1. CHAPTER 2
TRANSMISSION LINE MODELS
As we have discussed earlier in Chapter 1 that the transmission line parameters
include series resistance and inductance and shunt capacitance. In this chapter we shall
discuss the various models of the line. The line models are classified by their length. These
classifications are
Short line approximation for lines that are less than 80 km long.
Medium line approximation for lines whose lengths are between 80 km to 250 km.
Long line model for lines that are longer than 250 km.
These models will be discussed in this chapter. However before that let us introduce the
ABCD parameters that are used for relating the sending end voltage and current to the
receiving end voltage and currents.
2.1 ABCD PARAMETSRS
Consider the power system shown in Fig. 2.1. In this the sending and receiving end
voltages are denoted by VS and VR respectively. Also the currents IS and IR are entering and
leaving the network respectively. The sending end voltage and current are then defined in
terms of the ABCD parameters as
RRS BIAVV (2.1)
RRS DICVI (2.2)
From (2.1) we see that
0RIR
S
V
V
A (2.3)
This implies that A is the ratio of sending end voltage to the open circuit receiving end
voltage. This quantity is dimension less. Similarly,
0RVR
S
I
V
B (2.4)
i.e., B, given in Ohm, is the ratio of sending end voltage and short circuit receiving end
current. In a similar way we can also define
0RIR
S
V
I
C mho (2.5)

2. 1.31
0RVR
S
I
I
D (2.6)
The parameter D is dimension less.
Fig. 2.1 Two port representation of a transmission network.
2.2 SHORT LINE APPROXIMATION
The shunt capacitance for a short line is almost negligible. The series impedance is
assumed to be lumped as shown in Fig. 2.2. If the impedance per km for an l km long line is
z0 = r + jx, then the total impedance of the line is Z = R + jX = lr + jlx. The sending end
voltage and current for this approximation are given by
RRS ZIVV (2.7)
RS II (2.8)
Therefore the ABCD parameters are given by
0and,1 CZBDA (2.9)
Fig. 2.2 Short transmission line representation.
2.2 MEDIUM LINE APPROXIMATION
Medium transmission lines are modeled with lumped shunt admittance. There are two
different representations nominal- and nominal-T depending on the nature of the network.
These two are discussed below.
2.2.1 Nominal- Representation
In this representation the lumped series impedance is placed in the middle while the
shunt admittance is divided into two equal parts and placed at the two ends. The nominal-
representation is shown in Fig. 2.3. This representation is used for load flow studies, as we
shall see later. Also a long transmission line can be modeled as an equivalent -network for
load flow studies.

3. 1.32
Fig. 2.3 Nominal- representation.
Let us define three currents I1, I2 and I3 as indicated in Fig. 2.3. Applying KCL at
nodes M and N we get
RRs
Rs
IV
Y
V
Y
IIIIII
22
3121
(2.10)
Again
RR
RRRRs
ZIV
YZ
VI
Y
VZVZIV
1
2
2
2
(2.11)
Substituting (2.11) in (2.10) we get
RR
RRRRs
I
YZ
V
YZ
Y
IV
Y
ZIV
YZY
I
1
2
1
4
2
1
22
(2.12)
Therefore from (2.11) and (2.12) we get the following ABCD parameters of the
nominal- representation
1
2
YZ
DA (2.13)
ZB (2.14)
mho1
4
YZ
YC (2.15)
2.2.1 Nominal-T Representation
In this representation the shunt admittance is placed in the middle and the series
impedance is divided into two equal parts and these parts are placed on either side of the
shunt admittance. The nominal-T representation is shown in Fig. 2.4. Let us denote the
midpoint voltage as VM. Then the application of KCL at the midpoint results in

4. 1.33
Fig. 2.4 Nominal-T representation.
22 Z
VV
YV
Z
VV RM
M
MS
Rearranging the above equation can be written as
RSM VV
YZ
V
4
2
(2.16)
Now the receiving end current is given by
2Z
VV
I RM
R (2.17)
Substituting the value of VM from (2.16) in (2.17) and rearranging we get
RRs I
YZ
ZV
YZ
V 1
4
1
2
(2.18)
Furthermore the sending end current is
RMS IYVI (2.19)
Then substituting the value of VM from (2.16) in (2.19) and solving
RRR I
YZ
YVI 1
2
(2.20)
Then the ABCD parameters of the T-network are
1
2
YZ
DA (2.21)
1
4
YZ
ZB (2.22)
mhoYC (2.23)
2.3 LONG LINE MODEL

5. 1.34
For accurate modeling of the transmission line we must not assume that the
parameters are lumped but are distributed throughout line. The single-line diagram of a long
transmission line is shown in Fig. 2.5. The length of the line is l. Let us consider a small strip
x that is at a distance x from the receiving end. The voltage and current at the end of the
strip are V and I respectively and the beginning of the strip are V + V and I + I
respectively. The voltage drop across the strip is then V. Since the length of the strip is x,
the series impedance and shunt admittance are z x and y x. It is to be noted here that the
total impedance and admittance of the line are
lyYlzZ and (2.24)
Fig. 2.5 Long transmission line representation.
From the circuit of Fig. 2.5 we see that
Iz
x
V
xIzV (2.25)
Again as x 0, from (2.25) we get
Iz
dx
dV
(2.26)
Now for the current through the strip, applying KCL we get
xVyxVyxyVVI (2.27)
The second term of the above equation is the product of two small quantities and therefore
can be neglected. For x 0 we then have
Vy
dx
dI
(2.28)
Taking derivative with respect to x of both sides of (2.26) we get
dx
dI
z
dx
dV
dx
d
Substitution of (2.28) in the above equation results

6. 1.35
02
2
yzV
dx
Vd
(2.29)
The roots of the above equation are located at (yz). Hence the solution of (2.29) is of the
form
yzxyzx
eAeAV 21 (2.30)
Taking derivative of (2.30) with respect to x we get
yzxyzx
eyzAeyzA
dx
dV
21 (2.31)
Combining (2.26) with (2.31) we have
yzxyzx
e
yz
A
e
yz
A
dx
dV
z
I 211
(2.32)
Let us define the following two quantities
impedancesticcharacterithecallediswhich
y
z
ZC (2.33)
constantnpropagatiothecallediswhichyz (2.34)
Then (2.30) and (2.32) can be written in terms of the characteristic impedance and
propagation constant as
xx
eAeAV 21 (2.35)
x
C
x
C
e
Z
A
e
Z
A
I 21
(2.36)
Let us assume that x = 0. Then V = VR and I = IR. From (2.35) and (2.36) we then get
21 AAVR (2.37)
CC
R
Z
A
Z
A
I 21
(2.38)
Solving (2.37) and (2.38) we get the following values for A1 and A2.
2
and
2
21
RCRRCR IZV
A
IZV
A
Also note that for l = x we have V = VS and I = IS. Therefore replacing x by l and substituting
the values of A1 and A2 in (2.35) and (2.36) we get

7. 1.36
lRCRlRCR
S e
IZV
e
IZV
V
22
(2.39)
lRCRlRCR
S e
IZV
e
IZV
I
22
(2.40)
Noting that
l
ee
l
ee llll
cosh
2
andsinh
2
We can rewrite (2.39) and (2.40) as
lIZlVV RCRS sinhcosh (2.41)
lI
Z
l
VI R
C
RS cosh
sinh
(2.42)
The ABCD parameters of the long transmission line can then be written as
lDA cosh (2.43)
lZB C sinh (2.44)
CZ
l
C
sinh
mho (2.45)
Example 2.1: Consider a 500 km long line for which the per kilometer line impedance
and admittance are given respectively by z = 0.1 + j0.5145 and y = j3.1734 10 6
mho.
Therefore
5.54024.406
2
9079
101734.3
5241.0
90101734.3
795241.0
101734.3
5145.01.0
666
j
j
y
z
ZC
and
6419.00618.05.846448.0
2
9079
500101734.35241.0 6
j
lyzl
We shall now use the following two formulas for evaluating the hyperbolic forms
sincoshcossinhsinh
sinsinhcoscoshcosh
jj
jj
Application of the above two equations results in the following values
5998.00495.0sinhand037.08025.0cosh jljl

8. 1.37
Therefore from (2.43) to (2.45) the ABCD parameters of the system can be written as
0015.01001.2
72.2404.43
037.08025.0
5
jC
jB
jDA
2.3.1 Equivalent- Representation of a Long Line
The -equivalent of a long transmission line is shown Fig. 2.6. In this the series
impedance is denoted by Z while the shunt admittance is denoted by Y . From (2.21) to
(2.23) the ABCD parameters are defined as
1
2
ZY
DA (2.46)
ZB (2.47)
mho1
4
ZY
YC (2.48)
Fig. 2.6 Equivalent representation of a long transmission line.
Comparing (2.44) with (2.47) we can write
l
l
Z
yzl
l
zll
y
z
lZZ C
sinhsinh
sinhsinh (2.49)
where Z = zl is the total impedance of the line. Again comparing (2.43) with (2.46) we get
1sinh
2
1
2
cosh lZ
YZY
l C (2.50)
Rearranging (2.50) we get
2
2tanh
2
2
2tanh
2
2tanh2tanh
1
sinh
1cosh1
2
l
lY
yzl
lyl
l
z
y
l
Zl
l
Z
Y
CC
(2.51)
where Y = yl is the total admittance of the line. Note that for small values of l, sinh l = l and
tanh ( l/2) = l/2. Therefore from (2.49) we get Z = Z and from (2.51) we get Y = Y . This

9. 1.38
implies that when the length of the line is small, the nominal- representation with lumped
parameters is fairly accurate. However the lumped parameter representation becomes
erroneous as the length of the line increases. The following example illustrates this.
Example 2.2: Consider the transmission line given in Example 2.1. The equivalent
system parameters for both lumped and distributed parameter representation are given in
Table 2.1 for three different line lengths. It can be seen that the error between the parameters
increases as the line length increases.
Table 2.1 Variation in equivalent parameters as the line length changes.
Length of
the line
(km)
Lumped parameters Distributed parameters
Z ( ) Y (mho) Z Y (mho)
100 52.41 79 3.17 10 4
90 52.27 79 3.17 10 4
89.98
250 131.032 79 7.93 10 4
90 128.81 79.2 8.0 10 4
89.9
500 262.064 79 1.58 10 3
90 244.61 79.8 1.64 10 3
89.6
2.4 CHARACTERIZATION OF A LONG LOSSLESS LINE
For a lossless line, the line resistance is assumed to be zero. The characteristic
impedance then becomes a pure real number and it is often referred to as the surge
impedance. The propagation constant becomes a pure imaginary number. Defining the
propagation constant as = j and replacing l by x we can rewrite (2.41) and (2.42) as
xIjZxVV RCR sincos (2.52)
xI
Z
x
jVI R
C
R cos
sin
(2.53)
The term surge impedance loading or SIL is often used to indicate the nominal
capacity of the line. The surge impedance is the ratio of voltage and current at any point
along an infinitely long line. The term SIL or natural power is a measure of power delivered
by a transmission line when terminated by surge impedance and is given by
C
n
Z
V
PSIL
2
0
(2.54)
where V0 is the rated voltage of the line.
At SIL ZC = VR/IR and hence from equations (2.52) and (2.53) we get
xj
R
x
R eVeVV (2.55)
xj
R
x
R eIeII (2.56)

10. 1.39
This implies that as the distance x changes, the magnitudes of the voltage and current in the
above equations do not change. The voltage then has a flat profile all along the line. Also as
ZC is real, V and I are in phase with each other all through out the line. The phase angle
difference between the sending end voltage and the receiving end voltage is then = l. This
is shown in Fig. 2.7.
Fig. 2.7 Voltage-current relationship in naturally loaded line.
2.4.1 Voltage and Current Characteristics of an SMIB System
For the analysis presented below we assume that the magnitudes of the voltages at the
two ends are the same. The sending and receiving voltages are given by
SS VV and 
0RR VV (2.57)
where is angle between the sources and is usually called the load angle. As the total length
of the line is l, we replace x by l to obtain the sending end voltage from (2.39) as
sincos
22
RCR
jRCRjRCR
SS IjZVe
IZV
e
IZV
VV (2.58)
Solving the above equation we get
sin
cos
C
RS
R
jZ
VV
I (2.59)
Substituting (2.59) in (2.52), the voltage equation at a point in the transmission line that is at
a distance x from the receiving end is obtained as
sin
sinsin
sin
sin
cos
cos
xVxV
xjZ
jZ
VV
xVV RS
C
C
RS
R (2.60)
In a similar way the current at that point is given by
sin
coscos xVxV
Z
j
I RS
C
(2.61)

11. 1.40
Example 2.3: Consider a 500 km long line given in Example 2.1. Neglect the line
resistance such that the line impedance is z = j0.5145 per kilometer. The line admittance
remains the same as that given in Example 2.1. Then
6524.402
101734.3
5145.0
6
j
j
y
z
ZC
and
rad/km0013.0101734.35145.0 6
jjjyz
Therefore = l = 0.6380 rad. It is assumed that the magnitude of the sending and receiving
end voltages are equal to 1.0 per unit with the line being unloaded, i.e., VS = VR = 1 0 per
unit. The voltage and current profiles of the line for this condition is shown in Fig. 2.8. The
maximum voltage is 1.0533 per unit, while the current varies between –0.3308 per unit to
0.3308 per unit. Note that 1 per unit current is equal to 1/ZC.
Fig. 2.8 Voltage and current profile over a transmission line.
When the system is unloaded, the receiving end current is zero (IR = 0). Therefore we
can rewrite (2.58) as
cosRSS VVV (2.62)
Substituting the above equation in (2.52) and (2.53) we get the voltage and current for the
unloaded system as
x
V
V S
cos
cos
(2.63)

12. 1.41
x
V
Z
j
I S
C
sin
cos
(2.64)
Example 2.4: Consider the system given in Example 2.3. It is assumed that the system
is unloaded with VS = VR = 1 0 per unit. The voltage and current profiles for the unloaded
system is shown in Fig. 2.9. The maximum voltage of 1.2457 per unit occurs at the receiving
end while the maximum current of 0.7428 per unit is at the sending end. The current falls
monotonically from the sending end and voltage rises monotonically to the receiving end.
This rise in voltage under unloaded or lightly loaded condition is called Ferranti effect.
Fig. 2.9 Voltage and current profile over an unloaded transmission line.
2.4.2 Mid Point Voltage and Current of Loaded Lines
The mid point voltage of a transmission line is of significance for the reactive
compensation of transmission lines. To obtain an expression of the mid point voltage, let us
assume that the line is loaded (i.e., the load angle is not equal to zero). At the mid point of
the line we have x = l/2 such that x = /2. Let us denote the midpoint voltage by VM. Let us
also assume that the line is symmetric, i.e., VS = VR = V. We can then rewrite equation
(2.60) to obtain
sin
2sinVV
VM
Again noting that
22cos2
cos1
sin
tancos22
1sincos
1
VV
jVVV

13. 1.42
we obtain the following expression of the mid point voltage
22cos
2cosV
VM (2.65)
The mid point current is similarly given by
22sin
2sin
sin
2cos
CC
M
Z
VVV
Z
j
I (2.66)
The phase angle of the mid point voltage is half the load angle always. Also the mid point
voltage and current are in phase, i.e., the power factor at this point is unity. The variation in
the magnitude of voltage with changes in load angle is maximum at the mid point. The
voltage at this point decreases with the increase in . Also as the power through a lossless line
is constant through out its length and the mid point power factor is unity, the mid point
current increases with an increase in .
Example 2.5: Consider the transmission line discussed in Example 2.4. Assuming the
magnitudes of both sending and receiving end voltages to be 1.0 per unit, we can compute the
magnitude of the mid point voltage as the load angle ( ) changes. This is given in Table 2.2.
The variation in voltage with is shown in Fig. 2.10.
Table 2.2 Changes in the mid point voltage magnitude with load angle
in degree VM in per unit
20 1.0373
25 1.0283
30 1.0174
It is of some interest to find the Thevenin equivalent of the transmission line looking
from the mid point. It is needless to say that the Thevenin voltage will be the same as the mid
point voltage. To determine the Thevenin impedance we first find the short circuit current at
the mid point terminals. This is computed through superposition principle, as the short circuit
current will flow from both the sources connected at the two ends. From (2.52) we compute
the short circuit current due to source VS (= V ) as
2sin
1
C
SC
jZ
V
I (2.67)
Similarly the short circuit current due to the source VR (= V) is
2sin
2
C
SC
jZ
V
I (2.68)
Thus we have

14. 1.43
Fig. 2.10 Variation in voltage profile for a loaded line
22sin
2cos2
2sin
21
CC
SCSCSC
jZ
V
jZ
VV
III (2.69)
The Thevenin impedance is then given by
2
tan
2
C
SC
M
THTH
Z
j
I
V
jXZ (2.70)
2.4.3 Power in a Lossless Line
The power flow through a lossless line can be given by the mid point voltage and
current equations given in (2.66) and (2.67). Since the power factor at this point is unity, real
power over the line is given by
sin
sin
2
C
MMe
Z
V
IVP (2.71)
If V = V0, the rated voltage we can rewrite the above expression in terms of the natural power
as
sin
sin
n
e
P
P (2.72)
For a short transmission line we have
Xllc
c
lZZ CC sin (2.73)

15. 1.44
where X is the total reactance of the line. Equation (2.71) then can be modified to obtain the
well known power transfer relation for the short line approximation as
sin
2
X
V
Pe (2.74)
In general it is not necessary for the magnitudes of the sending and receiving end
voltages to be same. The power transfer relation given in (2.72) will not be valid in that case.
To derive a general expression for power transfer, we assume
SS VV and 
0RR VV
If the receiving end real and reactive powers are denoted by PR and QR respectively, we can
write from (2.52)
sincossincos
R
RR
CRSS
V
jQP
jZVjVV
Equating real and imaginary parts of the above equation we get
sincoscos
R
RC
RS
V
QZ
VV (2.75)
and
sinsin
R
RC
S
V
PZ
V (2.76)
Rearranging (2.76) we get the power flow equation for a losslees line as
sin
sinC
RS
RSe
Z
VV
PPP (2.77)
To derive expressions for the reactive powers, we rearrange (2.75) to obtain the
reactive power delivered to the receiving end as
sin
coscos
2
C
RRS
R
Z
VVV
Q (2.78)
Again from equation (2.61) we can write
sin
cos RS
C
S
VV
Z
j
I (2.79)
The sending end apparent power is then given by

16. 1.45
sinsin
cos
sin
cos
2
C
RS
C
SRS
C
SSSSS
Z
VV
j
Z
V
j
VV
Z
j
VIVjQP
Equating the imaginary parts of the above equation we get the following expression for the
reactive generated by the source
sin
coscos
2
C
RSS
S
Z
VVV
Q (2.80)
The reactive power absorbed by the line is then
sin
cos2cos
22
C
RSRS
RSL
Z
VVVV
QQQ (2.81)
It is important to note that if the magnitude of the voltage at the two ends is equal, i.e.,
VS = VR = V, the reactive powers at the two ends become negative of each other, i.e.,
QS = QR. The net reactive power absorbed by the line then becomes twice the sending end
reactive power, i.e., QL = 2QS. Furthermore, since cos 1for small values of , the reactive
powers at the two ends for a short transmission line are given by
R
C
S Q
X
V
Z
VV
Q cos1
sin
coscos 222
(2.82)
The reactive power absorbed by the line under this condition is given by
cos1
2 2
X
V
QL (2.83)
Example 2.5: Consider a short, lossless transmission line with a line reactance of 0.5
per unit. We assume that the magnitudes of both sending and receiving end voltages to be 1.0
per unit. The real power transfer over the line and reactive power consumed by the line are
shown in Fig. 2.11. The maximum real power is 2.0 per unit and it occurs for = 90 . Also
the maximum reactive power consumed by the line occurs at = 180 and it has a value of 8
per unit.

17. 1.46
Fig. 2.11 Real power flow and reactive power consumed by a transmission line.