Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Professional Engineers Ontario — 16-Elec-A6 Power Systems and Machines, May 2017. Closed-book; formula sheet supplied. Five questions of equal value; all five constitute a complete paper, and all are solved here. All ac quantities are rms; three-phase quantities are line-to-line and total real power unless noted.
Reference texts: S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (transformers ch. 2, DC machines ch. 8, synchronous machines ch. 4–5, induction machines ch. 6); J. D. Glover et al., Power System Analysis and Design, 6th ed. (per-phase and power-factor analysis, ch. 2).
Question 3: Wound-rotor induction motor (20 marks)
Given. The Y-connected six-pole machine and per-phase parameters below, operating at $s=0.033$ on a 480 V (line) 60 Hz supply with 2950 W of rotational loss.
Given data
Quantity
Value
Supply / poles
480 V (L-L), 60 Hz, 6-pole, Y
$R_1,X_1$
0.1 Ω, 0.205 Ω
$R_2',X_2'$
0.079 Ω, 0.186 Ω
$X_m$
7.5 Ω
Slip / rotational loss
0.033 / 2950 W
Find. Line current, the loss breakdown (stator copper, air-gap, rotor copper), developed and shaft mechanical power/torque, and efficiency.
Per-phase equivalent circuit with the R2'/s branch representing air-gap power transfer.
Approach. Build the per-phase circuit, find the stator current from the total input impedance, then walk the Chapman power flow $P_{in}\to P_{scl}\to P_{ag}\to P_{rcl}\to P_{dev}\to P_{out}$.
Speeds and phase voltage. $n_s=120(60)/6=1200$ rpm; $n_m=(1-0.033)1200=1160.4$ rpm. Phase voltage $V_1=480/\sqrt3=277.1$ V.
Stator current. Rotor branch $Z_2=R_2'/s+jX_2'=2.394+j0.186\,\Omega$; in parallel with $jX_m=j7.5$ gives $Z_f=2.078+j0.829\,\Omega$. Total $Z=R_1+jX_1+Z_f=2.178+j1.034\,\Omega$, so $I_1=V_1/Z=114.9\angle{-25.4^\circ}$ A at $\cos\theta=0.903$ lagging.
(a) Line current. For a Y connection the line current equals the phase current: $$\boxed{\,I_L=115.0\ \text{A}\,}$$
(b) Stator copper loss. $P_{scl}=3I_1^2R_1=3(114.9)^2(0.1)=3964$ W.
(c) Air-gap power. Input $P_{in}=3V_1I_1\cos\theta=3(277.1)(114.9)(0.903)=86{,}340$ W, so $P_{ag}=P_{in}-P_{scl}=82{,}376$ W (equal to $3I_2'^2R_2'/s$).
(d) Rotor copper loss. $P_{rcl}=sP_{ag}=0.033(82{,}376)=2718$ W.
(e) Developed mechanical power. $P_{dev}=(1-s)P_{ag}=0.967(82{,}376)=79{,}657$ W.
(f) Developed torque. $\omega_m=2\pi n_m/60=121.5$ rad/s, so $T_{dev}=P_{dev}/\omega_m=655.5\ \text{N}\cdot\text{m}$.
(g) Shaft torque. $P_{out}=P_{dev}-P_{rot}=79{,}657-2950=76{,}707$ W ($=102.8$ hp), so $T_{sh}=P_{out}/\omega_m=631.3\ \text{N}\cdot\text{m}$.
