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23 January, 11:40

A balanced three phase load is supplied over a three-phase, 60 hz, transmission line with each line have a series impedance of (12.84 j72.76) ohms, the load voltage is 132 kV and the load power is 55 MVA at 0.8 lagging power factor. Determine: a. The ABCD parameters assuming a short line

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  1. 23 January, 11:53
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    Given a three-phase system

    Frequency f=60Hz

    Line impedance Z = 12.84 + j72.76 Ω

    Then,

    The resistance is R=12.84Ω

    And reactance is X=72.76Ω

    Z=√ (12.84²+72.76²)

    Z=73.88

    Angle = arctan (X/R)

    Angle = arctan (72.76/12.84)

    Angle=80°

    Then, Z=73.88 < 80° ohms

    Load voltage is 132 kV

    Load power P=55 MWA

    Power factor = 0.8lagging

    the relation between the sending and receiving end specifications are given using ABCD parameters by the equations below.

    Vs = AVr + BIr

    Is = CVr + DIr

    Where

    Vs is sending Voltage

    Vr is receiving Voltage

    Is is sending current

    Ir is receiving current

    A is ratio of source voltage to received voltage A=Vs/Vr when Ir=0

    B is short circuit resistance

    B = Vs/Ir when Vr=0

    C is ratio of source current to received voltage C=Is/Vr when Ir=0

    D is ratio of source current to received current D=Is/Ir when Vr=0

    Now,

    The load at 55MVA at 132kV (line to line)

    Therefore, load current is

    Ir = P/V√3

    Ir=55*10^6 / (132*10^3*√3)

    Ir=240.56 Amps

    It has a power factor 0.8 lagging

    PF=Cosθ

    0.8=Cosθ

    θ=arcCos (0.8)

    θ=36.87°

    Therefore, Ir=240.56 <-36.87°

    Vr=V/√3

    Vr=132/√3

    Vr=76.21 kV. Phase voltage

    Vr = 76210 < 0° V

    For series impedance,

    Using short line approximation

    Vs = Vr + IrZ

    Vs = 76210 < 0° + (240.56 <-36.87° * 73.88 < 80°)

    Using calculator

    Vs=76210<0° + 17772.5728< (-36.87°+80°)

    Vs=76210<0° + 17772.5728<43.13°

    Vs=89970.67<7.7°

    Also

    Is = Ir = 240.56 <-36.87° Amps

    Therefore, the ABCD parameters is

    A=Vs/Vr

    A = 89970.67 <7.7° / 76210 <0°

    A=1.181 <7.7-0

    A=1.18 <7.7° no unit

    B = Vs/Ir

    B = 89970.67 < 7.7° / 240.56 <-36.87°

    B = 347.01 < 7.7+36.87

    B = 347.01 < 44.57° Ω

    C = Is/Vr = 240.56 <-36.87° / 76210 < 0°

    C = 0.003157 <-36.87-0

    C = 3.157 * 10^-3 < - 36.87° / Ω

    C = 3.157 * 10^-3 < - 36.87° Ω~¹

    D = Is/Ir

    Since Is=Ir

    Then, D = 1 no unit.
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Home » Engineering » A balanced three phase load is supplied over a three-phase, 60 hz, transmission line with each line have a series impedance of (12.84 j72.76) ohms, the load voltage is 132 kV and the load power is 55 MVA at 0.8 lagging power factor. Determine: a.
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