Chapter 6 Additional Problems
X6.1 The three-phase, wound-rotor induction motor of Fig. X6.1 is operating in the steadystate with a balanced set of three-phase voltage applied to the stator winding when the
two-pole switch is quickly moved from position 1 to position 2. Does the motor speed
increase or decrease in value?
Prior to movement of the switch, the rotor coils were shorted. After movement
of the switch, a balanced set of three-phase resistances have been added to the rotor
winding so that the rotor resistance per phase is now R2  R . Based on Example 6.14
and the resulting Fig. 6.23, it is concluded that the developed torque in the region of
operation is reduced; thus, the motor speed must decrease in value.
X6.2 The three-phase, wound-rotor induction motor of Fig. X6.2 has balanced three-phase
voltages applied. It is operating in the steady-state when the three-pole switch is
suddenly moved from position 1 to position 2. Does the motor reverse direction of
rotation?
1
Study of the switch arrangement shows that after the switch moves, the
impressed voltages are still connected in a  b  c phase sequence in a counter-clockwise
direction of the stator winding. The stator-produced mmf wave does not change
direction of rotation; thus, the rotor does not change rotation direction.
X6.3 Replace the three resistors for the wound-rotor induction motor of Fig. X6.1 by three
identical capacitors sized so that X C  1/  C  X 2 (the rotor leakage reactance).
Make a qualitative sketch of the developed torque-speed curve for the motor with the
switch in position 1 and in position 2.
Based on [6.52], [6.56], and [6.58], with the switch in position 2,
3VTh2
3Td 

 s  RTh 

smax  
R2
s
2

R2 
2
 X Th


s 

R2
1/ 2
2
2 
 RTh
 X Th


3Td max  
3VTh2
2
2 1/ 2 
2 s  RTh   RTh
 X Th
 

It is concluded that for the switch in position 2, the developed torque is larger at all
speeds, and the slip at which maximum developed torque occurs has a larger value
when compared to the case for the switch in position 1. Fig. X6.3 displays a
qualitative sketch of the torque-speed curves.
2
X6.4 Replace the three resistors for the wound-rotor induction motor of Fig. X6.1 by three
identical inductors. Assume the values of inductance are such that X L   L is about
the same order of magnitude as X 2 . Make a qualitative sketch of the developed
torque-speed curve for the motor with the switch in position 1 and in position 2.
Based on [6.52], [6.56], and [6.58], with the switch in position 2,
3Td 
3VTh
R2
s
2

R2 
2
 s  RTh     X Th  X 2  X L  
s 


smax  
R2
1/ 2
 R 2   X  X   X  2 
Th
2
L
 Th

3Td max  
3VTh2

2 1/ 2 
2
2 s  RTh   RTh
  X Th  X 2  X L   

 

It is concluded that for the switch in position 2, the developed torque is smaller at all
speeds, and the slip at which maximum developed torque occurs has a smaller value
when compared to the case for the switch in position 1. Fig. X6.4 shows a qualitative
sketch of the torque-speed curves.
X6.5 A three-phase, 6-pole, 10 HP, 400 Hz induction motor has a slip of 3% at rated output
power. Friction and windage losses are 300 W at rated speed. The rated condition total
core losses are 350 W. R1  R2  0.05  . X1  X 2  0.15  . If the motor is operating at
rated output power, speed, and frequency, find (a) rotor speed, (b) frequency of rotor
3
currents, (c) total power across the air gap, (d) efficiency, and (e) applied line voltage.
Use the approximate equivalent circuit for analysis.
(a)
ns 
120 f 120  400 

 8000 rpm
p
6
nm  1  s  ns  1  0.03 8000   7760 rpm
(b)
(c)
f r  s f   0.03 400   12 Hz
3Pd  Ps  PFW  10  746   300  7760 W
3Pg 
3Pd
7760

 8000 W
1  s  1  0.03
(d) The reflected secondary current is found by
 0.03 8000 / 3  40 A
1/ 2
 sPg 
I 2  

 R2 

Losses  3  I 2 
2
0.05
 R1  R2   3Pc  PFW
 3 40   0.05  0.05  350  300  1130 W
2

Ps 100 
Ps  losses

10 746 100   88.94%
10 746  1130
(e)
V1  I 2 R1 
R2
0.05
 jX eq  40 0.05 
 j 0.3  69.71 V
s
0.03
VL  3 V1  3  69.71  120.7 V
X6.6 The induction motor of Problem X6.5 operates from a 120 V, 400 Hz source. It is
known that due to rotor current distribution and saturation, the starting values of R2
and X 2 are, respectively, 1.3 and 0.8 times the values at full-load condition given in
Problem X6.5. Find the ratio of developed starting torque to developed full-load
torque for this motor if full-voltage is applied for both cases. Use the approximate
equivalent circuit for analysis.
From Problem X6.5, the rated-load reflected secondary current is I 2 R  40 A .
For start conditions,
4
I 2 s 
V1
120 / 3

 R1  R2   j  X1  X 2   0.5  1.3  0.05  j  0.15  0.8  0.15
I 2 s  236.09 A
 I 2 s  R2 s /1   236.09  1.3  0.05 
3Tds 3Pgs /  s


3TdR 3PgR /  s  I 2 R  2 R2 R / sR
 40 2  0.05 / 0.03
2
2
3Tds
 1.36
3TdR
X6.7 A 4-pole, 230 V, three-phase induction motor has a value of secondary resistance such
that the motor produces maximum developed torque at stall. Neglect core losses and
use the Thevenin equivalent circuit for analysis. (See Fig. 6.56). Known equivalent
circuit values are:
R1  0.2
R 2  1.1064
X1  0.5 
X m  20 
Find (a) the reflected value of X 2 and (b) the total developed torque at stall.
(a) Based on [6.50] with Rc   ,
ZTh 
 X m X1  jX m R1   20  0.5  j  20  0.2 

R1  j  X1  X m 
0.2  j  0.5  20 
ZTh  RTh  jX Th  0.19  j 0.49 
Since smax   1 , [6.56] leads to
2
RTh
  X Th  X 2   R2
2
Whence,
 X 2 2  2 X Th X 2   X Th2  RTh2   R2 2   0
Substitute known values and apply the quadratic formula.
 X 2 2  0.98 X 2  0.9479  0
X 2  0.6 
where the negative value was discarded as extraneous.
(b) By [6.49] for Rc   ,
VTh 
X mV1
R12   X1  X m 
2

 20   230 /
3

 0.2 2   0.5  20 2
5
 129.55 V
2
p
s   
2
 2  60  188.49 rad / s
4
By [6.58],
3Td max  
3VTh2
2
2
2 s  RTh  RTh
  X Th  X 2  



3 129.55 
2 188.49  0.19 

2
 0.19 2  1.09 2 

3Td max   103.02 N  m
X6.8 A three-phase,. 4-pole, 600 V, 60 Hz induction motor is modeled by
ZTh  0.6933  j1.933  , R2  4.5  , and X 2  2  . Find the shaft speed at which
maximum torque occurs if the motor is absorbing power from the three-phase lines at
rated frequency.
By [6.56],
smax  
ns 
R2
2
RTh
  X Th  X 2 
2

4.5
 0.6933
2
  3.933
2
 1.127
120 f 120  60 

 1800 rpm
p
4
nm  1  s  ns  1  1.127 1800  228.2 rpm
This motor is operating in the braking or plugging mode. See Fig. 6.22.
X6.9 A three-phase, 230 V induction motor is operating at no-load condition with rated
voltage applied. Equivalent circuit parameters are
R1  0.26 
X1  0.6 
R2  0.4 
X 2  1.4 
Rc  143 
X m  22.2 
It is known that the rated voltage core losses are equal to the rotational losses. Assume
that for this no-load condition the coil resistive voltage drops and the leakage
reactance voltage drops can be neglected. Determine (a) no-load slip, (b) no-load input
power factor, and (c) no-load line current.
(a) Under the stated assumption, the per phase equivalent circuit can be drawn as
shown by Fig. X6.5. Since 3Pc  PFW , and reasoning that the converted power must be
the rotational losses,
V12
V12

R c R2 1  s  / s
(1)
6
Whence,
s
R2
0.4

 0.00279

R2  R c 0.4  143
(b) From (1), it is seen that R c  R2 1  s  / s . Thus,
 Rc 
 143 
  j Xm  
  22.2 90 
2 
2 


Zin 

 21.2 72.75 
Rc
143
 j 22.2
 j Xm
2
2
PFNL  cos  Zin   cos  72.75   0.297 lagging
(c)
I1 
V1 230 / 3

 6.26 A
Z in
21.2
X6.10 A three-phase, 4-pole, NEMA design B, 3 HP, 440 V, 60 Hz induction motor is
known to have a per phase stator resistance of 2.57  . The following test data have
been recorded for the motor:
No-load
Blocked rotor
VL  440 V
I1  2.25 A
PT  220 W
VL  55 V
I1  5.92 A
PT  522 W
f  60 Hz
nm  1796
f  10 Hz
Determine the per phase equivalent circuit parameters for this motor.
7
Based on [6.30]-[6.32],
Req 
Pbr
Z br 
Vbr 55 / 3

 5.364 
I br
5.92
2
Ibr

552 / 3
 5.922
 5.25 
1/ 2
2
2 
X br   Zbr
 Req

1/ 2
2
2
  5.364    5.25  


 0.9456 
By [6.34] and [6.35],
R2  Req  R1  5.25  2.57  2.68 
X eq 
f
60
X br   0.9456   5.67 
fbr
10
By use of Table 6.2,
X1  0.4 X eq  0.4  5.67   2.268 
X 2  0.6 X eq  0.6  5.67   3.402 
Using [6.36]-[6.43],


 Pn 
552 / 3
1
  71.22
  cos 
V
I
440
/
3
2.25
  
 n n 

 n  cos 1 


E1  V10  I n    n  R1  jX1 
E1 
sn 
I 2 
440
3
0  2.25  71.22  2.57  j 2.268   247.370.78 V
ns  nn 1800  1796

 0.00222
ns
1800
E1
1/ 2
 R   2

 2    X 2 2 
 sn 

Pc  Pn  I12 R1   I 2 
Pc 
2

247.37
1/ 2
 2.68 2
2

3.402






 0.00222 
 0.205 A
R2
sn
220
2
2  2.68 
  2.25  2.57    0.205 
  22.60 W
3
 0.00222 
8
E12  247.37 

 2707.6 
Pc
22.60
2
Rc 
Qm  Vn I n sin  n
440
Qm 
3
  I n2 X1   I 2 2 X 2
 2.25 sin  71.22    2.252  2.268   0.2052  3.402 
Qm  529.54 VARs
E 2  247.37 
Xm  1 
 115.6 
Qm
529.54
2
X6.11 A three-phase, 50 HP, 460 V, 60 Hz, 865 rpm induction motor is operating at rated
conditions and has PT  43.75 kW and I1  61 A . It is known that R1  0.15  and
rotational losses at rated speed are 1050 W. Determine (a) full-load power factor,
(b) full-load efficiency, (c) total rotor coil ohmic losses, and (d) total core losses.
(a)
PT
PF 
(b)
3 VL I1

43,750
3  460  61
 0.9 lagging

100 Ps 100  50  746 

 85.26%
PT
43,750
s
ns  nm 900  865

 0.03889
ns
900
(c)
3Pg 
3Pd Ps  PFW 50  746   1050


 39,901.8 W
1 s
1 s
1  0.03889
 
3 I 2  R2  s 3Pg  0.03889  39,901.8  1551.8 W
2
(d)
3Pc  PT  3I12 R1  3Pg
3Pc  43,750  3 61  0.15  39,901.8  2173.7 W
2
X6.12 A three-phase induction motor is supplying 100 HP to a coupled mechanical load. At
this point of operation, the rotational losses are 2046 W, the total core losses are
3320 W, the total stator coil ohmic losses are 2690 W, and the slip is 3%. Determine
(a) the efficiency at this point of operation and (b) the total rotor coil ohmic losses.
9
(a)
3Pg 
3Pd Ps  PFW 100  746   2046


 76,907.2 W
1 s
1 s
1  0.03
PT  3I12 R1  3Pc  3Pg  2690  3320  76,907.2  82,917.2 W

100 Ps 100 100  746 

 89.97%
PT
82,917.2
(b)
 
3 I 2  R2  s 3Pg  0.03 76,907.2  2307.2 W
2
X6.13 A three-phase, 230 V, 60 Hz, 6-pole induction motor has the following equivalent
circuit parameters:
R1  0.13 
R 2  0.09 
R c  175 
X1  0.30 
X2  0.21 
Xm  8.8 
The rotational losses are known to be 332 W at synchronous speed and vary with
speed raised to the 2.8 power. Edit imdata.m to enter the equivalent circuit
parameter values. Edit im_perf.m to enter PFW value. Execute im_perf.m to
determine the rated speed and efficiency.
10
Fig. X6.6 shoes the resulting MATLAB plot of Ps vs. nm where it is seen that
the output power is 15 HP for a speed of 1172 rpm. Fig. X6.7 displays the resulting
efficiency vs. speed. For the rated speed of 1172 rpm,   89.2% .
X6.14 The stator phase coils of a three-phase squirrel-cage induction motor are reconnected
as shown in Fig. X6.8. The capacitor is added for current phase shift in coil b with
intent to build up a self-starting single-phase induction motor. Will the motor selfstart when the sinusoidal source vs is applied?
Yes. Coils a and b combined form a pulsating magnetic field along a vertical
axis. Coil c establishes a pulsating magnetic field along a horizontal axis. These two
fields are separated by 90º in space. Owing to the capacitor C, the two fields peak at
approximately 90º difference in time-phase. If C is properly sized, then the air gap
traveling mmf wave of [6.89] results.
X6.15 The stator phase coils of a three-phase squirrel-cage induction motor are reconnected
as shown in Fig. X6.9 with intent to build up a self-starting single-phase induction
motor. Will the motor self-start when source vs is applied?
11
No. Coils a and b combine to form a pulsating magnetic field along a horizontal
axis. Coil c also establishes a pulsating magnetic field along the same horizontal axis.
With no separation in space between these two magnetic fields, a revolving mmf
wave cannot be established.
12