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Inductance of 3-phase lines with unsymmetrical spacing In this case , the flux linkages and inductance of each phase are not the same . A different inductance in each phase results in an unbalance circuit . This case can be overcome by exchanging the positions of the conductors at regular intervals along the line so that each conductor occupies the original position of every other conductor over an equal distance . Such an exchange of conductor

Inductance of 3 Phase Line

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Page 1: Inductance of 3 Phase Line

Inductance of 3-phase lines with unsymmetrical spacingIn this case , the flux linkages and

inductance of each phase are not the same . A different inductance in each phase results in an unbalance circuit . This case can be overcome by exchanging the positions of the conductors at regular intervals along the line so that each conductor occupies the original position of every other conductor over an equal distance . Such an exchange of conductor positions is called "transpositiontransposition".

Page 2: Inductance of 3 Phase Line

A complete transposition cycle is shown as : The flux linkages of ( a ) in position ( 1 ) , when ( b ) is in position ( 2 ) and ( c ) in position ( 3 ) , is :

7a1 12 312 10 ln[ ln(1 ) + ln(1 ) + ln(1 )] /a b cr D DI I I wbt m

a

c b

D13 D12

D23

1

23

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The flux linkages of ( a ) in position ( 2 ) , when ( b ) is in position ( 3 ) and ( c ) in position ( 1 ) , is :

c

b a

D13 D12

D23

1

23

7a2 23 122 10 ln[ ln(1 ) + ln(1 ) + ln(1 )] /a b cr D DI I I wbt m

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The flux linkages of ( a ) in position ( 3 ) , when ( b ) is in position ( 1 ) and ( c ) in position ( 2 ) , is :

b

a c

D13 D12

D23

1

23

7a3 13 232 10 ln[ ln(1 ) + ln(1 ) + ln(1 )] /a b cr D DI I I wbt m

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The average value of the flux linkages of ( a ) is :

a1 a2 a3( ) 3a

7

12 23 13 12 23 132 10 ln[3 ln(1 ) + ln(1 ) + ln(1 )]

3 a a b br D D D D D DI I I

Assume that a balance current :

( )a b cI I I

Page 6: Inductance of 3 Phase Line

7

12 23 13

73

12 23 13

312 23 13

.

7

7

2 10 ln[3 ln(1 ) - ln(1 )]3

2 10 = ln[3 ln(1 ) - 3 ln(1 )] 3

= ln( )

= ln( )

2 102 10

a a a

a a

a

a eq

r D D D

r D D D

D D D r

D r

I I

I I

I

I

And the average inductance per phase is : 7

.

.

2 10 ln( )

log( )

=0.4711 m /a eq

eq

D r

D r

LH mile

However , a b cL L L

are the same of equation for equal spacing conductor 12 23 13D D D D

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Double Circuit Three Phase LineIt is a common practice to build double circuit lines so as to increase transmission reliability at somewhat enhanced cost.

Less Dm (GMD) and More Ds (Self GMR)

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Bundled conductors

• The trend toward ever higher voltages for T.L has stimulated interest in the use of two or more

conductors per phase.

• Such a line said to be composed of " bundled " conductors.

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• Usually the spacing of conductors of a phase is about ( 10 ) times the diameter of one conductor.

• The advantages of bundling are reduced reactance because of increased self GMD and reduce voltage drop and voltage gradient which result in reduced

radio interference.

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For First Section of the transposition cycle

For Second Section

For Third Section

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Effect of earth on the capacitance of 3-phase transmission lines

Let us imagine conductor of the same size and shape as the overhead conductor lying directly below the original conductor above the plane of the ground .

If the earth is removed and a charge equal and opposite to that on the overhead conductor is assumed on the imagine conductor .

The electric flux between the overhead conductor and this equipotential surface is the same as that which existed between the conductor and the earth .

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Bundle conductor

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Single Line Diagram and

PU System

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Per Unit CalculationsA key problem in analyzing power systems is

the large number of transformers. – It would be very difficult to continually refer

impedances to the different sides of the transformers

This problem is avoided by a normalization of all variables.

This normalization is known as per unit analysis.

actual quantityquantity in per unit base value of quantity

Page 44: Inductance of 3 Phase Line

Per Unit Conversion Procedure, 11. Pick a 1 VA base for the entire system, SB

2. Pick a voltage base for each different voltage level, VB. Voltage bases are related by transformer turns ratios. Voltages are line to neutral.

3. Calculate the impedance base, ZB= (VB)2/SB

4. Calculate the current base, IB = VB/ZB 5. Convert actual values to per unit

Note, per unit conversion affects magnitudes, not the angles. Also, per unit quantities no longer have units (i.e., a voltage is 1.0 p.u., not 1 p.u. volts)

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Three Phase Per Unit

1. Pick a 3 VA base for the entire system, 2. Pick a voltage base for each different

voltage level, VB. Voltages are line to line. 3. Calculate the impedance base

Procedure is very similar to 1 except we use a 3 VA base, and use line to line voltage bases

3BS

2 2 2, , ,3 1 1

( 3 )3

B LL B LN B LNB

B B B

V V VZ

S S S

Exactly the same impedance bases as with single phase!

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Three Phase Per Unit, cont'd4. Calculate the current base, IB

5. Convert actual values to per unit

3 1 13 1B B

, , ,

3I I3 3 3B B B

B LL B LN B LN

S S SV V V

Exactly the same current bases as with single phase!

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