Notes 10: Conductor Sizing & an Example

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 only appropriate 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

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: A C  (1) d 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

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 π- 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 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 Ia Ilinea Iam Node - n n Zaa Node - m + + Vag Ibn Ilineb Z Z Ibm n bb ab Zca Vag m + Ic Iline + Vbg n c Z Z Icm n cc bc Vbgm + + Vcgn 1 1 Vcg [Yabc] [Yabc] m [ICabc] 2 [ICabc] 2 ------n m

Fig. 1

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

Vagn  Vagm  Zaa Zab Zac Ilinea      V  V  Z Z Z Iline   bgn   bgm   ba bb bc  b  (2)        Vcgn  Vcgm  Zca Zcb Zcc Ilinec 

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  Iam  Yaa Yab Yac Vagm  1   Iline   I   Y Y Y  V  b   bm  2  ba bb bc  bgm  (3)        Ilinec  Icm  Yca Ycb Ycc Vcgm  Substituting eq. (3) into eq. (2) yields:

5 Vagn  Vagm      Vbgn   Vbgm      Vcgn  Vcgm    (4) Zaa Zab Zac  Iam  Yaa Yab Yac Vagm     1     Zba Zbb Zbc  Ibm  Yba Ybb Ybc Vbgm      2    Z Z Z I  Y Y Y V   ca cb cc  cm   ca cb cc  cgm  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 (5) L

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

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 2 intensity and Dq (coulomb/m ) as the electric flux density, both vectors directed radially outwards from the conductor such that

Dq  E (6) 12 Here,   0 r ,where  0  8.8510 f / meter . 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., D ° da  Q  qL  q (7) A Integrating about the cylindrical surface, we obtain: q D (2RL)  qL  D  (8) q q 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

Db v  v  v  E °dl ab a b  (10) Da

8 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 v  v  v  E °dl  dl ab a b   D D 2l a a (11) q Db 1 q D  dl  ln b 2  l 2 D Da a

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

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) . vab , the potential difference due to q1, and (2) . vab , 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

(1) q1 D1b vab  ln (12) 2 D1a Likewise, the potential difference between two points D2a and D2b away from conductor b is given by

(2) q2 D2b vab  ln (13) 2 D2a Then, by superposition,

10 (1) (2) q1 D1b q2 D2b vab  vab  vab  ln  ln 2 D1a 2 D2a (15) 1  D1b D2b   q1 ln  q2 ln  2  D1a D2a  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

1  D12 r2  vab  q1 ln  q2 ln  (16) 2  r1 D12 

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

1  D1b DNb  vab  q1 ln  ...  qN ln  (17) 2  D1a DNa  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

1  D1b Dib vab  q1 ln  ...qi ln  ... 2  D1a ri (18)  rj DNb   q j ln  ...  qN ln  Dja DNa 

But then Dib=Dja=Dij, and the voltage we are computing is vij. So,

1  D1 j Dij vij  q1 ln  ...qi ln  ... 2  D1i ri (19) rj DNj   q j ln  ...  qN ln  Dij DNi  A more compact version of eq. (19) is: N 1 Dkj vij   qk ln (20) 2 k 1 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.

Dij qj Dij qj

Sii/2 Sii/2

Sjj/2 Sjj/2

Sij Sii/2 Sjj/2

-qi -qi j’ i’ Fig. 4a Fig. 4b

13 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 1 Dki' Vii'   qk ln 2 k 1 Dki Note that k=1,…,4, with k=1i k=2i’ k=3j k=4j’ 1  D D D D  V   q ln ii'  q ln i'i'  q ln ji'  q ln j'i'  ii'  i i' j j'  (21) 2  Dii Di'i D ji D j'i  Also note that Dii’= Di’i=Sii, Dii=Di’i’=ri, Dji=Dj’i’=Dij, Dji’=Dj’i=Sij.

14 Then eq. (21) becomes: 1  S r S D  V   q ln ii  q ln i  q ln ij  q ln ij  ii'  i i' j j'  (22) 2  ri Sii Dij Sij 

Now we will use the fact that qi=-qi’ and qj=qj’, resulting in: 1  S r S D  V   q ln ii  q ln i  q ln ij  q ln ij  ii'  i i j j  (23) 2  ri Sii Dij Sij  Combining logarithms having the same charge out front, we have: 1  S S  V   2q ln ii  2q ln ij  ii'  i j  (24) 2  ri Dij  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: 1  S S  V  q ln ii  q ln ij  ig  i j  (25) 2  ri Dij  In the general case, we will have N conductors. In this case, the approach above can be applied, and it will result in:

15 1  S S S S  V   q ln i1  q ln i 2  ...  q ln ii  ...  q ln iN  ig  1 2 i N  (26) 2  Di1 Di 2 ri DiN 

Define the self and mutual potential coefficients: ˆ 1 Sii . Self: Pii  ln (27) 2 ri S ˆ 1 ij . Mutual: Pij  ln (28) 2 Dij With 12 2  air  8.85 10 F / meter  1.424  10 F / mile eqs. (27) and (28) become:

ˆ Sii . Self: Pii  11.17689ln (29) ri S ˆ ij . Mutual: Pij  11.17689ln (30) Dij 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: ˆ ˆ ˆ ˆ Vig  q1 Pi1  q2 Pi 2  ...  qi Pii  ...  qN PiN (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: ˆ ˆ ˆ ˆ Vag  qa Paa  qb Pab  qc Pac  qn Pan  (32) ˆ ˆ ˆ ˆ Vbg  qa Pba  qb Pbb  qc Pbc  qn Pbn  (33) ˆ ˆ ˆ ˆ Vcg  qa Pca  qb Pcb  qc Pcc  qn Pcn  (34) ˆ ˆ ˆ ˆ Vng  qa Pna  qb Pnb  qc Pnc  qn Pnn  (35) In matrix form, this is: V  ˆ ˆ ˆ ˆ   ag  Paa Pab Pac Pan qa       Vbg Pˆ Pˆ Pˆ Pˆ q     ba bb bc bn  b   ˆ ˆ ˆ ˆ  (37) Vcg  P P P P qc     ca cb cc cn   V ˆ ˆ ˆ ˆ q  ng  Pna Pnb Pnc Pnn  n  The primitive potential coefficient matrix is:  ˆ ˆ ˆ ˆ  Paa Pab Pac Pan  ˆ ˆ ˆ ˆ  ˆ Pba Pbb Pbc Pbn  Pprim  (38)  Pˆ Pˆ Pˆ Pˆ   ca cb cc cn  ˆ ˆ ˆ ˆ Pna Pnb Pnc Pnn  Define the primitive potential coefficient matrix in terms of its submatrices:

18  ˆ ˆ ˆ ˆ  Paa Pab Pac Pan  ˆ ˆ ˆ ˆ   ˆ ˆ  P P P P Ppp  Ppn  Pˆ   ba bb bc bn   prim  ˆ ˆ ˆ ˆ   ˆ ˆ  (39) Pca Pcb Pcc Pcn Pnp  Pnn      ˆ ˆ ˆ ˆ Pna Pnb Pnc Pnn  Then we can re-write eq. (37) as  ˆ ˆ  Vabc  Ppp  Ppn  qabc      V  ˆ ˆ  q  (40)   n   Pnp  Pnn   n  

But [Vn]=[0].

Then we can use our Kron reduction formula to eliminate [qn] as follows: ˆ ˆ ˆ 1 ˆ Pabc   Ppp  Ppn Pnn  Pnp  (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 1 1 Pabc   Cabc  Cabc   Pabc  (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: 1 Yabc   jCabc   jPabc  (44) This is the abc admittance matrix that we used in the KCL equation of eq. (3) above, repeated here for convenience.

Ilinea  Iam  Yaa Yab Yac Vagm  1   Iline   I   Y Y Y  V  b   bm  2  ba bb bc  bgm  (3)        Ilinec  Icm  Yca Ycb Ycc Vcgm  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 4.5 ft b a c

3.0 ft 4.0 ft

25.0 ft

Fig. 5 The distances are given as: 2 2 Saa=58 ft Sab=sqrt(58 +2.5 )=58.0539ft 2 2 Sbb=58 ft Sac=sqrt(58 +7 )=58.42 ft 2 2 Scc=58 ft Sbc=sqrt(58 +4.5 )=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 Sii . Self: Pii  11.17689ln (29) ri S ˆ ij . Mutual: Pij  11.17689ln (30) Dij The matrix elements are: 58 Pˆ  11.17689ln  84.56miles / F aa 0.03004 58.0539 Pˆ  11.17689ln  32.1522miles / F ab 2.5 58.42 Pˆ  11.17689ln  23.7147miles / F ac 7.0 ˆ ˆ Pba  Pab ˆ ˆ Pbb  Paa 58.1743 Pˆ  11.17689ln  28.6058miles / F bc 4.5 ˆ ˆ Pca  Pac ˆ ˆ Pcb  Pbc ˆ ˆ Pcc  Paa So the primitive potential coefficient matrix is:

22 84.5600 35.1522 23.7147 25.2469 35.1522 84.5600 28.6058 28.359  Pˆ    prim 23.7147 28.6058 84.5600 26.6131 miles/μF   25.2469 28.359 26.6131 85.6659 Now we do the Kron reduction, invert the matrix, multiply by j377, and we have it!

Kron reduction:

ˆ ˆ ˆ 1 ˆ Pabc   Ppp  Ppn Pnn  Pnp  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: 1 77.1194 26.7944 15.8715  0.0150  0.0049  0.0019 26.7944 75.1720 19.7957   0.0049 0.0159  0.0031     15.8715 19.7957 76.2923  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 Y  j(377.9911) 0.0049 0.0159  0.0031 abc    0.0019  0.0031 0.0143   j5.6712  j1.8362  j0.7034   j1.8362 j5.9774  j1.169 S / mile    j0.7034  j1.169 j5.3911 

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 N 1 Dkj vij   qk ln (20) 2 k 1 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.

k D12 2 Rb 012 Rb j d c ds 3

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 N 1 D j1 v p1   q j ln 2 j 1 D jp

1  Rb rS D21  q p ln  q1 ln  q2 ln (31) 2  rc Rb Rb

Di1 Dk1   ...  qi ln  ...  qk ln  Rb Rb 

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

1  Rb  q p ln  2  rc  q p  rS D21 Di1 Dk1   ln  ln  ...  ln  ...  ln  k  Rb Rb Rb Rb 

Factoring out qp, we obtain:

27 v p1

q p  R  ln b  2  rc  1  rS D21 Di1 Dk1   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  R 1  r D ...D ...D  (33)  p ln b   ln S 21 i1 k1    k  2  rc k  Rb 

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:

k D12 2 Rb 012 Rb j d c ds 3

Fig. 6 Question: how to compute D21 (=D12)? Answer: Use 2 trig identities. 1. Law of cosines:

a c a 2  b2  c 2  2bc cos α b

 1  cos 2. sin  2 2 From the Law of cosines, wrspt Fig. 6, 2 D  R 2  R 2  2R 2 cos 21 b b b 12 (34) 2  2Rb 1 cos12  Taking square roots of both sides, we have:

29 2 D21  2Rb 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 (35) 1  cos   2R 12 b 2 From trig identity #2 above, we recognize the square root term as sin(θ12/2), so:

12 D  2R sin ;θ12=2π/k (36) 21 b 2  D  2R sin (37) 21 b 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) D  2R sin (38) i1 b k Recall eq. (33): q  R 1  r D ...D ...D  v  p ln b  ln S 21 i1 k1  p1   k  (33) 2  rc k  Rb  We can now rewrite the numerator of eq. (33) using eq. (38), according to:

rS D21D31 ..Di1 ...Dk1

k 1   2 (i  1) (k  1)  (39)  rs Rb 2sin 2sin ...2sin ...2sin   k k k k 

Now here is where I pull something out of nowhere. The term inside the bracket happens to be….k. Another trig identity.

31 In that case, k 1 rS D21 D31 ..Di1 ...Dk1  rs Rb k (40) Substitution of eq. (40) into eq. (33) yields: q  R 1  r Rk1k  v  p ln b   ln s b  p1   k  (41) 2  rc k  Rb 

But we notice now that the Rb terms cancel, leaving q  R 1  r k  p b  s  v p1  ln   ln  (42) 2  rc k  Rb  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. q  R 1  r k  p b  s  v pg  ln   ln  (43) 2  rc k  Rb  So now recall that C=q/V, so….

32 q p 2 C pg   v pg Rb 1  rsk  (44) ln   ln  rc k  Rb  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.

Table 1: Relative permittivities Material Permittivity range

33 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 j2 Ypg  jC pg  S / mile Rb 1  rsk  (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.

13.8 Example

34 Three identical concentric neutral cables are buried 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 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 .

Substitution into eq. (45) yields:

35 j2 Ypg  jC pg  S / mile Rb 1  rsk  ln  ln  rc k  Rb  j2 (2.3)(0.0142)(2 )(60)   j96.5569S / mile 0.6132 1  0.03205(13)  ln   ln  0.2835 13  0.6132  The phase admittance is then  j96.5569 0 0  Y   0 j96.5569 0  abc    0 0 j96.5569

13.8 Tape shielded cables

Figure 7 illustrates a tape shielded cable with appropriate nomenclature.

36 AL or CU Phase R Conductor b Insulation

CU Tape Shield

Jacket 77.3619Fig. 7 y0  j Recallag eq. (45), for the Rconcentric neutral cable with k neutral lnstrands.b Equation (45) is repeated here for convenience:RD j2 c Ypg  jC pg  S / mile Rb 1  rsk  (45) ln   ln  rc k  Rb  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 j2 Y S / mile pg  lim  k Rb 1  rsk  (46) ln   ln  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: j2 Y  S / mile pg R ln b (47) rc 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  5 / 1000 R  s  0.4375inches b 2

Substitution into eq. (47), with   2.3 0 yields j2 (2.3)(0.0142)(2 )(60) Y   j89.3179S / mile pg 0.4375 ln 0.184