Lecture 2 Inductance of Transmission Lines EEEN 4372 Electrical Power Transmission and Distribution
Lecture 2
Inductance of
Transmission Lines
EEEN 4372
Electrical Power Transmission and Distribution
Outline of presentation
• Introduction
• Nature of inductance, resistance
• Analytical treatment of resistance and
inductance
Introduction
• A transmission line has four parameters:
resistance, inductance, capacitance and
conductance
• Conductance, accounting for insulator and
cable leakage currents can normally be
neglected
Nature of inductance
• Variation of current in conductors causes a
change in magnetic flux linking the circuit
• A change in flux induces a voltage in the
circuit and INDUCTANCE relates the
induced voltage to the rate of change of
current
Nature of resistance
• The longitudinal resistance of a
conductor is expressed in terms of the
resistivity of the conductor material, its
length and cross-sectional area.
Resistance
• Effective resistance of a conductor is
• Effective resistance equal to the dc resistance only if the current distribution in the conductor is uniform
• DC resistance is given by
• Spiralling of conductors makes the strands longer than the conductor route length (1-2%)
2
condutorin losspower
IR
A
lRo
Resistance as a function of temperature
• The variation of resistance with temperature is given by
» or
a is the temperature coefficient of the conductor, R20 is the resistance at 200C.
• where R1 and R2 and the resistances of the conductor at t1 and t2
respectively
and T is a constant dependent on the material
• Current distribution is only uniform for dc.
• As frequency of ac current increases, current density normally
increases from interior towards the surface (skin effect) TASK 1. Find
out the reason why
1
2
1
2
tT
tT
R
R
)]20(1[ 0
20 a TRRT
Temperature coefficient of
Aluminium and Copper Conductors
The percent conductivity relative to that of copper
Inductance of a conductor
• The inductance of a transmission line is calculated as the flux
linkages per unit current
• To determine the inductance of a conductor, both internal and
external flux linkage must be considered
• Ampere’s law
I
L
dsHI xx .
Inductance of a conductor due to internal flux
• Integrating around the path yields
• where Ix is the current enclosed.
• Assuming constant current density,
• Therefore
xx xHI 2
Ir
xI x 2
2
Ir
xH x 22
22 r
xIHB xx
x
IH x
x2
Inductance of a conductor due to internal flux• In the tubular element thickness dx, the flux is the flux density times the cross-
sectional area of the element normal to the flux lines. Hence;
• and the flux linkages per metre at distance x is given by:
• the total flux linkages inside the conductor is therefore
• For rel permeability of 1, Lint
Wb/m2 2
dxr
xId
dxr
Ixd
r
xd
4
3
2
2
2
rI
dxr
Ix
0
4
3
int82
H/m x102
1
8
7-
int
L
Flux linkages between two points external to an
isolated conductor
• Magnetic field at point x:
• Flux density is
• Flux d in the tubular element thickness dx
x
IH x
2
x
IBx
2
dxx
Id
2
Flux linkages between two points external to
an isolated conductor
• Flux linkages numerically equal to the flux
• For a relative permeability of 1,
• Therefore, the inductance due to only the flux included between P1
and P2 is
1
212 ln
22
2
1D
DIdx
x
ID
D
1
27
12 ln102D
DI
1
27
12 ln102D
DL
Inductance of a single-phase two-wire conductor
• Inductance of the CIRCUIT due to the current in conductor 1 only
• Noting that
• The inductance is
H/m 10ln22
1 7
1
1
r
DL
4/1ln 4/1
4/1
1
7
1 ln102
r
DL
H/m lnln1021
4
1
7
1
r
DL
Inductance of a single-phase two-wire
conductor
• Which can be re-written as
• Where
• The radius r’ is that of a fictitious conductor having no
internal flux but with the same inductance as a solid
conductor radius r
• is equal to 0.7788
4/1' rr
'ln102
1
7
1r
DL
4/1
Inductance of a single-phase two-wire conductor
• In the same way, the inductance due to the current in conductor 2
can be written as
• And the total inductance for the circuit is therefore
• If r1=r2, the inductance per ‘loop meter’ is given by
'ln102
2
7
2r
DL
'
1
'
2
7
21 ln104rr
DLLL
'
7 ln104r
DL
Flux linkages of one conductor in a group
• Consider a group of
conductors where the sum
of the currents is zero,
the flux linkages of
conductor 1 due to I1 are:
• The flux linkages of conductor 1
due to I2 up to point P are
7
1
11
111 10)ln2
2(
r
DI
I PP
'ln102
1
11
7
11r
DI P
P
12
22
7
21 102D
DI P
P ln
Flux linkages of one conductor in a group
• The flux linkages of conductor 1 by all group conductors is
• Which becomes, since
n
nPn
PPPP
D
DI
D
DI
D
DI
r
DI
113
33
12
221
1
11
7
1 ln..lnlnln102
0.....4321 nIIIII
nP
Pnn
nP
P
nP
P
nP
P
n
n
P
D
DI
D
DI
D
DI
D
DI
DI
DI
DI
rI
11
33
22
11
113
3
12
21
1
1
7
1
ln..lnlnln
1ln..
1ln
1ln
1ln
102
Flux linkages of one conductor in a group
• As P moves away to infinity
n
nPD
ID
ID
Ir
I113
3
12
2'
1
1
7
1
1ln..
1ln
1ln
1ln102
Inductance of composite-conductor linesStranded conductors are classified generally as composite conductors
Analysis will be limited to the case where all strands are identical and share current equally
The method of analysis is also applicable to multi-phase conductors and parallel lines
Consider conductor X, composed of n filaments and conductor Y (the return conductor) which consists of m filaments
• The flux linkages of filament a of conductor X are
Inductance of composite-conductor lines
amacabaa
anacaba
a
DDDDm
I
DDDrn
I
1ln...
1ln
1ln
1ln102
1ln...
1ln
1ln
1ln102
'''
7
'
7
• The average inductance of the filaments of conductor X is
Inductance of composite-conductor lines
• the inductance of all of the filaments i.e. the inductance
of conductor X is the average inductance /n
n
LLLLL ncba
av
....
2
....
n
LLLL
n
LL ncbaav
X
• Substituting for each expression of inductance and combining terms
Inductance of composite-conductor lines
2
'''''''''
)...)...(...)(...(
)...)...(...)(...(ln102 7
nnnncnbnabnbcbbbaanacabaa
mnnmncnbnabmbcbbbaamacabaa
XDDDDDDDDDDDD
DDDDDDDDDDDDL
Note that ra’, rb’, and rn’ have been replaced by Daa, Dbb and Dnn
The ‘mn’th root of the product of the ‘mn’ distances is called the
GEOMETRIC MEAN DISTANCE
The n2 root of the product of the distances from every filament in
conductor to itself and to every other filament is called the
SELF GEOMETRIC MEAN DISTANCE of the conductor
(Also called geometric mean radius)
Inductance of composite-conductor lines
In terms of GEOMETRIC MEAN DISTANCE Dm and the
SELF GEOMETRIC MEAN DISTANCE Ds
YX
sy
my
Y
sx
mxX
LLL
and
D
DL
D
DL
ln102
ln102
7
7
Use of tables• Inductive reactance is usually the desired parameter rather than
inductance
mD
DffLX
s
mL / ln10222 7
• Sometimes the logarithmic expression is expanded in order to
separate the Dm and Ds terms
mDfD
fX m
s
L / ln104 1
ln104 77
• The term involving Ds is called the reactance at 1m-spacing, Xa
which depends on GMR and frequency only
• The second term with Dm is called the inductive reactance spacing
factor Xd is independent of the type of conductor and depends only
frequency and spacing.
Inductance of three-phase lines with equilateral spacing
• Assuming balanced three-phase phasor
currents
• the flux linkages of conductor a ;
DI
DI
DI cb
s
aa
1ln
1ln
1ln102 7
• and since cba III
s
aa
s
aaD
DI
DI
DI ln102
1ln
1ln102 77
s
aD
DL ln102 7
Inductance of three-phase lines with
unsymmetrical spacing• With unsymmetrical spacing, the flux linkages of each phase are not the same
• A different inductance in each phase results in an unbalanced circuit
• BALANCE of the three phases can be restored by exchanging the positions of the conductors at regular intervals.
• Such an exchange is referred to as TRANSPOSITION
2331
7
3
1223
7
2
3112
7
1
1ln
1ln
1ln102
1ln
1ln
1ln102
1ln
1ln
1ln102
DI
DI
DI
DI
DI
DI
DI
DI
DI
bb
s
aa
bb
s
aa
bb
s
aa
Inductance of three-phase lines with
unsymmetrical spacing
• With cba III
3312312
7
33123127
312312
7
ln102
ln102
1ln
1ln3
3
102
DDDD
D
DL
D
DDDI
DDDI
DI
eq
s
eq
a
s
aa
a
s
aa
Summary
• The factor appears in all equations
• In the denominator, the logarithmic term is always the GMR of the conductor
• The numerator is
– (I) the distance between conductors of a 1-phase line
– (ii) the mutual GMD between sides of a composite-conductor single phase line
– (iii) the distance between conductors of an equilaterally-spaced three-phase line
– (iv) the equivalent equilateral spacing of an unsymmetrical 3-phase line
7102
Inductance calculations of bundled
conductors
• On EHV lines, corona losses can be excessive
• In order to reduce the electric filed strength at the conductor surface, each phase is made up of two, three or four conductors in close proximity and are said to be bundled
• Ref. Figure 4.13
• This type of arrangement also reduces line reactance considerably
4 316 42
3 29 32
24 2
09.1)2(
)(
)(
dDddDD
dDdDD
dDdDD
ss
b
s
ss
b
s
ss
b
s
Problems
1. Calculate the series impedance per km of a single-phase 11kV line
with 50mm2 ‘Hazel’ AAAC conductor (BS3242/BS215-1). The
conductor has 7 strands of aluminium alloy with diameter 3.3mm and
the inter-phase spacing is 50cm.
2. Calculate the per unit series reactance of a 400kV L12 tower strung
with 2x700mm2 Araucaria bundled conductor. Use 100MVA base.
Assume the GMR of a 61-stranded conductor to be 0.7720 times the
nominal radius of the conductor. Ref. BSEN 50182:2001, Page 65,
Table F.41 . Bundle spacing 300mm
Solution1. A= 50mm2 ‘Hazel’ AAAC conductor
7 strands of aluminium alloy with diameter 3.3mm
inter-phase spacing is 50cm.
radius) mean (geometric distance mean geometric elfs:D
distance mean geometric :D
D
DlnL
s
m
s
m 7102
77654321
50
ssssssss
m
DDDDDDDD
D
rDDDDDDDDDDDD 2675747372717165645342312 7
12
435
6
rDDD 4362514
r .reDD / 77880412211
50cm
rsinrDDDDDD )32(6022 0625146352413
Self-GMD of strand 1:
r.rrrrrrr.
DDDDDDDDs
2578222)32(4)32(2 77880
7171615141312111
Self-GMD of strands 2,3,4,5 and 6 are equal to self-GMD of strand 1 since they are
symmetrically placed on the conductor.
Self-GMD of strand 7:
r.r.rrrrrr
DDDDDDDDs
747917788022222 2
7777675747322717
Therefore:
r.r.)r.(DDDDDDDD ssssssss 176274791257827 677654321
mH/km 884μH/m 884331762
500102 7 ..
..lnL