Chapter 3. Capacitance of Transmission Lines

54 Electrical Power Systems



moving a unit charge of one coulomb from X2 to X1, through the electric field produced by the

charge on the conductor.

Therefore,

V12 =



z



D2

D1



\



Ey . dy =



z



D2

D1



q

dy

2p Îo y



FG IJ

H K



D2

q

V12 = 2p Î ln D Volts

o

1



...(3.2)



V12 is the voltage at X1 with respect to X2. V12 is positive, when q is positive and D2 > D1, i.e.,

X1 is at higher potential than X2. For alternating current, V12 is a phasor voltage and q is a

phasor representation of a sinusoidal charge.



3.2 POTENTIAL DIFFERENCE IN AN ARRAY OF SOLID

CYLINDRICAL CONDUCTORS



Fig. 3.2: Array of N conductors.



Neglecting the distortion effect and assuming that the charge is uniformly distributed

around the conductor, with the following constraint.

q1 + q2 + ... + qN = 0

...(3.3)

Now apply eqn. (3.2) to the multiconductor configuration shown in Fig. 3.2. Assume conductor

m has a charge qm coulomb/m. The potential difference VKi(qm) between conductors K and i due

to the charge qm alone is

VKi (qm) =



FG IJ Volt

H K



qm

D im

ln

2p Îo

DKm



...(3.4)



When K = m or i = m, Dmm = rm. Using superposition, the potential difference between

conductors K and i due to the all charges is



Capacitance of Transmission Lines 55



VKi =



1

2 p Îo



F D I Volts



å qm ln G D im J

H K

N



Km



m=1



...(3.5)



3.3 CAPACITANCE OF A SINGLE PHASE LINE

Figure 3.3 shows a single phase line consisting of two long solid conductors having raidus r1 and

r2 respectively. Conductor 1 carries a charge of q1 coulombs/m and conductor 2 carries a charge

of q2 coulombs/m. The field of the first conductor is disturbed due to the presence of the second

conductor and the ground. The distance between the conductors is D and D is much greater

than the raddi of the conductors. Also, the height of the conductors from the ground is much

larger than D. Therefore the effect of distortion is negligible and the charge is assumed to be

uniformly distributed.



Fig. 3.3: Single phase two-wire line.



The potential difference V12 can be obtained in terms of q1 and q2 by using eqn. (3.5). Thus,



F D I Volt



m

å qm ln G D2m J

H K

2



V12 =



\

Since



1

2 p Îo



V12 =



D

D

1

q1 ln 21 + q2 ln 22

D12

D11

2p Îo



m=1



LM FG IJ

NM H K



1



FG IJ OP Volt

H K QP



...(3.6)



q2 = –q1, D21 = D12 = D, D11 = r1 and D22 = r2



LM

N



V12 =



r

D

1

q1 ln - q1 ln 2

r1

D

2p Îo



\



V12 =



2q1

D

ln

Volts

2p Îo

r1r2



\



C12 =



q1

=

V12



OP Volt

Q



ln



p Îo

D



F/m



...(3.7)



r1r2



If r1 = r2 = r

\



C12 =



p Îo

D

ln

r



FG IJ F/m

H K



...(3.8)



56 Electrical Power Systems



0.0121

mF/km

...(3.9)

log ( D r )

Eqn. (3.9) gives the line-to-line capacitance between the conductors. For the purpose of

transmission line modeling it is convenient to define capacitance between each conductor and

a neutral as shown in Fig. 3.4.

\



C12 =



Fig. 3.4: (a) line-to-line capacitance



(b) line-to-neutral capacitance.



Since the potential difference to neutral is half of V12, therefore, C1n = C2n = 2C12

or

C1n = C 2 n =



0.0242

mF/km

D

log

r



The associated line charging current is



FG IJ

H K



...(3.10)



IC = jw C12V12 A/km



...(3.11)



3.4 CAPACITANCE OF THREE PHASE TRANSMISSION LINES

Figure 3.5 shows a three phase line, each with radius r and lines are transposed.



Fig. 3.5: Three phase transmission line (fully transposed).



For balanced three phase system

qa + qb + qc = 0



...(3.12)



Potential difference between phase a and b for the first transposition cycle can be obtained

by applying eqn. (3.5), Vab(I) is,

Vab(I) =



LM

N



D

D

r

1

+ qc ln 23

qa ln 12 + qb ln

r

D12

D31

2p Îo



OP

Q



...(3.13)



Capacitance of Transmission Lines 57



Similarly for the 2nd transposition cycle.

Vab(II) =



For the third transposition cycle

Vab(III) =



\



or



Where



Vab



...(3.14)



LM

N



OP

Q



...(3.15)



1

Vab (I) + Vab (II) + Vab (III)

3



1

=

2p Îo



Vab =



OP

Q



D

r

D

1

+ qc ln 12

qa ln 31 + qb ln

r

D31

D23

2p Îo



The average value of Vab is,

Vab =



LM

N



D

D

r

1

+ qc ln 31

qa ln 23 + qb ln

r

D23

D12

2p Îo



1

2p Îo



b



LM (D

MMq ln

N

LMq ln D

NM r



Similarly average value of Vac is

Vac =



12



a



eq



a



Deq = D12 D23 D31



1



g



D23 D31 ) 3

+ qb ln

r



+ qb ln



r

Deq



1



( D12 D23 D31 ) 3



...(3.17)



LM

MN



D eq

r

1

qa ln

+ qc ln

r

Deq

2p Îo



LM F I + bq

MN GH JK



Deq

1

qa ln

r

2p Îo



2



b



OP

PQ



...(3.18)



g



+ qc ln



r

Deq



From eqn. (3.12), substituting qb + qc = –qa in eqn. (3.19), we have,

Vab + Vac =



OP

PP

Q

...(3.16)



1

3



Adding eqns. (3.16) and (3.18), we get

Vab + Vac =



OP

QP



r



OP

PQ



FG IJ

H K



Deq

3qa

ln

2p Îo

r



...(3.19)



...(3.20)



Phasor diagram for balanced three phase

system is shown in Fig. 3.6.



Fig. 3.6: Phasor diagram for balanced

three phase system.



58 Electrical Power Systems



From Fig. 3.6, we can write

Vab = Van – Vbn



...(3.21)



Vac = Van – Vcn



...(3.22)



Vbn = Van -120º



...(3.23)



Vcn = Van -240º



...(3.24)



Also



Adding eqns. (3.21) and (3.22) and substituting

Vbn = Van -120º and Vcn = Van -240º , we have

Vab + Vac = 3Van



...(3.25)



From eqns. (3.20) and (3.25), we have,



FG

H

FD

q

ln G

2p Î

Hr



Deq

3qa

ln

r

2 p Îo

\



eq



a



o



IJ = 3V

K

IJ = V

K



an



...(3.26)



an



The capacitance per phase to neutral of the transposed transmission line is then given by

Can =



qa

=

Van



2 p Îo

D eq

ln

r



or

Can =



For equilateral spacing,



FG IJ F/m

H K



0.0242

Deq

log

r



FG IJ mF/km

H K



...(3.27)



...(3.28)



D12 = D23 = D31 = D, and Deq = D. Therefore,



Can =



0.0242

mF/km

D

log

r



FG IJ

H K



...(3.29)



The line charging current for a three phase transmission line

I (line charging) = jwCanVLN A/phase/km.



...(3.30)



3.5 BUNDLED CONDUCTORS

As mentioned in chapter-2 (section-2.13), the bundle usually comprises two, three or four

conductors. Geometric mean radius of the bundle conductor calculated earlier for the inductance



Capacitance of Transmission Lines 59



calculation with the exception that the radius r of each conductor is used; If d is the bundle

spacing, then for two conductor arrangement,



b g



Ds = rd



1

2



...(3.31)



For three conductor arrangement (equilateral triangle)



e j



Ds = r . d



1

2 3



...(3.32)



For a four conductor (quadruplex) arrangement,

Ds =



e 2 .r .d j



1

3 4



...(3.33)



Considering the line to be transposed, the capacitance per phase is given as

Can =



0.0242

Deq

log

Ds



FG IJ mF/km

H K



...(3.34)



3.6 CAPACITANCE OF THREE PHASE DOUBLE CIRCUIT LINES

Figure 3.7 shows three phase double circuit line with the three sections of the transposition

cycle.



Fig. 3.7: Three sections of three phase double circuit transposed line.



Each phase conductor is transposed within its groups. The effect of ground and shield wires

are considered to negligible. In this case per phase equivalent capacitance to neutral is



Where



0.0242

Deq

log

Ds



FG IJ mF/km

H K

D = bD D D g

D = bD D D g

Can =



eq



s



ab



sa



bc



sb



ca



sc



1

3



1

3



...(3.35)



...(3.36)

...(3.37)



60 Electrical Power Systems



b

= bd



g b

g b g

g = bD . d . D . d g = bDd g



Dab = dab . dab¢ . da¢ b¢ . da¢ b

Dbc



bc . dbc ¢ . d b¢ c ¢ . db¢c



\



\



Deq



Deq



= D . d2 . D . d2



1

4



2



2



1

4



1

4



= Dd2



2



1

2

1

2



b

g = bd . d . d . d g = bd d g

U

R

= Sb Dd g . b Dd g . bd d g V

W

T

U

R

|

|

= SDd . d . d V

|

|

W

T



Dca = dca . dca¢ . dc¢a¢ . dc ¢a



\



1

4



Deq =



2



1

2



2



1

2

1



2



1

1

3 .d3 .

D

2



1

2



5



1 5



1

2



1



5



1



1

4



1 5



1

2



1

3



1

3



1

2

5



1



1



6

d1 .



1

4



6

d5



...(3.38)



Now



b g

D = br d g

D = br d g

U

R

D = Sbr d g . br d g . br d g V

W

T

F

I

D = G r. d . r . d J

H

K

1

2



Dsa = r d3

sb



sc



\



\



1

2



4



3



1

2



3



s



s



4



1



1



1

2



3



1

2



1

3



1



1

2

4



1

2



3



1



1

2



3



1



\



3

6

Ds = r 3 . d3 . r 6 . d4



\



3

6

Ds = r 2 . d3 . d4



1



1



1



...(3.39)



Note that Deq and Ds will remain same for section-II and section-III of transposition cycle.

Substituting Deq and Ds in eqn. (3.34), we have,



Capacitance of Transmission Lines 61



Can =



0.0242



RD . d . d . d

|

log S

| r .d .d

T

1

3



1

2



1

3

2



1

3

3



1

6

1



1

6

4



1

6

5



U mF/km

|

V

|

W



...(3.40)



3.7 EFFECT OF EARTH ON THE CAPACITANCE

The effect of the presence of earth can be accounted for by the method of image charges

introduced by Kelvin. Figure 3.8(a) shows a single conductor with uniform charge distribution

and with height h above a perfectly conducting earth plane. Consider that the conductor has a

positive charge q coulomb/m, an equal amount of negative charge –q coulomb/m is induced on

the earth. The electric field lines will originate from the positive charge on the conductor and

terminate at the negative charge on the earth. Also, the electric field lines are perpendicular to

the surfaces of earth and the conductor. Figure 3.8 (b) shows that the earth is replaced by image

conductor, lies directly below the original conductor. The electric field above the plane (dashed

line) is the same as it is when the ground is present instead of image conductors. Therefore, the

voltage between any two points above the earth is the same in Fig. 3.8 (a) and Fig. 3.8 (b).



Fig. 3.8 (=): Single conductor and

earth plane.



Fig. 3.8 (>): Earth plane and image charge

of one conductor.



3.8 CAPACITANCE OF A SINGLE PHASE LINE CONSIDERING

THE EFFECT OF EARTH

Figure 3.9 shows a single phase line with flat horizontal spacing. The earth plane is replaced by

separate image conductor for each overhead conductor.



62 Electrical Power Systems



Fig. 3.9: Single phase transmission line with images.



Potential difference between conductors 1 and 2 can be easily obtained by applying eqn.

(3.5), i.e.

V12 =



\



V12 =



1

2 p Îo



LM FG IJ

NM H K



4



FD I



m =1



1



m

å qm lnG D2m J

H K



FG IJ

H K



FG IJ OP

H K QP



FG IJ

H K



D

D

D

D

1

q1 ln 21 + q2 ln 22 + q3 ln 23 + q4 ln 24

D14

D13

D12

D11

2 p Îo



...(3.41)



D11 = D22 = r, D12 = D21 = D.

D23 = D14 = 4h2 + D 2 , D13 = D24 = 2h

q1 = q, q2 = –q, q3 = –q and q4 = q



V12 =



LM F I

MN GH JK



FG IJ FG 4h + D IJ + q lnFG

H K GH 2h JK H

1 L

MM2q lnFGH D IJK + 2q ln FGH 4h 2h+ D IJK OPP

=

2p Î

N r

Q

L D . 2h OP

q

=

ln M

pÎ

MN r 4h + D PQ

2



D

r

1

q ln

- q ln

- q ln

r

D

2 p Îo



\



V12



\



V12



2



o



o



2



2



2



2



2h

4h2 + D 2



IJ OP

K PQ



Capacitance of Transmission Lines 63



\



V12



q

=

ln

p Îo



LM

MM D .

MM r . FG1 + D IJ

MMN H 4h K

2



1

2



OP

PP Volts

PP

PPQ



1

2



OP F/m

PP

PP

PP

Q



2



\



C12 =



q

=

V12



p Îo



LM

D.

ln M

MM F D I

MMr. GH1 + 4h JK

MN

2



2



\



C12 =



0.0121



LM

D

log M

MM F D I

MMr. GH1 + 4h JK

NM

2



2



1

2



OP mF/km

PP

PP

PP

Q



...(3.42)



...(3.43)



...(3.44)



F D IJ

From eqn. 3.44, it is observed that the presence of earth modifies the radius r to r G 1 +

H 4h K

2



2



However, the term



D2

4h 2



1

2



.



is small and hence the effect of earth on line capacitance is



negligible.

Reader is asked to derive the expression of capacitance considering the effect of earth for

three phase transposed transmission line.

[Hint: apply eqn. (3.5)]

Example 3.1: Fig. 3.10 shows a completely transposed 50 Hz, 250 km long three phase line has

flat horizontal phase spacing with 10 m between adjacent conductors. If the outside radius is 1.2

cm and the line voltage is 220 kV, determine the charging current, per phase and the total

reactive power in MVAr supplied by the line capacitance.



64 Electrical Power Systems



Fig. 3.10: Three phase bundled conductor line.



Solution. Outside radius

ro = 1.2 cm = 0.012 m., d = 0.4 m

ro . d =



Ds =



\



b

= ld



0.012 ´ 0.4 = 0.0693 m



Deq = Dab . D bc . Dca

Dab

\



g



1

3



q = b10 ´ 10.4 ´ 9.6 ´ 10g



ab . dab¢ . da¢ b . da¢ b¢



1

4



1

4



= 9.995 m



Dbc = Dab = 9.995 m



l



q = b20 ´ 19.6 ´ 20 ´ 20.4g = 19.997 m » 12.6 m

= b9.995 ´ 9.995 ´ 19.997 g = 12.594 m (exact).



Dca = d . d . d . d

ca

ca¢

c ¢a

c ¢a



1

4



1

4



1



\



Deq



3



However, approximate value of Deq can be calculated quickly, which is very very close to

exact value, i.e.



b



Deq = 10 ´ 10 ´ 20

Applying eqn. (3.34),

Can =



g



1

3



= 12.599 m » 12.6 m



0.0242

Deq

log

Ds



FG IJ mF/km

H K



0.0242

12.6

log

0.0693



FG

H



IJ mF/km

K



\



Can =



\



Can = 0.0107096 mF/km



or



Can = 0.0107096 × 10–6 × 250 Farad.



\



Can = 2.677 × 10–6 Farad.



Applying eqn. (3.30),