EEL303_L3_Inductance1

7/31/2019 EEL303_L3_Inductance1

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1A. R. Abhyankar, IIT Delhi (2012)

EEL303: Power Engineering - 1Transmission Line Inductance Calculation Part 1

Course Coordinator: A. R. Abhyankar


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Transmission Line Parameters

Modeling a transmission line is all about finding the values of above parameters

in per unit length (meter) and per phase

Parameters are distributed along the length of the line


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Resistance


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Resistance

AR

=Ohm per meter per phase

Resistivity of conductor material (ohm-meter)

Effective conductor area (meter2)

A


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Resistance

Formula is for DC resistance

Distribution of current is uniform

More the frequency of AC current, more

nonuniformity of current distribution Skin Effect

Current develops tendency to move towards

the surface Higher effective resistance


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Inductance


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Inductance

Most important line parameter

Has direct bearing on transmission capacityand voltage drop

Depends on line geometry (wire size andconfiguration)


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Flux Linkages of Infinite Straight Wire

Infinite wire is one-turn coil with return path

at infinity Straight infinitely long wire of radius r

Uniform current density in the wire. Total

current is i


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Flux lines form concentric circles

Assume current in the wire is coming out

Case 1: x > r (point outside the conductor)

Applying Amperes circuital law to path 1

H Magnetic Field Intensity (At/m)

dI differential path length (m)

i Total instantaneous current linked by closed path

ixHdIH == 21

x

iH

2=1


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Flux lines form concentric circles

Assume current in the wire is coming out

Case 2: x


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Calculation of flux linkages of the

wire per meter and flux withinfinite radius R

Flux linkages outside the wire

Substituting from 1

3


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Calculation of flux linkages of the

wire per meter and flux withinfinite radius R

Flux linkages inside the wire

4

Substituting from 2


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Total Flux Linkages per Meter

70 104 = permeability of free space

1r

outside conductor1r

inside conductor if non-magnetic (copper or aluminum)1r

5


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Flux Linkages: Multi-conductor Case


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Substituting from 3Flux linkage of conductor 1 due to current in k:

Total flux linkages of conductor 1 due to currents in all conductors:


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As 1

R 1

1

R

Rk

Add following to second term

021

=+++niii LSince

Second term becomes

6


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For non-magnetic wire 1=r

11

4/1

11 78.07788.0' rrerr ==


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Generic Expression Flux linkages of kth Conductor:

7


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Special Case of Generic Form of Equations: Two Conductor Single Phase Line

d12 = d21

r1 r2

+=

12

2

1

1

0

1

1ln

'

1ln

2 di

ri

21ii =since

=

1

12

1

0

1

'ln

2 r

di

=

1

120

1

'ln

2 r

dL

Special Case of Generic Form of Equations: Two Conductor Single Phase Line

Conductors for Single Phase


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The resulting flux for the two conductors is determined by the sum of the fluxlinkages of both the conductors.

Thus TOTAL INDUCTANCE of double conductor line:

+

=+=

2

210

1

120

21

'ln

2'ln

2 r

d

r

dLLL

mHrr

dL /

''ln

21

120

=

'''21

rrr ==If mHr

dL /

'ln104

127=8


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Special Case of Generic Form of Equations: Each Phase Inductance of Three Phase Line

D D

D

Conductors have equal radii r

0=++cbaiii


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++= Di

Di

ri cbaa

1

ln

1

ln'

1

ln2

0

=

D

i

r

iaaa

1ln

'

1ln

2

0

=

'ln

2

0

r

Diaa

==

'ln

2

0

r

D

iL

a

a

a

mHr

DL

a/

'ln102

7=9

EEL303_L3_Inductance1

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