Question

A 2000 V, 3-phase star-connected synchronous motor has synchronous impedance of (0.5 + j5) $\Omega$ per phase. For an excitation voltage of 3000 V, the motor takes an input of 900 kW at rated voltage. The power angle of motor would be around

• A. ${53}^{\circ }$
• B. ${49}^{\circ }$
• C. ${60}^{\circ }$
• D. ${40}^{\circ }$

Answer 1

The correct option is B ${49}^{\circ }$

Synchronous impedance, $Z_s=(0.5+j5)\Omega =5.025\angle {84.29}^{\circ }\Omega$

$I_a=\frac{V\angle 0^{\circ }-E\angle -\delta }{Z_s\angle \theta }$

$S=V{I}_a=V\angle 0\left[\frac{V\angle 0-E\angle -\delta }{Z_s\angle \theta }\right]$

$S=\frac{V^2}{Z_s}\angle \theta -\frac{EV}{Z_s}\angle \theta +\delta$

So from above equation,

$P=\frac{V^2}{Z_s}cos\theta -\frac{EV}{Z_s}cos(\theta +\delta )$

$900\times {10}^3=\frac{{2000}^2}{5.025}\left(cos{84.29}^{\circ }\right)-\frac{2000\times 3000}{5.025}cos({84.29}^{\circ }+\delta )$

${84.29}^{\circ }+\delta ={133.426}^{\circ }$

$\text{Power angle,}\delta ={49.13}^{\circ }$