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5th and 7th harmonic cancellation with phase shift

  1. Dec 2, 2016 #1
    Hi,

    I replied to this thread regarding a question asked by @BjornFanden , and I was asking if someone could verify my calculations so we do not cause any confusion. I think it is better to start a new thread for me, if that is a bad move please remove this thread again.

    It was asked if full cancellation of the 5th and 7th harmonic were possible when operating two identical non-linear loads in parallel were one load is connected in D-d and the other D-y, introducing a 30 deg phase shift.

    30 degree by itself is not enough for shifting either 5th or 7th harmonic 180 degrees, but because of the delta-wye configuration it seem to be possible and this is also stated in litterateur and internet pages (I do not have any references available at the moment), but none of them shows why complete cancellation is happening. Therefore I hope someone can verify my calculations below.
    test4.JPG
    So here are the calculations, demonstrating full cancellation:
    In the delta-delta the currents are the same at primary and secondary.
    [itex]
    a1 = x1 = sin(5 \omega t)\\b1 = y1 = sin(5 \omega t - 120^{\circ})\\c1 = z1 = sin(5 \omega t + 120^{\circ})
    [/itex]

    for the delta-wye we have on the secondary side:
    [itex]
    x2 = sin(5 \omega t+30^{\circ}\times 5) = sin(5 \omega t+150^{\circ})\\
    y2 = sin(5 \omega t - 120^{\circ}\times 5 +30^{\circ}\times 5) = sin(5 \omega t -90^{\circ})\\
    z2= sin(5 \omega t + 120^{\circ}\times 5 +30^{\circ}\times 5) = sin(5 \omega t +30^{\circ})
    [/itex]

    Going from secondary to primary we have to include the [itex]\sqrt3:1[/itex] ratio and thus:
    [itex]
    A2= (x2-z2)/\sqrt3 = (sin(5\omega t+150^{\circ})-sin(5\omega t +30^{\circ}))/\sqrt3 = -sin(5 \omega t)
    [/itex]

    test1.JPG
     
  2. jcsd
  3. Dec 6, 2016 #2
    I have to apologize for the delay. As usually I am late.

    Your approach is very interesting, indeed. However, I think there are a few corrections to state.

    First of all the actual formulae are:

    In the delta-delta the currents are the same at primary and secondary but has to be as it follows:

    sin(5*w*t)

    sin(5*(w*t-120))

    sin(5*(w*t-240))

    That means you have to multiply -120 and -240 by 5[or 7] also.

    For the delta-wye we have on the secondary side also:

    sin(5*(w*t+30))

    sin(5*(w*t+30-120))

    sin(5*(w*t+30-240))

    Remark: the phase shift it could be also -30.It depends on transformer connection type.

    Second:

    A2 is not (ia−ib)/√3 but =(ia+ib)/w2*w1*sqrt(3) where w1 is primary no.of turns and w2 is secondary no. of turns.

    The phase shift has to be so then for a certain harmonic this sum has to be

    be zero. This condition is the same for each phase.

    Imot*SIN(hn*(w*t)=-Imot*SIN(hn*(w*t-deff)) where hn=5 or 7 Imot it is motor current [rated or loaded] the same for both motors ,w=2*pi()*freq. freq=50 or 60 c/s and deff it is the phase shift.

    The result sum is zero for 180/5=36 degrees for harmonic 5 and 180/7=25.71 for harmonic 7.

    For 30 degrees will be remain 48.3% for 7 and 39.5% for 5.
     
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