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Estimating cost due to power

  1. Apr 11, 2009 #1
    1. The problem statement, all variables and given/known data

    A small city requires about 19 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 0.60cm -diameter copper wire, estimate the cost of the energy lost to heat per hour per meter. Assume the cost of electricity is about 8.5 cents/kWh.
    Cost = $ per hour per meter



    2. Relevant equations

    P=IV
    P=I^2*R
    R=(rhoe*L)/A

    3. The attempt at a solution

    I used P=IV to find the current (converting 19 MW to 19*10^3 to keep it in kW for the answer) and found I=158 A .. then when I go to find the resistance I figure to use pi*r^2 as area than multiply the whole thing by two for the two wires but I do not understand what to use for L because they provide no length after i find R i intend to plug it into the equation P=I^2*R .. i am so lost!
     
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  3. Apr 11, 2009 #2

    Redbelly98

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    Since they want cost per meter, you really just need R/L and P/L.

    p.s. welcome to PF.
     
  4. Apr 11, 2009 #3
    thanks but i do not understand what you are saying sorry :(
     
  5. Apr 11, 2009 #4

    Redbelly98

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    Okay, let's back up a little.

    From your 1st post, you said you tried to calculate R, but need L in order to do that. What equation for R are you using, that has an L in it?
     
  6. Apr 11, 2009 #5
    ohhh yeah thats what my problem is i am not given L .. i am trying to use R=(rho(resistivity)*L)/A
     
  7. Apr 11, 2009 #6
    L being the length of the wire and A being the cross sectional area of the wire
     
  8. Apr 11, 2009 #7

    Redbelly98

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    Okay.

    Since you need to calculate power and cost per meter, use L=1m.
     
  9. Apr 11, 2009 #8
    okay so then i use l=1 to find the resistance and i take the current that i found usuing p=iv and plug both into P=I^2R .. once i find that P how do i incorporate finding time?
     
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