Question

A 120V, 60 Hz, three phase, y-connected, round rotor, synchronous motor has an armature winding resistance of $(0.5+j3)\Omega /phase$. The motor takes a current of 10 A at 0.8 of leading when operating at a certain field current. The load is gradually increased until the motor develops the maximum torque while the field current is held constant. The power developed by motor is ___W.

• A. 4740

Answer 1

The correct option is A 4740
Initial operation,

Synchronous impedance, $Z_s=0.5+j3=3.04\angle {80.54}^{\circ }\Omega$

Voltage, $V_a=69.28\angle 0^{\circ }V$

$I_{a1}=10\angle {36.87}^{\circ }A$

$\overrightarrow{E_{a1}}=\overrightarrow{V_a}-\overrightarrow{I_{a1}}.\overrightarrow{Z_s}=87.54\angle -{17.96}^{\circ }V$

The power angle associated with maximum torque is

${\delta }_m=ta{n}^{-1}\left[\frac{3}{0.5}\right]={80.54}^{\circ }\left(lag\right)$

$\overrightarrow{E_{a2}}=87.54\angle -{80.54}^{\circ }$

Line current, $\overrightarrow{I_{a2}}=\frac{69.28\angle 0^{\circ }-87.54\angle -{80.54}^{\circ }}{0.5+j3}$

$=33.64\angle -{22.98}^{\circ }A$

The per phase maximum power developed by the motor is

$P_{dev}=Re\left[\right(87.54\angle -{80.54}^{\circ })\times (33.64\angle {22.98}^{\circ }\left)\right]=1580W/phase$

Total power = $3P_{dev}=4740W$