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9 June, 08:06

A three-phase line has an impedance of 0.4 j2.7 ohms per phase. The line feeds two balanced three-phase loads that are connected in parallel. The first load is absorbing 560.1kVA at 0.707 power factor lagging. The second load absorbs 132 kW at unity power factor. The line-to-line voltage at the load end of the line is 3810.5 volts. Determine: a. The magnitude of the line voltage at the source end of the line. b. Total real and reactive power loss in the line. c. Real power and reactive power supplied at the sending end of the line.

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  1. 9 June, 10:17
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    a. The magnitude of the line source voltage is

    Vs = 4160 V

    b. Total real and reactive power loss in the line is

    Ploss = 12 kW

    Qloss = j81 kvar

    Sloss = 12 + j81 kVA

    c. Real power and reactive power supplied at the sending end of the line

    Ss = 540.046 + j476.95 kVA

    Ps = 540.046 kW

    Qs = j476.95 kvar

    Explanation:

    a. The magnitude of the line voltage at the source end of the line.

    The voltage at the source end of the line is given by

    Vs = Vload + (Total current*Zline)

    Complex power of first load:

    S₁ = 560.1 < cos⁻¹ (0.707)

    S₁ = 560.1 < 45° kVA

    Complex power of second load:

    S₂ = P₂*1 (unity power factor)

    S₂ = 132*1

    S₂ = 132 kVA

    S₂ = 132 < cos⁻¹ (1)

    S₂ = 132 < 0° kVA

    Total Complex power of load is

    S = S₁ + S₂

    S = 560.1 < 45° + 132 < 0°

    S = 660 < 36.87° kVA

    Total current is

    I = S* / (3*Vload) ( * represents conjugate)

    The phase voltage of load is

    Vload = 3810.5/√3

    Vload = 2200 V

    I = 660 < - 36.87° / (3*2200)

    I = 100 < - 36.87° A

    The phase source voltage is

    Vs = Vload + (Total current*Zline)

    Vs = 2200 + (100 < - 36.87°) * (0.4 + j2.7)

    Vs = 2401.7 < 4.58° V

    The magnitude of the line source voltage is

    Vs = 2401.7*√3

    Vs = 4160 V

    b. Total real and reactive power loss in the line.

    The 3-phase real power loss is given by

    Ploss = 3*R*I²

    Where R is the resistance of the line.

    Ploss = 3*0.4*100²

    Ploss = 12000 W

    Ploss = 12 kW

    The 3-phase reactive power loss is given by

    Qloss = 3*X*I²

    Where X is the reactance of the line.

    Qloss = 3*j2.7*100²

    Qloss = j81000 var

    Qloss = j81 kvar

    Sloss = Ploss + Qloss

    Sloss = 12 + j81 kVA

    c. Real power and reactive power supplied at the sending end of the line

    The complex power at sending end of the line is

    Ss = 3*Vs*I*

    Ss = 3 * (2401.7 < 4.58) * (100 < 36.87°)

    Ss = 540.046 + j476.95 kVA

    So the sending end real power is

    Ps = 540.046 kW

    So the sending end reactive power is

    Qs = j476.95 kvar
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