Notes 5: Symmetrical Components

Inductance of Overhead Lines

1.0 Introduction

We will develop relations for computing series inductance of overhead lines. In computing series inductance for transmission circuits, it is usually possible to compute it under assumption that phases are arranged symmetrically, because transposition is used for transmission.

1.1 Self inductance of a single conductor

We consider first getting the inductance of a single, long conductor of length Λ carrying current i.

The flux linking a single conductor from a current i in that conductor can be broken into two parts: . λINT is the flux internal to the conductor linking only part of the current

1 . λEXT is the flux external to the conductor linking all the current Fig. 1a illustrates, where the solid line is the circumference of the conductor and the light lines are lines of flux.

Fig. 1a From Ampere’s law, we have that

H  dl  iEN (1)  where: . Γ is the (closed) integration path . dl is the differential path length . iEN is the current enclosed by the path . H is the magnetic field strength . H•dl=|H||dl|cosθ, the tangential component of H to dl. Fig. 1b illustrates.

2 Direction of dl and H

Fig. 1b Our strategy is to use eq. (1) to express the flux linkages and then obtain inductance from L=λ/i.

1.1.1 External flux

This is called “Case 1” on page 57 of your text.

Since at any point on the circle, dl and H are tangential to Γ (cosθ=1), then:

H EXT  dl  H EXT dl  2xH EXT  circle of radius x Since this equals the current enclosed, i H  EXT 2x This is eq. (3.9a) of your text.

3 Since magnetic flux density is B=μH, i B  (2) EXT 2x Now recall Gauss’s law for magnetic fields:   N  B  dA Surface A where . λ' is the flux linking the conductor (we reserve λ for the flux linkage/unit length) . N is the number of turns of conductor . A is the surface area and . dA is a vector normal to the surface element dA with magnitude equal to dA. Fig. 2 illustrates.

radius=r dx

Fig. 2

4 We assume the flux for x>R to be zero. This is because there will always be a distance R large enough such that all conductors comprising the circuit are encompassed. If the sum of the currents flowing in all conductors is zero, then the “current enclosed” is zero and by eq. (1), the flux beyond this point is also zero.

Note B and dA are in the same direction.

Also note dA=Λdx.

Then, with N=1, R   B (x)dx EXT  EXT (3) r Substituting (2) into (3) yields:

5 R i i R 1   dx  dx EXT  2x 2x  x r r (4) i R i  R   ln x  ln  2 r 2  r  Since for overhead lines, the medium is air, the permeability is μ0=4πE-7 h/m, so that:

0i  R  EXT  ln  (5) 2  r  This is the same as eq. (3.11) in your text except that here,

 We use nomenclature λ’EXT and the text uses λ1.

 We have Λ in the numerator. This is because eq. (5) gives flux linkage (units of weber-turns) whereas eq. (3.11) in the text gives flux linkages per unit length (weber- turns/meter).

Do problem 3.1 here (See 456 class notes directory of my hard disk).

6 1.1.2 Internal flux

This is called “Case 2” on page 58 of the text.

By “internal,” we mean internal to the conductor.

Equation (2), repeated here for convenience, is the same here as for the external flux case, except that the current enclosed will not be the entire current. i B  (2) EXT 2x Assuming the current is distributed uniformly over the cross-section of the conductor, the current enclosed will be proportional to the cross-sectional area enclosed, which depends on the distance from the center of the conductor as indicated in eq. (6).

7 x 2 x 2 i  i  i (6) EN r 2 r 2 Replacing i in eq. (2) with the expression of (6) yields: i x 2 i x B   (7) INT 2x r 2 2 r 2 In addition, this flux does not link with the entire conductor but only a part of it, the part of it up to x. So the effective flux seen by the conductor is given by x 2 i x 3 BEFF  2 BINT  4 (8) INT r 2 r Again using Gauss’ law, similar to eq. (3), we have (for N=1), r r i x 3   B (x)dx  dx INT  EFF  4 0 INT 0 2 r (9) i r ir 4 i  x 3dx   4  4 2r 0 2r (4) 8

8 Since the medium here is steel or aluminum, the permeability is μ=μrμ0, where μr is the relative permeability of steel and aluminum, so that eq. (9) becomes:   i   r 0 (10) INT 8

This is the same as eq. (3.12) in your text except that here,

 We use nomenclature λ’INT and the text uses λ2.

 We have Λ in the numerator. This is because eq. (10) gives flux linkage (units of weber-turns) whereas eq. (3.12) in the text gives flux linkages per unit length (weber-turns/meter).

1.1.3 Total flux

From eq. (5) (for λ’EXT) and eq. (10) (for λ’INT), we have

9 0i  R  r 0i EXT  ln  ; INT  2  r  8 So the total flux linking the conductor is

0i  R  r 0i   EXT  INT  ln   2  r  8 Factor out the common terms to get:  i  R     0 ln  r  (11) 2  r 4 

Eq. (11) is the same as eq. (3.13) in text.

We can get a summation of logarithms inside the bracket if we take the natural log of exp(μr/4), i.e.,  i  R r   0  4    ln  lne  (12) 2  r  This is advantageous because now we can combine the two logarithms (sum of two logarithms is the logarithm of the product):

10  i R r   0 ln e 4 (13) 2 r Taking the exponential into the denominator of the logarithm yields:  i R   0 ln  (14) 2  r re 4 Define: Geometric Mean Radius (GMR):   r r  re 4 (15) It is the radius of an equivalent hollow cylindrical conductor that would have the same flux linkages as the solid conductor.

Eq. (15) is on page 62 of your text.

Normally, μr is about 1.0 for a nonmagnetic material such as steel or aluminum. Thus, exp(μr/4)= exp(1/4)=0.7788 So the GMR is r  0.7788r (16) Substituting eq. (16) into eq. (14), we get

11  i R   0 ln (17) 2 r The flux linkage per unit length is then  i R   0 ln (18) 2 r

1.1.4 Inductance Recalling that L=λ/i, we substitute eq. (18) for λ to obtain the inductance per unit length for a single long conductor, assuming the flux linkage a distance R from the conductor is 0.  i R 0 ln  R (19) L  2 r  0 ln i 2 r This is a fictitious inductance since the only way we can have an R beyond which λ=0 is to have the return path at distance R. But in this case, we must account for the flux linkage of the second conductor. We do that next.

12 1.2 Two conductor case

Consider 2 parallel conductors in Fig. 3.

R1c R2c

radius=r1

i2 i1 d12

Fig. 3

We desire to compute the flux from conductor 2 linking with conductor 1 up to a point c in space. A cross-sectional picture of the situation is shown in Fig. 4.

R1c R2c

radius=r1 i2

i1 d12

Fig. 4 To compute the flux from conductor 2 linking with conductor 1 up to the point c, we will integrate the flux density (using Gauss’ law – see eq. (3)) from the inner circle of Fig. 4 to the outer circle, as explained below: . The flux from conductor 2 inside the inner circle does not link (encircle) conductor 1 and therefore is not included in the flux linkage calculation

14 . The flux from conductor 2 outside point c is ignored because of our problem statement – to compute the flux linkage up to the point c in space. We actually did a similar calculation for the single conductor case when we computed the external flux from the conductor that linked with the conductor. There, we integrated from r to R and obtained eq. (18), repeated here for convenience:  i R   0 ln (18) 2 r1

Now, integrating from d12 (inner circle) to R2c (outer circle), we obtain:

0i2 R2c 1,2,c  ln (20) 2 d12 The total flux linking conductor 1, up to point c, will be that from conductor 1 and that from conductor 2, i.e.,

15 0i1 R1c 0i2 R2c 1,c  ln  ln 2 r1 2 d12 (21)   R R  0  1c 2c    i1 ln  i2 ln  2  r1 d12 

1.3 Multi-conductor case

Now consider that we have n conductors, 1, …,n, each with current i1, …in, and we desire to obtain the total flux linking conductor 1 from the other n-1 conductors, up to our point p in space.

It should be easy to see that each conductor will simply add a term to eq. (21) similar to the second term that is there. Thus, we obtain:

0  R1c R2c Rnc  1,c  i1 ln  i2 ln  ...  in ln  2  r1 d12 d1n  (22) Expand the logarithms to obtain

16 0   1 1 1  1,c    i1 ln  i2 ln  ...  in ln     2   r1 d12 d1n    i1 ln R1c  i2 ln R2c  ...  in ln Rnc    (23) Now let’s assume that all of the conductors are part of an n-1 phase system (the nth phase is the neutral). Then it will be true that

I1  I 2  ...  I n  0 (24)

 In  I1  I2  ...  In1  (25) We will use eq. (25) only in the bottom part of eq. (23). Making this substitution,

0  1 1 1 1,c  i1 ln  i2 ln  ...  in ln 2  r1 d12 d1n

(26)  i1 ln R1c  i2 ln R2c  ...  in1 ln Rn1c   i1  i2  ... in1 ln Rnc    Now expanding the bottom part of eq. (26),

17 0  1 1 1 1,c  i1 ln  i2 ln  ...  in ln 2  r1 d12 d1n

(27)  i1 ln R1c  i2 ln R2c  ...  in1 ln Rn1c   i1 ln Rnc  i2 ln Rnc  ... in1 ln Rnc    Combining logarithms in the bottom terms (difference of two logarithms is the logarithm of their ratio) results in

0  1 1 1 1,c  i1 ln  i2 ln  ...  in ln 2  r1 d12 d1n (28)  R R R   1c 2c n1c    i1 ln  i2 ln  ...  in1 ln   Rnc Rnc Rnc  Now consider taking the point c far out in space, an infinite distance from any of the conductors. In this case, all Rkc are equal, and the logarithms in the bottom term of eq. (28) all become 0! Therefore we have:

0  1 1 1  1,c  i1 ln  i2 ln  ...  in ln  (29) 2  r1 d12 d1n 

18 Furthermore, since the point c is “far out” in space, we are getting all of the flux linking conductor 1, and therefore the expression of eq. (29) gives λ1. Therefore

0  1 1 1  1  i1 ln  i2 ln  ...  in ln  (30) 2  r1 d12 d1n 

The general case, for the kth conductor, is   1 1 1 1    0 i ln  i ln  ...  i ln  ...  i ln k  1 2 k n  (31) 2  dk1 dk2 rk dkn 

Equation (31) applies whenever all currents sum to 0, which is always the case for us.

Do Problem 3.5 and Example 3.2 from text.

19 Interesting note about using eq. (31) in calculations. Units for the distances dkj and rk’ do not matter, as long as they are consistent. Why…?

Proof: Recall that eq. (25), repeated here for convenience:

In  I1  I2  ...  In1  (25) Substitution into eq. (31) yields:

0  1 1 1 1 k  I1 ln  I 2 ln  ...  I k ln  ...  I n1 ln 2  Dk1 Dk 2 rk Dkn1 1   I1  I 2  ...  I n1 ln  Dkn 

0  Dkn Dkn Dkn Dkn   I1 ln  I 2 ln  ...  I k ln  ...  I n1 ln  2  Dk1 Dk 2 rk Dkn1  This proves that as long as the sum of the currents are zero, we can use eq. (31) with any units, as long as the units are consistent throughout the equation.