# A cylinder rotor generator delivers 0.5 pu power in the steady – state to an infinite bus through a transmission line of reactance 0.5 pu. The generat

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A cylinder rotor generator delivers 0.5 pu power in the steady – state to an infinite bus through a transmission line of reactance 0.5 pu. The generator no-load voltage is 1.5 pu and the infinite bus voltage is 1 pu. The inertia constant of the generator is 5 MW – s / MVA and the generator reactance in 1 pu. The critical clearing angle, in degrees, for a three – phase dead short circuit fault at the generator terminal is
1. 53.5
2. 60.2
3. 70.8
4. 79.6

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Correct Answer - Option 4 : 79.6

PS = Pe1 = 0.5

Before fault ${P_{m1}} = \frac{{{E_V}}}{X} = \frac{{1.5 \times 1.0}}{{1.5}} = 1.0$

During fault Pm2 = 0,

After the fault Pm3 = 1.0

$\begin{array}{l} \delta = {\sin ^{ - 1}}\left( {\frac{{{P_S}}}{{{P_{m1}}}}} \right) = {\sin ^{ - 1}}\left( {\frac{{0.5}}{{1.0}}} \right) = 30^\circ \\ {\delta _o} = \frac{{30 \times \pi }}{{180}} = 0.52\ rad\\ {\delta _{max}} = 180 - {\sin ^{ - 1}}\left( {\frac{{{P_S}}}{{{P_{m3}}}}} \right)\\ = 180 - {\sin ^{ - 1}}\left( {\frac{{0.5}}{1}} \right) = 150^\circ \\ {\delta _{\max \left( {radians} \right)}} = \frac{{150 \times \pi }}{{180}} = 2.618rad\\ {\delta _C} = {\cos ^{ - 1}}\left[ {\frac{{{P_S}\left( {{\delta _{max}} - {\delta _o}} \right) + {P_{m3}}\cos {\delta _{max}}}}{{{P_{m3}}}}} \right]\\ = {\rm{ }}79.45^\circ \end{array}$