ABCD Parameters of Nominal T method

Medium Transmission Line: Nominal T Method

The total capacitance of the transmission line is considered at the middle of the line in the nominal T method, considering half the impedance at the sending end side and another half at the receiving end side. 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_C = Capacitance current

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 T Method: ABCD Parameters

Here, the total impedance is divided into two parts, half on the sending end and another half on receiving side but both have same value.

Z/2 = R/2 + jX_L / 2

Voltage across capacitor V₁ = V_R + I_R ( Z/2 )…. ( 3 )

Sending end current is sum of capacitance current and receiving end current.

I_S = I_R + I_C

I_S = I_R + YV₁  

I_S = I_R + Y { V_R + I_R ( Z/2 ) } { from equation ( 3 ) }

I_S = I_R + YV_R + YI_R ( Z/2 )

I_S = YV_R + I_R + YI_R ( Z/2 )

I_S = YV_R + ( 1 + YZ/2 ) I_R ……( 4 )

Compare equation ( 2 ) and ( 4 )

I_S = CV_R + DI_R and

I_S = YV_R + ( 1 + YZ/2 ) I_R

Therefore C = Y and D = 1 + YZ/2

Now, sending end voltage

V_S = V₁ + I_S ( Z/2 )

     = V_R + I_R ( Z/2 ) + I_S ( Z/2 ) { from equation ( 3 ) }

Putting value of Is from equation ( 4 )

     = V_R + I_R ( Z/2 ) + ( Z/2 )( YV_R + ( 1 + YZ/2 ) I_R

     = V_R + I_R ( Z/2 ) + ( YZ/2 )V_R + ( Z/2 )( 1 + YZ/2 ) I_R

     = ( 1 + YZ/2 )V_R + I_R ( Z/2 ) + ( Z/2 )( 1 + YZ/2 ) I_R

     = ( 1 + YZ/2 )V_R + ( Z/2 + Z/2 + YZ²/4 ) I_R

     = ( 1 + YZ/2 )V_R + ( Z + YZ²/4 ) I_R

V_S = ( 1 + YZ/2 )V_R + Z( 1 + YZ/4 ) I_R ……( 5 )

Compare equation ( 1 ) and ( 5 )

V_S = AV_R + BI_R

V_S = ( 1 + YZ/2 )V_R + Z( 1 + YZ/4 ) I_R

Therefore

A = ( 1 + YZ/2 ) and  B = Z( 1 + YZ/4 )  

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

A = ( 1 + YZ/2 )

B = Z( 1 + YZ/4 )

C = Y

D = 1 + YZ/2

Therefore

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

                = 1 + YZ/2 + YZ/2 + Y²Z² /4 – YZ – Y²Z² / 4

                = 1 + YZ + Y²Z² /4 – YZ – Y²Z² / 4

                = 1

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