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Losses in power transmission lines

  1. Jan 24, 2009 #1
    Ok when they taught me about this they kept saying the power loss = I2*R

    Well isn't the power loss also = V2 / R

    Now before you give me the regular answer of "this V is not the same as this V", i understand. However, from what i know from the voltage divider rule is that the V across the lines is a fraction of the V across the source (the rest is across the consumer's end resistance). Therefore if V (from the source) say doubles, the V across the lines should also double. And this will lead to the same power loss in the lines as before.

    This is from what i know and i'm sure there's something wrong. Either i'm applying formulas incorrectly (eg. does the voltage divider rule even apply here?) or there's something i just don't know.

    ------------------

    I also have another question regarding the same thing:

    Here are the variables i used (just to make it clear):

    Ps - the power supplied by the generator
    V1 - the voltage in the primary coil of the transformer
    V2 - the voltage in the secondary coil of the transformer
    I1 - the current in the primary coil
    I2 - the current in the secondary coil
    RL - the resistance of the transmission lines
    Rc - the resistance of the consumer stuff
    VL - the voltage drop across the transmission lines
    Vc - the voltage drop across the consumer stuff


    This is what i did,

    Ps = V1*I1 = V2*I2

    V2 = VL + Vc
    Ps / I2 = I2*RL + I2*Rc
    Ps = (I2^2)*RL + (I2^2)*Rc

    Does this mean that if i decrease the current (as well as increase the voltage) by increasing the number of coils on the 2ndary coil, the power supplied by the generator decreases?? I'm really confused. Note that i'm in my 3rd semester in uni so there's still some stuff i didn't take (like voltage regulation...i just found out a few minutes ago that the output voltage of a transformer changes when the load changes...still don't understand it but that's not really my question here).
     
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  3. Jan 24, 2009 #2

    mgb_phys

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    This is isn't true if you have transformers at each end - it would only be true for a simple resistive load.

    There are a bunch of earlier threads explaining this - but basically.
    I want to send 1MW to a city, I can use transformers to change the voltage/current at each end. Lets say the transmission line has a resistance of 1 Ohm.
    I could just send it at 110V (above ground) but this would need 10,000A and so would give a voltage drop of 10,000V along the wire and nothing would come out of the other end - as well as wasting 100MW of power heating up the line.

    Or I could increase the voltage to 1MV and only send 1amp down the wire, then I would only lose 1W as heat and I would get 999,999V at the other end.
    The other transformer steps this back down to 100V for your house.
     
  4. Jan 24, 2009 #3
    You have some things confused. You speak of voltage across the line is a fraction of what it is across the source. I think you are thinking of the voltage from one end to the other of each individual wire. This voltage squared divided by the resistance in the individual wire will get the same watts that the current squared multiplied by the resistance in the individual wire. Of course this is the loss for only one wire. Total loss will be losses in all wires involved. You are not applying the ohms law correctly.
     
  5. Jan 27, 2009 #4
    @Averagesupernova: I don't see where i'm applying them incorrectly. What i mean by the voltage across is the lines is the TOTAL voltage across the lines so tht's the voltage across the 'send' wire + the voltage across the 'coming back' wire.

    @mgb_phys: My physics professor told me this too but how can i get 100MW dissipated in the wires (let's say they don't melt and stay intact) when i'm only sending 1MW? How can i even get 10000V when the source's voltage is 100 (i'm assuming that's the number u wanted to type right?)? I know you're going to say P=I2*R but shouldn't the other laws apply too? Or maybe not? I think I have a problem with which rules and formulas apply here and which don't.

    Thanks for clearing that up. It was really confusing me. :)
     
  6. Jan 27, 2009 #5
    You say this:
    Then you say this:

    Which is it? They are not the same thing. The voltage at the source minus the voltage at the load is the voltage from one end of the wire to the other end, figuring both wires additively. If you had a voltmeter with really long leads, (or just build the transmission line to loop back to where the source is so the load is right next to the source except being fed with a really long transmission line) you would be able to measure this voltage.
     
  7. Jan 28, 2009 #6

    mgb_phys

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    It was an example to show that you could't use 110V for such a transmission line - because nothing would come out of the other end, all the power would be dissipated as heat.

    Perhaps a more reasonable example, suppose you need to supply 1MW (enough for a city block) and again a 1 ohm line.
    At 4Kv (a typical trash can size pole transformer) current from I=P/V = 250A
    Vlost= I R = 250V so you get 3.75Kv out at the other end and waste I2R = 62KW as heat.

    At 25Kv (typical sub station) I = 40A, so only 1.6KW is wasted (about a tea-kettle)

    At 1MV only 1Watt would be wasted - however running million volt cables is difficult since they have to be kept a long way from the ground to prevent arcing and the switching gear is expensive.

    The power company makes a decision on what voltage to use based on the cost of high voltage equipment vs the power wasted from low voltage supplies.
     
  8. Jan 29, 2009 #7
    @Averagesupernova: Yes i understand and tht's wut i said. The voltage across the lines (as in both wires) is the voltage across the source minus the voltage across the consumer-end. I was referring to another rule that called the voltage divider rule which applies when resistors are in series. I thought it would apply here and tht's y i referred to it but i was wrong. It's basically:

    Vacross Resistor A = Vacross both resistors * (RA / [RA + RB])

    @mgb_phys: Thx for explaining. I get that part now. It was mainly the voltage divider thing tht confused me. How about 2nd question tho (the one in the 1st post)? Does it even make sense?
     
  9. Jan 29, 2009 #8

    stewartcs

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    You can't. The maximum power that can be dissipated as heat is limited to what the power supply is sending. If the power supply is sending 1MW, then the most you will dissipate as heat in the cable would be 1MW. Conservation of energy always applies.

    Use a step-up transformer.

    CS
     
  10. Jan 29, 2009 #9

    stewartcs

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    The generator will only supply what it needs for the load (plus any losses). The load determines how much power is sent by the generator. In other words, the power supplied by the generator will be equal (neglecting losses) to what the load draws. Any combination of voltage and current that equals the apparent power drawn by the load will work.

    CS
     
  11. Jan 31, 2009 #10
    I was referring to the voltage across the wires compared to the voltage across the 2ndary coil of the transformer. The voltage across the wires was higher and that wasnt possible because as you said the power dissipated as heat is limited by the power supplied by the generator.

    Hmmm...thx alot for your help :) It's all clear now. woohoo
     
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