Notes 12: Series Impedance of Cables
12.0 Introduction
We desire to determine methods of
computing the series impedance for cable
configurations. The main steps for doing so
are exactly the same as the steps for
computing series impedance of overhead
lines. These steps, given in Notes 10, are:
1.Determine resistance per mile and GMR
of each conductor.
2. Determine distance between conductors.
3. Compute primitive impedance matrix
using eqs. (39,40) of notes 9 (see below)
with appropriate value of resistivity ρ.
4. Perform Kron reduction to eliminate the
presence of the neutral and thus obtain the
phase impedance matrix.
1
Zˆ ii  Ri  0.00158836 f
 1
1 
 j 0.12134ln  7.6786  ln  (39)
2 f
 ri
Zˆ ij  0.00158836 f

1 
 1

 j 0.12134ln
 7.6786  ln  (40)
2 f

 Dij

Consider Fig. 1, and observe the fact that the
neutral is actually comprised of several
conductors.
Fig. 1
2
Now we could treat each neutral wire as a
separate conductor and build a very large
primitive impedance matrix. This would be
very tedious. A better approach is to obtain
an equivalent neutral.
12.1 Definitions, nomenclature, equations
Fig. 2 defines some distances that we need:
Phase Conductor
Insulation
Jacket
R
d od
dc
Concentric Neutral Strand
Insulation Screen
ds
Fig. 2
These distances are:
 dc: phase conductor diameter (inches)
 dod: nominal diameter over the concentric
neutrals of the cable (inches)
3
 ds: diameter of a concentric neutral strand
(inches)
 R: radius of a circle passing through the
center of the concentric neutral strands (ft)
We also need these additional definitions.
 GMRc: geometric mean radius of the
phase conductor (ft)
 GMRs: geometric mean radius of a single
neutral strand
 Rc: resistance of the phase conductor
(Ω/mile)
 Rs: resistance of a single neutral strand
(Ω/mile)
 k: the number of concentric neutral strands
All of the above information is obtained
from the tables given in “Notes 11,” with the
exception of R. This is computed as
R
d od  d s 1 d od  d s

2
12
24
(1)
4
Our approach is to convert the concentric
neutral configuration to an equivalent
configuration having a single neutral, Fig. 3.
R
Fig. 3
References [1-2] show how to derive
relations
for
such
an
equivalent
configuration. We assume phase conductors
are numbered first and concentric neutrals
numbered last. For example, for 3 cables,
each with concentric neutrals, we number
5
phase conductors 1,2,3 and concentric
neutrals 4,5,6.
The relations are:
 Concentric neutral geometric mean radius:
GMRcni  GMRsi kR
k
k 1
(2)
Note that GMRi= ri based on previously
defined nomenclature.
 Distance between a concentric neutral and
an adjacent phase conductor
k
Dij  k Dnm
 Rk
(3)
where Dnm is the center-to-center distance
between phase conductors corresponding
to the concentric neutrals ij. Equation (3)
gives the geometric mean distance
between all of the concentric neutral
strands of one cable and the phase
conductor of the other cable.
6
 Distance between a concentric neutral and
its own phase conductor:
Dij  R
(4)
where R is given by eq. (1) above.
 Distance between a concentric neutral of
one cable and a concentric neutral of
another cable:
Dij  Dnm
(5)
where, as before, Dnm is the center-tocenter distance between phase conductors.
 The equivalent resistance of the concentric
neutral is
Rsi
Rcni 
(6)
k
10.2 Example
Three concentric neutral cables are buried in
a trench with spacings as shown in Fig. 4.
The cables are 15kV, 250MCM stranded allaluminum with 13 strands of #14 annealed,
7
coated copper wires (1/3 neutral). Determine
the phase impedance matrix.
4
5
0.5 ft
R
0.5 ft
R
1
6
2
R
3
Fig. 4
We obtain the data for the phase conductor
and neutral strands from the conductor data
table given at the end of “Notes 11.”
 dod=1.29 inches
 250MCM AA phase conductor:
o GMRc=0.0171 ft
o Diameter dc=0.56 inches
o Resistance Rc=0.4100 Ω/mile
 #14 copper neutral strands:
o GMRs=0.00208 ft
o Diameter ds=0.0641 inches
o Resistance Rs=14.8722 Ω/mile
The radius of the circle passing through the
center of the strands, eq. (1), is:
R
d od  d s 1.29  0.0641

 0.0511 ft
24
24
8
The GMR of the equivalent concentric
neutral is:
GMRcni  k GMRsi kR k 1
 13 ( 0.00208)13( 0.0511)12  0.0486 ft
The resistance of the equivalent concentric
neutral is
Rsi 14.8722
Rcni 

 1.1438 / mile
k
13
Now assume that the phase conductors are
numbered 1,2, and 3, and the equivalent
concentric neutrals are numbered 4, 5, and 6.
The conductor-to-conductor and concentric
neutral to concentric neutral spacings are:
D12=D21=D45=D54=0.5 ft.
D23=D32=D56=D65=0.5 ft.
D31=D13=D64=D46=1.0 ft.
The spacings between conductors and their
concentric neutrals are
D14=D25=D36=R=0.0511 ft.
9
The distances between concentric neutrals
and adjacent phase conductors are given by
eq. (3), but because Dnm>> R, we may
obtain very good accuracy by just assuming
this distance is just Dnm.
D15=D51=0.5 ft
D26=D62=0.5 ft
D61=D16=1.0 ft.
The self impedance for the cable in position
1 is:
Zˆ 11  R1  0.00158836 f
 1
1 
 j 0.12134ln  7.6786  ln 
2 f
 r1
1
1 100 

 0.41  0.0953  j 0.12134ln
 7.6786  ln

2 60 
 0.0171
 0.5053  j1.4564 / mile
10
The self impedance for the concentric
neutral for Cable #1 is
Zˆ 44  R4  0.00158836 f
 1
1 
 j 0.12134ln  7.6786  ln 
2 f
 r4
1
1 100 

 1.144  0.0953  j 0.12134ln
 7.6786  ln

2 60 
 0.0486
 1.2393  j1.3296 / mile
The mutual impedance between Cable #1
and Cable #2 is
Zˆ 12  0.00158836 f
 1
1  
 j 0.12134ln
 7.6786  ln 
2 f 
 Dij
 1
1 
 0.0953  j 0.12134ln
 7.6786  ln 
2 f
 0.5
 0.0953  j1.0468 / mile
11
The mutual impedance between Cable #1
and its concentric neutral is
Zˆ 14  0.00158836 f
 1
1  
 j 0.12134ln
 7.6786  ln 
2 f 
 Dij

1
1 
 0.0953  j 0.12134ln
 7.6786  ln 
2 f
 0.0511
 0.0953  j1.3236 / mile
The mutual impedance between the
concentric neutral of Cable #1 and the
concentric neutral of Cable #2 is:
Zˆ 45  0.00158836 f
 1
1  
 j 0.12134ln
 7.6786  ln 
2 f 
 Dij
 1
1 
 0.0953  j 0.12134ln
 7.6786  ln 
2 f
 0.5
 0.0953  j1.0468 / mile
And so on for the rest of the terms.
12
The resulting primitive impedance matrix in
partitioned form is
  Zˆ 
  Zˆ 
 Zˆ pp
ˆ
Z
 Zˆ np
pn
nn
where
Zˆ 
 0.5053  j1.4564 0.0953  j1.0468 0.0953  j 0.9627 
  0.0953  j1.0468 0.5053  j1.4564 0.0953  j1.0468 
0.0953  j 0.9627 0.0953  j1.0468 0.5053  j1.4564 
Zˆ 
 0.0953  j1.3236 0.0953  j1.0468 0.0953  j 0.9627 
  0.0953  j1.0468 0.0953  j1.3236 0.0953  j1.068 
0.0953  j 0.9627 0.0953  j1.068 0.0953  j1.3236 
pp
pn
   
Zˆ np  Zˆ pn
Zˆ 
nn
T
 1.2391  j1.3296 0.0953  j1.0468 0.0953  j 0.9627 
  0.0953  j1.0468 1.2391  j1.3296 0.0953  j1.0468 
0.0953  j 0.9627 0.0953  j1.0468 1.2391  j .1.3296
13
Applying Kron reduction results in:
     Zˆ 
1
Z   Zˆ pp  Zˆ pn Zˆ nn

0.7981  j 0.4463
 0.3191  j 0.0328
0.2849  j 0.0143
np
0.3191  j 0.0328 0.2849  j 0.0143
0.7891  j 0.4041 0.3191  j 0.0328
0.3191  j 0.0328 0.7981  j 0.4463
10.3 Tape shielded cables
The tape shielded cable uses bare copper
tape helically applied around the insulation
screen. An insulation jacket encircles the
tape shield. Fig 5 illustrates.
AL or CU Phase
Conductor
Insulation
d od ds dc
CU Tape Shield
T
Fig. 5
14
Jacket
For a given ampacity, a tape shielded
conductor is less expensive than a concentric
neutral conductor because of the simpler
construction and manufacturing cost of the
tape shield relative to the concentric neutral.
Tape shielded cables are often called power
cables because they are more frequently
used in higher current applications such as
three-phase
mainlines,
since
such
applications typically see less unbalance and
the neutral conductor need not be designed
to carry full load.
Parameters of the taped shielded cables are:
 dc: phase conductor diameter (inches)
 dod: outside diameter over the jacket
(inches)
 ds: outside diameter of the tape shield
(inches)
 T: thickness of copped tape shield (mils)
 ρ: resistivity of tape shield (Ω-meters) at
50º C.
15
With this information we may compute the
resistance of the tape shield as:
rshield  7.9385  10 8

d sT
 / mile
(7)
where the constant 7.9385E8 converts from
meters per in-mil to per mile.
Since the shield is in fact a hollow thinwalled conductor, we need not correct for
the internal flux. Therefore, the number to
use as the shield’s geometric mean radius is
exactly its radius, given as half the
difference between its outside diameter ds
and its thickness T/1000 (where we divide
by 1000 to convert from mils to inches):
GMR shield 
d s  T / 1000 1 d s  T / 1000
 
ft
2
12
24
(8)
As in the concentric neutral, we number
phase conductors 1,2,3 and tape shields
4,5,6.
16
Then spacings between a tape shield and the
conductors and other tape shields are:
 Distance between a tape shield and an
adjacent phase conductor
Dij=Dnm
where Dnm is the center-to-center distance
between phase conductors corresponding
to tape shields i and j (ft). Note that with
the concentric neutrals, this distance was
given as
k
Dij  k Dnm
 Rk
but we concluded Dnm was usually a very
good approximation.
 Distance between a tape shield and its own
phase conductor:
Dij=GMRshield
which is the radius to midpoint of the
shield (ft). In the concentric neutral case,
this was R, the radius of a circle passing
through the center of the concentric
neutral strands.
17
 Tape shield to a tape shield of another
conductor:
Dij=Dnm (ft)
This is the same as in the concentric
neutral case.
References:
[1] W. Lewis and G. Allen, “Symmetrical component circuit
constants and neutral circulating currents for concentric neutral
underground distribution cables,” IEEE Trans. on PAS, Vol. PAS97, No. 1, pp 191-199, Jan/Feb 1978.
[2] W. Lewis, G. Allen, and J. Wang, “Circuit constants for
concentric neutral underground distribution cables on a phase
basis,” IEEE Trans. on PAS, Vol. PAS-97, no. 1, pp 200-207,
Jan/Feb 1978.
18