Chapter 9: Single-Phase and Special-Purpose Motors
9-1. A 120-V 1/4-hp 60-Hz four-pole split-phase induction motor has the following impedances:
R1 = 2.00 X 1 = 2.56 X M = 60.5
R2 = 2.80 X 2 = 2.56
At a slip of 0.05, the motors rotational losses are 51 W. The rotational losses may be assumed constant
over the normal operating range of the motor. If the slip is 0.05, find the following quantities for this
motor:
(a) Input power
(b) Air-gap power
(c) Pconv
(d) Pout
(e) ind
(f) load
(g) Overall motor efficiency
(h) Stator power factor
SOLUTION The equivalent circuit of the motor is shown below
I1
1 1 0.5 2
+ 1.28 ?
2.0 ? 2.56 ?
0.5 R2
0.5 0.5 Forward
30.25 ? s
V
2
1.28 ?
R2 Reverse
0.5 0.5XM 0.5
30.25 ?
2 s
The impedances Z F and Z B are:
R2 / s jX 2 jX M
ZF
R2 / s jX 2 jX M
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The slip s 0.05 , so R2 / s 2.80 / 0.05 56
56 j 2.56 j 60.5
ZF 28.82 j 28.05
56 j 2.56 j 60.5
R2 / 2 s jX 2 jX M
ZB
R2 / 2 s jX 2 jX M
The slip s 0.05 , so R2 / 2 s 2.80 / 2 0.05 1.436
1.436 j 2.56 j 60.5
ZB 1.321 j 2.486
1.436 j 2.56 j 60.5
(a) The input current is
V
I1
R1 jX 1 0.5Z F 0.5Z B
120 0 V
I1 4.862 46.2 A
2.00 j 2.56 0.5 28.82 j 28.05 0.5 1.321 j 2.486
PIN VI cos 120 V 4.862 A cos 46.2 403.8 W
(b) The air-gap power is
2
PAG,F I12 0.5RF 4.862 A 14.41 340.6 W
2 2
PAG,B I1 0.5 RB 4.862 A 0.661 15.6 W
PAG PAG,F PAG,B 340.6 W 15.6 W 325 W
(c) The power converted from electrical to mechanical form is
Pconv,F 1 s PAG,F 1 0.05 340.6 W 323 W
Pconv,B 1 s PAG,B 1 0.05 15.6 W 14.8 W
Pconv Pconv,F Pconv,B 323 W 14.8 W 308 W
(d) The output power is
POUT Pconv Prot 308 W 51 W 257 W
(e) The induced torque is
PAG 325 W
ind 1.72 N m
2 rad 1 min
sync
1800 r/min
1r 60 s
(f) The load torque is
POUT 257 W
load 1.44 N m
2 rad 1 min
m
0.95 1800 r/min
1r 60 s
(g) The overall efficiency is
POUT 257 W
100% 100% 63.6%
PIN 403.8 W
277
(h) The stator power factor is
PF cos 46.2 0.692 lagging
9-2. Repeat Problem 9-1 for a rotor slip of 0.025.
The impedances Z F and Z B are:
R2 / s jX 2 jX M
ZF
R2 / s jX 2 jX M
The slip s 0.025 , so R2 / s 2.80 / 0.025 112
112 j 2.56 j 60.5
ZF 24.81 j 46.53
112 j 2.56 j 60.5
R2 / 2 s jX 2 jX M
ZB
R2 / 2 s jX 2 jX M
The slip s 0.025 , so R2 / 2 s 2.80 / 2 0.025 1.418
1.418 j 2.56 j 60.5
ZB 1.305 j 2.485
1.418 j 2.56 j 60.5
(a) The input current is
V
I1
R1 jX 1 0.5Z F 0.5Z B
120 0 V
I1 3.874 60.9 A
2.00 j 2.56 0.5 24.81 j 46.53 0.5 1.305 j 2.485
PIN VI cos 120 V 3.874 A cos 60.9 226.1 W
(b) The air-gap power is
2
PAG,F I12 0.5RF 3.874 A 12.41 186.2 W
2
PAG,B I12 0.5RB 3.874 A 0.653 9.8 W
PAG PAG,F PAG,B 186.2 W 9.8 W 176.4 W
(c) The power converted from electrical to mechanical form is
Pconv,F 1 s PAG,F 1 0.025 186.2 W 181.5 W
Pconv,B 1 s PAG,B 1 0.025 9.8 W 9.6 W
Pconv Pconv,F Pconv,B 181.5 W 9.6 W 171.9 W
(d) The output power is
POUT Pconv Prot 171.9 W 51 W 120.9 W
(e) The induced torque is
PAG 176.4 W
ind 0.936 N m
2 rad 1 min
sync
1800 r/min
1r 60 s
(f) The load torque is
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POUT 120.9 W
load 0.658 N m
2 rad 1 min
m
0.975 1800 r/min
1r 60 s
(g) The overall efficiency is
POUT 120.9 W
100% 100% 53.5%
PIN 226.1 W
(h) The stator power factor is
PF cos 60.9 0.486 lagging
9-3. Suppose that the motor in Problem 9-1 is started and the auxiliary winding fails open while the rotor is
accelerating through 400 r/min. How much induced torque will the motor be able to produce on its main
winding alone? Assuming that the rotational losses are still 51 W, will this motor continue accelerating or
will it slow down again? Prove your answer.
SOLUTION At a speed of 400 r/min, the slip is
1800 r/min 400 r/min
s 0.778
1800 r/min
The impedances Z F and Z B are:
R2 / s jX 2 jX M
ZF
R2 / s jX 2 jX M
The slip s 0.778 , so R2 / s 2.80 / 0.778 3.60
3.60 j 2.56 j 60.5
ZF 3.303 j 2.645
3.60 j 2.56 j 60.5
R2 / 2 s jX 2 jX M
ZB
R2 / 2 s jX 2 jX M
The slip s 0.778 , so R2 / 2 s 2.80 / 2 0.778 2.291
2.291 j 2.56 j 60.5
ZB 2.106 j 2.533
2.291 j 2.56 j 60.5
The input current is
V
I1
R1 jX 1 0.5Z F 0.5Z B
120 0 V
I1 17.21 47.6 A
2.00 j 2.56 0.5 3.303 j 2.645 0.5 2.106 j 2.533
The air-gap power is
2
PAG,F I12 0.5RF 17.21 A 3.303 978.3 W
2
PAG,B I12 0.5 RB 17.21 A 2.106 623.8 W
PAG PAG,F PAG,B 978.3 W 623.8 W 354.5 W
The power converted from electrical to mechanical form is
279
Pconv,F 1 s PAG,F 1 0.778 978.3 W 217.2 W
Pconv,B 1 s PAG,B 1 0.778 623.8 W 138.5 W
Pconv Pconv,F Pconv,B 217.2 W 138.5 W 78.7 W
The output power is
POUT Pconv Prot 78.7 W 51 W 27.7 W
The induced torque is
PAG 354.5 W
ind 1.88 N m
2 rad 1 min
sync
1800 r/min
1r 60 s
Assuming that the rotational losses are still 51 W, this motor will still be able to speed up because Pconv is
78.7 W, while the rotational losses are 51 W, so there is more power than it required to cover the
rotational losses. The motor will continue to speed up.
9-4. Use MATLAB to calculate and plot the torque-speed characteristic of the motor in Problem 9-1, ignoring
the starting winding.
SOLUTION This problem is best solved with MATLAB, since it involves calculating the torque-speed
values at many points. A MATLAB program to calculate and display both torque-speed characteristics is
shown below. Note that this program shows the torque-speed curve for both positive and negative
directions of rotation. Also, note that we had to avoid calculating the slip at exactly 0 or 2, since those
numbers would produce divide-by-zero errors in Z F and Z B respectively.
% M-file: prob9_4.m
% M-file create a plot of the torque-speed curve of the
% single-phase induction motor of Problem 9-4.
% First, initialize the values needed in this program.
r1 = 2.00; % Stator resistance
x1 = 2.56; % Stator reactance
r2 = 2.80; % Rotor resistance
x2 = 2.56; % Rotor reactance
xm = 60.5 % Magnetization branch reactance
v = 120; % Single-Phase voltage
n_sync = 1800; % Synchronous speed (r/min)
w_sync = 188.5; % Synchronous speed (rad/s)
% Specify slip ranges to plot
s = 0:0.01:2.0;
% Offset slips at 0 and 2 slightly to avoid divide by zero errors
s(1) = 0.0001;
s(201) = 1.9999;
% Get the corresponding speeds in rpm
nm = ( 1 - s) * n_sync;
% Caclulate Zf and Zb as a function of slip
zf = (r2 ./ s + j*x2) * (j*xm) ./ (r2 ./ s + j*x2 + j*xm);
zb = (r2 ./(2-s) + j*x2) * (j*xm) ./ (r2 ./(2-s) + j*x2 + j*xm);
% Calculate the current flowing at each slip
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i1 = v ./ ( r1 + j*x1 + 0.5*zf + 0.5*zb);
% Calculate the air-gap power
p_ag_f = abs(i1).^2 .* 0.5 .* real(zf);
p_ag_b = abs(i1).^2 .* 0.5 .* real(zb);
p_ag = p_ag_f - p_ag_b;
% Calculate torque in N-m.
t_ind = p_ag ./ w_sync;
% Plot the torque-speed curve
figure(1)
plot(nm,t_ind,'Color','b','LineWidth',2.0);
xlabel('\itn_{m} \rm(r/min)');
ylabel('\tau_{ind} \rm(N-m)');
title ('Single Phase Induction motor torque-speed
characteristic','FontSize',12);
grid on;
hold off;
The resulting torque-speed characteristic is shown below:
9-5. A 220-V 1.5-hp 50-Hz six-pole capacitor-start induction motor has the following main-winding
impedances:
R1 = 1.30 X 1 = 2.01 X M = 105
R2 = 1.73 X 2 = 2.01
At a slip of 0.05, the motors rotational losses are 291 W. The rotational losses may be assumed constant
over the normal operating range of the motor. Find the following quantities for this motor at 5 percent
slip:
(a) Stator current
(b) Stator power factor
281
(c) Input power
(d) PAG
(e) Pconv
(f) Pout
(g) ind
(h) load
(i) Efficiency
SOLUTION The equivalent circuit of the motor is shown below
I1
1 1 0.5 2
+ 2.01 ?
1.30 ? 2.01 ?
0.5 R2
0.5 0.5 Forward
105 ? s
V = 220?0 V
2
2.01 ?
R2 Reverse
0.5 0.5XM 0.5
2 s
105 ?
The impedances Z F and Z B are:
R2 / s jX 2 jX M
ZF
R2 / s jX 2 jX M
The slip s 0.05 , so R2 / s 1.73 / 0.05 20.6
20.6 j 2.01 j105
ZF 19.12 j 5.654
20.6 j 2.01 j105
R2 / 2 s jX 2 jX M
ZB
R2 / 2 s jX 2 jX M
The slip s 0.05 , so R2 / 2 s 1.73 / 2 0.05 0.887
0.887 j 2.01 j105
ZB 0.854 j1.979
0.887 j 2.01 j105
(a) The input stator current is
282
V
I1
R1 jX 1 0.5Z F 0.5Z B
220 0 V
I1 17.32 27.3 A
1.30 j 2.01 0.5 19.12 j 5.654 0.5 0.854 j1.979
(b) The stator power factor is
PF cos 27.3 0.889 lagging
(c) The input power is
PIN VI cos 220 V 17.32 A cos 27.3 3386 W
(d) The air-gap power is
2
PAG,F I12 0.5RF 17.32 A 9.56 2868 W
2
PAG,B I12 0.5 RB 17.32 A 0.427 128 W
PAG PAG,F PAG,B 2868 W 128 W 2740 W
(e) The power converted from electrical to mechanical form is
Pconv,F 1 s PAG,F 1 0.05 2868 W 2725 W
Pconv,B 1 s PAG,B 1 0.05 128 W 122 W
Pconv Pconv,F Pconv,B 2725 W 122 W 2603 W
(f) The output power is
POUT Pconv Prot 2603 W 291 W 2312 W
(g) The synchronous speed for a 6 pole 50 Hz machine is 1000 r/min, so induced torque is
PAG 2740 W
ind 26.17 N m
2 rad 1 min
sync
1000 r/min
1r 60 s
(h) The load torque is
POUT 2312 W
load 23.24 N m
2 rad 1 min
m
0.95 1000 r/min
1r 60 s
(i) The overall efficiency is
POUT 2312 W
100% 100% 68.3%
PIN 3386 W
9-6. Find the induced torque in the motor in Problem 9-5 if it is operating at 5 percent slip and its terminal
voltage is (a) 190 V, (b) 208 V, (c) 230 V.
The impedances Z F and Z B are:
R2 / s jX 2 jX M
ZF
R2 / s jX 2 jX M
The slip s 0.05 , so R2 / s 1.73 / 0.05 20.6
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