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Question on power lost?

Discussion in 'Electronic Basics' started by henry chow, Nov 14, 2005.

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  1. henry chow

    henry chow Guest

    I have a simple question. The way the power company delivers power from the
    power plant to the homes is using stepped-up transformers and high voltage
    lines. The reason is that they want to minimize the power loss which is
    P=I * I * R. But P can also be equal to P=V*I. If you increase the
    voltage, why don't you have the same power loss which equals to P=I*I*R=V*I?
  2. John Fields

    John Fields Guest

    Because for the same line resistance, if V increases, I will go

    Say the power company was sending 120V through one mile of of #0 AWG
    into a 1000 watt load. Then the circuit would look like this:

    +-------+ <--8.333A-->
    | 120AC|-----[0.519R]---<<--+
    | | |
    | | [14R4]
    | | |
    | 120AC|-----[0.519R]---<<--+
    +-------+ <--8.333A-->

    Since the power company is putting out 120VRMS and the lines look
    like 0.519 ohm each, the total circuit resistance is:

    Rt = 0.516R + 14R4 + 0.516R = 15.432R

    and the current in the circuit is:

    E 120V
    It = --- = -------- ~ 7.78 amperes.
    R 15.43R

    And the power lost in each the lines would be:

    P = I²R = 7.78A² * 0.519R ~ 31.4 watts.

    For both lines, that would come to ~ 63 watts.

    Now, assuming that the power company has a 1:10 step-up transformer
    at the plant and a 10:1 step-down transformer at the load, the
    circuit will look like this:

    +-----+ +--------+ <--0.833A--> +--------+
    |120AC|--|120 1200|----[0.519R]----|1200 120|---<<--+
    | | | | | | |
    | | | | | | [14R4]
    | | | | | | |
    |120AC|--|120 1200|---[0.519R]-----|1200 120|---<<--+
    +-----+ +--------+- <--0.833A--> +--------+

    Assuming for a moment that the transformers are lossless, the
    current into our 1000 watt load, to a first approximation, will be:

    E 120V
    I = --- = ------ ~ 8.33 amperes.
    R 14R4

    Since the transformers are transforming power, if we increase the
    voltage out of the secondary to ten times that across the primary,
    then the current which the secondary can supply will be one tenth of
    that flowing in the primary. In this case, for a 1000 watt load,

    Now, since the line resistance hasn't changed, the power which the
    lines will dissipate in moving 1000 watts over ten miles of cable
    will be:

    P = I²R = 0.833A² * 1.038R = 0.72 watts,

    about an 88-fold reduction in power lost in the lines!

    There will, of course, be some power lost in the transformers, but I
    believe the above answers your basic question.
  3. John Fields

    John Fields Guest

    Oops... two
  4. John  Larkin

    John Larkin Guest

    The V in the equation is not the line voltage, it is the line voltage
    *loss*. That's equal to V(at generator) - V(at load). And that
    quantity goes down when you step up to high-voltage, low-current

  5. Power lost in the transmission lines is the current squared times the line
    resistance (called I squared R). Drop the current (by boosting the voltage)
    and you can eliminate much of the power lost in the transmission lines. Up
    the voltage so you can deliver the same power, at less current. 500 kV, or
    there abouts, is used for long-distance transmission.

    Some numbers: 120 volts at 100 amperes = 12 kW and 500,000 volts at 24 mA =
    12 kW. So, suppose it is your choice: send energy to your customers at 100
    amperes or 24 mA. It is an easy choice as you can also use smaller gage
    wires (beside saving the heat loss in your transmission lines). By the way,
    this is the basic point that Edison missed and he went so far as to publicly
    electrocute animals with ac to "prove" that ac was too dangerous. Tesla was
    way ahead of Edison on this particular point.

    Some loss numbers: Assuming 100 ohms of transmission line resistance with
    100 amperes = 1 MW! With 24 ma = 0.06 W!
  6. PeteS

    PeteS Guest

    Perhaps the OP is confused about the P = V*I part.

    Power is developed as V(ckt) * I(ckt). On a power line, the V(ckt) is
    quite small (actually derived from V = I*R). So on a power line, think
    of the voltage **across the cable** as being a result of the current,
    rather than the other way around. Certainly for any power line, I^2 R
    will equal V * I *for the power line*

    Looked at from that perspective, then by raising the transmission
    voltage, we lower the current (P remains roughly constant) so the
    voltage drop across the line decreases, thus reducing V * I (by a
    square function) at the same proportion as I^2 R.


  7. Bill Bowden

    Bill Bowden Guest

    The "V" in your equation is the voltage lost in the lines, not the line
    voltage. As you increase transmission voltage, both V and I will go
    down. In the limiting case, both V and I go to zero and there are no

  8. Jasen Betts

    Jasen Betts Guest

    you do, but the significant voltage in this case (for calculating the loss)
    is the difference between the voltage at each end of the transmission
    line not the difference between the line and ground.
  9. Rich Grise

    Rich Grise Guest

    Yes, it's the voltage drop from one end of the wire to the other, which
    will be constant, based on the resistivity of the wire and its length.
    By using a higher primary voltage, that portion dropped by the wires
    themselves is a smaller proportion of the whole loss set, so they save

  10. You will? ;-P ;-D ;-P
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