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3 PHASE CIRCUIT PROBLEM - Haaalllppp

  1. Mar 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Three load each having a resistance of 16 ohms and an inductive reactance of 12 ohms are wye connected to a 240V three phase supply. Determine:

    a) the impedance per phase
    b) the current per phase
    c) the kVA total
    d) the Power Factor
    e) the kW total

    2. Relevant equations

    impedance? (Z= √R^2 + XL^2)??
    wye: IPhase=ILine=ILoad
    kVA= 3 X VPhase X IPhase
    kVA= 1.732 X VLine X ILine
    PF= kW/kVA
    p= 3 X VPhase X IPhase X PF
    P= 1.732 X VLine X ILine X PF

    3. The attempt at a solution

    A) ?

    B) VLine= 240V

    IPhase= VPhase/RPhase

    VPhase= 240V/1.732
    VPhase= 138.5V

    IPhase= 138.5V/16Ω
    IPhase= 8.6A

    C) kVA= 3 X VPhase X IPhase

    kVA= 3 X 138.5V X 8.6A
    = 3.5kVA

    D) ?

    E) ?


    Not sure how to take these ones on. Any tips? formulas to use?
     
  2. jcsd
  3. Mar 16, 2013 #2
    http://1-media-cdn.foolz.us/ffuuka/board/wsg/image/1339/18/1339188551986.gif
     
  4. Mar 16, 2013 #3

    gneill

    User Avatar

    Staff: Mentor

    I did some browsing and came across this site:

    http://www.wisc-online.com/objects/ViewObject.aspx?ID=ACE302

    with the following graphic:

    attachment.php?attachmentid=56802&stc=1&d=1363469614.gif

    which seems to indicate that the voltage specification for a Wye generating system specifies the line-to-neutral (or phase) voltage of the generator. If that is true for this problem, then the line-to-line voltage will be in the neighborhood of 416V.

    A circuit diagram would look something like:

    attachment.php?attachmentid=56805&stc=1&d=1363469921.gif

    For the impedance you can use the reactance as you've shown. You can determine the power factor for the load from its real and reactive components.

    An alternative is to leave the impedance in complex form and do the calculations in complex form too (phasors). This would automatically take care of all the "1.732" and pf factors and give you the kVA, kW and kVAR values all at once. It helps if your calculator handles complex numbers :smile:
     

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