GATE Electrical Engineering (EE) 2021 Solved Papers

Question : 15 of 65

Marks: +1, -0

An alternator with internal voltage of

1 \angle \delta_{1}

p.u. and synchronous reactance of

0.4

p.u. is connected by 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

\left(\delta_{1}-\delta_{2}\right)

is _____________degrees. (Round off to 2 decimal places.) (Machines are of non-salient type. Neglect resistances.)

Solution:

Total power transferred,

P=\;{E_{f} V}/{X_{e q}} \sin (δ_{1}-δ_{2})

⇒ 0.866=\;{1 × 0.85}/{0.85} \sin (δ_{1}-δ_{2})

Total reactance of the system,

X_{e q}=j 0.4+j 0.1+j 0.35=j 0.85

p.u.

\sin (δ_{1}-δ_{2})=0.866

(δ_{1}-δ_{2})=60^{∘}

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