Three-Phase Induction Motor Performance: 22-Elec-A6 Power Systems and Machines, May 2014, Q6

Question 6: Three-Phase Induction Motor Performance (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, $V_L=440$ V so $V_1=254$ V/phase; stator $R_1=0.1$, $X_1=0.35\ \Omega$; rotor $R_2'=0.12$, $X_2'=0.4\ \Omega$; $P_{rot}=2150$ W; no-load 18 A at 0.089 pf; running speed 1755 rpm.

Given data
V₁	R₁, X₁	R₂′, X₂′	P_rot	No-load	n
254 V	0.1, 0.35 Ω	0.12, 0.4 Ω	2150 W	18 A, 0.089 pf	1755 rpm

Find. (a) line current, (b) developed torque, (c) output power, (d) efficiency at 1755 rpm.

Per-phase approximate equivalent circuit; the magnetizing branch is taken as a lossless $jX_m$ (core loss lumped into the rotational losses).

Check: No core-loss resistor is given, so the magnetizing branch is modelled as a lossless $jX_m$ obtained from the no-load reactive current; core loss is carried within the stated rotational loss and removed at the shaft. For a 60 Hz motor near 1755 rpm the synchronous speed is 1800 rpm (4 poles).

Approach. Find slip and the magnetizing reactance, compute the rotor-branch current and the air-gap power, then developed and output power and efficiency.

Slip and magnetizing reactance. $n_s=1800$ rpm, $s=(1800-1755)/1800=0.025$. From the no-load test the magnetizing current is $I_m=18\sin(\cos^{-1}0.089)=17.9$ A, so $$X_m=\frac{V_1}{I_m}=\frac{254}{17.9}=14.2\ \Omega.$$
Rotor-branch current. $$Z=\Big(R_1+\frac{R_2'}{s}\Big)+j(X_1+X_2')=(0.1+4.8)+j0.75=4.96\angle 8.7^\circ\ \Omega,$$ $$I_2'=\frac{V_1}{Z}=51.2\angle{-8.7^\circ}\ \text{A}.$$
(a) Line current. Adding the magnetizing current $I_m=17.9\angle{-90^\circ}$ A, $$I_1=I_2'+I_m = 56.8\angle{-26.9^\circ}\ \text{A}\ \Rightarrow\ \boxed{I_L=56.8\ \text{A}.}$$
(b) Developed torque. Air-gap power $P_{ag}=3I_2'^2\dfrac{R_2'}{s}=3(51.2)^2(4.8)=37.8$ kW, and with $\omega_s=188.5$ rad/s, $$T_{dev}=\frac{P_{ag}}{\omega_s}=\frac{37\,818}{188.5}=\boxed{200.6\ \text{N}\!\cdot\!\text{m}.}$$
(c) Output power. $P_{dev}=(1-s)P_{ag}=0.975(37\,818)=36.9$ kW, so $$P_{out}=P_{dev}-P_{rot}=36\,872-2150=\boxed{34.7\ \text{kW}\ (46.5\ \text{hp}).}$$
(d) Efficiency. Input $P_{in}=\sqrt3V_LI_L\cos\theta_1=3V_1\,\text{Re}(I_1)=38.6$ kW: $$\eta=\frac{P_{out}}{P_{in}}=\frac{34\,722}{38\,606}=\boxed{89.9\%.}$$

Final results — Question 6
Quantity	Value
Slip	0.025
Line current	56.8 A at 0.892 pf lag
Developed torque	200.6 N·m
Output power	34.7 kW (46.5 hp)
Efficiency	89.9%

Topic: Steady-state analysis of a three-phase induction motor from its per-phase equivalent circuit.

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

Why this works: The resistance $R_2'/s$ represents both rotor copper loss ($R_2'$) and mechanical output ($R_2'(1-s)/s$). Air-gap power crosses to the rotor as the torque-producing quantity referenced to synchronous speed, so dividing by $\omega_s$ gives developed torque directly. Subtracting rotational losses (which include core loss when no core-loss resistor is modelled) yields shaft output.

Common pitfall: Drawing core loss through the circuit as well as counting it in rotational loss (double counting), or using running speed instead of synchronous speed when computing torque from air-gap power.

Source: Chapman, ch. 6–7 (induction motors, equivalent circuit, power flow); Sen, ch. 6.

Question 6 of 6: Three-Phase Induction Motor Performance

Question 6: Three-Phase Induction Motor Performance (20 marks)