Power/Step-up and Step-down transformers?

Question

65kW is to arrive at a town over two 0.100 ohm lines. How much power is saved in the voltage is stepped up from 120V to 1200V and then down again, rather than simply transmitting at 120V. (It is assumed that the transformers are 99% efficient). The answer in book is 54kW.

Thanks in advance!

Answer 1

Could you please check precisely what the question sez again? As described, I believe it is impossible to deliver 65kw to the load from such a low voltage thru such a high impedance. I get a max power to load of 18kW, which will be when the load impedance equals the impedance of the lines,as in 0.2 ohms each, 300 amps, 18kW each. Any other load impedance will result in a lower power delivered. To get 65kW to the load, either raise the source voltage or lower the line resistance. Note the answers above and below do not account for voltage drop in the lines, which is more than just "substantial", it is the limiting factor.

Answer 2

Without transformers: Power arriving at the town is
E*I = 120*I = 65 kW. So I = 540 amps.
Power lost on the way is
I^2*R =540^2*(.1+.1) = 58.320 kW.

With transformers: Power arriving at the town is
E*I = 1200*I = 65 kW. So I = 54 amps.
Power lost on the way is
I^2*R =54^2*(.1+.1) = 583 W.
Power lost via transformer inefficiency: 65kW*.01 = 650 W
Total power lost = 583+650 = 1.233 kW
Hmm, maybe the 2 transformers (1 at each end) each loose 1%. Then
Total power lost = 583+650+650 = 1.883 kW. Still not getting what the book says.
58.320 kW - 1.233 kW = 57.1 kW saved is as close as I can come.

Answer 3

65Kw to arrive at the town.
Over 2 - 0.1 ohm lines (Rt=0.2 ohms)
Two transformers @ 99% efficiency.

How much power does the supplier have to generate to get 65000 watts to the town over a 0.2 ohm resistance?

Right off, the power company has to compensate for the transformers if he is to use 1500 volts - 65000 W will necessarily have to be 1.02*65000 to maintain 65Kw in the town. That's 66,300 W at the plant. Now he has to also compensate for the 0.2 Ohm resistance for either supply.

P=IE
E=IR
Substitute and P=I*I*R

First the necessary current:
65000=120*I = 541.7 A
66300=1200*I = 55.25 A

Then the power loss:
P=541.7^2*0.2 = 58687.78
P=55.25^2*0.2 = 610.51

At 120 volts, the power company will have to generate 65000 + 58687.78 watts to get 65000 into the town

At 1200 volts, the company will only have to generate 65000 + 610.51 watts to get 65000 into the town.

At 120 volts that's 123688 watts!
At 12000 volts, it's 65611 watts.

A savings of 58077 or 58Kv.

Answer 4

At 100 and twenty volts, the mandatory modern is 65000/100 and twenty = 541.667 A At 1200 volts, it must be fifty 4.1667 A At 100 and twenty volts, the ability loss is 541.667² * .2 = fifty 8,680 W At 1200 volts, fifty 4.1667² * .2 = 586.8 W Doug