Electrical Power Systems

Characteristics and Performance of Transmission Lines 129
Now note that, when x = 0, V(x) = V R and from eqn. (6.26), we get
VR = C1 + C2 also when x = 0, I(x) = IR and from eqn. (6.28), we get,
I R = 1
Z C
Solving eqns. (6.31) and (6.32), we obtain,
C 1 =
...(6.30)
(C 1 – C 2) ...(6.31)
VR+ ZCIR 2
( V - Z I )
C2 =
2
R C R
...(6.32)
...(6.33)
Substituting the values of C 1 and C 2 from eqns. (6.32) and (6.33) into eqns. (6.26) and (6.28),
we get
V(x) =
I(x) =
( V + Z I ) ( VRZ I )
e +
e
2 2
R C Rx g - C R -gx
( VR+ ZCIR) x ( VRZCIR) e -
e
2Z2Z C
g - -gx
The equations for voltage and currents can be rearranged as follows:
V(x) =
I(x) =
gx -gx gx -gx
( e + e ) ( e - e )
VR + ZC I
2 2
gx -gx gx -gx
( e - e ) ( e + e )
VR
+
I
2Z2 C
C
R
R
...(6.34)
...(6.35)
...(6.36)
...(6.37)
or V(x) = cosh(gx)V R + Z C sinh(gx) I R ...(6.38)
I(x) = 1
Z C
sinh(gx)V R + cosh(gx)I R
...(6.39)
Our interest is in the relation between the sending end and the receiving end of the line.
Therefore, when x = l, V(l) = V S and I(l) = I S. The result is
Therefore, ABCD constants are:
V S = cosh (gl)V R + Z C sinh(g l)I R
IS = 1
ZC sinh (gl ) V R + cosh(gl)I R
...(6.40)
...(6.41)
A = cosh(gl) ; B = Z C sinh(gl) ...(6.42)
C = 1
Z C
sinh(gl) ; D = cosh(gl ) ...(6.43)
It is now possible to find an accurate equivalent p model for long transmission line as shown
in ig. 6.5.