Chapter 7: Induction Motors
7-1. A dc test is performed on a 460-V Δ-connected 100-hp induction motor. If VDC = 21 V and I DC = 72 A,
what is the stator resistance R1 ? Why is this so?
SOLUTION If this motor’s armature is connected in delta, then there will be two phases in parallel with one
phase between the lines tested.
VDC R1 R1
R1
Therefore, the stator resistance R1 will be
VDC R (R + R1 ) 2
= 1 1 = R1
I DC R1 + (R1 + R1 ) 3
3 VDC 3 § 21 V ·
R1 = = ¨ ¸ = 0.438 Ω
2 I DC 2 © 72 A ¹
7-2. A 220-V three-phase six-pole 50-Hz induction motor is running at a slip of 3.5 percent. Find:
(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
SOLUTION
(a) The speed of the magnetic fields is
120 f e 120(50 Hz )
nsync = = = 1000 r/min
P 6
(b) The speed of the rotor is
nm = (1 − s ) nsync = (1 − 0.035)(1000 r/min ) = 965 r/min
(c) The slip speed of the rotor is
nslip = snsync = (0.035)(1000 r/min ) = 35 r/min
(d) The rotor frequency is
fr =
nslip P
=
( 35 r/min )( 6) = 1.75 Hz
120 120
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 0.025.
SOLUTION
(a) The speed of the magnetic fields is
114
120 f e 120 ( 60 Hz )
nsync = = = 1800 r/min
P 4
(b) The speed of the rotor is
nm = (1 − s ) nsync = (1 − 0.025)(1800 r/min ) = 1755 r/min
(c) The slip speed of the rotor is
nslip = snsync = ( 0.025)(1800 r/min ) = 45 r/min
(d) The rotor frequency is
fr =
nslip P
=
( 45 r/min )( 4) = 1.5 Hz
120 120
7-4. A three-phase 60-Hz induction motor runs at 715 r/min at no load and at 670 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?
SOLUTION
(a) This machine has 10 poles, which produces a synchronous speed of
120 f e 120 ( 60 Hz )
nsync = = = 720 r/min
P 10
(b) The slip at rated load is
nsync − nm 720 − 670
s= × 100% = × 100% = 6.94%
nsync 720
(c) The motor is operating in the linear region of its torque-speed curve, so the slip at ¼ load will be
s = 0.25(0.0694) = 0.0174
The resulting speed is
nm = (1 − s ) nsync = (1 − 0.0174 )( 720 r/min ) = 707 r/min
(d) The electrical frequency at ¼ load is
f r = sf e = ( 0.0174 )( 60 Hz ) = 1.04 Hz
7-5. A 50-kW 440-V 50-Hz two-pole induction motor has a slip of 6 percent when operating at full-load
conditions. At full-load conditions, the friction and windage losses are 520 W, and the core losses are 500
W. Find the following values for full-load conditions:
(a) The shaft speed nm
(b) The output power in watts
(c) The load torque τ load in newton-meters
(d) The induced torque τ ind in newton-meters
115
(e) The rotor frequency in hertz
SOLUTION
(a) The synchronous speed of this machine is
120 f e 120 ( 50 Hz )
nsync = = = 3000 r/min
P 2
Therefore, the shaft speed is
nm = (1 − s ) nsync = (1 − 0.06)( 3000 r/min ) = 2820 r/min
(b) The output power in watts is 50 kW (stated in the problem).
(c) The load torque is
POUT 50 kW
τ load = = = 169.3 N ⋅ m
ωm § 2π rad ·§ 1 min ·
(2820 r/min )¨ ¸¨ ¸
© 1 r ¹© 60 s ¹
(d) The induced torque can be found as follows:
Pconv = POUT + PF&W + Pcore + Pmisc = 50 kW + 520 W + 500 W = 51.2 kW
Pconv 51.2 kW
τ ind = = = 173.4 N ⋅ m
ωm § 2π rad ·§ 1 min ·
(2820 r/min )¨ ¸¨ ¸
© 1 r ¹© 60 s ¹
(e) The rotor frequency is
f r = sf e = ( 0.06)( 50 Hz ) = 3.00 Hz
7-6. A three-phase 60-Hz two-pole induction motor runs at a no-load speed of 3580 r/min and a full-load
speed of 3440 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 (4-57)]?
SOLUTION The synchronous speed of this machine is 3600 r/min. The slip and electrical frequency at no-
load conditions is
nsync − n nl 3600 − 3580
s nl = × 100% = × 100% = 0.56%
nsync 3600
f r ,nl = sf e = ( 0.0056)( 60 Hz ) = 0.33 Hz
The slip and electrical frequency at full load conditions is
nsync − n nl 3600 − 3440
sfl = × 100% = × 100% = 4.44%
nsync 3600
f r ,fl = sf e = ( 0.0444 )( 60 Hz ) = 2.67 Hz
The speed regulation is
n nl − nfl 3580 − 3440
SR = × 100% = × 100% = 4.1%
nfl 3440
116
7-7. A 208-V four-pole 60-Hz Y-connected wound-rotor induction motor is rated at 15 hp. Its equivalent
circuit components are
R1 = 0.220 Ω R2 = 0.127 Ω X M = 15.0 Ω
X 1 = 0.430 Ω X 2 = 0.430 Ω
Pmech = 300 W Pmisc ≈ 0 Pcore = 200 W
For a slip of 0.05, find
(a) The line current I L
(b) The stator copper losses PSCL
(c) The air-gap power PAG
(d) The power converted from electrical to mechanical form Pconv
(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
SOLUTION The equivalent circuit of this induction motor is shown below:
IA
R1 jX1 jX2 R2
+ j0.43 Ω 0.127 Ω
0.22 Ω j0.43 Ω
jXM §1 − s ·
Vφ j15 Ω R2 ¨ ¸
© s ¹
2.413 Ω
-
(a) The easiest way to find the line current (or armature current) is to get the equivalent impedance Z F
of the rotor circuit in parallel with jX M , and then calculate the current as the phase voltage divided by
the sum of the series impedances, as shown below.
IA
R1 jX1 jXF RF
+
0.22 Ω j0.43 Ω
Vφ
The equivalent impedance of the rotor circuit in parallel with jX M is:
1 1
ZF = = = 2.337 + j 0.803 = 2.47∠19° Ω
1 1 1 1
+ +
jX M Z 2 j15 Ω 2.54 + j 0.43
117
The phase voltage is 208/ 3 = 120 V, so line current I L is
Vφ 120∠0° V
IL = IA = =
R1 + jX 1 + RF + jX F 0.22 Ω + j 0.43 Ω + 2.337 Ω + j 0.803 Ω
I L = I A = 42.3∠ − 25.7° A
(b) The stator copper losses are
PSCL = 3I A R1 = 3 ( 42.3 A ) ( 0.22 Ω ) = 1180 W
2 2
2 R2 2
(c) The air gap power is PAG = 3I 2 = 3I A R F
s
2 R2 2
(Note that 3I A RF is equal to 3I 2 , since the only resistance in the original rotor circuit was R2 / s ,
s
and the resistance in the Thevenin equivalent circuit is RF . The power consumed by the Thevenin
equivalent circuit must be the same as the power consumed by the original circuit.)
R2
= 3I A RF = 3 ( 42.3 A ) ( 2.337 Ω ) = 12.54 kW
2 2 2
PAG = 3I 2
s
(d) The power converted from electrical to mechanical form is
Pconv = (1 − s ) PAG = (1 − 0.05)(12.54 kW ) = 11.92 kW
(e) The induced torque in the motor is
PAG 12.54 kW
τ ind = = = 66.5 N ⋅ m
ω sync § 2π rad ·§ 1 min ·
(1800 r/min)¨ ¸¨ ¸
© 1 r ¹© 60 s ¹
(f) The output power of this motor is
POUT = Pconv − Pmech − Pcore − Pmisc = 11.92 kW − 300 W − 200 W − 0 W = 11.42 kW
The output speed is
nm = (1 − s ) nsync = (1 − 0.05)(1800 r/min ) = 1710 r/min
Therefore the load torque is
POUT 11.42 kW
τ load = = = 63.8 N ⋅ m
ωm
(1710 r/min)§¨ 2π rad ·¸§¨ 1 min ·¸
© 1 r ¹© 60 s ¹
(g) The overall efficiency is
POUT POUT
η= × 100% = × 100%
PIN 3Vφ I A cos θ
11.42 kW
η= × 100% = 83.2%
3 (120 V )( 42.3 A ) cos 25.7°
(h) The motor speed in revolutions per minute is 1710 r/min. The motor speed in radians per second is
118
§ 2π rad · § 1 min ·
ω m = (1710 r/min ) ¨ = 179 rad/s
© 1 r ¸¹ ¨© 60 s ¸¹
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?
SOLUTION The slip at pullout torque is found by calculating the Thevenin equivalent of the input circuit
from the rotor back to the power supply, and then using that with the rotor circuit model.
jX M ( R1 + jX 1 ) ( j15 Ω )( 0.22 Ω + j0.43 Ω ) = 0.208 + j0.421 Ω = 0.470∠63.7° Ω
Z TH = =
R1 + j ( X 1 + X M ) 0.22 Ω + j ( 0.43 Ω + 15 Ω )
VTH =
jX M
Vφ =
( j15 Ω ) (120∠0° V ) = 116.7∠0.8° V
R1 + j ( X 1 + X M ) 0.22 Ω + j (0.43 Ω + 15 Ω )
The slip at pullout torque is
R2
smax =
RTH + ( X TH + X 2 )
2 2
0.127 Ω
smax = = 0.145
( 0.208 Ω ) + ( 0.421 Ω
2
+ 0.430 Ω )
2
The pullout torque of the motor is
119
2
3VTH
τ max =
2ω sync ª RTH + RTH + ( X TH + X 2 ) º
2 2
«¬ »¼
3 (116.7 V )
2
τ max =
2 (188.5 rad/s ) ª0.208 Ω + ( 0.208 Ω ) + ( 0.421 Ω + 0.430 Ω ) º
2 2
«¬ »¼
τ max = 100 N ⋅ m
7-9. (a) Calculate and plot the torque-speed characteristic of the motor in Problem 7-7. (b) Calculate and plot
the output power versus speed curve of the motor in Problem 7-7.
SOLUTION
(a) A MATLAB program to calculate the torque-speed characteristic is shown below.
% M-file: prob7_9a.m
% M-file create a plot of the torque-speed curve of the
% induction motor of Problem 7-7.
% First, initialize the values needed in this program.
r1 = 0.220; % Stator resistance
x1 = 0.430; % Stator reactance
r2 = 0.127; % Rotor resistance
x2 = 0.430; % Rotor reactance
xm = 15.0; % Magnetization branch reactance
v_phase = 208 / sqrt(3); % Phase voltage
n_sync = 1800; % Synchronous speed (r/min)
w_sync = 188.5; % Synchronous speed (rad/s)
% Calculate the Thevenin voltage and impedance from Equations
% 7-38 and 7-41.
v_th = v_phase * ( xm / sqrt(r1^2 + (x1 + xm)^2) );
z_th = ((j*xm) * (r1 + j*x1)) / (r1 + j*(x1 + xm));
r_th = real(z_th);
x_th = imag(z_th);
% Now calculate the torque-speed characteristic for many
% slips between 0 and 1. Note that the first slip value
% is set to 0.001 instead of exactly 0 to avoid divide-
% by-zero problems.
s = (0:1:50) / 50; % Slip
s(1) = 0.001;
nm = (1 - s) * n_sync; % Mechanical speed
% Calculate torque versus speed
for ii = 1:51
t_ind(ii) = (3 * v_th^2 * r2 / s(ii)) / ...
(w_sync * ((r_th + r2/s(ii))^2 + (x_th + x2)^2) );
end
% Plot the torque-speed curve
figure(1);
plot(nm,t_ind,'k-','LineWidth',2.0);
xlabel('\bf\itn_{m}');
ylabel('\bf\tau_{ind}');
120
V
IL =
(RS + RL ) 2
+ (X S + X L )
2
The power supplied to the load is
2 V 2 RL
P = I L RL =
(RS + RL )2 + ( X S + X L )2
∂P ¬( RS + RL ) + ( X S + X L ) ¼ V − V RL ¬ª2 ( RS + RL ) ¼º
ª 2 2
º 2 2
= 2
∂RL ª( RS + RL ) 2 + ( X S + X L ) 2 º
¬ ¼
To find the point of maximum power supplied to the load, set ∂P / ∂RL = 0 and solve for RL .
ª( RS + RL ) 2 + ( X S + X L ) 2 º V 2 − V 2 RL ª2 ( RS + RL ) º = 0
¬ ¼ ¬ ¼
ª( RS + RL ) 2 + ( X S + X L ) 2 º = 2 RL ( RS + RL )
¬ ¼
RS + 2 RS RL + RL + ( X S + X L ) = 2 RS RL + 2 RL
2 2 2 2
RS + RL + ( X S + X L ) = 2 RL
2 2 2 2
RS + ( X S + X L ) = RL
2 2 2
Therefore, for maximum power transfer, the load resistor should be
R L = RS + ( X S + X L )
2 2
7-14. A 440-V 50-Hz six-pole Y-connected induction motor is rated at 75 kW. The equivalent circuit
parameters are
R1 = 0.082 Ω R2 = 0.070 Ω X M = 7.2 Ω
X 1 = 0.19 Ω X 2 = 0.18 Ω
PF&W = 1.3 kW Pmisc = 150 W Pcore = 1.4 kW
For a slip of 0.04, find
(a) The line current I L
(b) The stator power factor
(c) The rotor power factor
(d) The stator copper losses PSCL
(e) The air-gap power PAG
(f) The power converted from electrical to mechanical form Pconv
(g) The induced torque τ ind
(h) The load torque τ load
(i) The overall machine efficiency η
127
(j) The motor speed in revolutions per minute and radians per second
SOLUTION The equivalent circuit of this induction motor is shown below:
IA
R1 jX1 jX2 R2
+
0.082 Ω j0.19 Ω
j0.18 Ω 0.07 Ω I2
jXM §1− s ·
Vφ j7.2 Ω R2 ¨ ¸
© s ¹
1.68 Ω
-
(a) The easiest way to find the line current (or armature current) is to get the equivalent impedance Z F
of the rotor circuit in parallel with jX M , and then calculate the current as the phase voltage divided by
the sum of the series impedances, as shown below.
IA
R1 jX1 jXF RF
+
0.082 Ω j0.19 Ω
Vφ
The equivalent impedance of the rotor circuit in parallel with jX M is:
1 1
ZF = = = 1.557 + j 0.550 = 1.67∠19.2° Ω
1 1 1 1
+ +
jX M Z 2 j 7.2 Ω 1.75 + j 0.18
The phase voltage is 440/ 3 = 254 V, so line current I L is
Vφ 254∠0° V
IL = IA = =
R1 + jX 1 + RF + jX F 0.082 Ω + j 0.19 Ω + 1.557 Ω + j 0.550 Ω
I L = I A = 141∠ − 24.3° A
(b) The stator power factor is
PF = cos 24.3° = 0.911 lagging
(c) To find the rotor power factor, we must find the impedance angle of the rotor
X2 0.18
θ R = tan −1 = tan −1 = 5.87°
R2 / s 1.75
Therefore the rotor power factor is
PFR = cos 5.87° = 0.995 lagging
(d) The stator copper losses are
PSCL = 3I A R1 = 3 (141 A ) ( 0.082 Ω ) = 4890 W
2 2
128
2 R2 2
(e) The air gap power is PAG = 3I 2 = 3I A R F
s
2 R2 2
(Note that 3I A RF is equal to 3I 2 , since the only resistance in the original rotor circuit was R2 / s ,
s
and the resistance in the Thevenin equivalent circuit is RF . The power consumed by the Thevenin
equivalent circuit must be the same as the power consumed by the original circuit.)
R2
= 3I A RF = 3 (141 A ) (1.557 Ω ) = 92.6 kW
2 2 2
PAG = 3I 2
s
(f) The power converted from electrical to mechanical form is
Pconv = (1 − s ) PAG = (1 − 0.04 )( 92.6 kW ) = 88.9 kW
(g) The synchronous speed of this motor is
120 f e 120(50 Hz )
nsync = = = 1000 r/min
P 6
§ 2π rad · § 1 min ·
ω sync = (1000 r/min ) ¨ = 104.7 rad/s
© 1 r ¸¹ ¨© 60 s ¸¹
Therefore the induced torque in the motor is
PAG 92.6 kW
τ ind = = = 884 N ⋅ m
ω sync § 2π rad ·§ 1 min ·
(1000 r/min)¨ ¸¨ ¸
© 1 r ¹© 60 s ¹
(h) The output power of this motor is
POUT = Pconv − Pmech − Pcore − Pmisc = 88.9 kW − 1.3 kW − 1.4 kW − 300 W = 85.9 kW
The output speed is
nm = (1 − s ) nsync = (1 − 0.04 )(1000 r/min ) = 960 r/min
Therefore the load torque is
POUT 85.9 kW
τ load = = = 854 N ⋅ m
ωm
(960 r/min )§¨ 2π rad ·¸§¨ 1 min ·¸
© 1 r ¹© 60 s ¹
(i) The overall efficiency is
POUT POUT
η= × 100% = × 100%
PIN 3Vφ I A cos θ
85.9 kW
η= × 100% = 87.7%
3 ( 254 V )(141 A ) cos 24.3°
(j) The motor speed in revolutions per minute is 960 r/min. The motor speed in radians per second is
§ 2π rad ·§ 1 min ·
ω m = (960 r/min )¨ ¸¨ ¸ = 100.5 rad/s
© 1 r ¹© 60 s ¹
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