Electric power question?

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 am going to repeat your questions and answer each one below:

P = IV (what does P represent here?-power from where?)
This P is the power delivered to the customer. (540 kW)

(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?
This P represents the "losses" through the distribution line.

Why are there 2 "P's" in the problem?
One is the power delivered to the customer, the other is the "line losses" serving the customer through the 2.5 Ω of the distribution line.

You computed 4970 watts which is 4.97 kW.

I find this problem amusing because so many professors in academia know nothing about the real world. It is not practical to serve individual customers of this size from an 87 kV transmission line.

Answer 2

P=IV is the power delivered to the load (the 540 KW). The P in P=I²R is the power that is lost in the transmission lines and which must -also- be supplied by the power company.

The difference between the two power loss numbers is the 4970 W (or 4.97KW) that do not have to be supplied by the power company if the delivery is at 87KV instead of 12KV. That is, the power company saves 4.97KW.

HTH

Doug

Answer 3

The 2 p's are the same,
they come from P=I*V and V=I*R
P=I*I*R=I^2*R