22-Elec-A6 Power Systems and Machines · December 2016

Question 3 of 6: Three-Phase Squirrel-Cage Induction Motor

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format: Professional Engineers Ontario — 07-Elec-A6 Power Systems and Machines, December 2016. Closed-book; formula sheet supplied. Six questions of equal value; candidates answer any five. All six are solved here as a study resource. All ac voltages and currents are rms; three-phase voltages are line-to-line unless noted.

Reference texts: S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (transformers, induction/synchronous machines, dc motors); J. D. Glover et al., Power System Analysis and Design, 6th ed. (three-phase power, two-wattmeter method, power-factor correction).

Question 3: 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. Y-connected, 6-pole, 60 Hz machine at slip $s=0.04$.

Given data (per phase, referred to stator)
R₁	X₁	R₂	X₂	X_m	P_core	P_rot
0.082 Ω	0.19 Ω	0.070 Ω	0.18 Ω	7.2 Ω	1400 W	1450 W

Find. Stator and rotor currents; input, air-gap (rotor input), developed and output power; efficiency; and output torque.

Per-phase equivalent circuit; the rotor resistance appears as $R_2/s$.

Approach. Reduce the per-phase circuit ($Z_2=R_2/s+jX_2$ in parallel with $jX_m$, in series with the stator impedance) to find the stator current; step through the power flow $P_{in}\to P_{ag}\to P_{conv}\to P_{out}$; and get torque from $P_{out}/\omega_m$. The magnetizing branch is lossless, so core loss is a rotational loss removed at the shaft.

Phase voltage and input impedance. $V_1=440/\sqrt3=254.0$ V. With $R_2/s=0.070/0.04=1.75\ \Omega$, $$Z_2=1.75+j0.18,\quad Z_2\parallel jX_m = 1.577+j0.550,\quad Z_{in}=(0.082+j0.19)+(1.577+j0.550)=1.816\angle 24.0^\circ\ \Omega.$$
Stator current (part a). $$\boxed{I_1=\frac{V_1}{Z_{in}}=\frac{254.0}{1.816\angle24.0^\circ}=139.9\angle{-24.0^\circ}\ \text{A}.}$$
Rotor current (part b). The air-gap voltage is $E=I_1(Z_2\parallel jX_m)=233.6\angle{-4.8^\circ}$ V, so $$\boxed{I_2=\frac{E}{Z_2}=\frac{233.6\angle{-4.8^\circ}}{1.759\angle 5.9^\circ}=132.8\ \text{A}.}$$
Input power and stator copper loss (parts c, d). $$P_{in}=3V_1 I_1\cos\theta_1=3(254.0)(139.9)\cos24.0^\circ = 97.35\ \text{kW},$$ $$P_{scl}=3 I_1^{2} R_1 = 3(139.9)^2(0.082)=4812\ \text{W}.$$
Air-gap (rotor input) power (part e). Since $X_m$ is lossless, all real power crossing the gap is dissipated in the rotor branch: $$\boxed{P_{ag}=3 I_2^{2}\,\frac{R_2}{s}=3(132.8)^2(1.75)=92.54\ \text{kW}}\;=\;P_{in}-P_{scl}\ \checkmark.$$
Developed power (part f). The rotor copper loss is $P_{rcl}=sP_{ag}=3701$ W, so $$\boxed{P_{conv}=(1-s)P_{ag}=0.96(92.54)=88.84\ \text{kW}.}$$
Output power (part g). Subtract rotational and core losses at the shaft: $$\boxed{P_{out}=P_{conv}-P_{rot}-P_{core}=88.84-1.45-1.40 = 85.99\ \text{kW}=115.3\ \text{hp}.}$$
Efficiency and output torque (parts h, j). $\eta=P_{out}/P_{in}=85.99/97.35 = 88.3\%$. The synchronous speed is $n_s=120f/P=1200$ rpm, the mechanical speed $n_m=(1-s)n_s=1152$ rpm, $\omega_m=120.6$ rad/s, so $$\boxed{T_{out}=\frac{P_{out}}{\omega_m}=\frac{85\,990}{120.6}=712.8\ \text{N}\cdot\text{m}.}$$

Final results — Question 3
Quantity	Value
(a) Stator current	139.9 A ∠ −24.0°
(b) Rotor current	132.8 A
(c) Input power	97.35 kW
(d) Stator copper loss	4812 W
(e) Rotor (air-gap) input power	92.54 kW
(f) Developed power	88.84 kW
(g) Output power	85.99 kW = 115.3 hp
(h) Efficiency	88.3 %
(j) Output torque	712.8 N·m

Topic: Induction-motor per-phase circuit and power-flow (Chapman) accounting.

Key relations: $P_{ag}=3I_2^2 R_2/s$; $P_{conv}=(1-s)P_{ag}$; $P_{rcl}=sP_{ag}$; $P_{out}=P_{conv}-P_{rot}-P_{core}$; $\eta=P_{out}/P_{in}$; $T_{out}=P_{out}/\omega_m$, $\omega_m=(1-s)\,2\pi n_s/60$.

Why this works: The load on the shaft appears electrically as $R_2/s$; the power delivered to that resistance is the air-gap power, of which a fraction $s$ is lost as rotor copper and $(1-s)$ is converted to mechanical form. Because the magnetizing branch here has no core-loss resistor, the stated core loss cannot flow through the circuit—it is grouped with friction and windage and removed at the shaft, exactly like the rotational loss. Adding it to the input would double-count it.

Common pitfall: Subtracting core loss twice (once in the circuit, once at the shaft), or computing air-gap power as $P_{in}-P_{scl}-P_{core}$ when the model has a lossless $X_m$. The consistency check $P_{ag}=P_{in}-P_{scl}$ guards against this.

Source: Chapman, Electric Machinery Fundamentals, 5th ed., §7 (induction motors, power-flow diagram).

← Question 2 Next: Question 4 →