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
VS = AVR + BI R
I S = CVR + DI R
(2.1)
(2.2)
From (2.1) we see that
A=
VS
VR
(2.3)
I R =0
This implies that A is the ratio of sending end voltage to the open circuit receiving end
voltage. This quantity is dimension less. Similarly,
B=
VS
IR
Ω
(2.4)
VR = 0
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
C=
IS
VR
mho
I R =0
(2.5)
1.31
D=
IS
IR V
(2.6)
R
=0
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
VS = VR + ZI R
IS = IR
(2.7)
(2.8)
Therefore the ABCD parameters are given by
A = D = 1, B = Z Ω and C = 0
(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-π and nominal-T depending on the nature of the network.
These two are discussed below.
2.2.1 Nominal-π 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-π
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 π-network for
load flow studies.
1.32
Fig. 2.3 Nominal-π 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
I s = I1 + I 2 = I1 + I 3 + I R
=
Y
Y
Vs + VR + I R
2
2
(2.10)
Again
 Y

Vs = ZI 2 + VR = Z VR + I R  + VR
 2

 YZ 
=
+ 1VR + ZI R
 2

(2.11)
Substituting (2.11) in (2.10) we get
Is =
 Y
Y  YZ

+ 1VR + ZI R  + VR + I R


2  2

 2
 YZ

 YZ

= Y
+ 1VR + 
+ 1 I R
 4

 2

(2.12)
Therefore from (2.11) and (2.12) we get the following ABCD parameters of the
nominal-π representation
 YZ 
+ 1
A= D=
 2

B=ZΩ
 YZ 
C = Y
+ 1 mho
 4

(2.13)
(2.14)
(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
1.33
Fig. 2.4 Nominal-T representation.
VS − VM
V − VR
= YVM + M
Z 2
Z 2
Rearranging the above equation can be written as
VM =
2
(VS + VR )
YZ + 4
(2.16)
Now the receiving end current is given by
IR =
VM − VR
Z 2
(2.17)
Substituting the value of VM from (2.16) in (2.17) and rearranging we get
 YZ 
 YZ 
Vs = 
+ 1 I R
+ 1VR + Z 

 4

 2
(2.18)
Furthermore the sending end current is
I S = YVM + I R
(2.19)
Then substituting the value of VM from (2.16) in (2.19) and solving
 YZ 
I R = YVR + 
+ 1 I R

 2
(2.20)
Then the ABCD parameters of the T-network are
 YZ 
A= D=
+ 1

 2
YZ


B = Z
+ 1 Ω
4


C = Y mho
(2.21)
(2.22)
(2.23)
1.34
2.3 LONG LINE MODEL
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
Z = z × l and Y = y × l
(2.24)
Fig. 2.5 Long transmission line representation.
From the circuit of Fig. 2.5 we see that
∆V = Iz ∆x ⇒
∆V
= Iz
∆x
(2.25)
Again as ∆x → 0, from (2.25) we get
dV
= Iz
dx
(2.26)
Now for the current through the strip, applying KCL we get
∆I = (V + ∆V ) y ∆x = Vy ∆x + ∆Vy ∆x
(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
dI
= Vy
dx
(2.28)
Taking derivative with respect to x of both sides of (2.26) we get
d  dV

dx  dx
dI

=z
dx

1.35
Substitution of (2.28) in the above equation results
d 2V
− yzV = 0
dx 2
(2.29)
The roots of the above equation are located at ±√(yz). Hence the solution of (2.29) is of the
form
V = A1e x
yz
+ A2e − x
yz
(2.30)
Taking derivative of (2.30) with respect to x we get
dV
= A1 yz e x
dx
yz
− A2 yz e − x
yz
(2.31)
Combining (2.26) with (2.31) we have
1  dV 
I= 
=
z  dx 
A1 x
e
z y
yz
−
A2 − x
e
z y
yz
(2.32)
Let us define the following two quantities
ZC =
z
Ω which is called the characteristic impedance
y
γ = yz which is called the propagation constant
(2.33)
(2.34)
Then (2.30) and (2.32) can be written in terms of the characteristic impedance and
propagation constant as
V = A1eγx + A2e −γx
A
A
I = 1 eγx − 2 e −γx
ZC
ZC
(2.35)
(2.36)
Let us assume that x = 0. Then V = VR and I = IR. From (2.35) and (2.36) we then get
VR = A1 + A2
A A
IR = 1 − 2
ZC ZC
(2.37)
(2.38)
Solving (2.37) and (2.38) we get the following values for A1 and A2.
A1 =
VR + Z C I R
V − ZC I R
and A2 = R
2
2
1.36
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
VR + Z C I R γl VR − Z C I R −γl
e +
e
2
2
V Z + I R γl VR Z C − I R −γl
IS = R C
e −
e
2
2
VS =
(2.39)
(2.40)
Noting that
eγl − e −γl
eγl + e −γl
= sinh γl and
= cosh γl
2
2
We can rewrite (2.39) and (2.40) as
VS = VR cosh γl + Z C I R sinh γl
sinh γl
I S = VR
+ I R cosh γl
ZC
(2.41)
(2.42)
The ABCD parameters of the long transmission line can then be written as
A = D = cosh γl
B = Z C sinh γl Ω
sinh γl
mho
C=
ZC
(2.43)
(2.44)
(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 Ω and y = j3.1734 × 10−6 mho.
Therefore
ZC =
z
=
y
0.1 + j 0.5145
0.5241∠79°
0.5241
 79° − 90° 
=
=
∠

−6
−6
−6
j 3.1734 × 10
3.1734 × 10 ∠90°
3.1734 × 10
2


= 406.4024∠ − 5.5° Ω
and
 79° + 90° 

2


γl = yz × l = 0.5241 × 3.1734 × 10− 6 × 500∠
= 0.6448∠84.5° = 0.0618 + j 0.6419
We shall now use the following two formulas for evaluating the hyperbolic forms
cosh (α + jβ ) = cosh α cos β + j sinh α sin β
sinh (α + jβ ) = sinh α cos β + j cosh α sin β
Application of the above two equations results in the following values
1.37
cosh γl = 0.8025 + j 0.037 and sinh γl = 0.0495 + j 0.5998
Therefore from (2.43) to (2.45) the ABCD parameters of the system can be written as
A = D = 0.8025 + j 0.037
B = 43.4 + j 240.72 Ω
C = −2.01 × 10− 5 + j 0.0015
∆∆∆
2.3.1 Equivalent-π Representation of a Long Line
The π-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
 Y′Z′ 
A= D=
+ 1
 2

B = Z′Ω
 Y′Z′ 
C = Y ′
+ 1 mho
 4

(2.46)
(2.47)
(2.48)
Fig. 2.6 Equivalent π representation of a long transmission line.
Comparing (2.44) with (2.47) we can write
Z ′ = Z C sinh γl =
z
sinh γl
sinh γl
Ω
sinh γl = zl
=Z
y
γl
l yz
(2.49)
where Z = zl is the total impedance of the line. Again comparing (2.43) with (2.46) we get
cosh γl =
Y′Z′
Y′
+1 =
Z C sinh γl + 1
2
2
(2.50)
Rearranging (2.50) we get
1 cosh γl − 1 1
Y′
=
=
tanh (γl 2 ) =
2 Z C sinh γl
ZC
Y tanh (γl 2 )
=
2 (γl 2)
y
yl tanh (γl 2 )
tanh (γl 2) =
2 (l 2) yz
z
(2.51)
1.38
where Y = yl is the total admittance of the line. Note that for small values of l, sinh γl = γl and
tanh (γl/2) = γl/2. Therefore from (2.49) we get Z = Z′ and from (2.51) we get Y = Y′. This
implies that when the length of the line is small, the nominal-π 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 (Ω)
Y (mho)
Z′ Ω
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 γ = jβ and replacing l by x we can rewrite (2.41) and (2.42) as
V = VR cos βx + jZ C I R sin βx
sin βx
I = jVR
+ I R cos βx
ZC
(2.52)
(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
SIL = Pn =
V02
ZC
(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
V = V R e γx = V R e − jβx
(2.55)
1.39
I = I R eγx = I R e − jβx
(2.56)
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 θ = βl. 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
VS = VS ∠δ and VR = VR ∠0o
(2.57)
where δ 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
VS = VS ∠δ =
VR + Z C I R jθ VR − Z C I R − jθ
e +
e = VR cosθ + jZ C I R sin θ
2
2
(2.58)
Solving the above equation we get
IR =
VS ∠δ − VR cosθ
jZ C sin θ
(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
V = VR cos βx +
VS ∠δ − VR cosθ
V ∠δ sin βx + VR sin (θ − βx )
( jZ C sin βx ) = S
jZ C sin θ
sin θ
(2.60)
In a similar way the current at that point is given by
I=
− j  VS ∠δ cos βx − VR cos(θ − βx ) 


ZC 
sin θ

(2.61)
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 Ω per kilometer. The line admittance
remains the same as that given in Example 2.1. Then
ZC =
z
=
y
j 0.5145
= 402.6524 Ω
j 3.1734 × 10− 6
and
γ = yz = jβ = j 0.5145 × 3.1734 × 10−6 = j 0.0013 rad/km
Therefore θ = βl = 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
VS = VS ∠δ = VR cosθ
(2.62)
Substituting the above equation in (2.52) and (2.53) we get the voltage and current for the
unloaded system as
V=
VS ∠δ
cos βx
cosθ
(2.63)
1.41
I=
j VS ∠δ
sin βx
Z C cosθ
(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 δ is not equal to zero). At the mid point of
the line we have x = l/2 such that βx = θ/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
VM =
(V∠δ + V )sin (θ 2)
sin θ
Again noting that
V∠δ + V = V (cos δ + j sin δ + 1)
 sin δ 
= V 2 + 2 cos δ tan −1 
 = 2V cos(δ 2)∠(δ 2)
 1 + cos δ 
1.42
we obtain the following expression of the mid point voltage
VM =
V cos(δ 2 ) δ
∠
2
cos(θ 2 )
( )
(2.65)
The mid point current is similarly given by
IM =
− j  (V∠δ − V ) cos(θ 2 )  V sin(δ 2 ) δ
 = Z sin(θ 2 ) ∠ 2
sin θ
Z C 

C
( )
(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 δ. 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 δ.
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 (δ) changes. This is given in Table 2.2.
The variation in voltage with δ is shown in Fig. 2.10.
Table 2.2 Changes in the mid point voltage magnitude with load angle
δ in degree
20
25
30
VM in per unit
1.0373
1.0283
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∠δ) as
I SC1 =
V∠δ
jZ C sin (θ 2 )
(2.67)
Similarly the short circuit current due to the source VR (= V) is
I SC 2 =
Thus we have
V
jZ C sin(θ 2 )
(2.68)
1.43
Fig. 2.10 Variation in voltage profile for a loaded line
I SC = I SC1 + I SC 2 =
2V cos(δ 2 ) δ
V∠δ + V
=
∠
2
jZ C sin(θ 2 ) jZ C sin(θ 2 )
( )
(2.69)
The Thevenin impedance is then given by
Z TH = jX TH =
( )
Z
VM
= j C tan θ
2
2
I SC
(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
Pe = V M I M∗ =
V2
sin δ
Z C sin θ
(2.71)
If V = V0, the rated voltage we can rewrite the above expression in terms of the natural power
as
Pe =
Pn
sin δ
sin θ
(2.72)
For a short transmission line we have
(
)
Z C sin θ ≅ Z Cθ =  l  ω lc τ = ω lτ = X
 c
(2.73)
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
Pe =
V2
sin δ
X
(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
V S = V S ∠δ and V R = V R ∠0 o
If the receiving end real and reactive powers are denoted by PR and QR respectively, we can
write from (2.52)
V S = V S (cos δ + j sin δ ) = V R cos θ + jZ C
PR − jQ R
sin θ
VR
Equating real and imaginary parts of the above equation we get
V S cos δ = V R cos θ +
Z CQR
sin θ
VR
(2.75)
and
V S sin δ =
Z C PR
sin θ
VR
(2.76)
Rearranging (2.76) we get the power flow equation for a losslees line as
Pe = PS = PR =
VS VR
Z C sin θ
sin δ
(2.77)
To derive expressions for the reactive powers, we rearrange (2.75) to obtain the
reactive power delivered to the receiving end as
2
QR =
V S V R cos δ − V R cos θ
Z C sin θ
(2.78)
Again from equation (2.61) we can write
IS =
− j  V S cos θ∠δ − V R 


sin θ
ZC 

The sending end apparent power is then given by
(2.79)
1.45
2
V S cos θ
V S V R ∠δ
j  V S cos θ∠ − δ − V R 
PS + jQ S = V S I = V S ∠δ
− j

= j
ZC 
sin θ
Z C sin θ
Z C sin θ

∗
S
Equating the imaginary parts of the above equation we get the following expression for the
reactive generated by the source
2
QS =
V S cos θ − V S V R cos δ
(2.80)
Z C sin θ
The reactive power absorbed by the line is then
QL = QS − QR
(V
=
2
S
+ VR
2
)cosθ − 2 V
S
V R cos δ
Z C sin θ
(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 θ ≈ 1for small values of θ, the reactive
powers at the two ends for a short transmission line are given by
QS =
V 2 cosθ − V 2 cos δ V 2
(1 − cos δ ) = −QR
≈
Z C sin θ
X
(2.82)
The reactive power absorbed by the line under this condition is given by
QL =
2V 2
(1 − cos δ )
X
(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 δ = 90°. Also
the maximum reactive power consumed by the line occurs at δ = 180° and it has a value of 8
per unit.
∆∆∆
1.46
Fig. 2.11 Real power flow and reactive power consumed by a transmission line.