Line Loss calculation

Discussion in 'General Electronics Chat' started by EEMajor, Sep 27, 2006.

  1. EEMajor

    Thread Starter Well-Known Member

    Aug 9, 2006
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    4
    Alright, I should be able to do this with no problem, but something about this particular question has me stumped.

    A 14KW load is supplied with wires that have a TOTAL resistance of .4ohms. The source voltage is 120VAC. How much power is lossed in the wires alone? The answer is somewhere between 3000-8000 watts.

    I have tried calculating the total current by (120V)/(.4ohms) = 300A, then taking that to find the total power, since we know that in a series circuit, the current is the same everywhere. So, Total Power = (300A)(120V) = 36000 watts. Then I subtracted 14000watts from that for 22000wats, but that's not right.

    Anyway, I have tried a few other things, but I am stumped on this simple question. I think I'm thinking too hard!

    Thanks for the help!

    Here is a schematic:
     
  2. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    What ohmic value would you give the 14 kW load, rated at 120 V?

    Then it becomes a simple series circuit.
     
  3. EEMajor

    Thread Starter Well-Known Member

    Aug 9, 2006
    67
    4
    Well, thats part of the deal, that value is not given. So, I would assume that we could say since the current is 300 amps (Given that 120V/.4ohms = 300A) that we could say the resistance of R2 is (14000W/300Asquared = .155555ohms). Not sure if I'm thinking this through correctly or not, but thats what I think it should be.

    So that would give us a total resistance of .5555556 ohms. If we do a check on that, (300A x .5555556ohms) we get around 166V, SO SOMETHING IS WRONG with my line of thought.....

    Thanks for the help!
     
  4. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    Forget about the 120V/0.4 ohms. You don't have the total resistance for the series circuit ... YET.

    What is the resistance of a load that consumes 14kW at 120 Volts?
     
  5. EEMajor

    Thread Starter Well-Known Member

    Aug 9, 2006
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    4
    Okay, thats E^2/P = 120^2/14000 ~ 1.03ohms. + the .4 ohms = 1.43ohms total for the circuit.
     
  6. nomurphy

    AAC Fanatic!

    Aug 8, 2005
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    The above calculation is incorrect, there isn't 120V across the 14KW load. Because of the R1/R2 divider, the voltage at R2 is LESS than 120V.

    This is a series circuit, therefore current is constant and assuming it's a purely resistive load, the equation would be more like:

    I = 120V / [.4 + (14KW / I^2)]

    If you can solve this in terms of "I", you may be able to reach an answer.
     
  7. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    The total resistance of the circuit is the characteristic resistance of the 14kW load plus the line loss.

    That makes his total resistance calculation correct and then his power loss in the lines will be correct.

    Naturally, the 14kW load will not draw 14kW because of the line loss.

    What figure did you get for the power loss in the line?
     
  8. richbrune

    Senior Member

    Oct 28, 2005
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    I think Joe Jester and nomurphy are both correct in some respects. In order to acheive 14000 watts from the circuit, you must either lower the "line" resistance to some number less than .256 ohms, or raise the voltage at the mains to above 120 volts. I'ts a quadradic equation that cannot be solved with the constants given. Anyone help me here?
    Rich
     
  9. Jon Borys

    New Member

    Aug 12, 2006
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    You can't determine the exact value with out measuring the current or the resistance of the load.

    A good approximation with the values given:
    14 Kw Load
    120 VAC Nomimal AC voltage
    .4 Ohm Round trip wire resistance

    14 Kw / 120 V = 11.67 A

    11.67 ^ 2 * .4 = 54.47 Watts
     
  10. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    Jon,

    You may want to revisit your calculations.
     
  11. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    Rich,

    The point of this problem is to point out that line losses don't "add" to the power consumption. The 14kW load, whose electrical characteristic of R [or Z] did not change with the introduction of the 400 milliohm line loss.

    Attached is a graphic of a 14kW load without
    and with
    a line loss.

    The 400 milliohm line loss, represented by R1 in the circuit, reduces the current applied to the box titled 14kW load. As we know with E constant and R increasing, I must decrease. Otherwise we would be discussing Ohm's theory and not Ohm's Law.

    I had this type of a problem, although line losses were very low on the suspect list since it was a new installation when testing my 300 kW three phase generators. The line currents were not within my calculated limits. After verifying the integrity of the connections at the various points, we went to verify the load characteristics. It turned out the 300kW load was constructed with a lot of parallel combinations, each group fused with about a 50 amp fuse. I had a few blown fuses causing the error. Had the load tested good, the next step in the troubleshooting process would be to measure the IR drop at each step of the path from Generator to Load.​
     
  12. Jon Borys

    New Member

    Aug 12, 2006
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    Ok I see the problem.

    14kw /120 V = 116.7 A 116.7 ^ 2 * .4 = 5544 W, 116.7 * .4 = 46.68 V drop across line. Which is too much to deliver 14kw so the approximation won't work and you won't be able to deliver the desired 14kw to the load.

    120 V ^2 /14kw = 1.028 Ohms = load resistance

    120 V / (1.028 + .4) = 84 A 84 A ^ 2 * .4 = 2824 Watts.
     
  13. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    Good job Jon.

    I see you found it while I was typing a response explaining what you had just found.
     
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