Question

A 3-phase , 2-pole, 50 Hz, synchronous generator has a rating of 250 MVA, 0.8 pf lagging. The kinetic energy of the machine at synchronous speed is 1000 MJ. The machine is running steadily at synchronous speed and delivering 60 MW power at a power angle of 10 electrical degrees. If the load is suddenly removed, assuming the acceleration is constant for 10 cycles, the value of the power angle after 5 cycles is electrical degrees is.

• A. 12.7

Answer 1

The correct option is A 12.7
$S=250MVA$
$cos\varphi =0.8$
$K.E=1000MJ$
$P_e=60MW$
${\delta }_0={10}^{\circ }$
$t=10cycles=\frac{10}{50}=0.2sec$
$t=5cycles\Rightarrow 0.1sec$
Load is removed,
$P_e=0$
$P_a=P_m-P_e$
$\Rightarrow P_m=60MW,$
$M=\frac{HS}{180f}=\frac{K.E.}{180f}$
$=\frac{1000}{180\times 50}=0.111$
$\frac{d^2\delta }{d{t}^2}=\frac{P_a}{M}=545.45ele.degree/s^2$
Integrating the above equation,
$\int \left(\frac{d^2\delta }{d{t}^2}\right)dt=\int \frac{P_a}{M}dt=545.45\times t$
Integrating it once again with $dt$
$=\int 545.45\times 6\left(dt\right)$
$\delta =\frac{545.45\times t^2}{2}={2.7}^{\circ }$
So, new value of power angle
$={10}^{\circ }+{2.7}^{\circ }={12.7}^{\circ }$