Power for 3 Phase Y Conn. Load

Discussion in 'Homework Help' started by jegues, Sep 29, 2011.

  1. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
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    43
    See figure attached for problem statement as well as my attempt.

    I'm having trouble finding the power dissapated across the transmission line.

    I thought this would be,

    P_{TL} = \sqrt{3}V_{L}I_{L}cos\phi, \quad \text{Where, } \quad cos\phi \quad \text{is the power factor across the transmission line}

    Where VL and IL are the line voltage and current across the transmission line.

    I_{L} = 150 \angle -36.87^{o}

    V_{L} = I_{L} \cdot Z_{TL}

    Where,

    Z_{TL} = (2.5 + j10.2) \Omega

    The solution gets the answer by doing the following,

    Re \left{ 3 I_{L}^{2} \cdot Z_{TL} \right}

    Can someone explain my confusion or clarify things?
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    There are several ways of coming up with an answer but having determined the line current, I would have thought the simplest way of expressing the transmission line power loss would be

    P_{TL}=3|{I_L}|^2R_{line}

    or using your values

    P_{TL}=3*150^2*2.5=56.25kW

    Sorry that was per phase

    it should obviously read

    P_{TL}=3*150^2*2.5=168.75kW

    BTW. Your formula would work fine as well provided you convert the voltage drop across the line impedance to a line-to-line equivalent voltage drop.
     
    Last edited: Sep 29, 2011
  3. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    735
    43
    Where does your formula come from?

    I can see you did P = IV, where V = I/R, but wheres does the 3 infront come from?

    Will the math show how it comes about?
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    No I didn't use P=IV - rather I used P=I^2R

    The RMS magnitude of the current in any line conductor is |I_line|.

    This current flows in the line resistance R_line of each of the 3-phase conductors.

    So the total power loss in the transmission line for all 3 conductors is

    3*|I_line|^2*R_line

    The phase angle of the current doesn't matter in this approach since current and voltage in the line resistances (as lumped elements) are always in phase.
     
  5. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    735
    43
    I mentioned after that V= I/R

    Hence, P=I^2R.

    Thank you for your other comments, this is much more clear now.
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    I guess you meant V=I*R ....
     
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