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Power loss in transmission lines

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  1. Apr 3, 2016 #1
    1. The problem statement, all variables and given/known data
    My textbook says that high voltages are used for power transmission because according to the equation P=VI and P= I2R, power loss can be reduced.

    What confuses me is why the equation P=V2/R cannot used here? And if used, would be contradictory? I know that R is the resistance of the transmission line and is constant, but the voltage or current supplied can be altered. For example, if for constant power supplied P, I can either transmit it at high V, low I or vice versa. If at high V, then if I is low then P=I2/R is low. On the other hand, P=V2/R is high.

    I don't know how to reconcile these two opposing results. Why can't P=V2/R be used?

    2. Relevant equations
    P=VI
    P=I2R
    P=V2/R

    3. The attempt at a solution
     
  2. jcsd
  3. Apr 3, 2016 #2
    Who knows what the value of R is? Also, there other types of losses that make R irrelevant.
     
  4. Apr 3, 2016 #3

    haruspex

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    That formula applies when V is the voltage drop across the resistance R. In a transmission line, the resistance of the wire, R, is, we hope, only a small portion of the whole load. The useful load has some other resistance, RL, say. We wish to supply a certain level of power PL to the useful load. If the source is at voltage V, the current is V/(R+RL), and we have PL=V2RL/(R+RL)2. The power lost is PLR/RL.
    If we increase V for the same PL, RL also increases, reducing the power lost.
     
  5. Apr 3, 2016 #4
    Hi thanks for answering! Maybe I'll include a question to make things clear.

    A power station generates electrical power at a rate of 10MW. This power is to be transmitted along cables whose total resistance is 10Ω. Calculate the power losses in the cable if the power is transmitted at 50kV and at 250kV.

    Now, according to the textbook, at 50kV, the ##I## is ##\frac{10M} {50k}## which gives 200A. And the power loss is P=I2R which is 400kW. What I want to know is why the other equation for power, P=V2/R cannot be used here to calculate power losses.
     
  6. Apr 3, 2016 #5
    i think the heat losses in power transmission is dependent on current- thermal energy=I^2 R .t where t is the time -so one tries to keep current smaller in high voltage transmission lines and any loss calculation is done involving current rather than voltage.
    this we learn't in school level physics-but i may be wrong. let us see what electrical tech. say!
     
  7. Apr 3, 2016 #6

    haruspex

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    @toforfiltum's error is taking the whole voltage drop V as being across the resistance of the transmission line only. Were that the case, a higher voltage would indeed generate more waste heat.
     
  8. Apr 3, 2016 #7
    Ok, I think I see it now. The resistance of useful load must be very much larger than resistance of transmission wire, so that voltage drop across wire is very small compared to voltage supplied to useful load. As such, using P=V2/R, it makes sense to transmit at high voltage. And since voltage drop across transmission wire is small, according to P=V2/R, power loss is also small. Because of high voltage supplied, current will be low, and P=I2R makes sense. Is this sort of right?
     
  9. Apr 3, 2016 #8

    haruspex

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    Yes.
     
  10. Apr 3, 2016 #9
    Phew, thanks! So the reason makes use of the concept of potential dividers. Btw, I'm still trying to figure out your steps in post #3.
     
  11. Apr 4, 2016 #10
    The resistance mentioned is more for DC current than for high Voltage AC.
    The P=V2/R formula assumes the current density is uniform in the conductor, it's not always true in practical situations. This formula is good for approximation for long thin conductors. However this formula is not exact for High Voltage alternating current (AC), because the skin effect inhibits current flow near the center of the conductor. The geometrical cross-section is different from the effective cross-section in which current actually flows, so the resistance is higher than expected. So the R in P=V2/R becomes less relevant and P=VA is more accurate.
     
  12. Apr 4, 2016 #11

    haruspex

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    Isn't that just equivalent to specifying a resistivity that takes into account the frequency and the physical characteristics of the wire? I see no definition of R in post 0.
     
  13. Apr 4, 2016 #12
    R in the formula is resistance. Transmission of high voltage alternating current involves skin effect and impedance too.
     
  14. Apr 4, 2016 #13

    haruspex

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    Yes, but I'm asking whether that can be allowed for, given the frequency and physical characteristics of the wire, by using an adjusted value for R.
     
  15. Apr 4, 2016 #14
    I would imagine so but it may be a negligible effect given that the high voltages of transmission lines negates the need.
     
  16. Apr 4, 2016 #15

    haruspex

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    My point is that if it can be handled that way then the discussion in the thread remains valid.
     
  17. Apr 4, 2016 #16
    I was focused on his question about why R was being ignored in the power equation in favor of an equation that ignores it.
     
  18. Apr 4, 2016 #17

    Merlin3189

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    I would have thought quite the contrary. Since the resistance is specified for a wire carrying high voltage AC, then presumably that is what it is for! The DC resistance may well be lower, but that is irrelevant in this context.
    Again, I would think in this context that P=V2/R (or rather I2R) applied exactly to the power, current and resistance quoted at the stated voltage..
    It does not assume any particular current distribution across the area of the conductor: provided the resistance is measured at the frequency used (which is not a term in the formula) then the formula is valid for any length of wire.
    I can't think of any mechanism, including skin effect, which changes the value of resistance for High Voltage.
    If the values (of power and resistance) are correct at low voltage, then the same value of resistance will give the correct result at high voltage.
    I'm not sure what the "expected" resistance is here. If someone has either measured or calculated the resistance at 50Hz, presumably that is what they expect and that is what it is.
    This is a puzzle. It may depend on how you define (and measure) V and A? It seems to me that I2R is the definitive quantity for power dissipated in the resistance of the wire, however I is measured. Then both V2/R and VI are equivalent to it when you define V as IR, rather than making an independent measurement of V.
    Certainly if you try to measure V, rather than calculating it from IR, you will have to determine the phase angle between voltage and current. But even that will not solve all problems when the current is not sinusoidal.
    I hope the designers have carefully calculated the resistance from the physical properties of the wire and verified that by measuring test specimens.
    I would be interested to hear what other losses there are and how they compare with I2R in the wire. I suppose like our water supply, maybe a significant part of the supplied current never reaches the load and power is lost to I2R in the earth? Is this why we no longer hear of attempts to engineer superconducting grid cables?
    My own impression had been that I2R losses in the cables were significant and increasing due to greater distortions in the load waveform.
     
  19. Apr 4, 2016 #18

    haruspex

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    But that was not the question. It was why I2R is used rather than V2/R, and the answer to that, as I wrote in post #3, is that those equations only apply when the other variable (V or I) applies to the whole of the resistance represented by R. If that resistance is the transmission line then I does apply right across it, but V does not.
     
  20. Apr 5, 2016 #19
    I recall that the original question was: "Why can't P=V2/R be used?" I thought it was because the power delivered was conditioned by the voltage applied to the load and power in a transmission is more accurate when VA is used because the R is bundled with the load.
     
  21. Apr 5, 2016 #20

    haruspex

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    That was the question, but the key answer is that the V referred to there included the voltage drop over the useful load, which is clearly inappropriate.
     
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