Question

A generator delivers power of 1.0 p.u. to an infinite bus through a purely reactive network. The maximum power that could be delivered by the generator is 2.0 p.u. A three-phase fault occurs at the terminals of the generator which reduces the generator output to zero. The fault is cleared after $t_c$ second. The original network is then restored. The maximum swing of the rotor angle is found to be ${\delta }_{max}=110$ electrical degree. Then the rotor angle in electrical degrees at $t=t_c$ is

• A. 55
• B. 70
• C. 69.14
• D. 72.4

Answer 1

The correct option is C 69.14

Power generated by the generator
$P_e=P_{max}sin\delta$
At ${\delta }_0$
$P_e=1p.u$
$P_{max}=2p.u.$
${\delta }_{cr}=$ rotor angle at $t=t_c$
$P_e=P_{max}sin\delta$
$1=2sin{\delta }_0$
${\delta }_0={30}^{\circ }$

• During fault, P_{e} become zero and the fault is cleared at ${\delta }_{cr}$. Mechanical input to the generator remains constant, $P_m=1p.u.$
• Applying equal area criterion,

$A_1=A_2$
$A_1={\int }_{\delta 0}^{{\delta }_{cr}}(P_m-0)d\delta =P_m({\delta }_{cr}-{\delta }_0)...\left(i\right)$
$A_2={\int }_{\delta cr}^{{\delta }_{max}}(P_{max}sin\delta -P_m)d\delta$
$=P_{max}(cos{\delta }_{cr}-cos{\delta }_{max})-P_m({\delta }_{max}-{\delta }_{cr})...\left(ii\right)$
Equating equation (i) and (ii),
$P_m({\delta }_{cr}-{\delta }_0)=P_{max}(cos{\delta }_{cr}-cos{\delta }_{max})-P_m({\delta }_{max}-{\delta }_{cr})$
$cos{\delta }_{cr}=\frac{P_m}{P_{max}}({\delta }_{max}-{\delta }_0)+cos{\delta }_{max}$
$cos{\delta }_{cr}=\frac{1}{2}\times (110-30)\times \frac{\pi }{180}+cos110$
${\delta }_{cr}={69.14}^{\circ }$