22-Elec-A6 Power Systems and Machines · December 2013
Question 6 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, Fall 2013 (closed book, formula sheet supplied). Six questions of equal value; any five constitute a complete paper. All ac voltages/currents are rms; three-phase voltages are line-to-line unless noted. All six questions are solved below as a complete study resource.
Reference texts: S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (McGraw-Hill); T. Wildi, Electrical Machines, Drives, and Power Systems, 6th ed. (Pearson); Fitzgerald, Kingsley & Umans, Electric Machinery, 6th ed.
Question 6: Three-Phase Squirrel-Cage Induction Motor (20 marks)
Given. Y-connected squirrel-cage induction motor, 4-pole, 230 V (line), 60 Hz, rated speed 1550 rpm. (The paper prints "1550 W"; W is a unit of power, so this is read as 1550 rpm.) Per-phase circuit parameters as tabulated. Friction/windage 5.9 W; core loss 10.72 W.
Per-phase equivalent circuit. The magnetizing branch is a pure reactance $X_m$; core loss is treated as a separate rotational loss.
Approach. Compute synchronous speed and slip; then the per-phase input impedance ($R_1+jX_1$ in series with $jX_m \Vert (R_2'/s+jX_2')$). Stator current from $V_\phi/Z_{in}$; the airgap voltage splits into the magnetizing and rotor branches. Power flows input → stator copper → airgap → rotor copper → developed → (minus rotational) output.
Synchronous speed and slip. $n_s=\dfrac{120(60)}{4}=1800$ rpm; $s=\dfrac{1800-1550}{1800}=0.1389$. Per-phase voltage $V_\phi=230/\sqrt3=132.79$ V.
Input impedance. Rotor branch $Z_2=\dfrac{R_2'}{s}+jX_2'=\dfrac{21.97}{0.1389}+j11.56=158.2+j11.56\ \Omega$. In parallel with $jX_m=j432.48$:
$$Z_2\,\Vert\,jX_m=133.2+j58.7\ \Omega,\qquad Z_{in}=(R_1+jX_1)+(Z_2\,\Vert\,jX_m)=143.3+j97.3=173.2\angle34.2^\circ\ \Omega$$
Stator current (part a).
$$\boxed{I_1=\frac{V_\phi}{Z_{in}}=\frac{132.79\angle0^\circ}{173.2\angle34.2^\circ}=0.767\angle{-}34.2^\circ\ \text{A}}$$
Airgap voltage and branch currents (parts b, c). $E_1=V_\phi-I_1(R_1+jX_1)$, $|E_1|=111.6$ V. Then
$$\boxed{I_m=\frac{E_1}{jX_m}=0.258\ \text{A}}\qquad \boxed{I_2'=\frac{E_1}{Z_2}=0.703\ \text{A}}$$
Input power (part d). $P_{in}=3V_\phi I_1\cos\theta_1=3(132.79)(0.767)\cos(34.2^\circ)=\boxed{252.7\ \text{W}}$.
Stator and rotor copper losses (parts e, f).
$$P_{scl}=3I_1^2R_1=3(0.767)^2(10.12)=\boxed{17.85\ \text{W}}$$
$$P_{rcl}=3I_2'^2R_2'=3(0.703)^2(21.97)=\boxed{32.62\ \text{W}}$$
Airgap and developed power. $P_{gap}=3I_2'^2\dfrac{R_2'}{s}=234.8$ W (check: $P_{scl}+P_{gap}=252.7$ W $=P_{in}$, since $X_m$ is lossless). Developed power $P_{dev}=(1-s)P_{gap}=202.2$ W.
Output power (part g). Subtract the rotational losses (friction/windage + core):
$$\boxed{P_{out}=P_{dev}-(P_{f\&w}+P_{core})=202.2-(5.9+10.72)=185.6\ \text{W}}$$
Shaft torque (part h). $\omega_m=2\pi(1550)/60=162.3$ rad/s, so $T_{out}=\dfrac{P_{out}}{\omega_m}=\dfrac{185.6}{162.3}=\boxed{1.14\ \text{N}\!\cdot\!\text{m}}$.
