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Working out power loss in power lines: Multiple wires

  1. Apr 8, 2015 #1
    Hey everyone,

    This is quite a simple question, I think I'm just getting confused.

    1. The problem statement, all variables and given/known data

    Given Data:
    Merlin: Resistance: 2 Ω / km, Maximum Current: 519
    Pelican: Resistance: 1.5Ω/km, Maximum Current: 646
    Penguin: Resistance: 4 Ω/km, Maximum Current: 357
    Sparrow: Resistance: 11Ω/km, Maximum Current: 184
    Swan: Resistance: 17Ω/km, Maximum Current: 140
    Note: Maximum current that will pass through the wire is 360 A
    State the maximum power loss per km for 10 possible combinations of wires.
    2. Relevant equations
    Ploss=I²R
    Total Resistance = 1/{(1/R1)+(1/R2)+(1/R3)..+(1/Rn)}
    3. The attempt at a solution
    I'm not sure how current is divided up depending on the combination of wires.
    First I tried to do 2x Merlin in parallel
    Current is the same in components in parallel so:
    Rtotal: [itex] (1/2 + 1/2)^-1 = 1 [/itex]
    Ploss= I²R
    [itex] = (360)^2 [/itex]
    = 129600
    Total Power loss; 129600x2= 259200 W
    This is the same as just figuring out Merlin x1. Would I divide the current by 2 if I did 2x Merlin in series?
    Also say I tried doing:
    1x swan, 1x sparrow, 1x penguin
    I have no idea what to do with currents. Add them together? Divide them up?


    Thanks for your help.



     
  2. jcsd
  3. Apr 9, 2015 #2

    BvU

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    Hello Roy, welcome to PF :smile: !

    For parallel wires, you know that the voltage is the same for each of the two. Uncle Ohm's law gives you the current ratios.

    For two Merlins in parallel that means each of the wires carries only 180 Amps ! A lot less max power loss, therefore.


    [edit] Ah, now I see: combinations in this exercise means parallel combimations !
     
  4. Apr 9, 2015 #3
    Oh! I thought current is the same in parallel rather than series, sorry my mistake.
    Not necessarily I think. It can be in series.
    So: Say I did 1x swan, 1x penguin in series.
    Total Resistance: 17+4 = 21 ohms
    Current: 360 A
    Power loss: I²R
    360 x 21
    = 7560 Watts
    Is this correct? I did not surpass the amount of maximum amount of current for each wire because together they can take a maximum current of 497. Is that correct? ie. if I have 2 wires in series with different max currents do I add them together to get the overall maximum current?
    Sorry for the bother!

    Thanks for the welcome ^_^

    Edit: This is all per km of course
     
  5. Apr 9, 2015 #4

    SammyS

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    From the above statement, if seems that you can answer your own question below.
     
  6. Apr 9, 2015 #5

    NascentOxygen

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    Swan maximum is 140A, this must not be exceeded
     
  7. Apr 9, 2015 #6
    oh? I thought that if I put it in series with penguin there will be a net maximum current of 140 + 357(penguin's max currrent) = 497 A

    So if current is the same in each point in the series circuit...the overall max current should be the sum of the two wires' max currents...I assume...
    I feel like I'm missing something here
     
  8. Apr 9, 2015 #7

    NascentOxygen

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    In a series circuit the same current passes through each, just as water pipes in series carry the same water flow.
     
  9. Apr 9, 2015 #8
    Okay, so the maximum current is different but the current flowing through each wire is the same. Gotcha. Okay say if I instead put Merlin in parallel with say pelican maybe I would not exceed the maximum current if current is split. How would I work out the current flowing through Merlin and in turn pelican? Extra information that probably will be helpful: 330, 000 V is supplied, may use up to 120 MW of power. These are the values I used to get the maximum current running through the wire. (I=P/V, 120 000 000/330 000= 363.6, simplified to 2 sig. figures; 360 A)
    Attempt:
    Current through Merlin: I = V/R
    =330, 000/2
    =165 000 A
    Obviously this is wrong. Maybe I can use another formula: I = √PR
    120 000 000 x 2 = 240 000 000
    √240 000 000= 15491.93338
    Obviously this is wrong as well. I'm just confusing myself even more now. This is something to do with the resistance since I didn't use resistance to work out the initial maximum current I think.
     
    Last edited: Apr 9, 2015
  10. Apr 9, 2015 #9

    NascentOxygen

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    Did you forget the 'squared' current? This would be the loss per 2km, so you'll probably express it as per km.

    Are you sure you can use Swan in series with Penguin? It won't meet the 360A specification, if I'm reading this correctly.
     
  11. Apr 9, 2015 #10

    NascentOxygen

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    The 300kV goes mostly to the distant load, whatever it is, and is not 'used up' by the resistance in these cables. So you don't use that figure to calculate a cable's power losses.

    When you have two cables in parallel, their voltage is common, so currents are
    I = I1 + I2 = V/R1 + V/R2

    and the current sharing can be expressed
    ∴ I1 / I2 = R2 / R1
     
  12. Apr 9, 2015 #11
    Sorry I forget to square it. So the answer would actually be 2721600 Watts. This is per 2km so I divide it by two; 1360800 W/km.

    Yes that's correct. I'm wrong.

    Wow thanks. So can I use the current divider rule: R1 current: Itotal * R2/(R1+R2)??
    So again: Merlin in parallel to pelican
    Current through Merlin:
    360 x 2 / (1.5+2)
    =154.3
    WHEW it works!! Doesn't overcome mass current
    SO: Total power loss
    I²R Total Resistance: [itex] (1/2 +1/1.5)^-1 [/itex]
    =6/7
    [itex] (360)^2 [/itex] x 6/7
    =111085.72
    For per km divide by 2:
    55542.86 W/km
    I hope that's correct.
     
    Last edited: Apr 9, 2015
  13. Apr 9, 2015 #12

    NascentOxygen

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    I doubt that final divde by 2. Probably leave it as per km of paralleled cable.
     
  14. Apr 9, 2015 #13
    Okay thank you so much.

    Edit: I have to put this on a graph cost per km (data is given) vs power loss per Km. Do I still leave it as power loss per km of paralleled cable or for the sake of comparison change it?
     
    Last edited: Apr 9, 2015
  15. Apr 10, 2015 #14

    NascentOxygen

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    I would leave it as per km of parallelled cable, but it's just a scaling factor so shouldn't complicate comparisons.
     
  16. Apr 11, 2015 #15
    Okay. Will do. Thanks!
     
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