Inductance of transmission line Flux linkages of one conductor in a group of conductors Inductance of composite conductor lines Inductance of 3-phase overhead line Bundled conductors

2.  Flux linkages of one conductor in a group of conductors
 Inductance of composite conductor lines
 Inductance of 3-phase overhead line
 Bundled conductors

3.  Theoretically , the flux due to a conductor extends from
the centre of the conductor to right upto infinity.
 Assuming that the flux linkages will extend upto a point P
very far from the group of the conductors , and the
distances are as shown in fig.
1
2
3
n
D1p
D2p
D3p
Dnp
P

4.  Consider a group of conductors 1, 2, 3,…… n such that
the sum of currents in all these conductors is zero.
 If the currents carried by respective strands are I1, I2, I3,
…In we have I1+I2+I3+…+In = 0
 ψ1p1 = All flux linkages of conductor 1 due to tis own
current I1 , internal and external , upto point P ….
 Ψ1p1 = 2 × 10 I1 ln D1p/r1‘ Wb.T/Mt
 Ψ1p2 = Flux linkages with conductor 1 due to current in
conductor-2
 Ψ1p2 = 2 × 10 × I2 ln D2p/D12
-7
-7

5.  Similarly ψ1p3…….ψ1pn
 Ψ1p=
2 × 10 { I1 ln D1p/r'1 + I2 ln D2p/D12 +...+ In ln Dnp/D1n}
 Flux linkages with conductor 1 due to I1 , I2…..In
 Net flux linkages ψ1p
 Ψ1p = 2 × 10 { I1 ln 1/r1'+ I2 ln 1/D12 +…+ In ln 1/D1n }
Wb-turns/m
-7
-7

6.  Two composite conductors consisting of number of strands,
each having ‘m’ and ‘n’ number of stands respectively is
shown in fig.
 The current is assumed to be equally divided amongst the
strands.
 One group act as a go conductor and the other as the return
for the 1-phase line.
 The current per strands is I/m ampere in one group and
-I/n ampere in the other.
1
m
3
2
n
3’2’
1’
(conductor ‘A’) (conductor ‘B’)

7.  Applying equation for net flux linkages in a group of
conductors.
 The flux linkage of strand 1 in conductor ‘A’ is given by :
 Ψ1 = 2 × 10 ln (D11,D12…D1n) /(r1' D12…D1m)
Wb-turns/mtr
 The inductance of strand 1.
L1 = ψ1/(I/m)
= 2m × 10 ln (D11 D12…D1n) /(r1’ D12…D1m)
Henry/m (r'= D11)
-7 1/n 1/m
-7 1/n 1/m

8.  Similarly the inductance of strand 2 in conductor A
L2 = ψ2/(I/m)
= 2m × 10 ln (D21 D22…D2n) /(D21 r2‘…D2m)
Henry/m (r2‘ =
D22)
 The average inductance of ‘m’ strands in conductor A.
Lav = (L1 + L2 +…+Lm)/m
 Since all the strands of conductor ‘A’ are electrically
parallel, the inductance of conductor A.
LA = Lav/m = (L1 +L2+…+Lm)/m substituting the values
of L1,L2…Lm
-7 1/n 1/m

9.  LA = 2 × 10 ln
(D11 D12…D1n) (D21 D22…D2n)…(Dm1 Dm2…Dmn)1/mn
(D11 D12…D1m) (D21 D22…D2m)…(Dm1 Dm2…Dmn)1/m
 The mn root of the product of the mn distance between ‘m’
strands of conductor A and ‘n’ strands of conductor B is called
geometric mean distance (GMD) denoted by Dm.
 The m th root of m distance of the various strands from one
of the strands and the radius of the same strand is called the
geometric mean radius (GMR) or self GMD and is denoted by
Ds.
 So LA = 2 × 10 LN Dm/Ds Henry/mtr
 The inductance of conductor “B” can also be similarly
obtained and the total inductance of the composite
conductors is
L = LA + LB
-7
2
th
2 2
-7

10.  In a 3-phase transmission line, the inductance of each
conductor is considered instead of loop inductance.
 The conductor of a 3-phase overhead line may be
placed symmetrically or unsymmetrically on the towers.
 With Symmetrical Spacing :
 A 3-phase line in which the space between any two
conductor is the same as shown in fig. the line is called
symmetrical line.
 Fig. shows the conductor of a 3-phase line conductor
has a radius r meters and spacing between the
conductors is D meters.

11. a c
b
D
D D

12.  Under balanced three-phase phasor currents, the
algebraic sum of the currents in the conductors is zero.
 Hence, Ia + Ib + Ic = 0
 The flux linkages of the conductor ‘a’ are
ψa = 2 × 10 [ Ia ln 1/Daa + Ib ln 1/Dab + Ic ln 1/Dac ] Wb-
T/m
 Inductance of conductor a,
La = ψa/Ia
= 2 × 10 ln D/r‘ H/m
-7
-7

13.  A 3-phase line in which the space between the
conductors is different as shown in fig., the line is called
unsymmetrical line.
 Consider 3-phase line with conductors a, b and c each of
radius r meters.
 Let the spacing between them be Dab, Dbc and Dca and
the currents flowing through them be Ia, Ib and Ic
respectively as shown in fig.
 From fig, the flux linkages of the conductor a
Ψa = 2 × 10 [Ia ln 1/r' + Ib ln 1/D12 + Ic ln 1/D31] Wb-T/m
-7

14.  Inductance of each line conductor is
L = 2 × 10 [ln √D12D23D31/r'] H/m
-7 3
a
3 2
c
D23
b
D31 D12
1

15.  A bundle conductor is a conductor made up of two or
more sub-conductors and is used as one phase
conductor.
 Lines of 400kv and higher voltages invariably use
bundled conductors.
 Sub-conductors of a bundled conductor are separated
from each other by a constant distance varying from 0.2
m to 0.6 m depending upon designed voltage and
surrounding conditions throughout the length of the line
with the help of spacers.

16.  It reduces corona loss.
 It reduces radio interference
 The bundled conductor lines transmit bulk power with
reduced losses, thereby giving increased transmission
efficiency
 Bundle conductor lines have a higher capacitance to
neutral so they have higher charging current, which
helps in improving power factor
 By bundling, the GMR is increased, the inductance per
phase is reduced. As a result reactance per phase is
reduced.