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Notes 13: Shunt Admittance
13.0 Introduction
Our motivation is to obtain a distribution
system branch model for use in “phaseframe” studies, in contrast to “sequenceframe” studies. Phase-frame models for
distribution system analysis are only
appropriate for computer studies. Before
computers, sequence-frame models were the
common approach. The advantage of phaseframe modeling is improved accuracy, since
we may easily represent variability in the
elements of the primitive impedance matrix
(rather than requiring equal diagonal
elements and equal off-diagonal elements in
order to achieve decoupling of the sequence
circuits and the resulting ability to perform
per-phase analysis of each sequence circuit).
Up until now, our interest has been entirely
focused on the series impedance part of the
1
distribution branch model. This is the part of
the distribution branch for which a voltage
drops as loading current increases, and the
voltage drop always leads the current that
caused it (or the current lags the voltage).
Another effect is that leading current is
produced as a function of the voltage at a
bus. This effect is caused by capacitance. In
the case of overhead lines, the capacitance is
between phases. In the case of underground
cables, the capacitance is between each
phase conductor and its outside shield.
Which do you think is larger, on a per mile
basis?
Recall the relation for capacitance of a
parallel plate configuration:
C
A
(1)
where ε is the permittivity of the medium
between the plates, A is the area of one of
the plates, and d is the distance of separation
d
2
between them. Note that as d decreases, C
increases. For overhead lines, the separation
is on the order of several feet. In contrast,
for cables, the separation is usually less than
an inch. As a result, underground cable
capacitance is typically much larger than
overhead line capacitance.
What does this do to currents? Recall that
Xc=1/ωC, so as C goes up, reactance goes
down. For a shunt capacitor, the current is
given by V/jXC, so for greater capacitance,
we get more charging currents.
So it is common to neglect capacitance for
overhead lines, but not for cables. And even
for lines, particularly if they are long, it can
be important to model capacitance.
In our case, however, since we intend to
embed our phase-frame models in computer
programs, there is no reason not to go ahead
3
and model the capacitance, despite the fact
that its effects are relatively small.
13.1 Distribution branch model
It is useful at this point, before we proceed
to describe how to obtain the distribution
branch capacitance model, to answer the
question, “What will we do with it when we
get it?”
To answer this question, Fig.1 shows a
distribution branch model. Note the
similarity between this model and the πequivalent model presented in EE 303 (and
EE 456) for transmission lines, with the
main difference being that here we see all
three phases.
Note the presence of the shunt components
characterized by the Yabc matrix (half on left
and half on right). Also note that n and m
indicate nodes (not neutral conductors).
4
Node - n
+
Vag n
+
Vbg n
Ia n
Iline a
Z aa
Ib n
Iline b
Z bb
Z ab
Z cc
Z bc
Ic n
Iline c
Ia m
-
-
[ICabc] n
+
Ib m
Z ca
Vag m
+
Ic m
Vbg m
+
+
Vcg n
Node - m
1
[Yabc]
2
[ICabc] m
-
1
[Yabc]
2
Vcg m
-
-
-
Fig. 1
We can relate voltages on the left to voltages
on the right using KVL according to:
Vagn  Vagm   Z aa

 
 
V

V
 bgn   bgm    Z ba
Vcgn  Vcgm   Z ca

 

Z ac   Ilinea 
Z bc   Ilineb 
Z cc   Ilinec 
Z ab
Z bb
Z cb
(2)
We can also relate currents to the left of
node m to currents to the right of node m
using KCL at node m according to:
 Ilinea   I am 
Yaa Yab Yac  Vagm 
 Iline    I   1 Y
 V 
Y
Y
b
bb
bc   bgm 

 bm  2  ba
 Ilinec   I cm 
Yca Ycb Ycc  Vcgm 
Substituting eq. (3) into eq. (2) yields:
5
(3)
Vagn  Vagm 

 

V

V
 bgn   bgm 
Vcgn  Vcgm 

 

 Z aa
  Z ba
 Z ca
Z ab
Z bb
Z cb
Z ac    I am 
Yaa Yab Yac  Vagm   (4)

1


Z bc    I bm   Yba Ybb Ybc  Vbgm  
2




Yca Ycb Ycc  Vcgm  
Z cc    I cm 

We can go further with this, and we will, but
this suffices, for the moment, to motivate
our work on capacitance, which we will
eventually convert to the Yabc matrix used in
eq. (4) above.
13.2 Infinite Uniform Charged Conductor
Consider a charge Q on infinitely long
charge on a conductor of length L and radius
r. Then the charge density is
Q
q   Q  Lq
L
(5)
6
L
R
+
+
+
+ + +
+ ++ +
+
+ +
+
+
+ +
Fig. 2
Now consider a cylindrical surface, Fig. 2,
with a cylinder enclosing the conductor
having surface area A. The radius of the
cylinder is R, where R>>r.
Define E (volts/meter) as the electric field
intensity and Dq (coulomb/m2) as the electric
flux density, both vectors directed radially
outwards from the conductor such that
Dq  E
(6)
12






8
.
85

10
f / meter .
Here,
0 r ,where
0
Also define da as a vector of differential
length normal and outwards to the surface of
the cylinder.
7
Then recall Gauss’ law for electric fields
which says that the surface integral of the
dot product of Dq and da equals the charge
enclosed, i.e.,
A Dq  da  Q  qL
(7)
Integrating about the cylindrical surface, we
obtain:
q
Dq ( 2RL)  qL  Dq 
(8)
2R
Substituting eq. (6) into eq. (8) results in
q
E
(9)
2R
Now recall that potential difference between
two points in space pa and pb that are located
a distance Da and Db, respectively, from the
conductor is obtained by
v ab  v a  v b 
Db
 E  dl
Da
8
(10)
Since E and dl are both in the radially
outwards direction, and using eq. (9), with R
replaced by l, we have
Db
Db
q
vab  va  vb   E  dl  
dl
2l
Da
Da
Db
Db
1
q

dl 
ln

2  Da l
2  Da
q
(11)
13.3 Capacitance of a 2-wire line
Consider two straight infinitely long
conductors separated in space by a distance
D12, having radii of r1 and r2, respectively.
Point a
Point b
q1
q2
D12
Fig. 3
9
We desire to obtain the potential difference
between two arbitrary points in space due to
the charges residing on the conductors.
Our approach will be to use superposition
and obtain
(1)
v
 ab , the potential difference due to q1, and
(2)
 v ab , the potential difference due to q2.
Then we will add to obtain the total potential
difference.
From eq. (11), we know that the potential
difference between two points D1a and D1b
away from conductor 1 is given by
D1b
q1
(1)
vab 
ln
(12)
2  D1a
Likewise, the potential difference between
two points D2a and D2b away from conductor
b is given by
v
( 2)
ab
D2b
q2

ln
2  D2a
(13)
Then, by superposition,
10
v ab  v
(1)
ab
v
( 2)
ab
D1b
D2b
q1
q2

ln

ln
2  D1a 2  D2a
D1b
D2b 
1 

 q2 ln
q1 ln

2  
D1a
D2a 
(15)
If we wanted to obtain the voltage drop
between the two conductors, we would
simply place point a on conductor 1 and
point b on conductor 2. The result would be
D12
r2 
1 
v ab 
 q2 ln
q1 ln

2  
r1
D12 
(16)
13.4 Voltage eq. for multi-wire configuration
Consider having N charged conductors
labeled 1,…,N. Then eq. (15) generalizes,
and we may obtain the voltage drop between
any two points a and b in space as
D1b
DNb 
1 
v ab 
 ...  q N ln
q1 ln

2  
D1a
DNa  (17)
Now let’s assume that we place the point a
on conductor i and the point b on conductor
11
j, where k=1,…,i,…,j,…N. Then eq. (17)
becomes
vab 
1 
D1b
Dib
q
ln

...
q
ln
 ...
 1
i
2  
D1a
ri
DNb 
 q j ln
 ...  qN ln

D ja
DNa 
rj
(18)
But then Dib=Dja=Dij, and the voltage we are
computing is vij. So,
D1 j
Dij
1 
vij 
 ...qi ln
 ...
q1 ln
2  
D1i
ri
DNj 

 q j ln
 ...  qN ln

Dij
DNi 

rj
(19)
A more compact version of eq. (19) is:
v ij 
1
N
q

2 
k 1
k
ln
Dkj
(20)
Dki
where
Dkj=distance between conductors k and j, ft.
Dki=distance between conductors k and i, ft.
Dkk=rk, the radius of conductor k.
12
13.5 Capacitance of overhead lines
Although the ground does not contribute
much capacitance for overhead lines, it does
contribute some. To account for this
influence, and the influence from nearby
conductors, we use the method of images.
The method of images proceeds from the
fact that a point or line charge above a
conducting plane, illustrated in Fig 4a, will
produce an electric field exactly the same as
the electric field produced by the same
configuration together with its image, but
without the conducting plane, illustrated in
Fig. 4b.
qi
qi
Dij
qj
Dij
Sii/2
qj
Sii/2
Sjj/2
Sjj/2
Sij
Sii/2
i’
Fig. 4a
13
Sjj/2
-qi
-qi
Fig. 4b
j’
Note that
 The image charges are the negative of the
actual charges
 We may apply eq. (20) to the
configuration of Fig. 4b to compute
potential difference between any two
conductors
So let’s apply eq. (20) to compute the
potential difference between conductor i and
its image. This will be:
N
Dki'
Vii' 
qk ln

2  k 1
Dki
1
Note that k=1,…,4, with
k=1i
k=2i’
k=3j
k=4j’
Vii' 
D ji'
D j 'i ' 
D
D
1 

qi ln ii'  qi ' ln i 'i '  q j ln
 q j ' ln

2  
Dii
Di 'i
D ji
D j 'i  (21)
Also note that
Dii’= Di’i=Sii, Dii=Di’i’=ri,
Dji=Dj’i’=Dij, Dji’=Dj’i=Sij.
14
Then eq. (21) becomes:
Vii ' 
S ij
Dij 
S
r
1 

qi ln ii  qi ' ln i  q j ln
 q j ' ln

2  
ri
S ii
Dij
S ij  (22)
Now we will use the fact that qi=-qi’ and
qj=qj’, resulting in:
S ij
Dij 
S ii
ri
1 

Vii' 
qi ln
 qi ln
 q j ln
 q j ln

2  
ri
S ii
Dij
S ij 
(23)
Combining logarithms having the same
charge out front, we have:
S ij 
S ii
1 

Vii' 
2qi ln
 2q j ln
2  
ri
Dij 
(24)
Equation (24) gives the total voltage drop
between conductor i and its image. The
voltage drop between conductor i and
ground will be one-half of that given in eq.
(24), that is:
S ij 
S ii
1 

Vig 
qi ln
 q j ln
2  
ri
Dij 
(25)
In the general case, we will have N
conductors. In this case, the approach above
can be applied, and it will result in:
15
Vig 
S
S
S
S
1 
 q1 ln i 1  q2 ln i 2  ...  qi ln ii  ...  q N ln iN
2  
Di 1
Di 2
ri
DiN

 (26)

Define the self and mutual potential
coefficients:
S ii
1
ˆ
Pii 
ln
 Self:
2  ri
Sij
1
ˆ
 Mutual: Pij  2  ln Dij
(27)
(28)
With
 air  8.85  10 12 F / meter  1.424  10 2 F / mile
eqs. (27) and (28) become:
Sii
ˆ
Pii  11.17689 ln
 Self:
ri
Sij
ˆ
 Mutual: Pij  11.17689 ln Dij
(29)
(30)
where eqs. (29) and (30) are given in units
of mile/μF.
Note that we must use consistent units
within the logarithm of eqs. (29) and (30).
16
With the notation of eqs. (29) and (30), eq.
(26) becomes:


Vig  q1 Pˆi 1  q2 Pˆi 2  ...  qi Pˆii  ...  q N PˆiN (31)
Equation (31) can be applied to any
configuration of overhead conductors.
Side question: Do you think eq. (31) is also
good for cables?
Answer: No.
Why not? Because eq. (26) assumes that the
electric field from the charged conductor is
not confined, i.e., it emanates in all
directions an infinite distance. Cables, on the
other hand, are purposely shielded to
confine the electric field to the area between
the phase conductor and the shield. If the
phase conductor charge induces equal and
opposite charge on the shield such that the
charge enclosed by a cylinder at the surface
of the cable is zero, then by Gauss’ Law for
electrostatic fields, E=0.
17
Let’s apply eq. (31) to a 4 wire, three phase
overhead line with phases a, b, c, and a
neutral.
We obtain:


 q Pˆ
 q Pˆ


 q Pˆ  q Pˆ 
 q Pˆ  q Pˆ 
Vag  qa Pˆaa  qb Pˆab  qc Pˆac  qn Pˆan
V  q Pˆ  q Pˆ  q Pˆ  q Pˆ
bg
Vcg
Vng
a
ba
b
a
ca
a
na
 qb Pˆcb
 q Pˆ
b
bb
nb
(32)
c
bc
n
bn
(33)
c
cc
n
cn
(34)
c
nc
n
nn
(35)
In matrix form, this is:
Vag   Pˆaa
V   ˆ
 bg    Pba
Vcg   Pˆca
  ˆ
Vng   Pna
Pˆab
Pˆbb
Pˆcb
Pˆnb
Pˆac
Pˆbc
Pˆcc
Pˆnc
Pˆan   q a 
 
ˆ
Pbn   q b 
Pˆcn   q c 
 
ˆ
Pnn  q n 
(37)
The primitive potential coefficient matrix is:
 Pˆaa
ˆ
P
Pˆ prim   ba
 Pˆ
 ca
ˆ
 Pna
Pˆab
Pˆbb
Pˆcb
Pˆnb
Pˆac
Pˆbc
Pˆcc
Pˆnc
Pˆan 

Pˆbn 
Pˆcn 

ˆ
Pnn 
(38)
Define the primitive potential coefficient
matrix in terms of its submatrices:
18
 Pˆaa
ˆ
P
Pˆ prim   ba
 Pˆ
 ca
 Pˆna
Pˆab
Pˆbb
Pˆcb
Pˆnb
Pˆac
Pˆbc
Pˆcc
Pˆnc
Pˆan 

Pˆbn  

ˆ

Pcn

 
Pˆnn 
Pˆ  Pˆ 
Pˆ  Pˆ 
pp
pn
np
nn
(39)
Then we can re-write eq. (37) as
Vabc  
 V    
 n  
Pˆ  Pˆ  q 
Pˆ  Pˆ   q  
pp
pn
abc
np
nn
n
(40)
But [Vn]=[0].
Then we can use our Kron reduction
formula to eliminate [qn] as follows:
     Pˆ 
Pabc   Pˆ pp  Pˆ pn Pˆnn
1
np
(41)
so that
Vabc   Pabc qabc 
(42)
Now recall that in the scalar case, C=q/V
 V=q/C V=C-1q. Comparing to eq. (42),
we see that
Pabc   Cabc 1  Cabc   Pabc 1 (43)
19
So we can obtain the abc capacitance matrix
by inverting the primitive potential
coefficient matrix.
The abc capacitance matrix may be
converted to an abc admittance matrix by
multiplying by jω according to:
Yabc   jCabc   jPabc 
1
(44)
This is the abc admittance matrix that we
used in the KCL equation of eq. (3) above,
repeated here for convenience.
 Ilinea   I am 
Yaa Yab Yac  Vagm 
 Iline    I   1 Y
 V 
Y
Y
b
bb
bc   bgm 

 bm  2  ba
 Ilinec   I cm 
Yca Ycb Ycc  Vcgm 
(3)
Hurray! 
13.6 Example
An overhead 3-phase distribution line is
constructed as in Fig. 5. Determine the
primitive potential coefficient matrix, the
abc potential coefficient matrix, the abc
shunt capacitive matrix, the abc admittance
matrix. The phase conductors are 336,400
20
26/7 ACSR (dc=0.721 inches, rc=0.03004 ft)
and the neutral conductor is 4/0 6/1 ACSR
(ds=0.563inches, rs=0.02346 ft).
2.5 ft
a
4.5 ft
b
c
3.0 ft
4.0 ft
n
25.0 ft
Fig. 5
The distances are given as:
Saa=58 ft
Sab=sqrt(582+2.52)=58.0539ft
Sbb=58 ft
Sac=sqrt(582+72)=58.42 ft
Scc=58 ft
Sbc=sqrt(582+4.52)=58.1743ft
Snn=50 ft
Dab=2.5 ft
Dac=7.0 ft
Dbc=4.5 ft
We use the above information in eqs. (29)
and (30), repeated here for convenience:
21
ˆ  11.17689 ln Sii
P
ii
 Self:
ri
Sij
ˆ
 Mutual: Pij  11.17689 ln Dij
(29)
(30)
The matrix elements are:
58
 84.56miles / F
0.03004
58.0539
ˆ
Pab  11.17689 ln
 32.1522miles / F
2.5
58.42
Pˆac  11.17689 ln
 23.7147miles / F
7.0
Pˆba  Pˆab
Pˆ  Pˆ
Pˆaa  11.17689 ln
bb
aa
58.1743
ˆ
Pbc  11.17689 ln
 28.6058miles / F
4.5
Pˆca  Pˆac
Pˆ  Pˆ
cb
bc
Pˆcc  Pˆaa
So the primitive potential coefficient matrix
is:
22
Pˆ prim
 84.5600 35.1522 23.7147 25.2469
 35.1522 84.5600 28.6058 28.359 


 23.7147 28.6058 84.5600 26.6131


 25.2469 28.359 26.6131 85.6659
miles/μF
Now we do the Kron reduction, invert the
matrix, multiply by j377, and we have it!
Kron reduction:
     Pˆ  
Pabc   Pˆpp  Pˆpn Pˆnn
1
np
84.5600 35.1522 23.7147  25.2469
 35.1522 84.5600 28.6058   28.359 85.66591 25.2469 28.359 26.6131

 

 23.2469 28.6058 84.5600  26.6131

77.1194 26.7944 15.8715
 26.7944 75.1720 19.7957


15.8715 19.7957 76.2923
The above is the primitive potential
coefficient matrix. We just need to invert it
to obtain the shunt capacitive matrix:
77.1194 26.7944 15.8715
 26.7944 75.1720 19.7957


15.8715 19.7957 76.2923
1
 0.0150  0.0049  0.0019
   0.0049 0.0159  0.0031


  0.0019  0.0031 0.0143 
23
And the above is the shunt capacitive
matrix. Now we multiply it by jω, with
ω=2π(60)=377.9911 rad/sec.
 0.0150  0.0049  0.0019
Yabc  j ( 377.9911)   0.0049 0.0159  0.0031


  0.0019  0.0031 0.0143 
 j 5.6712  j1.8362  j 0.7034
   j1.8362 j 5.9774
 j1.169  S / mile


  j 0.7034  j1.169
j 5.3911 
13.7 Concentric Neutral Cable
We recall that eq. (31), and even eq. (26), is
only applicable to overhead lines. To assess
cable capacitance, we must go back before
the point where we used the method of
images (Section 13.5), because it was in
using the method of images that we were
implicitly assuming that the electric field
was not confined.
This would be eq. (20), repeated here for
convenience:
24
v ij 
1
N
q

2 
k 1
k
ln
Dkj
(20)
Dki
where
Dkj=distance between conductors k and j, ft.
Dki=distance between conductors k and i, ft.
Dkk=rk, the radius of conductor k.
Now one thing to remember about eq. (20).
It gives the potential difference between two
points in space, given the presence of any
number of charged conductors. The
individual terms in the summation require
the distances between the two points (point i
and point j) and the various charged
conductors k=1,…,N.
Now let’s consider carefully the case of the
concentric neutral cable. Fig. 6 illustrates.
We assume that the entire electric field
created by the charge on the phase
conductor is confined to the boundary of the
concentric neutral strands.
25
1
k
D12
Rb
2
012 R b
dc
j
ds
3
4
i
5
Fig. 6
Define the following:
 Rb=radius of a circle passing through the
centers of the neutral strands.
 dc=diameter of the phase conductors=2rc
 ds=diameter of a neutral strand=2rs
 k=total number of neutral strands
We use these definitions to compute the
voltage between the conductor and strand #1
in the presence of the other strands.
26
v p1 
1
N
q

2 
j 1
j
ln
D j1
D jp
1 
Rb
rS
D21

 q1 ln
 q2 ln
q p ln
2  
rc
Rb
Rb
D
D 
 ...  qi ln i 1  ...  qk ln k 1 
Rb
Rb 
(31)
Assume that the charge on each of the
neutral strands is 1/k of the charge on the
phase conductor and opposite in sign.
Therefore,
q1= q1= qi= qk= -qp/k
(32)
Substitution of eq. (32) into (31) yields
v p1
1 
Rb


q p ln
2  
rc
q p  rS
D21
Di 1
Dk 1  
 ln
 
 ln
 ...  ln
 ...  ln
k  Rb
Rb
Rb
Rb  
Factoring out qp, we obtain:
27
v p1
q p  Rb


ln
2   rc
1  rS
D21
Di 1
Dk 1  
 ln
 
 ln
 ...  ln
 ...  ln
k  Rb
Rb
Rb
Rb  
Recalling that the sum of logarithms is the
logarithm of the products, we rewrite the
above as:
v p1
q p  Rb 1  rS D21 ... Di 1 ... Dk 1  
 

  ln
ln
k
2   rc k 
Rb

(33)
Now we come to the following question….
What are the distances D21, …,Di1, …Dk1?
These are the distances between strand 1 and
all of the other strands (strand 2, 3, …, k).
How to obtain them? Go back to Fig. 6,
repeated here for convenience:
28
1
k
D12
Rb
2
012 R b
dc
j
ds
3
4
i
5
Fig. 6
Question: how to compute D21 (=D12)?
Answer: Use 2 trig identities.
1. Law of cosines:
a
c
a 2  b 2  c 2  2bc cos 
α
sin

b

1  cos 
2
2.
2
From the Law of cosines, wrspt Fig. 6,
2
D21  Rb2  Rb2  2 Rb2 cos  12
(34)
 2 Rb2 1  cos  12 
Taking square roots of both sides, we have:
29
D21  2 Rb2 1  cos12   Rb 21  cos12 
Now multiply top and bottom inside the
square root by 2.
2  21  cos 12 
D21  Rb
 2 Rb
2
1  cos12 
(35)
2
From trig identity #2 above, we recognize
the square root term as sin(θ12/2), so:
12
D21  2 Rb sin
(36)
2 ; θ12=2π/k

D21  2 Rb sin
(37)
k
Everyone is happy about this.
But now we have another small problem.
What is D31, D41,… ,Di1, …Dk1?
30
There is no reason why eq. (37) will not
apply for the other distances as well, if we
use the right angle.
But the angle is easy, it will just be a
multiple of π/k.
And the multiplier, for computing Di1, will
just be one less than i. So….
(i  1)
Di 1  2 Rb sin
(38)
k
Recall eq. (33):
v p1
q p  Rb 1  rS D21 ... Di 1 ... Dk 1  
 

  ln
ln
k
2   rc k 
Rb
  (33)
We can now rewrite the numerator of eq.
(33) using eq. (38), according to:
rS D21 D31 .. Di 1 ... Dk 1

2
( i  1)
( k  1)  (39)

 rs Rbk 1 2 sin 2 sin
...2 sin
...2 sin
k
k
k
k 

31
Now here is where I pull something out of
nowhere. The term inside the bracket
happens to be….k. Another trig identity.
In that case,
rS D21 D31 .. Di 1 ... Dk 1  rs Rbk 1 k
(40)
Substitution of eq. (40) into eq. (33) yields:
v p1
q p  Rb 1  rs Rbk 1k  
 

  ln
ln
k
2   rc k 
Rb  
(41)
But we notice now that the Rb terms cancel,
leaving
v p1
q p  Rb 1  rs k  
 

  ln
ln
2   rc k  Rb  
(42)
Eq. (42) gives the voltage drop from the
phase conductor to neutral strand #1.
 Since all the neutral strands are at the
same potential, this is the voltage drop from
the phase conductor to each and every
neutral strand.
 Since all neutral strands are grounded,
this is the voltage drop to ground.
32
v pg
q p  Rb 1  rs k  
 

  ln
ln
2   rc k  Rb  
(43)
So now recall that C=q/V, so….
C pg 
qp
v pg

2 
Rb 1  rs k 

ln
  ln
rc k  Rb 
(44)
Last issue: what value of permittivity to use?
First of all, recall that    r  0 , where
12
  0  8.85  10 F / m  0.0142F / mile
  r is the relative permittivity of the
medium in which the E-field exists.
The medium in which the E-field exists is,
for cables, not air, but rather the insulation
material. Table 1 provides typical values of
relative permittivity for standard insulating
materials.
33
Table 1: Relative permittivities
Material
Permittivity
range
Polyvinyl Chloride (PVC)
3.4-8.0
Ethylene-Propylene Rubber
2.5-3.5
(EPR)
Polyethylene (PE)
2.5-2.6
Cross-linked
Polyethlyene
2.3-6.0
(XLPE)
The admittance then becomes
Y pg  jC pg 
j 2  
S / mile
Rb 1  rs k 
(45)

ln
  ln
rc k  Rb 
Question:
Three identical concentric neutral cables are
buried in a trench spaced 6 inches apart. If
the admittance of one of them is Ypg, write
down the shunt admittance matrix Yabc.
34
13.8 Example
Three identical concentric neutral cables are
buried in a trench spaced 6 inches apart. The
cables are 15 kV, 250 MCM stranded allaluminum with 13 strands of #14 annealed,
coated copper wires, 1/3 neutral. The outside
diameter of the cable over the neutral
strands is dod=1.29 inches, and the neutral
diameter is ds=0.0641 inches. Determine the
shunt admittance matrix.
The radius is
Rb=(dod-ds)/24=(1.29-0.0641)/2=0.6132in
The neutral radius is rs=0.0641/2=0.03205in
The diameter of the 250MCM phase
conductor is rc=0.567/2=0.2835 in.
We will assume a relative permittivity of
2.3, with  0  0.0142F / mile .
35
Substitution into eq. (45) yields:
Y pg  jC pg 

j 2  
S / mile
Rb 1  rs k 

ln
  ln
rc k  Rb 
j 2 ( 2.3)( 0.0142)( 2 )( 60)
 j 96.5569S / mile
0.6132 1  0.03205(13) 
ln
  ln

0.2835 13 
0.6132 
The phase admittance is then
Y abc
 j 96.5569
 
0

0
0
j 96.5569
0
0


0

j 96.5569
13.8 Tape shielded cables
Figure 7 illustrates a tape shielded cable
with appropriate nomenclature.
36
R
AL or CU Phase
Conductor
b
Insulation
CU Tape Shield
Jacket
Fig. 7
77.3619
y

0

j
ag
Recall eq. (45), for the R
concentric neutral
cable with k neutral ln
strands.bEquation (45) is
repeated here for convenience:
RD
c
j 2  
Y pg  jC pg 
Rb 1  rs k 

ln
  ln
rc k  Rb 
S / mile
(45)
The tape-shielded cable may be thought of
as a concentric neutral cable with an infinite
number of strands, i.e., k=∞. Applying this
idea to eq. (45) results in:
37
Y pg
j 2  
 lim
S / mile
R 1 rk
k 
(46)
ln b   ln s 
rc k  Rb 
Eq. (46) is not hard to evaluate because we
know that
 as k gets big
 the natural log gets big but
 much more slowly than 1/k gets small
So the second term in the denominator of eq.
(46) is dominated by 1/k. Therefore this
second term goes to 0 as k gets big.
So we are left with:
Y pg 
j 2  
S / mile
R
ln b
rc
(47)
Equation (47) is what we will use for
computing the shunt admittance for a tape
shielded cable.
Values of permittivity should still come
from Table 1.
38
13.9 Example
Determine the shunt admittance of the
single-phase tape-shielded cable having
outside diameter of 0.88 inch with 1/0 AA
phase conductor. The thickness of the tape
shield is 5 mils.
First, we may obtain the diameter of the
phase conductor from the table of conductor
data. This is read off as 0.368 inches, so that
the radius is rc=0.184 inches.
The radius of a circle passing through the
center of the tape shield is
d s  5 / 1000
Rb 
 0.4375inches
2
Substitution into eq. (47), with   2.3 0
yields
Y pg
j 2 ( 2.3)( 0.0142)( 2 )( 60)

 j 89.3179S / mile
0.4375
ln
0.184
39