ABCD Parameters of Nominal π method

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

V_S = Sending end voltage to neutral

V_R = Receiving end voltage to neutral

I_S = Sending end current

I_R = Receiving end current

I_C1 = Capacitance current at the receiving end side

I_C2 = 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

Y = jX_C = Admittance

A, B, C and D = Constant

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

V_S = AV_R + BI_R……. ( 1 )

I_S = CV_R + DI_R……....( 2 )

Nominal π method: ABCD Parameters

Apply KCL at point Q,

I₁ = I_R + I_C1

I₁ = I_R + ( Y/2 ) V_R ……( 3 )

Apply KCL at point P,

I_s = I₁ + I_C2

I_s = I₁ + ( Y/2 ) V_s ………( 4 )

Substitute value of  I₁ from equation ( 3 )

I_s = I_R + ( Y/2 ) V_R + ( Y/2 ) V_s ………( 5 )

Sending end voltage V_S = V_R + I₁( Z )

Substitute value of  I₁ from equation ( 3 )

V_S = V_R + { I_R + ( Y/2 ) V_R } ( Z )

      = V_R + Z I_R + ( ZY/2 ) V_R

      = V_R + ( ZY/2 ) V_R  + Z I_R

      = ( 1 + ZY/2 ) V_R  + Z I_R

V_S = ( 1 + ZY/2 ) V_R  + Z I_R ….. ( 6 )

Compare equation ( 6 ) and ( 1 )

V_S = AV_R + BI_R

V_S = ( 1 + ZY/2 ) V_R  + Z I_R

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

From equation ( 5 )

I_s = I_R + ( Y/2 ) V_R + ( Y/2 ) V_s

Substitute value of  V_s from equation ( 6 )

I_s = I_R + ( Y/2 ) V_R + ( Y/2 ) {( 1 + ZY/2 ) V_R  + Z I_R }

I_s = I_R + ( Y/2 ) V_R + ( Y/2 + Y²Z/4 ) V_R  + ( YZ/2 ) I_R

I_s = ( Y/2 ) V_R + ( Y/2 + Y²Z/4 ) V_R  + I_R + ( YZ / 2 ) I_R +   

I_s = { Y/2 + Y/2 + Y²Z/4 ) V_R  + { 1 + ( YZ / 2 ) } I_R

I_s = { Y + Y²Z/4 ) V_R  + { 1 + ( YZ / 2 ) } I_R

I_s = { 1 + ( YZ / 2 ) } I_R + Y ( 1 + Y²Z/4 ) V_R ….. ( 7 )

Compare equation ( 7 ) and ( 2 )

I_S = CV_R + DI_R

I_s = Y ( 1 + YZ/4 ) V_R + { 1 + ( YZ / 2 ) } I_R

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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