Delta connected loads question

Discussion in 'Homework Help' started by matt_t, Sep 9, 2012.

  1. matt_t

    Thread Starter Member

    Aug 18, 2012
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    Hi I have the following question:

    The following are connected in delta to a 600 Volt three-phase supply.
    • A-B has a 20 Amp load with a power factor of .9 lag.
    • B-C has a 30 Amp load with a power factor of 1
    • C-A has a 40 Amp load with a power factor of .9 lag.
    Calculate the current in line A:
    Calculate the current in line B:
    Calculate the current in line C:

    I have drawn out the circuit in delta form to try and understand where the current is going but without the neutral of a star system I am confused when using kirchoffs law.

    I know that two of the currents are not in phase so I have:

    A-B 20A /_ 25.85 deg
    B-C 30A /_ 0 deg
    C-A 40A /_ 25.85 deg

    I can't use the formula I line = I phase X √3 as the loads are not balanced. So I am unsure how the current is distributed between the phases.
     
  2. t_n_k

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    Mar 6, 2009
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    Any of the line currents would be the vector summation of the two associated delta currents. So Ia=Iab-Ica and so forth.
     
  3. matt_t

    Thread Starter Member

    Aug 18, 2012
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    So to find the current in line B I would do the following?

    Ib= Iba-Ibc

    20/_ 35.85 deg - 30/_ 0 deg
    18+j8.717 - 30+j0 = -12-j8.717 = 14.83/_ -144 which isnt the correct answer.

    Think im missing something here.
     
  4. t_n_k

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    But

    Ib=Ibc-Iab

    etc....
     
  5. t_n_k

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    Also how could a lagging pf current give a +ve imaginary part.
     
  6. matt_t

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    Aug 18, 2012
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    I don't understand, the correct answers I have are B = 37.2A and C = 67A.

    Would you be able to show me the vector sums so I can see clearly how i've gone wrong?
     
  7. matt_t

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    Aug 18, 2012
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    Am I missing the fact that all of the phases are 120 deg out of phase with eachother? Do I need to take this into account?
     
  8. t_n_k

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    Yes - absolutely you must remember the individual line voltage phase displacements are an important consideration.
     
  9. matt_t

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    Aug 18, 2012
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    But how do I add currents which are normally 120 deg out of phase with eachother when two of them are out of phase with themselves?
     
  10. mlog

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    Feb 11, 2012
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    I'm giving my solution since we've hashed it out enough. Don't you think?

    Here's the way I see it. Given Iab, Ibc, & Ica, find Ia, Ib, & Ic.

    Given:

    Iab = 20A @ pf 0.9 lag
    Ibc = 30A @ pf 0.0
    Ica = 40A @ pf 0.9 lag

    Since all currents will be in Amperes, I will drop the "A."

    Iab = 20.0 ang(-25.85 deg)
    Ibc = 30.0 ang(0.00 deg)
    Ica = 40.0 ang(-25.85 deg)

    I shall define Ia as the current entering node A, etc. As t_n_k said (at least for Ib):

    Ia = Iab+Iac = Iab-Ica
    Ib = Ibc+Iba = Ibc-Iab
    Ic = Ica+Icb = Ica-Ibc

    Ia = Iab-Ica = -18.00 + j8.72 = 20.0 ang(154.2 deg)
    Ib = Ibc-Iab = +12.00 + j8.72 = 14.83 ang(36.0 deg)
    Ic = Ica-Ibc = +6.00 - j14.44 = 18.00 ang(-71.0 deg)

    I think you'll find if you sum Ia, Ib, & Ic, you will get zero.

    If you draw a phasor diagram, you'll see the 3 currents, Ia, Ib, & Ic are almost 120 degrees apart. I say "almost," because A & B are ~118 deg apart, B & C are ~107 deg apart, and C & A are ~135 deg apart.
     
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  11. matt_t

    Thread Starter Member

    Aug 18, 2012
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    Thanks for the reply this is what I understand according to hat tnk has said. I do not understand why the textbook answers are line currents B=37.2A and C=67A?
     
  12. mlog

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    Feb 11, 2012
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    Is that the entire problem? I noticed the supply voltage was given as 600 volts. Is there something else given? What does it say for the A current?
     
  13. matt_t

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    Aug 18, 2012
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    That is the entire problem in the first post the question only asked to work out lines A and B so I've been spending alot of time trying to figure out the answers given.
     
  14. t_n_k

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    Unfortunately this solution is incorrect - the Op's answers are good.
     
  15. mlog

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    Feb 11, 2012
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    What did I do wrong?
     
  16. matt_t

    Thread Starter Member

    Aug 18, 2012
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    I have been on this same question for 2 days now and still cannot come up with the answers. This is as close as I can get:

    B = 20/_ -25.84 - 30/_ 120 = 18+j8.717 - -15+j25.98 = 33+j34.697 = 47.88 /_46.435
     
  17. matt_t

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    Aug 18, 2012
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    OK so i've finally realised what I did wrong.

    B = 20/_ -25.84 - 30/_ 120 = 18+j8.717 - -15+j25.98 = 33-j17.26 = 37.24/_-27.61A

    FINALLY.
     
  18. matt_t

    Thread Starter Member

    Aug 18, 2012
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    C= 30/_ 0 - 40/_ 145.84 = 30+j0 - -33.09+j22.46 = 63.09-j22.46 = 66.96/_19.59A
     
  19. t_n_k

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    Correct answers are

     I_a=52.92 \angle{-66.74^o} \ A

     I_b=37.2 \angle{ -152.5^o} \ A

     I_c=66.9 \angle{ 79.58^o} \ A
     
    Last edited: Sep 12, 2012
  20. matt_t

    Thread Starter Member

    Aug 18, 2012
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    how did you get those different angles?
     
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