Question

A 208 V, 45 kVA, 0.8 pf leading, $\Delta -$ connected, 60 Hz synchronous machine has a synchronous reactance of $2.5\Omega$ and a negligible armature resistance. Its friction and windage losses are 1.5 kW and its core losses are 1.0 kW. The shaft is supplying a 15-hp load and the motor's pf is 0.80 leading. The value of induced voltage is______ (1 hp = 746 W)

• A. $186\angle -{12.4}^{\circ }V$
• B. $255\angle -{12.4}^{\circ }V$
• C. $190\angle -{17.1}^{\circ }V$
• D. $255\angle -{17.1}^{\circ }V$

Answer 1

The correct option is B $255\angle -{12.4}^{\circ }V$

Initially the motor's output power is 15 hp. This corresponds to an output of $P_{\text{out}}=\left(15hp\right)\left(0.746kW/hp\right)=11.19kW$

The electric power supplied to machine is:
$P_{\text{in}}=P_{\text{out}}+P_{\text{mech loss}}+P_{\text{coreloss}}+P_{\text{elec loss}}$

$=11.19kW+1.5kW+1.0kW+0kW=13.69kW$

$\therefore$ The motor's power factor is 0.80 leading, resultant line current flow:
$I_L=\frac{P_{\text{in}}}{\sqrt{3}{V}_T\cos \theta }=\frac{13.69\times {10}^3}{\sqrt{3}\times 208\times 0.80}=47.5A$

The armature current, $I_A=\frac{I_L}{\sqrt{3}}$

$I_A=27.4\angle {36.87}^{\circ }A$

$E_A=V-j{I}_A{X}_s$

$=208\angle 0^{\circ }-j\left(2.5\right)\left(27.4\angle {36.87}^{\circ }\right)$

$=255\angle -{12.4}^{\circ }V$