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DC Transmission Line Problem

  1. Apr 10, 2013 #1
    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. jcsd
  3. Apr 10, 2013 #2

    haruspex

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    You've applied the 2% in the wrong way (backwards) leading to 1/50th of the correct answer.
     
  4. Apr 10, 2013 #3
    Thanks for the response. How are you saying I should change the values in my equation? Should power be equal to .98 * 300MW?
     
  5. Apr 10, 2013 #4

    haruspex

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    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.).
     
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