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
Similar Threads
Forum Title Date
Homework Help Power engineering help(Automatic Voltage Regulator) Jun 8, 2010
Homework Help Power Engineering Unbalanced 3 Phase Aug 25, 2007
Homework Help Power Spectrum Problem Mar 24, 2015
Homework Help Power Supply Mar 2, 2015
Homework Help DC power supply 12V 2A Feb 26, 2015

Share This Page