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Welcome to lesson 5 on Power System Analysis
course. In this course we will talk about
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the Transmission Line Capacitance, before
we get into the calculation of transmission
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line capacitance. I would like to answer those
questions that I asked in lesson 4. First
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question was why bundled conductors are used
in EHV lines? Well the answer to this
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question is bundling of conductors that is
instead of using one single conductor; use
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of a
number of conductor connected by conducting
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frames, reduces electric field strength on
conductor surface, which in effect reduces
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the corona losses, which result in power loss
as well as radio interference and audible
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noise in the system. Bundling also increases
the
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effective radius of the conductor. And there
by reduces the inductance of the
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transmission line. This in effect well improve
the regulation of the transmission line.
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As seen from here, the effective radius for
a three conductor bundle which are spaced
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at
a distance d from the center or d from each
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other with a radius r is given by 9th root
of r
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dash into d into d whole cube, which is equal
to cube root of r dash d square. This is
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much larger than r dash, which is used when
a single conductor is used. And therefore,
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.bundling helps in reducing the inductance
as it increases the effective radius. Second
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question was, what is transposition?
(Refer Slide Time: 0:03:12)
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Well transposition means, exchanging the position
of the three phase conductors along
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the length of the line. That is when we have
three phase conductors, which are not
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equilaterally spaced. That is equal spacing
between them is not possible due to physical
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constraints of that transmission line construction.
There in such a case we will get
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unequal flux linkages with the three conductors.
Most of the time if you have seen the power
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line transmission towers. You will find that
the three phase conductors are either placed
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in a horizontal configuration, if it is a
single
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circuit line. Or they may be placed in a vertical
configuration if it is a double circuit line.
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Therefore the equilateral spacing is not there
and the distance between the conductors
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phase conductors are not equal. This results
in different difference in flux linkages with
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the three conductors, Rresulting in difference
in inductances; in order to have a balance
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three phase system.
What we do is we make these conductors a phase
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conductors go through all the three
positions. That is what we do in transposition.
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We exchange the position of the three
conductors along the length of the line. So,
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that each conductor or each phase conductor
is in all the three positions for equal length
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of the line, which means that the average
flux
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linkage of all the conductors or for all the
three phase conductors has same.
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.So, if the spacing between three conductors
are in unequal. Unequal flux linkages will
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occur when the three phase conductors are
even when the three phase currents has
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balanced.
.
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Here we show the transposition the phase conductor
a is in position 1 for 1 3rd length of
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the line. It occupies position 2 for the next
1 3rd length of the line. And position 3 for
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the
last 1 3rd portion of the line. Similarly,
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phase b conductor is in position 2 for 1 3rd
of
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length of the line next 1 3rd length of the
line it is in position 3.
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And the last 1 3rd length of the line it is
in position 1. In this way each phase conductor
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is occupying all the three positions, which
results in average flux linkages to be equal.
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So, we say that by transposition each phase
conductor occupies the three positions for
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1
3rd length of the line. Therefore the average
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flux linkage and inductance become equal.
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..
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The third question was how the effect of earth
return current is taken into account in
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inductance calculation? Well J R Carson in
his paper a 1923 said that the earth return
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currents can be taken into account by means
of considering earth return conductance.
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That is we assume imaginary conductors, which
are placed below the ground. These
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conductors are suppose to have the same radius
are the GMR as the overhead conductors
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and they are placed directly below them in
the ground.
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The distance between the overhead conductor
and these imaginary conductors is given
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by the distance D e. Where D e or is given
by an empirical relation 658.5 square root
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of
rho by f, where rho is the earth resistivity
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normally if earth resistivity is not known.
Then
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we assume it to be around 100 ohm meter. And
f is the system frequency. So, having
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answer these questions, now we will going
to the main part of the lesson 5.
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..
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Now, this lessons as I said earlier we will
be talking about transmission line capacitance
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and it is calculation. So, here we will first
start with electric field and voltage calculation.
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Next we will take up the transmission line
capacitance calculation for a single phase
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line
with solid conductors. Then we will take up
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three phase line with equal spacing. And
finally, we will talk about three phase line
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with bundle conductors and unequal spacing.
Well as we have learnt in the lessons 3 and
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4 about calculating the resistance and
inductance of the transmission line. The transmission
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line also has capacitance, this is
mainly because the line conductors have a
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voltage difference between the phases. That
is
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phase conductors have voltage difference and
there is a difference of voltage between the
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phase conductor and the ground.
And these conductors are separated by a dielectric
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medium for overhead line this is air.
And in case of cables, this can be some kind
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of a dielectric, which can be impregnated
paper or it can be XLPE that is cross linked
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polyethylene insulation or some other kind
of insulation. Therefore we have two conductors
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at different voltages and there is a
dielectric between them, which results into
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a capacitance between the conductors or
between the conductors and the ground.
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..
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Now, for calculating the transmission line,
capacitance we have to go through first using
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Gauss’s law to calculate the electric field
strength. Once we are calculated the electric
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field strength, we find out the voltage between
the conductors and then we find out the
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capacitance by relationship C equal q by V.
So, will start with the Gauss’s law and
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calculation of electric field strength.
.
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Well Gauss’s law states that, total electric
flux leaving a close surface is equal to total
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charge within the volume enclose by the close
surface, which basically leads to that
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.statement that normal electric flux density
integrated over the closed surface will be
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equal to the charge enclosed by thus closed
surface. Mathematically we can write this
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as
the surface integral over the closed surface
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of electric flux density D or the norm the
electric flux density which is normal to the
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surface into d s will be equal to or integrated
over the surface. Will be equal to
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the surface integral of epsilon E, which is
again normal
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to the surface into d s that is over the whole
surface.
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Now, this epsilon is the permittivity of the
medium and E is the electric field strength
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at
the surface. So, that is what we have done
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is electric flux density D is replaced by
epsilon
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E, because D equal epsilon E, that we know
from electric field theory. So, this is equal
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to
the total charge enclosed by the volume, enclose
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by this close surface, this we can see in
this figure very clearly.
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.
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Let us say we have a conductor, which is of
very long conductor of radius r and this is
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charge with q Coulomb’s per meter. Here
we have assume q to be a positive charge it
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can be positive or negative a does not matter.
So, here we have assume there is conductor
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of radius r is having a charge of q Coulomb’s
per meter and it is very long conductor.
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Now, since this is perfect conductor electric
field inside is going to be zero. Now, for
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finding out the electric field, outside the
conductor what we need to do is we take up
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concentric cylindrical volume of 1 meter long
length. So, this volume shown here this is
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.a concentric cylinder of 1 meter length at
with a radius of X. Now since the electric
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field
lines will be radial to this surface, which
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is of this concentric cylinder.
Therefore, there will be no tangential component
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of the electric field. Now, we can find
out with this, what is the fields strength
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at the surface of the this concentric cylinder.
Now, using Gauss’s law, we will use the
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integration surface integration of epsilon
E d s.
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So, epsilon E x that is field strength at
this surface of this concentric cylinder integrated
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over this close surface. Closed surface area
is to going be equal to twice pi X into 1
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meter length of this cylinder. So, epsilon
Ex into twice pi X into 1 is equal to the
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charge
enclosed which is q into 1 meter length of
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the line. Because q coulomb per meter is the
charge on the conductor from this we can calculate
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the electric field strength, at the
surface of the concentric cylinder of radius
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X.
.
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So, E x equal to q by twice pi epsilon X volt
per meter. Now, using this Ex or the
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relationship for the field strength at any
point we can always find out the voltage
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difference between two points. So, if you
want to find out the voltage difference between
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2 points P 1 and P 2, which are at a distance
D 1 and D 2. From the center of the
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conductor, what we need to do is integrate
from D 1 to D 2 of Ex d x over D 1 to D 2.
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So, E x the integral Ex to E x d x to from
D 1 to D 2 will give me a the voltage difference
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V 1 2. So, V 1 2 equal to integral D 1 integral
of q by twice pi epsilon X d x from D 1 to
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D 2 this is equal to q by twice pi epsilon
log n D 2 by D 1 volts. Here epsilon is as
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I said
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.earlier is the permittivity of the medium
and this equal epsilon r into epsilon 0, where
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epsilon 0 is the permittivity of the free
space and for here also it is almost the same,
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epsilon 0 equal to 8.854 into 10 to the power
minus 12 Farad per meter. Now, real take
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since we have now found out the electric field
strength and the voltage between any 2
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points for any charge conductor. We can extended
to a multi conductor system.
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Now, let us a that we have a system of n conductors
with any conductor k having radius r
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k and has a charge q k per meter length of
the conductor. Then we can find out the
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voltage between 2 conductors i and j due to
charge on this conductor k. We write this
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as
V i j k. That is voltage between conductor
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i and j due to charge on conductor k. This
will
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be equal to q k by twice pi epsilon log n
the distance d of k to j, k to j and divided
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by
distance of k to i.
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As we can see using the old relationship,
that we have found here; q by twice pi epsilon
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log n D 2 by D 1. Now, if you want to find
out the voltage difference due to charge from
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all the conductors, which are in this system.
Then what we need to do is use the
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superposition theorem. That is add the voltages
due to the charge from each of this
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conductors. So, V i j in that case due to
charges on all the conductor will be equal
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to
summation k equal to 1 2 n of q k by twice
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pi epsilon log n D j k by D i k volts. So,
this
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way we can find out the voltage between any
two conductors or any two points due to
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charge on various conductors in a multi conductor
system.
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.
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.Now, let us try to apply this to find out
the voltage and then the capacitance of a
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single
phase line. Let us say we have a single phase
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line with two conductors are with
conductor x and conductor y. Conductor x has
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a radius r x and has a charge q Coulomb
per meter. Conductor y has a radius r y with
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a charge of minus q Coulomb per meter
because conductor y will be the written conductor.
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So, it will be carrying the written currents.
So, the charge will be negative after charge
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on
the conductor x. So, now we using the same
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formula we can find out V x y will be given
by one twice pi epsilon q log n D y x by D
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x x. That is because of the charge on this
conductor the voltage difference between these
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two will be given by this relationship 1
by twice pi epsilon q log n D y x by D x x.
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And the voltage between these two conductors
that is V x to V y. So, V x y will be equal
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to minus 1 by twice pi epsilon minus q log
n D y y by D x y. So, this can be reduce to
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q
by twice pi epsilon log n D y x into D x y
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divided by D x x into D y y. That is this
minus
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is converted to plus by changing reversing
these D x y and D y y; that is taking the
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inverse of this. So, we get D y x into D x
y divided by log n D x x by D y y. So, finally,
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we get the voltage between to conductors by
the relation q by twice pi epsilon log n D
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y
x into D x y divided by D x x into D y y.
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.
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Which if we substitute the values we get V
x y equal to q by pi epsilon log n D by square
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root of r x, r y. Because, this would have
become D square and this would be r x r y.
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So,
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.this square is taken out that becomes 2 into
q and this 2 pi epsilon. So, this 2 well cancel
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out and we get finally, q by pi epsilon log
n D by square root of r x into r y.
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Now, once we have found out the voltage, we
can find out the capacitance very easily.
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Capacitance C x y that is capacitance between
the two conductors x and y is equal to q
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by voltage between the two conductor. So,
this will be equal to pi epsilon divided by
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log
n D by square root r x r y. That is what we
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have done is divided q by this term. And then
we get this C x y is pi epsilon divided by
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log n D by square root of r x r y.
.
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So, this if we consider the radius of the
two conductors to be equal then r x will be
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equal
to r y; and therefore, square root of r x
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into r y equal r. Therefore we can write C
x y
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equal to pi epsilon divided log n D by r Farad
per meter this is a line capacitance
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between line to line. Now, if we see the system,
in the center of this, we will get 0
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voltage because, this will be positive potential,
this will be equal negative potential.
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So, by symmetry at the center we will get
a 0 potential or the potential of the ground
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or
the neutral. Therefore we can find out the
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voltage to the neutral as V x n equal to V
y n
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equal to V x y divided by 2. This will be
half the voltage because the voltage of one
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conductor is positive another conductor is
negative. So, the zero potential line will
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be in
between at the half, so V x n equal to V y
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n equal to V x y divided 2. Therefore we can
find out the capacitance to the neutral or
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capacitance to ground as C n which will be
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.equal to C x n. And which will be all again
equal to C y n that is capacitance from to
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ground for conductor x will be same as capacitance
to ground for conductor y.
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And this will be equal to q by V x n, which
is equal to 2 times C x y. As we can see here
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between x and y we have a capacitance C x
y. And if we have taken the neutral at the
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center, then we have now capacitance C x n,
between x and n. And capacitance C y n
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between n and y. And these capacitance is
will be equal to 2 times the capacitance
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between x and y, because they are in series.
So, 2 x y and 2 x, the 2 C x y and 2 C x y
203
00:26:24,240 --> 00:26:27,840
in
series will give me C x y.
204
00:26:27,840 --> 00:26:38,110
So, C y n or C x n which is equal to C n capacitance
to ground for any conductor, after
205
00:26:38,110 --> 00:26:45,559
for any of the 2 phase conductor will be equal
2 times C x y. This equal to twice pi
206
00:26:45,559 --> 00:26:56,309
epsilon divided by log n D by r Farad per
meter for line to neutral capacitors. Now,
207
00:26:56,309 --> 00:27:02,409
we
will take up the case of 3 phase system, because
208
00:27:02,409 --> 00:27:08,909
we know that post of the power system
that we have to the specially transmission
209
00:27:08,909 --> 00:27:14,809
systems are 3 phase systems. Therefore will
take the case of a three phase system.
210
00:27:14,809 --> 00:27:15,809
.
211
00:27:15,809 --> 00:27:21,110
Will start with the three phase line with
equilateral spacing, because as we have seen
212
00:27:21,110 --> 00:27:25,549
for
inductance calculation. We can always convert,
213
00:27:25,549 --> 00:27:31,750
if the line is transpose, we can always
convert any system into the equivalent 3 phase
214
00:27:31,750 --> 00:27:40,630
equivalent equilateral spacing of the
conductors by finding out the equivalent distance
215
00:27:40,630 --> 00:27:41,900
d e q.
216
00:27:41,900 --> 00:27:50,929
.So, here we have equilateral spacing conductor
a b and c each with a distance d from
217
00:27:50,929 --> 00:27:57,679
each other. We also assume that this system
consist of only three conductors. So, the
218
00:27:57,679 --> 00:28:10,850
some of the total charges will be equal to
0. That is q a plus q b plus q c equal 0.
219
00:28:10,850 --> 00:28:16,669
Again
as earlier we can find out the voltage between
220
00:28:16,669 --> 00:28:18,020
any two conductors.
.
221
00:28:18,020 --> 00:28:25,710
So, let us find out the voltage between a
and b. So, V a b will be equal to again using
222
00:28:25,710 --> 00:28:28,760
the
same relationship that we are used earlier.
223
00:28:28,760 --> 00:28:32,230
It will be equal to 1 by twice pi epsilon
into
224
00:28:32,230 --> 00:28:45,380
the q to charge on conductor a q a log n D
b a by D a a. Due to charge q b it will be
225
00:28:45,380 --> 00:28:52,190
plus
q b log n D b b by D a b that distance of
226
00:28:52,190 --> 00:28:59,220
conductor b to conductor b divided by distance
of the conductor b to conductor a. And due
227
00:28:59,220 --> 00:29:08,470
to charge q c log n D b c by D a c, that is
distance from conductor c to conductor b and
228
00:29:08,470 --> 00:29:17,430
distance from conductor c to conductor a.
Now, this if we substitute the values of D
229
00:29:17,430 --> 00:29:23,130
b a D a b or D b a and D a a D b b and so
on.
230
00:29:23,130 --> 00:29:32,909
This can be written as V a b equal to one
twice pi epsilon into q a log n D by r D b
231
00:29:32,909 --> 00:29:38,299
a is
equal to D a b, which is equal to D. And D
232
00:29:38,299 --> 00:29:43,809
a a is its self distance that is r the radius
of the
233
00:29:43,809 --> 00:29:51,691
conductor plus q b into log n D b b again.
Since the conductor radius we have assume
234
00:29:51,691 --> 00:30:01,100
to be same that is 3 conductors are same
radius or diameter. Therefore, D b b is equal
235
00:30:01,100 --> 00:30:06,770
to r Dab we will be equal to D plus qc log
n
236
00:30:06,770 --> 00:30:13,910
D b c which is equal to D and D a c, which
is equal to D. So, this is q c log and D by
237
00:30:13,910 --> 00:30:17,789
D
which will be q c log and 1; therefore, this
238
00:30:17,789 --> 00:30:21,640
term will go to 0. Therefore we have V a b
is
239
00:30:21,640 --> 00:30:31,330
equal to 1 by twice be epsilon into q a log
n D by r plus q b log n r by D.
240
00:30:31,330 --> 00:30:32,330
..
241
00:30:32,330 --> 00:30:40,110
Similarly, we can find out the voltage between
conductor a and conductor c. So, V a c
242
00:30:40,110 --> 00:30:49,159
will be equal to 1 by twice by epsilon into
due to charge q a, q a log n the distance
243
00:30:49,159 --> 00:30:59,490
between a and c D c a the distance between
a to a. So, D a plus q b log n distance of
244
00:30:59,490 --> 00:31:06,230
b to
c that is D c b divided by D distance of b
245
00:31:06,230 --> 00:31:14,700
to a. So, that is D a b plus q c log n distance
of
246
00:31:14,700 --> 00:31:25,590
c to c divide by distance of c to a, that
is equal to D c c by D c a.
247
00:31:25,590 --> 00:31:34,470
This will be equal to 1 by twice pi epsilon
q a log n D by r that substituting the distances
248
00:31:34,470 --> 00:31:44,970
D by r plus q b log n D by D plus qc log n
r by D, again this term since it is D by D
249
00:31:44,970 --> 00:31:48,590
log n
one. So, this term goes to 0, so we have V
250
00:31:48,590 --> 00:31:53,419
a c is equal to 1 by twice pi epsilon into
q a
251
00:31:53,419 --> 00:32:00,240
log n D by r plus qc log n r by D.
252
00:32:00,240 --> 00:32:01,240
..
253
00:32:01,240 --> 00:32:15,669
Now, if we look at this equilaterally space
conductors, here the voltage V a b is equal
254
00:32:15,669 --> 00:32:25,790
to
root 3 times V a n at an angle of plus 30
255
00:32:25,790 --> 00:32:31,059
degrees from the x axis. So, this is equal
to root
256
00:32:31,059 --> 00:32:44,769
3 times V a n into root 3 by 2 plus j half.
That is we have converted this from polar
257
00:32:44,769 --> 00:33:00,870
coordinate to rectangular. Similarly V a c
if you see is equal to minus of V c a. That
258
00:33:00,870 --> 00:33:05,290
is
negative of V c a, this is equal to again
259
00:33:05,290 --> 00:33:12,010
root 3 times V a n. And this if you see V
a c’s at
260
00:33:12,010 --> 00:33:20,150
minus 30 degree from the x axis. So, root
3 V a n with an angle of minus 30 degrees.
261
00:33:20,150 --> 00:33:28,500
This is equal to route 3 V a n into root 3
by 2 minus j half.
262
00:33:28,500 --> 00:33:29,500
.
263
00:33:29,500 --> 00:33:38,640
.So, now if we add these two voltages V a
b plus V a c we get this as equal to root
264
00:33:38,640 --> 00:33:45,179
3 times
V a n. That is if we see if we add these two
265
00:33:45,179 --> 00:33:55,370
this j plus j half and minus j half will cancel
out and we have got two times root 3 by 2
266
00:33:55,370 --> 00:34:00,559
into times V a n. So, that two will cancel
out
267
00:34:00,559 --> 00:34:07,950
and root 3 into root 3 will give us 3. So,
this will be equal to 3 V a n. Therefore,
268
00:34:07,950 --> 00:34:12,570
we will
get V a n that is voltage of phase a to the
269
00:34:12,570 --> 00:34:23,100
neutral will be equal to 1 3rd of the V a
b plus
270
00:34:23,100 --> 00:34:27,240
V a c.
So, substituting the values from V a b for
271
00:34:27,240 --> 00:34:34,270
V a b and V a c as we have calculated earlier.
We will get this as 1 3rd of 1 by twice pi
272
00:34:34,270 --> 00:34:41,460
epsilon into 2 times q a log n D by r plus
q b
273
00:34:41,460 --> 00:34:51,550
plus q c log n r by D. That is if we see here
V a c is q a log n D by r plus qc log n r
274
00:34:51,550 --> 00:35:00,610
by D
and V a b is 1 by twice by epsilon q a log
275
00:35:00,610 --> 00:35:05,590
n D by r plus q b log n r by D. So, if we
add we
276
00:35:05,590 --> 00:35:16,280
will get twice q a log n D by r and q b log
n r by D plus q c log n r by D that is what
277
00:35:16,280 --> 00:35:17,460
we
have got here.
278
00:35:17,460 --> 00:35:26,730
So, we have two times q a log n r by D plus
q b plus qc log n r by D. So, from this
279
00:35:26,730 --> 00:35:37,099
relationship, we can find out now V a n is
equal to 1 by twice pi epsilon into; now q
280
00:35:37,099 --> 00:35:42,760
b
plus qc is equal to minus q a, because q a
281
00:35:42,760 --> 00:35:52,570
plus q b plus qc equal to 0. So, this becomes
twice q a log n r D by r minus q a log n r
282
00:35:52,570 --> 00:36:00,410
by D, which can be put as plus q a log n r
D by
283
00:36:00,410 --> 00:36:06,579
r. And therefore, we get 3 q a log n D by
r.
284
00:36:06,579 --> 00:36:12,630
And this three will cancel with this 1 by
3. So, we have got 1 by twice pi epsilon q
285
00:36:12,630 --> 00:36:19,520
a log
n D by r. So, once we have calculated the
286
00:36:19,520 --> 00:36:26,810
voltage between phase conductor a and the
neutral, which is at zero potential. Then
287
00:36:26,810 --> 00:36:34,530
we have got C a n or the voltage or the
capacitance between conductor a and the neutral
288
00:36:34,530 --> 00:36:40,710
or the ground will be equal to q a by V
a n, which comes out to be equal to twice
289
00:36:40,710 --> 00:36:49,990
pi epsilon divided by log n D by r Farad per
meter. This is line to neutral capacitance
290
00:36:49,990 --> 00:36:57,690
or line to ground capacitance.
291
00:36:57,690 --> 00:36:58,690
..
292
00:36:58,690 --> 00:37:07,550
In general, we instead of having D as we have
seen earlier, we if the conductor spacing
293
00:37:07,550 --> 00:37:16,280
are not equal. That is we do not have a equilateral
configuration of conductors, then other
294
00:37:16,280 --> 00:37:27,460
configurations can always be converted into
an equivalent equilateral
295
00:37:27,460 --> 00:37:35,710
configuration there
the distance D equal to D e q that is the
296
00:37:35,710 --> 00:37:39,010
equivalent equilateral configuration distance.
So,
297
00:37:39,010 --> 00:37:46,400
D e q is equal to cube root of D a b into
D b c into D a c; that is what we had seen
298
00:37:46,400 --> 00:37:51,349
in
earlier lecture on 3 phase line that is in
299
00:37:51,349 --> 00:37:52,980
lessons 4.
.
300
00:37:52,980 --> 00:38:00,880
.Now, let us take a more general case, where
we have considered a 3 phase bundle
301
00:38:00,880 --> 00:38:09,790
conductor line with unequal spacing. We have
here a 3 phase line with 2 conductors in a
302
00:38:09,790 --> 00:38:17,260
bundle for each phase. And this configuration
for the three phase system is now
303
00:38:17,260 --> 00:38:23,900
horizontal configuration. That is distance
from a to b that is centre of conductor bundle
304
00:38:23,900 --> 00:38:27,640
a
to conductor bundle b is D a b.
305
00:38:27,640 --> 00:38:34,500
The distance between centre of conductor bundle
b and conductor bundle c is D b c. And
306
00:38:34,500 --> 00:38:41,140
distance from the centre of the conductor
bundle a to centre of conductor bundle c is
307
00:38:41,140 --> 00:38:48,660
D a
c. Whereas each conductor has a radius r,
308
00:38:48,660 --> 00:38:54,640
here we are assuming that all the conductors
have the same radius. And the bundle distance
309
00:38:54,640 --> 00:38:58,660
that is distance between the centre of the
two conductors is d.
310
00:38:58,660 --> 00:39:06,349
As we had said earlier this d is generally
much larger than r, it is normally 10 times
311
00:39:06,349 --> 00:39:11,561
or
more r. In general we have around 30 centimeter
312
00:39:11,561 --> 00:39:18,280
or 40 centimeter distance between the
conductors for bundle conductor lines. So,
313
00:39:18,280 --> 00:39:25,070
d can be of the order of 30 to 40 centimeters
where as the conductor radius will be of the
314
00:39:25,070 --> 00:39:31,940
order of 2 to 3 centimeter. So, now again
we
315
00:39:31,940 --> 00:39:36,970
need to find out the voltage difference between
the two phase conductors.
316
00:39:36,970 --> 00:39:37,970
.
317
00:39:37,970 --> 00:39:43,280
So, we find out V a b, now here what we have
assumed ((Refer Time: 39:41)) is that
318
00:39:43,280 --> 00:39:50,930
conductor a is having a total charge of q
u. And since we have two conductor bundle.
319
00:39:50,930 --> 00:39:54,890
So,
each conductor or the sub conductor in the
320
00:39:54,890 --> 00:39:59,021
bundle will be having a charge q by 2. So,
it
321
00:39:59,021 --> 00:40:04,290
will be having q a by 2, this will be having
q a by 2. This conductor will be having
322
00:40:04,290 --> 00:40:12,130
.charge q b by 2, this will be having charge
q b by 2, because all of them are of the same
323
00:40:12,130 --> 00:40:18,880
kind and same radius.
So, c will be having charge q c by 2 this
324
00:40:18,880 --> 00:40:26,160
sub conductor and this subsequently conductor
will be having charge q c by 2. So, now in
325
00:40:26,160 --> 00:40:29,280
order to find out V a b again what we need
to
326
00:40:29,280 --> 00:40:36,970
do is find out the voltage between a and b
due to the charges on each of these conductors
327
00:40:36,970 --> 00:40:45,040
or sub conductors.
So, V a b will be equal to 1 by twice pi epsilon.
328
00:40:45,040 --> 00:40:52,690
Now q 2 charge on conductor sub
conductor one of phase a it will be q a by
329
00:40:52,690 --> 00:40:55,260
2, that is the charge on that sub conductor
into
330
00:40:55,260 --> 00:41:06,690
log n. D b a by D a a that is the distance
between a to b divided by distance of a to
331
00:41:06,690 --> 00:41:10,619
a that
will be again the radius of the conductor.
332
00:41:10,619 --> 00:41:22,800
Plus q a by 2 due to the sub conductor the
second sub conductor in phase a that is, this
333
00:41:22,800 --> 00:41:27,080
conductor this sub conductor it will be q
a
334
00:41:27,080 --> 00:41:40,170
by 2 into log n D b a dash by D a a dash.
Now, due to this sub conductor the voltage
335
00:41:40,170 --> 00:41:46,710
V a b will be given by q b by 2 log n D b
b
336
00:41:46,710 --> 00:41:53,970
divided by Dab again for the other sub conductor
of phase b this will be plus q b by 2 log
337
00:41:53,970 --> 00:42:02,690
n D b b divided Dab dash. Similarly, for the
two sub conductors charge on the two sub
338
00:42:02,690 --> 00:42:10,650
conductors of phase c. We will have components
qc by 2 log n D b c by D a c plus qc by
339
00:42:10,650 --> 00:42:17,060
two log n D b c dash by D a c dash.
.
340
00:42:17,060 --> 00:42:30,410
.This after substituting the values will be
given by q a by 2 log n Dab by r. That is
341
00:42:30,410 --> 00:42:40,310
that a a
plus log n D a b by D, which is the distance
342
00:42:40,310 --> 00:42:51,030
D a a dash. Now, here what we are saying
that D b a and D b a dash are almost same,
343
00:42:51,030 --> 00:42:57,010
because the distance between the two phase
conductors is much larger compared to the
344
00:42:57,010 --> 00:43:00,819
distance between two sub conductors of a
bundle.
345
00:43:00,819 --> 00:43:10,290
So, this small approximation we have made,
that is D a b is the distance between the
346
00:43:10,290 --> 00:43:18,470
centre of the two conductors. So, D this is
q a by 2 log n D a b by r plus log n D a b
347
00:43:18,470 --> 00:43:23,430
by
D. Similarly for conductor b or the charge
348
00:43:23,430 --> 00:43:27,290
due to conductor b will be q b by 2 into log
n r
349
00:43:27,290 --> 00:43:41,891
by D a b plus log n D by D a b plus due to
charge on conductor c q c by 2 log n D b c
350
00:43:41,891 --> 00:43:45,001
by
D a c plus log n D b c by D a c.
351
00:43:45,001 --> 00:43:54,630
Now, this when we add these terms finally,
can be written as 1 by twice pi epsilon q
352
00:43:54,630 --> 00:44:01,560
a
log n D a b by r d square root of r d plus
353
00:44:01,560 --> 00:44:08,540
q b by log n square root of r d by D a b plus
q c
354
00:44:08,540 --> 00:44:16,740
log n D b c by D a c. Now, this if we remember
the inductance relationship is very
355
00:44:16,740 --> 00:44:19,430
similar to that relationship.
.
356
00:44:19,430 --> 00:44:28,550
And therefore this can be again written in
terms of capacitance is equal to, capacitance
357
00:44:28,550 --> 00:44:36,240
between neutral can be written as twice pi
epsilon log n D e q by D s, where we have
358
00:44:36,240 --> 00:44:50,960
already defined D e q, as cube root of Dab
into D b c into D a c.
359
00:44:50,960 --> 00:45:01,310
.Therefore, capacitance to neutral is given
by the same relationship except that we have
360
00:45:01,310 --> 00:45:12,349
now replaced r by D s, Where D s is the effective
radius of the bundle conductor and D e
361
00:45:12,349 --> 00:45:20,130
q is the equivalent equilateral spacing for
that conductor. So, for two conductor bundle
362
00:45:20,130 --> 00:45:24,099
as
we had see earlier or in this case the D s
363
00:45:24,099 --> 00:45:28,460
is equal to the self distance. D s is equal
to
364
00:45:28,460 --> 00:45:36,470
square root of r into D for a 3 conductor
bundle the self D s is equal to cube root
365
00:45:36,470 --> 00:45:40,329
of r into
D square and for a 4 conductor bundle.
366
00:45:40,329 --> 00:45:53,640
It is equal to 1.0491 4th root of rd cube
this we had seen in the last lesson. Now,
367
00:45:53,640 --> 00:46:01,970
once we
have calculated the capacitance. Now, we can
368
00:46:01,970 --> 00:46:08,349
find out the charging current, because
there is voltage and the there is capacitance
369
00:46:08,349 --> 00:46:14,430
between the conductor and the ground. So,
there will be some current, which will be
370
00:46:14,430 --> 00:46:18,330
flowing in the system and this current, which
is
371
00:46:18,330 --> 00:46:24,230
flowing in the capacitance is called charging
current for the system.
372
00:46:24,230 --> 00:46:25,230
..
373
00:46:25,230 --> 00:46:36,290
Now, this transmission line has a capacitance
C a n for phase a C b n for phase b n c, c
374
00:46:36,290 --> 00:46:40,290
n
for phase c. Now, this capacitance is we can
375
00:46:40,290 --> 00:46:43,420
find out that admittance, admittance will
be
376
00:46:43,420 --> 00:47:00,530
equal to twice pi f into c. That is we will
have y will be equal to minus j omega c.
377
00:47:00,530 --> 00:47:09,190
So, we can find out the admittance between
the phase conductor and neutral or the
378
00:47:09,190 --> 00:47:18,730
ground. So, if we are finding between two
conductors x and y. So, then we will find
379
00:47:18,730 --> 00:47:24,069
out
the admittance for C x y as Y x y and the
380
00:47:24,069 --> 00:47:27,960
charging current will be Y x y into V x y.
This
381
00:47:27,960 --> 00:47:36,050
is equal to j omega C x y into V x y Amperes.
So, once we have calculated the charging
382
00:47:36,050 --> 00:47:45,660
current, we can also find out how much is
the volt Ampere reactive, which is supplied
383
00:47:45,660 --> 00:47:49,940
by
this charging capacitance of the transmission
384
00:47:49,940 --> 00:47:54,460
line.
So, the capacitance of the transmission line
385
00:47:54,460 --> 00:47:58,970
will. In fact, be supplying reactive power
to
386
00:47:58,970 --> 00:48:06,700
the system. And this will be given by Q c,
which is equal to V x y square divided by
387
00:48:06,700 --> 00:48:14,440
X c
or Y x y into V square x y or V x y square.
388
00:48:14,440 --> 00:48:26,680
So, we can find out the charging volt Ampere
reactive as omega C x y into V x y square.
389
00:48:26,680 --> 00:48:33,860
So, this will provide as the volt Ampere
reactive which is generated by the transmission
390
00:48:33,860 --> 00:48:43,220
line.
And if we are looking for the charging current
391
00:48:43,220 --> 00:48:46,100
to the ground, then this will be equal to
y
392
00:48:46,100 --> 00:48:54,540
into V a n. This will be equal to j omega
Can into the voltage between line and the
393
00:48:54,540 --> 00:49:04,640
neutral. So, once we know this current we
can find out by multiplying this by another
394
00:49:04,640 --> 00:49:11,470
V
that is j omega C a n into V l n square will
395
00:49:11,470 --> 00:49:17,700
give us the reactive power generated by this
line.
396
00:49:17,700 --> 00:49:18,700
..
397
00:49:18,700 --> 00:49:28,190
So, for a single phase system we have the
charging reactive power supplied by the line
398
00:49:28,190 --> 00:49:35,410
given by y into V a n square. This is equal
to omega C a n into V l n square volt ampere
399
00:49:35,410 --> 00:49:45,349
reactive of the 3 phases system this will
be equal to 3 times the charging volt ampere
400
00:49:45,349 --> 00:49:52,720
reactive generated by a single phase line
or between each phase and neutral of the line.
401
00:49:52,720 --> 00:50:01,349
So, this will be equal to three times omega
C a n into V l n square, which is equal to
402
00:50:01,349 --> 00:50:09,460
omega C a n into V l l square that is if you
take line to line voltage or the system voltage
403
00:50:09,460 --> 00:50:15,810
for a 3 phase system. Then charging reactive
power generating by the line will be given
404
00:50:15,810 --> 00:50:28,640
by omega C a n V l l square. Now, that we
have seen have we can calculate the
405
00:50:28,640 --> 00:50:35,020
capacitance and also find out the charging
current and the reactive power generated by
406
00:50:35,020 --> 00:50:45,180
the line. Let us take one example to see have
we calculate these quantities for a
407
00:50:45,180 --> 00:50:46,190
Transmission system.
408
00:50:46,190 --> 00:50:47,190
..
409
00:50:47,190 --> 00:50:54,570
So, let us this example we have a three phase
400 kV 50 Hertz 350 kilometer overhead
410
00:50:54,570 --> 00:51:01,630
transmission line that has flat horizontal
spacing with three identical conductors. That
411
00:51:01,630 --> 00:51:05,450
is
we have three identical conductors placed
412
00:51:05,450 --> 00:51:12,950
in a flat horizontal spacing. The conductors
have on outside diameter of 3.28 centimeter
413
00:51:12,950 --> 00:51:20,579
that is the diameter of the conductor is 3.28
centimeter and the distance between the adjacent
414
00:51:20,579 --> 00:51:26,510
conductors is 12 meter. Now, for this
system determine the capacitive reactance
415
00:51:26,510 --> 00:51:34,319
to neutral in Ohms per meter per phase and
the capacitive reactance for the line in Ohms
416
00:51:34,319 --> 00:51:37,059
per phase.
.
417
00:51:37,059 --> 00:51:49,380
.So, for solving this, we will take case of
a fully transpose 3 phase line. So, we have
418
00:51:49,380 --> 00:51:53,750
the
fully transpose 3 phase line. The voltage
419
00:51:53,750 --> 00:51:57,800
is given by the relationship V a b equal to
1 by
420
00:51:57,800 --> 00:52:07,420
twice pi epsilon this is epsilon into q a
into log n D e q by r plus q b into log n
421
00:52:07,420 --> 00:52:14,280
r by D e q,
where D e q is equal to D 1 2 into D 2 3 into
422
00:52:14,280 --> 00:52:23,450
D 3 1 into the power 1 by 3 that is cube root
of D 1 2 D 2 3 into D 3 1. Now, substituting
423
00:52:23,450 --> 00:52:32,120
this values we get this as equal to 12 meters
into 12 meters into 24 meters. That is the
424
00:52:32,120 --> 00:52:40,050
distance between the conductor a and c, the
cube root of this will give us equal to 15.119
425
00:52:40,050 --> 00:52:49,420
meter. Similarly, we can write the
relationship for V a c and V a c will be equal
426
00:52:49,420 --> 00:52:54,450
to 1 by twice pi epsilon into q a into log
n
427
00:52:54,450 --> 00:52:59,849
D e q by r plus q c into log n r by D e q.
.
428
00:52:59,849 --> 00:53:07,450
So, substituting again the values for the
system.
429
00:53:07,450 --> 00:53:08,450
..
430
00:53:08,450 --> 00:53:19,470
We have
substituting the values we will get V a b
431
00:53:19,470 --> 00:53:25,570
and V a c and we know V a b plus V a
c equal to 3 V a n. And also we know that
432
00:53:25,570 --> 00:53:33,010
q b plus qc equal to minus q a therefore,
adding V a b and V a c we will get V a n equal
433
00:53:33,010 --> 00:53:43,220
to q a divided by twice pi epsilon log n D
e q by r.
434
00:53:43,220 --> 00:53:44,220
.
435
00:53:44,220 --> 00:53:50,200
Therefore the capacitance to neutral will
be equal to q a by V a n, which will be equal
436
00:53:50,200 --> 00:53:56,790
to
twice pi epsilon log n D e q by r. And this
437
00:53:56,790 --> 00:53:59,921
will after substituting value comes out to
be
438
00:53:59,921 --> 00:54:07,940
8.163 into 10 to the power minus 6 micro Farad
per meter. And we can find out the
439
00:54:07,940 --> 00:54:14,240
.admittance the charging admittance of the
line to neutral, which will be equal to twice
440
00:54:14,240 --> 00:54:17,140
pi
f into C n.
441
00:54:17,140 --> 00:54:23,630
So, after substituting the value, we will
get this as equal to 2.565 into 10 to the
442
00:54:23,630 --> 00:54:27,460
power
minus 9 Siemens per meter per phase. Now,
443
00:54:27,460 --> 00:54:36,700
we given the length of the line as 350
kilometers therefore, Y n equal to 8.978 into
444
00:54:36,700 --> 00:54:42,290
10 to the power minus 4 Siemens per phase.
That is multiplying this by 350 kilometer
445
00:54:42,290 --> 00:54:51,740
which is 350 into 10 to the power 3 meters.
So, this gives as this much and the reactance
446
00:54:51,740 --> 00:54:59,530
will be equal to 1 by admittance, which
comes out be equal to 1.1138 into 10 to the
447
00:54:59,530 --> 00:55:10,800
power 3 Ohms per phase. That is 1.1138 kilo
Ohms per phase. That is all per this lecture,
448
00:55:10,800 --> 00:55:18,079
we will continue with capacitance
calculation in the next lessons. Where will
449
00:55:18,079 --> 00:55:27,920
talk about how we take into account the effect
of earth for capacitance calculation.
450
00:55:27,920 --> 00:55:34,160
Thank you.
Preview of next lecture
451
00:55:34,160 --> 00:55:41,450
Lecture No. #06
Transmission Line Capacitance (Contd.)
452
00:55:41,450 --> 00:55:49,000
Welcome to lessons 6 on Power System Analysis.
In this lesson we are going to discuss
453
00:55:49,000 --> 00:55:54,369
Transmission Line Capacitance, which we were
discussing in lessons 5, we are
454
00:55:54,369 --> 00:56:02,630
continuing with it.
455
00:56:02,630 --> 00:56:03,630
..
456
00:56:03,630 --> 00:56:09,180
Here what we will do is we will consider the
effect of earth on transmission line
457
00:56:09,180 --> 00:56:20,640
capacitance. Actually when we have this transmission
lines, the phase conductors are in
458
00:56:20,640 --> 00:56:27,040
above the ground in over transmission system
and the distance between the phase
459
00:56:27,040 --> 00:56:33,670
conductors. And the distance between phase
conductor and ground are of the same
460
00:56:33,670 --> 00:56:42,130
magnitude. And therefore, the earth which
acts as equi-potential surface thus affect
461
00:56:42,130 --> 00:56:45,980
the
electric field lines and there by the capacitance.
462
00:56:45,980 --> 00:56:52,320
So, we need to consider the effect of earth
and calculating the capacitance for
463
00:56:52,320 --> 00:57:02,859
transmission lines. So, in these lessons we
will discuss how the earth or the ground
464
00:57:02,859 --> 00:57:10,570
affects the capacitance of a single phase
transmission line and how it affects the 3
465
00:57:10,570 --> 00:57:19,099
phase
transmission line.
466
00:57:19,099 --> 00:57:20,099
..
467
00:57:20,099 --> 00:57:28,799
Then we will talk about a double circuit line
and we will talk about, how we calculate the
468
00:57:28,799 --> 00:57:33,490
series and shunt impedance of the double circuit
line.
469
00:57:33,490 --> 00:57:34,490
.
470
00:57:34,490 --> 00:57:43,390
Such a system we are showing here, here we
have a three phase line. This is one circuit
471
00:57:43,390 --> 00:57:52,089
and this another circuit which is placed here.
So, this is a double circuit system, here
472
00:57:52,089 --> 00:57:55,690
we
are showing this double circuit line. This
473
00:57:55,690 --> 00:58:02,200
is one circuit, this is another circuit connecting
two bus bars or two substations.
474
00:58:02,200 --> 00:58:09,099
.Now, the current flowing in one circuit is
I p 1, current flowing in the other circuit
475
00:58:09,099 --> 00:58:17,550
is I p
2 where I p is basically a three element vector,
476
00:58:17,550 --> 00:58:28,690
which consists of I a, I b and I c. So, we
have I a 1, I b 1, I c 1 after circuit 1 and
477
00:58:28,690 --> 00:58:33,930
I a 2, I b 2, I c 2 for the circuit 2. Now,
this
478
00:58:33,930 --> 00:58:42,299
current flowing here from one circuit will
be I p 1 plus I p 2, which gets divide into
479
00:58:42,299 --> 00:58:46,569
I p 1
I p 2 and may be x getting connected on this
480
00:58:46,569 --> 00:59:08,660
side. This is the system that we are showing
we may we have learnt about we will do a lessons
481
00:59:08,660 --> 00:59:19,039
7.
Thank you.
482
00:59:19,039 --> 00:59:19,039
.