Question

A 50 Hz 4-pole, 500 MVA, 22 kV turbo-generator is delivering rated MVA at 0.8 power factor. Suddenly a fault occurs reducing its electric power output by 40%. Neglect losses and assume constant power input to the shaft. The accelerating torque in the generator in MN-m at the time of the fault will be

• A. 1.528
• B. 1.018
• C. 0.848
• D. 0.509

Answer 1

The correct option is B 1.018

Before fault, mechanical input$\left(P_m\right)$= Electrical output $\left(P_e\right)$

$=500\times 0.8=400MW$

While electrical output reduces by 40%.

So, electrical output at the time of fault
$(P_e{)}_f=0.6\times 400=240MW$

Accelerating Power $=P_a=P_m-(P_e{)}_f$

$=400-240=160MW$

Synchronous speed (in rpm) $=N_s$

$\frac{120\times f}{P}=\frac{120\times 50}{4}$

$=1500rpm$

Synchronous speed (in rad/sec)$={\omega }_s$

$\frac{2\pi N_s}{60}=\frac{2\pi \times 1500}{60}=157rad/sec$

$P_a=T_a{\omega }_s$

$\Rightarrow T_a\left(acceleratingTorque\right)=\frac{P_a}{\omega s}$

$=\frac{160}{157}MN-m=1.018MN-m$