Question

1 MW power is to be delivered from a power station to a town 10 km away. One uses a pair of Cu wires of radius 0.5 cm for this purpose. Calculate the fraction of ohmic losses to power transmitted if

1. power is transmitted at 220 V. Comment on the feasibility of doing this.
2. a step-up transformer is used to boost the voltage to 11000 V, power transmitted, then a step-down transfomer is used to bring voltage to 220 V. (ρ_Cu = 1.7 × 10^–8 SI unit)

Answer 1

i. When power is transmitted at 220 V.

Power lost in transmission P = I²R

P = VI

P = 1 MV = 10⁶ W

V = 220 Volt

`I = P/V = 1000000/220`

`I = 50000/11 A`

R = `ρ I/A`

⇒ R = `(ρl)/(pir^2)` 

I = 10 km × 2 = 20,000 M

∴ A = πr²

r = 0.5 cm = 0.5 cm × 10^–2 = 5 × 10^–3 cm

ρ_Cu = 1.7 × 10^–8 Ωm

R = `(ρI)/A`

∴ R = `(1.7 xx 10^-8 xx 20,000)/(3.14 xx 5 xx 10^-3 xx 5 xx 10^-3)`

= `(170 xx 20000 xx 10^(-8+6))/(314 xx 5 xx 5)`

= `(170 xx 20000)/(314 xx 25 xx 100)` Ω

R = `(170 xx 800)/(314 xx 100) = (170 xx 4)/157 = 680/157` ≅ 4Ω

∴ Power loss = I²R

= `50000/11 xx 50000/11 xx 4 = (100 xx 10^8)/121 = 8.26 xx 10^7`

Power loss in heating = 82.6 MW

As 82.6 MW > 1 MW

So this method cannot be used to transmit the power.

ii. When power is transmitted at 11000 V

P = 10⁶ W

VI = 1000000

11000 I = 1000000

`I = 1000000/11000 = 1000/11`

∴ R_Cu = 4Ω as already calculated in part (i)

∴ Power loss = P = I²R

P = `1000/11 xx 1000/11 xx 4 = 4000/121 xx 4^4`

P = `3.3 xx 10^4` Watt

Fractional power loss = `(3.3 xx 10^4)/10^6 = 3.3/1000` = 0.033

Power loss in 3.3%