The Line Parameters

CHAPTER

THE LINE PARAMETERS

2

An electrical transmission line has four parameters that influence its operation. They are the series resistance, series inductance, shunt-connected capacitance, and shunt-connected conductance. Line conductors possess resistance, and this resistance for a specified conductor material, size, and shape that is in practice is supplied in standard tables and hand books as ohms per unit length. The magnetic field around a conductor-carrying current is in the form of closed loops linking the conductor. The electric lines of flux originate on positive charges and are radial. They terminate on negative charges. The electric and magnetic field distributions for a single-conductor and two-conductor system are shown in Fig. 2.1. In case of alternating currents the changing current produces change in flux linking the conductor or circuit. This induces a voltage in the conductor or circuit and is proportional to the rate of change of flux linkages. The flux linkages per ampere of changing current is called inductance of the line. The capacitance of the line is the charge per unit potential difference between conductors or conductor and earth. Through these line charging capacitors, small leakage currents will flow to ground. To consider the effect of this leakage current in high and extra high voltage lines, the parameter conductance (G) is used.

2.1 THE LINE RESISTANCE The effective resistance of a conductor R5

Power loss in the conductor ðohmsÞ I2

(2.1)

The DC resistance Rdc 5

ρl ðΩÞ A

(2.2)

where ρ is the resistivity of conductor; l is the length of conductor; and A is the cross-sectional area of the conductor. Due to nonuniform distribution of current density on account of skin effect, the resistance to alternating currents is slightly higher than that due to direct currents. Furthermore the variation of resistivity of conductor material with temperature is also important. For accurate calculations these effects are to be included. In general the ac resistance Rac C1:2 Rdc Electrical Power Systems. DOI: http://dx.doi.org/10.1016/B978-0-08-101124-9.00002-4 Copyright © 2017 BSP Books Pvt. Ltd. Published by Elsevier Ltd. All rights reserved.

(2.3)

5

6

CHAPTER 2 THE LINE PARAMETERS

FIGURE 2.1 Electric and magnetic fields. (A) Single-conductor system and (B) two-conductor system.

FIGURE 2.2 Conductor and internal flux.

2.2 THE LINE INDUCTANCE The inductance of a transmission line conductor is composed of two parts, one due to the flux internal to the conductor and the other due to the flux external to the conductor.

2.2.1 INDUCTANCE DUE TO INTERNAL FLUX Consider the conductor flux system shown in Fig. 2.2. Let H be the magnetic field intensity in AT/m. I is the total current enclosed by the conductor. Hx is the field intensity at a radial distance x from the center of the conductor O. we have, the relation for magneto motive force I

mmf 5

H  ds 5 I ðATÞ

(2.4)

2.2 THE LINE INDUCTANCE

7

Then, I Hx  ds 5 Ix

(2.5)

where Ix is the current enclosed up to radial distance x by the conductor. Assuming uniform current density πx2  I ðAÞ πr 2

(2.6)

x  I ðAT=mÞ 2πr2

(2.7)

Ix 5

Substituting Eq. (2.6) into Eq. (2.5) Therefore, Hx 5

The flux density at a radial distance x Bx 5 μHx

μxI ðwb=m2 Þ 2πr 2

(2.8)

In the tubular element of width dx the flux enclosed dφ 5 Bx 3 ðarea normal to flux linesÞ 5

μxI 3 ðdx 3 1Þ ðwb=m-lengthÞ 2πr 2

The flux linkages dψ per meter length due to dφ dψ 5

πx2 μIx3  dφ 5  dx ðwb-term=mÞ πr 2 2πr 4

The total internal flux linkages ψInt: 5

ðr

μIx3 μI ðwb-turns=mÞ dx 5 4 8π 2πr 0

(2.9)

If permeability of air is taken as 4π 3 1027 h/m ψInt: 5

I I 3 4π 3 107 5 3 107 ðwb-turns=mÞ 8π 2

(2.10)

Inductance due to internal flux linkages LInt: 5

which is constant.

ψint 1 5 3 107 h=m I 2

(2.11)

8

CHAPTER 2 THE LINE PARAMETERS

2.2.2 INDUCTANCE DUE TO EXTERNAL FLUX LINKAGES Consider two points P1 and P2 external to a conductor-carrying current I. It distant S1 and S2 from the center of the conductor as shown is Fig. 2.3. Consider the flux enclosed between the two concentric cylindrical surfaces with radii S1 and S2. At a radial distance x, so that S1 , x , S2, the tubular element of width dx will have a field intensity Hx given by 2πxHx 5 I; by Ampere theorem

(2.12)

The flux density correspondingly is Bx 5

μI  dx 3 1 ðwb=m2 Þ 2πx

(2.13)

The differential flux dφ in the tubular element per meter length dφ 5

μI  dx 3 1 2πx

(2.14)

The differential flux dφ in the tubular element per meter length of the conductor dψ 5

μI dx ðwb-t=mÞ 2πx

(2.15)

The total flux linkages between P1 and P2 linking the conductor ψ12 5

ð S2 S1

μI μI S2 ðwb-t=mÞ  dx 5 ln S1 2πx 2π

(2.16)

Since μ 5 1 and μr 5 4π 3 1027 for free space ψ12 5 2 3 1027 I ln

S2 ðwb-t=mÞ S1

The inductance due to flux between P1 and P2 L12 5

FIGURE 2.3 Flux linkages due to external flux.

2 3 107  I S2 S2 ln 5 2 3 1027 ln ðh=mÞ I S1 S1

(2.17)

2.2 THE LINE INDUCTANCE

9

2.2.3 INDUCTANCE OF A SINGLE-PHASE LINE Consider a two-conductor system shown in Fig. 2.4. Using Eqs. (2.11) and (2.17) the total inductance of conductor 1 

L1 5

 1 D 1 2 ln 3 1027 ðh=mÞ 2 r1

(2.18)

1

But ln A4 5 14. Using this relation

 1 D 1 ln ðh=mÞ 4 r1   1 D 27 4 5 2 3 10 ln A 1 ln ðh=mÞ r1   D ðh=mÞ 5 2 3 1027 ln 1 r1 A 4

L1 5 2 3 1027



(2.19)

1

Let r1 A 4 5 r11 Then L1 5 2 3 1027 ln

D ðh=mÞ r11

(2.20)

Given 1

A 4 5 0:7788 r11 5 0:7788 r l is called the geometric mean radius ðGMRÞ L2 5 2 3 1027

(2.21)

D ln 1 ðh=mÞ r2

The total inductance D ffi ðh=mÞ L 5 L1 1 L2 5 4 3 1027 ln pffiffiffiffiffiffiffiffiffi r11 r21

(2.22)

In general for practical systems r1 5 r2 5 r1 Hence the inductance per loop per meter length of single-phase system (two-conductor system) L 5 4 3 1027 ln

FIGURE 2.4 Two-conductor system.

D ðh=mÞ r1

(2.23)

10

CHAPTER 2 THE LINE PARAMETERS

2.2.4 FLUX LINKAGES IN A GROUP OF CONDUCTORS Consider a group of n-conductors 1, 2, 3, . . ., n as shown in Fig. 2.5 carrying currents I1, I2, I3,. . ., In, respectively. Consider a point P external to the group. The distances of the point P from the group of n conductors are given, respectively, by D1p, D2p, D3p, . . ., Dnp. The flux linkages of conductor 1 due to I1 from Eq. (2.18) excluding all flux beyond point P. ψ1p1 5 I1

    D1p D1p 1 1 2 ln 3 107 5 2 3 107 I1 ln 2 r1 r11

(2.24)

where r1 is the radius of conductor 1. In a similar way the flux-linking conductor 1 due to current I2 in conductor 2 between point P and conductor 1 ψ1p2 5 2 3 107 I2 ln

  D2p D1p

(2.25)

Extending the argument, the total flux linkages ψ1p with conductor 1 due to currents in all the n-conductors in the group, up to point P   D1p D2p D3p Dnp I ψ1p 5 2 3 10 I1 ln 1 1 ln 1 I3 ln 5 ? 1 In ln 2 D12 D13 D1n r1 7

(2.26)

Rearranging the terms   1 1 1 1 ψ1p 5 2 3 107 I1 ln 1 1 I2 ln 1 I3 ln 1 ? 1 In ln D12 D13 D1n r1

(2.27)

1½I1 ln D1p 1 I2 ln D2p 1 I3 ln D3p 1 ?In ln Dnp 

The sum of all currents I1 1 I2 1 I3 1 ? 1 In 5 0 Hence In 5 2 ½I1 1 I2 1 I3 1 ? 1 In21 Þ

FIGURE 2.5 Conductor group near point P.

(2.28)

2.2 THE LINE INDUCTANCE

11

Substituting Eq. (2.28) in (2.27) 7

(

ψ1p 5 2 3 10

I1 ln

1 1 1 1 1 I2 ln 1 I3 ln 1 ? 1 In ln 1 D D D r1 12 13 1n



1I1 ln D1p 1 I2 ln D2p 1 I3 ln D3p 1 ? 1 In21 ln Dðn21Þ Þ )

(2.29)

2ðI1 1 I2 1 I3 1 ? 1 In21 Þ ln Dnp 7

ψ1p 5 2 3 10 " 1

"

1 1 1 1 I1 ln 1 1 I2 ln 1 I3 ln 1 ? 1 In ln D12 D13 D1n r1

D1p D2p Dðn21Þp D3p 1 I2 ln 1 I3 ln 1 ? 1 In21 ln I1 ln Dnp Dnp Dnp Dnp

# #

(2.30)

If the point P is now moved to infinity, so that conductor 1 links with all the flux generated by the n-conductors, then the ratios D1p D2p D3p Dðn21Þp ; ; ; . . .; Dnp Dnp Dnp Dnp

tend to unity and the contribution of second term in Eq. (2.30) becomes negligible. Therefore   1 1 1 1 ðwb-t=mÞ ψ1p 5 ψ1 5 2 3 107 I1 ln 1 1 I2 ln 1 I3 ln 1 ? 1 In ln D12 D13 D1n r1

(2.31)

2.2.5 INDUCTANCE BETWEEN CONDUCTOR GROUPS: SELF AND MUTUAL GEOMETRIC MEAN DISTANCE Consider a single-phase line consisting of two groups of conductors. Group A has n-conductors and group B has m-conductors. Conductors in each group are identical. Each conductor in group A carries a current of I=n and each conductor in group B, the return conductor, carries a current of (I/m) (Fig. 2.6).

FIGURE 2.6 Two groups of conductor systems.

12

CHAPTER 2 THE LINE PARAMETERS

The distances are marked D11, D12, . . ., D1n and D110 ; D120 ; . . .D1m . Conductor 1 of group A has flux linkages

  I 1 1 1 1 1 ln 1 ? 1 ln ln 1 1 ln n D12 D13 D1n r1   I 1 1 1 1 ln 2 3 107 ðwb-t=mÞ 1 ln 1 ln 1 ? 1 ln m D110 D120 D130 D1m

ψ1 5 2 3 107

Simplifying 7

ψ1 5 2 3 10

I ln

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi! m D110 D120 D130 . . .D1m pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðwb-t=mÞ n r11 D12 D13 . . .D1n

(2.32)

(2.33)

The inductance of conductor 1 due to these flux linkages L1 5

ψ1 5 2n 3 1027 ln I=n

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi! m D110 D120 D130 . . .D1m pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðh=mÞ n r11 D12 D13 . . .D1n

(2.34)

Similarly the inductance L2 of conductor 2 in group A. ψ L2 5 2 5 2n 3 107 ln I=n

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ! m D210 D220 D230 . . .D2m pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðh=mÞ n r21 D21 D22 D23 . . .D2n

(2.35)

The average inductance of all the conductors in group A Laverage 5

L1 1 L2 1 L3 1 ? 1 Ln n

(2.36)

All the n conductors in group A are connected in parallel and carry equal currents (I/n). The inductance of the group or line A LA 5

Laverage L1 1 L2 1 L3 1 ?Ln 5 n n2

(2.37)

The inductance of line A containing n-conductors in parallel LA 5 2 3 107 ln

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðD110 D120 . . .D1m ÞðD210 D220 . . .D2m Þ. . .ðDn10 Dn20 . . .Dnm Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi n2 ðD11 D12 . . .D1n ÞðD21 D22 . . .D2n Þ. . .ðDn1 Dn2 . . .D2n Þ

mn

(2.38)

where r11 ; r21 ; . . .; rn1 are replaced by D11, D22, . . ., Dnn, respectively. The numerator of Eq. (2.38) contains the product of m 3 n distances and its mn-th root. This mn-th root of m 3 n distances is called mutual geometric mean distance (GMD) between the groups A mean B, containing n and m conductors each in their group. It is denoted by Dm. The denominator of Eq. (2.38) contains the product of n 3 n distances and its n2-th, root. The n2-th, root of n2 distances is called self-GMD or self-geometric mean radius (GMR) and is denoted by DS. Hence, LA 5 2 3 1027 ln

DmA ðh=mÞ DSA

(2.39)

2.2 THE LINE INDUCTANCE

13

Similarly, the inductance of the group B or line B LB 5 2 3 1027 ln

DmB ðh=mÞ DSB

(2.40)

The loop inductance L 5 LA 1 LB

(2.41)

2.2.6 INDUCTANCE OF THREE-PHASE LINES SYMMETRICAL SPACING Consider that the three-phase conductors a, b, and c are at the corners of an equilateral triangle of side Dm. Let the conductor radius be r (m) for each phase (Fig. 2.7). At any instance Ia 1 Ib 1 Ic 5 0. The flux linkages of conductor a due to current Ia, Ib, and Ic are given from Eq. (2.32) by   1 1 1 ðwb-t=mÞ ψa 5 2 3 107 Ia ln 1 1 Ib ln 1 Ic ln r D D   1 1 ðwb-t=mÞ 5 2 3 107 Ia ln 1 1 ðIb 1 Ic Þ ln r D

Since Ib 1 Ic 5 2 Ia   D ψa 5 2 3 107 Ia ln 1 ðwb-t=mÞ r

(2.42)

ψa D 5 2 3 1027 ln 1 ðh=mÞ Ia r

(2.43)

Inductance of conductor a La 5

Due to symmetry La 5 Lb 5 Lc 5 2 3 1027 ln

FIGURE 2.7 Three-phase line—symmetrical spacing.

D ðh=mÞ r1

(2.44)

14

CHAPTER 2 THE LINE PARAMETERS

2.2.7 THREE-PHASE LINE WITH UNSYMMETRICAL SPACING Let the three-phase conductors a, b, and c be placed at the corners of an unsymmetrical triangle of sides Dab, Dbc, and Dca, respectively, as shown in Fig. 2.8. The flux linkages ψa of conductor a are obtained from   1 1 1 ψa 5 2 3 107 Ia ln 1 1 Ib ln 1 Ic ln r Dab Dac

Inductance of phase a La 5

  ψa 1 Ib 1 Ic 1 5 2 3 1027 ln 1 1 ln 1 ln r Dab Dac Ia Ia Ia

Let Ia be chosen as reference phasor. I b = Ia – 120o and Ic = Ia 120° Ib 1 3 = 1 –120° = − − j Ia 2 2 Ic 1 3 = 1120° = − + j Ia 2 2

Hence,

pffiffiffi pffiffiffi     1 1 1 1 1 3 3 ln ln La 5 2 3 1027 ln 1 1 2 2 j 1 2 1j r 2 Dab 2 Dac 2 2   pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 1 Dab ðh=mÞ 5 2 3 1027 ln 1 1 ln Dab Dac 1 j 3 ln r Dac

In a similar way,

and

(2.45)

  pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi Dbc 1 Lb 5 2 3 1027 ln 1 1 ln Dbc Dba 1 j 3 ln r Dba

(2.46)

  pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi Dca 1 Lc 5 2 3 1027 ln 1 1 ln Dca Dcb 1 j 3 ln Dcb r

(2.47)

The individual phase inductances of an unsymmetrical line are complex is nature. The complex nature indicates exchange of energy between phases.

FIGURE 2.8 Three-phase line—unsymmetrical spacing.

2.4 INDUCTANCE OF A TRANSPOSED THREE-PHASE LINE

15

FIGURE 2.9 Transposition of three-phase conductors.

2.3 TRANSPOSITION OF TRANSMISSION LINES Unequal flux linkages due to asymmetrical spacing causes energy exchange between phases. To eliminate this effect, in practice, transmission lines are transposed at regular intervals as shown in Fig. 2.9. Transposition means that each phase conductor is made to occupy the original position of every other conductor over an equal distance. This is explained in Fig. 2.9. This process is repeated for the entire length of the line. This exchanging the positions of the conductors at regular intervals balances the flux linkages of each phase conductors. However, the error in inductance due to asymmetry is negligible. The average distance between conductors can be taken as p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 ðDab Dbc Dca Þ 5 Dequivalent

(2.48)

and inductance may be calculated as in the symmetric case taking Dequivalent in place of D.

2.4 INDUCTANCE OF A TRANSPOSED THREE-PHASE LINE Let the phase conductor a occupy in one cycle of transposition all the three positions 1, 2, and 3. The flux linkages are then with conductor a in position 1.   1 1 1 ðwb-t=mÞ ψa1 5 2 3 107 Ia ln 1 1 Ib ln 1 Ic ln r D12 D13

Similarly when conductor a occupies position 2   1 1 1 ðwb-t=mÞ ψa2 5 2 3 107 Ia ln 1 1 Ib ln 1 Ic ln r D23 D21

16

CHAPTER 2 THE LINE PARAMETERS

Likewise,

  1 1 1 ðwb-t=mÞ ψa3 5 2 3 107 Ia ln 1 1 Ib ln 1 Ic ln r D31 D32

The average flux linkages are ψa 5

  ψa1 1 ψa2 1 ψa3 2 3 1027 1 1 1 1 Ic ln 5 3Ia ln 1 1 Ib ln ðwb-t=mÞ r D12 D23 D31 D12 D23 D31 3 3

But Ib 1 Ic 5 2 Ia Therefore,   2 3 1027 1 1 3Ia ln 1 2 Ia ln r D12 D23 D31 3 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi p  3 D12 D23 D31 ðwb-t=mÞ 5 2 3 107 Ia ln r1

ψa 5

(2.49)

The average inductance per phase a ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p  3 D12 D23 D31 ðh=mÞ r1   Deq ðh=mÞ ln r1

La 5 2 3 107 ln 7

5 2 3 10

(2.50)

where r1 is the self-GMD or self-GMR of each phase conductor.

2.5 BUNDLE CONDUCTORS In transmission lines operating at voltages greater than 220 KV, corona power loss and interference with communication lines will be excessive, if only one conductor is used per phase. The potential gradient at the conductor surface can be reduced considerably if the self-GMD or self-GMR of the conductor can be increased. This is done by using two or more conductors in close proximity per phase compared to the spacing between phases and connected in parallel. The spacing between these same phase conductors is much smaller than the spacing between inter phases. Such a line is called bundled conductor line. Two or three conductor bundles are common. The current in each conductor of the bundle will be the same, if the conductors within the bundle are also transposed. However, it is not generally required. Increase of GMR reduces reactance of the line also (Fig. 2.10). Let r be the radius of each conductor of the bundle and d be the spacing between conductors in the bundle. The GMR of a two-conductor bundle is D2b 5

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffi 4 ð0:7788Þ2  d 2  r 2 5 r 1 d

(2.51)

2.6 INDUCTANCE OF A THREE-PHASE, DOUBLE-CIRCUIT LINE

17

FIGURE 2.10 Different types of bundle conductors.

For a three-conductor bundle it is D3b 5

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi 3 9 ð0:77 δδÞ3 3 d 3 3 r 3 5 r 1 d 2

(2.52)

For a four-conductor bundle the GMR is D4b 5

ffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  p ffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 4 4 4 r 1 d:d  2d 5 1:09 r 1 3 d 3

(2.53)

2.6 INDUCTANCE OF A THREE-PHASE, DOUBLE-CIRCUIT LINE WITH UNSYMMETRICAL SPACING AND TRANSPOSITION Consider a double-circuit line with conductors arranged in a vertical line for both the circuits. Let a, b, c be the conductors is one circuit and a1, b1 and c1 be the conductors in another circuit. If the conductors are transposed regularly, then for one cycle of transposition the conductor positions are shown in Fig. 2.11. The distance between conductors are marked in Fig. 2.11. With reference to Fig. 2.11 the flux linkages of phase a in position 1, 2, 3 are written down. The flux linkages of phase a in position 1        1 1 1 1 1 1 1 Ib ln 1 ln 1 Ic ln 1 ln ψa1 5 2 3 107 Ia ln 1 1 ln r m h l 2h D

Similarly in position 2 ψa2 is obtained as

       1 1 1 1 1 1 ψa2 5 2 3 107 Ia ln 1 1 ln 1 Ib ln 1 ln 1 Ic ln 1 ln r D h l h l

and ψa3 is position 3 is obtained as

       1 1 1 1 1 1 1 Ib ln 1 ln 1 Ic ln 1 ln ψa3 5 2 3 107 Ia ln 1 1 ln r m 2h D h l

18

CHAPTER 2 THE LINE PARAMETERS

FIGURE 2.11 Double-circuit unsymmetrical transposed three-phase line.

The total average flux linkages for phase a ψa1 1 ψa2 1 ψa3 3 "    # 27 2 3 10 1 1 1 1 1 U U 1 Ib ln 3Ia ln 1 1 Ia ln 5 r D m m 2h3 3    #   1 1 1 1 1 1 1   1 Ic ln   1 Ic ln 1Ib ln l l D 2h3 D l l " # 2 3 1027 1 1 1 1 5 3Ia ln 1 1 Ia ln 2 1 Ib ln 3 1 Ib ln 2 r m D 2h l D 3

ψa 5

1 1 1 Ic ln 2 2h3 Dl   2 3 1027 1 1 1 1 5 3Ia ln 1 1 Ia ln 2 1 ðIb 1 Ic Þ ln 3 1 ðIb 1 Ic Þ ln 2 r m D 2h l D 3 1Ic ln

Since Ia 1 Ib 1 Ic 5 0 ðIb 1 Ic Þ 5 2 Ia

Substituting the above

 2 3 1027 1 3Ia ln 1 1 Ia ln 3 r  2 3 1027 1 5 3Ia ln 1 1 Ia ln 3 r 2 0 1 2 13 23 hl3 A5 5 2 3 1027 4 ln @ 2 r 1 m3 2 1 3  2 3h 3 2 l 5 5 2 3 1027 Ia ln 4 1 r m

ψa1 5

1 1 1 2 Ia ln 3 2 Ia ln 2 m2 D 2h l D  2h3 l2 D m2 D

ðwb-t=mÞ



2.7 CAPACITANCE OF TRANSMISSION LINES

The inductance of phase a, La 5

" 1  2 # ψa 23 h l 3 ðh=mÞ 5 2 3 1027 ln 1 Ia r m

If D .. h then l=m tends to unity. In this case 7

La 5 2 3 10

ln

! 1 23 h r1

19

(2.54)

(2.55)

Since there are two conductors per phase inductance per conductor is La/2, i.e., "  1  1 # h 2 l 3 1 L 5 2 3 10 ln 26 1 U ðh=mÞ r m "  1 # h 2 1 27 ðh=mÞ LC2 3 10 ln 26 U 1 r 7

(2.56)

(2.57)

2.7 CAPACITANCE OF TRANSMISSION LINES Capacitance always exists between conductors carrying charges in a medium of permittivityA. Calculation of capacitances between line conductors and between conductor to neutral or earth is based on determining: 1. electric field strength E using Gauss’s law 2. utilizing the known potential difference between conductors, and 3. by dividing the total charge with the potential difference

2.7.1 POTENTIAL DIFFERENCE BETWEEN TWO POINTS DUE TO AN ELECTRIC CHARGE The electric flux density (D) at a point distance x (m) from a current-carrying conductor having a charge of q (c/m) is given by (Fig. 2.12)

FIGURE 2.12 Electric flux lines and equipotential lines.

20

CHAPTER 2 THE LINE PARAMETERS

D5

q ðc=m2 Þ 2πx

(2.58)

E5

q ðv=mÞ 2πεx

(2.59)

The electric field intensity

where ε is the permittivity of the medium. The potential difference between any two points P1 and P2 at distance D1 and D2 from the center of the conductor (Fig. 2.13). ð D2

v12 5

EUdx 5

D1

5

ð D2 D1

q Udx 2πxε

q D2 ln ðVÞ 2πε D1

(2.60)

2.7.2 CAPACITANCE OF A TWO-CONDUCTOR LINE Capacitance between the two conductors of a transmission line is defined as the charge on the conductors per unit potential difference between them. Thus the capacitance per unit length of the line is (Fig. 2.14) C5

q ð f=mÞ v12

(2.61)

The potential difference between the two conductors due to charges q1 and q2 on conductors 1 and 2 is the sum of the values due to charges q1 and q2. V12 5

FIGURE 2.13 Potential difference.

FIGURE 2.14 Capacitance of single-phase line.

q1 D q2 r2 ln ln 1 r1 2πε 2πε D

(2.62)

2.7 CAPACITANCE OF TRANSMISSION LINES

21

where ν 12 due to q1 5

q1 D ln 2πε r1

ν 112 due to q2 5

q2 r2 ln 2πε D

and

and V12 5 ν 12 1 ν 112

However, since q1 5 2q2

  q1 D r2 ln ðVÞ  r1 D 2πε  2 q1 D 5 ln ðVÞ 2πε r1 r2

V12 5

(2.63)

The capacitance between the two conductors C12 5

q1 q1 2πε   25  2 5 v12 q1 D D ln U ln 2πε r1 r2 r1 r2

(2.64)

In general for a two-conductor system q 5 q1 5 2 q2 and r 5 r1 5 r2

So that, C12 5

2pε πε   ðf=mÞ  2 5 D D ln ln r r2

(2.65)

Capacitance to neutral or ground of each phase conductor is double this value, since the potential to neutral is half the potential difference between the two conductors. Cn 5

2πε   ðf=mÞ D ln r

(2.66)

2.7.3 CAPACITANCE OF A THREE-PHASE LINE WITH SYMMETRIC SPACING Consider the three-phase system shown in Fig. 2.15. The three-phase conductors are placed at the corners of an equilateral triangle of side D (m). The radius of each conductor is r (m). The potential difference between conductors a and b due to charges qa, qb, and qc is obtained as   1 D r D 5 qa ln 1 qb ln 1 qc ln ðVÞ Vab 5 2πε r D D

22

CHAPTER 2 THE LINE PARAMETERS

FIGURE 2.15 Symmetrically spaced three-phase line.

Similarly the potential difference between a and c is given by Vac 5

  1 D D r qa ln 1 qb ln 1 qc ln ðVÞ 2πε r D D

Then, Vab 1 Vac 5

  1 D r r 2qa ln 1 qb ln 1 qc ln ðVÞ 2πε r D D

(2.67)

Assuming that there are only qa, qb, and qc and no other charges are present in the vicinity. qa 1 qb 1 qc 5 0 or ðqb 1 qc Þ 5  qa

Substituting this result is Eq. (2.67)

  1 D r 2qa ln  qa ln 2πε r D 1 D 5 3qa ln 2πε r

Vab 1 Vac 5

Further, in a balanced system Vab 1 Vac 5 3Van Hence, 1 D 3qa ln 2πε r 1 D qa ln ðVÞ Van 5 2πε r 3Van 5

(2.68)

Capacitance to neutral or ground of each phase conductor Cn 5

qa 5 Van

2πε   ðf=mÞ D ln r

(2.69)

It is to be noted that the radius of the conductor is not affected in capacitance calculations, while for inductance calculations the effective radius is 0.7788 3 r (GMR) The capacitive reactance per meter length Xc 5

1 ðΩÞ 2πfc

(2.70)

2.7 CAPACITANCE OF TRANSMISSION LINES

23

The line charging current Ic 5

V 5 2πfc V ðAÞ xc

(2.71)

2.7.4 CAPACITANCE OF A THREE-PHASE LINE WITH UNSYMMETRICAL SPACING Consider a three-phase line spaced at distances D12, D23, and D31 as shown in Fig. 2.16. Such lines are generally transposed so that each conductor occupies every other position over a cycle of transposition. The potential difference Vab between conductors a and b in position 1 is   D12 r D23 ðVÞ Vab 5 qa ln 1 qc ln 1 qb ln D12 r D31

(2.72)

when conductor a is in position 2 Vab 5

  1 D23 r D31 qa ln 1 qb ln ðVÞ 1 qc ln 2πε D23 r D12

(2.73)

Vab 5

  1 D31 r D12 qa ln ðVÞ 1 qb ln 1 qc ln 2πε D31 r D23

(2.74)

Likewise, in position 3

The average potential difference Vab over a cycle of transposition

  1 1 D12 D23 D31 r3 D23 D31 D12 qa ln ðVÞ Vab 5 U 1 q ln 1 q ln b c 3 2πε r3 D12 D23 D31 D31 D12 D23 p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  3 1 r D12 D23 D31 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðVÞ 1 qb ln p qa ln 5 3 r 2πε D12 D23 D31

Let

p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 D12 D23 D31 5 Deq

FIGURE 2.16 Three-phase conductor over a cycle of transposition.

(2.75)

24

CHAPTER 2 THE LINE PARAMETERS

then Vab 5

  1 Deq r qa ln 1 qb ln ðVÞ 2πε Deq r

(2.76)

In a similar manner the potential difference across conductors a and c over a transposition cycle can be proved to be   Deq 1 r qa ln 1 qc ln ðVÞ 2πε Deq r   Deq 1 r 2qa ln 1 ðqb 1 qc Þ ln Vab 1 Vac 5 3Van 5 2πε Deq r Vac 5

(2.77)

Since qa 1 qb 1 qc 5 0   Deq 1 3qa ln 3Van 5 2πε r Van 5

Deq q ðVÞ ln r 2πε

(2.78)

(2.79)

Capacitance to neutral or ground Cn 5 qa =Van Therefore, Cn 5

2πε   ðf=mÞ Deq ln r

(2.80)

2.8 CAPACITANCE OF A THREE-PHASE, DOUBLE-CIRCUIT LINE WITH SYMMETRICAL SPACING Consider a double-circuit line spaced as shown in Fig. 2.17. The phase-to-phase distance is D (m). The potential difference Vab can be written down using Fig. 2.17 as pffiffiffi  pffiffiffi       1 D r 2D D 3D 3D qa ln 1 ln 1 qb ln 1 ln pffiffiffi 1 qc ln pffiffiffi 1 ln 2πε r D 2D D 3D 3D pffiffiffi   1 2r 3D 5 qa ln 1 qb ln pffiffiffi 2πε 2r 3D

Vab 5

(2.81)

In a similar manner Vac can be written as Vac 5

pffiffiffi   3D 1 2r 1 qc ln pffiffiffi qa ln 2r 2πε 3D

(2.82)

2.9 EFFECT OF EARTH ON THE CAPACITANCE OF TRANSMISSION LINES

25

FIGURE 2.17 Capacitance of three-phase, double-circuit line-symmetrical spacing.

Vab 1 Vac 5 3Van 5

pffiffiffi   1 2r 3D 2qa ln 1 ðqb 1 qc Þ ln pffiffiffi 2πε 2r 3D

Using the relation that qa 1 qb 1 qc 5 0 3Van 5

So that

3qa ln 2πε

pffiffiffi  3D 2r

pffiffiffi qa 3D ln Van 5 2r 2πε

(2.83)

(2.84)

(2.85)

Capacitance to neutral of each phase conductor Can 5

qa 5 Van

2πε pffiffiffi  ðf=mÞ 3D ln 2r

(2.86)

Capacitance to neutral of each phase Ca 5 2  Can 5

2U2πε 2πε pffiffiffi  5 pffiffiffi 1=2 ðf=mÞ  3D 3D ln ln 2r 2r

(2.87)

2.9 EFFECT OF EARTH ON THE CAPACITANCE OF TRANSMISSION LINES The electric lines of force and equipotential lines are orthogonal to each other for an isolated charged conductor. The presence of earth alters the electric field of a charged conductor. If earth

26

CHAPTER 2 THE LINE PARAMETERS

FIGURE 2.18 Conductor and its image.

is assumed as a perfect conductor in the form of a horizontal plane of infinite extent, then we notice that the electric field of the charged conductors is forced to conform to the presence of earth’s equipotential surface. The potential distribution in space will remain the same if we imagine an isolated opposite charged conductor at the same depth h as the conductor is above the earth (Fig. 2.18). The capacitance between the conductor and its fictitious image Caa0 5

πε ðf=mÞ 2h ln r

(2.88)

where r is the radius of the conductor. Then, capacitance to earth of an isolated conductor Cn 5

πε   ðf=mÞ 2h ln r

(2.89)

2.10 EFFECT OF EARTH CAPACITANCE OF A SINGLE-PHASE LINE Consider Fig. 2.19 where a and b are the conductors carrying charges qa and qb. As earth is at zero potential, consider a1 and b1, the images of a and b carrying charges 2 qa and 2 qb at a depth of h meters. The spacing between the conductors is D (m). The potential difference 1 Vab 5 2πε

" # pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  D r 2h D2 1 4h2  qb ln pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qa ln 1 qb ln  qa ln 2h r D D2 1 4h2

(2.90)

2.10 EFFECT OF EARTH CAPACITANCE OF A SINGLE-PHASE LINE

27

FIGURE 2.19 Effect of earth capacitance of line.

But qa 5 2qb, substituting this

" pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi# 1 D D D2 1 4h2 D2 1 4h2 qa ln 1 qa ln  qa ln  qa ln Vab 5 2πε r r 2h 2h   1 D 2h 2qa ln 1 2qa ln pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 5 2πε r D2 1 4h2 3 2   D 1 qa D 2h qa 6 U pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi7 ln U pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 5 5 4 ln r D2 1 4h2 5 r πε πε D2 1 4h2 4h2 2 3 D 1 q 6 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi7 5 ln 4 r 1 1 D2 5V πε 4h2 qa πε 1 ðf=mÞ 0 5 Cab 5 Vab D 1 C B U rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ln @ r D2 A 11 2 4h

(2.91)

(2.92)

The line capacitance is slightly increased due to the presence of earth. As h tends to infinity D2 =4h2 tends to zero and 0

1 πε   Cab 5 @ D A ln r

28

CHAPTER 2 THE LINE PARAMETERS

2.11 CAPACITANCE OF THREE-PHASE LINE INCLUDING EFFECT OF EARTH Consider a three-phase transmission line with unsymmetrical spacing above the ground level and the images of the three-phase conductors below the earth at the same depth as the original conductors are above the ground level. Let the conductor carry charges qa, qb, and qc. Their images carry charges 2 qa, 2 qb, and 2 qc, respectively. The distances are all marked as shown in Fig. 2.20. The numbers 1, 2, and 3 indicate transposition cycle positions.

FIGURE 2.20 Capacitance of three-phase line in the phase presence of earth.

2.12 SKIN EFFECT AND PROXIMITY EFFECT

29

With conductor a in position 1 Vab1 5

       1 D12 H12 r H2 D23 H23 qa 2 ln 1 qb ln 1 qc ln 2 ln 2 ln 2πε D12 r H1 H12 D31 H31

(2.93)

Similarly when conductor a is in position 2

       1 D23 H23 r H3 D31 H31 qa ln 2 ln 1 qb ln 1 qc ln Vab2 5  ln  ln 2πε D23 r H2 H23 D12 H12

and likewise, Vab3 5

       1 D31 H31 r H1 D12 H12 qa ln 2 ln 1 qb ln 1 qc ln 2 ln 2 ln 2πε D31 r H3 H31 D23 H23

1 Vab 5 ½Vab1 1 Vab2 1 Vab3  3     1 1 D23 D31 D12 H12 H23 H31 r3 H3 H1 H2 1q  ln ln 2 ln qa ln 5 U b r3 H1 H2 H3 D23 D12 D31 H23 H12 H31 3 2πε   D31 D12 D23 H23 H31 H12 1qc ln 2 ln D31 D12 D23 H31 H12 H23      1 D23 D31 D12 H12 H23 H31 r3 H1 H2 H3 qa ln 1 q 2 ln ln 2 ln 5 b 6πε r3 H1 H2 H3 D12 D23 D31 H12 H23 H31

(2.94)

(2.95)

(2.96)

In a similar manner Vac can be obtained as Vac 5

     1 D12 D23 D31 H12 H23 H31 r3 H12 H23 H31 qa ln 1 q 2 ln ln 2 ln c 6πε r3 H1 H2 H3 D12 D23 D31 H1 H2 H3   3 r H12 H23 H31 1qc ln 2 ln D12 D23 D31 H1 H2 H3

(2.97)

We know that Vab 1 Vac 5 3Van and qb 1 qc 5 2qa. Substituting both,

     1 D12 D23 D31 H12 H23 H31 D12 D23 D31 H12 H23 H31 1 q 2 ln ln  ln 2qa ln a r3 H1 H2 H3 r3 H1 H2 H3 6πε      qa D12 D23 D31 H12 H23 H31 ln 2 ln 5 2πε r3 H1 H2 H3

3Van 5

Capacitance of conductor to neutral with Deq 5

(2.98)

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p 3 D12 D23 D31 ðf=mÞ

p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 2 3 2πε H12 H23 H31 Van 4   2 ln p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 5 3 Cn 5 5 Deq H1 H2 H3 qa ln r

(2.99)

2.12 SKIN EFFECT AND PROXIMITY EFFECT Consider a conductor of cross section A. Imagine that this area A is divided into small elemental areas uniformly as shown in Fig. 2.21.

30

CHAPTER 2 THE LINE PARAMETERS

FIGURE 2.21 Conductor section divided into elemental areas.

FIGURE 2.22 Electric field near two conductors carrying current (A) in opposite directions and (B) in the same direction.

Each elemental area carries a current I/n where I is the conductor current and n is the number of elemental areas. Each area current I/n produces an alternating flux of frequency determined by supply frequency. The central elemental areas link all the flux produced by inner areas (X) and also by area currents. The peripheral areas (Y) link much less flux than the inner elemental areas. Treating these elemental areas along the conductor length as filaments, the central filaments link more flux and possess more inductance and inductive reactance compared to the peripheral filaments (Y). Hence the current flowing in the central part will be less compared to current flowing in the outer or periphnal filaments, the variation in current distribution being nonuniform. Higher the frequency higher will be the reactance of central filaments, and less will be the current flowing in the central part of the conductor. This tendency to limit the flow of current at higher frequencies to only the outer most part of the conductor is called skin effect. The apparent resistance of the conductor is increased due to this effect. In case of alternating currents, which does not happen if direct

WORKED EXAMPLES

31

current flows through the conductor. If two conductors are placed near each other and they carry currents in opposite directions, adjacent sides of the conductors carry more flux than the far sides. Similarly, if both conductors carry currents in opposite direction, the adjacent sides carry less flux than the far sides (Fig. 2.22). The result in either case is that due to this effect the resistance of conductors becomes a little more than for uniform distribution of flux just as in the case of skin effect. Thus the effective increase in resistance due to the proximity of two alternating current-carrying conductors is called proximity effect. Due to these two effects, viz., skin effect and proximity effect, the effective ac resistance is obtained as about 1.2 times the resistance for a direct current-carrying conductor.

WORKED EXAMPLES E.2.1 Calculate the loop inductance per kilometer of a single-phase circuit comprising two parallel conductors 1 m apart and 1 cm in diameter. Solution: D 5 1 m 0:5 3 0:7788 m 100   1 3 100 L 5 2 3 ð2 3 107 Þ ln 0:5 3 0:7788 r1 5

5 4 3 1027 ln 256:8 5 22:194 3 1027 h=m 5 2:21 mh=km

E.2.2 A single-phase line has two pairs of conductors. Each pair comprises two 1.25 cm diameter conductors in parallel spaced vertically and 75 cm apart. The two parallel pairs are spaced laterally by a distance of 1.5 m. Calculate the total inductance of the line per kilometer assuming the current to be equally divided. Solution: The self-GMD or self-GMR of each pair of conductors sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1:25 6:042 4 1:25 3 75 3 3 75 3 ð0:7788Þ2 5 6:042 cm 5 m D5 5 2 2 100

FIGURE E.2.2

32

CHAPTER 2 THE LINE PARAMETERS

The mutual GMD between the two pairs

p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 1:5 3 1:677 3 1:5 3 1:677 5 1:58 m   1:56 3 100 5 2 3 107 3 ln 25:819 L 5 2 3 107 3 loge 6:042 Dm 5

5 2 3 10 3 7 3 3:2511 h=m

Total inductance per kilometer 5 2 3 2 3 107 3 3:2511 h=m 5 1:3 mh=km

E.2.3 A wire of 4 mm in diameter is suspended at a constant height of 10 m above the sea level which constitutes the return conductor. Calculate the inductance of the system per kilometer. Solution: Consider the conductor 10 m above the sea level. The image of the conductor at a depth of 10 m below the sea water level (a0 ). Then the inductance of the system L 5 2 3 107 ln



 20 3 1000 2 3 0:7788

5 18:92 3 107 h=m 5 1:892 mh=km

FIGURE E.2.3

E.2.4 Calculate the loop inductance per kilometer of a single-phase transmission line when the line conductors are spaced 1.2 m apart and each conductor has a diameter of 1.3 cm. Also, calculate the reactance of line, if its length is 50 km and the line!is operating at 50 Hz. Solution: Loop inductance 5 2 3 ð2 3 107 Þ ln

1:2

ð1:3=2Þ 3

5 20:8 3 107 h=m 5 2:08 mh=km

Line reactance per kilometer 5 2π 3 50 3 2:08 5 653:12 3 1023 Ω 5 0:653 Ω

Total reactance 5 50 3 0:653 5 32:65 Ω

1 100

WORKED EXAMPLES

33

FIGURE E.2.5

E.2.5 A single-phase transmission line consists of three conductors in the phase and two conductors in the return as shown in Fig E.2.5. The radius of each phase conductor is 0.22 cm and that of return is 0.45 cm. Phase conductors are a, b, c and are in one vertical line with a spacing of 5 m and the return conductors d and e are also in a vertical line spaced 5 m apart and separated from the phase group by a horizontal spacing of 7 m. Determine the inductance of both the phase and return conductors and the inductance of the complete line in h/m. Solution: The GMD of phase and return groups Dm is computed: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Dad 5 7 m; Dbd 5 Dae 5 52 1 72 5 8:6 m pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Dcd 5 502 1 72 5 12:201 m pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Dm 5 6 Dad Dae Dbd Dbe Dcd Dce pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Dm 5 6 7 3 8:6 3 7 3 8:6 3 8:6 3 12:201 5 6 380; 265:44 5 8:51 m

The GMR for phase conductor group

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p 9 Daa Dab Dac Dbb Dbc Dba Dca Dcb Dcc rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p 0:22 0:22 9 0:22 9 3 3 3 5 3 10 3 5 3 5 3 5 3 10 3 ð0:7788Þ3 5 0:0006655 5 0:408155 m 5 100 100 100

Dph 5

Dreturn

5

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:45 4 0:45 3 3 5 3 5 3 ð0:7788Þ2 5 4 0:0003071 5 0:1323794 m 100 100

Self-inductance of phase conductor group Lph 5 2 3 107 ln

8:51 5 6:074 3 107 h=m 0:408155

Self-inductance of return conductor group Lreturn 5 2 3 107 ln

8:51 5 8:3324 3 107 h=m 0:132

34

CHAPTER 2 THE LINE PARAMETERS

Total inductance of line per meter 5 ð6:0742 1 8:3324Þ107 h=m 5 14:4066 3 104 m h=m 5 1:44066 mh=km

E.2.6 A three-phase unsymmetrical circuit has the arrangement shown in Fig E.2.6. The conductor radius is 1.5 cm. Determine the inductance per kilometer. If f 50 HZ find the line reactance. Solution: 1:5 3 0:7788 5 0:011682 m 100 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Dm 5 3 6 3 6 3 6:63 5 6:203 m r1 5

Inductance, L 5 2 3 107 ln

6:203 0:011682

5 2 3 107 ln 530:988 5 2 3 107 3 6:27474 h=m 5 12:549 3 107 h=m 5 1:2549 mh=km Line reactance xL 5

2π 3 50 3 1:2549 5 0:394 Ω=km 1000

FIGURE E.2.6

E.2.7 A two-conductor, three-phase transmission line is arranged horizontally as shown in Fig E.2.7. The spacing between conductors of the bundle is 40 cm. The phase-to-phase

FIGURE E.2.7

WORKED EXAMPLES

35

spacings are 8, 8, and 16 m, respectively. Determine the inductance of the line per phase per kilometer. The conductor diameter is 3 cm. Solution: D2b 5

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 4 ð0:7788 3 1:5 3 40Þ2 5 2184:5 5 6:8365 cm Deq 5

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p 3 8 3 8 3 16 5 10:08 m 27

Inductance per phase 5 2 3 10



10:08 ln 0:06836



5 2 3 107 3 4:99 5 9:987 3 107 h=m 5 0:9987 mh=km

The line reactance, X L 5 2π 3 50 3 0:9987 3 1023 Ω=km=phase 5 0:3136 Ω=km=phase

E.2.8 Determine the inductance per km of a double-circuit, three-phase line shown in Fig E.2.8. Each circuit is transposed perfectly. The diameter of the conductor is 2.0 cm. Solution: Self-GMD of each conductor 5 1 3 0:7788 5 0:7788 3 1022 m

FIGURE E.2.8

36

CHAPTER 2 THE LINE PARAMETERS

Dab 5 Dbc 5 Db1 c1 5 Db1 a1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 5 0:752 1 3:52 5 3:58 m qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Dab1 5 3:52 1 ð8:5  7:5Þ2 5 3:52 1 7:752 5 8:5 m pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Daa1 5 72 1 72 5 9:9 m ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p 4 Da1 5 4 Dab UDac UDab1 UDac1 5 3:58 3 7 3 7 3 8:5 5 4 1491 5 38:613 5 6:21 m Da2 5

p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi 4 3:58 3 3:58 3 8:5 3 8:5 5 4 925:985 5 2 30:43 5 5:516 m Da3 5 Da 5

p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffiffi 4 7 3 3:58 3 7 3 8:5 5 4 1491 5 6:21 m

p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 6:21 3 6:21 3 5:516 5 5:96947 m 5 Db 5 Dc

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Ds1 5 Self GMD of each phase 5 0:007788 3 9:9 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 5 0:0771 5 0:2777 m 5 Ds3 Ds2 5 Ds 5

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:007788 3 8:5 5 0:2573 m

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p p 3 3 Ds1 UDs2 UDs3 5 0:2573 3 0:2777 3 0:2777 5 0:271 m 5:96947 0:271 5 2 3 107 3 3:0923

Inductance per phase 5 2 3 1027 ln

5 6:1846 3 107 h=m 5 0:61846 mh=km

E.2.9 The 2-cm diameter conductors of a three-phase, three-wire (line to neutral) of a three-phase system are located at the corners of a triangle, giving conductor spacings of 3.6, 5.4, and 8.1 m. The conductors are transposed at regular intervals and the load is balanced. Calculate the inductance per kilometer per phase. Solution: The self-GMD of phase conductors Ds 5

1 3 0:7788 m 100

The mutual GMD of the three-phase conductors Dm 5

p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 3:6 3 5:4 3 8:1 5 5:4 m

Inductance per phase L 5 2 3 1027 ln

5:4 3 100 5 2 3 1027 ln 693:3749 0:7788

5 13:08 3 107 h=m 5 1:308 mh=km

WORKED EXAMPLES

37

E.2.10 A three-phase, three-wire system consisting of 2.5-cm diameter conductors spaced 3 m apart in a horizontal plane supplies a balanced load. Calculate the inductance per kilometer of each phase.

FIGURE E.2.10

Solution:

 La 5 2 3 1027 ln

 pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 3 100 3 2 1 ln 3 3 6 1 j 3 ln 0:7788 3 2:5 6 h i pffiffiffi 27 5 2 3 10 4:632 1 ln 4:24264 1 j 3ð0:3467Þ 5 2 3 1027 ½4:632 1 1:4452  j 0:6005

5 2 3 1027 ½6:077  j 0:6 h=m 5 ½1:2154  j 0:12mh=km  Lb 5 2 3 1027 ln

pffiffiffiffiffiffiffiffiffiffiffi 100 3 2 1 ln 3 3 3 0:7788 3 2:5



5 2 3 1027 ½4:632 1 1:09861 5 11:46 3 107 h=m 5 1:146 mh=km sffiffiffi# " pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 18 3 100 3 2 6 Lc 5 2 3 10 ln 1 j 3 ln 3 2:5 3 0:7788 p ffiffi ffi 5 2 3 1027 ½6:077213 1 j 3 0:34642 27

5 2 3 1027 ½6:077 1 j 0:6 h=m 5 ½1:2154 1 j 0:12 mh=km

E.2.11 Calculate the capacitance of a pair of parallel conductors of 5 mm diameter spaced 1 uniformly 20 cm apart in air. Ao 5 8:854 3 10212 5 36π 3 109 . Solution: D 2 3 20 3 10 5 5 80 r 5   5 D r5 cm ln 5 ln 80 5 4:382 2 3 10 r D 5 20 cm

38

CHAPTER 2 THE LINE PARAMETERS

C5

5

2πA 1 1   5π3 f=m U 9 4:382 D 36π 3 10 ln r 1 1 3 106 3 103 μf=km 3 4:382 36 3 109

5 0:00634 μf=km

E.2.12 A singlephase overhead line 32 km long consists of two parallel wires each 0.5 cm diameter, 1.5 m apart. If the line voltage is 50 kv at 50 Hz, calculate the charging current when the line is open circuited. Solution: C5

πA 1 1 1 1   5π3  5 f=m 3 3 D 1:5 3 100 6:3969 36π 3 109 36 3 109 ln ln r 0:25 Xc 5

1 1 36 3 109 3 6:3969 5 3 5 0:7334 3 109 Ω 2πfc 2π 3 50 1

Charging current V 50 3 103 3 32 3 1000 l5 XC 0:7334 3 109 5 2:18 A

Ic 5

E.2.13 Determine the capacitance per phase of a three-phase system of conductors arranged in horizontal configuration with a spacing of 3 m. The diameter of the conductors is 30 mm.

FIGURE E.2.13

Solution: Deq 5

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p 3 3 3 3 3 6 5 3:78 m

Cm 5

2π 3 109 3 36π

5

1 109 1 109 5 5 3 f=m  3 ln 252 18 3 5:529 18 3:78 3 10 ln 15

109 3 103 3 106 μf=km 5 0:01 μf=km 18 3 5:529

E.2.14 A three-phase, double-circuit line is composed of 2.6-cm diameter conductors spaced vertically at a distance of 2.5 m and spaced 5.5 m apart as shown in Fig. E.2.14. Determine the capacitance of the line per kilometer.

WORKED EXAMPLES

FIGURE E.2.14

Solution: GMD between phases in position 1 Dab1 5

p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 6:04 3 2:5 3 6:04 3 2:5 5 3:8858 m

Dbc1 5

p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 2:5 3 6:04 3 2:5 3 6:04 5 3:8058 m

Dca1 5

p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 5 3 5:5 3 5 3 5:5 5 5:244 m

DM1 5

p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 5:244 3 3:8058 3 3:8058 5 4:294 m

It can be proved that DM1 5 DM2 5 DM3 5 DM 5 4:294 m The self-GMD

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1:3 1:3 6 1:3 3 7:433 3 3 7:433 3 3 5:5 Ds 5 100 100 100 1:14 3 1:951 3 1:3286 5 5 0:2955 10

Therefore C5

5

2π 3 109 3 36π

1 1 1 1 5 9 5 4:294 10 3 18 ln 14:5373 109 3 18 3 2:6762 ln 0:2955 

0:202076 f=m 5 0:202076 μf=km 109

E.2.15 Determine the capacitance per phase of a three-phase, double-circuit line, the conductors of which are arranged in hexagonal spacing, the distance between conductors being 2.5 m. The diameter of each conductor is 3 cm. Total length of the line is 120 km.

39

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CHAPTER 2 THE LINE PARAMETERS

Solution: Ca 5 Cb 5 Cc 5 2Can 5

5 2 3 2π 3

2 3 2πA pffiffiffi  f=m 3D ln 2r

1 3 36π 3 109

1 pffiffiffi  3 3 2:5 3 100 ln 2 3 1:5

5

1 1 1 5 3 f=m ln 144:333 9 3 109 3 4:972 9 3 109

5

106 3 103 μf=m 9 3 109 3 4:972

5

106 3 103 3 120 μf 5 2:68 f 9 3 109 3 4:972

E.2.16 A single-phase line constructed 13.5 m above ground has spacing between the conductors 3.9 m. The radius of the conductor is 1.78 cm. Determine the capacitance of the line per km length, considering the effect of earth and neglecting it. Solution: sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi D2 3:92 11 2 5 11 5 1 1 0:20864 5 1:09938 2 4h 4 3 13:5 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi D D2 ln 11 2 r 4h 5 ln

3:9 3 100 1 5 ln 199:295 5 5:2947 1:78 1:09938 C5

πA π31 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 5 f=m 9 3 5:2947 36π 3 10 2 D D ln 11 2 4h r

5 0:005246 μf=km

Without considering the effect of earth πA π31 1 1 5 3 f=m 5 9 D 3:9 3 100 5:389528 36 3 10 ln 36π 3 109 ln r 1:78 109 μf=km 5 0:005154 μf=km 5 36 3 109 3 5:389528

C5

The effect of earth on capacitance is to increase it by 0.0052460.005154 5 0.000092 μf/km

PROBLEMS

41

PROBLEMS P.2.1 A single-phase supply is effected by conductors arranged as shown in Fig. P.2.1, the current being equally divided between both conductors forming a pair. Find the inductance per kilometer of the system.

FIGURE P.2.1

P.2.2 Find the inductance per kilometer per conductor (line to neutral) of a three-phase system which are placed at the corners of an equilateral triangle of side 1.49 m. The diameter of the conductor is 1.24 cm. P.2.3 A three-phase system has its conductors placed at the corners of a triangle with spacings 3.6, 5.4, and 8.1 m. The conductors are transposed at regular intervals. Calculate the inductance per kilometer of the line, if each conductor has a diameter of 2 cm. P.2.4 A symmetrical double-circuit line has its conductors arranged as shown in Fig. P.2.4. If the radius of each conductor is 1.25 cm, determine the effective inductance in mh/km of the line.

FIGURE P.2.4

P.2.5 Calculate the capacitance per kilometer of a pair of parallel wires 5 mm is diameter spaced uniformly 25 cm apart in air. P.2.6 A wire of 6 mm in diameter and 1 km in length is suspended at a constant height of 12 m above the sea water. Calculate the capacitance between conductor and earth.

42

CHAPTER 2 THE LINE PARAMETERS

P.2.7 A three-phase overhead line with conductors having diameter of 1.2 cm is operating at 66 kv and 50 Hz. Its conductors are arranged horizontally with a spacing of 3.2 m. Calculate the charging current, if the length of the line is 50 km. P.2.8 Determine the capacitance per kilometer of the three-phase double-circuit line shown in Fig. P.2.8. The line is perfectly transposed. The diameter of each conductor is 2 cm. If the line operates at 200 kv, determine the charging current per kilometer.

FIGURE P.2.8

QUESTIONS Q.2.1 Q.2.2 Q.2.3 Q.2.4 Q.2.5 Q.2.6 Q.2.7 Q.2.8 Q.2.9

Derive an expression for the inductance of a single-phase line. Derive an expression for the inductance per phase of a three-phase unsymmetrical line. Explain the terms: (1) GMR and (2) GMD. What is transposition? Explain. What are bundle conductors? Why are they used? Obtain an expression for the capacitance of a single-phase line. Derive an expression for the capacitance of a three-phase line. What is the effect of earth on the capacitance of a transmission line? Explain the following: (1) Skin effect and (2) Proximity effect.