Capacitance of Transmission Line:
The capacitance between the conductors is the charge
per unit of potential difference .
q
C
V
F /m
……………(25)
Capacitance between parallel conductors is a constant
depending on the size and spacing of the conductors .
For the length of transmission line less than 80 Km ,
the effect of capacitance is usually neglected .
The flow of charge is a current , and this current is
called the charging current of the line .
The charging current effects the voltage drop along the
line , efficiency , power factor and stability of the
power system .
Gauss theorem states that at any instant of time , the
total electric flux through any closed surface (A) is
equal to the total charge enclosed by that surface .
X
+
٧٤
q
D
coulomb / m 2
………..(26)
2 x
Where q - charge on the conductor coulomb /m
x - distance , from the conductor to the point
where the electric flux density is computed.
D - the electric flux density .
The electric flux density
The electric field intensity = The permittivity of the medium
q
D

V /m
E
……….(27)
2 x

Where  - actual permittivity of material .
r   /0
r - relative permittivity ;  0 - permittivity of free space
 0  8 . 85  10
 12
1- coulomb

1
4   9  10
9
F /m
E-force on unit
charge
Electric flux lines
q- coulomb/meter
on line conductor
equipotential
surface
Electric
flux lines
Electric field between two line conductors
٧٥
equipotential surface
P1
D1
q
P2
D2
Therefore , instantaneous voltage drop between P1 and
P2 is :
v 12 
D2
 E dx
D1

D2

D1
q
dx
2 x 

q
D2
ln
2 
D1
volts
…..(28)
Capacitance of a two wire line :
qa
qb
ra
rb
D
٧٦
From eq. ( 28)
v ab due to q a 
And voltage due to
v ba
v ab
qa
D
ln
2 
ra
qb , calculated by :
qb
D

ln
2 
rb
  v ba
 v ab
 v ab
volt
volt
 D 
qb
D
qb

 

ln
ln 
rb
2 
2 
 rb 
qb
rb
due to q b 
ln
volt
2 
D
1
By the principle of superposition the voltage drop from
conductor (a) to conductor (b) due to charges on both
conductors is the sum of the voltage drop caused by
each charge alone .
 v ab
1

2 

rb 
D
 q a ln
 volt
 q b ln
D 
ra

…..(29)
For two-wire line
qa = - qb , so that :

qa
D
qa


ln
ln 

2 
2 
ra rb

2
v ab
D
ra rb




2
٧٧

qa

ln 



qa

C ab 
v ab
ra  rb  r
If

C ab




D
ra rb
volts


ln 


D
ra rb
Farads




( radius of two conductors are same )


 D 
ln 

 r 
F /m
………(30)
1
If ,  r  1 , and  0 
4  9  10 9

C ab 
/ meter
10
F /m
9
 D
36 ln 
 r
1

D
36 ln
r



F /m
 F / Km
…(31)
٧٨
a
C ab
b
a
n
b
Ca n
Cb n
C ab - capacitance between conductors (a) and (b) .
C a n - capacitance between conductor (a) and neutral .
Cb n - capacitance between conductor (b) and neutral .
C a n = Cb n = C n = 2 C ab

C

n
Cn
1
 2 
36 ln
0 . 0555

D
ln
r
D
r

1
18 ln
D
r
 F / Km
……(32)
Capacitive reactance between one conductor and
neutral is :
X
c
1

2 f C n
٧٩
X
c

18 ln
D
 10
r
2 f
6
6 .6
 10
f

Where , ln  2 . 303 log
6
log
D
r
10
6 .6  10 6
1 6 .6  10 6
Xc 
log 
log D  / Km
f
r
f
Capacitive reactance Capacitive reactance
at 1 meter spacing
spacing factor
( X d )
( X a )
Potential difference between two conductors of
a group of charged conductors :
The voltage drop between the two conductors is the
sum of the voltage drops due to each charged
conductor .
b
Dab
Dbc
The voltage drop
Dac
between (a) to (b) :
a
c
Dam
Dbm
Dcm
m
٨٠
D
r
D


 q a ln ab  q b ln b  q c ln cb  ... 
D ca
ra
D ba
1 

volts
v ab 


D
2 

......  q m ln mb 
D ma 

....…(33)
In similar manner :
D
D
r


 q a ln ac  q b ln bc  q c ln c  ... 
D ca
ra
D ba
1 

volts
v ac 


D
2 

......  q m ln mc 
D ma 

Capacitance of three – phase line:
a- Equilateral spacing :
b
D
By using eq. ( 33)
D
a
0
D
r
D

ln
ln
ln


q
q
q
 a

b
c
r
D
D


r 
D
D

 qa ln  qb ln  qc ln 
r
D
D

1 
D
r 
vab  vac 
 2 qa ln  ( qb  qc ) ln 
2 
r
D
c
D
1
2
1
vac 
2
vab 
Volts
٨١
But q b  qc   qa
1 
D
r 
 2 qa ln  qa ln 
2 
D
r
2
3
r 
1 
1 
 D  
D
qa ln    qa ln
qa ln  





2 
D  2 
 r  
r 
3q
D
 a ln
volts
2
r
 vab  vac 
Phasor diagram of voltages v ab , v bc and v ca :
b

v ab  v ab  30 
v ab
2
 V an cos 30 


v ab
a
3
V an
2
v ab 
v ca
3 V an
c

v ab  3V an  30 


v ac   v ca  3V an   30 


 v ab  v ac 

v bc
30  n
3V an  30   3V an   30 
3V an  2  cos 30  
3V an  3
 3 V an
٨٢


3q a
D
ln
2 
r
q
D
V an  a ln
volt
2 
r
q
2 
F /m
Cn  a 
D
V an
ln
r
For  r  1
3 V an 
to neutral
1
0 . 0555
Cn 

D
D
18 ln
ln
r
r
 F / Km
(as in the case of single-phase lines , eq(32)
b) Unsymmetrical spacing but transposed :
1
D 31
D 12
2
3
v ab 1
1

2
D 23

D 23 
D12
r
 q b ln
 q c ln
 q a ln
 ….(34)
r
D
D
12
31 

٨٣
v ab 2
1

2

D 23
D 31 
r
 q b ln
 q c ln
 q a ln

r
D
D
23
12 

v ab 3
1

2

D 31
D12 
r
 q b ln
 q c ln
 q a ln

r
D
D
31
23 

v ab 1  v ab 2  v ab 3
3


D12 D 23 D 31
r3
 q b ln

 q a ln
3
r
D12 D 23 D 31 
1 

6 
D12 D 23 D 31 
 q c ln


D
D
D
12
23
31


v ab 
v ab
v ab
1

2

 q a ln

3
D12 D 23 D 31
r
 q b ln
3
r
D12 D 23 D 31
3
If , D eq  D12 D 23 D 31
 v ab
1

2
Similarly ,
v ac

D eq
r 
 q b ln
 q a ln

r
D

eq 

1

2

D eq
r 
 q c ln
 q a ln

r
D

eq 

٨٤



As in section (a) :
v ab  v ac  3 V an
and
qb  qc   qa

D eq
r
q
q
 v an
2
ln
ln



a
a
r
D eq


D eq2
1
r 
v an 
 q a ln 2  q a ln

r
3  2 
D eq 

D eq3 
1
v an 
 q a ln 3 
3  2 
r 
D eq
qa
v an 
ln
volts
2
r
1

3  2
Cn 
 Cn 
qa
2

D
v an
ln eq
r
1
18  ln
D eq
r




F /m
0 .0555
D
ln eq
r
 F / Km
…(35)
٨٥
Effect of earth on the capacitance of a line :
The presence of earth will change the electric flux lines
and equipotential surfaces considerably , which in effect
will change the effective capacitance between the wires
,as shown in fig. below :
equipotential surface
Electric flux
lines
h
D
influence of ground on the electric field picture
٨٦
a
b
qa
qb
conductor
h
H
Ground
h
a
b
 qa
 qb
D
v ab 
1
2
but
 v ab 
Images of
conductors
(a) and (b)
1 
2h 
r
H
D

ln
ln
ln
ln
q
q
q
q




b
b
 a
 2  a 2 h
r
D
H 



H 
D 2  (2h) 2 
D 2  4h 2
D
r

q
ln
q
ln


b
 a

r
D

1 
D 2  4h 2
 q b ln
  q a ln
2 
2h
1
2


2
2
D  4 h 
2h
٨٧
v ab
1

2
but
 v ab

r D 2  4h 2
2 hD
 q b ln
 q a ln
2
2
2 hD
r D  4h




q a   qb
1

2


4h 2 D 2
q
ln
 a
2
2
2 
r
(
D
4
h
)




1 
D
q a ln 

 r 1  ( D 2 4h 2 )
2 






2





1 
D
 2 q a ln


2
2
2 
r 1  ( D 4 h ) 

qa  D
1
 ln  ln

2
2
  r
1
(
4
)
D
h


 C ab 
qa

v ab



volts

D 
ln
 ln

r

1
1  (D 2


4 h 2 ) 
F /m
effect of earth
term
٨٨
Charging current due to capacitance :
Single-phase line :
v
I chg 
 j c v
 j / c
Where ,   2 f
( amp
c-capacitance between lines (Farads)
v- phase voltage (volts)
Three – phase line :
I chg 
v
 j c v
 j /c
Where ,
( amps )
v- voltage to neutral (volts)
c-capacitance to neutral (Farads)
Capacitance of bundle conductors
b
Instead of (r) , put D sc , where :
D scb - is the GMR of a bundle conductor, and can be
calculated as in inductance (page 67) .
1- For two – conductor bundle :
b
DSC
 2 ( DS  d ) 2  4 ( DS  d ) 2  ( DS  d )
2
٨٩
2- For three – conductor bundle :
b
DSC
 3 ( DS  d  d )3  9 ( DS  d 2 )3  3 DS  d 2
2
3- For four – conductor bundle :
b
DSC
 4 (DS  d  d  d 2)4  16 (DS  d 3 2)4  4 DS  d 3 2
2
1.09 4 DS  d 3
IF conductor is solid , DS  r in above three condition
 Capacitance of 3-phase line , bundle conductor with
equilateral spacing :
C nb
0 . 0555

D
ln
D scb
 F / Km
And for unsymmetrical spacing but transposed , the
capacitance is :
C nb
0 .0555

D eq
ln b
D sc
 F / Km
٩٠
EX.1 A 3-phase , 50Hz , 132 KV overhead transmission lines has
conductor diameter of 4 cm each , are arranged in a horizontal
plane as shown in fig . supplies a balanced load , assume the line
is completely transposed . Find the capacitance to neutral per
phase per Km.
Phase A
Phase B
4m
Phase C
4m
Solution :
D m  Deq  3 D AB D BC DCA
 3 4  4  8  5.04 m
DS  r  2 cm
C
0.0555
0.0555
0.0555


 0.01 F / Km
D
5.04
5.529
ln m
ln
DS
2  10  2
EX.2 A 3-phase , 50Hz , 400 KV overhead transmission lines are
arranged in a horizontal plane , each phase has two – strand
bundle conductors , the diameter of each strand is 25mm , as
shown in the fig. below . Find the capacitance to neutral per
phase per Km .
0.3m
25mm
6m
Phase A
6m
Phase B
Phase C
٩١
Solution :
D m  Deq  3 D AB D BC DCA
 3 6  6  12  7.56 m
25
DS  r 
 12 .5 mm
2
b
DSC
 DS  d
 12 .5  10 3  0.3  0.0612 m
C nb 
0.0555
0.0555
0.0555


 0.0115 F / Km
Deq
7.56
4.816
ln
ln b
0.0612
DSC
Discussion Questions
1 – Discuss the effect of earth on the capacitance of a line :
2 – Derive in expression for the capacitance to neutral per phase per
Km of a single phase overhead transmission line , taking into
account the effect of earth .
3 – Derive in expression for the capacitance to neutral per phase per
Km of a 3- phase overhead transmission line when conductors
are of equilateral spacing .
٩٢
Tutorial Problems
Q1. A 3-phase , 50 Hz , 110 KV , overhead transmission line consists
of three solid conductors of 3 cm diameter positioned on the
corners of triangle with sides of 2 m , 2.5 m , 3.125 m .
If the conductor of each phase of this line is replaced by threestrand bundle conductor has the same equivalent area of one solid
conductor , and the space between the strands of bundle is 0.2 m .
Find the capacitance to neutral per phase per Km for two
conditions . Assume the line is transposed.
Q2. A 3-phase , 132 KV , 50 Hz , 100 Km , single circuit bundle
conductor transmission line as shown in the figure below . If the
diameter of each strand is 1 cm , and the conductors are
regularly transposed . Determine , the capacitance to neutral per
phase per Km .
6m
5m
0.18 m
٩٣
Q3. By using the method of images , prove that the capacitance
between conductor and earth , taking into account the effect of
earth for 3- phase equilateral spacing (as shown in fig. below) is :
C an 
2
H 1  H 2
D
ln  ln 3
r
H 3 2 H 4
Where
2
H 4 2

H1H 3
H3
H2
and ( r) is the radius of the conductor.
b
D
a
D
D
c
H2
H1
Ground level
H3
a
H4
c
D
D
D
b
٩٤
Performance of Transmission lines
For balance 3-phase system only 1-phase need be
considered for analysis .
For purpose of analysis , transmission lines may be
classified as :
a- Short line - The length of line is less than 80 Km.
b- Medium line - The length of line is between 80 to
250 Km.
c- Long line - The length of line is more than 250 Km.
Short Transmission line
 The Fig. below , shown the equivalent circuit of a
short line .
 The effect of shunt capacitance is neglected .
 The series impedance can be taken a lumped .
Ir
IL
Vr
Load
VS &Vr - phase voltage (rms value)
For short line :
   
I S  I r  I L  I …………………….(1)
٩٥
 

VS  Vr  Z I r
Eq. (1) can be written in matrix form as :
VS  1 Z  Vr 
I   0 1  I 
 r
 S 
…………………(2)
Phasor diagram :
Vr - as reference .
 - Power angle ( load angle ) , it is the angle between
VS &Vr , and is small ( 1 to 7 for short line ) .
VS
IZ
s


IX
Vr
IR
Ir  IS  I
Phasor diagram per phase , any value taken from it is
per phase .
Vr 0 ; VS 
VS  (Vr cos  IR)2  (Vr sin  IX )2
٩٦
VS  Vr2  2Vr I ( R cos  X sin )  ( IR)2  ( IX )2
Vo  Vr
 100
Percent voltage regulation of line (V.R) = V
r
(at rated current and at a given power factor )
Vo - The magnitude of receiving end voltage at no-load.
Vr - The magnitude of receiving end voltage at fullload.
With VS - constant .
For short line Vo  VS
 V.R (for short line) 
VS  Vr
100
Vr
f
VS
o


s
Ir  IS  I

IX
Vr a b
c
IR
90  
From phasor diagram :
VS  Vr  of  oa
٩٧
 is very small ( of  oc )
 VS  Vr  oc  oa  ac
 ab  bc
 I R cos   I X sin 
(approximately)
+ For lagging P.F.
- For leading P.F.
 V.R 
VS  Vr
IR cos  IX sin 

100
Vr
Vr
V.R , depends on I , R ,X and P.F.
Condition for zero voltage regulation ( approximately)
IR cos   IX sin   0
V.R becomes zero at leading P.F.
 IR cos   IX sin 
R
tan 
X
Also from phasor diagram :
 Vr sin   IX 

s  tan 
 Vr cos   IR 
1
  s  
٩٨
Medium Transmission line
The effect of shunt capacitance becomes more and more
pronounced with the increase in the length of line .
For medium length T.L , the shunt capacitance can be
considered as lumped , by two type of equivalent circuit :
a) Nominal π circuit .
b) Nominal T circuit .
Nominal π circuit
Ics
I cr
Ir
Vr
I L  I r  I cr
Y
 I r  Vr
2
VS  Vr  I L Z
Y

 Vr   I r  Vr  Z
2

٩٩
 Vr  I r Z  Vr
YZ
2
 YZ 
 Vr  1 
  IrZ
2 

...............................(3)
I S  Ics  I L  Ics  Icr  I r
Y
Y
 Vr  I r
2
2
  YZ 
 Y
Y
 Vr 1    I r Z    Vr  I r
2 
2
 
 2
 VS

Y 2Z 
 YZ 
  I r 1  
 Vr Y 
4 
2 


 YZ 
 YZ 
 Vr Y 1    I r 1  
4 
2 


..........................(4)
Eq. (3) and (4) can be written in the matrix form as :
VS    YZ 
 Vr 
Z
   1  
 
2


 
   .....................(5)
    YZ   YZ   
  Y 1   1    
4  
2  I r 
I S   
١٠٠
Phasor diagram for nominal π circuit
VS
Ics
Icr

IZ
IX
Vr
Is
IR
I  Ics
Ir Icr
Icr and Ics are leads the Vr and VS by 90
Angles :
 - between Vs and Vr
 - between Vr and I r
 s - between Vs and I s
Nominal T circuit
In the nominal T circuit , the total capacitance of each
conductor is concentrated at the center of the line ,
while the series impedance is split into two equal parts.
١٠١
VS
VC
Vr
I s  I r  I c  I r  VcY
Z

 I r  Vr  I r  Y
2

YZ 

.......... .......( 6 )
 I r 1 
  Vr Y
2 

Z
Z
VS  Vr  I r  I s
2
2
 Z
Z  
YZ 
 Vr  I r   I r 1 
  VrY  
2  
2 
 2
 YZ
 V r 1 
2

YZ 


I
Z
1




r 
4



.......... ........( 7 )
Eq. (6) and (7) can be written in the matrix form as :
VS    YZ 

   1 
2 
  
  
Y
  
I S  
 YZ  Vr 
Z 1 
  
4   


...............(8)


 YZ  
1 
  
2  Ir 

١٠٢
Phasor diagram for nominal T circuit
IS Z / 2
Ir Z / 2
VS
IS
Ic
VC
IS
Vr
Is
Ir
Ir
X
2
R
2
X
2
R
2
Ic
Ir
Ic - Leads the phasor VC by 90
Angles :  - between Vs and Vr
 - between Vr and I r
s - between Vs and I s
Long Transmission line
The line parameters are distributed uniformly over
entire length .
Assumption of lumped circuit analysis tails in the case
of long length T.L.
١٠٣
Fig. below , shows distribution parameter line with
series impedance (z) per unit length and shunt
admittance (y) per unit length .
Ir
Vr
( Where z and y per unit length )
From Krichhoff 's voltage law :
V ( x   x )  V ( x )  z x  I ( x )
V ( x   x)  V ( x)
 zI ( x )
x
V ( x   x)  V ( x)
lim
 zI ( x )
x  0
x
or
dV ( x )
 zI ( x )
dx
.......... .......... ........( 9 )
١٠٤
By Krichhoff 's current law :
I ( x   x )  I ( x )  y x  V ( x   x )
Where y  x  V ( x   x ) is the current flowing in
the shunt admittance y x of the element .
I ( x  x)  I ( x)
 yV ( x   x )
x
lim
 x0
I ( x  x )  I ( x)
 lim yV ( x   x )
 x0
x
dI ( x )
or
 yV ( x )
.......... .......... .......( 10 )
dx
Differentiating eq. (9) and (10) with respect to x :
d 2V ( x )
dI ( x )

z
dx 2
dx
d 2 I ( x)
dV ( x )
y

dx 2
dx
.......... .......... .(11)
dV ( x)
dI ( x)
Substituting values of dx and dx from eq.(9)
and (10) into eq.(11) :
d 2V ( x )
2

zy

V
(
x
)


V ( x)
2
dx
.......... .....(12 )
١٠٥
d 2 I ( x)
2

zy

I
(
x
)


I ( x)
2
dx
Where  2  z y or  
.......... .........( 13)
zy
 - Propagation constant , it is complex quantity .
    j
Where   Attenuatio n factor / Km
  phase shift rad / Km
Linear differential equation of the type
d2y
ky
2
dz
y  ae
, has its complet solution
kz
 be
kz
Where a and b are constant to be evaluated
Therefore , solution of eq. (12) is :
V ( x)  a e x  b e  x
.......... ...( 14 )
dV ( x)
x
 x

a

e

b

e
Differentiating eq. (14) , dx
and substituting the result in eq. (9) to give :
z I ( x )  a e x  b e  x
١٠٦
a   x b   x
e 
or I ( x ) 
e
z
z
a zy  x b zy   x

e 
e
z
z
b
1
a
x
 x

e 
(a e x  b e x )
e 
z/ y
z/ y
z/ y
1
x
 x

(a e  be )
Zc
.......... ..... (15 )
Where Z c  z / y ohm , is called characteristic
impedance , and it is complex quantity .
For lossless line Z c is called Surge impedance and
equal to L / C ohm .
For Over Head Transmission Line , Z c  400  600
, and for underground cable Z c  40  60
For eq. (14) and (15) , at x = 0 :
V ( x)  Vr
; I ( x)  I r
Vr  a  b
Ir 
1
a
b


(a  b)
Zc
Zc
Zc
١٠٧
V  Ir Zc
 a r
2
V  Ir Zc
; b r
2
V  I r Z c  x V r  I r Z c  x

e
V ( x)  r
e
2
2
V
 x I r Z c  x V r  x I r Z c  x
 r e



e
e
e
2
2
2
2
 e x  e  x


2

1
I ( x) 
Z
c
 e x  e  x

 Vr  


2


 V  I Z
r c
 r
2


 Z c I r .........(16)


  x  Vr  I r Z c
e



2


  x
e




  x  x 

 e x  e  x 
1  e  e
 Z I  ......(17)
V 

 c r
 r 
2
2
Z 
c 




In term of hyperbolic functions , eq. (16) and (17) are:
V ( x )  (cosh  x ) V r  Z c (sinh  x ) I r
I ( x) 
1
(sinh  x ) V r  (cosh  x ) I r ……(18)
Zc
To obtain sending end values of voltage and current ,
we set (x) equal to (l) in eq. (18) :
١٠٨
VS  (cosh  l ) Vr  Z c (sinh  l ) I r
1
IS 
(sinh  l ) Vr  (cosh  l ) I r
Zc
……(19)
Eq. (19) can be written in matrix form :
Zc sinh  l  Vr 
VS   cosh l
 
   ...............(20)
1
  
sinh  l cosh l   
 I S   Zc
  I r 
The equivalent circuit of along line :
Ir
Ir
Vr
Vr
Nominal  circuit
Equivalent  circuit
( approximately )
( Exactly )
١٠٩
Nominal  circuit is approximate equivalent circuit ,
generally used for medium length T.L.
The results from nominal  circuit are approximate ,
but the results from equivalent  circuit are exact as
by the long line equations .
For equivalent  circuit :
 Y Z  
VS  Vr 1 
  Ir Z
2 

Compare this equation with equation (19) :
1) Z   ZC sinh l
sinh l
 z / y sinh l  zl
zy l
sinh l
 Z  Z
........................................(21)
l
Where, Z = z l - Total series impedance of the line .
Y Z 
 cosh l
2
cosh l  1 cosh l  1
Y / 2 

Z
ZC sinh l
2) 1 
1
l
tanh

ZC
2
 l cosh  l  1
Where, tanh 
2
sinh  l
١١٠

Y  Y  tanh  l / 2 

 
2 2  l /2 
...........................(22)
Where Y = y l - Total shunt admittance of the line .
Similar , an equivalent T circuit can be found for T.L.
Transmission efficiency : (  TL )
The ratio of receiving end power to the sending power
of transmission line .
Transmission efficiency (TL ) 
Vr I r cos
100 .....(23)
VS I S cosS
Generalized Constants
In any four terminal network , the input voltage and input
current can be expressed in term of output voltage and
output current . T.L. is a 4-terminal network as shown :
Is
Ir
A
B
Vr
Vs
C D
2 – port , 4 terminals network
The network should be :
1- Passive ; 2- Linear ; 3- Bilateral
This condition is fully met in T.L.
١١١
Therefore , equations ( 2,5,8,20 ) can be written in
general form as :
VS   A B  Vr 
 I   C D  I 
 r
 S 
.................(24)
Vs  A Vr  B I r
I s  C Vr  D I r
………………….(25)
The following points may be kept in mind :
1- Constants A,B,C, and D are all Phasors



A  A   ; B  B  ; C  C  
 
2- D  A for all symmetrical line .
;
 
A D

3- A - Dimensionless .

B - Ohms .

C - Siemens ( mhos ) .

D - Dimensionless .
 
4- AD  BC  1
The values of the generalized constants depend upon
the particular method adopted for solving a T.L. , as
shown in below table .
١١٢
Comparison of ( A,B,C,D ) constant for T.Ls :
A
Short line
1

Medium
line
B
1
C
Z
ZY
2
0
Z

A
 ZY 
Y 1 

4  A

T 1  ZY
 ZY 
Z 1 

2
D
Y
4 
A
Length <80Km(66Kv) ,
Capacitance can be
ignored ,
Parameters can be
taken as lumped
Length 80-250Km ,
Capacitance - is line to
neutral per Km ,
Parameters can be
taken as lumped ,
Z- Total series
impedance of line ,
Y – Total shunt
admittance of line .
Length >250Km ,
  zy
Long line
cosh  l Z c sinh  l
1
sinh  l
Zc
 l  zlyl  ZY
A
Zc 
z
Z

y
Y
Z  zl ; Y  yl
z–series impedance
per unit length
.
 = propagation constant .
=   j
 = Attenuation constant .
 = Phase constant ( rad. / unit length )
Zc = Characteristic impedance , for lossless line it
is referred to as the surge impedance .
١١٣
Also , can be determined receiving end condition in
term of sending end conditions :
Vr   D  B  VS 
 I    C A   I 
 S
 r 
...............( 26)
Summary for long line :
 
zy
;
l 
zy l 
Zc 
z

y
A  cosh ( l )
C
Z  zl
zlyl 
; Y  yl
ZY
Z
Y
;
1
sinh ( l ) ;
Zc
B  Z c sinh ( l )
DA
V S  cosh ( l ) Vr  Z c sinh ( l ) I r
IS 
1
sinh ( l ) V r  cosh ( l ) I r
Zc
cosh( l )  cosh(a  jb)  cosh(a) cos(b)  j sinh(a) sin(b)
sinh( l )  sinh(a  jb)  sinh(a) cos(b)  j cosh(a) sin(b)
Or
( l ) 2 ( l ) 4
cosh ( l )  (1 

 .......... )
2!
4!
( l ) 3 ( l ) 5
sinh ( l )  {(  l ) 

 .......... }
3!
5!
١١٤
Two transmission Lines in Cascade ( Series):
I S1
IS2
I r1
Ir2
A1 B1
A2
Vr1
VS1
C1 D1
B2
Vr 2
VS12
C2 D2
VS1  A1 Vr1  B1 I r1
and
VS 2  A2 Vr 2  B 2 I r 2
I S1  C1 Vr1  D1 I r1
and
I S 2  C 2 Vr 2  D 2 I r 2
VS1  A1 B1 Vr1 
VS 2  A2 B2 Vr 2 
I   C1 D1 I  and I   C2 D2 I 
  r1 
  r2 
 S1  
 S2  
Vr1  VS 2 and I r1 I S 2
VS1  A1 B1 A2 B2 Vr 2 
I   C1 D1 C2 D2 I 

  r2 
 S1  
VS1  A1A2 B1C2
I   C1A2 C2D1
 S1  
VS1  Ao
I   C
 S1   o
Bo  Vr 2 
Do  I r 2 
A1B2 B1D2 Vr 2 
B2C1 D1D2 I r 2 
…………………….(27)
١١٥
Two transmission Lines in Parallel:
IS
I S1
I r1
Ir
A1 B1
Vr
VS
C1 D1
IS2
Ir2
A2 B2
Vr
VS
C2 D2
VS  VS1  VS 2
I S  I S1  I S 2
;
Vr  Vr1  Vr 2
;
I r  I r1  I r 2
VS  A1 Vr  B1 I r1
and
I S1  C1 Vr  D1 I r1
and
VS  A2 Vr  B 2 I r 2 .....( 28)
I S 2  C 2 Vr  D2 I r 2 .....( 29)
From eq.(28) :
A1 Vr  B1 I r1  A2 Vr  B 2 I r 2
Vr ( A1  A2)  B 2 I r 2  B1 I r1
 B2 ( I r  I r1 )  B1 I r1
 B2 I r  I r1 ( B1  B 2)
١١٦
 I r1 
B 2 I r  Vr ( A1  A2)
............(30)
B1  B2
Put eq.(30) in eq.(28)
 B 2 I r  Vr ( A1  A2) 
VS  A1 Vr  B1 

B1  B2


B1( A1  A2) 
 B1B 2 

Vr
Ir

  A1 



B1  B 2 
 B1  B 2 

 B1B2 
 A1B2  A2 B1
VS  
Vr
Ir




 B1  B2 
 B1  B2 
Further, I S  I S1  I S 2
Substituting values of I S1 and I S 2 from eq. (29)
I S  C1 Vr  D1 I r1  C 2 Vr  D2 I r 2
 C1  C 2 Vr  D1 I r1  D 2 ( I r  I r1 )
(D1 D2)B2I r Vr ( A1 A2)
 C1 C2 Vr 
 D2 I r
B1 B2
( A1  A2)(D1 D2) 
 B1 D2  D1 B2 

IS  C1 C2 
Vr  
Ir


B1  B2
 B1  B2 


VS   Ao
 I   C
 S  o
Bo  Vr 
Do   I r  ………………(31)
١١٧
Experimental Determination of Generalized
Constants ( A , B , C , D ) :

IS0
1- Open circuit test :

Ir  0
A

VS 0 
  Z S 0  Z S 0   S 0
IS0

VS 0
B

Vr 0
C D
 


VS 0  A Vr 0
( I r 0  0)
 


I S 0  C Vr 0
( I r 0  0)



V
A
 Z S 0  S 0  
...........(32)
IS0 C
2- Short circuit test :

VSS 
  Z SS  Z SS    SS
I SS


VSS  B I r

 
I SS  D I r


V
 Z SS  SS 
I SS

B

D

I SS

VSS

Ir
A
B
C
D

Vr  0

(Vr  0)

(Vr  0)
...........(33)
١١٨
From eq. (32) and (33) :
   


A B AD  BC
1
  
Z S 0  Z SS     
C D
DC
DC
...........(34)
 

2

ZS 0
A/ C

     AD  A
Z S 0  Z SS 1 / DC
...........(35)

 A
 
A D

Z
 S 0
Z S 0  Z SS
and
..........(36)
 
AD  BC
..........(37)
From equations (34 , 36 and 37
) can be
   
determined the constants ( A , B , C , D) .
١١٩
Power Flow and Power Circle Diagram of T.L.
Receiving End power Circle Diagram :
   
Vs  A Vr  B I r
    
I s  C Vr  D I r

A  A 

; B  B 
………………….(38)

; C  C 



Vr  Vr 0 ; VS  VS 
;
 
AD
From eq.(38) :
  
 VS  A Vr  VS
  A Vr


       
Ir 
   
B

B
  B

The conjugate of I r is :
   VS
  A Vr

   
I r        

B
  B
The complex power at the receiving end ( The voltamperes delivered to the load ) is given by :

 
S r  Vr I r  Pr  j Qr
١٢٠
2

 VSVr
  A Vr
   
     
 Pr  j Qr  
 B
  B

……………..(39)
2

 VSVr
  A Vr

Pr  
cos (    )   
cos (    ) 
 B
  B

2

 VSVr
  A Vr
Qr  
sin (    )   
sin (    ) 
 B
  B

…………………(40)
Where Pr - Real power at receiving end .
Qr - Reactive power at receiving end .
j
reactive

Sr  Pr  j Qr
power axis
(Var)
AVr2
B
VS Vr
B
 
 
0
real power axis
(Watt)
١٢١

Sr  S r  Vr I r r
 Pr  j Qr
 Vr I r cosr  j Vr I r sin r
Var
Vr I r
 
AVr2
B
n
(x , y)
r
O
 
K
Watt
L
VS Vr
B
d
f
Sr  Vr I r r  OK r
Pr  Vr I r cosr  OL
Qr  Vr I r sin r  LK
١٢٢
Coordinates of the center of a receiving end circle :
x (Horizontal)  
y (Vertical)

A
B
A
B
2
Vr cos (    ).... Watt
2
Vr sin (    ).... Var
………………(41)
VS Vr
Radius of a receiving end  circle (nk) 
..(VA)
B
Units of Circle :
1- If voltages are phase to neutral in volts , P and Q
( and coordinates x , y ) are in Watts and Vars per
phase.
2- If voltages are line-to-line in volts , the Watts and
Vars on the diagram are total three-phase
quantities .
3- If voltages are in Kilovolts from line-to-line , the
coordinates given by above equation are total in
MW , MVar for three-phase .
From circle diagram , maximum power is
delivered if :
   0
or
 
١٢٣
2

 VS Vr   A Vr
 Pr (max.)  
cos (    ) 
  
 B   B

Pr (max.)  df
( For maximum power , the load must draw a
large leading current . )
 If Vr - constant , and VS -Variable , the
following receiving-end power circles are
obtained :
Q (Var)
K3
Load line
a
K2
K1
b
O
P(Watt)
r
n
(x , y)
VS1
VS 2 VS 3
١٢٤
EX. (1) : A 3–phase , 50Hz . medium transmission line , the load at
the receiving end is 75MVA at 0.8 power factor lagging with
132KV between lines . If at the sending end , the voltage is
165 .56.9  KV , and the power is 68.44 MW at 0.7791 power
factor lagging , find the total resistance , inductance and
capacitance per phase of line .
Solution : We can be use nominal T circuit method :

IS

Z /2

Z /2

Ir

IC

VS

Y

VC

Vr
132
Vr / phase 
 76.21 KV
3

Vr  76 .21 0  KV

165 .5
VS / phase 
 95 .5 6.9  KV
3
 94 .9  j11 .5 KV

Ir 

IS 
75  10 6
1


328
.
1


cos
0
.
8

328
.
1


36
.
9
A
3
3  132  10
 262 .5  j196 .9 A
68 .44  10 6
1

306
.
5


cos
0 .779
3
3  165 .5  10  0 .779
 306 .5  31 .9  A
 260 .3  j162 .46 A
١٢٥


 Z 
VC  Vr  I r
2

Z
 76 . 21  0   10 3   328 . 1  36 . 9  Volt
2



Z 
VS  VC  I S
2

Z
95 .6 6.8  10 3  76 .21 0   10 3   328 .1  36 .9  
2

Z
  306 .5  31 .9 
2

Z
(94.9  j11.5)  103  76.21 103  (262.5  j196.9  260.3  j162.4)
2

Z
 34 .666 .1  14  j 31 .6 
2

Z  28  j 63 
Therefore , the resistance per phase of line is = 28  ,
and the reactance per phase of line is = 63 
 the inductance per phase of line 




I
I  I
Y  C  S  r
VC
VC

XL
63

 0 .2 H
2 f
2  50
( 260 . 3  j162 . 4 )  ( 262 . 5  j196 . 9 )
76 . 21  0   10 3  34 . 6  66 . 1  328 . 1  36 . 9 
34 . 57   86 . 35 

 4  10  4   90  mho

86297 . 5  3 . 68
١٢٦
 1
Y
XC
1
4

0
.
25

10

4
4 10
1
1
C

 1.2 F
2 f X C 2  50  0.25 104
XC 
EX. (2)
A 3-phase , 50 Hz transmission line 160 Km long . delivers a load of
90 MW at 0.85 power factor lagging . The line voltage at the receiving
end is 230 KV . The generalized circuit constants for the line are as
follows :





A  D  0 . 9785  0 . 3 ; B  85 . 2  77 . 47 ; C  0 . 000503  9 . 01 
Calculate :
1- The sending end voltage , current , and power factor .
2- Efficiency of transmission .
Solution :
Receiving end voltage / phase (V r ) 
Receiving end current , I r 
230
 132 .8 KV
3
90  10 6
3  230  10  0.8

Vr  132 .80  KV

I r  282 .4   cos 1 0.8  282 .4   36 .9  A
3
 282 .4 A
Sending end voltage per phase :
 

 
Vs  A Vr  B I r
 0.97850.3  132.80  103  85.277.47  282.4  36.9
 130  0.3  10 3  24060 .48  40 .57 
١٢٧
 129 .998  j 680 .6  18276 .6  j15648 .373
 148274 .6  j16328 .97
 149 .171 6.28  KV
Sending end current :
 

 
I s  C Vr  D I r
 0.000503 9.01  132 .8  10 3 0   0.9785 0.3  282 .4  36 .9 
 66 .89.01  276 .32   36 .6 
 65 .97  j10 .46  221 .83  j164 .74
 287 .8  j154 .28
 326 .54   28 .19  A
 S  6 .28  28 .19  34 .47 

Power factor of sending end = cos S  cos 34.47  0.824 lagging
Sending end power = 3VS I S cos S
 3  149.171  326.54  0.824  120 .47 MW
Receiving end power
 100
Sending end power
90

 100  74 %
120 .47
Efficiency of transmission 
١٢٨
Discussion Questions
1- Draw the phasor diagram of a nominal  -circuit of transmission line
.
2- Prove that the voltage regulation (V.R) for short line is :
V.R 
IR cos   IX sin 
Vr
Where , I –load current , Vr and cos  are the receiving end
voltage and power factor , R and X are resistance and reactance of
line .
3- Prove that the voltage regulation (V.R) for short transmission line ,
unity power factor is :
V.R 
IR
Vr
Where , I –load current , Vr - receiving end voltage , R-resistance
of line .
4- State the advantage of bundle conductor lines over single conductor
lines .
5- What is the effect of unsymmetrical spacing of conductor in a 3phase transmission line .
Tutorial Problems
Q1. A 3-phase , 132 KV , 50 Hz transmission line , 200 Km long has
constants per phase per Km as follows : resistance = 0.26 Ω , inductive
6
reactance = j0.38 Ω , shunt susceptance = j 2.75  10 mho.
The power received is 1.8 MW at 0.8 power factor lagging ,
calculate : 1) The sending end power . 2) The percentage regulation.
Q2. A 3-phase load of 5 MW at a power factor of 0.9 lagging is to be
transmitted over a distance of 100 Km with a receiving end voltage of
66 KV , 50 Hz . Each conductor has a resistance 0f 0.8 Ω /Km and
radius of 26.1mm , the conductors have an equilateral spacing of 2 m
١٢٩
. Find the sending end voltage , current and power factor , by using :
1) Nominal  circuit method . 2) Nominal T circuit method .
Q3, A 3-phase transmission line , 10 Km long consists of three solid
copper conductors are arranged at corner of an equilateral triangle of
0.5 m side . Load conditions at receiving end are 5 MW at 0.8 power
factor lagging , 11 KV , 50 Hz . The efficiency of transmission is
6
90.9% and the resistivity of copper is 1.7774  10 Ω cm . Calculate
:
i) The sending end voltage and power factor .
ii) The voltage regulation of the transmission line.
Q4. A 3-phase , 50 Hz , 100 Km long overhead transmission line has
the following line constants :
Resistance per phase per Km = 0.153 Ω
Inductance per phase per Km = 1.21 mH
Capacitance per phase per Km = 0.00958 F
The line supplies a load of 20 MW at 0.9 power factor lagging at a
line voltage of 132 KV . Calculate :
1- Sending end voltage and current .
2- The voltage regulation of line .
3- Efficiency of transmission .
Q5. A 3-phase , 50 Hz , 100 Km transmission line has conductors
composed of 12 strands of Aluminum wound around a core of 7
steel strands . Each of the aluminum and steel strands has a
diameter of 0.28 cm . The resistivity and T constant for the
6
aluminum are given by 2.85  10 Ω m at 20 C , and 228
respectively . The steel strands carry no current due to the skin
effect . If the line conductors are placed at corners of an equilateral
triangle of 2 m a side . Determine :


1- The conductor resistance per Km of the line at 40 C
2- The line inductance and capacitance .
3- The voltage and current at the sending end , if the receiving end
voltage is 132 KV .
١٣٠
Mechanical Design of Overhead System
Type of line supports :
The supports for an overhead line ( O.H.L) must be
capable of carrying the load due to the conductors
and insulators ( including the ice and wind loads on
the conductors ) .
The supports are of the following types :
1- Poles :
a - Wooden poles .
b - Reinforced concrete poles .
c – Steel pipes poles .
Poles used for high voltage (H.V) distribution and
for low voltage (L.V) transmission line up to 33 KV.
There are separate types of poles designed for single
circuit and double circuit lines , figures below
illustrate the various types of poles :
Fig
Single circuit pole
double circuit pole
2- Steel Towers :
Steel towers are extremely useful for long distance
transmission lines running across open country
where long spans are a decided advantage . There
are separate types of towers designed for single
circuit and double circuit lines , figures below
illustrate the types of tower which are used for 66
KV and 132 KV :
Fig
Single circuit tower
Double circuit tower
Figures below illustrate the types of tower which
are used for voltage levels above 220 KV :
Fig
Calculation of Length , Sag , and Tension :
A. Line Supported at equal levels :
Fig
d- Maximum Sag at mid span , in metres .
S- length of section OP.
W- Weight/unit length of conductor.
l - Span ( distance between supports A and B ).
L- Actual length of conductor between A and B
( AOB).
T0 - Ultimate strength ( horizontal tension in the
section OP at point O in Kg .
 - Angle between the horizontal axis at point P
with the tangent line at point P .
TX , Ty - Horizontal and vertical components of
tension in the section OP at point P .
TA , TB - Tension at the supporting points A and B.
dS
dy Ty
tan 

dx TX
dy

dx
The part OP (S) under equilibrium :
TX  T0
Ty  W  S
dy Ty W  S
tan  

dx TX
T0
................(1)
And for the infinitesimal part dS
dS 2  dx 2  dy 2
2
 dy 
 dS 
1


 


 dx 
 dx 
2
W  S 

 1  
 T0 
2
W  S 
dS

 1  
dx
 T0 
2
dS
dx 
dx 
W  S 

1  
 T0 
T0
W
dS
W
T0
2
W  S 

1  
 T0 
2
Integrating both sides :
T0
1  W  S 
  A
sinh 
X 
W
 T0 
Where , A – Constant of integration .
At X = 0 ( i.e. at point 0 ) , S = 0
 A0
T0
1  W  S 

sinh 
X 
W
 T0 
S
W  S 
T0

sinh 
W
 T0 
.......... ........( 2)
Length (L) of the conductor :
l
L
; S
2
2
Wl 
L T0

sinh 

2 W
 2T0 
At

X
Wl 
2T0

sinh 
W
 2T0 
 Wl 
 can be represented as follows :
sinh 
 2T0 
L
3
5

 W l  W l  W l 
 Wl 
  
 3!  
 5!  ........
 
sinh 

 2T0 
 2T0   2T0  2T0 
5
 Wl 
 5! etc.
Ignorting 
 2T0 
3


2T0 W l  W l 
 3!

 
 L
W  2T0  2T0 


3!  1 2  3  6
 W 2 l2 

L  l 1 
2 
 24 T0 
........................(3)