Answer:
Explanation:
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.
