( S )  S 
( Ys  35mm to 25mm)
Ys
Z
Ys  35mm  S 
  0.3556
3
 32
35 10
  0.3556
Ys  25mm  S 
 45
3
25 10
No. of Slots S  32 to 45
No. of Slots / Pole( S / p )  (9 to 16)
S
 (9 to 16)  S  36 to 64
p4
Combining above two slot ranges, we get
S  36 to 45
Exercise 35: Page: 9.112
In Lap winding no. slots should be multiples of pole pairs
Pole pair ( p / 2)  4 / 2  2
ie no. of slots S  36,38, 40, 42, 44
To avoid flux pulsation loss slot / pole should be Integer  1/ 2
S
1
 Integer 
p
2
36

90  
4

38
1

 38 
9

4
2

40

 40 
 10  0   Choo sin g S  38, 42
4

42
1 
 42 
 10 
4
2 

44
 44 
 11  0  

4
S  36 
S
S
S
S
Exercise 35: Page: 9.112
Further to avoid flux pulsation loss ( S / p )  Integer  0.5(nearly )
38

S  38  0.7  6.65 

4
 finally choose S  38
42
S  42  0.7
 7.35

4
Z 228
No. of Conductor per slot Z s  
6
S
38
Total no of conductor Z  Z s  S
Z  6  38
Z  228
Example 9.15 page 9.56


A 500KW, 460V, 8 pole, 375 rpm compund generator has an
armature diameter of 1.1m and a core length of 0.33m.The ampere
conductors per meter are 34000. Internal voltage drop is 4% of
terminal voltage and the field current is 1% of output current.
The ratio of pole arc to pole pitch is 0.7. The voltage between
adjacent segments at no load should not exceed 15V and the slot
loading should not exceed 1500A. The diameter of commutator is
0.65 of armature diameter and the minimum allowable pitch of
segments is 4mm. Make other suitable assumptions.
Internal voltage drop I a Ra  4% of V  0.04  460  18.4V
Generated EMF Eg  V  I a Ra  460  18.4  478.4V
P 500 KW

 1086.9565 A
V
460V
Field current I f  1% of I L  0.011086.9565  10.8696 A
Line current I L 
Armature current I a  I L  I f  1086.9565  10.8696  1097.8261A
Type of Winding
Ia
Current per parallel path I z 
( A  2 for wave w in ding )
A
I a 1097.8261
Iz  
 548.9130 A
A
2
Current per brush arm I b  2 I z  2  548.9130  1097.8261A
I b  1097.8261A which exceeds 400 A, so wave winding is not possible
Example 9.15 page 9.56
Ia
Current per parallel path I z 
( A  p  8 for lap w in ding )  Given
A
I a 1097.8261
Iz  
 137.2283 A
A
8
Current per brush arm I b  2 I z  2 137.2283  274.4565 A
I b  274.4565 A which does not exceeds 400 A, so lap winding is possible
Number of Conductors ( S ) Slot pitchYs var ies from 35mm to 25mm
Ys 
D
S
S
D
Ys
Ys  35mm  S 
 1.1
35 103
 1.1
Ys  25mm  S 
25 103

99 

 No. of slots S  99 to 138
138

Example 9.15 page 9.56
No. of slots per pole( S / p ) Should be 9 to 16
S
 (9 to 16)  S  p (9 to 16)
( p  8 given)
p
S  8  (9 to 16)  72 to 128 in comparing with S  99 to 138
Now no. of slots S  99 to 128
Further No. of slots should be multiples of pole pair ( p / 2  8 / 2  4)
ie
Now no. of slots S  100,104,108,112,116,120,124,128
S
1
In order to reduce flux pulsation loss  Integer 
p
2
100
1
104
s  100 
 12   correct : s  104 
 13  Wrong
8
2
8
108
1
112
s  108 
 13   correct : s  112 
 14  Wrong
8
2
8
116
1
120
s  116 
 14   correct : s  120 
 15  Wrong
8
2
8
124
1
128
s  124 
 15   correct : s  128 
 16  Wrong
8
2
8
Now No of Slots ( S )  100,108,116,124
Example 9.15 page 9.56
S
1
further reducing flux pulsation loss   Integer  (nearly )
p
2
100
108
S  100  0.7
 8.75  : S  108  0.7
 9.45 correct
8
8
116
124
S  116  0.7
 10.15  : S  124  0.7
 10.85 
8
8
So choose No. of slots S  108
No. of armature conductor  Z 
IzZ
 D  ac  1.1 34000
ac 
Z 

 Z  856
D
Iz
137.2283
Z 856
No. of conductors per slot Z s  
 7.9259
S 108
No. of conductor Z  Z s  S  8 108  Z  864
8  Zs
Example 9.15 page 9.56
No. of Coils (C ) :
Minimum no. of coils C 
E p 478.4  8

15
15
255
1
No of Coils C  US  forU  2, 4, 6,8,10.....
2
1
U  2  C  2 108  108 
2
1
U  4  C  4 108  216 
2
Zs 8
1
U  6  C  6 108  324 
 (not divisable) 
2
U 6
Zs 8
1
U  8  C  8 108  432 
 ( Divisable)
2
U 8
No. of coils C  432
Example 9.15 page 9.56
Z
864
Turns per Coil Tc 

 1 (Sin gle turn coils )
2C 2  432
Checks :
Slot loading I z Z s  137.2283  8  1097.8264  1500
Commutator Diameter Dc  0.65  D  0.65  1.1  Dc  0.715m
Commutator segment pitch(  c ) 
 c  5.2mm  4mm (acceptable)
Lap winding , Z  864,
S  108, Z s  8
C  432, U  8, Tc  1
 Dc
C

  0.715
432
 5.2mm  4mm
Exercise: 33, Page: 9.111


Find the main dimensions of a 200KW, 250V, 6 pole generator.
The maximum value of flux density in the gap is 0.87T, and
ac=31000. The ratio of pole arc to pole pitch is 0.67 and the
efficiency is 91%. Assume the ratio of length of the core to
pole pitch is 0.75.
Given data:
Output Power P  200 KW ;V  250V ; poles p  6;
Speed N  1000rpm; Bg  0.87 tesla; ac  31000;
b

 0.67;  0.91;
L

 0.75
Exercise: 33, Page: 9.111
L  0.75  0.75
D
p
 0.75
D
6
 L  0.3927 D
P
200
Power developed in armature Pa  
 Pa  219.78 KW
 0.91
N 1000
Speed n 

 n  16.6667 rps
60
60
B
b
   0.67  av  Bav   Bg  0.67  0.87  Bav  0.5829T

Bg
Output co  efficient Co   2 Bav ac  103   2  0.5829  31000
Co  178.3428
Pa  Co D 2 Ln  D 2 L 
Pa
219.78

 0.0739 m3
Co n 178.3428 16.6667
D 2 L  0.0739 m3  D 2 ( L  0.3927 D)  0.0739
D 3  0.1883 m3  D  0.5732m & L  0.2251m
Exercise 9.34: page: 9.112

Find the main dimensions and the number of poles of a 37KW,
230V, 1400rpm, shunt motor so that a square pole face is
obtained. The average flux density is 0.5wb/m2, and the
ac=22000. The ratio of pole arc to pole pitch is 0.7 and the full
load efficiency is 90%.
N 1400
Speed n 

 23.3333rps
60
60
pn
No. of poles ( p )  frequency f  25Hz to 50 Hz 
2
2  23.3333

p 2 f 
 23.3333 Hz  

2
 So choose p  4
4  23.3333
p 4 f 
 46.6667 Hz 

2
Exercise 9.34: page: 9.112
37 KW
 41.1111KW

0.9
Total loss  Pi  P  41.1111  37  4.1111KW
Input power Pi 
P

At full load cons tan t loss  Copper loss
Total Loss 4.1111KW

 2.0556 KW
2
2
P 37000
Line current I L  
 160.8696 A
V
230
Field current I f  1% of I L  0.01160.8696 A  1.6087 A
ie Cons tan t loss 
Armature current I a  I L  I f  160.8696  1.6087  159.2609 A
Cons tan t loss  Field cu  loss  friction & windage loss
Field cu loss  VI f  230 1.6087  370W
friction & windage loss  Cons tan t loss  Field cu  loss  2.0556 KW  370W
friction & windage loss  1.6856 KW
Armature power Pa  output power ( P )  friction & windage loss
Armature power Pa  37 KW  1.6856 KW  38.6856 KW
Exercise 9.34: page: 9.112
Co   2 Bav ac  103   2  0.5  22000  103  Co  108.5656
Power developed in Armature Pa (Output equation)
Pa
38.6856
Pa  Co D Ln  D L 

 0.0153m3
Co n 108.5656  23.3333
2
D L  0.0153 m
2
2
3
but L  0.7  0.7
D
p
 0.7
D 2 (0.5498 D)  0.0153 m3  D 3  0.0278
D  0.3028m and L  0.1665m
D
4
 L  0.5498D
Example 3.2, Page 3.12


Calculate the MMF required for the air gap of a machine having
core length=0.32m, including 4 ducts of 10mm each, pole
arc=0.19m; slot pitch=65.4mm; slot opening=5mm; air gap
length=5mm; flux per pole=52mWb. Given Carter’s co-efficient is
0.18 for opening/gap=1, and is 0.28 for opening / gap=2.
Solution:
Example 3.2, Page 3.12
Slot opening 5
  1  Carter ' s coefficient K cs  0.18
gap length
5
This is a salient pole machine with semienclosed slots.
Ys
Gap contraction factor for slots K gs 
Ys  K csWo
65.2
K gs 
 1.014
65.2  0.18  5
Duct width 10

 2  K cd  0.28
Gap length 5
Gap contraction factor for Ducts K gd 
L
L  K cd ndWd
0.32
K gd 
 1.036
0.32  0.28  4 10mm
Total gap contraction factor k g  K gs  K gd
k g  1.014  1.036  1.05
Example 3.2, Page 3.12
Flux per pole
Flux density at the centre of pole Bg 
Area
Flux per pole
52 103
Bg 

 0.854 Tesla
Pole arc  Core length 0.19  0.32
MMF required for air gap ATg  800000 K g Bg lg
ATg  800000 1.05  0.854  5 10
ATg  3587 A
3
TUTORIAL PROBLEMS