:
7-1.
A dc test is performed on a 460-V resistance R
1
Δ
-connected 100-hp induction motor. If V
DC
= 24 V and I
DC
= 80 A, what is the stator
? Why is this so? (If this motor’s armature is connected in delta, then there will be two phases in parallel with one phase between the lines tested.
Therefore, the stator resistance R
(a)
The speed of the magnetic fields in revolutions per minute
(b) The speed of the rotor in revolutions per minute
(c) The slip speed of the rotor
(d) The rotor frequency in hertz
1
will be 0.45
Ω
)
7-2.
A 220-V, three-phase, two-pole, 50-Hz induction motor is running at a slip of 5 percent. Find:
((a) n sync
= 3000 r/min, (b) n m
=2850 r/min , (c) n slip
= 150 r/min , (d) f r
= 2.5 Hz)
7-3.
Answer the questions in Problem 7-2 for a 480-V, three-phase, four-pole, 60-Hz induction motor running at a slip of
0.035.
((a) n sync
= 1800 r/min, (b) n m
=1737 r/min , (c) n slip
= 63 r/min , (d) f r
= 2.1 Hz)
7-4.
A three-phase, 60-Hz induction motor runs at 890 r/min at no load and at 840 r/min at full load.
(a) How many poles does this motor have?
(b)
What is the slip at rated load?
(c)
What is the speed at one-quarter of the rated load?
(d) What is the rotor’s electrical frequency at one-quarter of the rated load?
((a) 8 poles, (b) s=6.67%, (c) n = 885 r/min , (d) m f = 1.00 Hz ) r
7-5.
A 50-kW, 440-V, 50-Hz, six-pole induction motor has a slip of 6 percent when operating at full-load conditions. At fullload conditions, the friction and windage losses are 300 W, and the core losses are 600 W. Find the following values for full-load conditions:
(a) The shaft speed n m
(b)
The output power in watts
(c) The load torque τ load in newton-meters d) The induced torque τ ind
in newton-meters
(e) The rotor frequency in hertz
((a) n m
= 940 r/min , (b) The output power in watts is 50 kW (stated in the problem)., (c) τ load
=508 N
⋅ m , (d) τ ind
= 517
N
⋅ m , (d) f r
= 3.00 Hz )
7-5.
A three-phase, 60-Hz, four-pole induction motor runs at a no-load speed of 1790 r/min and a full-load speed of 1720 r/min. Calculate the slip and the electrical frequency of the rotor at no-load and full-load conditions. What is the speed regulation of this motor [Equation (4-68)]? (at no-load s nl
=0.56%, f =0.33 Hz , at full-load s fl
= 4.44%, f = 2.67
Hz, SR = 4.1% )
7-7. A 208-V, two-pole, 60-Hz Y-connected wound-rotor induction motor is rated at 15 hp. Its equivalent circuit components are
R
1
X
1
= 0.200 Ω
= 0.410 Ω
P mech
= 250 W
For a slip of 0.05, find
R
X
2
= 0.410 Ω
P misc
≈ 0
Ω X
P
M core
= 15.0 Ω
= 180 W
(a)
(b)
The line current I
L
The stator copper losses
(c) The air-gap power P
AG
P
SCL
(d) The power converted from electrical to mechanical form P conv
(e) The induced torque τ ind
(f) The load torque τ load
(g) The overall machine efficiency
(h) The motor speed in revolutions per minute and radians per second
((a) I
L
= 44.8
∠ − 25.5
o
A , (b) P
SCL
=1205 W , (c) P
AG
= 13.4 kW , (d) P conv
= 12.73 kW , (e) τ ind
= 35.5 N
⋅ m ,

(f) τ load
= 34.3 N
⋅ m , (g)
η
=84.5% , (h)
ω
=3420 r/min = 358 rad/s ) m
7-8. For the motor in Problem 7-7, what is the slip at the pullout torque? What is the pullout torque of this motor? (s max
=0.144,
τ max
= 53.1
N m )
7-10. For the motor of Problem 7-7, how much additional resistance (referred to the stator circuit) would it be necessary to add to the rotor circuit to make the maximum torque occur at starting conditions (when the shaft is not moving)? Plot the torque-speed characteristic of this motor with the additional resistance inserted. (an additional 0.713 Ω must be added to the rotor circuit. The resulting torque-speed characteristic is:
)
7-11. If the motor in Problem 7-7 is to be operated on a 50-Hz power system, what must be done to its supply voltage? Why?
What will the equivalent circuit component values be at 50 Hz? Answer the questions in Problem 7-7 for operation at 50
Hz with a slip of 0.05 and the proper voltage for this machine.( the input frequency is decreased to 50 Hz, then the applied voltage must be decreased by 5/6 also. If this were not done, the flux in the motor would go into saturation, since
φ =
1
N
T
vdt and the period T would be increased. At 50 Hz, the resistances will be unchanged, but the reactances will be reduced to 5/6 of their previous values. The equivalent circuit of the induction motor at 50 Hz is shown below:
((a) I
L
= 38.0
∠ − 24.4
o
A , (b) P
SCL
=886 W , (c) P
AG
= 9.52 kW , (d) P conv
= 9.04 kW , (e) τ ind
= 30.3 N
⋅ m ,
(f) τ load
= 28.8 N
⋅ m , (g)
η
=82.9% , (h)
ω
=2850 r/min = 298.5 rad/s ) m
7-14. A 440-V 50-Hz two-pole Y-connected induction motor is rated at 75 kW. The equivalent circuit parameters are
R
1
= 0.075 Ω
X
1
= 0.17 Ω
P mech
= 1.0 kW
For a slip of 0.04, find
R Ω
X
2
= 0.17 Ω
P misc
= 150 W
X
M
= 7.2 Ω
P core
= 1.1 kW

(a)
The line current
I
L
(b) The stator power factor
(c) The rotor power factor
(d) The stator copper losses P
SCL
(e)
The air-gap power
P
AG
(f) The power converted from electrical to mechanical form P conv
(g)
(h)
The induced torque
The load torque
τ ind
τ load
(i) The overall machine efficiency
(j) The motor speed in revolutions per minute and radians per second
((a) I
L
= 149.4
∠ − 18.3
o A , (b) ) PF = 0.949 lagging , (c) PF
R
= 0.995 lagging , (d) P
SCL
= 1675 W , (e) P
AG
(f) P conv
= 98.9 kW , (g) τ ind
=327.9
N
⋅ m , (h) τ load
= 327.6 N
⋅ m , (i) η = 89.4% , (j)
= 103 kW ,
ω
=2880 r/min = 301.6 rad/s ) m
7-15. For the motor in Problem 7-14, what is the pullout torque? What is the slip at the pullout torque? What is the rotor speed at the pullout torque? (s max
=0.189,
τ max
= 704
N m , )
7-16. If the motor in Problem 7-14 is to be driven from a 440-V 60-Hz power supply, what will the pullout torque be? What will the slip be at pullout? (s max
=0.159,
τ max
= 507
N m )
7-18. A 208-V, 60 Hz, six-pole Y-connected 25-hp design class B induction motor is tested in the laboratory, with the following results:
No load:
Locked rotor:
DC test:
208 V, 22.0 A, 1200 W, 60 Hz
24.6 V, 64.5 A, 2200 W, 15 Hz
13.5 V, 64 A
Find the equivalent circuit of this motor, and plot its torque-speed characteristic curve.
(The resulting equivalent circuit is shown below:
The resulting plot is shown below:
)

7-19.
A 460-V, four-pole, 50-hp, 60-Hz, Y-connected three-phase induction motor develops its full-load induced torque at 3.8 percent slip when operating at 60 Hz and 460 V. The per-phase circuit model impedances of the motor are
R
1
X
1
= 0.33 Ω
= 0.42 Ω
= 30 Ω
X
2
= 0.42 Ω
Mechanical, core, and stray losses may be neglected in this problem.
(a) Find the value of the rotor resistance R
2
.
(b) Find
τ
,s max max
and the rotor speed at maximum torque for this motor.
(c) Find the starting torque of this motor.
(d) What code letter factor should be assigned to this motor?
((a) R
2
= 0.0059 Ω or 0.172 Ω These two solutions represent two situations in which the torque-speed curve would go through this specific torque-speed point. The two curves are plotted below. As you can see, only the 0.172 Ω solution is realistic, since the 0.0059 Ω solution passes through this torque-speed point at an unstable location on the back side of the torque-speed curve.
(b)
τ
= 448 N
⋅ m and s max max
= 0.192, (c)
τ
=199 N
⋅ m , (d) The phase voltage is 460 / 3 = 266 V, ind
L start
= 274 ∠ − 59.2
o
A
, S = 218 kVA, kVA/hp = 4.36 kVA/hp
D covers the range 4.00-4.50.)
7-20. Answer the following questions about the motor in Problem 7-19.
(a) If this motor is started from a 460-V infinite bus, how much current will flow in the motor at starting?
(b) If transmission line with an impedance of 0.35 + j 0.25 . per phase is used to connect the induction motor to the infinite bus, what will the starting current of the motor be? What will the motor’s terminal voltage be on starting?
(c) If an ideal 1.4:1 step-down autotransformer is connected between the transmission line and the motor, what will the current be in the transmission line during starting? What will the voltage be at the motor end of the transmission line
((a) during starting?
I
L
= 273 ∠ − 59.2
o
A , (b) I
L
= 193.2
∠ − 52.0
o
A ,
, This motor would have starting code letter D, since letter terminal voltage will be 325 V . Note that the terminal voltage sagged byabout 30% during motor starting, which would be unacceptable.
, (c) the starting current referred to the primary side of the transformer I
'
L
= 115.4
∠ − 54.9
o .
The line voltage at the motor end of the transmission line will be
380.5 V . Note that this voltage sagged by 17.3% during motor starting, which is less than the 30% sag with case of across-the-line starting.
)
7-21. In this chapter, we learned that a step-down autotransformer could be used to reduce the starting current drawn by an induction motor. While this technique works, an autotransformer is relatively expensive. A much less expensive way to reduce the starting current is to use a device called
Y
Δ starter
. If an induction motor is normally Δ -connected, it is possible to reduce its phase voltage V
φ
(and hence its starting current) by simply re-connecting the stator windings in Y during starting, and then restoring the connections to Δ when the motor comes up to speed. Answer the following questions about this type of starter.
(a) How would the phase voltage at starting compare with the phase voltage under normal running conditions?
(b)
How would the starting current of the Y-connected motor compare to the starting current if the motor remained in a
Δ -connection during starting?

((a) The phase voltage at starting would be 57.7% of the phase voltage under normal running conditions. (b) the line current is reduced by a factor of 3 in a Y-
Δ starter. I
L
, Δ
= 3 I )
7-22. A 460-V 100-hp four-pole Δ -connected 60-Hz three-phase induction motor has a full-load slip of 5 percent, an efficiency of 92 percent, and a power factor of 0.87 lagging. At start-up, the motor develops 1.9 times the full-load torque but draws
7.5 times the rated current at the rated voltage. This motor is to be started with an autotransformer reduced voltage starter.
(a) What should the output voltage of the starter circuit be to reduce the starting torque until it equals the rated torque of the motor?
(b) What will the motor starting current and the current drawn from the supply be at this voltage?
((a) V
T
2
= 334 V , (b) I line
= 463 A )
7-23. A wound-rotor induction motor is operating at rated voltage and frequency with its slip rings shorted and with a load of about 25 percent of the rated value for the machine. If the rotor resistance of this machine is doubled by inserting external resistors into the rotor circuit, explain what happens to the following:
(a) Slip s
(b) Motor speed n m
(c)
The induced voltage in the rotor
(d) The rotor current
(e)
τ max
(f) P out
(g) P
RCL
(h) Overall efficiency η
((a) The slip s will increase. (b) The motor speed n m
will decrease.(c) The induced voltage in the rotor will increase. (d)
The rotor current will increase. (e) The induced torque will adjust to supply the load’s torque requirements at the new speed. This will depend on the shape of the load’s torque-speed characteristic. For most loads, the induced torque will decrease. (f) The output power will generally decrease:
P
OUT
= τ ↓ ω m
↑ , (g) The rotor copper losses (including the ind external resistor) will increase. , (h) The overall efficiency η will decrease.)
7-25.
When it is necessary to stop an induction motor very rapidly, many induction motor controllers reverse the direction of rotation of the magnetic fields by switching any two stator leads. When the direction of rotation of the magnetic fields is reversed, the motor develops an induced torque opposite to the current direction of rotation, so it quickly stops and tries to start turning in the opposite direction. If power is removed from the stator circuit at the moment when the rotor speed goes through zero, then the motor has been stopped very rapidly. This technique for rapidly stopping an induction motor is called plugging . The motor of Problem 7-19 is running at rated conditions and is to be stopped by plugging.
(a) What is the slip s before plugging?
(b) What is the frequency of the rotor before plugging?
(c) What is the induced torque
τ
before plugging? ind
(d)
What is the slip s immediately after switching the stator leads?
(e) What is the frequency of the rotor immediately after switching the stator leads?
(f) What is the induced torque
τ
immediately after switching the stator leads?
ind
((a) The slip before plugging is 0.038
, (b) The frequency of the rotor before plugging is f r
= 2.28 Hz, (c) The induced torque
τ before plugging is 205.7 N ⋅ m in the direction of motion, (d) s = 1.962, (e) The frequency of the rotor after ind plugging is f r
= 117.72 Hz , (f) The induced torque immediately after switching the stator leads is
τ
=110 N ⋅ m, ind opposite the direction of motion )