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

Question 5 of 7: 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, Spring 2016. Closed book (formula sheets supplied); do any five of seven questions, all of equal value (20 marks each). Three hours. All ac quantities are rms; three-phase voltages are line-to-line and power is total real power unless noted. All seven questions are solved here as a complete study resource.

Reference texts. S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (DC machines ch. 8, transformers ch. 2, induction motors ch. 7, synchronous machines ch. 4–5, parallel operation ch. 5); J. D. Glover et al., Power System Analysis and Design, 6th ed. (per-phase and power-factor analysis ch. 2); T. Wildi, Electrical Machines, Drives, and Power Systems. Canadian frame: EGBC/PEO national exam; SI units.

Question 5: 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.

Given data — Question 5
Ratings	4-pole, 208 V, 3φ, 60 Hz, 10 hp, Y
Stator	R₁ = 0.4 Ω, X₁ = 0.35 Ω
Rotor (referred)	R₂′ = 0.14 Ω, X₂′ = 0.35 Ω
Magnetizing reactance	X_m = 16 Ω
Rotational losses	360 W
Speed	1746 rpm

Find. (a) stator current; (b) rotor current; (c) input power; (d) stator copper loss; (e) air-gap (rotor input) power; (f) developed power; (g) output power (W and hp); (h) efficiency; (j) output torque.

Figure 5 — per-phase equivalent circuit with the slip-dependent rotor resistance R₂′/s and the magnetizing branch jX_m.

Approach. Compute slip from speed, form the per-phase circuit with R₂′/s, find the input impedance, then peel the power flow: input → stator copper loss → air-gap power → developed power (×(1−s)) → shaft output (minus rotational loss).

Slip and rotor resistance. $n_s=\dfrac{120(60)}{4}=1800$ rpm, so $s=\dfrac{1800-1746}{1800}=0.03$ and $\dfrac{R_2'}{s}=\dfrac{0.14}{0.03}=4.667\ \Omega.$
Stator current (a). With $V_{\text{ph}}=\dfrac{208}{\sqrt3}=120.1$ V, the input impedance is $Z_{\text{in}}=R_1+jX_1+\big(jX_m\parallel(\tfrac{R_2'}{s}+jX_2')\big)=4.533+j1.871=4.904\angle 22.42^\circ\ \Omega$, so $$\boxed{I_1=\frac{V_{\text{ph}}}{Z_{\text{in}}}=24.49\angle{-22.42^\circ}\text{ A}.}$$
Rotor current (b). By the current divider, $I_2'=I_1\dfrac{jX_m}{\tfrac{R_2'}{s}+j(X_2'+X_m)}=23.05\angle{-6.51^\circ}$ A, i.e. $|I_2'|=23.05$ A.
Stator input power (c). $P_{\text{in}}=3V_{\text{ph}}I_1\cos\theta=3(120.1)(24.49)\cos 22.42^\circ=8155\text{ W}=8.16\text{ kW}.$
Stator copper loss (d). $P_{\text{scl}}=3I_1^2R_1=3(24.49)^2(0.4)=720\text{ W}.$
Air-gap (rotor input) power (e). $P_{\text{ag}}=P_{\text{in}}-P_{\text{scl}}=8155-720=7435\text{ W}$ (check: $3I_2'^2\tfrac{R_2'}{s}=7435$ W ✓).
Developed power (f). $P_{\text{dev}}=(1-s)P_{\text{ag}}=0.97(7435)=7212\text{ W}.$
Output power (g). Subtracting the rotational loss, $$\boxed{P_{\text{out}}=P_{\text{dev}}-P_{\text{rot}}=7212-360=6852\text{ W}=9.18\text{ hp}.}$$
Efficiency (h) and output torque (j). $\eta=\dfrac{6852}{8155}=84.0\%$; with $\omega_m=1746\cdot\dfrac{2\pi}{60}=182.8$ rad/s, $T_{\text{out}}=\dfrac{6852}{182.8}=37.47\text{ N}\cdot\text{m}.$

Final results — Question 5
Quantity	Value
(a) Stator current	24.49 A
(b) Rotor current	23.05 A
(c) Input power	8155 W
(d) Stator copper loss	720 W
(e) Air-gap power	7435 W
(f) Developed power	7212 W
(g) Output power	6852 W (9.18 hp)
(h) Efficiency	84.0%
(j) Output torque	37.47 N·m

Topic: Induction-motor power-flow accounting from the per-phase equivalent circuit.

Key relations: $s=(n_s-n_m)/n_s$; $P_{\text{ag}}=3I_2'^2R_2'/s$; $P_{\text{dev}}=(1-s)P_{\text{ag}}$; $P_{\text{out}}=P_{\text{dev}}-P_{\text{rot}}$; $T_{\text{out}}=P_{\text{out}}/\omega_m$; $\eta=P_{\text{out}}/P_{\text{in}}$.

Why this works: The rotor resistance appears as R₂′/s, which splits into R₂′ (real rotor copper loss) and R₂′(1−s)/s (developed mechanical power). Working down the power tree — input, stator copper loss, air-gap power, developed power, shaft output — guarantees every loss is counted exactly once. The rotational (friction/windage/core) loss is a mechanical loss and is subtracted at the shaft, not drawn through the circuit.

Common pitfall: Treating the 360 W as rotor copper loss (that is $sP_{\text{ag}}$, computed separately) or adding it into the electrical input — it is a rotational loss removed at the shaft only.

Source: Chapman, Electric Machinery Fundamentals §7.4–7.6 (induction-motor equivalent circuit and power flow).

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