Question

Suppose 55 kW is to be transmitted over two 0.150 Ω lines.
Estimate how much power is saved if the voltage is stepped up from 160 V to 1600 V and then down again, rather than simply transmitting at 160 V . Assume the transformers are each 99% efficient.
Psaved = __________kW

Answer 1

given

pdeli=55000w

vout=160v

current i=pdeli/vout=343.75 A

plost=i2r=(343.75)2 2( 0.150 Ω)

=36.45 kw

thus there must be 55 kw+35.45kw=90.45 kw

i=vout*iout/0.99*v =160*343.75/0.99*1600 =34.72 A plost=I2*R =(34.72A)22*0.15Ω=361.64 w the power that must be delivered to the step down transformeris 55000/0.99=55555.56 w the power that must be present at the start of thetransmission must be 55555.56w+361.64w=55917.21 w the power that must enter the transmission lines fromthe99% efficient step up transformer is 55917.21/0.99 =56482.03 w so the power saved is 90.45kw-56.482kw=33.96kw = 34 Kw