Lecture 2

Inductance of

Transmission Lines

EEEN 4372

Electrical Power Transmission and Distribution

Outline of presentation

• Introduction

• Nature of inductance, resistance

• Analytical treatment of resistance and

inductance

Introduction

• A transmission line has four parameters:

resistance, inductance, capacitance and

conductance

• Conductance, accounting for insulator and

cable leakage currents can normally be

neglected

Nature of inductance

• Variation of current in conductors causes a

change in magnetic flux linking the circuit

• A change in flux induces a voltage in the

circuit and INDUCTANCE relates the

induced voltage to the rate of change of

current

Nature of resistance

• The longitudinal resistance of a

conductor is expressed in terms of the

resistivity of the conductor material, its

length and cross-sectional area.

Resistance

• Effective resistance of a conductor is

• Effective resistance equal to the dc resistance only if the current distribution in the conductor is uniform

• DC resistance is given by

• Spiralling of conductors makes the strands longer than the conductor route length (1-2%)

2

condutorin losspower

IR

A

lRo

Resistance as a function of temperature

• The variation of resistance with temperature is given by

» or

a is the temperature coefficient of the conductor, R20 is the resistance at 200C.

• where R1 and R2 and the resistances of the conductor at t1 and t2

respectively

and T is a constant dependent on the material

• Current distribution is only uniform for dc.

• As frequency of ac current increases, current density normally

increases from interior towards the surface (skin effect) TASK 1. Find

out the reason why

1

2

1

2

tT

tT

R

R

)]20(1[ 0

20 a TRRT

Temperature coefficient of

Aluminium and Copper Conductors

The percent conductivity relative to that of copper

Inductance of a conductor

• The inductance of a transmission line is calculated as the flux

linkages per unit current

• To determine the inductance of a conductor, both internal and

external flux linkage must be considered

• Ampere’s law

I

L

dsHI xx .

Inductance of a conductor due to internal flux

• Integrating around the path yields

• where Ix is the current enclosed.

• Assuming constant current density,

• Therefore

xx xHI 2

Ir

xI x 2

2

Ir

xH x 22

22 r

xIHB xx

x

IH x

x2

Inductance of a conductor due to internal flux• In the tubular element thickness dx, the flux is the flux density times the cross-

sectional area of the element normal to the flux lines. Hence;

• and the flux linkages per metre at distance x is given by:

• the total flux linkages inside the conductor is therefore

• For rel permeability of 1, Lint

Wb/m2 2

dxr

xId

dxr

Ixd

r

xd

4

3

2

2

2

rI

dxr

Ix

0

4

3

int82

H/m x102

1

8

7-

int

L

Flux linkages between two points external to an

isolated conductor

• Magnetic field at point x:

• Flux density is

• Flux d in the tubular element thickness dx

x

IH x

2

x

IBx

2

dxx

Id

2

Flux linkages between two points external to

an isolated conductor

• Flux linkages numerically equal to the flux

• For a relative permeability of 1,

• Therefore, the inductance due to only the flux included between P1

and P2 is

1

212 ln

22

2

1D

DIdx

x

ID

D

1

27

12 ln102D

DI

1

27

12 ln102D

DL

Inductance of a single-phase two-wire conductor

• Inductance of the CIRCUIT due to the current in conductor 1 only

• Noting that

• The inductance is

H/m 10ln22

1 7

1

1

r

DL

4/1ln 4/1

4/1

1

7

1 ln102

r

DL

H/m lnln1021

4

1

7

1

r

DL

Inductance of a single-phase two-wire

conductor

• Which can be re-written as

• Where

• The radius r’ is that of a fictitious conductor having no

internal flux but with the same inductance as a solid

conductor radius r

• is equal to 0.7788

4/1' rr

'ln102

1

7

1r

DL

4/1

Inductance of a single-phase two-wire conductor

• In the same way, the inductance due to the current in conductor 2

can be written as

• And the total inductance for the circuit is therefore

• If r1=r2, the inductance per ‘loop meter’ is given by

'ln102

2

7

2r

DL

'

1

'

2

7

21 ln104rr

DLLL

'

7 ln104r

DL

Flux linkages of one conductor in a group

• Consider a group of

conductors where the sum

of the currents is zero,

the flux linkages of

conductor 1 due to I1 are:

• The flux linkages of conductor 1

due to I2 up to point P are

7

1

11

111 10)ln2

2(

r

DI

I PP

'ln102

1

11

7

11r

DI P

P

12

22

7

21 102D

DI P

P ln

Flux linkages of one conductor in a group

• The flux linkages of conductor 1 by all group conductors is

• Which becomes, since

n

nPn

PPPP

D

DI

D

DI

D

DI

r

DI

113

33

12

221

1

11

7

1 ln..lnlnln102

0.....4321 nIIIII

nP

Pnn

nP

P

nP

P

nP

P

n

n

P

D

DI

D

DI

D

DI

D

DI

DI

DI

DI

rI

11

33

22

11

113

3

12

21

1

1

7

1

ln..lnlnln

1ln..

1ln

1ln

1ln

102

Flux linkages of one conductor in a group

• As P moves away to infinity

n

nPD

ID

ID

Ir

I113

3

12

2'

1

1

7

1

1ln..

1ln

1ln

1ln102

Inductance of composite-conductor linesStranded conductors are classified generally as composite conductors

Analysis will be limited to the case where all strands are identical and share current equally

The method of analysis is also applicable to multi-phase conductors and parallel lines

Consider conductor X, composed of n filaments and conductor Y (the return conductor) which consists of m filaments

• The flux linkages of filament a of conductor X are

Inductance of composite-conductor lines

amacabaa

anacaba

a

DDDDm

I

DDDrn

I

1ln...

1ln

1ln

1ln102

1ln...

1ln

1ln

1ln102

'''

7

'

7

• The average inductance of the filaments of conductor X is

Inductance of composite-conductor lines

• the inductance of all of the filaments i.e. the inductance

of conductor X is the average inductance /n

n

LLLLL ncba

av

....

2

....

n

LLLL

n

LL ncbaav

X

• Substituting for each expression of inductance and combining terms

Inductance of composite-conductor lines

2

'''''''''

)...)...(...)(...(

)...)...(...)(...(ln102 7

nnnncnbnabnbcbbbaanacabaa

mnnmncnbnabmbcbbbaamacabaa

XDDDDDDDDDDDD

DDDDDDDDDDDDL

Note that ra’, rb’, and rn’ have been replaced by Daa, Dbb and Dnn

The ‘mn’th root of the product of the ‘mn’ distances is called the

GEOMETRIC MEAN DISTANCE

The n2 root of the product of the distances from every filament in

conductor to itself and to every other filament is called the

SELF GEOMETRIC MEAN DISTANCE of the conductor

(Also called geometric mean radius)

Inductance of composite-conductor lines

In terms of GEOMETRIC MEAN DISTANCE Dm and the

SELF GEOMETRIC MEAN DISTANCE Ds

YX

sy

my

Y

sx

mxX

LLL

and

D

DL

D

DL

ln102

ln102

7

7

Use of tables• Inductive reactance is usually the desired parameter rather than

inductance

mD

DffLX

s

mL / ln10222 7

• Sometimes the logarithmic expression is expanded in order to

separate the Dm and Ds terms

mDfD

fX m

s

L / ln104 1

ln104 77

• The term involving Ds is called the reactance at 1m-spacing, Xa

which depends on GMR and frequency only

• The second term with Dm is called the inductive reactance spacing

factor Xd is independent of the type of conductor and depends only

frequency and spacing.

Inductance of three-phase lines with equilateral spacing

• Assuming balanced three-phase phasor

currents

• the flux linkages of conductor a ;

DI

DI

DI cb

s

aa

1ln

1ln

1ln102 7

• and since cba III

s

aa

s

aaD

DI

DI

DI ln102

1ln

1ln102 77

s

aD

DL ln102 7

Inductance of three-phase lines with

unsymmetrical spacing• With unsymmetrical spacing, the flux linkages of each phase are not the same

• A different inductance in each phase results in an unbalanced circuit

• BALANCE of the three phases can be restored by exchanging the positions of the conductors at regular intervals.

• Such an exchange is referred to as TRANSPOSITION

2331

7

3

1223

7

2

3112

7

1

1ln

1ln

1ln102

1ln

1ln

1ln102

1ln

1ln

1ln102

DI

DI

DI

DI

DI

DI

DI

DI

DI

bb

s

aa

bb

s

aa

bb

s

aa

Inductance of three-phase lines with

unsymmetrical spacing

• With cba III

3312312

7

33123127

312312

7

ln102

ln102

1ln

1ln3

3

102

DDDD

D

DL

D

DDDI

DDDI

DI

eq

s

eq

a

s

aa

a

s

aa

Summary

• The factor appears in all equations

• In the denominator, the logarithmic term is always the GMR of the conductor

• The numerator is

– (I) the distance between conductors of a 1-phase line

– (ii) the mutual GMD between sides of a composite-conductor single phase line

– (iii) the distance between conductors of an equilaterally-spaced three-phase line

– (iv) the equivalent equilateral spacing of an unsymmetrical 3-phase line

7102

Inductance calculations of bundled

conductors

• On EHV lines, corona losses can be excessive

• In order to reduce the electric filed strength at the conductor surface, each phase is made up of two, three or four conductors in close proximity and are said to be bundled

• Ref. Figure 4.13

• This type of arrangement also reduces line reactance considerably

4 316 42

3 29 32

24 2

09.1)2(

)(

)(

dDddDD

dDdDD

dDdDD

ss

b

s

ss

b

s

ss

b

s

Problems

1. Calculate the series impedance per km of a single-phase 11kV line

with 50mm2 ‘Hazel’ AAAC conductor (BS3242/BS215-1). The

conductor has 7 strands of aluminium alloy with diameter 3.3mm and

the inter-phase spacing is 50cm.

2. Calculate the per unit series reactance of a 400kV L12 tower strung

with 2x700mm2 Araucaria bundled conductor. Use 100MVA base.

Assume the GMR of a 61-stranded conductor to be 0.7720 times the

nominal radius of the conductor. Ref. BSEN 50182:2001, Page 65,

Table F.41 . Bundle spacing 300mm

Solution1. A= 50mm2 ‘Hazel’ AAAC conductor

7 strands of aluminium alloy with diameter 3.3mm

inter-phase spacing is 50cm.

radius) mean (geometric distance mean geometric elfs:D

distance mean geometric :D

D

DlnL

s

m

s

m 7102

77654321

50

ssssssss

m

DDDDDDDD

D

rDDDDDDDDDDDD 2675747372717165645342312 7

12

435

6

rDDD 4362514

r .reDD / 77880412211

50cm

rsinrDDDDDD )32(6022 0625146352413

Self-GMD of strand 1:

r.rrrrrrr.

DDDDDDDDs

2578222)32(4)32(2 77880

7171615141312111

Self-GMD of strands 2,3,4,5 and 6 are equal to self-GMD of strand 1 since they are

symmetrically placed on the conductor.

Self-GMD of strand 7:

r.r.rrrrrr

DDDDDDDDs

747917788022222 2

7777675747322717

Therefore:

r.r.)r.(DDDDDDDD ssssssss 176274791257827 677654321

mH/km 884μH/m 884331762

500102 7 ..

..lnL