Medium Transmission Line: Nominal π method

The total capacitance of the transmission line is divided into two parts, half at the sending end side and another half at the receiving end side in the nominal π method. All the parameters are given considering three phase transmission line.

Let

VS = Sending end voltage to neutral

VR = Receiving end voltage to neutral

IS = Sending end current

IR = Receiving end current

IC1 = Capacitance current at the receiving end side

IC2 = Capacitance current at the sending end side

R = Resistance of transmission line, resistance to the neutral in the three-phase line

X = Reactance of transmission line, reactance to the neutral in the three-phase line

Z = R + j X = Total impedance

A, B, C and D = Constant

Generalized Constant of the transmission line is given by sending end voltage and sending end current equation

VS = AVR + BIR……. ( 1 )

IS = CVR + DIR……....( 2 )

Nominal π method: ABCD Parameters

Apply KCL at point Q,

I1 = IR + IC1

I1 = IR + ( Y/2 ) VR ……( 3 )

Apply KCL at point P,

Is = I1 + IC2

Is = I1 + ( Y/2 ) Vs ………( 4 )

Substitute value of  I1 from equation ( 3 )

Is = IR + ( Y/2 ) VR + ( Y/2 ) Vs ………( 5 )

Sending end voltage VS = VR + I1( Z )

Substitute value of  I1 from equation ( 3 )

VS = VR + { IR + ( Y/2 ) VR } ( Z )

= VR + Z IR + ( ZY/2 ) VR

= VR + ( ZY/2 ) V+ Z IR

= ( 1 + ZY/2 ) V+ Z IR

VS = ( 1 + ZY/2 ) V+ Z IR ….. ( 6 )

Compare equation ( 6 ) and ( 1 )

VS = AVR + BIR

VS = ( 1 + ZY/2 ) V+ Z IR

Therefore A = ( 1 + YZ/2 ) and B = Z

From equation ( 5 )

Is = IR + ( Y/2 ) VR + ( Y/2 ) Vs

Substitute value of  Vs from equation ( 6 )

Is = IR + ( Y/2 ) VR + ( Y/2 ) {( 1 + ZY/2 ) V+ Z IR }

Is = IR + ( Y/2 ) VR + ( Y/2 + Y2Z/4 ) V+ ( YZ/2 ) IR

Is = ( Y/2 ) VR + ( Y/2 + Y2Z/4 ) V+ IR + ( YZ / 2 ) IR +

Is = { Y/2 + Y/2 + Y2Z/4 ) V+ { 1 + ( YZ / 2 ) } IR

Is = { Y + Y2Z/4 ) V+ { 1 + ( YZ / 2 ) } IR

Is = { 1 + ( YZ / 2 ) } IR + Y ( 1 + Y2Z/4 ) VR ….. ( 7 )

Compare equation ( 7 ) and ( 2 )

IS = CVR + DIR

Is = Y ( 1 + YZ/4 ) VR + { 1 + ( YZ / 2 ) } IR

Therefore C = Y ( 1 + YZ/4 ) and D = 1 + ( YZ / 2 )

From the above value of A, B, C and D

A = ( 1 + YZ/2 )

B = Z

C = Y ( 1 + YZ/4 )

D = 1 + ( YZ / 2 )

Therefore

AD – BC = ( 1 + YZ/2 ) × ( 1 + YZ/2 )  – Z × Y ( 1 + YZ/4 )

= ( 1 + YZ + Y2Z2/4 ) – YZ – Y2Z2/4

= 1

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