Faculty of electrical and computer engineering

POWER SYSTEM I
EEng-3132

Characteristics and Performance of Transmission

Lines

Compiled by: Biniyam Z.

1 4/9/2019
 Chapter Five
Characteristic and performance of power transmission
lines
Outline
 Introduction
 ABCD constants
 Representation of transmission lines
 Short transmission line
 medium transmission line and
 Long transmission line

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 INTRODUCTION
 This chapter deals with the representation and performance of transmission
lines under normal operating conditions
 Transmission lines are represented by an equivalent model with appropriate
circuit parameters on a per-phase basis
 The terminal voltages are expressed from one line to neutral, the current
for one phase and, thus, the three-phase system is reduced to an equivalent
single phase system
 This model used to calculate voltages, currents, and power flows depends
on the length of the line

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ABCD Constants
 It is convenient to represent a transmission line by the two port
network, where in the sending-end voltage Vs and current Is are related
to the receiving-end voltage VR and IR through A, B, C and D
parameters (ABCD constants) as
VS = AVR + BI R Volts
IS = CVR + DI R Amp

or, in matrix form

Figure: Two-port representation ofTx lines

 VS   A B   VR 
 I  = C D   I R 
(5.1)
 S  
 A, B, C and D are the parameters that depend on the transmission line
constants R, L, C and G
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 The ABCD parameters are, in general complex numbers
 A and D are dimensionless
 B has units of Ohms and
 C has units of Siemens
 Also the following identity holds for ABCD constants
AD  BC  1 (5.2)

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SHORT TRANSMISSION LINES
 Capacitors may be ignored without much error if the lines are less than
80 km long or if the voltage is not over 66 kV
 The short line model is obtained by multiplying the series impedance
per unit length by line length
Z   r  j L  l
 R  jX (5.3)
where: r - per phase resistance per unit length
L - per phase inducatnce per unit length
l - the line length
 The short line model on per-phase basis is shown in figure 5.2 below

Figure 5.2 Short line model

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 If a three phase load with apparent power S R (3 ) is connected at the end
of the transmission line, the receiving end current is obtained by:
S R* (3 )
IR  *
(5.4)
3V R

The phase voltage at the sending end is

VS  VR  ZI R (5.5)
since the shunt capacitance is neglected, the sending end and the receiving end
currents are equal
IS  IR (5.6)

 The relationship between sending-end, receiving-end voltages and

currents can be written as in the form of equation 5.1
A  1, B  Z , C  0 and D  1
VS  1 Z  VR 
 I   0 1   I  (5.7)
 S   R
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 The phasor diagram for the short line is shown in figure 5.3

Figure 5.3 Phasor diagram for short transmission line

Voltage Regulation and Efficiency

 Voltage regulation of the transmission line may be defined as the
percentage change in voltage at the receiving end of the line ( expressed
as percent of full load voltage) in going from no-load to full-load
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 VR ( NL )  VR ( FL )
Percent VR  x100 (5.8)
VR ( FL )
At no load, I R  0, and from eqn. (5.1)
VS
VR ( NL )  (5.9)
A

 For a short line, A  1 ( VR ( NL )  VS ) and VR ( FL )  VR

VS  VR
 Percent voltage regulation  x100 (5.10)
VR

 Voltage regulation is a measure of line voltage drop and depends on the

load power factor
 Voltage regulation will be poorer at low lagging power factor loads.
With capacitive loads (leading P.f.) regulation may become negative

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 Once the sending end voltage is calculated the sending end power is
obtained by
S S (3 )  3VS I S* (5.11)

The total line loss is then given by

S L (3 )  S S (3 )  S R (3 ) (5.12)

and the transmission line efficiency is given by

PR (3 )
 (5.13)
PS (3 )

where PR (3 ) and PS (3 ) are the total real power at the receiving end and sending end
of the line, respectively

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 Example
 A 220 kV, three phase transmission line is 40 km long. The resistance
per phase is 0.15 Ω per km and the inductance per phase is 1.3263
mH per km. The shunt capacitance is negligible. Use the short line
model to find the voltage and power at the sending end and the
voltage regulation and efficiency when the line is supplying a three
phase load of
a). 381 MVA at 0.8 power factor lagging at 220 kV
b). 381 MVA at 0.8 power factor leading at 220 kV
Solution
(a).The series impedance per phase is
Z  (r  j L)l
 (0.15  j 2 x60 x1.3263 x103 )40
 6  j 20 

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 The receiving end voltage per phase is
22000
VR   12700 kV
3
The apparent power is
SR (3 ) =381cos -1 0.8 =38136.870
 304.8  j 228.6 MVA
From eqn. (5.5), the sending end voltage is
VS  VR  ZI R
 12700   6  j 20  1000  36.87 0 103 
 144.334.930 kV
The sending end line-to-line voltage magnitude is
VS ( L  L )  3 VS  250 kV

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The sending end power is
S S (3 )  3VS I S*  3x144.334.930 x100036.87 0 x10 3
 322.8 MW  j 288.6 MVar
 433 41.80 MVA

Voltage regulation is
250 - 220
Percent VR  x100  13.6%
220

Transmission line efficieny is

PR (3 ) 304.8
  x100  94.4%
PS (3 ) 322.8

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(b) The current for 381 MVA with o.8 leading power factor is
S R* (3 ) 38136.87 0 x103
IR    100036.87 0
3VR* 3 x1270 0

The sending end voltage is

VS  VR  ZI R  12700   6  j 20  100036.87 0 10 3 
 121.399.290 kV
The sending end line-to-line voltage magnitude is
VS ( L  L )  3VS  210.26 kV
The sending end power is
S S (3 )  3VS I S*  3 x121.399.290 x1000  36.87 0 x10 3
 322.8 MW  j168.6 Mvar
 364.18  27.580 MVA
Voltage regulation is
210.26  220
Percent VR  x100  4.43%
220
Transmission line efficiency is
PR (3 ) 304.8
  x100  94.4%
14 PS (3 ) 322.8 4/9/2019
 MEDIUM TRANSMISSION LINE
 For the lines more than 80 km long and below 250 km in length are
treated as medium lines, and the line charging current becomes
appreciable and the shunt capacitance must be considered
 For medium length lines, half of the shunt capacitance may be
considered to be lumped at each end of the line. This is referred to as
the nominal π model and is shown in figure 5.4.

Figure 5.4 nominal π model for medium length line

 The sending end voltage and current for the nominal π model are
obtained as follows:

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 From KCL the current in the series impedance designated by IL is
Y
IL  IR VR (5.14)
2
From KVL the sending end voltage is
VS  VR  ZI L (5.15)
substituting for I L from (5.14), we obtain
 ZY 
VS   1   VR  ZI R (5.16)
 2 
The sending end current is
Y
IS  IL  VS (5.17)
2
Substituting for I L and VS
 ZY   ZY 
I S  Y 1   VR  1   IR (5.18)
 4   2 
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 Eqns. (5.16) and (5.18) can be written in matrix form
 ZY  
 1   Z
VS   2   V 
I     
R
 (5.19)
 S  Y  1  ZY   ZY    I R 
    1+ 
  4   2 
Therefore, the ABCD constants for the nominal  model are given by
 ZY 
A  1  , BZ
 2 
 ZY   ZY 
C  Y 1   , D   1  
 4   2 
Solving (5.1), the receiving end quantity can be expresses in terms
of the sending end quantities by
VR   D  B  VS 
 I    C A   I  (5.20)
 R   S 
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 Example 1
 Determine the efficiency and regulation of a 3-phase, 150 km long, 50
Hz transmission line delivering 20 MW at a power factor of 0.8 lagging
and 66 kV to a balanced load. Resistance of the line is 0.075 Ω/km,
inductance of the line and capacitance per phase is 1.117 mH/km and
9.933 nF/km respectively. Use nominal π method.
 Solution
R  0.075 x150  11.25
Inductive reactance per phase is
 since, L  1.117 mH per phase/per km
 X  2 x x50 x150 x0.1.117 x10 3  52.62
and the capacitance per phase per unit length is given as
C  9.933 nF/km
The capacitance per phase is
18
C  9.933 nF/km x 150 =1.49  F 4/9/2019
 Y  jC  j 2 x50 x1.49 x106 mho
 j 468.1x106 mho
Y
  j 234 x10 6 mho
2
Z  11.25  j 52.62   53.80977.90 
Now

 cos 
*
P
S*R(3 ) 20MW-36.87 0
IR = = =
3V*R 3V*R
3x0.8x
66
kV 00
3
= 218.7  36.87 0 Amp
 I R  218.7 Amp. at 0.8 pf lagging

Receiving end phase voltage

66
VR   38.104 kV
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From eqn. (5.19), we have
 ZY 
VS  1   VR  ZI R
 2 
I R  218.7  36.870
VR  38.10400

 53.809 x 234 x106   77.90  900 

ZY
2
 0.01259167.90  0.0123  j 0.00264

 VS  1  0.0123  j 0.00264  x38.10400 

 53.80977.9 x 218.7  36.87 
0 0

1000
  0.9877  j 0.00264  x38.10400  11.7641.030
 37.63  j 0.1  8.87  j 7.72
  46.5  j 7.82   47.159.540 kV
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 VS ( L  L )  3 x 47.159.540 kV  81.669.540 kV
VS
 VR 81.66
 66
A
Voltage regulation VR   0.9877  25.26%
VR 66
Power loss per phase  I R   218.7  x11.25 x10 6 MW  0.538 MW
2 2

Per phase receiving end power

20
PR  MW
3
Per phase sending end power
20
PS   0.538  7.204 MW
3
Transmission efficiency
20
= 3 x100  92.54%
21 7.204 4/9/2019
Example 2
 Determine the voltage, current and power factor at the sending end of a 3
phase, 50 HZ, overhead transmission line 160 km long delivering a load of
100 MVA at 0.8 pf lagging and 132 kV to a balanced load. Resistance per
km is 0.16 Ω, inductance per km is 1.2 mH and capacitance per km per
conductor is 0.0082 μF . Use nominal π method.
 Solution
R  0.16 x160  25.6 
X  1.2 x103 x 2 x50 x160  60.3
Y  j 2 x50 x 0.0082 x106 x160  j 4.12 x104 mho
Z  R  jX  25.6  j 60.3  65.5167 0 
 ZY 
VS   1   VR  ZI R
 2 
Phase voltage at the receiving end,
132
VR  00 kV  76.2100 kV
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 Receiving end current
100 x106
IR  3
Amp  437.38 Amp
3 x132 x10
Load has lagging power factor of 0.8, i.e.,  R  36.87 0
 I R  437.38  36.870 Amp
ZY
 65.51670 x 4.12 x104 900
2
 0.0124  j 0.0053
65.5167 0 x 437.38  36.87 0
 VS  1  0.0124  j 0.0053 x 76.210  0

1000
 VS  101.078.180 kV
Sending end line to line voltage
VS ( L  L )  3x101.078.180 kV
 175.058.180 kV
From equation
 ZY   ZY 
I S  Y 1  
 R 
V 1   IR
 4   2 
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 ZY
 0.00675157 0
4
ZY
1  0.99380.150
4
 ZY  4
Y 1    4.12 x10 x 0.993890.15 0

 4 
 4.094 x104 90.150

 4
I S  4.094 x10 90.15 x76.210
0 0

1  0.0124  j 0.0053  x 437.38  36.87 0
1000
I S  0.031190.150  0.432  36.560
 0.414  33.060 kAmp
 414  33.060 Amp
Sending end power factor angle  8.180   33.06 0   41.240
Sending end power factor  cos  41.24 0   0.752

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 LONG TRANSMISSION LINE
 For the short and medium length lines reasonably accurate model were
obtained by assuming the line parameters to be lumped
 In case the lines are more than 250 km long, for accurate solutions the
parameters must be taken as distributed uniformly along the length as a
result of which the voltage and currents will vary from point to point on
the line
 In this section, expressions for voltage and current at any point on the
line are derived. Then, based on these equations, an equivalent π model
is obtained for long transmission line.
 Figure 5.6 shows one phase of a distributed line of length l km, the
series impedance per unit length is shown by the lowercase letter z, and
the shunt admittance per phase is shown by the lowercase letter y,
where z  r  j L and y  g  jC

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 Figure 5.5: Long line with distributed parameters

 Consider a small segment of line ∆x at a distance x from the receiving

end of the line
 From Kirchhoff's voltage law
V ( x  x )  V ( x )  z xI ( x ) (5.21)
or
V ( x  x )  V ( x )
 zI ( x ) (5.22)
x
Taking the limit as x  0, we have
dV ( x )
 zI ( x ) (5.23)
dx
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 Also, from KCL
I ( x  x )  I ( x )  y xV ( x  x ) (5.24)
or
I ( x  x)  I ( x)
 yV ( x  x ) (5.25)
x
Taking the limit as x  0, we have
dI ( x )
 yV ( x ) (5.26)
dx

Differentiating (5.23) and substituting from (5.26), we get

d 2V ( x) dI ( x )
 z
dx 2 dx
 zyV ( x ) (5.27)
Let
 2  zy (5.28)

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The following second-order diffrent equation will result
d 2V ( x) 2
2
  V ( x)  0 (5.29)
dx
The solution of the above equation is
V ( x )  A1e x  A2e  x (5.30)
where  , known as the propagation constant, is a complex expression given by (5.28) or
    j   zy   r  j L  g  jC  (5.31)
The real part  is known as the attenuation constant, and the imaginary component
 is known as the phase constant,  is measured in radian per unit length
From (5.23), the current is
1 dV ( x) 
I ( x)    A1e x  A2e  x 
z dx z


y
z
 A1e x  A2 e  x  (5.32)

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or

I ( x) 
1
Zc
 A1e x  A2e  x  (5.33)

Where Z c is known as the characteristic impedance, given by

z
Zc  (5.34)
y
To find the constants A1 and A2 we note that when x  0, V ( x)  VR , and
I ( x)  I R . From (5.30) and (5.33) these constants are found to be
VR  Z c I R
A1 =
2
V  Zc I R
A2 = R (5.35)
2
upon substitution in (5.30) and (5.33), the general expression for voltage and current along a long
trnsmission line become
VR  Z c I R  x VR  Z c I R - x
V ( x)  e + e (5.36)
2 2
VR VR
 IR  IR
Z Z
I ( x)  c e x - c e - x (5.37)
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 The equations for voltage and currents can be rearranged as follows:
e x  e   x e rx  e  rx
V ( x)  VR  Z c IR (5.38)
2 2
1 e x  e   x e x  e   x
I ( x)  VR  IR (5.39)
Z 2 2
Recognizing the hyperbolic functions sinh, and cosh, the above equations are
written as follows:
V ( x)  cosh  xVR  Z c sinh  xI R (5.40)
1
I ( x)  sinh  xVR  cosh  xI R (5.41)
Zc
we are particularly intersted in the relation b/n the sending end and the
receiving end of the line. Setting x  l , V (l )  VS and I (l )  I S , the result is
VS  cosh  lVR  Z c sinh  lI R (5.42)
1
IS  sinh  lVR  cosh  lI R (5.43)
Zc
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Rewriting the above equations in terms of the ABCD constants as before, we have
VS   A B  VR 
 I   C D   I 
 S   R
Where
A  cosh  l B  Z c sinh  l (5.44)
1
C sinh  l D  cosh  l (5.45)
Zc
Note that, as before, A  D and AD  BC  1
It is now possible to find an accurate equivalent  model, shown in
figure 5.6, to replace the ABCD constants of the two-port network
for the equivalent  model we have
 Z 'Y ' 
VS   1   VR  Z ' I R (5.46)
 2 
 Z 'Y '   Z 'Y ' 
I S  Y ' 1   R 1 
V   IR (5.47)
 4   2 
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 Comparing (5.46) and (5.47) with (5.42) and (5.43), respectively,
and making use of the identity
l cosh  l  1
tanh  (5.48)
2 sinh  l
the parameters of the equivalent  model are obtained
sinh  l
Z '  Z c sinh  l  Z (5.49)
l
l
Y' 1 l Y tanh
2
 tanh  (5.50)
2 Zc 2 2 l
2

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 Q?
Thank you
33 4/9/2019