SOLUTION

For Loss-less line at no-load:

V_s = V_r.coshγl……………..(1)

where

γl = jωl√LC

ω = 2πf

∴ coshγl = cos(jωl√L C) = cos(wl√L C)

= cos(2πfl√L C)…………..(2)

In the above question

frequency is given = 60 Hz
Length of transmission i.e l = 320 km

The time take for the surge to travel distance T =(x/v) s

Where

x is the distance

v is the velocity of the surge which is equal to 3 x 10⁸ m/s

Distance given 600km which is equal to 6 x 10⁵ m therefore,

(320 × 10³ / 3 × 10⁸ ) = 0.001 s

Putting all the value in equation number 2 we get

coshγl = cos(2π × 60 × 320 × 1000 / 3 × 10⁸) = 0.92

⇒Putting the above value in equation number 1 we get

= Vr = Vs/coshγl

= 1 / 0.92 = 1.087 pu