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

VA (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

VB (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

IC mho (2.5)

1.31

0RVR

S

I

ID (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

0 and ,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.

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

IVY

VY

IIIIII

22

3121

(2.10)

Again

RR

RRRRs

ZIVYZ

VIY

VZVZIV

12

22

(2.11)

Substituting (2.11) in (2.10) we get

RR

RRRRs

IYZ

VYZ

Y

IVY

ZIVYZY

I

12

14

21

22 (2.12)

Therefore from (2.11) and (2.12) we get the following ABCD parameters of the

nominal- representation

12

YZDA (2.13)

ZB (2.14)

mho 14

YZYC (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.

22 Z

VVYV

Z

VV RMM

MS

Rearranging the above equation can be written as

RSM VVYZ

V4

2 (2.16)

Now the receiving end current is given by

2Z

VVI RM

R (2.17)

Substituting the value of VM from (2.16) in (2.17) and rearranging we get

RRs IYZ

ZVYZ

V 14

12

(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 IYZ

YVI 12

(2.20)

Then the ABCD parameters of the T-network are

12

YZDA (2.21)

14

YZZB (2.22)

mho YC (2.23)

2.3 LONG LINE MODEL

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

Izx

VxIzV (2.25)

Again as x 0, from (2.25) we get

Izdx

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

Vydx

dI (2.28)

Taking derivative with respect to x of both sides of (2.26) we get

dx

dIz

dx

dV

dx

d

Substitution of (2.28) in the above equation results

1.35

02

2

yzVdx

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

yzxyzxeyzAeyzA

dx

dV21 (2.31)

Combining (2.26) with (2.31) we have

yzxyzxe

yz

Ae

yz

A

dx

dV

zI 211

(2.32)

Let us define the following two quantities

impedancestic characteri thecalled is which y

zZC (2.33)

constant npropagatio thecalled is which yz (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

eZ

Ae

Z

AI 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

RZ

A

Z

AI 21 (2.38)

Solving (2.37) and (2.38) we get the following values for A1 and A2.

2 and

221

RCRRCR IZVA

IZVA

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

1.36

lRCRlRCRS e

IZVe

IZVV

22 (2.39)

lRCRlRCRS e

IZVe

IZVI

22 (2.40)

Noting that

lee

lee llll

cosh2

and sinh2

We can rewrite (2.39) and (2.40) as

lIZlVV RCRS sinhcosh (2.41)

lIZ

lVI R

C

RS coshsinh

(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

lC

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 106 mho.

Therefore

5.54024.406

2

9079

101734.3

5241.0

90101734.3

795241.0

101734.3

5145.01.0666j

j

y

zZC

and

6419.00618.05.846448.0

2

9079500101734.35241.0 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

5998.00495.0sinh and 037.08025.0cosh jljl

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

12

ZYDA (2.46)

ZB (2.47)

mho 14

ZYYC (2.48)

Fig. 2.6 Equivalent representation of a long transmission line.

Comparing (2.44) with (2.47) we can write

l

lZ

yzl

lzll

y

zlZZ C

sinhsinhsinhsinh (2.49)

where Z = zl is the total impedance of the line. Again comparing (2.43) with (2.46) we get

1sinh2

12

cosh lZYZY

l C (2.50)

Rearranging (2.50) we get

2

2tanh

2

2

2tanh

22tanh2tanh

1

sinh

1cosh1

2

l

lY

yzl

lyll

z

yl

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

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 3.17 104

90 52.27 79 3.17 104

89.98

250 131.032 79 7.93 104

90 128.81 79.2 8.0 104

89.9

500 262.064 79 1.58 103

90 244.61 79.8 1.64 103

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)

xIZ

xjVI R

C

R cossin

(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

nZ

VPSIL

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)

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

sincos22

RCR

jRCRjRCR

SS IjZVeIZV

eIZV

VV (2.58)

Solving the above equation we get

sin

cos

C

RS

RjZ

VVI (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

sinsinsin

sin

coscos

xVxVxjZ

jZ

VVxVV RS

C

C

RS

R (2.60)

In a similar way the current at that point is given by

sin

coscos xVxV

Z

jI RS

C

(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 per kilometer. The line admittance

remains the same as that given in Example 2.1. Then

6524.402101734.3

5145.06j

j

y

zZC

and

rad/km 0013.0101734.35145.0 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

xV

V S coscos

(2.63)

1.41

xV

Z

jI S

C

sincos

(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

2sinVVVM

Again noting that

22cos2cos1

sintancos22

1sincos

1 VV

jVVV

1.42

we obtain the following expression of the mid point voltage

22cos

2cosVVM (2.65)

The mid point current is similarly given by

22sin

2sin

sin

2cos

CC

MZ

VVV

Z

jI (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

2sin1

C

SCjZ

VI (2.67)

Similarly the short circuit current due to the source VR (= V) is

2sin2

C

SCjZ

VI (2.68)

Thus we have

1.43

Fig. 2.10 Variation in voltage profile for a loaded line

22sin

2cos2

2sin21

CC

SCSCSCjZ

V

jZ

VVIII (2.69)

The Thevenin impedance is then given by

2tan

2

C

SC

M

THTH

Zj

I

VjXZ (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

sinsin

2

C

MMeZ

VIVP (2.71)

If V = V0, the rated voltage we can rewrite the above expression in terms of the natural power

as

sinsin

n

e

PP (2.72)

For a short transmission line we have

Xllcc

lZZ CC sin (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

sin2

X

VPe (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)

sincossincosR

RR

CRSSV

jQPjZVjVV

Equating real and imaginary parts of the above equation we get

sincoscosR

RC

RSV

QZVV (2.75)

and

sinsinR

RC

SV

PZV (2.76)

Rearranging (2.76) we get the power flow equation for a losslees line as

sinsinC

RS

RSeZ

VVPPP (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

coscos2

C

RRS

RZ

VVVQ (2.78)

Again from equation (2.61) we can write

sin

cos RS

C

S

VV

Z

jI (2.79)

The sending end apparent power is then given by

1.45

sinsin

cos

sin

cos2

C

RS

C

SRS

C

SSSSSZ

VVj

Z

Vj

VV

Z

jVIVjQP

Equating the imaginary parts of the above equation we get the following expression for the

reactive generated by the source

sin

coscos2

C

RSS

SZ

VVVQ (2.80)

The reactive power absorbed by the line is then

sin

cos2cos22

C

RSRS

RSLZ

VVVVQQQ (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 QX

V

Z

VVQ cos1

sin

coscos 222

(2.82)

The reactive power absorbed by the line under this condition is given by

cos12 2

X

VQL (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.

1.46

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