Question

A 4-pole, 50 Hz generating unit has H-constant of 2 MJ/MVA. The machine is initially operating under steady state condition at synchronous speed and producing 1 p.u. of real power. The initial value of rotor angle $\delta$ is $5^o$, when a bolted 3-phase to ground fault occurs at the terminals of generator. Assuming the input mechanical power to remain at 1 p.u., the speed of rotor in rpm, at the end of 10 cycles of fault is ______. (Assume acceleration is constant for 10 cycles)

• A. 1575

Answer 1

The correct option is A 1575
Given, P = 4

$f_0=50Hz$ $H=2MJ/MVA$

${\delta }_0=5^o$ $P_s=1p.u.$

Swing equation,

$\frac{H}{\pi f}.\frac{d^2\delta }{d{t}^2}=P_s-P_e=P_a=1-0+1p.u.$

But, $P_e=0$ after the 3- phase to grounded fault

$\Rightarrow$ Acceleration constant $\alpha =\frac{d^2\delta }{d{t}^2}=\frac{P_a\times \pi f}{H}elec-rad/se{c}^2$

$=\frac{1\times \pi \times 50}{2}=25\pi$

Time of 10 cycles $=\frac{10}{50}=\frac{1}{5}sec$

Frequency at the end of 10 cycles = f
Since acceleration is constant for 10 cycles, then

$f=f_0+\frac{\alpha }{2\pi }.t$

$f=50+\frac{25\pi }{2\pi }.\left(\frac{1}{5}\right)$

$f=50+\frac{12.5}{5}=52.5Hz$

Speed of the end of 10 cycles = N

$N=\frac{120.f}{P}=\frac{120\times 52.5}{4}=1575rpm$