Question

A 10 hp, 230 V, 60 Hz, 3-phase, star connected, synchronous motor delivers full load at a power factor of 0.707 leading. The synchronous reactance of the motor is $j5\Omega /phase$. The rotational loss is 230 W and the filed winding loss is 70 W. The generated voltage per phase is ______(take $1hp=746W$)

• A. $132.79\angle -0^{\circ }V$
• B. $248.8\angle -{22.8}^{\circ }V$
• C. $168.15\angle -{11.8}^{\circ }V$
• D. $221.10\angle -{11.2}^{\circ }V$

Answer 1

The correct option is B $248.8\angle -{22.8}^{\circ }V$

The power output : $P_o=10\times 746=7460W$

Power developed : $P_d=P_o+P_r$

$=7460+230=7690W$

Since there is no copper loss in armature winding, the power supplied by 230 V alternative current source is 7690 W.

$\overrightarrow{I_a}=\frac{7690}{\sqrt{3}\times 230\times 0.707}=27.3A$

Total power input :$P_{in}=7690+70=7760W$

$\overrightarrow{V_a}=\frac{230\angle 0^{\circ }}{\sqrt{3}}=132.79\angle 0^{\circ }V$

$\overrightarrow{I_a}=27.3\angle {45}^{\circ }A$

Generated voltage per phase is
$\overrightarrow{E_a}=132.79\angle 0^{\circ }-\left(27.3\angle {45}^{\circ }\right)j5$

$=248.8\angle -{22.8}^{\circ }V$