Question

A 400 V, 4-pole, 50 Hz, star connected synchronous motor has a shaft load of 25 kW, 0.8 leading power factor operating at full load. Its synchronous reactance is 7 $\Omega$. A sudden disturbance causes its rotor angle to slip by 0.25 mechanical degree. The synchronizing current is

• A. 1.41 A
• B. 0.611 A
• C. 0.919 A
• D. 0.306 A

Answer 1

The correct option is B 0.611 A

$\text{Full load current}=\frac{25\times {10}^3}{\sqrt{\times }400\times 0.8}=45.11A$

$\text{Excitation emf,}\overrightarrow{E_f}=\overrightarrow{V}-j\overrightarrow{I_a}X$

$=\frac{400}{\sqrt{3}}-\left(45.11\angle {36.87}^{\circ }\right)\left(j7\right)=490.5\angle -{31}^{\circ }V$

$\text{Rotor angle slip by 0.25 mechanical degree,}$

${\theta }_e=\frac{P}{2}{\theta }_m$

$\Delta \delta =\frac{4}{2}\times 0.25={0.5}^{\circ }$

$\text{Synchronizing emf =}2E_f\sin \frac{\Delta \delta }{2}$

$=2\times 490.5\sin \left(\frac{0.5}{2}\right)=4.28V$

$\text{Synchronizing current}=\frac{4.28}{7}=0.611A$