Power in a balanced 3 phase system

Discussion in 'Homework Help' started by p75213, Mar 4, 2012.

  1. p75213

    Thread Starter Member

    May 24, 2011
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    See attached for problem.
    Part a) of the question asks to find the complex power of the combined load. Following is my attempt at the answer.

    1. Find complex power for load 1.
    S1=\frac{|Vl|^{2}\sqrt{3}}{Z^{*}}
    S1=\frac{840^{2}\sqrt{3}}{30-j40}
    S1=14.665+j19.554kVA

    2. Find complex power for load 2.
    S2=P+jQ
    S2=48+j48tan36.87
    S2=48+j36kVA

    3. Total Complex Power
    S=S1+S2
    S=62.665+j55.554kVA
    The answer is 56.47+j47.29kVA.
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Perhaps I'm half asleep. I fail to understand how two lagging pf loads produce a combined leading complex power. I believe the solution is incorrect.
     
  3. p75213

    Thread Starter Member

    May 24, 2011
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    The correct answer is - according to the book - 56.47+j47.29kVA. I thought that was lagging.
     
  4. t_n_k

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    Mar 6, 2009
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    The book is wrong - only for part (a). Your method for finding the complex power is incorrect.
     
  5. p75213

    Thread Starter Member

    May 24, 2011
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    Ok - Can you show the workings for the correct answer?
     
  6. t_n_k

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    Mar 6, 2009
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    The complex power for load 1 is

    S1=|Vline|^2/(30+j40)

    The complex power for the motor is

    48-j36 kVa

    Sum=56.47-j47.29 kVa
     
  7. p75213

    Thread Starter Member

    May 24, 2011
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    So the formula is S=VI?
    The book shows the total complex power as 3VpIp* or √3VlIl*.
    Vp=Phase Voltage
    Ip=Phase Current
    Vl=Line Voltage
    Il=Line Current.

    I'm confused.
     
  8. t_n_k

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    Mar 6, 2009
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    I'll use the method I'm familiar with.

    For the Y connected load 1.

    Vp=Vl / √3
    Il=Vp/Z
    Stot=3*Vp*Il=3*(Vl/√3)*(Vl/√3)/Z=|Vl|^2/Z
     
  9. t_n_k

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    Mar 6, 2009
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    I've never seen the use of the complex conjugate term in the context of complex power calculations. Perhaps something different is meant by the term than what I would regard as conventional. Are we talking about what is often referred to as apparent power?
    S=P+jQ ....?
     
  10. t_n_k

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    Mar 6, 2009
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    Actually I lie. I've seen this before. I've actually posted on the matter in another thread. The form you use is correct. My apologies.
     
  11. p75213

    Thread Starter Member

    May 24, 2011
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    I can see my error now.
    The load on load1 is per phase but the voltage is line voltage. Therefore convert line voltage to phase voltage to get the complex power. The formula then becomes 3VpIp* or (3|Vp|^2)/Z*.

    Thanks for helping me sort that out t_n_k
     
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