What Is the Minimum Diameter for a 600kV DC Transmission Line with 2% Loss?

[colink96]

Homework Statement

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?

Homework Equations

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

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 that's 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/questions-and-answers/design-a-dc-transmission-line-that-can-transmit-300-mw-ofelectricity-200-km-with-only-a-2--q75618)

 

[haruspex]

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

 

[colink96]

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

 

[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.).

 

[Nickie]

As a scientist, it is important to always double check your calculations and assumptions to ensure accuracy. In this case, it seems there may be some discrepancies in the calculations and assumptions made by both yourself and the online sources you referenced.

Firstly, in the equation P=V^2/R, the power (P) represents the total power being transmitted, not just the power lost. The power lost can be calculated using the equation P_loss=I^2*R, where I is the current and R is the resistance.

Additionally, it is important to consider the skin effect, which is the tendency of high frequency currents to flow mainly near the surface of a conductor. This can affect the resistance and therefore the diameter of the wire needed for a given power transmission.

Without access to the online sources you referenced, it is difficult for me to pinpoint exactly where the discrepancies may lie. However, I would suggest double checking your calculations and assumptions, and perhaps seeking guidance from a colleague or teacher to ensure accuracy in your solution.