Delta Power System

Discussion in 'Homework Help' started by camino, Dec 9, 2009.

  1. camino

    Thread Starter New Member

    Sep 17, 2009
    8
    0
    [​IMG]
    Here's what I have so far:

    I1 = 49.68A<-68.6 degrees (top loop)
    I2 = 49.68A<-128.6 degrees (bottom loop)
    I3 = 28.68<-98.6 degrees (triangle loop)

    a)
    PA12 = (49.68A)^2(12) = 29,617.23 W
    PB12 = (49.68A-49.68A)^2(12) = 0 W
    PC12 = (49.68A)^2(12) = 29,617.23 W
    Ptotal12 = 29,617.23W + 0W + 29,617.23W = [59,234 W] answer

    b)
    Pab400 = (49.68A-28.68A)^2(400) = 176,400 W
    Pbc400 = (49.68A-28.68A)^2(400) = 176,400 W
    Pca400 = (28.68A)^2(400) = 329,017 W
    Ptotal400 = 176,400W + 176,400W + 329,017W = [681,817 W] answer

    Now c) and d) I am unsure of. I was thinking power factor would be 59,234 W/681,817W which would give me a p.f. of 0.087 but that doesn't seem right, and efficiency I have no clue how to calculate.

    I'm not 100% sure I even did parts a) and b) right. Any help/explanations would be greatly appreciated!!
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    All phase loads and line impedances are balanced [or equal] so the individual line current and load current magnitudes will be equal. In that case the power in each 12 Ω line resistor and 400 Ω load resistor will be the same - which is not what your solution indicates.
     
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