anon - shunt admitance lines and cables notes 0000.pdf

7/28/2019 Anon - Shunt admitance lines and cables Notes 0000.pdf

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Notes 13: Shunt Admittance

13.0 Introduction

Our motivation is to obtain a distribution

system branch model for use in phase-

frame studies, in contrast to sequence-

frame studies. Phase-frame models for

distribution system analysis are onlyappropriate for computer studies. Before

computers, sequence-frame models were the

common approach. The advantage of phase-

frame 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


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2

distribution branch model. This is the part of

the distribution branch for which a voltage

drops as loading current increases, and thevoltage 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. Inthe 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:

d

AC

====

(1)where 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


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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 thatXc=1/C, 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 toembed our phase-frame models in computer

programs, there is no reason not to go ahead


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4

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 weget it?

To answer this question, Fig.1 shows a

distribution branch model. Note the

similarity between this model and the -

equivalent 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 componentscharacterized by the Yabc matrix (half on left

and half on right). Also note that n and m

indicate nodes (not neutral conductors).


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5

+

+

++

+

+

---

[Yabc]2

1[Yabc]

2

1

- - -

Vagn

Vcgm

Vbgm

m

Node - m

Vagm

Iam

Ibm

Icm

Zaa

Zbb

Zcc

Zab

Zbc

Zca

aIline

bIline

cIline

Ian

Ibn

Icn

Node - n

Vbgn

Vcgn[ICabc]n

[ICabc]

Fig. 1

We can relate voltages on the left to voltageson the right using KVL according to:

++++

====

c

b

a

cccbca

bcbbba

acabaa

cgm

bgm

agm

cgn

bgn

agn

Iline

Iline

Iline

ZZZ

ZZZ

ZZZ

V

V

V

V

V

V

(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:

++++

====

cgm

bgm

agm

cccbca

bcbbba

acabaa

cm

bm

am

c

b

a

V

V

V

YYY

YYY

YYY

I

I

I

Iline

Iline

Iline

2

1

(3)

Substituting eq. (3) into eq. (2) yields:


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6

++++

++++

====

cgm

bgm

agm

cccbca

bcbbba

acabaa

cm

bm

am

cccbca

bcbbba

acabaa

cgm

bgm

agm

cgn

bgn

agn

V

V

V

YYY

YYY

YYY

I

I

I

ZZZ

ZZZ

ZZZ

V

V

V

V

V

V

2

1 (4)

We can go further with this, and we will, but

this suffices, for the moment, to motivate

our work on capacitance, which we willeventually 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

LqQ

L

Qq ========

(5)


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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 fieldintensity and Dq (coulomb/m2) as the electric

flux density, both vectors directed radially

outwards from the conductor such that

EDq ==== (6)

Here, r 0==== ,where meterf /1085.812

0

==== .Also define da as a vector of differential

length normal and outwards to the surface of

the cylinder.


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Then recall Gauss law for electric fields

which says that the surface integral of thedot product of Dq and da equals the charge

enclosed, i.e.,

qLQdaDA

q ======== o (7)Integrating about the cylindrical surface, we

obtain:

R

qDqLRLD qq

2)2( ======== (8)

Substituting eq. (6) into eq. (8) results in

R

qE

2====

(9)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

========b

a

D

D

baab dlEvvv o(10)


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Since E and dl are both in the radially

outwards direction, and using eq. (9), with R

replaced by l, we have

a

b

D

D

D

D

D

D

baab

D

Dqdl

l

q

dll

qdlEvvv

b

a

b

a

b

a

ln

2

1

2

2

========

============

o

(11)

13.3 Capacitance of a 2-wire line

Consider two straight infinitely long

conductors separated in space by a distanceD12, having radii of r1 and r2, respectively.

q2

q1

D12

Point a Point b

Fig. 3


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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(abv , the potential difference due to q1, and )2(abv , 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

a

bab

D

Dqv

1

11)1( ln2

====(12)

Likewise, the potential difference between

two points D2a and D2b away from conductor

b is given by

a

b

abD

Dqv

2

22)2( ln2

====(13)

Then, by superposition,


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++++====

++++====++++====

a

b

a

b

a

b

a

b

ababab

D

DqD

Dq

D

Dq

D

Dqvvv

2

22

1

11

2

22

1

11)2()1(

lnln2

1

ln2

ln2

(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

++++====12

22

1

121 lnln

2

1

D

rq

r

Dqvab

(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

++++++++====Na

Nb

N

a

b

abD

DqD

Dqv ln...ln2

1

1

11

(17)

Now lets assume that we place the point a

on conductor i and the point b on conductor


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j, where k=1,,i,,j,N. Then eq. (17)

becomes

++++++++++++

++++++++====

Na

NbN

ja

j

j

i

ibi

a

bab

D

Dq

D

rq

rDq

DDqv

ln...ln

...ln...ln2

11

11

(18)

But then Dib=Dja=Dij, and the voltage we are

computing is vij. So,

++++++++++++

++++++++====

Ni

Nj

N

ij

j

j

i

ij

i

i

j

ij

D

Dq

D

rq

r

Dq

D

Dqv

ln...ln

...ln...ln2

1

1

1

1

(19)

A more compact version of eq. (19) is:

====

====N

k ki

kj

kijD

Dqv

1

ln2

1

(20)

where

Dkj=distance between conductors k and j, ft.Dki=distance between conductors k and i, ft.

Dkk=rk, the radius of conductor k.


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13

13.5 Capacitance of overhead lines

Although the ground does not contributemuch 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 thefact 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.

i

Sjj/2

Sjj/2

Sii/2

Sii/2Sii/2

-qi

Fig. 4b

-qi

qjqj

qiqi

Sij

Dij

Sjj/2

Dij

Fig. 4a

j


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14

Note that

The image charges are the negative of theactual chargesWe may apply eq. (20) to theconfiguration of Fig. 4b to compute

potential difference between any two

conductors

So lets apply eq. (20) to compute thepotential difference between conductor i and

its image. This will be:

====

====N

k ki

ki

kiiD

DqV

1

'' ln

2

1

Note that k=1,,4, withk=1i

k=2i

k=3j

k=4j

++++++++++++==== ij

ij

jji

ji

jii

ii

iii

ii

iii D

D

qD

D

qD

D

qD

D

qV'

''

'

'

'

''

'

'

' lnlnlnln2

1

(21)Also note that

Dii= Dii=Sii, Dii=Dii=ri,

Dji=Dji=Dij, Dji=Dji=Sij.


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Then eq. (21) becomes:

++++++++++++====

ij

ij

j

ij

ij

j

ii

ii

i

iiiii

S

Dq

D

Sq

S

rq

r

SqV lnlnlnln

2

1'''

(22)

Now we will use the fact that qi=-qi and

qj=qj, resulting in:

++++====

ij

ij

j

ij

ij

j

ii

ii

i

iiiii

S

Dq

D

Sq

S

rq

r

SqV lnlnlnln

2

1'

(23)

Combining logarithms having the same

charge out front, we have:

++++====

ij

ij

j

i

iiiii

D

Sq

r

SqV ln2ln2

2

1'

(24)

Equation (24) gives the total voltage drop

between conductor i and its image. The

voltage drop between conductor i andground will be one-half of that given in eq.(24), that is:

++++====

ij

ij

j

i

iiiig

D

Sq

r

SqV lnln

2

1

(25)

In the general case, we will have Nconductors. In this case, the approach above

can be applied, and it will result in:


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++++++++++++++++++++====

iN

iNN

i

iii

i

i

i

iig

D

Sq

r

Sq

D

Sq

D

SqV ln...ln...lnln

2

1

2

22

1

11

(26)

Define the self and mutual potentialcoefficients:

Self:i

ii

iir

SP ln

2

1

==== (27)

Mutual: ijij

ij

D

SP ln

2

1

====(28)

With

mileFmeterFair /10424.1/1085.8212

========

eqs. (27) and (28) become:

Self: i

iiii

r

SP ln17689.11 ====

(29)

Mutual:ij

ij

ijD

SP ln17689.11 ==== (30)

where eqs. (29) and (30) are given in units

of mile/F.

Note that we must use consistent units

within the logarithm of eqs. (29) and (30).


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17

With the notation of eqs. (29) and (30), eq.

(26) becomes:

iNNiiiiiig PqPqPqPqV

...

...2211 ++++++++++++++++++++==== (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 alldirections 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.


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Lets apply eq. (31) to a 4 wire, three phase

overhead line with phases a, b, c, and a

neutral.

We obtain:

annaccabbaaaag PqPqPqPqV ++++++++++++==== (32)

bnnbccbbbbaabg PqPqPqPqV ++++++++++++==== (33)

cnnccccbbcaacg PqPqPqPqV ++++++++++++==== (34)

nnnnccnbbnaang PqPqPqPqV ++++++++++++==== (35)

In matrix form, this is:

====

n

c

b

a

nnncnbna

cncccbca

bnbcbbba

anacabaa

ng

cg

bg

ag

q

q

q

q

PPPP

PPPP

PPPP

PPPP

V

V

V

V

(37)

The primitive potential coefficient matrix is:

====

nnncnbna

cncccbca

bnbcbbba

anacabaa

prim

PPPPPPPP

PPPP

PPPP

P

(38)

Define the primitive potential coefficient

matrix in terms of its submatrices:


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[[[[ ]]]] [[[[ ]]]]

[[[[ ]]]] [[[[ ]]]]

====

====

nnnp

pnpp

nnncnbna

cncccbca

bnbcbbba

anacabaa

prim

PP

PP

PPPPPPPP

PPPP

PPPP

P

(39)

Then we can re-write eq. (37) as

[[[[ ]]]][[[[ ]]]]

[[[[ ]]]] [[[[ ]]]][[[[ ]]]] [[[[ ]]]]

[[[[ ]]]][[[[ ]]]]

====

n

abc

nnnp

pnpp

n

abc

q

q

PP

PP

V

V

(40)

But [Vn]=[0].

Then we can use our Kron reduction

formula to eliminate [qn] as follows:

[[[[ ]]]] [[[[ ]]]] [[[[ ]]]][[[[ ]]]] [[[[ ]]]]npnnpnppabc PPPPP 1

====(41)so that

[[[[ ]]]] [[[[ ]]]][[[[ ]]]]abcabcabc qPV ==== (42)

Now recall that in the scalar case, C=q/V

V=q/C

V=C

-1

q. Comparing to eq. (42),we see that

[[[[ ]]]] [[[[ ]]]] 1==== abcabc CP [[[[ ]]]] [[[[ ]]]]1

==== abcabc PC (43)


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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 j according to:

[[[[ ]]]] [[[[ ]]]] [[[[ ]]]] 1========abcabcabc

PjCjY (44)This is the abc admittance matrix that we

used in the KCL equation of eq. (3) above,repeated here for convenience.

++++

====

cgm

bgm

agm

cccbca

bcbbba

acabaa

cm

bm

am

c

b

a

V

V

V

YYY

YYY

YYY

I

I

I

Iline

Iline

Iline

2

1

(3)

Hurray!

13.6 ExampleAn overhead 3-phase distribution line is

constructed as in Fig. 5. Determine the

primitive potential coefficient matrix, theabc potential coefficient matrix, the abc

shunt capacitive matrix, the abc admittance

matrix. The phase conductors are 336,400


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21

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).

25.0 ft

4.0 ft

3.0 ft

2.5 ft 4.5 ft

n

cb

a

Fig. 5

The distances are given as:

Saa=58 ft Sab=sqrt(582+2.52)=58.0539ftSbb=58 ft Sac=sqrt(58

2+7

2)=58.42 ft

Scc=58 ft Sbc=sqrt(582+4.5

2)=58.1743ft

Snn=50 ft

Dab=2.5 ft

Dac=7.0 ftDbc=4.5 ft

We use the above information in eqs. (29)

and (30), repeated here for convenience:


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22

Self:i

iiii

r

SP ln17689.11 ==== (29)

Mutual:ij

ijij

DSP ln17689.11 ==== (30)

The matrix elements are:

FmilesPaa /56.8403004.0

58ln17689.11 ========

FmilesPab /1522.325.2

0539.58

ln17689.11 ========

FmilesPac /7147.230.7

42.58ln17689.11 ========

abba PP ====

aabb PP ====

FmilesPbc /6058.285.4

1743.58ln17689.11 ========

acca PP ====

bccb PP ====

aacc PP ====

So the primitive potential coefficient matrixis:


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23

====

6659.856131.26359.282469.25

6131.265600.846058.287147.23

359.286058.285600.841522.35

2469.257147.231522.355600.84

primP miles/F

Now we do the Kron reduction, invert the

matrix, multiply by j377, and we have it!

Kron reduction:

[[[[ ]]]] [[[[ ]]]] [[[[ ]]]][[[[ ]]]] [[[[ ]]]]======== npnnpnppabc PPPPP 1

[[[[ ]]]] [[[[ ]]]]

====

2923.767957.198715.15

7957.191720.757944.26

8715.157944.261194.77

6131.26359.282469.256659.85

6131.26

359.28

2469.25

5600.846058.282469.23

6058.285600.841522.35

7147.231522.355600.841

The above is the primitive potential

coefficient matrix. We just need to invert it

to obtain the shunt capacitive matrix:

====

0143.00031.00019.0

0031.00159.00049.0

0019.00049.00150.0

2923.767957.198715.15

7957.191720.757944.26

8715.157944.261194.771


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And the above is the shunt capacitive

matrix. Now we multiply it by j, with

=2(60)=377.9911 rad/sec.

mileS

jjj

jjj

jjj

jYabc

/

3911.5169.17034.0

169.19774.58362.1

7034.08362.16712.5

0143.00031.00019.0

0031.00159.00049.0

0019.00049.00150.0

)9911.377(

====

====

13.7 Concentric Neutral Cable

We recall that eq. (31), and even eq. (26), is

only applicable to overhead lines. To assesscable 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:


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25

====

====N

k ki

kj

kijD

Dqv

1

ln2

1

(20)

whereDkj=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 chargedconductors k=1,,N.

Now lets 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.


26/39


27/39

27

++++++++++++++++

++++++++====

==== ====

b

kk

b

ii

bb

S

c

bp

N

j jp

j

jp

R

Dq

R

Dq

R

Dq

R

rq

r

Rq

D

Dqv

11

2121

1

1

1

ln...ln...

lnlnln2

1

ln2

1

(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

++++++++++++++++++++

====

b

k

b

i

bb

Sp

c

bp

p

RD

RD

RD

Rr

k

q

r

Rq

v

1121

1

ln...ln...lnln

ln2

1

Factoring out qp, we obtain:


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28

++++++++++++++++++++

====

b

k

b

i

bb

S

c

bp

p

R

D

R

D

R

D

R

r

k

r

Rq

v

1121

1

ln...ln...lnln1

ln2

Recalling that the sum of logarithms is the

logarithm of the products, we rewrite the

above as:

====

k

b

kiS

c

bp

p

R

DDDr

kr

Rq

v

1121

1

......ln

1ln

2(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:


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29

D12

i

5

j

4

3

1

2k

012

dc

Rb

ds

Rb

Fig. 6Question: how to compute D21 (=D12)?

Answer: Use 2 trig identities.

1. Law of cosines:

b

ac

2. 2

cos1

2sin

====

From the Law of cosines, wrspt Fig. 6,

(((( ))))122

12

222

21

cos12cos2

2

====++++====

b

bbb

RRRRD

(34)

Taking square roots of both sides, we have:

cos2222

bccba ++++====


30/39

30

(((( )))) (((( ))))12122

21 cos12cos12 ======== bb RRD

Now multiply top and bottom inside the

square root by 2.

(((( ))))

(((( ))))

2

cos12

2

cos122

12

1221

====

====

b

b

R

RDo

(35)

From trig identity #2 above, we recognize

the square root term as sin(12/2), so:

2sin2 1221

bRD ==== ;12=2/k (36)

kRDb

sin221 ==== (37)

Everyone is happy about this.

But now we have another small problem.

What is D31, D41, ,Di1, Dk1?


31/39

31

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 /k.

And the multiplier, for computing Di1, will

just be one less than i. So.

k

iRD bi

)1(sin21

====

(38)

Recall eq. (33):

====

k

b

kiS

c

bp

p

R

DDDr

kr

Rqv 11211

......ln

1ln

2 (33)We can now rewrite the numerator of eq.

(33) using eq. (38), according to:

====

k

k

k

i

kkRr

DDDDr

k

bs

kiS

)1(sin2...

)1(sin2...

2sin2sin2

.....

1

113121

(39)


32/39

32

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,

kRrDDDDrk

bskiS

1

113121 .....==== (40)

Substitution of eq. (40) into eq. (33) yields:

====

k

b

k

bs

c

bpp

RkRr

krRqv

1

1 ln1ln2 (41)

But we notice now that the Rb terms cancel,

leaving

====

b

s

c

bp

p

R

kr

kr

Rqv ln

1ln

2

1

(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.


33/39

33

====

b

s

c

bp

pgR

kr

kr

Rqv ln

1ln

2 (43)

So now recall that C=q/V, so.

========

b

s

c

bpg

p

pg

R

kr

kr

Rv

qC

ln1

ln

2

(44)

Last issue: what value of permittivity to use?

First of all, recall that 0 r==== , where

mileFmF /0142.0/1085.8 120 ======== 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.


34/39

34

Table 1: Relative permittivities

Material Permittivity

rangePolyvinyl Chloride (PVC) 3.4-8.0

Ethylene-Propylene Rubber

(EPR)

2.5-3.5

Polyethylene (PE) 2.5-2.6

Cross-linked Polyethlyene

(XLPE)

2.3-6.0

The admittance then becomes

mileS

R

kr

kr

R

jCjY

b

s

c

b

pgpg /

ln1

ln

2

========(45)

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.


35/39

35

13.8 Example

Three identical concentric neutral cables areburied in a trench spaced 6 inches apart. The

cables are 15 kV, 250 MCM stranded all-

aluminum 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 neutraldiameter is ds=0.0641 inches. Determine the

shunt admittance matrix.

The radius is

Rb=(d

od-d

s)/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 mileF/0142.00 ==== .


36/39

36

Substitution into eq. (45) yields:

mileSjj

mileS

R

kr

kr

R

jCjY

b

s

c

b

pgpg

/5569.96

6132.0

)13(03205.0ln

13

1

2835.0

6132.0ln

)60)(2)(0142.0)(3.2(2

/

ln1

ln

2

====

====

========

The phase admittance is then

====

5569.9600

05569.960

005569.96

j

j

j

Yabc

13.8 Tape shielded cables

Figure 7 illustrates a tape shielded cable

with appropriate nomenclature.


37/39

37

AL or CU PhaseConductor

Insulation

Jacket

CU Tape Shield

Rb

Fig. 7

Recall eq. (45), for the concentric neutral

cable with k neutral strands. Equation (45) is

repeated here for convenience:

mileS

R

kr

kr

R

jCjY

b

s

c

b

pgpg /

ln1

ln

2

========(45)

The tape-shielded cable may be thought of

as a concentric neutral cable with an infinite

number of strands, i.e., k=. Applying thisidea to eq. (45) results in:


38/39

38

mileS

R

kr

kr

R

jY

b

s

c

bkpg /

ln1

ln

2lim

==== (46)

Eq. (46) is not hard to evaluate because we

know that

as k gets bigthe natural log gets big butmuch more slowly than 1/k gets smallSo 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:

mileS

r

RjY

c

bpg /

ln

2 ====(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.


39/39

13.9 Example

Determine the shunt admittance of thesingle-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 thephase 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

inchesd

Rs

b 4375.02

1000/5====

====

Substitution into eq. (47), with 03.2 ==== yields

mileSjj

Ypg /3179.89

184.0

4375.0ln

)60)(2)(0142.0)(3.2(2

========

anon - shunt admitance lines and cables notes 0000.pdf

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