Line Capacitance
Transmission line conductors exhibit capacitance with respect to each other due to the
potential difference between them. The amount of capacitance between conductors is a
function of conductor size, spacing, and height above ground. By definition, the capacitance
C is the ratio of charge q to the voltage V, given by
Consider a long round conductor far away from other conductors and earth with radius r,
carrying a charge of q coulombs per meter length as shown in Figure 1.
The charge on the conductor gives rise to an electric field with radial flux lines. The total
electric flux is numerically equal to the value of charge on the conductor. The intensity of the
field at any point is defined as the force per unit charge and is termed electric field intensity
designated as E. Concentric cylinders surrounding the conductor are equipotential surfaces
and have the same electric flux density. From Gauss‘s law, for one meter length of the
conductor, the electric flux density at a cylinder of radius x is given by
( )
The electric field intensity E may be found from the relation
For air
,
F/m
is the permittivity of free space.
The potential difference between cylinders from position D1 to D2 is defined as
the work done in moving a unit of charge of one coulomb from D 2 to D1
through the electric field produced by the charge on the conductor. This is given
by
∫
∫
∫
∫
[
]
Or,
The notation
implies the voltage drop from 1 relative to 2, that is, 1 is understood to be
positive relative to 2. The charge q carries its own sign.
Capacitance of Single-Phase Lines
Consider one meter length of a single-phase line consisting of two long solid round
conductors each having a radius r as shown in Figure 2. The two conductors are separated by
a distance D. Conductor 1 carries a charge of q1 coulombs/meter and conductor 2 carries a
charge of q2 coulombs/meter. The presence of second conductor and ground disturb the field
of the first conductor. If the distance of separation of the conductors D is great with respect to
r and the height of conductors above earth are much larger compared with D, then the
distortion effect will be small and the charge is assumed to be uniformly distributed on the
surface of the conductors.
q1+q2 =0, Let q2 = - q1 = - q
Voltage between conductor 1 and conductor 2 due to charge on conductor 1 only,
(
)
Voltage between conductor 2 and conductor 1 due to charge on conductor 2 only,
(
)
Now,
(
So.
)
(
(
(
)
&
)
(
)
(
)
)
From the principle of superposition, the potential difference due to presence of both charges
is
(
)
(
)
The Capacitance between Conductors is
F/m
The above equation gives the line to line Capacitance between the conductors. For the
purpose of transmission line modeling, we find it convenient to define a capacitance C
between each conductor and a neutral. Since the voltage to neutral is half of
, the
capacitance to neutral , C=2 C12
Or,
F/m
Potential Difference in a multiconductor configuration
Assume that the distortion effect is negligible and the charge is uniformly distributed around
the conductor, with the following constraint
Assume conductor m has a charge
conductors i and j due to the charge
(
)
(
)
coulomb/m. The potential difference
alone is
(
)
between
When i = m or j = m,
. Using superposition theorem, the potential difference
between i and j due to all charges is
∑
(
) Volts
Capacitance of a Single-phase Line
Figure 6 shows a single phase line consisting of two long solid conductors having radius r1
and r2 respectively. Conductor 1 carries a charge of q1 coulomb/meter and conductor 2 carries
a charge of q2 coulomb/meter. The field of the first conductor is disturbed due to the presence
of the second conductor and the ground. Let us assume that the distance between the
conductors D is much greater than the radius of the conductors. Also the height of the
conductors from the earth is much larger than D. Therefore, the effect of distortion is
negligible and the charge is assumed to be uniformly distributed.
Now, we have,
∑
(
∑
But
)
(
)
(
, So,
(
Generally,
*
*
( )
(
)
(
)+
)
)
,
( )+
*
( )
( )+
*
( )+
( )
The Capacitance between the conductors is
( )
F/m
is the line to line capacitance between the conductors.
Capacitance per phase (Capacitance between conductor and neutral)
C=2
( )
F/m
F/m,
F/m
Capacitance of a three-phase line with equilateral spacing
Figure 8 shows a three-phase line composed of three identical conductors of radius r placed
in equilateral configuration. Balanced three phase currents are flowing in them. The charge
on conductor A is qA coulomb/meter, that on B is qB and on C is qC coulomb/meter. Clearly,
The radius of cross-section of each conductor is r.
∑
Now,
∑
Hence,
Or,
Now,
(
)
(
*
(
*
( )
)
)
(
(
(
*
( )
(
( )
∑
*
)
)+
( )+ --------(1)
)
)
(
( )
Adding equations (1) and (2), we get
)
(
)+
( )+ -------(2)
)
( )
*
( )
( )+
*
( )
( )+
*
( )+ --------(4)
√
Hence,
(
*
( )+
√
√
√
√
[
√
[
√
*
]
(
√
√
+
-------(5)
)
(
)]
From equations (4) and (5)
*
( )+
*
Or,
( )+
Capacitance of line ‗A‘ to neutral is
( )
----(6)
Capacitive Susceptance,
Hence, line charging current of phase A is
(
)
Capacitance of three-phase line with unsymmetrical spacing
Figure 9 shows three identical conductors of radius r of a three-phase line with
unsymmetrical spacing. It is assumed that the line is fully transposed.
∑
Now,
(
)
When the conductors are in 1st section of the transposition cycle, we have
*
+ -----(1)
When the conductors are in 2nd section of the transposition cycle, we have
*
+ ----(2)
When the conductors are in 3rd section of the transposition cycle, we have
*
+ ------(3)
Taking Average
[
]
*
,
-
,
-
,
-+
*
(
)
(
)
(
)+
)
Let √(
Hence,
[
(
)
(
)]
[
(
)
(
)]
[
Similarly,
(
)
[
(
(
)] -------(5)
)
(
)
(
)] ----(4)
Adding equations (4) and (5)
[
(
But,
)
(
)
(
)]
&
(
[
(
)
[
(
)
*
(
)+
*
(
)
(
(
)]
)]
*
(
)
(
)+
)+
Hence, the capacitance of line to neutral is given by
(
)
--------(6)
It is obvious that for equilateral spacing
the exact result presented earlier.
, the above (approximate) formulae gives
The line charging current for a 3 phase line in phasor form is given by
(
)
Ampere/meter.
Effect of earth on Transmission line Capacitance
So, far in calculating the capacitance of transmission lines, the presence of earth was ignored.
The effect of earth on capacitance can be conveniently taken into account by the method of
images.
Method of Images
The electric field of transmission line conductors must conform to the presence of the earth
below. The earth for this purpose may be assumed to be a perfectly conducting horizontal
sheet of infinite extent which therefore acts like an equipotential surface.
Figure 10 shows the electric field of two long, parallel conductors charged +q and –q per
unit. It is such that it has a zero potential plane midway between them. If a conducting sheet
of infinite dimensions is placed at the zero potential plane, the electric field remains
undisturbed. Further, if the conductor carrying charge –q is now removed, the electric field
above the conducting sheet stays intact, while that below it vanishes. Using these well-known
results in reverse, we may equivalently replace the presence of ground below a charged
conductor by a fictitious conductor having equal and opposite charge and located as far below
the surface of the ground as the overhead conductor above it—such a fictitious conductor is
the mirror image of the overhead conductor. This method of creating the same electric field
as in the presence of the earth is known as the ―Method of Images‖ originally suggested by
Lord Kelving.
Capacitance of a single-phase line considering the effect of earth
Figure 11 shows a single-phase line. The conductors have radius ‗r‘ and they are separated by
a distance ‗D‘. A‘ and B‘ are images of conductors A and B. The conductors are at height ‗h‘
from the earth.
Now,
∑
∑
(
(
)
)
[
(
)
(
(
[
Or,
)
(
)
(
)
]
(
0
0
0
2
3
(
)
(
)
2
2
)
2
3
)]
31
(
)
(
)
2
31
31
(
0
)
1 --------(1)
(
)
Hence, line to line capacitance,
F/m -------(2)
(
)
{
Line to neutral capacitance,
F/m
{
.
/
.
/
}
--------(3)
}
It is observed from equations (2) and (3) that the presence of earth modifies the radius r to
{
} . For h >>D (this is normally the case), the effect of earth on line capacitance or
(
)
line to neutral capacitance is of negligible order. Under this condition,
Line to line capacitance is
, -
F/m
Capacitance of a three-phase line (considering the effect of earth)
The method of images can similarly be applied for the calculation of capacitance of a 3-phase
line, shown in Figure 12. The line is considered to be fully transposed. The conductors A, B
and C carry charges qA , qB , qC col/m and occupy positions 1, 2 and 3 respectively, in the first
section of the transposition cycle. The effect of earth is simulated by image conductors with
charges -qA , - qB , -qC respectively as shown in Figure 12.
For first 1/3rd of the line (i.e. 1st section of transposition cycle),
*
+ ----(1)
*
+
For 2nd section of transposition cycle,
*
In last section of transposition cycle,
+ ---(2)
*
+ ---(3)
Average value of voltage between conductors A & B is
[
]
[
{
}
{
}
{
}
{
}
{
}
{
}]
Or,
(
0
)
(
.
(
)
(
0
.
)
/
(
)
(
)
)1
/
(
)
(
)
.
/
(
)1
Where, √
0
(
0
. (
)
)
(
(
)/
)
2 (
(
)
)1
(
) 31 ----(4)
Similarly,
0
. (
)
(
)/
2 (
)
) 31 –(5)
(
Adding equations (4) and (5)
0
Or, 3
. (
)
0
0
(
. (
. (
)/
)
)
(
(
(
)2 (
)/
)/
)
2 (
2 (
(
)
)
) 31
(
(
) 31
) 31
0
Or,
. (
)
(
) /1
0
. (
)
(
) /1
0
. (
)
(
) /1 ----(6)
Capacitance to neutral
F/m -----(7)
(
)
(
)
It is evident that the effect of earth is to increase the capacitance of a line. If the conductors
are high above earth compared to the distances among them, the effect of earth on the
capacitance of 3-phase lines can be neglected.
Method of GMD (Modified)
A comparison of various expressions for inductance and capacitance of transmission lines
brings out the fact that the two are similar except that in inductance expressions, we have to
use the fictitious conductor radius
while in the expressions for capacitance
actual conductor radius r is used. This fact suggests that the method of GMD would be
applicable in the calculations for capacitance as well provided it is modified by using the
outer conductor radius for finding
, the self-geometric mean distance.
Problem 1: The six conductors of a double-circuit 3-phase line having an overall radius of
0.865
m are arranged as shown in Figure 15. Find the capacitive reactance to neutral
and charging current per Km per conductor at 110 KV, 50
. Assume full transposition.
Solution :
(
(
(
(
)
)
(
))
(
By symmetry,
(
))
(
(
)
(
)
)
)
(
)
)+
*(
Now,
(
(
)
(
))
(
)
(
)
(
(
)
(
))
(
)
(
)
(
By symmetry,
(
)
*(
)
)+
*(
)+
So, total capacitance to neutral per phase (there are two conductors per phase in parallel),
(
F/m
)
[{
[{
} ]
} ]
Now, h = 6 m, d = 8 m, j = 8 m, radius, r =
m
Referring to Figure (15), we have,
[( )
(
) ]
(
)
(
(
)
√
[( )
( ) ]
)
m
√
m
m
Substituting these values, we have,
F/m
[{
} ]
[2
3 ]
(
)
Hence, Capacitive reactance to neutral,
Charging current per phase
Charging current per conductor
(
√
)
A/Km
A/Km
Problem 2: A 3 phase, 50 Hz, 110 KV, transmission line has flat horizontal spacing with 3.5
m between adjacent conductors. The conductor diameter is 1.05 cm. Find the capacitance to
neutral and the charging current per Km of the line. Assume full transposition.
Solution:
The equivalent spacing,
(
)
(
Radius of conductor, r
Diameter of conductor, d
Capacitance to neutral,
(
(
)
Capacitive reactance to neutral,
Charging current,
⁄
√
)
)
(
)