El-Hawary M.E. Electrical Energy Systems
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133
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×=
−
GMR
GMD
)102(
7
lnL
(5.23)
In many instances, the subconductor spacing S in the bundle circle is
given. It is easy to find the radius A using the formula
=
N
AS
π
sin2
(5.24)
which is a consequence of the geometry of the bundle as shown in Figure 5.4.
Example 5.2
Figure 5.5 shows a 1000-kv, single-phase, bundle-conductor line with eight
subconductors per phase. The phase spacing is D
1
= 18 m, and the subconductor
spacing is S = 50 cm. Each subconductor has a diameter of 5 cm. Calculate the
line inductance.
Solution
We first evaluate the bundle radius A. Thus,
=
8
sin25.0
π
A
Figure 5.4 Conductor Geometry.
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Figure 5.5 1000-kV Single-Phase Bundle-Conductor Line.
Therefore,
A = 0.6533 m
Assume that the following practical approximation holds:
GMD = D
1
= 18 m
The subconductor’s geometric mean radius is
m10947.1
10
2
5
7788.0
2
2
1
−
−
×=
×=
′
r
Thus we have
[]
[]
×
×=
′
×=
−
−
−
−
81
72
7
1
1
1
7
)6533.0)(10947.1)(8(
18
)102(
)(
GMD
)102(
ln
ArN
lnL
N
N
The result of the above calculation is
terhenries/me1099.6
7
−
×=
L
135
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The following MATLAB
script implements Example 5.2 based on Eqs. (5.21)
to (5.24)
The answers obtained from MATLAB
are as follows:
Inductance of a Balanced Three-Phase Single-Circuit Line
We consider a three-phase line whose phase conductors have the
general arrangement shown in Figure 5.6. We use the voltage drop per unit
length concept. This is a consequence of Faraday’s law. In engineering practice
we have a preference for this method. In our three-phase system, we can write
)(
)(
)(
3332231133
3232221122
3132121111
ILILILjV
ILILILjV
ILILILjV
++=
++=
++=
ω
ω
ω
Here we generalize the expressions of Eqs. (5.8) and (5.10) to give
′
×=
−
i
ii
r
lnL
1
)102(
7
(5.25)
% Example 5-2
%
N=8
S=0.5
d=0.05
r=d/2
r_prime=0.7788*r
GMD=18
A=(S/2)/sin(pi/N)
GMR=(N * r_prime *(A)^(N-1))^(1/N)
L=2*1e-7*log(GMD/GMR)
EDU»
N = 8
S = 0.5000
d = 0.0500
r = 0.0250
r_prime = 0.0195
GMD = 18
A = 0.6533
GMR = 0.5461
L = 6.9907e-007
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Figure 5.6 A Balanced Three-Phase Line.
×=
−
kj
kj
D
lnL
1
)102(
7
(5.26)
We now substitute for the inductances in the voltage drops equations and use the
condition of balanced operation to eliminate one current from each equation.
The result is
′
+
=
′
+
′
=
′
2
23
2
12
23
12
12
13
2
1
13
11
r
D
lnI
D
D
lnIV
D
D
lnI
r
D
lnIV
′
+
=
′
3
13
3
23
13
23
r
D
lnI
D
D
lnIV
(5.27)
Here,
)102(
7
−
×
=
′
ω
j
V
V
i
i
We note that for this general case, the voltage drop in phase one, for example,
depends on the current in phase two in addition to its dependence on I
1
. Thus
the voltage drops will not be a balanced system. This situation is undesirable.
Consider the case of equilaterally spaced conductors generally referred
to as the delta configuration; that is
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rrrr
DDDD
′
=
′
=
′
=
′
===
321
231312
The voltage drops will thus be given by
′
=
′
′
=
′
r
D
lnIV
r
D
lnIV
22
11
′
=
′
r
D
lnIV
33
(5.28)
And in this case the voltage drops will form a balanced system.
Consider the so often called H-type configuration. The conductors are
in one horizontal plane as shown in Figure 5.7. The distances between
conductors are thus
DD
DDD
2
13
2312
=
==
and the voltage drops are given by
′
=
′
+
′
=
′
r
D
lnIV
lnI
r
D
lnIV
22
211
2
2
′
+=
′
r
D
lnIlnIV
2
2
323
(5.29)
We note that only conductor two has a voltage drop proportional to its current.
Figure 5.7 H-Type Line.
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Figure 5.8 Transposed Line.
Transposition of Line Conductors
The equilateral triangular spacing configuration is not the only
configuration commonly used in practice. Thus the need exists for equalizing
the mutual inductances. One means for doing this is to construct transpositions
or rotations of overhead line wires. A transposition is a physical rotation of the
conductors, arranged so that each conductor is moved to occupy the next
physical position in a regular sequence such as a-b-c, b-c-a, c-a-b, etc. Such a
transposition arrangement is shown in Figure 5.8. If a section of line is divided
into three segments of equal length separated by rotations, we say that the line is
“completely transposed.”
Consider a completely transposed three-phase line. We can
demonstrate that by completely transposing a line, the mutual inductance terms
disappear, and the voltage drops are proportional to the current in each phase.
Define the geometric mean distance GMD as
31
231312
)(GMD DDD
=
(5.30)
and the geometric mean radius GMR as
r
′
=
GMR
(5.31)
we attain
139
© 2000 CRC Press LLC
terhenries/me
GMR
GMD
)102(
7
×=
−
lnL
(5.32)
Example 5.3
Calculate the inductance per phase of the three-phase solid conductor line shown
in Figure 5.9. Assume that the conductor diameter is 5 cm and the phase
separation D
1
is 8 m. Assume that the line is transposed.
Figure 5.9 A Three-Phase Line.
Solution
The geometric mean distance is given by
()
[]
m 08.10
2599.1
2GMD
1
31
111
=
=
=
D
DDD
The geometric mean radius is
m 0195.0
2
105
)(
2
41
=
×
=
′
−
−
er
Therefore,
terhenries/me1025.1
0195.0
08.10
)102(
6
7
−
−
×=
×=
lnL
Inductance of Multiconductor Three-Phase Systems
Consider a single-circuit, three-phase system with multiconductor-
configured phase conductors as shown in Figure 5.10. Assume equal current
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Figure 5.10 Multiconductor Single-Circuit Three-Phase Line.
distribution in the phase subconductors and complete transposition. We can
show that the phase inductance for the system is the following expression:
×=
−
GMR
GMD
)102(
7
lnL
(5.33)
In this case, the geometric mean distance is given by
31
)(GMD
CABCAB
DDD
=
(5.34)
where D
AB
, D
BC
, and D
CA
are the distances between phase centers. The
geometric mean radius (GMR) is obtained using the same expression as that for
the single-phase system. Thus,
()
N
N
i
si
D
1
1
GMR
=
∏
=
(5.35)
For the case of symmetrical bundle conductors, we have
[]
N
N
ArN
1
1
)(GMR
−
′
=
(5.36)
The inductive reactance per mile per phase X
L
in the case of a three-
phase, bundle-conductor line can be obtained using
daL
XXX
+=
(5.37)
where as before for 60 Hz operation,
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GMR
1
log2794.0
=
a
X
(5.38)
GMDlog2794.0
=
d
X
(5.39)
The GMD and GMR are defined by Eqs. (5.34) and (5.36).
Example 5.4
Consider a three-phase line with an eight subconductor-bundle delta
arrangement with a 42 in. diameter. The subconductors are ACSR 84/19
(Chukar) with 0534.0
=
′
r ft. The horizontal phase separation is 75 ft, and the
vertical separation is 60 ft. Calculate the inductive reactance of the line in ohms
per mile per phase.
Solution
From the geometry of the phase arrangements, we have
ft97.69
96.30cos
60
96.30
60
36
tan
=
=
=
=
$
$
AB
D
θ
θ
Thus,
[]
ft577.71)75)(97.69)(97.69(GMD
31
==
For Chukar we have 0534.0
=
′
r ft. The bundle particulars are N = 8 and A =
(42/2) in. Therefore,
ft4672.1
12
21
)0534.0(8GMR
81
7
=
=
Thus,
518.0
577.71log2794.0
0465.0
4672.1
1
log2794.0
=
=
−=
=
d
a
X
X
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© 2000 CRC Press LLC
As a result,
mileperohms4715.0
=+=
daL
XXX
Inductance of Three-Phase, Double-Circuit Lines
A three-phase, double-circuit line is essentially two three-phase circuits
connected in parallel. Normal practice calls for identical construction for the
two circuits. If the two circuits are widely separated, then we can obtain the line
reactance as simply half that of one single-circuit line. For the situation where
the two circuits are on the same tower, the above approach may not produce
results of sufficient accuracy. The error introduced is mainly due to neglecting
the effect of mutual inductance between the two circuits. Here we give a simple
but more accurate expression for calculating the reactance of double-circuit
lines.
We consider a three-phase, double-circuit line with full line
transposition such that in segment I, the relative phase positions are as shown in
Figure 5.11.
The inductance per phase per unit length is given by
()
×=
−
GMR
GMD
102
7
lnL
(5.40)
where the double-circuit geometric mean distance is given by
(
)
31
eqeqeq
GMD
ACBCAB
DDD
=
(5.41)
with mean distances defined by
()
()
41
32323223
41
21212112
eq
eq
′′′′
′′′′
=
=
DDDDD
DDDDD
BC
AB
()
41
31313113
eq
′′′′
=
DDDDD
AC
(5.42)
is
Figure 5.11 Double-Circuit Conductors’ Relative Positions in Segment I of Transposition.