Question

Suppose 59 kW is to be transmitted over two 0.115 Ω lines.

Part A

Estimate how much power is saved if the voltage is stepped up from 110 V to 1100 V and then down again, rather than simply transmitting at 110 V . Assume the transformers are each 99% efficient.

Express your answer using two significant figures.

Psaved =

kW

Answer 1

Pout = V*I

I= P/V = 59*10^3 /110 =536.363636 A

so power loss in line = 536.363636^2 *2*0.115 =66167.7685 Watt

step down

0.99* I* 1100 = 536.363636*110

I = 54.178 A

hence power loss = 54.178^2*2*0.115 =675.108807 Watt

for the 1 % lossess in the transformer power in each transformer is

P = 0.01* ( 66167.7685+675.108807) = 668.428773 watt

Total power loss = 668.428773+675.108807 =1343.53758 watt

therfore power saved = 66167.7685 - 1343.53758 =64824.2309 Watt

your answer = 64.8 kWatt = 65 Watt

let me know in comment if there is any problem or doubt