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

Question

Engineering

Clarence Deleon

20 November, 10:15

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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Brennen Wallace · Answer 1

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.