EM - Problems

First Semester
2017-2018

REFERENCE:
Electric Machinery Fundamentals, Stephen J. Chapman, Fifth Edition

1
DC Motors and Generators
Problems 8-1 to 8-12 refer to the following dc motor:
Prated = 30 hp IL,rated = 110 A
VT = 240 V N F = 2700 turns per pole
nrated = 1200 r/min NSE = 14 turns per pole
RA = 0.19  RF = 75 
R S = 0.02  R adj = 100 to 400 

Rotational losses = 3550 W at full load Magnetization curve as shown in Figure P8-1

241V

n rated = 1200 r/min

0.96 A

2
In Problems 8-1 through 8-7, assume that the motor described above can be connected in shunt.
The equivalent circuit of the shunt motor is shown in Figure P8-2.

8-1. If the resistor Radj is adjusted to 175  what is the rotational speed of the motor at no-load conditions?

SOLUTION At no-load conditions, E A VT  240 V . The field current is given by

VT 240 V 240 V
IF     0.960 A
Radj  RF 175   75  250 

From Figure P9-1, this field current would produce an internal generated voltage E Ao of 241 V at a speed
no of 1200 r/min. Therefore, the speed n with a voltage E A of 240 V would be
EA n

E Ao no
 E   240 V 
n   A  no   1200 r/min   1195 r/min
 EAo   241 V 

8-2. Assuming no armature reaction, what is the speed of the motor at full load? What is the speed regulation of
the motor?
SOLUTION At full load, the armature current is
VT 240 V
I A  IL  I F  IL   110 A   109 A
Radj  RF 250 

The internal generated voltage EA is

EA  VT  I A RA  240 V  109 A 0.19   219.3 V

The field current is the same as before, and there is no armature reaction, so EAo is still 241 V at a speed
no of 1200 r/min. Therefore,
 E   219.3 V 
n   A  no    1200 r/min   1092 r/min
 EAo   241 V 
The speed regulation is
nnl  nfl 1195 r/min  1092 r/min
SR  100%  100%  9.4%
nfl 1092 r/min

3
8-3. If the motor is operating at full load and if its variable resistance Radj is increased to 250 , what is the
new speed of the motor? Compare the full-load speed of the motor with Radj = 175  to the full-load
speed with Radj = 250 . (Assume no armature reaction, as in the previous problem.)

SOLUTION If Radj is set to 250  , the field current is now

VT 240 V 240 V
IF     0.739 A
Radj  RF 250   75  325 

Since the motor is still at full load, E A is still 218.3 V. From the magnetization curve (Figure P8-1), the
new field current I F would produce a voltage EAo of 212 V at a speed no of 1200 r/min. Therefore,
 E   218.3 V 
n   A  no    1200 r/min   1236 r/min
 Ao 
E  212 V 

Note that Radj has increased, and as a result the speed of the motor n increased.

8-4. Assume that the motor is operating at full load and that the variable resistor Radj is again 175 . If the
armature reaction is 1000 Aturns at full load, what is the speed of the motor? How does it compare
to the result for Problem 8-2?
SOLUTION The field current is again 0.96 A, and the motor is again at full load conditions. However, this
time there is an armature reaction of 1200 Aturns, and the effective field current is
AR 1000 A  turns
IF*  IF   0.96 A   0.59 A
NF 2700 turns

From Figure P9-1, this field current would produce an internal generated voltage E Ao of 185 V at a speed
no of 1200 r/min. The actual internal generated voltage E A at these conditions is
E A  VT  I A RA  240 V  109 A  0.19    219.3 V

Therefore, the speed n with a voltage of 240 V would be

 E   219.3 V 
n   A  no    1200 r/min   1423 r/min
 E Ao   185 V 
If all other conditions are the same, the motor with armature reaction runs at a higher speed than the
motor without armature reaction.

8-5. If Radj can be adjusted from 100 to 400 , what are the maximum and minimum no-load speeds possible
with this motor?
SOLUTION The minimum speed will occur when Radj = 100  , and the maximum speed will occur when
Radj = 400  . The field current when Radj = 100  is:

VT 240 V 240 V
IF     1.37 A
Radj  RF 100   75  175 

From Figure P9-1, this field current would produce an internal generated voltage E Ao of 271.5 V at a
speed no of 1200 r/min. Therefore, the speed n with a voltage of 240 V would be

4
 EA n

E Ao no
 E   240 V 
n   A  no    1200 r/min   1061 r/min
 EAo   271.5 V 

The field current when Radj = 400  is:

VT 240 V 240 V
IF     0.505 A
Radj  RF 400   75  500 

From Figure P8-1, this field current would produce an internal generated voltage E Ao of 167 V at a speed
no of 1200 r/min. Therefore, the speed n with a voltage of 240 V would be

EA n

E Ao no
 E   240 V 
n   A  no   1200 r/min   1725 r/min
 EAo   167 V 

8-6. What is the starting current of this machine if it is started by connecting it directly to the power
supply VT ? How does this starting current compare to the full-load current of the motor?
SOLUTION The starting current of this machine (ignoring the small field current) is
VT 240 V
I L,start    1260 A
RA 0.19 

The rated current is 110 A, so the starting current is 11.5 times greater than the full-load current. This
much current is extremely likely to damage the motor.
8-7. Plot the torque-speed characteristic of this motor assuming no armature reaction, and again assuming a
full-load armature reaction of 1200 A turns. (Assume that the armature reaction increases linearly with
increases in armature current.)
SOLUTION This problem is best solved with MATLAB, since it involves calculating the torque-speed
values at many points.
The resulting plot is shown below:

5
For Problems 8-8 and 8-9, the shunt dc motor is reconnected separately excited, as shown in Figure P8-3. It
has a fixed field voltage V F of 240 V and an armature voltage V A that can be varied from 120 to 240 V.

8-8. What is the no-load speed of this separately excited motor when Radj = 175  and (a) V A = 120 V, (b)
V A = 180 V, (c) V A = 240 V?
SOLUTION At no-load conditions, E A  VA . The field current is given by

VF 240 V 240 V
IF     0.96 A
Radj  RF 175   76  250 

From Figure P9-1, this field current would produce an internal generated voltage E Ao of 241 V at a speed
no of 1200 r/min. Therefore, the speed n with a voltage of 240 V would be

EA n

E Ao no
 E 
n   A  no
 E Ao 

(a) If VA = 120 V, then E A = 120 V, and

 120 V 
n  1200 r/min   598 r/min
 241 V 
(b) If VA = 180 V, then E A = 180 V, and
 180 V 
n  1200 r/min   986 r/min
 241 V 
(c) If VA = 240 V, then E A = 240 V, and
 240 V 
n  1200 r/min   1195 r/min
 241 V 

6
8-9. For the separately excited motor of Problem 8-8:
(a) What is the maximum no-load speed attainable by varying both V A and Radj ?

(b) What is the minimum no-load speed attainable by varying both V A and Radj ?

(c) What is the motor’s efficiency at rated conditions? [Note: Assume that (1) the brush voltage drop
is 2 V; (2) the core loss is to e determined at an armature voltage equal to the armature voltage under
full load; and (3) stray load losses are 1 percent of full load.]
SOLUTION
(a) The maximum speed will occur with the maximum VA and the maximum Radj . The field current
when Radj = 400  is:

VT 240 V 240 V
IF     0.505 A
Radj  RF 400   75  475 

From Figure P8-1, this field current would produce an internal generated voltage E Ao of 167 V at a speed
no of 1200 r/min. At no-load conditions, the maximum internal generated voltage E A  VA = 240 V.
Therefore, the speed n with a voltage of 240 V would be
EA n

E Ao no
 E   240 V 
n   A  no   1200 r/min   1725 r/min
 E Ao   167 V 

(b) The minimum speed will occur with the minimum VA and the minimum Radj . The field current
when Radj = 100  is:

VT 240 V 240 V
IF     1.37 A
Radj  RF 100   75  175 

From Figure P8-1, this field current would produce an internal generated voltage E Ao of 271 V at a speed
no of 1200 r/min. At no-load conditions, the minimum internal generated voltage E A  VA = 120 V.
Therefore, the speed n with a voltage of 120 V would be
EA n

E Ao no
 E   120 V 
n   A  no    1200 r/min   531 r/min
 E Ao   271 V 
For Problems 8-10 to 8-11, the motor is connected cumulatively compounded as shown in Figure P8-4.

7
 8-10. If the motor is connected cumulatively compounded with Radj = 175 :
(a) What is the no-load speed of the motor?
(b) What is the full-load speed of the motor?
(c) What is its speed regulation?
(d) Calculate and plot the torque-speed characteristic for this motor. (Neglect armature effects in this
problem.)
SOLUTION At no-load conditions, EA  VT  240 V . The field current is given by

VT 240 V 240 V
IF     0.960 A
Radj  RF 175   75  250 

From Figure P8-1, this field current would produce an internal generated voltage E Ao of 241 V at a speed
no of 1200 r/min. Therefore, the speed n with a voltage of 240 V would be
EA n

E Ao no
 E   240 V 
n   A  no   1200 r/min   1195 r/min
 EAo   241 V 
At full load conditions, the armature current is
VT
IA  IL IF  IL   110 A  0.96 A  109 A
Radj  RF

The internal generated voltage EA is

E A  VT  I A  RA  RS   240 V  109 A  0.21    217.1 V

The equivalent field current is

NSE 14 turns
IF*  IF  I A  0.96 A  109 A   1.53 A
NF 2700 turns

From Figure P8-1, this field current would produce an internal generated voltage E Ao of 279 V at a speed
no of 1200 r/min. Therefore,
 E   217.1 V 
n   A  no    1200 r/min   934 r/min
 Ao 
E  279 V 
The speed regulation is
nnl  nfl 1195 r/min  934 r/min
SR   100%   100%  27.9%
nfl 934 r/min
The resulting plot is shown below:

Compare this torque-speed curve to that of the shunt

motor in Problem 8-7. (Both curves are plotted on

the sam e scale to facilitate comparison.)

8
8-11. The motor is connected cumulatively compounded and is operating at full load. What will the new speed
of the motor be if Radj is increased to 250 ? How does the new speed compared to the full-load speed
calculated in Problem 8-10?
SOLUTION If Radj is increased to 250  , the field current is given by

VT 240 V 240 V
IF     0.739 A
Radj  RF 250  75 250 

At full load conditions, the armature current is

I A  I L  I F  110 A  0.739 A  109.2 A

The internal generated voltage E A is

E A  VT  I A  RA  RS   240 V  109.2 A  0.21    217.1 V

The equivalent field current is

NSE 14 turns
IF*  IF  I A  0.739 A  109.2 A   1.305 A
NF 2700 turns

From Figure P9-1, this field current would produce an internal generated voltage E Ao of 268 V at a speed
no of 1200 r/min. Therefore,
 E   217.1 V 
n   A  no    1200 r/min   972 r/min
 E Ao   268 V 
The new full-load speed is higher than the full-load speed in Problem 8-10.

For Problem 8-12, the motor is now connected differentially compounded as shown in Figure P8-4.
8-12. The motor is now connected differentially compounded.
(a) If Radj = 175  , what is the no-load speed of the motor?

(b) What is the motor’s speed when the armature current reaches 20 A? 40 A? 60 A?
(c) Calculate and plot the torque-speed characteristic curve of this motor.
SOLUTION
(a) At no-load conditions, E A  VT  240 V . The field current is given by

VF 240 V 240 V
IF     0.96 A
adj
R  R F 175   75  275 

From Figure P8-1, this field current would produce an internal generated voltage E Ao of 241 V at a speed
no of 1200 r/min. Therefore, the speed n with a voltage of 240 V would be
EA n

EAo no
 E   240 V 
n   A  no    1200 r/min   1195 r/min
 EAo   241 V 

(b) At I A = 20A, the internal generated voltage E A is

EA  VT  I A  RA  RS   240 V   20 A  0.21    235.8 V

The equivalent field current is 223

N SE 14 turns
IF *  IF  I A  0.96 A   20 A   0.856 A
NF 2700 turns

10
From Figure P8-1, this field current would produce an internal generated voltage E Ao of 229 V at a speed
no of 1200 r/min. Therefore,
 E   235.8 V 
n   A  no    1200 r/min   1236 r/min
 Ao 
E  229 V 

At I A = 40A, the internal generated voltage E A is

EA  VT  I A  RA  RS   240 V   40 A  0.21    231.6 V

The equivalent field current is

NSE 14 turns
IF*  IF  I A  0.96 A   40 A   0.753 A
NF 2700 turns
From Figure P8-1, this field current would produce an internal generated voltage E Ao of 214 V at a speed
no of 1200 r/min. Therefore,
 E   231.6 V 
n   A  no    1200 r/min   1299 r/min
 EAo   214 V 
At I A = 60A, the internal generated voltage E A is

EA  VT  I A  RA  RS   240 V   60 A  0.21    227.4 V

The equivalent field current is

NSE 14 turns
IF*  IF  I A  0.96 A   60 A   0.649 A
NF 2700 turns
From Figure P8-1, this field current would produce an internal generated voltage E Ao of 196 V at a speed
no of 1200 r/min. Therefore,
 E   227.4 V 
n   A  no    1200 r/min   1392 r/min
 EAo   196 V 
(c) The torque-speed characteristic can best be plotted with a MATLAB program.

The resulting plot is shown below:

Compare this torque-speed curve to that of the shunt motor in Problem 8-7 and the cumulatively-
compounded motor in Problem 8-10.

11
8-13. A 7.5-hp 120-V series dc motor has an armature resistance of 0.1  and a series field resistance of 0.08 .
At full load, the current input is 56 A, and the rated speed is 1050 r/min. Its magnetization curve is shown
in Figure P8-5. The core losses are 220 W, and the mechanical losses are 230 W at full load.
Assume that the mechanical losses vary as the cube of the speed of the motor and that the core losses are
constant.

(a) What is the efficiency of the motor at full load?

(b) What are the speed and efficiency of the motor if it is operating at an armature current of 40 A?
(c) Plot the torque-speed characteristic for this motor.

SOLUTION
(a) The output power of this motor at full load is
POUT   7.5 hp  746 W/hp   5595 W

The input power is

PIN  VT I L  120 V  56 A   6720 W

Therefore the efficiency is

POUT 5595 W
  100%   100%  83.3%
PIN 6720 W

12
(b) If the armature current is 40 A, then the input power to the motor will be
PIN  VT I L  120 V  40 A   4800 W

The internal generated voltage at this condition is

E A2  VT  I A  R A  R S   120 V   40 A  0.10   0.08    112.8 V

and the internal generated voltage at rated conditions is

E A 2  VT  I A  RA  RS   120 V   56 A  0.10   0.08    109.9 V

The final speed is given by the equation

E A2 K 2  2 E Ao ,2 n2
 
E A1 K 2  2 E Ao ,1 n1

since the ratio E Ao ,2 / E Ao ,1 is the same as the ratio 2 / 1 . Therefore, the final speed is

E A2 E Ao ,1
n2  n1
E A1 E Ao ,2

From Figure P8-5, the internal generated voltage E Ao ,2 for a current of 40 A and a speed of no = 1200
r/min is E Ao ,2 = 120 V, and the internal generated voltage E Ao ,1 for a current of 56 A and a speed of no =
1200 r/min is E Ao ,1 = 133 V.

E A2 E Ao ,1  112.8 V  133 V 
n2  n1     1050 r/min   1195 r/min
E A1 E Ao ,2  109.9 V  120 V 
The power converted from electrical to mechanical form is
Pconv  E A I A  112.8 V  40 A   4512 W

The core losses in the motor are 220 W, and the mechanical losses in the motor are 230 W at a speed of
1050 r/min. The mechanical losses in the motor scale proportionally to the cube of the rotational
speed so the mechanical losses at 1326 r/min are
3
n 
3
 1195 r/min 
Pmech   2   230 W      230 W   339 W
 n1   1050 r/min 
Therefore, the output power is
POUT  Pconv  Pmech  Pcore  4512 W  339 W  220 W  3953 W
and the efficiency is
POUT 3953 W
  100%   100%  82.4%
PIN 4800 W

13
 (c) A MATLAB program to plot the torque-speed characteristic of this motor is shown below:

The resulting torque-speed characteristic is shown below:

8-14. A 20-hp 240-V 76-A 900 r/min series motor has a field winding of 33 turns per pole. Its
armature resistance is 0.09 , and its field resistance is 0.06 . The magnetization curve expressed in terms
of magnetomotive force versus EA at 900 r/min is given by the following table:

EA , V 95 150 188 212 229 243

F, A  turns 500 1000 1500 2000 2500 3000

Armature reaction is negligible in this machine.
(a) Compute the motor’s torque, speed, and output power at 33, 67, 100, and 133 percent of full-load
armature current. (Neglect rotational losses.)
(b) Plot the terminal characteristic of this machine.

14
SOLUTION

(a) Since full load corresponds to 76 A, this calculation must be performed for armature currents of
25.3 A, 50.7 A, 76 A, and 101.3 A.
If I A = 23.3 A, then

E A  VT  I A  RA  RS   240 V  25.3 A  0.09   0.06   236.2 V

The magnetomotive force is F  NI A   33 turns  25.3 A   835 A  turns , which produces a voltage E Ao
of 134 V at no = 900 r/min. Therefore the speed of the motor at these conditions is

n
EA
no 
236.2 V
134 V
 900 r/min  1586 r/min
E Ao

The power converted from electrical to mechanical form is

Pconv  E A I A   236.2 V 25.3 A  5976 W

Since the rotational losses are ignored, this is also th e output power of the motor. The induced torque is
Pconv 5976 W
 ind    36 N  m
m 1586 r/min   21 rad   1 min 
r   60 s 
If I A = 50.7 A, then

E A  VT  I A  RA  RS   240 V  50.7 A 0.09   0.06   232.4 V

The magnetomotive force is F  NI A   33 turns  50.7 A   1672 A  turns , which produces a voltage
E Ao of 197 V at no = 900 r/min. Therefore the speed of the motor at these conditions is

n
EA
no 
232.4 V
 900 r/min   1062 r/min
E Ao 197 V
The power converted from electrical to mechanical form is
Pconv  E A I A   232.4 V 50.7 A  11,780 W

Since the rotational losses are ignored, this is also th e output power of the motor. The induced torque is
Pconv 11,780 W
 ind    106 N  m
m  2 rad   1 min 
1062 r/min   1 r   60 s 
If I A = 76 A, then

E A  VT  I A  RA  RS   240 V  76 A 0.09   0.06    228.6 V

The magnetomotive force is F  NI A   33 turns  76 A   2508 A  turns , which produces a voltage E Ao

of 229 V at no = 900 r/min. Therefore the speed of the motor at these conditions is
EA 228.6 V
n no 
229 V
 900 r/min  899 r/min
E Ao

15
 The power converted from electrical to mechanical form is
Pconv  E A I A   228.6 V  76 A   17,370 W

Since the rotational losses are ignored, this is also the output power of the motor. The induced torque is
Pconv 17,370 W
 ind    185 N  m
m  2 rad   1 min 
899 r/min    
1 r   60 s 
If I A = 101.3 A, then

E A  VT  I A  RA  RS   240 V  101.3 A  0.09   0.06    224.8 V

The magnetomotive force is F  NI A   33 turns 101.3 A   3343 A  turns , which produces a voltage
E Ao of 252 V at no = 900 r/min. Therefore the speed of the motor at these conditions is
EA 224.8 V
n no   900 r/min   803 r/min
E Ao 252 V
The power converted from electrical to mechanical form is
Pconv  E A I A   224.8 V 101.3 A   22,770 W

Since the rotational losses are ignored, this is also the output power of the motor. The induced torque is
Pconv 22,770 W
 ind    271 N  m
m  2 rad   1 min 
 803 r/min    
1 r   60 s 

The resulting torque-speed characteristic is shown below:

16
8-15. A 300-hp 440-V 560-A, 863 r/min shunt dc motor has been tested, and the following data were taken:
Blocked-rotor test:
VA  14.9 V exclusive of brushes VF  440 V

I A  500 A I F  7.52 A
No-load operation:
VA  440 V includi ng brushes I F  7.50 A
I A  23.1 A n  863 r/min

What is this motor`s efficiency at the rated condition?

Assume that:-
1- the brush voltage drop is 2V
2- the core loss is be determined at an armature voltage equal
to the armature under full load
3- stray losses are 1 percent of full load.

SOLUTION The armature resistance of this motor is

V ,br 14.9 V
RA  A   0.0298 
I A,br 500 A

Under no-load conditions, the core and mechanical losses taken together (that is, the rotational losses ) of
this motor are equal to the product of the internal generated voltage E A and the armature current I A ,
since this is no output power from the motor at no-load conditions. Therefore, the rotational losses at
rated speed can be found as
E A  V A  Vbrush  I AR A  440 V 2 V  23.1 A  0.0298   437.3 V

Prot  Pconv  EA I A   437.3 V  23.1 A  10.1 kW

The input power to the motor at full load is

PIN  VT I L  440 V  560 A   246.4 kW

The output power from the motor at full load is

POUT  PIN  PCU  Prot  Pbrush  Pstray

The copp er losses are

PCU  I A RA  VF IF   560 A   0.0298     440 V  7.52 A   12.65 kW

2 2

The brush losses are

Pbrush  Vbrush I A   2 V 560 A  1.12 kW
Therefore ,
POUT  PIN  PCU 

Prot 

Pbrush 

Pstray
POUT  246.4 kW  12.65 kW  10.1 kW  1.12 kW  2.46 kW  220.1 kW
The motor’s efficiency at full load is
POUT
  100% 
220.1 kW 100%  89.3%

PIN 246.4 kW

17
Problems 8-16 to 8-19 refer to a 240-V 100-A dc motor which has both shunt and series windings.
Its characteristics are
RA = 0.14  N F = 1500 turns

R S = 0.05  N SE = 15 turns

RF = 200  nm = 3000 r/min

R adj = 0 to 300 , currently set to 120 

This m otor has compensating windings and interpoles. The magnetization curve for this motor at 3000 r/min is
shown in Figure P8-6.

245 V

0.75 A

18
8-16. The motor described above is connected in shunt.
(a) What is the no-load speed of this motor when Radj = 120 ?

(b) What is its full-load speed?

(c) What is its speed regulation?
(d) Plot the torque-speed characteristic for this motor.
(e) Under no-load conditions, what range of possible speeds
can be achieved by adjusting Radj ?
S OLUTION
(a) If Radj = 120 , the total field resistance is 320 , and the resulting field current is

VT 240 V
IF    0.75 A
RF  Radj 200   120 

This field current would produce a voltage EAo of 245 V at a speed of no = 3000 r/min. The actual E A
is 240 V, so the actual speed will be
EA 240 V
n no  3000 r/min   2939 r/min
E Ao 245 V
(b) At full load, I A  I L  I F  100 A  0.75 A  99.25 A , and

E A  VT  I AR A  240 V   99.25 A  0.14    226.1 V

The refore, the speed at full load will be
EA 226.1 V
n no  3000 r/min   2769 r/min
EAo 245 V
(c) The speed regulation of this motor is
nnl  nfl 2939 r/min  2769 r/min
SR   100%   100%  6.16%
nfl 2769 r/min

The resulting torque-speed curve is shown below:

19
 (e) If Radj is maximum at no-load conditions, the total resistance is 500 , and

VT 240 V
IF    0.48 A
RF  Radj 200   300 

This field current would produce a voltage E Ao of 195 V at a speed of no = 3000 r/min. The actual E A is
240 V, so the actual speed will be
EA 240 V
n no  3000 r/min   3692 r/min
EAo 195 V

If Radj is minimum at no-load conditions, the total resistance is 200  , and

VT 240 V
IF    1.2 A
RF  R adj 200   0 

This field current would produce a voltage E Ao of 282 V at a speed of no = 3000 r/min. The actual E A is
240 V, so the actual speed will be
EA 240 V
n no   3000 r/min   2553 r/min
EAo 282 V

8-17. This machine is now connected as a cumulatively compounded dc motor with Radj = 120 .

(a) What is the no-load speed of this motor?

(b) What is its full-load speed?
(c) What is its speed regulation?
(d) Plot the torque-speed characteristic for this motor.
S OLUTION

Th e field current will be

VT 240 V
(a) IF    0.75 A
RF  R adj 200   120 

At no load, I A  0 A , and

E A VT  I A  RA  RS   240 V

and the effective field current will be

NSE 15 turns
I F*  I F  I A  0.75 A   0 A   0.75 A
NF 1500 turns

This field current would produce a voltage E Ao of 245 V at a speed of no = 3000 r/min. The actual E A
is 240 V, so the actual speed at full load will be
EA 240 V
n no   3000 r/min   2939 r/min
EAo 245 V

20
 (b) The field current will be
VT 240 V
IF    0.75 A
RF  R adj 200   120 

At full load, I A  I L  I F  100 A  0.75 A  99.25 A , and

E A  VT  I A  R A  RS   240 V   99.25 A  0.14   0.05    221.1 V

and the effective field current will be

NSE 15 turns
I F*  I F  I A  0.75 A   99.25 A   1.74 A
NF 1500 turns

This field current would produce a voltage E Ao of 292 V at a speed of no = 3000 r/min. The actual E A
is 240 V, so the actual speed at full load will be
EA 221.1 V
n no   3000 r/min   2272 r/min
EAo 292 V
(c) The speed regulation of this motor is
nnl  nfl 2939 r/min  2272 r/min
SR   100%   100%  29.4%
nfl 2272 r/min

The resulting torque-speed curve is shown below. Note that is curve is plotted on the same scale as the
shunt motor in Problem 8-16.

21
8-18. The motor is reconnected differentially compounded with Radj = 120 . Derive the shape of its torque-speed
characteristic.
The resulting torque-speed curve is shown belo
w.

8-19. A series motor is now constructed from this machine by leaving the shunt field out entirely. Derive the
torque-speed characteristic of the resulting motor.
SOLUTION This motor will have extremely high speeds, since there are only a few series turns, and the
flux in the motor will be very small.
The resulting torque-speed characteristic is shown below:

The extreme speeds in this characteristic are due to the very light flux in the machine. To make a
practical series motor out of this machine, it would be necessary to include 20 to 30 series turns instead of 15.

22
8-20. An automatic starter circuit is to be designed for a shunt motor rated at 20 hp, 240 V, and 75 A. The
armature resistance of the motor is 0.12 , and the shunt field resistance is 40 . The motor is to start
with no more than 250 percent of its rated armature current, and as soon as the current falls to rated value,
a starting resistor stage is to be cut out. How many stages of starting resistance are needed, and how big
should each one be?
S OLUTION The rated line current of this motor is 75 A, and the rated armature current is
I A  I L  I F = 75A – 6 A = 69 A.
The maximum desired starting current is (2.5)(69 A) = 172.5 A.

Therefore, the total initial starting resistance must be

240 V
RA  Rstart,1   1.391 
172.5 A
Rstart,1  1.391   0.12   1.271 

The current will fall to rated value when E A rises to

E A  240 V  1.391   69 A   144 V

At that time, we want to cut out enough resistance to get the current back up to 172.5 A. Therefore,
240 V  144 V
RA  Rstart,2   0.557 
172.5 A
Rstart,2  0.557   0.12   0.437 
With this resistance in the circuit, the current will fall to rated value when E A rises to
E A  240 V   0.557   69 A   201.6 V

At that time, we want to cut out enough resistance to get the current back up to 172.5 A. Therefore,
240 V  201.6 V
RA  Rstart,3   0.223 
172.5 A
Rstart,3  0.223   0.12   0.103 
With this resistance in the circuit, the current will fall to rated value when E A rises to
E A  240 V   0.223   69 A   224.6 V

If the resistance is cut out when E A reaches 224,6 V, the resulting current is

240 V  224.6 V
IA   128 A  172.5 A ,
0.12 
so there are only three stages of starting resistance. The three stages of starting resistance can be found
from the resistance in the circuit at each state during starting.
Rstart,1  R1  R2  R3  1.217 
Rstart,2  R2  R3  0.437 
Rstart,3  R3  0.103 

Therefore, the starting resistances are

R1  0.780 
R2  0.334 
R3  0.103 

23
8-21. A 10-hp 120-V 1000 r/min shunt dc motor has a full-load armature current of 70 A when operating at
rated conditions. The armature resistance of the motor is RA = 0.12 , and the field resistance RF is 40 

The adjustable resist ance in the field circuit Radj may be varied over the range from 0 to 200  and is
currently set to 100 . Armature reaction may be ignored in this machine. The magnetization curve for
this motor, taken at a speed of 1000 r/min, is given in tabular form below:
EA , V 5 78 95 112 118 126 130

IF , A 0.00 0.80 1.00 1.28 1.44 2.88 4.00

(a) What is the speed of this motor when it is running at the rated conditions specified above?
(b) The output power from the motor is 10 hp at rated conditions. What is the output torque of the
motor?
(c) What are the copper losses and rotational losses in the motor at full load (ignore stray losses)?
(d) What is the efficiency of the motor at full load?
(e) If the motor is now unloaded with no changes in terminal voltage or Radj , what is the no-load
speed of the motor?
(f) Suppose that the motor is running at the no-load conditions described in part (e). What would
happen to the motor if its field circuit were to open? Ignoring armature reaction, what would the final
steady-state speed of the motor be under those conditions?
(g) What range of no-load speeds is possible in this motor, given the range of field resistance
adjustments available with Radj ?

SOLUTION
(a) If Radj = 100 , the total field resistance is 140 , and the resulting field current is

VT 120 V
IF    0.857 A
RF  Radj 100   40 

This field current would produce a voltage E Ao of 82.8 V at a speed of no = 1000 r/min. The actual E A
is
E A  VT  I A RA  120 V   70 A  0.12    111.6 V

so the actual speed will be

EA 111.6 V
n no  1000 r/min   1348 r/min
E Ao 82.8 V
(b) The output power is 10 hp and the output speed is 1000 r/min at rated conditions, therefore, the
torque is

 out 
Pout

10 hp  746 W/hp   71.2 N  m
m 2 rad  1 min 
1000 r/min    
 1 r  60 s 
(c) The copper losses are

24
 PCU  I A 2 RA  VF I F   70 A   0.12    120 V  0.857 A   691 W
2

The power converted from electrical to mechanical form is

Pconv  EA I A  111.6 V  70 A   7812 W
The output power is
POUT  10 hp  746 W/hp   7460 W
Therefore, the rotational losses are
Prot  Pconv  POUT  7812 W  7460 W  352 W
(d) The input power to this motor is
PIN  VT  IA  IF   120 V  70 A  0.857 A   8503 W

Therefore, the efficiency is

POUT 7460 W
  100%   100%  87.7%
PIN 8503 W

(e) The no-load EA will be 120 V, so the no-load speed will be

EA 120 V
n no  1000 r/min   1450 r/min
EAo 82.8 V

(f) If the field circuit opens, the field current would go to zero   drops to  res  E A   I A 
 ind   n to a very high speed. If I F = 0 A, E Ao = 8.5 V at 1800 r/min, so
EA 230 V
n no  1000 r/min   46,000 r/min
EAo 5V
(In reality, the motor speed would be limited by rotational losses, or else the motor will destroy itself
first.)
(g) The maximum value of R adj = 200 , so

VT 120 V
IF    0.500 A
RF  Radj 200   40 

This field current would produce a voltage E Ao of 50.6 V at a speed of n o = 1000 r/min. The actual EA
is 120 V, so the actual speed will be
EA 120 V
n no  1000 r/min   2372 r/min
EAo 50.6 V

The minimum value of Radj = 0  , so

VT 120 V
IF    3.0 A
RF  Radj 0   40 

This field current would produce a voltage EAo of about 126.4 V at a speed of n o = 1000 r/min. The
actual E A is 120 V, so the actual speed will be

EA 120 V
n no  1000 r/min   949 r/min
E Ao 126.4 V

25
8-22. The magnetization curve for a separately excited dc generator is shown in Figure P8-7. The generator is
rated at 6 kW, 120 V, 50 A, and 1800 r/min and is shown in Figure P8-8. Its field circuit is rated at 5A.
The following data are known about the machine:
R A  0.18  VF  120 V
Radj  0 to 40  RF  20 
N F  1000 turns per pole
Answer the following questions about this generator, assuming no armature reaction.
(a) If this generator is operating at no load, what is the range of voltage adjustments that can be
achieved by changing Radj ?

(b) If the field rheostat is allowed to vary from 0 to 30  and the generator’s speed is allowed to vary
from 1500 to 2000 r/min, what are the maximum and minimum no-load voltages in the generator?

26
SOLUTION
(a) If the generator is operating with no load at 1800 r/min, then the terminal voltage will equal the
internal generated voltage E A . The maximum possible field current occurs when Radj = 0 . The
current is
VF 120 V
I F ,max   6 A
RF  Radj 20   0 

From the magnetization curve, the voltage E Ao at 1800 r/min is 135 V. Since the actual speed is 1800
r/min, the maximum no-load voltage is 135 V.
The minimum possible field current occurs when Radj = 40 . The current is

VF 120 V
I F ,max    2.0 A
RF  Radj 20   40 

From the magnetization curve, the voltage E Ao at 1800 r/min is 79.5 V. Since the actual speed is 1800
r/min, the minimum no-load voltage is 79.5 V.
(b) The maximum voltage will occur at the highest current and speed, and the minimum voltage will
occur at the lowest current and speed. The maximum possible field current occurs when Radj = 0 . The
current is
VF 120 V
I F ,max   6 A
RF  Radj 20   0 

From the magnetization curve, the voltage E Ao at 1800 r/min is 135 V. Since the actual speed is 2000
r/min, the maximum no-load voltage is
EA n

E Ao no

n 2000 r/min
EA  E Ao  135 V   150 V
no 1800 r/min

27
 The minimum possible field current occurs and minimum speed and field current. The maximum
adjustable resistance is Radj = 30 . The current is

VF 120 V
I F ,max    2.4 A
RF  Radj 20   30 

From the magnetization curve, the voltage E Ao at 1800 r/min is 93.1 V. Since the actual speed is 1500
r/min, the maximum no-load voltage is
EA n

E Ao no

n 1500 r/min
EA  E Ao   93.1 V   77.6 V
no 1800 r/min
8-23. If the armature current of the generator in Problem 8-22 is 50 A, the speed of the generator is 1700 r/min,
and the terminal voltage is 106 V.

- how much field current must be flowing in the generator?

S OLUTION The internal generated voltage of this generator is

E A  VT  I ARA  106 V  50 A   0.18    115 V

at a speed of 1700 r/min. This corresponds to an E Ao at 1800 r/min of

EA n

E Ao no
n 1800 r/min
E Ao  o EA 
1700 r/min
115 V   121.8 V
n
From the magnetization curve, this value of E Ao requires a field current of 4.2 A.

8-24. Assuming that the generator in Problem 8-22 has an armature reaction at full load equivalent to 400 A.turns
of magnetomotive force, what will the terminal voltage of the generator be when
I F = 5 A, nm = 1700 r/min, and I A = 50 A?
S OLUTION When I F is 5 A and the armature current is 50 A, the magnetomotive force in the generator is

Fnet  NI F  FAR  1000 turns  5 A   400 A  turns  4600 A  turns

I F  Fnet / N F  4600 A  turns / 1000 turns  4.6 A

*
or
The equivalent internal generated voltage EAo of the generator at 1800 r/min would be 126 V. The actual
voltage a t 1700 r/min would be

EA 
n
EAo 
1700 r/min
126 V  119 V
no 1800 r/min
Therefore, the terminal voltage would be
VT  E A  I A RA  119 V  50 A  0.18   110 V

28
8-25. The machine in Problem 8-22 is reconnected as a shunt generator and is shown in Figure P8-9. The shunt
field resistor Radj is adjusted to 10 , and the generator’s speed is 1800 r/min.

(a) What is the no-load terminal voltage of the generator?

(b) Assuming no armature reaction, what is the terminal voltage of the generator with an armature
current of 20 A? 40 A?
(c) Assuming an armature reaction equal to 300 Aturns at full load, what is the terminal voltage of
the generator with an armature current of 20 A? 40 A?
(d) Calculate and plot the terminal characteristics of this generator with and without armature
reaction.
SOLUTION
(a) The total field resistance of this generator is 30 , and the no-load terminal voltage can be found
from the intersection of the resistance line with the magnetization curve for this generator. The
magnetization curve and the field resistance line are plotted below. As you can see, they intersect at a
terminal voltage of 121 V.

29
(b) At an armature current of 20 A, the internal voltage drop in the armature resistance is
20 A 0.18    3.6 V .
As shown in the figure below, there is a difference of 3.6 V between E A and
VT at a terminal voltage of about 116 V.

At an armature current of 40 A, the internal voltage drop in the armature resistance is

 40 A  0.18    7.2 V . As shown in the figure below, there is a difference of 7.2 V between E A and
VT at a terminal voltage of about 110 V. The program to create this plot is identical to the one shown
above, except that the gap between E A and VT is 7.2 V. The resulting terminal voltage is about 110 V.

30
 (c) The rated current of this generator is 50 A, so 20 A is 40% of full load. If the full load armature
reaction is 300 Aturns, and if the armature reaction is assume d to change linearly with armature current,
then the armature reaction will be 120 A turns. The demagnetizing effect of armature reaction is
equivalent to a reduction in field current of
120 A t
I F   0.12 A
1000 t
The figure below shows that a triangle consisting of 3.6 V and (120 Aturns)/(1000 turns) = 0.12 A fits
exactly between the E A and VT lines at a terminal voltage of 114 V.

The rated current of this generated is 50 A, so 40 A is 80% of full load. If the full load armature reaction
is 300 Aturns, and if the armature reaction is assumed to change linearly with armature current, then the
armature reaction will be 240 Aturns. The demagnetizing effect of armature reaction is equivalent to a
reduction in field current of
240 A t
IF   0.24 A
1000 t
The figure below shows that a triangle consisting of 7.2 V and (240 Aturns)/(1000 turns) = 0.24 A fits
exactly between the E A and VT lines at a terminal voltage of 105 V.

31
 The resulting terminal characteristic is shown below:

The resulting terminal characteristic is shown below. Note that the armature reaction reduces the terminal
voltage for any given load current relative to a generator without armature reaction.

32
8-26. If the machine in Problem 8-25 is running at 1800 r/min with a field resistance Radj = 10  and an armature
current of 25 A.
- What will the resulting terminal voltage be?
- If the field resistor decreases to 5 ,while the armature current remains 25 A, what will the new terminal voltage be?
(Assume no armature reaction.)
SOLUTION

If I A = 25 A, then I A RA   25 A  0.18   = 4.5 V. The point where the distance between the
E A and VT curves is exactly 4.5 V corresponds to a terminal voltage of 114 V, as shown below.

If Radj decreases to 5  , the total field resistance becomes 25 , and the terminal voltage line gets
shallower. The new point where the distance between the E A and VT curves is exactly 4.5 V
corresponds to a terminal voltage of 125 V, as shown below.

Note that decreasing the field resistance of the shunt generator increases the terminal voltage.

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8-27. A 120-V 50-A cumulatively compounded dc generator has the following characteristics:
R A  RS  0.21  N F  1000 turns

RF  20  N SE  25 turns
Radj  0 to 30 , set to 10  nm  1800 r / min
The machine has the magnetization curve shown in Figure P8-7. Its equivalent circuit is shown in Figure
P8-10. Answer the following questions about this machine, assuming no armature reaction.

(a) If the generator is operating at no load, what is its terminal voltage?

(b) If the generator has an armature current of 20 A, what is its terminal voltage?
(c) If the generator has an armature current of 40 A, what is its terminal voltage'?
(d) Calculate and plot the terminal characteristic of this machine.
SOLUTION
(a) The total field resistance of this generator is 30 , and the no-load terminal voltage can be found
from the intersection of the resistance line with the magnetization curve for this generator. The
magnetization curve and the field resistance line are plotted below. As you can see, they intersect at a
terminal voltage of 121 V.

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 (b) If the armature current is 20 A, then the effective field current contribution from the armature
current
N SE 25
IA   20 A   0.5 A
NF 1000

and the I A  R A  RS  voltage drop is I A  R A  RS    20 A  0.21   4.2 V . The location where the
N SE
triangle formed by I A and I A RA exactly fits between the E A and VT lines corresponds to a terminal
NF
voltage of 121 V, as shown below.

(c) If the armature current is 40 A, then the effective field current contribution from the armature
current
NSE 25
IA   40 A   1.0 A
NF 1000

and the I A  RA  RS  voltage drop is I A  RA  RS    40 A   0.21    8.4 V . The location where

N SE
the triangle formed by I A and I A RA exactly fits between the E A and VT lines corresponds to a
NF
terminal voltage of 121 V, as shown below.

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 The resulting terminal characteristic is shown below. Compare it to the terminal characteristics of the
shunt dc generators in Problem 8-25 (d).

8-28. If the machine described in Problem 8-27 is reconnected as a differentially compounded dc generator,
what will its terminal characteristic look like? Derive it in the same fashion as in Problem 8-27.

The resulting terminal characteristic is shown below. Compare it to the terminal characteristics of the
cumulatively compounded dc generator in Problem 9-28 and the shunt dc generators in Problem 9-25 (d).

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