A 60 hz, 320 km lossless line has sending end voltage 1.0 p.U. The receiving end voltage on no-load is
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For Loss-less line at no-load:
Vs=Vr.coshγl............................(1)
where,
γl = jwl√LC
so,
coshγl = cos(jwl√L C) = cos(wl√L C) = cos(2∏x60x320x1000x√L C).......(2)
But,
1╱√(LC) = v = c = velocity of light
Putting this in equation (2),
coshγl = cos(2∏x60x320x1000/3*10^8) = 0.92
Putting this value in equation (1)
Vr = Vs/coshγl = 1/0.92 = 1.087 pu
Vs=Vr.coshγl............................(1)
where,
γl = jwl√LC
so,
coshγl = cos(jwl√L C) = cos(wl√L C) = cos(2∏x60x320x1000x√L C).......(2)
But,
1╱√(LC) = v = c = velocity of light
Putting this in equation (2),
coshγl = cos(2∏x60x320x1000/3*10^8) = 0.92
Putting this value in equation (1)
Vr = Vs/coshγl = 1/0.92 = 1.087 pu
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