Electric power???

Question

Suppose a company delivers 540kW of power at 12kV thru wires that have a total resistance of 2.5 ohms. How much power is saved by delivering the power at 87kV rather than 12kV?

I read in my textbook that {power lost due to heat} is proportional to R*I^2. I also know how to get the right answer. I just don't know why (I messed around with the numbers and somehow got the right answer).
The procedure to get the right answer is:
P = IV (what does P represent here?-power from where?)
(540kW) = I (12kV)
I = 45 A for the delivery at 12kV
Then: P = I^2*R = (45^2)(2.5) = 5062.5
Another question here: what does this P represent? Why are there 2 "P's" in the problem?

Repeat procedure for 87kV
P = IV
Solve for I.
Then: P = I^2*R = 96 A
Then the correct answer is obtained by subtracting:
5063 - 96 = 4967 = 4970 W
They give the answer as 4.97 kW.

Answer 1

I think an explanation is in order. Any power company has two main tasks:

a) Generating power.
b) Delivering generated power to customers, wherever they are.

Both processes involve, inevitably, some loss; nothing is perfect in real life. Your problem refers specifically to power delivery. Power is to be carried over from one location to another, and in the process, some of the delivered power is lost. Power is delivered from one place to other by means of "transmission lines". Theoretically, you could reduce transmission losses to naught. In practice, you can do this only at a cost. As could be expected, reduced loss comes along with increasing costs.

Specifically, transmission loss is –mainly– caused by resistance in the lines. Using heftier lines (wires) will decrease line resistance and indeed reduce loss, but heavier wire costs more. As in any other engineering endeavor, you have to compromise. So much for briefing. Let’s get into the problem.

Power available to be transmitted is 540 kW. Thus, 540 kW are fed into the “sending” or “transmitting” end of the line. At 12 kV, this means 540/12 = 45 A circulating thru the transmission line, and, of course, at the load, which will be connected to the “receiving” end of the line, usually at hundreds of miles from the opposite end. Resistance of line is given as 2.5 Ω. Thus, P’ = I²R = 45² ×2.5 = 5 062.5 W = 5.0625 kW are lost in the line. Consequently, actual power delivered is 540 − 5.0625, or slightly less than 535 kW.

Consider now transmitting this same power at 87 kV –instead of 12 kV– over the same line. In order to transmit this much power, I = P/V = 540/87 = 6.207 A have to flow thru the line. Power lost in the line is now P’ = I²R = 6.2² × 2.5 = 96.31 W, or just 1.9% of the power lost when transmitting at 12 kV. Clearly, power saved at 87 kV –as compared to 12 kV– is 5.0625 − 0.09631 = 4.9662 kW. Rounded to 3 significant figures, power saved = 4.97 kW.

Hope this has clarified the issue to you.

Answer 2

Normally the "P" symbols for power would have subscripts indicating the meaning, for example "P sub d" for "power delivered to the load" and "P sub h" for "power lost as heat". It is confusing to use the same symbol for two different quantities, even if they both are power.

Answer 3

Electric power is the rate at which electric energy is transferred. Electric power is measured by capacity and is commonly expressed in megawatts (MW). A megawatt (MW) is one million watts.

The electric power P is equal to the energy consumption E divided by the consumption time t:

P is the electric power in watt (W).
E is the energy consumption in joule (J).
t is the time in seconds (s).

https://www.electrikals.com/

Answer 4

p(1)= power supplied through wire=540kW=IV
V=12kV
p(2)=I^2*R is power lost as heat