Effect of Bundling and Transposition on
Series Inductance of Overhead Lines
1.0 Bundling
The text, pp 64-65, considers a four
conductor bundle in an equilateral
configuration. We will consider a twoconductor bundle in an equilateral
configuration, as shown in Fig. 1.
c-phase
5
6
D
1
D
2
a-phase
3
D
4
b-phase
Fig. 1
We assume:
 i1=i2=ia/2, i3=i4=ib/2, i5=i6=ic/2
 r’k=r’ for k=1,…,6
1
Recall our eq. (31) which is the same as eq.
3.22 in the text:
k 
0
2

1
1
1
1 
 i2 ln
 ...  ik ln  ...  in ln
i1 ln
 (31)
d k1
dk 2
rk
d kn 

Apply this to the configuration of Fig 1, for
conductor #1.
0
2
 ia
ib
ib
ic
ib
1 ia
1
1
1
1
1 
ln

ln

ln

ln

ln

ln


 2 r  2 d 12 2 d 13 2 d 14 2 d 15 2 d 16 
 i
i
i
1
1
1 
 0  a ln
 b ln
 c ln

2  2 r d 12 2 d 13 d 14 2 d 15 d 16 
1 
Bring ½ into the logarithm.
1 
0
2


1
1
1
i
ln

i
ln

i
ln
a

b
c
( r d 12 )1 / 2
(d 13 d 14 )1 / 2
(d 15 d 16 )1 / 2 

Define:
 Geometric Mean Radius (GMR) of bundle
Rb=(r’d12)1/2
 Geometric Mean Distance (GMD)
between phase positions
o From phase a to phase b
D1b=(d13d14)1/2 ≈(D2)1/2=D
o From phase a to phase c
D1c=(d15d16)1/2 ≈(D2)1/2=D
2
Substituting D into the last expression for λ1
1 
0
2

1
1
1
 i b ln  i c ln 
i a ln
Rb
D
D

(1)
Factor out the ln(1/D):
1 
0
2


1
1
 ln i b  i c 
i a ln
Rb
D


(2)
Assuming balanced conditions, so that
ib+ic=-ia, we have
1 
0
2

1
1  0 ia
D
i
ln

i
ln
ln
a

a
Rb
D  2
Rb

Use l=λ/i to get inductance of conductor 1:
l1 
0 ia
i
D
D
ln
 2 0 a ln
2 ( ia / 2) Rb
2
Rb
We can repeat the process for conductor 2,
which is the other conductor in the phase a
bundle. We will obtain:
l2  2
 0 ia
D
ln
2
Rb
Since we have two equal conductors in
parallel, the phase a inductance is just:
la 
l1  0 i a
D

ln
2
2
Rb
Since we have an equilateral configuration
of the phases,
3
0 D
lb  lc  la 
ln
2 Rb
(3)
Recall the solution to example 3.2 in the text
where we had equilateral configuration but
no bundling, i.e., only 1 conductor per
phase. In this case,
0 D
lb  lc  la 
ln
2 r 
(4)
Note that the only difference between the
inductance expression for bundled (3) and
non-bundled (4) equilateral configurations is
the use of Rb for bundled case vs. the use of
r’ for non-bundled case, where Rb=(r’d12)1/2.
How does Rb related to r’?
To answer this question, note that it will
always be the case that the distance of
separation between conductors in a bundle
4
will exceed the radius of a conductor, i.e.,
d12>r’ (Even Fig. 1, which is not drawn to
scale, would suggest this).
The implication of this fact on Rb=(r’d12)1/2
is that Rb>r’.
Therefore, D/Rb<D/r’.
Therefore ln (D/Rb)<ln (D/r’).
Therefore the inductance of a bundled
conductor configuration is less than the
inductance of a non-bundled conductor
configuration, given the same D.
Therefore the reactance of a bundled
conductor configuration is less than the
reactance of a non-bundled conductor
configuration, given the same D.
In the text’s treatment of the case of 4
conductors per bundle, they get the same eq.
(3), but Rb=(r’d12d13d14)1/4.
5
In general for bundled conductors with b
conductors per bundle, Rb=(r’d12…d1b)1/b.
Question: As we increase b, what does
geometry tend to become?
b=2
b=5
b=3
b=6
b=4
b=8
Fig. 2
We observe that as b gets large, the bundle
begins to take on the configuration of a
circle, with all current flowing at the
6
circumference. A hollow conductor!!!! So
the denominator Rb tends from r’ (for the
case of b=1) to the radius of the circle
comprised by the bundle as b gets large.
Exercise 1:
It asks you to look at the case of b=31 when
conductors are located around the
circumference of a circle of radius=1 ft, with
conductor radius=0.1 ftr’=0.078 ft.
The circle would have a circumference of
2π. Therefore, the spacing between 31
conductors would be (along the arc of the
circle) 2π/31≈0.2.
7
1 ft
Fig. 3
The problem is asking you to graphically
measure the distance from the dark
conductor to every other conductor. Then
compute Rb=(r’d12d13d14…d131)1/31. You
should get ~1.0 ft. Compare to r’=0.078 ft
for single conductor without bundling.
The basic concept to be absorbed here is that
increasing Rb of the bundle lowers the
inductance. The reason for this is that, as b
gets large, there is less and less flux interior
8
to the “circle” and therefore less flux linking
the conductors. When b is very large, there
is no flux interior to the circle, which is the
case of the hollow conductor.
Reference [1] indicates the following usages
of bundling:
345 kV: 1 or 2 conductors per phase
500 kV: 1, 2, 3, or 4 conductors per phase
765 kV: 3, 4, 6, or 8 conductors per phase
2.0 Stranding
The procedure used above to obtain Rb may
also be applied to the case of a stranded
conductor, where a single conductor consists
of several individual strands.
See problem 3.7 on how to compute RS, the
GMR of a stranded conductor in a threephase equilateral geometry. Problem 3.8
gets a numerical answer for this case. Can
work these problems in-class (see notes).
9
The bottom of page 70 in your text discusses
the difference in computing GMR for a
conductor considering stranding (as done in
Problems 3.7 and 3.8) vs. not considering
stranding (simply using r’=0.778r, where r is
the radius of the conductor), concluding
“even if we ignore the stranding, our results
are quite accurate.” Comparison of the
answers to Problems 3.9 and 3.10, and of the
answers to Problems 3.12 and 3.13, confirm
this conclusion.
3.0 Transposition
From structural point of view, it is more
economic to arrange the phases of a
transmission line in a vertical or horizontal
configuration, as shown in Fig. 4.
10
Fig. 4
These and other arrangements are not
symmetric like the equilateral configuration
is. This means that la=lb=lc is not valid. If
you have such an arrangement, and you
apply balanced voltages to the sending end
of this transmission line, and connect it to a
balanced load, you will not see a balanced
current flow. This has effects which are not
desirable (e.g., neutral current is not zero,
also generation of negative and zero
sequence currents causes increased heating
in loads).
11
To address this, line designers use
transposition where line positions are
interchanged. Position interchange is
typically done at substations.
We want to determine the influence of
transposition.
Let’s make the following assumptions:
1.The line is “completely” transposed. This
means that each phase occupies each
position for the same fraction of total
length. This assumption allows us to
compute the flux linkages for a phase by
averaging flux linkages over all three
positions. That is:
a 

1 ( 1)
 a   (a2)   (a3)
3

(3)
where the overbar indicates “average” and
(k )
the nomenclature  a , k=1,2,3, is the flux
linkages computed for phase a when
located at position k.
2.Currents are balanced, i.e., ia+ib+ic=0.
12
3.The line is bundled.
Note that assumption #3 is a departure from
the text. The text (page 68) first does the
non-bundled case and then uses that result to
get the bundled case. In contrast, we will do
both cases at the same time.
Recall eq. (1), repeated here for
convenience, which is the flux linkages for a
bundled conductor with dab=dbc=dca=D.
1 
0
2

1
1
1
 i b ln  i c ln 
i a ln
Rb
D
D

(1)
This equation also applies to our case, but
we need to change the denominators of the
second two terms.
(ak ) 
0
2

1
1
1 
i
ln

i
ln

i
ln
a
b
c
(k )
(k )  ,
Rb
d ab
d ac


k=1,2,3
(5)
where Rb is the GMR of the bundle given
by:
Rb  r d12  ,
1/ 2
1/ 3
 r d12d13  ,
for 2 conductor bundle
for 3 conductor bundle
1/ 3
 r d12d13d14  ,
for 4 conductor bundle
13
So substitution of eq. (5) for k=1,2,3 into eq.
(3) to get the average flux linkages, yields:


1 (1)
a  (a2 )  (a3 )
3
1 0 
1
1
1

i
ln

i
ln

i
ln
a
b
c
(1)
(1)
3 2 
Rb
d ab
d ac
a 
 ia ln
1
1
1
 ib ln ( 2 )  ic ln ( 2 )
Rb
d ab
d ac
(6)
1
1
1 
 ia ln
 ib ln ( 3 )  ic ln ( 3 ) 
Rb
d ab
d ac 
Adding terms that are multiplied by the
same current, we get:
a 

1
3
i
ln
 a
Rb


1
1
1 
 ib  ln (1)  ln ( 2 )  ln ( 3 ) 
d ab
d ab 
 d ab
1 0
3 2

1
1
1 

 ic  ln (1)  ln ( 2 )  ln ( 3 )  
d ac
d ac  
 d ac
(7)
Now combine logarithms in the second 2
terms:
14

1 0 
1
1
1
a 
 ib ln (1) ( 2 ) ( 3 )  ic ln (1) ( 2 ) ( 3 ) 
3ia ln
3 2 
Rb
d ab d ab d ab
d ac d ac d ac 
(8)
Now bring the 1/3 into the logarithms:

a  0
2

1
1

i
ln

i
ln
a
b
(1) ( 2 ) ( 3 )
R

d ab
d ab d ab
b



1/ 3
 ic ln
d
1
(1)
ac
( 2) ( 3)
d ac
d ac
(9)
(1) ( 2 ) ( 3 )
(1) ( 2 ) ( 3 )
d
d
d

d
ac d ac d ac
Note that ab ab ab
because
this is just the product of the distances
between the positions.
So we define:
 d

(1) ( 2 ) ( 3 ) 1 / 3
ab ab ab
 d

(1) ( 2 ) ( 3 ) 1 / 3
ac ac ac
Dm
d d
d d
(10)
where Dm is the GMD between positions.
Substitution of eq. (10) into eq. (9) yields:
0 
1
1
1 
a 
 ib ln
 ic ln
ia ln

2 
Rb
Dm
Dm 
But ia=-(ib+ic), so that eq. (11) becomes:
15
(11)


1/ 3 



a 
0
2

 0 
1
1
1
1 


i
ln

ln
i

i

i
ln

i
ln
a
a

b
c 
a
R
D
2

R
D
b
m
b
m




(12)
Factoring out ia and combining logarithms,
we get:
a 
0 ia Dm
ln
2
Rb
(13)
The inductance is then obtained by dividing
by the current ia to get:
la 
a
ia

 0 Dm
ln
2
Rb
(14)
where
 Dm is the GMD between phase positions
 Rb is the GMR of the bundle
Note that
 if the phase configuration is equilateral,
then Dm=D
 if the conductor is not bundled, then Rb=r’
We will skip section 3.6. You should take
EE 455 to learn about getting inductance for
the unbalanced line.
16
[1] “Transmission Line Reference Book: 345 kV and Above,”
EPRI, second edition, 1982.
17