Power Loss, Power Factor and Voltage Drop

Discussion in 'General Electronics Chat' started by ben22, Oct 8, 2014.

  1. ben22

    Thread Starter New Member

    Jul 10, 2013
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    I am having trouble grasping how these three concepts can be reconciled in a given circuit. I’ll use the following two examples with two different power factors to illustrate:

    Two-wire, single phase circuit; Load of 1,000 watts @ 100 volts, connected to load by wire that has a resistance of 0.2 ohms. The voltage of the generator can be adjusted to make sure that the load sees 100 V.

    I will calculate the generator voltage set point and the resulting power that the generator must produce when the load has a power factor of 1.0 and 0.5. Then I can calculate power loss in two different ways: using the formula PL = 2*r* I^2, and also subtracting the power seen by the load from the power required of the generator.

    For the voltage drop formula I’ll skip reactance to simplify, so VD = 2*r*I

    For current required by load I’ll use I = watts / (V*PF)

    When PF = 1.0, I = 10 amps

    VD = 2*10*0.2 = 4 volts

    PL = 2*0.2*100 = 40 watts

    So we set generator at 104 volts and generator power = 104*10 = 1,040 watts; load sees 100 V, so power at load is 100*10 = 1,000 watts; power loss is 1,040 – 1,000 = 40 watts, same as calculated above.


    When PF = 0.5, I = 20 amps

    VD = 2*20*.2 = 8 volts

    PL = 2*0.2*400 = 160 watts

    Keeping in mind that the generator generates watts not VA, we set generator at 108 volts and generator power = 108*10=1,080 watts; load sees 100 V so power at load is still 1,000 watts; power loss is 1,080 – 1,000 = 80 watts, which is not equal to the 160 watts calculated above (its actually 160*PF). Where happened to the extra 80 watts of power loss??? Does the generator somehow not have to produce this power?

    Can someone please help clarify where the inconsistency / problem is with the reasoning here?

    Thank you!
     
  2. studiot

    AAC Fanatic!

    Nov 9, 2007
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    What sort of load do you have that seems to draw a constant 1000 watts, regardless of powerfactor, and if the load is drawing a constant wattage, how does the power factor change?

    Why are you calculating the voltage drop as 2IR ?

    Edit OK I see you are allowing for the double cable run (2wire) in the resistance.
     
    Last edited: Oct 8, 2014
  3. ben22

    Thread Starter New Member

    Jul 10, 2013
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    Its not a real load, its more for the thought experiment. What is the problem with the voltage drop formula? each length of wire has resistance r, so since there are two lengths of wire the voltage drop is 2*I*r...?
     
  4. ben22

    Thread Starter New Member

    Jul 10, 2013
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    lets say the load in example one is a bank of 10 100-watt incandescent bulbs, in example 2 its a bank of 20 50-watt CFLs
     
  5. crutschow

    Expert

    Mar 14, 2008
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    You generator power calculation is not correct. A generator can indeed generate VA (consisting of both real and reactive power). You stated the current is 20A so the generator has to output 20A. It's output power is then the 100V * 10A through the load plus 8V *20A loss in the line. The other 10A is reactive current going through the parallel reactance of the load. The generator output power can be calculated from the voltage, current, and power factor at the generator. (I don't know if the PF of 0.5 you stated is for the load or at the generator output, the difference being the line resistance factor).
     
  6. ben22

    Thread Starter New Member

    Jul 10, 2013
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    The PF of 0.5 is the load PF. Will the line resistance cause there to be a different PF at the generator? Does this account for the different in power loss?

    So then the generator output is 108 x 20 x 0.5 = 1,080 w ...?
     
  7. crutschow

    Expert

    Mar 14, 2008
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    It will change the power factor since it changes the ratio of real to reactive power. Don't know if that's sufficient to account for the difference since I haven't cranked through the math but it should (I'm somewhat allergic to hand math ;)).
    That's not the correct power factor as noted in the previous paragraph.
     
  8. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Suppose (purely for convenience) the load voltage phase angle is 0 deg.
    If the load power factor is lagging then the load current is

    20 \angle{-60^o} \ A

    The conductor voltage drop will therefore be

    8  \angle{-60^o} \ V

    meaning the generator voltage will be

    100 \angle{0^o}+8 \angle{-60^o} \ V =104.231 \angle{-3.8113^o} \ V

    The phase difference between generator voltage and current is therefore 56.1887 deg giving the generator pf as cos(56.1887) lagging.

    The generator power is therefore

    104.231*20*cos(56.1887) or 1160W.
     
    Last edited: Oct 8, 2014
  9. ben22

    Thread Starter New Member

    Jul 10, 2013
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    Ok thanks t n k for doing that math. Let me see if I can extend it to see if doing it this way accounts for the full 160 watts of power loss given by PL=2*r*I^2

    if the load is PF 0.5, so draws 20 amps @-60 deg (lagging as you assumed) with a voltage of 100 V @0 deg, then the Z for the load = 5 ohms @60 deg. Z for the wire should be 0.4 ohms @0 deg, adding these two resistances together then gives a total Z of the circuit of 5.2115 ohms @56.19 deg. this gives a current of 19.1883 amps @-56.19 deg at the generator, which gives a generator power factor of 0.5564.

    So 0.5564 * 104.231*19.1883 = 1,113...closer but still not at 1,160...
     
  10. crutschow

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    There is an error somewhere. The generator current has to be the same as the load current or 20A, not 19.1883A.
     
  11. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Yes the error occurred when ben22 divided 100V by the total impedance to give a current of 19.1883A. The generator voltage is applied to the total impedance and the generator voltage, as shown earlier, is not 100V.

    I'd make the point here to ben22 that he seems to be trying to show there is an anomaly which defies the laws of nature. The reality is that if there appears not to be conservation of energy in the system under consideration, then the anomaly always boils down either to mathematical errors or incorrect assumptions on our part as we do the analysis. Ensuring no such anomaly exists at the completion of our calculations is in fact a very useful check on our working.
     
    Last edited: Oct 9, 2014
  12. ben22

    Thread Starter New Member

    Jul 10, 2013
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    Well...that is quite simply and clearly put. Thank you.

    Thank you as well for your help crutschow.
     
  13. ben22

    Thread Starter New Member

    Jul 10, 2013
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    Yes indeed that is why I posted this - to figure out where I went wrong.

    I suppose to have a fully accurate example I should, among other things, include the effect on current draw of the resistance of the wire as well, which would lower total current draw of the circuit.

    So if I am trying to figure out how much voltage I need out of a generator to serve a bunch of loads in a very "branchy" circuit, all with the same power factor, I guess I will then need to add the complex form of the voltage drops together to arrive at the actual complex voltage at the generator, giving me the set point I need...
     
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