Three-Phase Squirrel-Cage Induction Motor: 22-Elec-A6 Power Systems and Machines, May 2013, Q2

Question 2: Three-Phase Squirrel-Cage Induction Motor (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A wye-connected cage induction motor with the per-phase circuit parameters below, running at 5 % slip.

Given data (per phase, wye)
Line voltage / phase voltage	$V_L = 208\ \text{V}$, $\ V_\phi = 208/\sqrt{3} = 120.1\ \text{V}$
Stator	$R_1 = 0.210\ \Omega,\ X_1 = 0.442\ \Omega$
Rotor (referred)	$R_2' = 0.137\ \Omega,\ X_2' = 0.442\ \Omega$
Magnetizing	$X_m = 13.2\ \Omega$
Poles / frequency / slip	$P = 4,\ f = 60\ \text{Hz},\ s = 0.05$
Rotational & core losses	$P_{rot} = 300\ \text{W},\ P_{core} = 200\ \text{W}$

Find. Speed, line current, stator copper loss, air-gap power, output torque, and efficiency (parts a–g; (g) repeats (e)).

Figure 2.1 — per-phase equivalent circuit used for the power-flow analysis.

Check: The stated “rotor losses = 300 W” are taken as the machine’s rotational (friction, windage and stray-load) losses. The rotor copper loss cannot be a given here — it is $s\,P_{ag}$, computed from the circuit — so the 300 W is the mechanical loss subtracted after conversion. Core loss (200 W) is likewise a fixed loss removed from the developed power.

Approach. Reduce the rotor and magnetizing branches to one input impedance, obtain the stator (line) current, then walk the power-flow diagram: $P_{in}\!\to\!P_{scl}\!\to\!P_{ag}\!\to\!P_{conv}\!\to\!P_{out}$.

Synchronous and shaft speed (a). $n_s = \dfrac{120 f}{P} = \dfrac{120(60)}{4} = 1800\ \text{rpm}$, so $$n_m = (1-s)\,n_s = 0.95 \times 1800.$$ $$\boxed{n_m = 1710\ \text{rpm}}$$
Input impedance. The rotor branch is $Z_2 = \dfrac{R_2'}{s} + jX_2' = \dfrac{0.137}{0.05} + j0.442 = 2.74 + j0.442\ \Omega$. In parallel with $jX_m = j13.2$ this gives $Z_f = 2.47 + j0.92\ \Omega$, so $$Z_{in} = R_1 + jX_1 + Z_f = 0.210 + j0.442 + 2.47 + j0.92 = 2.68 + j1.37 = 3.00\,\angle 27.0^\circ\ \Omega.$$
Line current (b). With $V_\phi = 120.1\,\angle 0^\circ$, $$I_1 = \frac{V_\phi}{Z_{in}} = \frac{120.1\,\angle 0^\circ}{3.00\,\angle 27.0^\circ} = 40.0\,\angle{-27.0^\circ}\ \text{A}.$$ $$\boxed{I_L = |I_1| = 40.0\ \text{A}}$$
Stator copper loss (c). $P_{scl} = 3\,I_1^2 R_1 = 3(39.98)^2(0.210)$, hence $$\boxed{P_{scl} = 1007\ \text{W}}$$
Air-gap power (d). The rotor-branch current is $I_2' = I_1\dfrac{jX_m}{Z_2 + jX_m} = 37.9\ \text{A}$, and the air-gap power is the power into $R_2'/s$: $$P_{ag} = 3\,I_2'^{\,2}\frac{R_2'}{s} = 3(37.9)^2(2.74).$$ $$\boxed{P_{ag} = 11\,820\ \text{W} \approx 11.82\ \text{kW}}$$
Developed power, output power and torque (e, g). Rotor copper loss $P_{rcl} = s\,P_{ag} = 591\ \text{W}$; converted (developed) power $P_{conv} = (1-s)P_{ag} = 11\,232\ \text{W}$. Removing the rotational and core losses, $$P_{out} = P_{conv} - P_{rot} - P_{core} = 11\,232 - 300 - 200 = 10\,732\ \text{W}.$$ With $\omega_m = \dfrac{2\pi n_m}{60} = 179.1\ \text{rad/s}$, $$T_{out} = \frac{P_{out}}{\omega_m} = \frac{10\,732}{179.1}.$$ $$\boxed{T_{out} = 59.9\ \text{N}\cdot\text{m}}$$ (Parts (e) and (g) ask the same quantity; the answer is 59.9 N·m.)
Efficiency (f). The real power drawn is $P_{in} = 3V_\phi I_1\cos\theta = P_{scl}+P_{ag} = 12\,830\ \text{W}$. Therefore $$\eta = \frac{P_{out}}{P_{in}} = \frac{10\,732}{12\,830}.$$ $$\boxed{\eta = 83.6\%}$$

The machine is running close to its rated operating point: a per-unit slip of 5 % gives a healthy power factor of about 0.89 lagging and an efficiency in the mid-80 % range typical of a small integral-horsepower cage motor. Note how the air-gap power splits: 95 % is converted to mechanical form and 5 % (the slip fraction) is dissipated as rotor copper loss.

Question 2 — results
Quantity	Value
(a) Shaft speed	1710 rpm
(b) Line current	40.0 A
(c) Stator copper loss	1007 W
(d) Air-gap power	11.82 kW
(e), (g) Output torque	59.9 N·m
(f) Efficiency	83.6 %

Topic. Steady-state analysis of a polyphase induction motor via the per-phase equivalent circuit and power-flow diagram.

Key relations. $n_s = 120f/P$; $Z_2 = R_2'/s + jX_2'$; $P_{ag} = 3 I_2'^2 R_2'/s$; $P_{conv} = (1-s)P_{ag}$; $T_{out} = P_{out}/\omega_m$; $\eta = P_{out}/P_{in}$.

Why this works. All of the rotor’s real power enters through the single resistance $R_2'/s$, so multiplying the rotor current squared by that resistance gives the total air-gap power directly. The slip cleanly partitions that power: a fraction $s$ becomes rotor $I^2R$ heating and the remaining $(1-s)$ becomes shaft mechanical power — which is exactly why a low slip means high efficiency.

Common pitfall. Using the stator current $I_1$ (rather than the rotor-branch current $I_2'$) in $P_{ag}=3I^2R_2'/s$; the magnetizing branch diverts several amps, so the two currents differ. Also, do not double-count core loss: it is drawn from the line but is not in this circuit model (there is no $R_c$), so it is subtracted from the converted power exactly once and is not also added to $P_{in} = P_{scl} + P_{ag}$.

Source. Chapman, Electric Machinery Fundamentals, 5th ed., Ch. 7 (induction-motor power-flow and torque).

Question 2 of 6: Three-Phase Squirrel-Cage Induction Motor

Question 2: Three-Phase Squirrel-Cage Induction Motor (20 marks)