Question 3.AE.1: A Pout=100 hp, VL-L_rat=480 V, f=60 Hz, p=6 pole, three-phas......
A P_{out}=100 hp, V_{L-L\_rat}=480 V, f=60 Hz, p=6 pole, three-phase induction motor runs at full load and rated voltage with a slip of 3%. Under conditions of stress on the power system, the line-to-line voltage drops to V_{L-L\_rat}=430 V. If the load is of the constant torque type, compute for the lower voltage:
a) Slip s_{low} (use small-slip approximation).
b) Shaft speed n_{m\_low} in rpm.
c) Output power P_{out\_low}.
d) Rotor copper loss P_{cur\_low}= \left(I_r^{\prime}\right)^2 R_r^{\prime} in terms of the rated rotor copper loss at ratedvoltage.
One concludes that a decrease of the terminal voltage increases the rotor loss (temperature) of an induction motor.a) Based on the small-slip approximation s \propto \frac{1}{V^2}, the new slip at the low voltage is
s_{\text {low }}=0.03(480 / 430)^2=0.0374.
b) With the synchronous speed n_s=\frac{120 f}{p}=\frac{120(60)}{6}=1200 \mathrm{rpm} one calculates the shaft speed at low voltage as
n_{m_{-} \text {low }}=\left(1-s_{\text {low }}\right) n_s=(1-0.0374)(1200)=1155.14 \mathrm{rpm} .c) The output power at low voltage is P_{\text {out }}=\omega_m \cdot T_m. For constant mechanical torque one gets P_{out \ _low}=\frac{(1-0.0374)}{(1-0.03)}(100 h p)=99.23 h p
d) Due to the relation T=\frac{P_g}{\omega_{s 1}} \text { (where } P_g \text { is the air-gap power), } the torque is proportional to P_g, and therefore, for constant torque operation the air-gap power is constant. Thus the rotor loss is P_{\text {cur } \_ \text {low }}=s_{\text {low }} P_g=\frac{0.0374}{0.03} P_{\text {out } \_ \text {rat }}=1.247 P_{\text {out } \_ \text {rat }}
One concludes that a decrease of the terminal voltage increases the rotor loss (temperature) of an induction motor.