El-Hawary M.E. Electrical Energy Systems
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meterper farads
586.5
2
075.0
20
2
0
0
2
πε
πε
=
=
ln
C
AN
The relative error involved if we neglect earth effect is:
0014.0
1
21
=
−
AN
ANAN
C
CC
which is clearly less than 1 %.
Capacitance of a Single-Circuit, Three-Phase Line
We consider the case of a three-phase line with conductors not
equilaterally spaced. We assume that the line is transposed and as a result can
assume that the capacitance to neutral in each phase is equal to the average
value. This approach provides us with results of sufficient accuracy for our
purposes. This configuration is shown in Figure 5.17.
We use the three-phase balanced condition
0
=++
cba
qqq
The average potential on phase A
=
r
DDD
ln
q
V
a
A
31
132312
0
)(
2
πε
(5.70)
The capacitance to neutral is therefore given by
Figure 5.17 Three-Phase Line with General Spacing.
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© 2000 CRC Press LLC
=
=
r
D
ln
V
q
C
A
a
AN
eq
0
2
πε
(5.71)
where
3
132312eq
DDDD
=
(5.72)
Observe that D
eq
is the same as the geometric mean distance obtained in the case
of inductance. Moreover, we have the same expression for the capacitance as
that for a single-phase line. Thus,
meterper farad
GMD
2
0
=
r
ln
C
AN
πε
(5.73)
If we account for the influence of earth, we come up with a slightly
modified expression for the capacitance. Consider the same three-phase line
with the attendant image line shown in Figure 5.18. The line is assumed to be
transposed. As a result, the average phase A voltage will be given by
()()
=
)(
)2(3
231312
3
321132312
0
HHHr
HHHDDD
ln
q
V
a
A
πε
(5.74)
From the above,
Figure 5.18 Three-Phase Line with Ground Effect Included.
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© 2000 CRC Press LLC
=
31
231312
321
eq
0
2
HHH
HHH
r
D
ln
C
AN
πε
or
31
231312
321
eq
0
2
+
=
HHH
HHH
ln
r
D
ln
C
AN
πε
(5.75)
We define the mean distances
()
31
321
HHHH
s
= (5.76)
()
31
132312
HHHH
m
= (5.77)
Then the capacitance expression reduces to
−
=
s
m
AN
H
H
ln
r
D
ln
C
eq
0
2
πε
(5.78)
We can thus conclude that including the effect of ground will give a higher
value for the capacitance than that obtained by neglecting the ground effect.
Figure 5.19 Conductor Layout for Example 5.8.
Example 5.8
Find the capacitance to neutral for the signal-circuit, three-phase, 345-kV line
with conductors having an outside diameter of 1.063 in. with phase
configuration as shown in Figure 5.19. Repeat including the effect of earth,
assuming the height of the conductors is 50 ft.
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© 2000 CRC Press LLC
Solution
()()()
[]
meterperfarads105404.8
GMD
2
ft0443.0
)12)(2(
063.1
ft61.29
475.235.23GMD
12
0
31
−
×=
=
==
=
=
r
ln
C
r
AN
πε
()()
() ( )
0512.0
)49.110)(72.102)(72.102(
)100)(100)(100(
49.11010047
72.1021005.23
ft100502
31
22
13
22
2312
321
−=
=
=+=
=+==
=×===
ln
H
H
ln
H
HH
HHH
m
s
Thus,
meterperfarads106082.8
)0512.0505.6)(1018(
1
12
9
−
×=
−×
=
AN
C
The following MATLAB
script implements Example 5.8.
% Example 5-8
% data
D12=23.5; % ft
D23=23.5; % ft
D13=47; % ft
r=0.0443; % ft
eo=(1/(36*pi))*10^-9;
% To find the capacitance to neutral in
farads/m
GMD=(D12*D23*D13)^(1/3)
CAN=(2*pi*eo)/(log(GMD/r))
% To calculate the capacitance to
neutral,
% including the effect of earth
H1=2*50; % ft
H2=H1;
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© 2000 CRC Press LLC
MATLAB
con’t.
The results of running the script are as shown below:
Capacitance of Double-Circuit Lines
The calculation of capacitance of a double-circuit line can be quite
involved if rigorous analysis is followed. In practice, however, sufficient
accuracy is obtained if we assume that the charges are uniformly distributed and
that the charge q
a
is divided equally between the two phase A conductors. We
further assume that the line is transposed. As a result, capacitance formulae
similar in nature to those for the single-circuit line emerge.
Consider a double-circuit line with phases, A, B, C, A
′
, B
′
, and C
′
placed in positions 1, 2, 3, 1
′
, 2
′
, and 3
′
, respectively, in segment I of the
transposition cycle. The situation is shown in Figure 5.20.
The average voltage of phase A can be shown to be given by
Figure 5.20 Double-Circuit Line Conductor Configuration in Cycle Segment I of Transposition.
EDU»
GMD = 29.6081
CAN = 8.5407e-012
CAN = 8.6084e-012
H3=H1;
H12=(D12^2+H1^2)^.5;
H23=H12;
H13=(D13^2+H3^2)^.5;
Hs=(H1*H2*H3)^(1/3);
Hm=(H12*H23*H13)^(1/3);
CAN=(2*pi*eo)/(log(GMD/r)-log(Hm/Hs))
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()
()( )()
()( )
=
′′′
′′′′′′′′′′′′
2
33
2
22
2
11
6
313131133232322321212112
0
212
DDDr
DDDDDDDDDDDD
ln
q
V
a
A
πε
(5.79)
As a result,
=
GMR
GMD
2
0
ln
C
AN
πε
(5.80)
As before for the inductance case, we define
(
)
31
eqeqeq
GMD
ACBCAB
DDD
=
(5.81)
()
41
21212112
eq
′′′′
=
DDDDD
AB
(5.82)
()
41
32323223
eq
′′′′
=
DDDDD
BC
(5.83)
()
41
31313113
eq
′′′′
=
DDDDD
AC
(5.84)
The GMR is given by
()
31
CBA
rrrGMR
=
(5.85)
with
()
21
11
′
=
rDr
A
(5.86)
()
21
22
′
=
rDr
B
(5.87)
()
21
33
′
=
rDr
C
(5.88)
If we wish to include the effect of the earth in the calculation, a simple
extension will do the job.
As a result, we have
α
πε
+
=
GMR
GMD
2
0
ln
C
AN
(5.89)
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Figure 5.21 Double-Circuit Line with Ground Effect.
where GMD and GMR are as given by Eqs. (5.81) and (5.85). Also, we defined
=
m
s
H
H
ln
α
(5.90)
()
31
321
ssss
HHHH
=
(5.91)
with
()
41
2
1111
1
′′
=
HHHH
s
(5.92)
()
41
2
2222
2
′′
=
HHHH
s
(5.93)
()
41
2
3333
3
′′
=
HHHH
s
(5.94)
and
()
31
233212
mmmm
HHHH
=
(5.95)
()
41
21212112
12
′′′′
=
HHHHH
m
(5.96)
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© 2000 CRC Press LLC
()
41
31313113
13
′′′′
=
HHHHH
m
(5.97)
()
41
32323223
23
′′′′′
=
HHHHH
m
(5.98)
Capacitance of Bundle-Conductor Lines
It should be evident by now that it is sufficient to consider a single-
phase line to reach conclusions that can be readily extended to the three-phase
case. We use this in the present discussion pertaining to bundle-conductor lines.
Consider a single-phase line with bundle conductor having N
subconductors on a circle of radius A. Each subconductor has a radius of r.
We have
()
[]
meterperfarads
2
1
1
0
=
−
N
N
AN
ArN
D
ln
C
πε
(5.99)
The extension of the above result to the three-phase case is clearly
obtained by replacing D by the GMD. Thus
()
[]
=
−
N
N
AN
ArN
ln
C
1
1
0
GMD
2
πε
(5.100)
with
()
31
GMD
ACBCAB
DDD
=
(5.101)
The capacitive reactance in megaohms calculated for 60 Hz and 1 mile
of line using the base 10 logarithm would be as follows:
()
[]
=
−
N
N
c
ArN
X
1
1
GMD
log0683.0
(5.102)
dac
XXX
′
+
′
=
(5.103)
This capacitive reactive reactance can be divided into two parts
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()
[]
=
′
−
N
N
a
ArN
X
1
1
1
log0683.0
(5.104)
and
)GMDlog(0683.0
=
′
d
X
(5.105)
If the bundle spacing S is specified rather than the radius A of the circle
on which the conductors lie, then as before,
1for
sin2
>
=
N
N
S
A
π
(5.106)
5.5 TWO-PORT NETWORKS
A network can have two terminals or more, but many important
networks in electric energy systems are those with four terminals arranged in
two pairs. A two-terminal pair network might contain a transmission line model
or a transformer model, to name a few in our power system applications. The
box is sometimes called a coupling network, or four-pole, or a two-terminal
pair. The term two-port network is in common use. It is a common mistake to
call it a four-terminal network. In fact, the two-port network is a restricted four-
terminal network since we require that the current at one terminal of a pair must
be equal and opposite to the current at the other terminal of the pair.
An important problem arises in the application of two-port network
theory to electric energy systems, which is called the transmission problem. It is
required to find voltage and current at one pair of terminals in terms of
quantities at the other pair.
The transmission problem is handled by assuming a pair of equations
of the form
rrs
BIAVV
+=
(5.107)
rrs
DICVI
+=
(5.108)
to represent the two-port network. In matrix form, we therefore have
=
r
r
s
s
I
V
DC
BA
I
V
(5.109)
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For bilateral networks we have
1
=−
BCAD
(5.110)
Thus, there are but three independent parameters in the ABCD set as well.
Symmetry of a two-port network reduces the number of independent
parameters to two. The network is symmetrical if it can be turned end for end in
a system without altering the behavior of the rest of the system. An example is
the transmission line, as will be seen later on. To satisfy this definition, a
symmetrical network must have
A = D
(5.111)
We consider an important two-port network that plays a fundamental
role in power system analysis – this is the symmetrical
π
-network. Figure 5.22
shows a symmetrical
π
-network. We can show that
+=
2
1
ZY
A
(5.112)
ZB
=
(5.113)
+=
4
1
ZY
YC
(5.114)
AD
=
(5.115)
One of the most valued aspects of the ABCD parameters is that they are
readily combined to find overall parameters when networks are connected in
cascade. Figure 5.23 shows two cascaded two-part networks. We can write
Figure 5.22 A
π
-Network.