Date post: | 23-Jan-2018 |
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Engineering |
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Presented By:
Anisur Rahman,ID: 1402EEE00033
Department of EEEManarat International University (MIU)
Ashulia, Khagan, Savar, Dhaka, Bangladesh
Definition Of Inductance
Flux Linkages of Conductors
1. Flux linkages inside the conductor
2. Flux linkages outside the conductor
Flux linkages of one conductor in a group of
conductors
Inductance of a single phase two wire line
Inductance of 3-phase overhead line
Bundled conductors
•
4
•
Consider a conductor of radius r carrying a
current I. At a distance x from the center of
this conductor, the magnetic field intensity Hx
The internal inductance per meter is
intint
8H ml
I
7
70int
4 10
8 8H ml
If the relative permeability of the conductor is 1 (non-ferromagnetic
materials, such as copper and aluminum), the inductance per meter
reduces to
The external inductance per meter is :
The total external flux linkages per meter can be found via integration
To find the flux linkages external to a
conductor, we will consider the portion of
flux between two points P1 and P2 that lie
at distances D1 and D2 from the center of
the conductor.
Theoretically , the flux due to a conductor extends from
the center of the conductor to right unto infinity.
Assuming that the flux linkages will extend unto a point P
very far from the group of the conductors , and the
distances are as shown in fig.
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 , unto 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
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
The inductance of a single-phase line
consisting of two conductors of radii r
spaced by a distance D and both carrying
currents of magnitude I flowing into the
page in “A” conductor and out of the page
in the “B“ conductor.
x xH dl I Ñ
Since the path of radius x2 encloses both
conductors and the currents are equal and
opposite, the net current enclosed is 0 and,
therefore, there are no contributions to the
total inductance from the magnetic fields
at distances greater than D.
A B
The total inductance of a line per unit length in this transmission line is a sum of
the internal inductance and the external inductance between the conductor surface
(r) and the separation distance (D):
int
1ln
2 4ext H m
Dl l l
r
By symmetry, the total inductance of the other line is the same, therefore, the total
inductance of a two-wire transmission line is
1
ln4
H mD
lr
Where r’=is GMR (Geometric mean radius)
For a solid conductor G.M.R = 0.7788 times the radius of conductor.
D is the distance between conductors
‘
‘
‘
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.
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
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
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.
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.