inductance calculation

1. EE 8402
Transmission and distribution
Transmission Line Inductance
1

2. Substation Bus
2

3. Inductance Example
•Calculate the inductance of an N turn coil wound
tightly on a toroidal iron core that has a radius of
R and a cross-sectional area of A. Assume
1) all flux is within the coil
2) all flux links each turn
3) Radius of each turn is negligible compared to R
Circular path Γ of radius R within the
iron core encloses all N turns of the
coil and hence links total enclosed
current of Ie = NI.
Since the radius of each turn is
negligible compared to R, all circular
paths within the iron core have
radius approximately equal to R.
3

4. Inductance Example, cont’d
0
2 (path encloses ; path length is 2 ,)
( varies somewhat with , but ignore,)
2
(linear magnetic material,)
(assuming and therefore constant,)
(ea
e
e
r
I d
NI H R I NI R
NI
H H R
R
B H H
AB H B
N
π π
π
µ µ µ
φ
λ φ
Γ
=
= =
=
= =
=
=
∫ H lgÑ
0
2
0
ch of the turns link flux ,)
2
/ H
2
r
r
N
NI
NAB NA
R
N A
L I
R
φ
λ µ µ
π
µ µ
λ
π
= =
= =
4

5. Inductance of a Single Wire
•To develop models of transmission lines, we first
need to determine the inductance of a single,
infinitely long wire. To do this we need to
determine the wire’s total flux linkage, including:
1. flux linkages outside of the wire
2. flux linkages within the wire
•We’ll assume that the current density within the
wire is uniform and that the wire is solid with a
radius of r.
•In fact, current density is non-uniform, and
conductor is stranded, so our calculations will be
approximate.
5

6. Flux Linkages outside of the wire
We'll think of the wire as a single loop "closed" by
a return wire that is infinitely far away. Therefore
= since there is = 1 turn. The flux linking
a length of wire outside it to a distance of
N
R
λ φ
0A
from
the wire center is derived as follows:
d length
2
R
r
I d
I
dx
x
φ µ
π
Γ
=
= =
∫
∫ ∫
H l
B a
g
g
Ñ
6

7. Flux Linkages outside, cont’d
0A
0
0
d length
2
Since length = we'll deal with per unit length values,
assumed to be per meter.
ln
meter 2 2
Note, this quantity still goes to infinity as
R
r
R
r
I
dx
x
I R
dx I
x r
R
λ φ µ
π
µλ µ
π π
= = =
∞
= =
→ ∞
∫ ∫
∫
B ag
7

8. Flux linkages inside of wire
Current inside conductor tends to travel on the outside
of the conductor due to the skin effect. The penetration
of the current into the conductor is approximated using
1
the skin depth = where isf
fπ µσ
the frequency in Hz
and is the conductivity in siemens/meter.
0.066 m
For copper skin depth 0.33 inch at 60Hz.
For derivation we'll assume a uniform current density.
f
σ
≈ ≈
8

9. Flux linkages inside, cont’d
Wire cross section
x
r
2
2
2
Current enclosed within distance
from center
2 2
e
e
x
x
x I I
r
I Ix
H
x rπ π
= =
= =
2
inside 2 2A 0
3
40
However, situation is not as simple as outside wire
case since flux only links part of wire (need Biot-Savart law
to derive): d (length) d
2
(length) d (length)
2
r
r
Ix x
x
r r
Ix
x
r
λ µ
π
µ
π
= = ×
= × = ×
∫ ∫
∫
B ag
0
.
8
r
I
µ µ
π
9

10. Line Total Flux & Inductance
0 0
0
0
(per meter) ln
2 8
(per meter) ln
2 4
(per meter) ln
2 4
Note, this value still goes to infinity as we let
go to infinity.
Note that inductance depends on
r
Total
r
Total
r
R
I I
r
R
I
r
R
L
r
R
µ µ µ
λ
π π
µ µ
λ
π
µ µ
π
= +
 = + ÷
 
 = + ÷
 
logarithm
of ratio of lengths.
10

11. Inductance Simplification
0 0 4
0 4
Inductance expression can be simplified using
two exponential identities:
ln( )=ln + ln ln ln ln ln( )
ln ln ln ln
2 4 2
ln ln
2
r
r
a
r
a
ab a b a b a e
b
R
L R r e
r
L R re
µ
µ
µ µµ
π π
µ
π
−
−
= − =
   = + = − + ÷  ÷ ÷
    
  
= −  ÷
  
0
4
r
ln
2 '
Where ' 0.78 for 1
We call ' the "effective radius" of the conductor.
r
R
r
r re r
r
µ
µ
π
µ
−
=÷
≈ =@
11

12. Two Conductor Line Inductance
•Key problem with the previous derivation is we
assumed no return path for the current. Now
consider the case of two wires, each carrying the
same current I, but in opposite directions; assume
the wires are separated by distance D.
D
Creates counter-
clockwise field
Creates a
clockwise field
To determine the
inductance of each
conductor we integrate
as before. However
now we get some
field cancellation.
12

13. Two Conductor Case, cont’d
Key Point: Flux linkage due to currents in each conductor tend
to cancel out. Use superposition to get total flux linkage.
For a distance from left conductor that is greater than 2 ,
flux due to the right
Consider flux linked by
conductor from distanc
left conduc
e 0 to
can
tor from distance 0
cels the flux due t
to
o t e r
.
h
D
R
D
0 0
left
ight conductor from to 2 :
Summing the fluxes yields: ln ln
2 ' 2
D D
R R D
I I
r D
µ µ
λ
π π
− = −  ÷
 
Direction of integration
D
D
R
Left Current Right Current13

14. Two Conductor Inductance
( )
0
left
0
0
0
0
Simplifying (with equal and opposite currents)
ln ln
2 '
ln ln ' ln( ) ln
2
ln ln
2 '
ln as
2 '
ln H/m
2 '
left
R R D
I
r D
I R r R D D
D R
I
r R D
D
I R
r
D
L
r
µ
λ
π
µ
π
µ
π
µ
π
µ
π
−  = −  ÷ ÷
  
= − − − +
 = + ÷
− 
 = → ∞ ÷
 
 =  ÷
 
14

15. Many-Conductor Case
Now assume we now have n conductors,
with the k-th conductor having current ik, and
arranged in some specified geometry.
We’d like to find flux linkages of each conductor.
Each conductor’s flux
linkage, λk, depends upon
its own current and the
current in all the other
conductors.
For example, to derive the flux linkage for conductor 1, λ1, we’ll be integrating from
conductor 1 (at origin) to the right along the x-axis.
15

16. Many-Conductor Case, cont’d
At point b the net
contribution to λ1
from ik , λ1k, is zero.
We’d like to integrate the flux crossing between b to c.
But the flux crossing between a and c is easier to
calculate and provides a very good approximation of λ1k.
Point a is at distance d1k from conductor k.
Rk is the
distance
from con-
ductor k
to point
c.
16

17. Many-Conductor Case, cont’d
[ ]
0 1 2
1 1 2'
12 11
0
1 2'
12 11
0
1 1 2 2
1 1 2
0
1
ln ln ln ,
2
1 1 1
ln ln ln
2
ln ln ln .
2
As goes to infinity, so the second
term from above can be written =
2
n
n
n
n
n
n n
n
j
j
RR R
i i i
d dr
i i i
d dr
i R i R i R
R R R R
i
µ
λ
π
µ
π
µ
π
µ
π =
 
= + + + 
 
 
= + + + 
 
+ + + +
= =
L
L
L
1ln .
n
R
 
 ÷
 
∑
17

18. Many-Conductor Case, cont’d
1
0
1 1 2'
12 11
11 1 12 2 1
Therefore if 0, which is true in a balanced
three phase system, then the second term is zero and
1 1 1
ln ln ln ,
2
System has self and mutual inductan
n
j
j
n
n
n n
i
i i i
d dr
L i L i L i
µ
λ
π
=
=
 
= + + + 
 
= + +
∑
L
L
ce.
However, the mutual inductance can be canceled for
balanced 3 systems with symmetry.φ 18

19. Symmetric Line Spacing – 69 kV
19

20. Line Inductance Example
Calculate the reactance for a balanced 3φ, 60Hz
transmission line with a conductor geometry of an
equilateral triangle with D = 5m, r = 1.24cm (Rook
conductor) and a length of 5 miles.
0 1 1 1
ln( ) ln( ) ln( )
2 '
a a b ci i i
r D D
µ
λ
π
 = + +  
Since system is assumed
balanced
a b ci i i= − −
20

21. Line Inductance Example, cont’d
0
0
7
0
3
6
Substituting , obtain:
1 1
ln ln
2 '
ln .
2 '
4 10 5
ln ln
2 ' 2 9.67 10
1.25 10 H/m.
Again note logarithm of ratio of distance between
p
a b c
a a a
a
a
i i i
i i
r D
D
i
r
D
L
r
µ
λ
π
µ
π
µ π
π π
−
−
−
= − −
    = − ÷  ÷     
 =  ÷
 
×   = = ÷  ÷
   ×
= ×
hases to the size of the conductor. 21

22. Line Inductance Example, cont’d
6
6
4
Total for 5 mile line
1.25 10 H/m
Converting to reactance
2 60 1.25 10
4.71 10 /m
0.768 /mile
3.79
(this is the total per phase)
The reason we did NOT have mutual inductance
was because
a
a
L
X
X
π
−
−
−
= ×
= × × ×
= × Ω
= Ω
= Ω
of the symmetric conductor spacing
22

23. Conductor Bundling
To increase the capacity of high voltage transmission
lines it is very common to use a number of
conductors per phase. This is known as conductor
bundling. Typical values are two conductors for
345 kV lines, three for 500 kV and four for 765 kV.
23

24. Bundled Conductor Flux Linkages
•For the line shown on the left,
define dij as the distance between
conductors i and j.
•We can then determine λk for conductor k.
•Assuming ¼ of the phase current flows
in each of the four conductors in
a given phase bundle, then for conductor 1:
18
12 13 14
0
1
15 16 17
19 1,10 1,11 1,12
1 1 1 1
ln ln ln ln
4 '
1 1 1 1
ln ln ln ln
2 4
1 1 1 1
ln ln ln ln
4
a
b
c
i
r d d d
i
d d d d
i
d d d d
µ
λ
π
  
+ + + +  ÷
  
  
 = + + + + ÷ ÷  
 
  
+ + + ÷ 
    24

25. Bundled Conductors, cont’d
1
4
12 13 14
0
1 1
4
15 16 17 18
1
4
19 1,10 1,11 1,12
Simplifying
1
ln
( ' )
1
ln
2 ( )
1
ln
( )
a
b
c
i
r d d d
i
d d d d
i
d d d d
µ
λ
π
  
  ÷+
  ÷
  
  
  ÷= +  ÷
   
 
  
 ÷ 
 ÷   
25

26. Bundled Conductors, cont’d
1
4
12 13 14
1
12 1
1
1
4
15 16 17 18 2 3 4
1 19 1
geometric mean radius (GMR) of bundle
( ' ) for our example
( ' ) in general
geometric mean distance (GMD) of
conductor 1 to phase b.
( )
(
b
b
b
b
b b b ab
c
R
r d d d
r d d
D
d d d d D D D D
D d d
=
=
= ≈ ≈ ≈ ≈
=
@
K
@
1
4
,10 1,11 1,12 2 3 4) c c c acd d D D D D≈ ≈ ≈ ≈
26

27. Inductance of Bundle
0
1
0 0
1
0
1
If and
Then
1 1
ln ln
2
ln 4 ln
2 2
4 ln , which is the
2
self-inductance of wire 1.
ab ac bc a b c
a a
b
a
b b
b
D D D D i i i
i i
R D
D D
I I
R R
D
L
R
µ
λ
π
µ µ
π π
µ
π
= = = = − −
    = −  ÷  ÷    
   
= = ÷  ÷
   
 
= × ×  ÷
 
27

28. Inductance of Bundle, cont’d
0
1
But remember each bundle has conductors
in parallel (4 in this example).
So, there are four inductances in parallel:
/ ln .
2
Again note that inductance depends on the
logarithm of t
a
b
b
D
L L b
R
µ
π
 
= =  ÷
 
he ratio of distance between phases
to the size of bundle of conductors.
Inductance decreases with decreasing distance between
phases and increasing bundle size. 28

29. Bundle Inductance Example
0.25 M0.25 M
0.25 M
Consider the previous example of the three phases
symmetrically spaced 5 meters apart using wire
with a radius of r = 1.24 cm. Except now assume
each phase has 4 conductors in a square bundle,
spaced 0.25 meters apart. What is the new inductance
per meter?
( )
2 3
1
3 4
70
1.24 10 m ' 9.67 10 m
9.67 10 0.25 0.25 ( 2 0.25)
0.12 m (ten times bigger than !)
5
ln 7.46 10 H/m
2 0.12
Bundling reduces inductance.
b
a
r r
R
r
L
µ
π
− −
−
−
= × = ×
= × × × × ×
=
= = ×
29

30. Transmission Tower Configurations
•The problem with the line analysis we’ve done
so far is we have assumed a symmetrical tower
configuration.
•Such a tower configuration is seldom practical.
Typical Transmission Tower
Configuration
• Therefore in
general Dab ≠
Dac ≠ Dbc
•Unless something
was done this would
result in unbalanced
Phases.
30

31. Transposition
To keep system balanced, over the length of
a transmission line the conductors are
“rotated” so each phase occupies each
position on tower for an equal distance.
This is known as transposition.
Aerial or side view of conductor positions over the length
of the transmission line. 31

32. Line Transposition Example
32

33. Line Transposition Example
33

34. Transposition Impact on Flux
Linkages
0
a
12 13
0
13 23
0
23 12
For a uniformly transposed line we can
calculate the flux linkage for phase "a"
1 1 1 1
ln ln ln
3 2 '
1 1 1 1
ln ln ln
3 2 '
1 1 1 1
ln ln ln
3 2 '
a b c
a b c
a b c
I I I
r d d
I I I
r d d
I I I
r d d
µ
λ
π
µ
π
µ
π
 
= + + + 
 
 
+ + + 
 
 
+ + 
 
“a” phase in
position “1”
“a” phase in
position “3”
“a” phase in
position “2”
34

35. Transposition Impact, cont’d
( )
( )
1
3
1
3
12 13 230
1
3
12 13 23
Recognizing that
1
(ln ln ln ) ln ( )
3
We can simplify so
1 1
ln ln
'
2
1
ln
a b
a
c
a b c abc
I I
r d d d
I
d d d
µ
λ
π
 + + =  ÷
 
  
  ÷+ +
 ÷  
=  
  
 ÷ 
 ÷   
35

36. Inductance of Transposed Line
( )
1
3
12 13 23
0 0
a
70
Define the geometric mean distance (GMD)
Then for a balanced 3 system ( - - )
1 1
ln ln ln
2 ' 2 '
Hence
ln 2 10 ln H/m
2 ' '
Again, logarithm of ratio
m
a b c
m
a a a
m
m m
a
D d d d
I I I
D
I I I
r D r
D D
L
r r
φ
µ µ
λ
π π
µ
π
−
=
 
= − = 
 
= = ×
@
of distance between phases
to conductor size. 36

37. Inductance with Bundling
0
70
If the line is bundled with a geometric mean
radius, , then
ln
2
ln 2 10 ln H/m
2
b
m
a a
b
m m
a
b b
R
D
I
R
D D
L
R R
µ
λ
π
µ
π
−
=
= = ×
37

38. Inductance Example
Calculate the per phase inductance and
reactance of a balanced 3φ, 60 Hz, line with:
– horizontal phase spacing of 10m
– using three conductor bundling with a spacing
between conductors in the bundle of 0.3m.
Assume the line is uniformly transposed and
the conductors have a 1cm radius.
38

39. Inductance Example
( )
1
3
12 13 23
1/3
4
1
3
bundle bundle bundle
1/3
0
,
(10 (2 10) 10) 12.6m,
'= 0.0078m,
( ' ) , where is the
distance between conductors in bundle
( ' 0.3 0.3) 0.0888m,
ln
2
9.9 10
r
m
b
m
a
b
D d d d
r re
R r d d d
r
D
L
R
µ
µ
π
−
−
=
= × × × =
=
=
= × × =
=
= × 7
H/m,
2 (1600m/mile) = 0.6 /mile.a aX fLπ= Ω 39