Three phase ac newtork

Discussion in 'Homework Help' started by SilverKing, Jan 31, 2015.

  1. SilverKing

    Thread Starter Member

    Feb 2, 2014
    72
    0
    Hi everyone,

    I need some help in solving this problem:

    Three inductors with 0.1 H each are delta connected to three phase ac newtork with 220V/50Hz. Determine Iph, IL, P, Q and Theta.

    This how I tried to solve it:
    ______________________
    XL=2*pi*f*L=31.4 ohms
    Iph=Uph/XL=220/31.4=7 A
    IL=1.73*7=12.12 A

    Q=(Iph)^2 XL=Uph Iph sin(theta)

    Since I don't have theta and I need to determine it in thte first place, I shall use he first law.
    Q=(7)^2 31.4=1538.6 VAR
    QT=3*Q=4615 VAR

    Then, I shall substitute the value of Q in the second law to get theta.
    sin^-1 (theta)=Q/Uph Iph=1538.6/220*7=87.55 ْ

    P=Uph Iph cos(theta)=65.65 W
    PT=3*P=196.95 W

    ______________________


    Is that right?
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    If the load comprises purely inductive components where will any real load power be dissipated?
     
  3. WBahn

    Moderator

    Mar 31, 2012
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    4,796
    Always ask if the answer makes sense!
     
  4. SilverKing

    Thread Starter Member

    Feb 2, 2014
    72
    0
    t_n_k
    So, any purly inductor would disspate zero power? I didn't know that before. Thanks

    WBahn

    ???
     
  5. WBahn

    Moderator

    Mar 31, 2012
    17,748
    4,796
    Yep. If the voltage and current are 90 degrees out of phase, then no real power is delivered or received. Energy flows from the source and is temporarily stored in the magnetic field of the inductor (or electric field of a capacitor) and then returned to the source. This happens twice each cycle.
     
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