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Welcome to lesson 6 on Power System Analysis.
In this lesson, we are going to discuss
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Transmission Line Capacitance which we were
discussing in lessons 5. We are
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continuing with it.
.
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Here, what we will do is, we will consider
the effect of earth on transmission line
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capacitance. Actually when we have this transmission
lines, the phase conductors are in
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above the ground in over a transmission system.
And the distance between the phase
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conductors and ground are of the same magnitude.
And therefore, the earth which acts as
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a equipotential surface. Thus effect the electric
field lines and thereby the capacitors.
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So, we need to consider the effect of earth.
And calculating the capacitance for
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transmission lines. So, in these lessons we
will discuss, how the earth or the ground
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affects the capacitance of a single phase
transmission line, and how it affects the
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three
phase transmission line.
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..
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Then, we will talk about a double circuit
line. And we will talk about how we calculate
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the series, and shunt impedance of the double
circuit line. In fact, when we have a double
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circuit line the two circuits are somewhat
close to each other. That is the three phase
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conductors of one circuit and the three phase
conductors of the other circuit, are most
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of
the time placed on the same tower. And therefore,
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they are in quite close proximity.
And because of which the both the magnetic
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field lines, as well as electric field line
do
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get effected by the currents flowing in the
other circuit. And the voltages present in
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the
phase conductors in the other circuit. Therefore,
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for a double circuit line, we have the
mutual effects. Because of the two circuits
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and that needs to be considered. And we will
see how we take care of this, in this lesson.
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Well, we will first start with the effect
of
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earth on transmission line capacitance.
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..
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Now, as I said earlier, we consider the earth
surface as an equipotential surface. Now,
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the
earth surface may not be a plane. It may be
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undulating, it may not be horizontal, it may
incline. But, for our purpose of calculation
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we will always consider that to be a
horizontal plane surface.
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.
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Now, if we see that there is a conductor as
shown here. The phase conductor, which is
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charged with positive charge q per meter length,
then the lines of electrical lines or field
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line should be radiating from this in radial
direction. But, because the earth is acting
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as
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.an equipotential surface. So, these field
lines will bend and go to the negative potential
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on the earth. So, the effect of earth is to
change the configuration of the electric field
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lines. And thereby, this will affect the capacitance
of the conductor system. Now, how do
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we take care of this effect. Well, what we
do is, we use what we call as method of
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images.
.
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Now, in this what we do is, we consider a
conductor directly below the overhead
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conductor in the ground at a distance; which
is equal to the height of the conductor above
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the ground. That is we are considering an
image of this conductor below the ground.
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So,
this is a mirror image of this conductor.
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And the charge on the image conductor is
opposite of that of the overhead conductor.
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Now, if you see, if you have such a system,
then the field lines will look like as shown
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here.
Now, these field lines if you see are very
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much similar to if we had as just this
conductor, and the ground and consideration.
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Therefore, what we find as that, that if we
take another conductor with an opposite charge.
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And place it below the ground, just
below the ground of this conductor at a distance,
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which is equal to the height of the
conductor above the ground.
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Then, the field lines created are exactly
same as that, if this conductor as above that
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earth
surface. That is the effect of earth can be
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taken care of by this kind of a situation.
So, the
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method of images basically consist of considering
image conductors below the ground at
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.a depth; equal to the height of the overhead
conductor above the ground. And the charge
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on this image conductor is of the opposite
polarity. So, if we do this then we have a
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system which is consistent of the same effect
as that of the earth. Now, we will try to
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use
this for calculating the capacitance of a
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single phase line.
.
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So, capacitance of a single phase line considering
the effect of earth.
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.
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Now, here we have a single phase line with
two conductors x and y. Placed at a distance
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D, the conductors have a radius of r x and
r y. The charge on conductor x is q coulombs
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.per meter and charge on conductor is minus
q coulombs per meter, because conductor y
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in a single phase system will be acting as
a return conductor. That is current flowing
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in x
will be returning through conductor y. Now,
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this system of conductors single phase 2
conductors are above the ground at a height
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of H meters.
So, if we have to now take the effect of ground
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or earth. Then, what we have to do is, we
have to consider image conductor x dash, which
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is placed directly below the conductor x.
At a distance equal to H below the ground,
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that is the distance between conductor x.
And
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it is image H x x will be equal to 2 H. And
the charge on this image conductor x dash
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will be minus q coulombs per meter.
Similarly, we will consider the image conductor
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for conductor y, as y dash which will be
placed directly below the conductor y at a
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distance of 2 H. That is at a distance H below
the ground directly under conductor y. And
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the charge on this conductor y dash will be
equal to plus q per coulomb per meter. So,
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now with this system we would like to find
out the voltage between the two conductors
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x and y.
.
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We can write this voltage from the earlier
lesson, that we have had. We can write this
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voltage V x y is equal to q by twice pi epsilon
into log n D y x. That is the distance of
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the
conductor b to conductor a divide by D x x.
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That is the distance of conductor a to
conductor a, that is conductor x to conductor
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x. So, in similar way we will write for the
charge on conductor y, considering the charge
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on conductor y.
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.We will have the relationship q by twice
pi epsilon minus q by twice pi epsilon D y
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y.
That is the distance of conductor y to conductor
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y itself divided by distance of conductor
x from y. Now, due to the image conductor
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x dash, we will have the effect on the voltage
given by q by minus q by twice pi epsilon
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log n H y x. That is distance of image
conductor x to conductor y. And H x x that
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is divided H x x, that is the distance between
x and the image conductor x.
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For the charge on the conductor y dash, we
will have the term q by twice pi epsilon log
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n
H y y divided by H x y. Now, if we consider
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all these terms together. And arrange them,
then we will get this as equal to q by twice
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pi epsilon log n D y x into D x y divided
by D
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x x into D y y. That is this minus log n D
y y by D x y can be written as plus log n
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D x y
by D y y. And then we can add these two together
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that will get me log n D y x into D x y
divided by D x x into D y y.
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Similarly, considering these two conductors,
this if we make it as negative then this
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becomes H x y by H y y. So, this will be minus
log n H x y by H y y. Therefore, if we
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add these two, then we will get minus log
n H y x into H x y divided by H x x into H
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y y.
Now, considering the symmetry of the figure,
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that we have H x x will be equal to H y y
and H x y will be equal to H y x. Also D x
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x is equal to r x and D y y is equal to r
y and if
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both the conductor have the same radius, then
we will have that equal to r.
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Therefore, this can become as D y x and D
x y are basically equal to D. So, this becomes
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q by twice pi epsilon log n D square by r
square, which can be written as this square
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can
be taken out. So, this will have 2 here and
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2 will cancel with this 2. So, it will become
q
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by pi epsilon log n D by r. In the similar
H x y is equal to H y x. So, this becomes
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equal
to H x y square and H x x is equal to H y
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y which is equal to H x x square, therefore
this
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square term can be taken here. So, 2 will
cancel with this 2 here, so we get q by pi
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epsilon minus log n H x y by H x x. So, this
way we have calculated the voltage between
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the 2 phase conductors x and y for a single
phase line.
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..
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Once we have got this voltage, we can calculate
the line to line capacitance as C x y is
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equal to q by V x y. So, now this is simply
q divided by V x y, that we have calculated
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this will come out to be pi epsilon by log
n D by r minus log n H x y by H x x farad
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per
meter. Now, as we had seen earlier, the capacitance
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to neutral or capacitance to ground
will be equal to twice C x y, because the
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earth surface or the ground will have a potential
which will be half that of between x and y.
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So, we can write C n is equal to 2 C x y,
which is equal to twice pi epsilon divided
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by log
n D by r minus log n H x y by H x x. So, in
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this way we can calculate the capacitance
considering the effect of ground. Now, here
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what we are finding that, what how this
effect of ground is taken into account. If
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you see this expression and compare it with
the
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expression for the single phase capacitance
of a single phase transmission line, then
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we
are saying seeing that, it is this term which
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is an addition or for taking the effect of
earth.
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And here, we can see that if the conductors
are very much above the ground compared to
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the distance between them. Then, D x y and
D x x will be almost equal, that is this
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distance and this distance will be almost
equal. If the conductors are very much above
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the ground that is H is very large. And in
that case, we find that the effect of ground
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will
be negligible. So, if the conductors are not
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far off above the ground, as compared to the
distance between the phase conductors.
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.Then, the earth does affect the capacitance
and the effect as in increasing the capacitance
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as compared to when we have not considered
the effect of earth, because this total term
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is now getting reduced.
.
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Now, we will talk about the capacitance of
3 phase line considering the effect of earth.
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Now, here we have 3 conductors a, b and c
placed at position 1, 2 and 3. The distance
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between these conductors are D 1 2, which
is the distance between a and b. D 2 3 which
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is distance between b and c, and D 3 1 which
is distance between c and a. The
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conductors are carrying a charge of q a, the
phase a conductor has a charge q a per meter.
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Phase b conductor has charge q b per meter
and phase c conductor has charge q c per
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meter.
Now, these conductors are placed above ground.
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And if we have to take into account the
effect of ground. Then we have to consider
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the image conductors. The image conductors
will be for the phase a conductor. Here which
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will be again at a distance, which will be
below the ground which will be equal to the
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height of this conductor above the ground.
So, this will be an image conductor here with
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a charge, which will be negative of that of
the conductor here.
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Similarly, for phase b conductor, we have
a image conductor here with a charge minus
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q
b. And for phase conductor c, we have an image
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conductor with a charge minus q c.
Now, the distance is from the phase conductors
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to the image conductors are given as
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.from conductor a to it is image a dash it
will be h 1 1. That is considering this is
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position
one. Similarly, for phase a conductor and
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00:20:54,919 --> 00:21:03,309
image of b, we will have the distance h 1
2. For
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00:21:03,309 --> 00:21:12,720
the distance from phase b conductor to image
of a, it will be h 2 1 and so on. So, we have
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all these distances which are shown in this
figure.
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.
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Now, we can calculate the voltage between
2 phase conductors. So, we will calculate
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that
voltage V a b, the voltage between conductor
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a and b. And this will be given by 1 twice
pi epsilon n 2 q 2 charge on conductor a,
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that is q a. It will be log n D 1 2 by r minus
log
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n h 1 2 by h 1 1, . that is log n D 1 2 by
r. That is from this
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conductor distance of this from phase b conductor
D 1 2.
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And distance of this conductor from itself
that is D 1 2 and divided by the distance
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from
itself that is D 1 1, which is equal to r.
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So, we have this first term taking into account
that
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part. Now, because of the image conductor,
that is placed here. Again we will have
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minus q a is the charge on this conductor.
And the distance is from this to b, it will
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be h 2
1 and distance of from this to a will be h
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1 1. Therefore, we have this minus log n,
because minus q a is the charge, so minus
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log n h 1 2 by h 1 1.
Similarly, for charge on conductor b, and
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00:23:13,070 --> 00:23:16,860
it is image we will get the term 1 by twice
pi
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epsilon q b log n r by D 1 2 minus log n h
2 2 by h 1 2. Plus due to the charge on
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00:23:29,830 --> 00:23:37,380
conductor c and it is image we will have the
term 1 by twice pi epsilon. Q c into log n
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00:23:37,380 --> 00:23:47,619
D
2 3 by D 3 1 minus log n h 2 3 by h 3 1.
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00:23:47,619 --> 00:23:48,619
..
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So, just as we have calculated the voltage
between the conductor a and b. We can also
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calculate in the same way the voltage between
conductors a and c. So, we will get the
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00:24:02,340 --> 00:24:10,090
voltage V a c in that case, if we add these
two voltages V a b plus V a c, we had seen
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00:24:10,090 --> 00:24:16,210
this
is equal to 3 V a n. In this way we can calculate
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00:24:16,210 --> 00:24:22,620
the voltage V a n, that is voltage
between phase a conductor and the ground or
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the neutral, and since we know of the
charge on conductor a as q a. Therefore, we
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can calculate the capacitance C n for or C
a
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n for the conductor a. That is the capacitance
for phase a conductor to ground will be
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coming out to be equal to twice pi epsilon
divided by log n D e q by r. Minus log n cube
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00:24:53,149 --> 00:25:03,429
root of h 1 2 into h 2 3 into h 3 1 divided
by cube root of h 1 1 into h 2 2 into h 3
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00:25:03,429 --> 00:25:08,980
3. Now,
here again what we are seeing is that, the
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00:25:08,980 --> 00:25:20,870
term log n cube root of h 1 2 h 2 3 h 3 1
divided by cube root of h 1 1 h 2 2 h 3 3
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is the effect of earth on the capacitance.
Now, this is getting subtracted in the denominator.
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So, the effect of ground when we
consider for calculating the capacitance,
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the capacitance will be higher than if we
neglect
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the effect of ground. Again here we see if
the conductors of the phase conductors a,
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00:25:48,820 --> 00:25:53,929
b, c
are vary for above the ground compared to
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the distance between them. Then h 1 2, h 2
3
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and h 3 1 will be very much nearly equal to
h 1 1, h 2 2 and h 3 3. And therefore, this
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term will come out to be 0, that is the effect
of ground will be negligible.
202
00:26:17,929 --> 00:26:25,550
However, since these distances or between
the phase conductors, and the distance
203
00:26:25,550 --> 00:26:30,990
between phase conductor on the ground are
of the same magnitude. The effect of ground
204
00:26:30,990 --> 00:26:39,950
.is basically to increase the capacitance
of the transmission line. Now, we will talk
205
00:26:39,950 --> 00:26:44,730
about
3 phase line with earth return.
206
00:26:44,730 --> 00:26:45,730
.
207
00:26:45,730 --> 00:26:53,760
So, again we take into consideration a 3 phase
system, with neutral wires of the ground
208
00:26:53,760 --> 00:27:00,740
wires on top of the phase conductors. This
is the configuration that we have for a 3
209
00:27:00,740 --> 00:27:08,679
phase
EHV system. Now, if the phase conductors are
210
00:27:08,679 --> 00:27:16,399
fully transposed and are carrying balance
current, then there will be no current flowing
211
00:27:16,399 --> 00:27:24,159
in the earth conductors or in the ground.
However, when there is unbalanced current
212
00:27:24,159 --> 00:27:29,220
flowing through the phase conductors. There
is going to be some current which will be
213
00:27:29,220 --> 00:27:38,580
flowing in the earth conductors or the ground
conductors, and also in the ground itself.
214
00:27:38,580 --> 00:27:50,720
As we had seen earlier we can take care of
this by means of using the effect of images
215
00:27:50,720 --> 00:27:54,769
as
we had done for calculating inductance, and
216
00:27:54,769 --> 00:28:03,299
calculating the resistance. So, here again
since, these conductors are above the ground
217
00:28:03,299 --> 00:28:10,610
what we are doing is? We are considering
the image conductors below the overhead conductors.
218
00:28:10,610 --> 00:28:21,220
Again at a distance which is equal
to the distance of the conductor above the
219
00:28:21,220 --> 00:28:26,450
ground.
So, we are taking an image conductor and using
220
00:28:26,450 --> 00:28:32,610
the method of images. Now, we will
have instead of 3 conductors and n neutral
221
00:28:32,610 --> 00:28:40,179
conductors. We will have now 6 conductors
and 2 n neutral conductors in the system,
222
00:28:40,179 --> 00:28:48,419
because we have 3 plus n conductors for the
neutral as the image conductors also.
223
00:28:48,419 --> 00:28:49,419
..
224
00:28:49,419 --> 00:29:00,019
Now, for this system, again we can find out
the voltage between any conductor and it is
225
00:29:00,019 --> 00:29:08,289
image. That will be given by V k k dash. For
any conductor k, we can write the voltage
226
00:29:08,289 --> 00:29:14,340
between the conductor k and its image k dash.
That will be equal to 1 by twice by
227
00:29:14,340 --> 00:29:21,279
epsilon, summation of m is equal to a to n
N, where n N is the total number of
228
00:29:21,279 --> 00:29:32,220
conductors in the system into q m, the charge
on conductor m log n.
229
00:29:32,220 --> 00:29:44,509
The distance from the image conductor k to
the m th conductor above the ground,
230
00:29:44,509 --> 00:29:53,899
divided by D k m the distance between the
k th conductor and the m th conductor. Minus
231
00:29:53,899 --> 00:30:03,769
summation m is equal to a to n N, that is
to for all the conductors in the system. Q
232
00:30:03,769 --> 00:30:09,730
m the
charge on m th conductor into log n D k m
233
00:30:09,730 --> 00:30:24,330
by H k m. That is again the distance between
the image conductor k to the conductor m divided
234
00:30:24,330 --> 00:30:33,990
by, distance between the conductor k
and conductor m divided by the distance between
235
00:30:33,990 --> 00:30:42,779
the k th conductor.
And the image conductor m, that is this part
236
00:30:42,779 --> 00:30:50,370
is coming because of the image conductor
m. This part is coming because of the overhead
237
00:30:50,370 --> 00:30:58,200
conductor m. So, because the image
conductor carries a negative charge to that
238
00:30:58,200 --> 00:31:06,059
of the overhead conductor above. So, this
negative sign is coming here. So, if we add
239
00:31:06,059 --> 00:31:10,120
this we will get this as equal to 2 by twice
pi
240
00:31:10,120 --> 00:31:20,919
epsilon summation m is equal to a to n N q
m log n H k m by D k m. That is we can
241
00:31:20,919 --> 00:31:33,770
multiplied with by minus 1, then we can inward
this. So, this will be plus sigma m is
242
00:31:33,770 --> 00:31:42,869
equal to a to n N q m log n H k m by D k m
and then we can add them.
243
00:31:42,869 --> 00:31:43,869
..
244
00:31:43,869 --> 00:31:53,190
Now, the voltage between the k th conductor
and the neutral, or the earth will be equal
245
00:31:53,190 --> 00:31:59,309
to
half of that between the conductor k and it
246
00:31:59,309 --> 00:32:07,269
is image k dash. Therefore, V k n will be
equal to half V k k dash which will be equal
247
00:32:07,269 --> 00:32:12,059
to 1 by twice pi epsilon. Because, that 2
is
248
00:32:12,059 --> 00:32:19,130
cancelled out with this 2 here. So, it is
1 by twice pi epsilon summation m is equal
249
00:32:19,130 --> 00:32:27,809
to a to
n N of q m log n H k m by D k m.
250
00:32:27,809 --> 00:32:42,940
Now, since the ground conductors which are
on top of the phase conductors. Since, they
251
00:32:42,940 --> 00:32:50,789
are grounded at regular intervals by means
of tower footing. Therefore, the voltage of
252
00:32:50,789 --> 00:32:56,119
these conductors with respect to neutral or
ground is going to be 0, because these
253
00:32:56,119 --> 00:33:07,309
conductors are at ground potential. Therefore,
V k n is equal to 0 for k is equal to n plus
254
00:33:07,309 --> 00:33:15,840
1 n plus 2 up to n N. That is expect for the
phase conductors, all these ground conductors
255
00:33:15,840 --> 00:33:20,119
than ground conductors their voltage is going
to be 0
256
00:33:20,119 --> 00:33:21,119
..
257
00:33:21,119 --> 00:33:33,730
Now, we can put this whole equation into a
matrix form as shown here. V a n, V b n, V
258
00:33:33,730 --> 00:33:40,369
c
n and V n 1, V n 2 and up to V n N, these
259
00:33:40,369 --> 00:33:48,299
are 0’s is equal to a matrix of P’s into
q a the
260
00:33:48,299 --> 00:33:56,169
charge on conductor a. Q b the charge on conductor
b, q c the charge on conductor c. Q n
261
00:33:56,169 --> 00:34:03,539
1 up to q n N the charge on neutral, conductors
or the ground conductors which are
262
00:34:03,539 --> 00:34:14,480
placed above the phase conductors. That is
we are talking about the charges on these
263
00:34:14,480 --> 00:34:29,639
. conductors, n 1 n 2 up to n N.
Now, here we see, we can divide this P matrix
264
00:34:29,639 --> 00:34:44,399
which will be 3 plus n N matrix, 3 plus n
N by 3 plus n N matrix. This matrix we can
265
00:34:44,399 --> 00:34:55,119
divide into 4 sub-matrices P A, P B, PC and
P D, where P A is a 3 by 3 matrix of which
266
00:34:55,119 --> 00:35:01,540
relates the voltage with respect to charge
on
267
00:35:01,540 --> 00:35:18,790
the phase conductors. P B is going to be are
3 by n matrix, P C is a n rows and 3 column
268
00:35:18,790 --> 00:35:29,050
that is n by 3 matrix. And P D is a n rows
and n column, that is n by n matrix for all
269
00:35:29,050 --> 00:35:32,590
the
ground conductors.
270
00:35:32,590 --> 00:35:39,920
And see P A is relating the phase conductors,
P B is relating the phase conductors with
271
00:35:39,920 --> 00:35:49,420
the charges on the ground conductors. P C
is relating the voltage on the neutral
272
00:35:49,420 --> 00:35:56,100
conductors, with respect to the charges on
the phase conductors. And P D is relating
273
00:35:56,100 --> 00:35:58,730
the
voltage on the neutral conductors, with respect
274
00:35:58,730 --> 00:36:05,910
to the charges on the neutral conductors.
275
00:36:05,910 --> 00:36:06,910
..
276
00:36:06,910 --> 00:36:16,140
Here P k m, that is any of these terms P k
P a b P b b or P n 1 b. Any of these terms,
277
00:36:16,140 --> 00:36:22,530
is
given by the relationship 1 by twice pi epsilon
278
00:36:22,530 --> 00:36:34,150
log n H k m by D k m meters per farad.
Now, this matrix which we had this big matrix
279
00:36:34,150 --> 00:36:43,800
relationship can be written in a shorter
form as V p which is the voltage for the phase
280
00:36:43,800 --> 00:36:53,660
conductors. And 0 which is the voltage for
all the n ground conductors is equal to P
281
00:36:53,660 --> 00:36:58,810
A, P B, P C, P D which we had already defined
earlier.
282
00:36:58,810 --> 00:37:06,080
And q p is the charges on the phase conductors.
That is q a, q b, q c and q n is the
283
00:37:06,080 --> 00:37:14,640
charges on the neutral conductors q n 1, q
n 2, up to q n N. Now, from this relationship
284
00:37:14,640 --> 00:37:26,380
we can write this as V p is equal to P A into
q p plus P B into q n. Similarly, we can
285
00:37:26,380 --> 00:37:33,660
write 0 is equal to P C into q p plus P D
into q n.
286
00:37:33,660 --> 00:37:34,660
..
287
00:37:34,660 --> 00:37:44,080
So, 0 is equal to P C into q p plus P D into
q n, these results send to q n is equal to
288
00:37:44,080 --> 00:37:49,410
that is
we can take this part on the other side. So,
289
00:37:49,410 --> 00:37:52,680
this becomes minus P C q p is equal to P D
q
290
00:37:52,680 --> 00:38:01,780
n. And again pre multiplying both sides by
PD inverse we will get q n is equal to minus
291
00:38:01,780 --> 00:38:14,940
of P D inverse into P C into q p. Now, if
we substitute the value of q n into this equation,
292
00:38:14,940 --> 00:38:26,450
then we will get V p is equal to P A minus
P B into P D inverse P C into q p. We can
293
00:38:26,450 --> 00:38:34,450
write this relationship as q p is equal to
C p into V p.
294
00:38:34,450 --> 00:38:41,660
We know the charge is product of capacitance
and voltage. So, q p will be equal to C p
295
00:38:41,660 --> 00:38:50,900
into V p and where C p will be equal to the
inverse of this term. That is inverse of PA
296
00:38:50,900 --> 00:39:01,700
minus P B, P D inverse P C. Now, this C p
will be a 3 by 3 matrix which relates the
297
00:39:01,700 --> 00:39:13,450
capacitances between the phases or the 3 phase
conductors. Now, here what we have
298
00:39:13,450 --> 00:39:24,431
done by doing all this, is we have eliminated
the effect of the ground conductors. That
299
00:39:24,431 --> 00:39:29,060
is
the ground currents which were flowing that
300
00:39:29,060 --> 00:39:37,950
effect is included by modifying the phase
conductor capacitances.
301
00:39:37,950 --> 00:39:38,950
..
302
00:39:38,950 --> 00:39:51,480
So, C p is a 3 by 3 matrix, C a a the capacitance
of a with respect to a C a b between the
303
00:39:51,480 --> 00:40:00,300
a and b. C a c between a and c, similarly
C a b between a and b, C b b the capacitance
304
00:40:00,300 --> 00:40:08,550
between b. And it is image C c c is between
C b c is between b and c and so on. So, C
305
00:40:08,550 --> 00:40:13,970
p
is giving us the capacitance for the 3 phase
306
00:40:13,970 --> 00:40:24,240
conductor system. Now, if the line is fully
transposed, then we will have the same capacitances
307
00:40:24,240 --> 00:40:30,750
for all the 3 phase conductors. As
well as the same capacitance between the 2
308
00:40:30,750 --> 00:40:38,580
phase conductors. So, phase to neutral
conductor capacitance, as well as the capacitance
309
00:40:38,580 --> 00:40:45,700
between 2 phases will be same, for all
the 3 conductors.
310
00:40:45,700 --> 00:40:46,700
.
311
00:40:46,700 --> 00:40:57,770
.Therefore, C a a hat is equal to C b b hat
and will be equal to C c c. So, the three
312
00:40:57,770 --> 00:41:02,080
will
have the same capacitance to neutral, whereas
313
00:41:02,080 --> 00:41:11,220
C a b and C b c will also be same. So, we
are for a transpose line system, we are representing
314
00:41:11,220 --> 00:41:18,040
them with hat which shows that. This
line is a this capacitance is for transpose
315
00:41:18,040 --> 00:41:22,860
line, not to confuse with the capacitance
that we
316
00:41:22,860 --> 00:41:24,380
had calculated earlier.
.
317
00:41:24,380 --> 00:41:34,610
Now, the relationship which governs this capacitance
is C a a hat is the average of C a a,
318
00:41:34,610 --> 00:41:44,500
C b b and C c c. Because, each phase conductor
is occupy the 3 positions for 1 3rd length
319
00:41:44,500 --> 00:41:57,030
of the line. Similarly, Cab which will be
equal to C b c and C a c, for the transposed
320
00:41:57,030 --> 00:42:02,080
line
this C a b will be equal to 1 3rd of C a b
321
00:42:02,080 --> 00:42:07,030
plus C b c plus C a c, that is the average
value of
322
00:42:07,030 --> 00:42:15,660
the capacitance between the phases a b, b
c and c a. Once we have calculated the
323
00:42:15,660 --> 00:42:24,190
capacitance, we can calculate the admittance
very easily by just multiplying it by j
324
00:42:24,190 --> 00:42:31,470
omega.
So, Y p or the admittance for the 3 phase
325
00:42:31,470 --> 00:42:37,820
system with ground wires will be given by
this
326
00:42:37,820 --> 00:42:46,240
Y p which is a 3 by 3 matrix. Again this will
have the terms j omega into C p, which will
327
00:42:46,240 --> 00:42:54,990
be j into twice pi f C p, where C p again
as we have seen earlier is a 3 by 3 matrix.
328
00:42:54,990 --> 00:43:00,720
Same
thing will happen for the transposed line.
329
00:43:00,720 --> 00:43:05,250
That is Y p hat equal to j omega C p hat plus
j
330
00:43:05,250 --> 00:43:15,880
into twice pi f C p hat Siemens per meter.
So, in this way we have seen that, we can
331
00:43:15,880 --> 00:43:24,550
calculate the capacitance for any general
system of 3 phase conductors. That is 3 phase
332
00:43:24,550 --> 00:43:27,340
conductors with any number of neutral
333
00:43:27,340 --> 00:43:37,770
.wires or the ground wires on top of that.
Next, we will talk about double circuit line.
334
00:43:37,770 --> 00:43:48,930
Now, in double circuit line as we know most
of the time when the power demand to an
335
00:43:48,930 --> 00:43:57,310
area increases, we would use instead of 1
circuit, 2 circuits. And both these circuits
336
00:43:57,310 --> 00:44:01,730
are
running parallel to each other mostly on the
337
00:44:01,730 --> 00:44:06,850
same tower.
So, since these two circuits are running parallel
338
00:44:06,850 --> 00:44:12,650
to each other. And are on the same tower
the distance between the conductors of the
339
00:44:12,650 --> 00:44:21,420
two circuit, is not large. Because, this distance
is not large, the current flowing in them
340
00:44:21,420 --> 00:44:27,760
is going to effect the magnetic field of the
current flowing in the other circuit. And
341
00:44:27,760 --> 00:44:33,510
similarly, also because the voltage of 1 circuit
is
342
00:44:33,510 --> 00:44:39,520
going to produce electric filed, which is
going to effect the electric field produced
343
00:44:39,520 --> 00:44:43,560
by the
voltages in the other circuit. That means,
344
00:44:43,560 --> 00:44:49,201
there is mutual effect which takes place
between these two circuits. Because, they
345
00:44:49,201 --> 00:44:53,920
are physically not very far off. So, they
have
346
00:44:53,920 --> 00:44:58,880
effect from one on the other.
.
347
00:44:58,880 --> 00:45:08,770
Such a system we are showing here, here we
have a three phase line this is one circuit
348
00:45:08,770 --> 00:45:17,270
and this is another circuit, which is placed
here. So, this is a double circuit system,
349
00:45:17,270 --> 00:45:21,120
here
we are showing this double circuit line. This
350
00:45:21,120 --> 00:45:26,470
is one circuit this is another circuit
connecting two bus-bars or two sub-stations.
351
00:45:26,470 --> 00:45:31,330
Now, the current flowing in one circuit is
I
352
00:45:31,330 --> 00:45:41,180
p 1, the current flowing in the other circuit
is I p 2. Where I p is basically a three element
353
00:45:41,180 --> 00:45:47,650
vector which consists of I a, I b and I c.
354
00:45:47,650 --> 00:45:57,700
.So, we have I a 1, I b 1, I c 1 of the circuit
1 and I a 2, I b 2, I c 2 for the circuit
355
00:45:57,700 --> 00:46:04,580
2. Now,
this current flowing here from one circuit
356
00:46:04,580 --> 00:46:07,680
will be I p 1 plus I p 2 which gets divided
into
357
00:46:07,680 --> 00:46:13,270
I p 1 I p 2. And maybe it is getting connecting
on this side this is this is shown that we
358
00:46:13,270 --> 00:46:19,110
are showing. We are talking about the effect
of the two currents flowing in these two
359
00:46:19,110 --> 00:46:26,050
circuits. How they are going to effect the
series impedance of the transmission line.
360
00:46:26,050 --> 00:46:29,070
As
well as how they are going effect the shunt
361
00:46:29,070 --> 00:46:37,360
admittance of the transmission system. So,
first we will take up the series impedance.
362
00:46:37,360 --> 00:46:38,360
.
363
00:46:38,360 --> 00:46:49,330
Now, for the two circuits we can write, the
relationship E p 1 is equal to Z p 1 into
364
00:46:49,330 --> 00:46:55,520
I p 1
and E p 2 is equal to Z p 2 into I p 2. But,
365
00:46:55,520 --> 00:46:59,860
since these two circuits are close together
there
366
00:46:59,860 --> 00:47:08,740
is mutual coupling. So, it will not be right
to write separate equations for the two. In
367
00:47:08,740 --> 00:47:12,350
fact,
what we have here is we are considering a
368
00:47:12,350 --> 00:47:19,300
system of 6 conductors, instead of 3
conductors.
369
00:47:19,300 --> 00:47:27,880
Again since we see that, . these two circuits
are connected under
370
00:47:27,880 --> 00:47:35,930
two ends. So, the voltage drop between them
is going to be same. So, the voltage drop
371
00:47:35,930 --> 00:47:39,831
E
p 1 is same as E p 2 which is equal to E p.
372
00:47:39,831 --> 00:47:44,190
So, we can write E p for circuit 1 and E p
for
373
00:47:44,190 --> 00:47:53,280
circuit 2 is equal Z p into I p for circuit
1 into I p for circuit 2. And Z p is going
374
00:47:53,280 --> 00:48:01,400
to be a 6
by 6 matrix of the series impedance.
375
00:48:01,400 --> 00:48:12,130
Now, from this relationship, we can write
I p 1, I p 2 is equal to Z p inverse into
376
00:48:12,130 --> 00:48:18,450
E p E p.
Because, E p 1, E p 2 are same, so we are
377
00:48:18,450 --> 00:48:21,680
writing E p for that. So, we are from here
we
378
00:48:21,680 --> 00:48:28,410
.are calculating this I p 1 and I p 2. So,
we are pre multiplying both sides by Z p inverse.
379
00:48:28,410 --> 00:48:37,750
So, we get I p 1, I p 2 is equal to now Z
p inverse E p E p. Now, Z p inverse will be
380
00:48:37,750 --> 00:48:45,321
admittance and this will be again a 6 by 6
matrix which we can divide into 4 sub
381
00:48:45,321 --> 00:48:52,970
matrices Y A, Y B, Y C, Y D. Where each of
these sub matrices are going to be a 3 by
382
00:48:52,970 --> 00:48:55,370
3
matrix.
383
00:48:55,370 --> 00:49:06,240
So, this is equal to Y A Y B Y C Y D E p E
p. And this is equal to Y A into E p plus
384
00:49:06,240 --> 00:49:11,630
Y B
into E p, so Y A plus Y B into E p and Y C
385
00:49:11,630 --> 00:49:15,300
into E p plus Y D into E p, so Y C plus Y
D
386
00:49:15,300 --> 00:49:24,400
into E p. So, now we can write if we add the
two currents I p 1 and I p 2.
387
00:49:24,400 --> 00:49:25,400
.
388
00:49:25,400 --> 00:49:33,710
Then, we get I p 1 plus I p 2 is equal to
Y A plus Y B plus Y C plus Y D into E p. That
389
00:49:33,710 --> 00:49:37,600
is
if you see from here, if we have I p 1 is
390
00:49:37,600 --> 00:49:41,820
equal to Y A plus Y B into E p and I p 2 is
equal
391
00:49:41,820 --> 00:49:50,200
to Y C plus Y D into E p. So, if we add them
we are going to get this I p 1 plus I p 2
392
00:49:50,200 --> 00:49:54,360
is
equal to Y A plus Y B plus Y C plus Y D into
393
00:49:54,360 --> 00:50:04,300
E p. And this equation we can again solve
for E p. Then, E p will be equal to Z p dash
394
00:50:04,300 --> 00:50:13,340
into I p 1 plus I p 2, where Z p dash is
nothing but the inverse of this matrix. Now,
395
00:50:13,340 --> 00:50:18,110
here if you look at this matrix, this is sum
of
396
00:50:18,110 --> 00:50:24,400
this 3 by 3 matrix plus this 3 by 3 matrix,
solve these matrixes are 3 by 3. So, Z p dash
397
00:50:24,400 --> 00:50:31,660
is
going to be a 3 by 3 matrix.
398
00:50:31,660 --> 00:50:32,660
..
399
00:50:32,660 --> 00:50:45,170
Now, we will talk about the Shunt Admittance
for this double circuit line. Again in the
400
00:50:45,170 --> 00:50:50,780
same way as we have done for a single circuit
line. We can write down the relationship
401
00:50:50,780 --> 00:51:02,070
for q p for circuit 1 and q p for circuit
2. Where, q p will be the charges on phase
402
00:51:02,070 --> 00:51:15,420
conductors a, b and c. So, q p 1 is q a 1
q b 1 q c 1, q p 2 is q a 2 q b 2 and q c
403
00:51:15,420 --> 00:51:27,680
2. So, this
is a vector having 6 rows C p will be a 6
404
00:51:27,680 --> 00:51:35,970
by 6 matrix.
And V p again is the voltage of the phase
405
00:51:35,970 --> 00:51:40,590
conductors a, b and c for circuit 1 and circuit
2.
406
00:51:40,590 --> 00:51:49,860
Since, these two circuits are in parallel.
Therefore, the voltages are going to be same,
407
00:51:49,860 --> 00:52:00,460
therefore instead of V p 1 and V p 2 we are
writing same as V p and V p. Now, this C p
408
00:52:00,460 --> 00:52:09,230
which is a 6 by 6 matrix can be divided into
4 sub-matrices C A, C B, C C, C D. So, we
409
00:52:09,230 --> 00:52:20,500
have got C A into V p plus C B into V p, so
C A plus C B into V p. C C into V p plus C
410
00:52:20,500 --> 00:52:27,790
D into V p, that is C C plus C D into V p
is equal to q p 1 q p 2.
411
00:52:27,790 --> 00:52:34,980
From here again we get q p 1 is equal to CA
plus C B V p, q p 2 is equal to C C plus C
412
00:52:34,980 --> 00:52:38,520
D
into V p. Therefore, if we add these two we
413
00:52:38,520 --> 00:52:47,950
get q p 1 plus q p 2 is equal to C p equivalent
into V p. Now, is this C P is going to be
414
00:52:47,950 --> 00:52:57,220
C P equivalent is going to be a 3 by 3 matrix
instead of a 6 by 6 matrix C P. So, where
415
00:52:57,220 --> 00:53:01,260
C P equivalent is nothing but equal to C A
plus
416
00:53:01,260 --> 00:53:12,680
C B plus C C plus C D, and this is a 3 by
3 matrix and therefore, we can get the shunt
417
00:53:12,680 --> 00:53:22,380
admittance for this double circuit line. In
terms of 3 phase system, for at which will
418
00:53:22,380 --> 00:53:28,561
gives
us a 3 by 3 matrix for the phase a, phase
419
00:53:28,561 --> 00:53:35,800
b and phase c, which includes the mutual effect
between the two circuits. So, we will get
420
00:53:35,800 --> 00:53:43,430
Y P dash is equal to j omega C P e q. So,
now
421
00:53:43,430 --> 00:53:54,680
.that we have seen how to calculate the capacitance
taken into account, the effect of earth.
422
00:53:54,680 --> 00:54:02,510
And we have also seen how we can calculate
the series, or shunt admittance for a double
423
00:54:02,510 --> 00:54:06,460
circuit line, taking into the account of mutual
effects.
424
00:54:06,460 --> 00:54:07,460
.
425
00:54:07,460 --> 00:54:13,210
Well, let us take an example to illustrate
some of the concepts that we learn today.
426
00:54:13,210 --> 00:54:19,130
We
consider a 3 phase 400 kV, 50 Hertz, 350 kilometer
427
00:54:19,130 --> 00:54:26,109
overhead transmission line with flat
horizontal spacing with three identical conductors.
428
00:54:26,109 --> 00:54:31,541
The conductors have an outside
diameter of 3.28 centimeters with 12 meter
429
00:54:31,541 --> 00:54:38,580
between the adjacent conductors. The
conductors are placed 20 meter above the ground,
430
00:54:38,580 --> 00:54:44,170
we consider the earth plane to be
perfect conductor. Determine the capacitive
431
00:54:44,170 --> 00:54:49,580
admittance to neutral in Siemens per meter
per phase and the capacitive var generated
432
00:54:49,580 --> 00:54:52,460
by the line. So, this is the problem that
we
433
00:54:52,460 --> 00:54:53,460
have.
434
00:54:53,460 --> 00:54:54,460
..
435
00:54:54,460 --> 00:55:00,050
And we can show this configuration graphically
as shown here. We have the 3
436
00:55:00,050 --> 00:55:06,701
conductors, a, b and c placed here, we will
say that these are positions 1, 2 and 3. There
437
00:55:06,701 --> 00:55:12,420
are 12 meters apart from each of the conductors.
Since, we are considering the ground to
438
00:55:12,420 --> 00:55:20,820
be a perfect conductor, we take the effect
of image conductors. So, we have a image
439
00:55:20,820 --> 00:55:28,430
conductor of a, as a dash which is again 20
meters below the earth. Similar for b and
440
00:55:28,430 --> 00:55:34,400
c
and we have the distances h i j which is showing
441
00:55:34,400 --> 00:55:42,150
the distance between the conductor I
with the image conductor j. So, h 1 3 is the
442
00:55:42,150 --> 00:55:50,220
distance between conductor a and the image
of conductor c and so on.
443
00:55:50,220 --> 00:55:51,220
.
444
00:55:51,220 --> 00:55:59,660
.So, we have seen that h 1 1, h 2 2, h 3 3
is the distance between the conductors and
445
00:55:59,660 --> 00:56:04,430
their
mirror images. That is h 1 1, this should
446
00:56:04,430 --> 00:56:08,500
be h 1 1, from this conductor to this a to
a dash
447
00:56:08,500 --> 00:56:15,210
is the distance between image conductor this
will be a 40 meters. Similarly, b b dash will
448
00:56:15,210 --> 00:56:23,850
be forty meter and c c dash will be 40 meters.
So, and h i j as we have said is the distance
449
00:56:23,850 --> 00:56:31,280
shown in the figure.
Now, we have for such a system, following
450
00:56:31,280 --> 00:56:42,280
the same procedure as we have shown
earlier. That is if we go back . here we have
451
00:56:42,280 --> 00:56:50,890
the capacitance to
neutral for 3 phase system with image conductors.
452
00:56:50,890 --> 00:57:01,230
This is given by this relationship, we
write the same relationship here. So, we have
453
00:57:01,230 --> 00:57:09,660
C n is equal to twice pi epsilon divided log
n D e q by r minus log n h 1 1 h 2 3 h 3 1
454
00:57:09,660 --> 00:57:14,950
divided h 1 1 h 2 2 h 3 3 a cube root of that.
.
455
00:57:14,950 --> 00:57:23,910
So, now what we do is we substitute all the
values, we can calculate D e q as 15.119
456
00:57:23,910 --> 00:57:28,980
meter which is basically the cube root of
the distance between the conductors
457
00:57:28,980 --> 00:57:36,500
themselves. R is the radius of the conductor
which is equal to 1.64, it is 3.28 divided
458
00:57:36,500 --> 00:57:40,580
by
2. From the diagram, we have the values by
459
00:57:40,580 --> 00:57:45,480
h 1 1 h 2 2 h 3 3 as 40 meters, h 1 2 h 2
3 are
460
00:57:45,480 --> 00:57:54,650
shown here is equal to 41.761 meters, h 3
1 is equal to 46.648 meters. Now, putting
461
00:57:54,650 --> 00:57:59,170
these
values in the equations we get C n is equal
462
00:57:59,170 --> 00:58:06,550
to 8.2595 into 10 to power minus 6
microfarad per meter.
463
00:58:06,550 --> 00:58:07,550
..
464
00:58:07,550 --> 00:58:13,970
Hence, we can calculate the admittance Y n
is equal to twice pi f which is 50 into the
465
00:58:13,970 --> 00:58:18,890
C.
This is C n that is 8.2595 into 10 to power
466
00:58:18,890 --> 00:58:24,630
minus 12 Siemens per meter per phase, this
is
467
00:58:24,630 --> 00:58:32,180
equal to 2.595 into 10 to power minus 9 Siemens
per phase. Now, length of the line is
468
00:58:32,180 --> 00:58:39,700
given as three 50 kilometers, therefore total
admittance will be simply multiply this by
469
00:58:39,700 --> 00:58:43,400
350, this conserved to be 908 micro Siemens
per phase.
470
00:58:43,400 --> 00:58:52,320
Now, the var generated by the line Q c will
be equal to Y n into V L L square or Q c is
471
00:58:52,320 --> 00:58:59,410
equal to 908 into 10 to the power minus 6
into 400 into 10 to power minus 3 square,
472
00:58:59,410 --> 00:59:02,060
that
is 400 into 10 to power minus 10 to power
473
00:59:02,060 --> 00:59:07,520
3 into 400 into 10 to power 3, there is a
400
474
00:59:07,520 --> 00:59:15,330
kilo volts. So, this conserved to be 145.358
M v a r. So, that is all discuss about the
475
00:59:15,330 --> 00:59:27,560
capacitance calculation. In the next class
on lessons 7, we will discuss about modeling
476
00:59:27,560 --> 00:59:29,090
of
transmission lines.
477
00:59:29,090 --> 00:59:34,920
Once that we have we need we have learnt about
how to calculate the parameters of the
478
00:59:34,920 --> 00:59:40,000
transmission line. That is the resistance,
the inductance and the capacitance for the
479
00:59:40,000 --> 00:59:48,870
transmission line. Now, we would like to model
the transmission line for analysis. So,
480
00:59:48,870 --> 00:59:49,870
that we will do in lesson 7.
Thank you.
481
00:59:49,870 --> 00:59:49,870
.