Calculating neutral currents with varying power factors

Discussion in 'Homework Help' started by smithbrian, Sep 3, 2012.

  1. smithbrian

    Thread Starter New Member

    Sep 3, 2012
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    Hello I am having trouble calculating the neutral currents for the following system
    Phase A = 10A @ PF of 1
    Phase B = 9A @ PF of 0.9
    Phase C = 8A @ PF of 0.8
    The answer is meant to be 4.55A but I get 1.73

    I used the equation from the following website.
    http://www.electrician2.com/electa1/electa4htm.htm
    Here they seem to not take power factor into account there can someone please point out to me how to calculate this taking power factor into account.
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    If all pf's are lagging then

    Phase A current

    I_A=10\angle{0^o} \ Amps

    Phase B current

    I_B=9\angle(-120^o-{\arccos(0.9)}^o) \ Amps

    Phase C current

    I_C=8\angle(-240^o-{\arccos(0.8)}^o) \ Amps

    Then

    I_N=I_A+I_B+I_C \ Amps
     
  3. smithbrian

    Thread Starter New Member

    Sep 3, 2012
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    tnk Thank you for your answer.

    But I still don't get the right answer. I don't understand some of your symbols I am guessing you meant.

    Ia = 10* cos (0)
    Ib = 9* cos (-120 - 25.84)
    Ic = 8* cos (-240 - 36.87)

    If I do the above I get Ia=10, Ib=-7.45 and Ic= 0.957

    In=3.507A (not 4.55A which is the correct answer)

    Can you please explain further
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Hi there - remember you have to think in terms of complex quantities.

    So

    I_A=10+j0 \ Amps

    I_B=9 \angle (-145.84^o)=-7.447-j5.054 \ Amps

    I_C=8 \angle (-276.87^o)=0.957+j7.942 \ Amps

    And finally

    I_N=I_A+I_B+I_C=3.51+j2.89 \ Amps=I_x \angle(\theta^o) \ Amps

    Where Ix is your unknown.
     
  5. mlog

    Member

    Feb 11, 2012
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    36
    You're only finding the real components. You can't ignore the imaginary components. The answer you want is the vector magnitude of the real and imaginary parts. In other words, take the square root of the sum of the squares.
     
  6. smithbrian

    Thread Starter New Member

    Sep 3, 2012
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    Yes I understand it now. I realised that I was making that mistake.Thank you guys for your help.
     
  7. smithbrian

    Thread Starter New Member

    Sep 3, 2012
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    I repeated the same question but this assuming the sequence to be 0,120,-120. But I get a different answer.
    Ia = 10* cos(0)
    = 10 + J0 A
    Ib= 9* cos(120-25.84)
    = -0.65 + J8.98 A
    Ic= 8*(-120-36.87)
    = -7.36 -J3.14 A
    In = Ia + Ib +Ic
    = 1.99 +J5.84
    =6.17A
    I don't get it why is the answer different?
    Shouldn't it be the same answer?
    Can someone please explain.
     
  8. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Once again it boils down to correctly tracking phase rotation and power factor angles.

    If you make A phase the reference [0°], B at 120° and C at -120° I would assume a clockwise rotation sequence A-B-C-A-B ....etc.

    With lagging power factors for the non-unity cases [B & C phases] then the three phasors would be ..

    I_A=10\angle{0^o} \ Amps \\ I_B=9\angle{145.84^o} \ Amps \\ I_C=8\angle{-83.13^o} \ Amps

    Adding these three gives the correct answer.
     
  9. smithbrian

    Thread Starter New Member

    Sep 3, 2012
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    Well why did you do the following can you please explain.

    Ib= 9* cos(120+25.84)

    and

    Ic= 8* cos(-120 + 36.87)

    when previously you said

    Ib = 9* cos (-120 - 25.84)
    Ic = 8* cos (-240 - 36.87)

    my question is why in one instance we are subtracting and in one instance we are adding?

    Can someone please explain.
     
  10. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    782
    In that earlier case you quote I was using the phase rotation convention I normally adopt - a counter-clockwise convention. The A phase is reference at 0°. The B phase lags the A phase by 120° which I denote as -120°. The C phase lags the A phase by 240° which I denote as -240°. The current must therefore either lead or lag those values depending upon whether the power factor is leading or lagging. For a counter-clockwise convention a lagging current of θ degrees on the B phase would be interpreted as a phasor angle of -120°-θ°.

    Your proposed convention differs to mine so the angles would differ.

    If you were taught how to draw phasor diagrams then this all becomes second nature.
     
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