Power Engineering Help (Star-Delta)

Discussion in 'Homework Help' started by mksa, May 6, 2006.

  1. mksa

    mksa Thread Starter New Member

    Joined:
    May 6, 2006
    Messages:
    2
    1) An unbalanced delta-connected load on a 400 V (line-line) 50 Hz system has
    impedances of:
    Za = 40 ohm; Zb = j 40 ohm; Zc = 20 – j 20 ohm.
    Calculate the phase and line currents.

    2) A star-connected 400 V resistive load is connected to a 3-wire system. The
    load has impedances of:
    Za = 20 ohm; Zb = 40 ohm; Zc = 40 ohm.
    Calculate the voltage of the neutral point.


    3) An unbalanced star-connected load on a 220 V 60Hz 4-wire system. The legs of the load have impedances of:
    Za = 5 ohm; Zb = 5 +j 5 ohm; Zc = 5– j 5ohm.
    To insure correct phase voltage the star is connected to the neutral line
    Calculate the phase and neutral currents.

    4) An 400 V (line-line) star-connected unbalanced load has an earthed start point.
    The load on phase 1 is 10 ohm, the load on phase 2 is 10 + j5 ohm and the load on
    phase 3 is 10 – j5 ohm. Calculate the phase currents and the star to earth current
    and sketch the current phasor diagram.

    I can do the balance system but the unbalanced I can't. and I would like to ask what is the diffrenece between xxx V (line-line) and xxxV xxHz and the difference between the 4-wire system and 3-wire system.

    thank you.
  2. mksa

    mksa Thread Starter New Member

    Joined:
    May 6, 2006
    Messages:
    2
    the phase currnts
    for Iab=Vab/Za=400/40=10A
    for Ibc =Vbc/Zb=400 phase-120/40 phase90 = 10 phase -210 A
    for Ica=Vca/zc=400 phase 120 / 28.28 phase -45 = 14.144 phase 165 A
    the line currents
    for IAa = Iab-Ica=10 phase 0 - 14.144 phase 165 = 23.94 phase -8.79 A
    for IBb = Ibc-Iab=10 phase -210 - 10 phase 0 = 218.43 phase -23.55 A
    for ICc = Ica - Ibc = 14.144 phase 165 - 10 phase -210 = 191.793 phase 24.69 A

    Vsn=Ya*Van+Yb*Vbn+Yc*Vcn/Ya+Yb+Yc
    Vsn=1/20 * (400/ √ 3 phase 0) + 1/40 * 230.94 phase -120 + 1/40 * 230.94 phase 120 / (1/20 + 1/40 + 1/40) = 95.6751 phase 180.013 V

    first we find Van , Vbn and Vcn which 220/ √ 3
    Van=127 phase 0 V
    Vbn= 127 phase -120 V
    Vcn=127 phase 120 V
    Ian = Van/Za = 127 phase 0 / 5 = 25.4 A
    Ibn=Vbn/Zb = 127 phase -120 / 7.07 phase 45 = 17.96 phase 165 A
    Icn=Vcn/Zc = 127 phase 120 / 7.07 phase -45 = 17.96 phase 165 A
    In = Ian + Ibn + Icn = 25.4 phase 0 + 17.96 phase 165 + 17.96 phase 165
    In = 13.15 phase 134.998 A
    first we find Van , Vbn and Vcn which 400/ √ 3
    Van=231 phase 0 V
    Vbn= 231phase -120 V
    Vcn=231phase 120 V
    Ian = Van/Za = 231phase 0 / 10 = 23.1 A
    Ibn=Vbn/Zb = 231phase -120 / 11.18 phase 26.565 = 20.66 phase 146.565 A
    Icn=Vcn/Zc = 231 phase 120 / 11.18 phase -26.565 = 20.66 phase 146.565 A
    In = Ian + Ibn + Icn = 23.1 phase 0 + 20.66 phase 146.565 A+ 20.66 phase 146.565 A
    In = 25.45 phase 116.56 A
  3. mksa

    mksa Thread Starter New Member

    Joined:
    May 6, 2006
    Messages:
    2
    the phase currnts
    for Iab=Vab/Za=400/40=10A
    for Ibc =Vbc/Zb=400 phase-120/40 phase90 = 10 phase -210 A
    for Ica=Vca/zc=400 phase 120 / 28.28 phase -45 = 14.144 phase 165 A
    the line currents
    for IAa = Iab-Ica=10 phase 0 - 14.144 phase 165 = 23.94 phase -8.79 A
    for IBb = Ibc-Iab=10 phase -210 - 10 phase 0 = 218.43 phase -23.55 A
    for ICc = Ica - Ibc = 14.144 phase 165 - 10 phase -210 = 191.793 phase 24.69 A

    Vsn=Ya*Van+Yb*Vbn+Yc*Vcn/Ya+Yb+Yc
    Vsn=1/20 * (400/ √ 3 phase 0) + 1/40 * 230.94 phase -120 + 1/40 * 230.94 phase 120 / (1/20 + 1/40 + 1/40) = 95.6751 phase 180.013 V

    first we find Van , Vbn and Vcn which 220/ √ 3
    Van=127 phase 0 V
    Vbn= 127 phase -120 V
    Vcn=127 phase 120 V
    Ian = Van/Za = 127 phase 0 / 5 = 25.4 A
    Ibn=Vbn/Zb = 127 phase -120 / 7.07 phase 45 = 17.96 phase 165 A
    Icn=Vcn/Zc = 127 phase 120 / 7.07 phase -45 = 17.96 phase 165 A
    In = Ian + Ibn + Icn = 25.4 phase 0 + 17.96 phase 165 + 17.96 phase 165
    In = 13.15 phase 134.998 A
    first we find Van , Vbn and Vcn which 400/ √ 3
    Van=231 phase 0 V
    Vbn= 231phase -120 V
    Vcn=231phase 120 V
    Ian = Van/Za = 231phase 0 / 10 = 23.1 A
    Ibn=Vbn/Zb = 231phase -120 / 11.18 phase 26.565 = 20.66 phase 146.565 A
    Icn=Vcn/Zc = 231 phase 120 / 11.18 phase -26.565 = 20.66 phase 146.565 A
    In = Ian + Ibn + Icn = 23.1 phase 0 + 20.66 phase 146.565 A+ 20.66 phase 146.565 A
    In = 25.45 phase 116.56 A
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