EE 334: Power Systems
Inductance, Bundling and Transposition
Anupama Kowli
Indian Institute of Technology Bombay, India
Outline of Today’s Lecture
Composite conductors
Inductance of three-phase transmission lines
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Bundled Conductors
Conductors may be bundled
together to increase the current
carrying capacity (skin effect!)
Bundling is desirable from
mechanical design aspect; it also
reduces the effect of corona and
increases effective radius
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Composite Conductors
Composite conductors are usually composed of two or more strands
electrically in parallel – stranded or bundled conductors
Consider the case of a single phase, two wire system with arbitrary
number of cylindrical conductors in each line
Assumption: strands in each line share current equally
I
n
−I
Cond. Y – m identical parallel strands, each carrying current
m
Cond. X – n identical parallel strands, each carrying current
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Inductance Of Individual Strands
The flux linkages of strand a in cond. X are
1
1
1
1
−7 I
ln 0 + ln
+ ln
+ . . . + ln
λa = 2 × 10
n
ra
Dab
Dac
Dan
1
1
1
1
−7 I
− 2 × 10
ln
+ ln
+ ln
+ . . . + ln
m
Daa0
Dab0
Dac0
Dam
!
√
m
Daa0 Dab0 Dac0 . . . Dam
−7
p
= 2 × 10 I ln
Wbt/m
n
ra0 Dab Dac .....Dan
 v

uY
m
u
t
m
Daq 


0


q=a
λa
−7

 H/m with Daa = ra0
∴ La =
= 2n × 10 ln  v

n
I/n
 u

Y
 u
n
t
Daq 
q=a
Similarly, inductances of strands b, c and so on can be computed
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Inductance Of Composite Conductors
Let the average inductance of the strands of conductor X be Lav
Since the n filaments of conductor X are electrically parallel, we
can obtain the inductance of conductor X as
La + Lb + Lc + . . . + Ln
Lav
=
LX =
n
n2
Substituting the expression for inductance of each filament,
 v

uY
n Y
m
u
t
 mn
Dpq 


p=a q=a0


−7
 H/m

LX = 2 × 10 × ln  v

n Y
n

 u
Y
2
 nu
t
Dpq 
p=a q=a
where Daa = ra0 , Dbb = rb0 and so on
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Geometrical Mean Distances
The numerator of the expression for LX represents the mnth root
of the products of distances from all the n filaments of conductor
X to all the m filaments of conductor Y
The mnth root product of mn distances is termed as geometrical
mean distance (GMD) or mutual GMD between the two conductors
th
The denominator represents the n2 root of product of the
distances from every filament in the conductor to itself and to every
other filament in conductor X
The denominator is termed as geometrical mean radius (GMR) of
conductor X or its self GMD
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Inductance Of Composite Conductors
In terms of GMD and GMR,
LX = 2 × 10−7 ln
Dm
H/m
Ds
where Dm and Ds are the mutual and self GMDs respectively
The above equation is similar (in form) to the one we derived for
the inductance of one conductor in a 1-φ two-wire configuration
(L = 2 × 10−7 ln(D/r0 ))
The inductance of the line with both conductor X and Y is,
L = LX + LY
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Example: Inductance Of 1-φ Transmission Line
One circuit of a 1-φ transmission line is composed of three solid
conductors of 0.2 cm radius. The return circuit is composed of two
0.4 cm radius conductors. Find the inductance due to currents in
each circuit, and the inductance of the complete line.
4
2
2
2
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Example: Inductance Of 1-φ Transmission Line
√
√
Dad = Dbe = 4 m; Dae = Dbd = Dce = 20 m; Dcd 32 m
p
Mutual GMD, Dm = 6 Dad Dae Dbd Dbe Dcd Dce = 4.4810 m
Self GMD for group A,
q
9
Ds = Daa Dab Dac Dba Dbb Dbc Dca Dcb Dcc
q
9
= (0.25 × 0.7788 × 10−2 )3 × 64 × 122 = 0.2147 m
Self GMD for group B,
q
4
Ds = (0.4 × 0.7788 × 10−2 )2 × 62 = 0.1116 m
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Example: Inductance Of 1-φ Transmission Line
Inductance due to currents in group A,
4.4810
0.2147
= 6.0767 × 10−7 H/m
LA = 2 × 10−7 ln
Inductance due to currents in group B,
4.4810
0.1116
= 7.3854 × 10−7 H/m
LB = 2 × 10−7 ln
Inductance of the complete line (per unit length) is
L = LA + LB = 13.4621 × 10−7 H/m
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Inductance Of 3-φ Line With Equilateral Spacing
The equations derived for the inductance of 1-φ can be adapted for
3-φ lines when Ia + Ib + Ic = 0 (i.e., there is no neutral wire, or the
currents are balanced)
The flux linkages of conductor a is,
1
1
1
λa = 2 × 10−7 Ia ln
Wbt/m
+ Ib ln + Ic ln
Ds
D
D
Since Ia = −(Ib + Ic ),
1
1
D
−7
λa = 2 × 10
− Ia ln
= 2 × 10−7 Ia ln
Wbt/m
Ia ln
Ds
D
Ds
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Inductance Of 3-φ Line With Equilateral Spacing
Inductance of conductor a,
La = 2 × 10−7 ln
D
H/m
Ds
The equation is similar to that of 1-φ lines with r0 replaced by Ds ,
the self GMD for conductor of phase a (which could be a single
solid conductor or stranded one)
Because of symmetry, inductance of conductors b and c are same as
that of a
What happens if the spacing is not symmetrical?
unbalanced circuit due to unequal inductances
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Transposing Of 3-φ Lines
Transposing of lines: each phase conductor sees the same average
topology
For the arrangement shown, conductor a sees configurations 1, 2,
and 3 equal number of times and hence, its inductance per km is
average of the inductance of the three configurations
a
c
b
b
a
c
c
b
Config−1
Config−2
a
Config−3
If the lines are not transposed then due to unsymmetrical geometry,
the phase inductances will not be equal and network will be not be
balanced
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Line Transposition
15 / 19
Aerial View Of Transposition
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Flux Linkages In A Transposed Line
The total flux linkages for phase a conductor under the three
configuration (superscript reflects the config.) and their average are
1
1
1
1
−7
Ia ln 1 + Ib ln 1 + Ic ln 1
λa = 2 × 10
Daa
Dac
Dab
1
1
1
−7
2
λa = 2 × 10
Ia ln 2 + Ib ln 2 + Ic ln 2
Daa
Dac
Dab
1
1
1
3
−7
λa = 2 × 10
Ia ln 3 + Ib ln 3 + Ic ln 3
Daa
Dac
Dab
j
Plug-in known values of the inter-phase distances and use Daa
= ra0
for j = 1, 2, 3;
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Inductance Of 3-φ Transposed Lines
Average flux linkages on phase a can be computed
λav
a =
λ1a + λ2a + λ3a
3 1
1
1
+ Ic ln √
+ Ib ln √
3
3
ra0
D12 D23 D13
D13 D12 D23
√
Using Ib + Ic = −Ia , Dm = 3 D12 D23 D13 = mutual GMD
between the three phase conductors and Ds = ra0 = self GMD for
phase a conductor gives
Dm
−7
λav
a = 2 × 10 Ia ln
Ds
= 2 × 10−7 Ia ln
The per phase inductance is then calculated as
λav
Dm
a
av
av
Lav
= 2 × 10−7 ln
a = Lb = Lc =
Ia
Ds
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References
Power System Analysis by J. J. Grainger and W. D. Stevenson
Power System Analysis and Design by J. Duncan Glover, M. S.
Sarma and Thomas J. Overbye
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