Question 3 of 5: Three-Phase Induction Motor Performance
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Professional Engineers Ontario — 07-Elec-A6 Power Systems and Machines, Spring 2015. Closed-book; formula sheets supplied. FIVE questions constitute a complete paper and all are of equal value (20 marks each). All ac quantities are rms; three-phase voltages are line-to-line and power is total unless noted.
Reference texts: S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (McGraw-Hill) — Ch. 2 transformers, Ch. 6–7 induction machines, Ch. 5 synchronous machines, Ch. 8–9 DC machines; A. E. Fitzgerald, C. Kingsley & S. Umans, Electric Machinery, 6th ed.; P. C. Sen, Principles of Electric Machines and Power Electronics, 3rd ed.
Question 3: Three-Phase Induction Motor Performance (20 marks)
Given. A two-pole, Y-connected, 480 V (line) squirrel-cage induction motor at slip $s=0.03$, with constant rotational losses of 1850 W and the per-phase stator-referred parameters below. No core-loss resistor is given, so core loss is accounted with the rotational losses at the shaft.
Given data (Ω per phase, referred to stator)
$R_1$
$X_1$
$R_2$
$X_2$
$X_m$
$s$
$P_{rot}$
0.322
0.675
0.196
0.510
12.5
0.03
1850 W
Find. Line current, developed (converted) power, developed torque, efficiency, and reactive power drawn.
Figure Q3. Per-phase equivalent circuit of the 2-pole squirrel-cage induction motor. The magnetizing branch is purely reactive (no core-loss resistor).
Approach. Build the per-phase equivalent circuit with $R_2/s$, reduce it to an input impedance to get the stator (line) current, split off the rotor current to find the air-gap power, then apply the standard power flow $P_{ag}\to P_{conv}\to P_{out}$ and the reactive balance.
Per-phase voltage and rotor branch. For the Y connection $V_\phi=480/\sqrt3=277.1$ V. The rotor branch carries the slip-dependent resistance $R_2/s=0.196/0.03=6.533\ \Omega$, so $Z_2=6.533+j0.510\ \Omega$, in parallel with $jX_m=j12.5\ \Omega$.
Input impedance and line current (part a). Combining the branches, $Z_{2}\parallel jX_m=4.82+j2.91\ \Omega$, so$$Z_{in}=R_1+jX_1+(Z_2\parallel jX_m)=5.14+j3.58=6.27\angle 34.9^\circ\ \Omega,$$$$I_1=\frac{V_\phi}{Z_{in}}=\frac{277.1}{6.27\angle 34.9^\circ}=\boxed{44.2\ \text{A}}\ (\text{pf}=\cos 34.9^\circ=0.820\ \text{lag}).$$ For the Y connection the line current equals the phase current.
Air-gap and developed power (part b). The rotor current follows by current division, $I_2=I_1\,\dfrac{jX_m}{Z_2+jX_m}=37.98\ \text{A}$, and the air-gap power is the power into $R_2/s$:$$P_{ag}=3I_2^2\frac{R_2}{s}=3(37.98)^2(6.533)=28.27\ \text{kW},$$$$P_{conv}=(1-s)P_{ag}=0.97(28.27)=\boxed{27.4\ \text{kW}}.$$
Developed torque (part c). With $n_s=120f/P=3600$ rpm ($\omega_s=376.99$ rad/s), the developed torque is most cleanly written from the air-gap power:$$\tau_{dev}=\frac{P_{ag}}{\omega_s}=\frac{28.27\times10^3}{376.99}=\boxed{75.0\ \text{N}\cdot ext{m}}=\frac{P_{conv}}{\omega_m}.$$
Efficiency (part d). Subtract the rotational losses at the shaft and divide by the electrical input:$$P_{out}=P_{conv}-P_{rot}=27.4-1.85=25.6\ \text{kW},$$$$P_{in}=\sqrt3\,V_LI_L\cos\theta=\sqrt3(480)(44.2)(0.820)=30.15\ \text{kW},\quad \eta=\frac{25.6}{30.15}=\boxed{84.8\%}.$$
Reactive power drawn (part e).$$Q=\sqrt3\,V_LI_L\sin\theta=\sqrt3(480)(44.2)\sin 34.9^\circ=\boxed{21.0\ \text{kVAR}}.$$
