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Homework Help: City requires about 15 MW of power problem

  1. May 25, 2015 #1
    1. The problem statement, all variables and given/known data
    A small city requires about 15 MW of power. Suppose that instead of using high-voltage lines to supply the power, the power is delivered at 120 V. Assuming a two-wire line of 0.50-cm-diameter copper wire, estimate the cost of the energy lost to heat per hour per meter. Assume the cost of electricity is about 12 cents per kWh.

    2. Relevant equations
    P=V^2 / R, R=r x l /A, P=I x V

    3. The attempt at a solution
    Can I have a hint at solving the problem? I know the formulas but I'm not sure how to approach. What does it mean by "energy lost to heat"?
  2. jcsd
  3. May 25, 2015 #2


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    The reason why it gives you the diameter of the copper wire is to probably use the resistance of a copper wire to figure out how much is lost. Think of the copper being a very small small resistor. Unless it stated that this was a ideal wire, but I don't believe that's the case.
  4. May 25, 2015 #3
    R=r x l /A or R/l=0.0086 ohm/per meter.
    I assume that means the wire will result that much resistance per meter. ( 2 times l and below 2 times A invalidates the multiple 2)

    If I put this value to P= V^2 / R it gives me 16.83MW
    Is this the power delivered per meter??

    If city requires 15MW, the difference 1.83MW is what?? power lost per meter?? It can't be.
    Where do I go from here?
  5. May 25, 2015 #4
    Why it cant be? According to me its correct.
    You are given resistance in the question. So you HAVE to put that in the formula. Potential difference too is given. So after putting values you get to know how much power is actually SUPPLIED. But since the village needs 15MW only then where will the rest power go? It will cause heating.
  6. May 25, 2015 #5
    1.83MW power loss per meter seemed way too much!!
  7. May 25, 2015 #6


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    Well also remember that a low resistance wire will heat up more then a higher resistance wire.
  8. May 25, 2015 #7
    I think you mean 0.00086 ohm/m.
  9. May 25, 2015 #8


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    I didn't check the numbers, did OP forget to convert cm to m?
  10. May 25, 2015 #9
    No, I think it's a typo. Also, that's ohm/m for each wire, so double that for ohm/m of 2-wire line, assuming basis is one meter of electric line.
  11. May 25, 2015 #10
    First calculate the amps needed to deliver 15MW at 120V.
  12. May 26, 2015 #11


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    2017 Award

    Pretty much all losses will be as heat
  13. May 27, 2015 #12
    It's a typo sorry it's 0.00086 ohm/m.
  14. May 27, 2015 #13
    Why? Why do I need to calculate the current? Why is it that P=V^2/R does not give the same power value as I^2 x R ?

    I calculated the difference of two power values, and got 1.83MW as the power lost per meter, So multiplying that to 12 cents gave me 219.6$.
    But it's not right according to solution manual. I'm lost.
  15. May 27, 2015 #14


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    Because the voltage at the end of the line is not the same as the voltage dropped across the line. Sketch the circuit.

    Power is consumed in both the load and the line. You're interested in the power lost in the line. The straightforward approach then is to determine the current required by the load (which you should know from the load power and voltage), which must also be the current in the line since it's a series circuit.
  16. May 27, 2015 #15
    Thank you.
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