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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
- power is transmitted at 220 V. Comment on the feasibility of doing this.
- 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. (ρCu = 1.7 × 10–8 SI unit)
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Solution
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
ρCu = 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`
∴ RCu = 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%
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