Question 2 of 5: Three-Phase Squirrel-Cage Induction Motor

Question 2: 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. A six-pole 60 Hz induction motor with a full per-phase equivalent circuit (including a core-loss branch $R_c$) running at 1164 rpm.

Find. Line current; input P, Q, S; air-gap power; rotor copper loss; developed mechanical power; developed and shaft torques; and the speed of maximum torque.

Approach. Compute synchronous speed and slip, reduce the circuit to a single input impedance to get $I_1$, then find the air-gap voltage and rotor current to split the power flow ($P_{ag}$, rotor copper, developed) and convert to torque; the maximum-torque slip follows from the Thevenin equivalent.

1. Synchronous speed and slip. $n_s=120f/p = 120(60)/6 = 1200\ \text{rpm}$; $\omega_s = 2\pi(1200)/60 = 125.66\ \text{rad/s}$. Slip $s=(n_s-n_m)/n_s=(1200-1164)/1200 = 0.030$. Phase voltage $V_\phi = 575/\sqrt3 = 331.98\ \text{V}$.
2. Input impedance. Rotor branch $Z_2 = R_2'/s + jX_2' = 13.00 + j2.151\ \Omega$. Excitation branch $Z_\phi = R_c \parallel jX_m = 1.983 + j26.44\ \Omega$. Their parallel combination $Z_{ab}=Z_\phi\parallel Z_2 = 9.102 + j5.857\ \Omega$, and adding the stator series impedance $$Z_{in}=R_1+jX_1+Z_{ab}=9.474+j7.291 = 11.955\,\angle 37.58^\circ\ \Omega.$$
3. Line current. With $V_\phi$ as reference, $$I_1=\frac{V_\phi}{Z_{in}}=\frac{331.98}{11.955\,\angle 37.58^\circ}=\boxed{27.77\,\angle{-}37.58^\circ\ \text{A}}.$$ For the Y connection the line current equals the phase current, $I_L=27.77\ \text{A}$.
4. Input power (P, Q, S). The pf is $\cos 37.58^\circ = 0.792$ lagging. $$S=\sqrt3\,V_LI_L=\sqrt3(575)(27.77)=\boxed{27.66\ \text{kVA}},$$ $$P=S\cos\theta=\boxed{21.92\ \text{kW}},\qquad Q=S\sin\theta=\boxed{16.87\ \text{kvar (lagging)}}.$$
5. Air-gap power. The air-gap (magnetizing-node) voltage is $E=V_\phi-I_1(R_1+jX_1)=300.6\,\angle{-}4.82^\circ\ \text{V}$, giving rotor current $I_2'=E/Z_2 = 22.81\,\angle{-}14.21^\circ\ \text{A}$. The air-gap power is the power crossing into the rotor branch: $$P_{ag}=3\,I_2'^{\,2}\frac{R_2'}{s}=3(22.81)^2(13.00)=\boxed{20.29\ \text{kW}}.$$ (Check: the core-loss branch dissipates $3E^2/R_c = 764\ \text{W}$, matching the stated $P_{core}=764.2\ \text{W}$ — so $R_c$ correctly models the core loss.)
6. Rotor copper loss. $$P_{rcl}=3\,I_2'^{\,2}R_2'=sP_{ag}=(0.03)(20.29\ \text{kW})=\boxed{608.7\ \text{W}}.$$
7. Developed mechanical power. $$P_{dev}=(1-s)P_{ag}=(0.97)(20.29\ \text{kW})=\boxed{19.68\ \text{kW}}.$$
8. Developed torque. With $\omega_m=2\pi(1164)/60 = 121.9\ \text{rad/s}$, $$T_{dev}=\frac{P_{dev}}{\omega_m}=\frac{P_{ag}}{\omega_s}=\frac{19680}{121.9}=\boxed{161.5\ \text{N}\cdot\text{m}}.$$
9. Shaft (output) torque. Because the core loss is already accounted for electrically by $R_c$, only friction and windage are removed at the shaft: $P_{out}=P_{dev}-P_{fw}=19682-345.8=19337\ \text{W}$, so $$T_{out}=\frac{P_{out}}{\omega_m}=\frac{19337}{121.9}=\boxed{158.6\ \text{N}\cdot\text{m}}.$$
10. Speed of maximum torque. From the stator Thevenin equivalent $Z_{th}=jX_m(R_1+jX_1)/(R_1+j(X_1+X_m)) = 0.335 + j1.365\ \Omega$, $$s_{maxT}=\frac{R_2'}{\sqrt{R_{th}^2+(X_{th}+X_2')^2}}=\frac{0.390}{\sqrt{0.335^2+(1.365+2.151)^2}}=0.1104,$$ hence $n_{maxT}=n_s(1-s_{maxT})=1200(0.8896)=\boxed{1067\ \text{rpm}}.$