A windmill park produces energy at a tension level of 11kV. The park is going to

Question

A windmill park produces energy at a tension level of 11kV. The park is going to be coupled up with a 11kV net not far from the park. To reduce the loss in transer the tension is tranfomred p at the parka nd down again right before coupling to the 11 kV net. The losses in the transfer reduces by a degree of 8/9 in relation to 11kV. What is the tension level of the transfer line?

A 16kV
B 22kV
C 33kV

1-8/9=1/9
So Tension level in the transfer line=11kV/?1/9
=11kV x 3
=33kV

Could you please explain the logic behind this calculation?

Explanation / Answer

losses are in terms of power which is the product of current and voltage

Power is directly proportional to (current)2 and (voltage)2

as copper losses in transmission

P = I2R = V2/R

=>P1 / P2 = (V1)2 / (V2)2

now the losses in the first transmission are given by 8/9 of voltage

so remaining power P1 = 1-(8/9) = 1/9

there are no losses in the second transmission because there is no transmission

so P2 = 1 - 0 = 1

=> (1/9) / 1 = (11000)2 / (V2)2

=>(V2)2 = (11000)2 / (1/9)

=>V2 = 11000 /(1/9) = 11000 x 3 = 33000 = 33kV

please ask me if you've any doubt..

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