The electricity supplied in exercise 34 is transmitted over a line 80.0km long w

Question

The electricity supplied in exercise 34 is transmitted over a line 80.0km long with a resistance of 0.80ohms/km.

a.) how many kilowatt-hours are saved in 5.00 hours by stepping up the voltage?

b.) At $0.15/kWh, how much of a savings (to the nearest $10) is this to all the consummers the line supplies in a 30 day month, assuming that the energy is supplied continuously?

Exercise 34 reads: "An ac generator supplies 20A at 440V to a 10,000V power line through a step-up transformer that has 150 turns in its primary coil. Transformer is 95% efficient". Only use this to answer my first question. Please show work, thanks

answers are a.) 128kWh and b.)$1840

Explanation / Answer

resistance of wire of length 80km = 80*.8=64 ohm

current in power line=I'=IV*.95/V'=0.836 A

energy dissipated when current is 20 A =W= 20^2*64 watt=25.6 kwatt

energy dissipated when corrent transfer is at .836 A=W'=.836^2*64= .0447 kwatt

a) energy saved in 5 h = (25.6-0.0447)x5 watt=127.7 kwatt h

b) energy saved in 30 days=(25.6-.0447)*30*24=18399.8kwatt h

savins in 30 days= eaergy saved X cost per uni energy(.15$)

18399.8*.15=2759.97 $

( answer that u have given for (b0 is wrong )

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