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Homework Help: High voltage distribution to reduce power loss

  1. Mar 11, 2013 #1
    1. The problem statement, all variables and given/known data
    In Australia, New South Wale's final year of high school, a common question is asking something like:

    How does distributing electricity at high voltages reduce power loss in the transmission lines.

    2. Relevant equations

    P = VI = (I^2)R = (V^2)/R. Not very clear without subscripts.

    3. The attempt at a solution

    I attached an image http://imageshack.us/f/16/image0001lmg.jpg/
    R_wire is the resistance of the transmission lines
    R_load is the load

    All the textbooks and teachers will teach it as shown in Case1 which essentially says:

    P_input = V_supply * I = constant
    If we distribute at high V_supply, current I must fall (since P_input is constant).
    Since current I falls, P_loss = I^2 * R_wire must also fall.

    But what about Case 2? Using P_loss = V_wire^2/R_wire seems to end up with a contradictory result.
  2. jcsd
  3. Mar 11, 2013 #2
    You are not modelling this correctly. You neglect the fact that voltage (and current) are transformed at the consumer.

    The voltage equation: Vs = Vw + Vc, where Vw is the voltage drop in the wire, and Vc is the voltage drop at the high-voltage cascade of the transformer at the consumer.

    The resistance of the consumer (in the low-voltage cascade) is R, so the current in the low-voltage cascade is I = U/R; consequently, the current in the high-voltage cascade is J = I/k = (U/R)/k.

    On the other hand, Vc = kU, so the "effective" resistance of the consumer, as seen in the high-voltage cascade, is Rc = Vc/J = kU/((U/R)/k) = Rk^2.
  4. Mar 11, 2013 #3
    So you're saying that the value of the load impedance is wrong because I didn't perform a impedance transformation from the secondary (low voltage) coil to the primary? R_load in the diagram is an arbitrary number so it doesn't matter if it's been scaled by the constant k^2 right? (k = turns in primary/turns in secondary). So I'm assuming you're saying the circuit layout is completely incorrect? What would a more correct model be?

    And before we continue to dive too deeply into this, wouldn't the contradiction I mentioned in my diagram still exist if the question simply: gave you that circuit diagram, stated that the power supplied by the voltage supply is constant and asked whether P_wire would increase or decrease if you increased V_supply.
  5. Mar 11, 2013 #4
    R_load (which I denoted simply as R) is arbitrary, but the power loss in the wire is governed by the effective resistance, which is Rk^2.

    The contradiction is resolved if you start with the consumer voltage and regard that as a given; then you should find that the losses decrease as k increases.
  6. Mar 14, 2013 #5
    So since no one else commented, does everyone agree with voko's explanation? From what I understand, voko is saying:
    1) Take consumer voltage as a constant value with their load R
    2) High voltage distribution side of it will see the consumer load as Rk^2
    3) If k increases, Rk^2 increases so total resistance increases -> less current -> less power loss
    4) k increasing means N_p/N_s increases and since we took consumer voltage to be constant, increasing k means increasing distribution voltage

    I can't really find a way to dispute that.

    I decided to sub some values into the 2 cases I drew in the diagram and realised that the circuit model is inconsistent with the constraint of keeping delivered power constant.

    eg. Take V_s = 5V, R_w = 1Ω, R_l = 4Ω. Apply Ohm's law and you'll find I = 1A
    So power delivered by power supply, P_s = V_s * I = 5W

    Now try it again with a higher supply voltage.
    Take V_s = 10V and keep the other resistances constant.
    If we want to keep power delivered constant, I = P_s/V_s = 5/10 = 0.5A
    However, if we apply Ohm's law to the circuit, we find I = V_s/(R_w + R_l) = 10/5 = 2A.

    So out of curiosity, can anyone supply a simplified circuit model of what the distribution system looks like that would describe the transmission losses.
  7. Mar 15, 2013 #6
    The simplified circuit model should use the effective resistance of the load, which is Rk^2. In the second case, k is close to 2, so it should be 16 Ω, not 4 Ω. This gives ~0.6 A, which is close to 0.5 A you got by considering the power.
  8. Mar 15, 2013 #7


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    Staff: Mentor

    PF allows you to create super- and sub- scripts, e.g., P = VI = (I[sup]2[/sup])R = (V[sup]2[/sup]))/R.
    Correct, because I is the current in the transmission cable.
    http://img441.imageshack.us/img441/3331/nooo1.gif [Broken]
    The voltage drop across the transmission cable is not the V in the formula P_input = V_supply * I = constant
    Last edited by a moderator: May 6, 2017
  9. Mar 15, 2013 #8
    I was wondering if there was a better model where the all the simple ohmic equations and power equations will be entirely consistent and not just close.

    lol, I subscripted V_supply and V_wire to highlight that I was aware of the difference between the two. Though it's understandable that you skimmed the post since this topic may have been brought up countless times before.
  10. Mar 15, 2013 #9
    Yes, you can get them to agree exactly. But first you have to agree on what you regard as a given. One possible set: the voltage in the consumer cascade; the consumer's resistance; the transformation factor; the wire resistance. Another possible set: the voltage in the consumer cascade; the consumer's resistance; the source voltage; the wire resistance. Yet another: the source voltage; the consumer's resistance; the transformation factor; the wire resistance.

    Then all it takes is a bit of algebra.
  11. Mar 15, 2013 #10


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    Staff: Mentor

    I didn't fail to notice that. It was the fact that, despite your recognition of the difference, you still wrote of an apparent contradiction.

    Perhaps I could have more succinctly answered your OP this way: There is no contradiction, because the total supply voltage is not dropped across the transmission line! :smile:
  12. Mar 15, 2013 #11
    Did you look at the image I linked in the OP? I wrote:

    Vwire = Vsupply (Rwire/(Rwire + Rload) )

    implying that the voltage drop across the wire was directly proportional to the total voltage supply
  13. Mar 15, 2013 #12
    "But first you have to agree on what you regard as a given."
    What do you mean by this? By given, do you mean a fixed value has been assigned to it? If yes, then the paragraph below has relevance. If not, then I'm not sure what's happening anymore.

    So in the first set, are you implying that we're treating the source voltage as a variable? However with all the other parameters being "given"/set, the source voltage can be calculated directly from the other parameters specified so it's not actually a variable. So the set will describe the entire system with fixed numbers leaving no room for arguments like "if we increase the supply voltage, ... will happen" because we can't increase the supply voltage without changing the other values which happen to be "given".

    Now I'm starting to confuse myself. Is anyone kind enough to make a quick sketch on Paint or something to explain what's happening?
  14. Mar 15, 2013 #13
    This could have been a poor choice of words. What I meant by "given" is free parameters that we may set or vary, and the values of the other parameters are then deduced.

    In the first set, we could analyze the source power as we change, for example, k, which has a direct bearing on your original question.
  15. Mar 15, 2013 #14


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    Staff: Mentor

    I had not noticed that, but I'll develop that idea ....

    Let's increase the supply voltage by a factor , but keep the transmission line fixed at R_wire.

    If the load is a resistance R_load, it now becomes 2R_load to account for the same power (the voltage now being V and the current I/)

    With the original supply voltage, V_supply,
    V_wire/V_supply = R_wire/(R_wire + R_load)

    and this is approximately,
    V_wire/V_supply R_wire/R_load

    Now, determine the same fraction with the higher voltage, V_supply, and a load 2R_load


    Remember, V_supply is a constant.
    Comparing the above fractions, we see with the higher operating voltage the voltage lost across the cable is lower by a factor .....

    (I'll leave you to fill in the dots.)
    Last edited: Mar 15, 2013
  16. Mar 15, 2013 #15
    You lost me at this step. Wasn't the constraint to keep the power delivered by source constant? So I would agree with this bit: "the voltage now being V and the current I/"

    But saying: "If the load is a resistance R_load, it now becomes 2R_load to account for the same power" means you're trying to keep the power consumed by the consumers constant as well? I'm not sure where the reasoning for this comes from.

    I mean if P_source = P_loss + P_load and you forcibly changed the R_load to keep P_load the same, that would mean P_loss stays the same (P_source is still constant by constraint)?

    Apart from my lack of understanding of what happened there, I see that you're trying to say that with supply voltage going up by factor of , the voltage drop across wire goes down by ≈.

    I'm still thinking about this. Though I don't think I'm getting anywhere. Here's what I have so far: http://imageshack.us/f/109/image0002zr.jpg/
  17. Mar 15, 2013 #16
    V: source voltage (unknown)
    I: current (unknown)
    P: power at the source (unknown)
    Q: useful power
    R: load resistance ("given")
    Rw: wire resistance ("given")
    U: load's (low) voltage ("given")
    k: transformation factor ("given")

    I = (U/R)/k

    V = IRw + kU = Rw(U/R)/k + kU = U(k + Rw/(kR))

    P = VI = U(k + Rw/(kR))(U/R)/k = U^2(1/R + Rw/(kR)^2)

    Q = kUI = U^2/R

    => P = Q + U^2 Rw/(kR)^2

    The second term, U^2 Rw/(kR)^2, is the power wasted.
  18. Mar 15, 2013 #17
    Thanks for the reply. Let me just check my understanding

    So U^2 Rw/(kR)^2 shows that as k increases, the power wasted decreases.

    V = U(k + Rw/(kR)) shows that as k increases from 0, V will first decrease to a stationary point and then increase. The k used in practice is large so it's past the stationary point and will therefore V will increase with increasing k.

    P = U^2(1/R + Rw/(kR)^2) shows that the power provided by source will decrease with increasing k. So I've been going about this the wrong way from the start by trying to tackle it from the angle that the power provided by the source was constant? Q = kUI = U^2/R shows that Q is kept constant instead since U and R are given.

    Would you still agree with the textbook answer given in the OP?
    P_input = V_supply * I = constant
    If we distribute at high V_supply, current I must fall (since P_input is constant).
    Since current I falls, P_loss = I^2 * R_wire must also fall.
  19. Mar 16, 2013 #18
    Don't forget, this is an idealization. This is what we would like to have. In practice, neither R nor U are constant. But for the sake of the analysis we have to make certain assumptions.

    That looks OK. P_input is the sum of the used and wasted power. By increasing the source voltage, we decrease the waste.
  20. Mar 17, 2013 #19


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    Staff: Mentor

    It wasn't, as far as I'm aware, though you could make it so if you choose. A perfectly realistic scenario is: "To deliver X MVA to an industrial plant (or town), determine the losses to be accounted for in the transmission system if ...."
    That is what I illustrated, to show there is no contradiction where you originally claimed there may be one.
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