Question

A 50 Hz generator is delivering rated power of 1.0 p.u. to an infinite bus through a lossless network. A three phase fault under this condition reduces $P_{\text{max}}$ to 0 per unit. If the fault is cleared in 0.05 sec, then the rotor angle at this instant is
(Assume $P_{\text{max}}$ before fault is 2.0 p.u. and 1.5 p.u. after fault clearing and H= 7.5 MJ/MVA)

• A. 30 electrical degrees
• B. 31.5 electrical degrees
• C. 34.2 electrical degrees
• D. 32.6 electrical degrees

Answer 1

The correct option is B 31.5 electrical degrees

Before the fault occurs,
$P_e=P_{\text{max}}\sin {\delta }_0$

$1=2\sin {\delta }_0$

${\delta }_0={\sin }^{-1}\left(\frac{1}{2}\right)={30}^{\circ }$

$M=\frac{GH}{180\times f}=\frac{1\times 7.5}{180\times 50}=\frac{1}{1200}p.u.$

$Afterthefault,Acceleration=\frac{d^2\delta }{d{t}^2}=\frac{P_a}{M}=\frac{1}{\left(\frac{1}{1200}\right)}=1200{\text{electrical degree/sec}}^2$

Integrating the above expression two times,
$\delta =\frac{1200t^2}{2}=\frac{1200\times (0.05{)}^2}{2}$

$\delta =1.5electricaldegrees$

$\therefore \text{After the fault at}t=0.05s,\delta ={30}^{\circ }+{1.5}^{\circ }=31.5electricaldegrees$