Unbalanced 3-phase Y connection

Discussion in 'Homework Help' started by Jess_88, Aug 5, 2011.

  1. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
    1
    Hey guys :)

    I have been writing up my lab report and I'v run into a sang.

    I constructed and analysed a three-phase Y connection similar to the one shown below. Only with resistance values
    Red line = 75 + j345.58
    White line = 85 + j345.58
    Blue line = 95 + j345.58

    [​IMG]

    My theoretical calculated values are (by mesh analysis)
    IL(red) = Ip(red) = 0.681<-76.84 A
    IL(blue) = Ip(white) = 0.676<44.52 A
    IL(whie) = Ip(blue) = 0.664<163.47 A

    where Ip = phase current and IL = line current
    Vp(red) = 240<0
    Vp(white) = 240<-120
    Vp(blue) = 240<120

    Sp(red) = 37.21 + j159.15 VA
    Sp(blue) = 40.68 + j157.06 VA
    Sp(white) = 37.12 + j154.9 VA

    Now I know how to calculate the Volt Amp reactance and power for each line but I think I need to determine the total power, total volt amps and total volt amp reactance. Can I do this??? how would I go about doing it?

    thanks guys :)
     
    Last edited: Aug 5, 2011
  2. subtech

    Senior Member

    Nov 21, 2006
    123
    4
    Let me make sure I'm clear on something before I proceed.

    I can see that the load is connected wye, but did you intend for the source to be connected Delta?
     
    Last edited: Aug 5, 2011
  3. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
    1
    ah sorry.
    I'm refering to the "Y" load connection.
    Is that ok???
     
  4. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
    1
    I'm pretty sure the source was delta for my laboratory.
    I live in australia... is this maybe the cause for some confusion?
    or is it my calculations maybe?
     
  5. subtech

    Senior Member

    Nov 21, 2006
    123
    4
    I am probably the one who is confused. It has been ages since I've used the maths...

    In answer to your question, I would treat each line as a single phase source.
    Use your calcs to establish VA for each. Sum the three lines for a total VA.
    Determine the Watt and VAr component from the angles and VA for each line.
    Again, sum the three Watt components for the 3 phase total, and the do the same for the VAr.

    I'm hesitant to say more at this point because my quantities are somewhat different than yours. However, that is very likely MY mistake and I am trying to find where I've went wrong.
    I'm not familiar with mesh analysis, but I'm going to investigate it.
    My approach to this problem is more from the "old school" divide and simplify method.

    I hope that I've not confused you.
     
    Jess_88 likes this.
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    You need mesh analysis since the load star point isn't the same as a system with a neutral connected. That is it can't be treated as three individual phases in isolation.
     
    Jess_88 likes this.
  7. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
    1
    ah ok.
    Yeah I was thinking I should do it that way to, but my calculated values have been different from my measurements. Maybe I made an error.
    Here is my working

    (75 + j345.58)I1 + (85 + j345.58)(I1 - I2) = 415<30
    (160 + j691.16)I1 - (85 + j345.58)I2 = 359.4 + j207.5

    (95 + j345.58)I2 + (85 + j345.58)(I2 - I1) = 415<270
    -(85 + j345.58)I1 + (180 + j691.16)I2 = -j415

    Matrix
    [160 + j641 -85 - j345.58][I1] = [359.4 + j207.5]
    [-85 - j345.58 180 + j691.16][I2] [ 0 - j415 ]

    delta = 336701.61 + j176245.8
    delta1 = 64692 + j250477.904
    delta2 = 245672.55 + j75438.952

    I1 = delta1/delta = 0.155 - j0.663 = 0.68<-76.84
    I2 = delta2/delta = -0.482 - j0.474 = 0.676<-44.52

    I1 = I(red line)
    I2 = I(blue line)
    I(red line) = I2 - I1 = - 0.637 + j0.189 = 0.664<163.47

    Hows that look?
     
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