GATE 2021 EE – Question 29
An alternator with internal voltage of $1\angle\delta_1$ p.u and synchronous reactance of 0.4 p.u is connected through a transmission line of reactance 0.1 p.u to a synchronous motor having synchronous reactance 0.35 p.u and internal voltage of $0.85\angle\delta_2$ p.u. If the real power supplied by the alternator is 0.866 p.u, then $(\delta_1-\delta_2)$ is ________ degrees. (Round off to 2 decimal places.)
(Machines are of non-salient type. Neglect resistances.)
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Correct answer: 60
Explanation
The total reactance between the two internal voltages is $0.4+0.1+0.35=0.85$ p.u. The power transferred is $P=\frac{E_1E_2}{X}\sin(\delta_1-\delta_2)=\frac{1\times0.85}{0.85}\sin(\delta_1-\delta_2)=0.866$, so $\delta_1-\delta_2=60^\circ$.