Question

An electric power station that operates at 15 kV and uses a 10:1 step-up ideal transformer is producing 360 MW (Mega-Watt) of power that is to be sent to a big city which is located 260 km away with only 2.0% loss. Each of the two wires are made of copper (resistivity = 1.68 × 10^ -8 Ω.m). What is the diameter of the wires?

Answer 1

From Transformer turns ratio

V_S/V_P=N_S/N_P

V_S/15 =(10/1)

V_s=150 KV (transmitted voltage)

Current transmitted

I=P/V_s=(360*10⁶)/(150*10³)=2400 A

Given Transmission line losses

P_loss=0.02*360 =7.2 MW

Since P_loss=I²R

2400²*R =7.2*10⁶

R=1.25 ohms

Since Resistance is given by

R=pL/A

=>A=pL/R =(1.68*10^-8)(2*260*10³)/1.25

A=6.9888*10^-3 m²

A=pi*D²/4=6.9888*10^-3

D=0.09433 m =9.433 cm