Question

A 50 Hz generator is delivering 60% of power that it is capable of delivering through a transmission line to an infinite bus. A fault occurs that increases the reactance between the generator and the infinite bus to 400% of the value before the fault. When fault is isolated, the maximum power that can be delivered is 75% of the original maximum value. For this condition critical clearing angle (in degres) is

• A. ${67.33}^o$
• B. ${54.17}^o$
• C. ${36.86}^o$
• D. ${51.32}^o$

Answer 1

The correct option is B ${54.17}^o$

Given, before fault,

$0.6P_{m1}=P_{m1}sin{\delta }_0$
$\therefore$ ${\delta }_0={36.86}^o\left(or\right)0.643$ radian

During fault, $P_{m2}=0.25P_{m1}as{X}_2=4X_2$

After fault,

$P_{m3}=0.75P_{m1}$ (given)

$\therefore$ ${\delta }_{max}=180-si{n}^{-1}\left(\frac{0.6P_{m1}}{0.75P_{m1}}\right)$

$={126.86}^o$ (or) 2.214 radian

Since, $cos{\delta }_{cr}=\frac{P_s({\delta }_{max}-{\delta }_0)+P_{m3}cos{\delta }_{max}-P_{m2}cos{\delta }_0}{P_{m3}-P_{m2}}$

$cos{\delta }_{cr}=\frac{0.6P_{m1}[2.214-0.643]+0.75P_{m1}cos\left(126.86\right)-0.25P_{m1}cos\left(36.86\right)}{0.75P_{m1}-0.25P_{m1}}$

$=\frac{0.6(2.214-0.643)+0.75cos\left(126.86\right)-0.25cos\left(36.86\right)}{0.75-0.25}$


$cos{\delta }_{cr}=0.585$
${\delta }_{cr}={54.2}^o$