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Question

The terminal voltage of the generator is held at 1.20 p.u. and the voltage of the infinite bus is 1.0 p.u. The value of steady state power limit of a system consisting of a generator equivalent reactance 0.5 p.u. connected to an infinite bus through a series reactance of 1.0 p.u. is

A
2.512 p.u.
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B
0.591 p.u
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C
1.151 p.u.
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D
1.615 p.u.
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Solution

The correct option is C 1.151 p.u.

I=(VtVj1)=(1.2θ10190)=(1.2(θ90)190)

Eδ=Vt+j0.5I

Eδ=1.2θ+(1.2(90+θ)+j)0.5j

=1.2θ+1.2×0.5θ0.5

=1.2θ+0.6θ0.5

At steady state power, δ=90

1.2cosθ+0.6cosθ0.5=0

1.8cosθ=0.5

θ=73.87

I=1.273.871j1=1.331630.1A

E=V+j1.5I

=1+j1.5×1.331630.1

=1.72790p.u.

Pmax=1.727×11.5=1.151p.u.


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