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Electric power transmission - what is different at a particle level?

  1. Mar 11, 2008 #1
    Hello!
    1 MW can be transmitted , for example, in 2 ways:

    1MW=100V*10000A;

    or

    1MW=10000V*100A.

    In the second case less energy is lost because P=RI^2. But my question is: what is different at a particle level to deeply understand what is going on?

    Thanks
     
  2. jcsd
  3. Mar 11, 2008 #2

    vanesch

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    Staff Emeritus
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    Gold Member

    Take the water analogy: voltage can be compared to difference in height, and current can be compared to water flow. You can obtain the same power transmission by a small tube mounted 2 km above the surface, which will transport a small water flow a long way, until it arrives at the point of where you want to use the power. There you let it flow downward, and, say, drive a water mill. The high pressure of the water (2 km water column!) makes that a small amount of water can deliver a lot of power.
    Or you can have a big tube, at 10 meters above ground, with a big flow of water. This time the pressure isn't very high, but the big flow compensates, and the water mill will deliver the same amount of power.

    Why is it interesting to use high voltages and small currents ? For two reasons: our insulators are better than our conductors, and they are cheaper too.

    A high voltage over an insulator causes almost no losses. A big current through a conductor does. If we would have had more leaky insulators (say, in the sea), and superconductors, then the decision might have been different.

    The power is transported both by the conductors and the insulators.
     
  4. Mar 12, 2008 #3
    Ok, thanks. I understood, but not at a particle level.
     
  5. Mar 12, 2008 #4
    The loss is only lower using AC transmission.

    It was the primary cause of the falling out between Edison (a DC proponent) and Faraday (an AC proponent) who showed that AC transmission had much lower losses than DC transmission for the same Power delivered to the customer. It's a big part of the reason Westinghouse gave him 20 million for the patent around the 1920's. Imagine, a brand new Ford is $80 and you just got a cheque for 20 million. Good idea that AC.
     
  6. Mar 12, 2008 #5

    berkeman

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    Not true. For a given voltage and current, the losses are the same. The difference is that high AC voltages are much easier to make, since all you need are transformers for the step-up at the generating station, and the step-down at the distribution stations and at the neighborhood distribution transformers. Especially back in Edison's time, DC-DC voltage conversion was not very well developed.

    To the OP -- at the atomic level, higher voltage means less current, which means fewer collisions between the electrons and the lattice of atoms in the conducting wires. Fewer collisions means less power lost to phonons in the lattice (vibrations making heat).
     
  7. Mar 12, 2008 #6
    Very good post. Let me just add a few comments along those lines. Insulators are so good that we just take them for granted. To get a perfect conductor, we must super-cool certain types of materials, which is expensive and unwieldy. Until high-temperature superconductors become available and affordable, conductors will be much much very much lossier than insulators.

    We don't use the term "super insulator" because insulators are very good over a vast range of temperature. No super-cooling or heating is needed at all to get super-insulativity.

    With power transmission, there are two losses, namely I^2*R (conductor), and V^2*G (insulator). For affordable common materials used as conductors and insulators I^2*R greatly exceeds V^2*G. Hence there is a great advantage to using high V with low I as it minimizes losses.

    Also, I used to wonder why power generators are always operated at constant speed to output constant voltage. Why not turn the generator at a constant torque to get a constant current source? The answer is along the same line as above.

    With constant current source (CCS) operation, the load receives a constant current, and the voltage varies with resistance of the load. To shut off power, let's say we're turning off a lamp, a switch is placed across the lamp, in parallel, and CLOSED. The constant current is shunted and the lamp turns OFF. Opening the switch turns the lamp on.

    So, with CCS power distribution, when we shut off power, a constant current is transported from the power generator all the way to our homes only to be shunted by switches in parallel with the loads. This results in huge power losses. With CVS (constant voltage source) generators, we turn off power by opening a series switch. With CCS we incur I^2*R loss under no load. With CVS we only incur V^2*G. CVS incurs MUCH LOWER power losses than CCS.

    The same holds with batteries. All battery producers for many decades have focused their effort on CVS operation, NOT CCS. If we had CCS batteries in our flashlight, and we turn them off by shunting the current through a switch, we get I^2*R losses continuously. A week later, when you attempt to turn on the flashlight, your batteries are gone.

    Something to think about. BR.

    Claude
     
  8. Mar 12, 2008 #7
    Thank you berkeman. I was trying to figure out how to explain the collisions part and couldn't come up with a good analogy. I too like the water analogy and will happily watch the thread develop.

    Thanks again.
     
  9. Mar 13, 2008 #8
    Ok, but with lower voltage means more current, which means more collisions between the electrons and the lattice of atoms in the conducting wires.

    I think that in both cases the collision are not energetically equal, and the question is: which one contribute with less power lost to phonons in the lattice?

    Thanks!
     
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