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DC transmission lower loss on higher voltage?

  1. Apr 9, 2013 #1
    Does a DC transmission line for a given amount of power, do better at higher voltage? I know power is the product of the Voltage and current. The logic given to me by one of my peers was that I2R losses are less if the current is low. And so for a given power, it would imply V to be higher.

    But power can also be given as V2 / R, right? So by this logic wouldn't increasing the voltage increase the power dissipation? Which reasoning is right?
     
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  3. Apr 10, 2013 #2

    jambaugh

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    The [itex] I^2R [/itex] formula gives line loss because power is [itex]I\cdot V [/itex] and [itex]V=IR [/itex] is the voltage difference of the line. That is not the same as the supplied voltage which is [itex] V = I(R +R_{load}) [/itex].

    The [itex]V^2/R [/itex] formula with V the supplied voltage is assuming the line resistance is the ONLY resistance, i.e. it is the power dissipation if you have a short circuit.

    To express line dissipation in terms of supplied voltage you must know load resistance and calculate current to get voltage drop from line resistance, then subtract load power, from total power...or just use first formula once you know current.
     
  4. Apr 10, 2013 #3

    FOIWATER

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    maybe not think about it in terms of fixing the power.
    Take for example a 120 watt light bulb. If you apply anything other than its rated voltage, it won't operate at its rated wattage. It draws current and dissipates power based on its resistance at a fixed temperature, Not based on some value of power that is fixed. The power is determined from the product of the other two, not the other way around.
    In discussing a current decrease across a line, this would be done with a decrease in voltage. So discussing a decrease in current and an increase in voltage to keep the power constant really isn't correct. Since you then change the ratio of V/I and that would imply a resistance change which we know won't happen
    Any corrections welcomed
     
  5. Apr 12, 2013 #4

    NascentOxygen

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    Most likely.

    Yes.

    No, because you have already defined V to be the supply voltage. The equation for line losses will involve the voltage drop across the length of the transmission line, and this is not that V.
     
  6. Apr 13, 2013 #5

    anorlunda

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    Applying the formulas

    P=VI is the correct formula for power if V is the voltage between line and ground and I is the current through the conductors. P is the power transmitted by the line.

    P=V*V/R is the correct formula if V is the voltage difference from one end of the line to the other, not the voltage to ground. But in this case power P is the power lost during transmission, not the power transmitted. If the line resistance R is small, then the voltage difference from one end to the other is nearly zero and the power losses are nearly zero.

    jambaugh mentioned short circuits. It is the same thing. If you have a short circuit on one end of the line, then the voltage to ground is V on one end and zero on the other end. In that case the voltage difference is V, and V*V/R is the power delivered to the line, but 100% of that power goes to losses and 0% is transmitted. That's one of the reasons why short circuits are bad.
     
  7. Apr 13, 2013 #6

    sophiecentaur

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    You need to be considering the right "V" in your comparison. The V dropped would be less because the I would be lower for a higher transmission voltage.
     
  8. Apr 13, 2013 #7

    rbj

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    this is the case. but it is the case for either AC or DC. the reason why, for very long distances, DC transmission lines have less transmission loss is because of the the skin effect. for AC current, most of the current flow closer to the surface of the cylindrical conductor. that means that the inside of the conductor is not used much and this skin effect effectively reduces the cross section of the conductor and increases its resistance per unit length.

    so DC does better than AC because of decreased R in I2R losses assuming the same conductor diameter. it's the same I (at least in terms of r.m.s.).

    this is the case when the lossy R is in parallel with the load. this is leakage loss. any parallel path that causes leakage, you want that R to be as high as possible. but series R, you want that to be as low as possible.
     
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