# Homework Help: DC Transmission Line Problem

1. Apr 10, 2013

1. The problem statement, all variables and given/known data

You have been asked to develop a cross-country DC electrical transmission line to deliver 300MW of power to a town 200km away. This design is to only have a 2% loss (98% efficiency). If the line voltage is 600kV, what is the minimum diameter that the aluminum wire can be?

2. Relevant equations

R = p * l / A
P = I * V
P = (I ^ 2) * R
P = (V ^ 2) / R

3. The attempt at a solution

This problem seems simple enough to me, yet I heard the answer I got was wrong.

P = (V^2)/R
---R = p * l / A
------A = pi * d^2 / 4
---R = 4 * p * l / (pi * d^2)
P = (V^2) / (4 * p * l / (pi * d^2))
P = pi * d^2 * V^2 / (4 * p * l)
d = sq(4 * p * l * P / (pi * V^2))
---P = 2% of total power = .02 * 300e6 Watts
d = sq(4 * 2.7e-8 * 200e3 * .02 * 300e6 / (pi * (600e3)^2)
d = 3.460e-4 m = 0.000346 meters

However something is wrong, because thats not the right answer (and that would be a REALLY small wire!).
From what I understand power in the P=V^2*R equation is equal to the power dissipated, or the power lost. Is that correct?

I found someones attempt online, though some parts seem off to me, and I don't understand why they did the two things they noted. Also, we used different resistivities, but that is not a big deal.

(http://physics.hivepc.com/eminduct.html)

p of aluminum is 2.65x10^-8

I = P/V.
I = 300x10^6W/600x10^3V = 500A.

P_Loss = I^2R.
(300x10^6*.02*1.02 ) = 500^2*R (PAY CLOSE ATTENTION TO .02*1.02*P_input)
24.48 = R
R = pL/A, 24.48= 2.65x10^-8 * 2*200x10^3/pi*r^2
*NOTE* in a dc line there is a to and fro, so two times the distance.
2r = d.
d = 2.348cm

It also appeared on chegg, but I can't view it without the subscription... (http://www.chegg.com/homework-help/...mw-ofelectricity-200-km-with-only-a-2--q75618)

2. Apr 10, 2013

### haruspex

You've applied the 2% in the wrong way (backwards) leading to 1/50th of the correct answer.

3. Apr 10, 2013

Thanks for the response. How are you saying I should change the values in my equation? Should power be equal to .98 * 300MW?

4. Apr 10, 2013

### haruspex

It's a bit hard to follow your working because you don't discriminate clearly between the total power the line is carrying and the power lost to resistance. P = (V^2)/R is valid if R is the total resistive load across which the voltage drop V occurs. So if R includes the useful load on the circuit, not just the losses, then P is the 300MW, but if R is the resistance of the wire then you need to break this into two stages:
Total power = Voltage * Current
Lost power = Current2 * Resistance
The net effect of your confusion was to multiply by the 2% instead of dividing by it (or v.v.).