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Total Resistance of Transmission Cables

  1. Oct 24, 2011 #1
    1. The problem statement, all variables and given/known data
    A mining company use an electric pump with and operating voltage in the range 1.25kV - 1.5kV. There is only a 240 Vrms supply available. A transformer is used to step up the output voltage to 1.5 kVrms. The secondary winding of the transformer has 2000 turns of wire.

    The first part of this question was to find the number of turns on the Primary winding of the transformer, which I have calculated correctly to be 320 turns.

    The second part is thus;
    The transformer has an electrical power output of 6.45kW. The underground pump is connected by 1.1km of cables to the surface. The potential difference across the pump is 1.45kV. Calculate the total resistance of the cables.

    2. Relevant equations
    P = (V^2) / R
    This is the equation I tried but it didn't work obviously.


    3. The attempt at a solution

    P = (V^2) / R
    R = (V^2) / P
    = (1.45x10^3)^2 / 6.45x10^3
    R = 325.97Ω

    This result is obviously quite outlandish for the situation, and I'm really stuck as to how to progress in this question.
     
  2. jcsd
  3. Oct 24, 2011 #2

    Delphi51

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    You are applying this formula to the cable. The trouble is that 6.45x10^3 is the power coming out of the transformer, not the power lost in the cable. You could subtract the power arriving at the pump to get the power lost in the cable and use it in this formula.

    Alternatively, since you know the power and voltage of the transformer, you could find the current. The current is the same throughout the circuit, so you can apply V = IR to the cable with that current and the known voltage drop in the cable to find the resistance.

    You should get the same answer both ways!
     
  4. Oct 24, 2011 #3

    cepheid

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    Welcome to PF!

    Is the result really that outlandish for the situation? I have no real intuition for this sort of thing, but assuming that the cable was just a solid copper wire of circular cross-section, and using the resistivity of copper, the length of the cable, and the total resistance you derived, I worked out that the wire would have to have a diameter of 0.27 mm, which does seem pretty ridiculous at these current/power levels. :biggrin:

    Of course, I did all of this as an academic exercise for fun, because in reality you DID make a mistake. Your real issue seems to be your interpretation of the word problem. If the output at the transformer is 1.5 kV, and the input to the pump is 1.45 kV, then what is the loss across the cable? Hint: the answer is certainly not 1.45 kV. :wink:
     
  5. Oct 24, 2011 #4

    gneill

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    Why do you suppose that the voltage at the pump end of the cables is different from the voltage at the transformer end?
     
  6. Oct 24, 2011 #5
    Ah, I think I see now. Instead of using the power output used in the transformer, I need to use the power loss in the cables (which is 1.5kv - 1.45kv according to the info), so using .05kv in the same equation I get a value for the resistance at ~42050Ω. This is still a really large answer, what am I missing?
     
  7. Oct 24, 2011 #6

    gneill

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    Voltage is not power. Voltage times current is power.

    Can you determine the current delivered by the transformer?
     
  8. Oct 24, 2011 #7

    cepheid

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    As Delphi51 pointed out, not only was your V wrong, but your P was also wrong. 6.45 kW was the total power delivered by the transformer (which includes BOTH the power dissipated as heat in the cable, and the power consumed by the electric pump). You should listen to Delphi51's tips on how to solve for the power dissipated in the cable.
     
  9. Oct 24, 2011 #8
    The power output is 6.45kW, so P=VI dictates the current coming out of the transformer is 4.3A (I think), so using this same equation on the other end, instead finding P, I get 6235W or 6.235kW. So the Power loss is .21kW or 210W, which we can then put into P=(V^2)/R. Is this right?
     
  10. Oct 24, 2011 #9

    gneill

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    If you've got the current and the voltage drop, why not just use Ohm's law to find the resistance :wink:
     
  11. Oct 24, 2011 #10

    cepheid

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    215 W actually, but yeah, that's right.
     
  12. Oct 24, 2011 #11
    Well that just makes it easy. V = IR, so R = V/I, so R = 11.63Ω. Correct?
     
  13. Oct 24, 2011 #12

    gneill

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    Looks like a reasonable result!
     
  14. Oct 24, 2011 #13
    Thanks to everyone for the help, this has really enhanced my knowledge. :)
     
  15. Oct 24, 2011 #14

    Delphi51

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    It is worth working out with P = V²/R as a check!
     
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