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  • A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V

Q7.25 (b)  A small town with a demand of 800 kW of electric power at 220V is situated 15km away from an electric plant generating power at 440V. The resistance of the two wire line carrying power is 0.5\Omega  per km.The town gets power from the line through a 4000-220V step-down transformer at a sub-station in the town.

How much power must the plant supply, assuming there is negligible power loss due to leakage?

Answers (1)

best_answer

Power required 

P=800kW=800*10^3kW

The total resistance of the two-wire line

R=2*15*0.5=15\Omega

Input voltage 

V_{input}=4000V

Output voltage:

V_{output}=220V

RMS current in the wireline

I=\frac{P}{V_{input}}=\frac{800*10^3}{4000}=200A

Now,

Total power delivered by plant = line power loss + required electric power = 800 + 600 = 1400kW.

Posted by

Pankaj Sanodiya

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