3-phase delta circuit help

Discussion in 'Homework Help' started by dan_man555, Sep 1, 2010.

  1. dan_man555

    Thread Starter New Member

    Sep 1, 2010
    18
    0
    I need help working this one out. Would really appreciate the help.

    A 3-phase, Δ-connected source has an internal impedance of 0.9 + 9j Ω/phase. The terminal voltage is equal to 13,200 Volts when no load is connected. The generator feeds a Δ-connected load with a per-phase impedance of 645 + 171j Ω/phase. The impedance of the transmission line connecting the generator to the load is 0.7 + 3j Ω/phase. The a-phase
    internal voltage of the generator has been specified as the reference.

    a) Construct a single phase equivalent circuit of the 3-phase system
    b) Calculate the line currents: laA, IbB, and IcC
    c) Calculate the magnitude of the line voltages at the terminals of the load
    d) Calculate the magnitude of the line voltages at the terminals of the source
    e) Calculate the magnitude of the phase current in the load
     
  2. dan_man555

    Thread Starter New Member

    Sep 1, 2010
    18
    0
    Here is a scan of the question page.
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    The first challenge is to convert the delta connected source & it's associated source impedance to a wye equivalent.

    Can you do this?
     
  4. dan_man555

    Thread Starter New Member

    Sep 1, 2010
    18
    0
    Here's my attempt. Let me know what you think.
    [​IMG]
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Looks like you left out the equivalent source impedances for the star (wye) transformation. Apart from that it looks OK.
     
  6. dan_man555

    Thread Starter New Member

    Sep 1, 2010
    18
    0
    This is my answer for part a. I really hope I got the values right.

    [​IMG]
     
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    That's pretty good. Why have you got the 30 degree phase lag on the source impedance - it's not relevant. The source impedance is just 0.3+j3 Ω.
     
  8. dan_man555

    Thread Starter New Member

    Sep 1, 2010
    18
    0
    Ok.
    So for part b:
    Z total = 216 + 63j
    = 225 < 16.26 Ω
    IaA = 7621<-30 / 225 < 16.26
    =33.87<-46.26 A
    Therefore:
    IbB= 33.87<-166.26 A
    IcC= 33.87<73.74 A
     
  9. dan_man555

    Thread Starter New Member

    Sep 1, 2010
    18
    0
    If that's all fine, then for part D:

    [​IMG]
    Z=3.015<84.29
    Vz= (33.87<-46.26)(3.015<84.29)
    = 102.12<38.08v
    Van = (6599.98-3810.5j)-(80.38+62.98j)
    =6519.6-3873.48j
    =7583.47<-30.72 V
    Vab=13135.42<-0.72V
    Vbc=13135.42<-120.72V
    Vca=13135.42<-150.72V

    Am I close?
     
  10. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    All looks fine except for the Vca phase angle.
     
  11. dan_man555

    Thread Starter New Member

    Sep 1, 2010
    18
    0
    Oh, I see.
    Vca=13135.42<-240.72V

    c: line voltages at the terminals of the load
    [​IMG][​IMG]
    Ztotal=215+57j+0.7+3j
    =215.7+60j
    =223.89<15.54
    Van=(223.89<15.54)(33.87<-46.26)
    =7583.15<-30.72
    Vab=13135.42<-0.72V
    Vbc=13135.42<-120.72V
    Vca=13135.42<-240.72V
    I hope both sides were supposed to have the same line voltages.

    How do you 'calculate the magnitude of the phase current in the load' for part e?
     
  12. dan_man555

    Thread Starter New Member

    Sep 1, 2010
    18
    0
    I'd really like to know if I'm on the right track for part c, and how to tackle the last part.
     
  13. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    The load terminal-to-neutral voltage on a given phase must simply be the phase line current times the equivalent phase line-to-neutral load. You would not include the line impedance - which you did.

    So
    Ia_line=33.87<-42.26°
    Z_load_a=215+57j

    Hence V_load_a (line-to-neutral) =33.87<-42.26°*(215+57j)=7533.62<-27.41°

    From which you can find the line-to-line value....etc

    The load phase currents are (presumably) the values of current flow in the load delta arms based on the previously calculated line currents. Essentially a star-to-delta transformation of the line currents.
     
  14. dan_man555

    Thread Starter New Member

    Sep 1, 2010
    18
    0
    Would that make the line voltages at the terminals of the load:
    Vab=13048<2.59v, Vbc=13048<-117.4v, Vca=13048<-237.41v

    And for the phase current:
    I=33.87*sqrt(3)
    I=58.66A
     
  15. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Voltages look OK w.r.t magnitude - I haven't carefully checked your phase angles throughout your solution. I have Vab at the load leading the phase-neutral supply (your Va'n) by 28.59° whereas you have the corresponding phase lead as 32.59°. So we have a discrepancy there - looks like a simple algebraic sign issue. I'll re-check my overall solution when I have some time.

    The phase current (in the delta arms) magnitude would be 33.87/sqrt(3)=19.55A.
     
  16. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Had another look at the solution - attached a pdf copy of some of my working.

    You'll note we have different starting phase angles for the single phase equivalent. I've based it all on the no-load line-to-line (source) voltage Vab having a zero phase shift. Then I worked everything off that basis. The real issue is the related phase shifts of the various quantities given some assumed initial phase at the source.

    I'll still disagree with your overall phase relationship as mentioned in my earlier post.
     
    Last edited: Sep 6, 2010
    dan_man555 likes this.
  17. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Did it again as a check with phase angle of source the same as yours. Still have the discrepancy. Sorry the copy isn't particularly legible.
     
Loading...