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 V_{1} = 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
}+ YV_{1 }
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} = V_{1}
+ 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 + YZ^{2}/4 ) I_{R}
= ( 1 + YZ/2 )V_{R} + ( Z + YZ^{2}/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 + Y^{2}Z^{2} /4 –
YZ – Y^{2}Z^{2} / 4
= 1 + YZ
+ Y^{2}Z^{2} /4 – YZ – Y^{2}Z^{2} / 4
= 1
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