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Welcome to lesson 4 on Power System Analysis.
Before we start this lesson 4, first I
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would like to take up the questions that we
asked in lessons 3.
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.
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First question was, what are the types of
conductors used for over head transmission
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line? Well, some of the types of conductors
used for over head transmission lines are
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copper conductors, which is very rarely use
nowadays. ACSR, that is Aluminum
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Conductors Steel Reinforced conductors. Then,
all aluminum conductors are ACAR
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conductors or expanded ACSR conductors, which
are used.
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As I have already said in lesson 3. Sometimes,
we want to increase the effective radius of
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the conductor. And for that, we used expanded
ACSR conductor in ESB transmission
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lines.
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..
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Next question was, how does conductor diameter
affect inductance of a transmission
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line? Well, if you remember the equation for
inductance, which was L, is equal to 2 into
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10 to power minus 7 log n D by r dash Henry’s
per meter. Now, from this expression it is
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very clear. That inductance is inversely proportional
to r dash or the effective radius of
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the conductor. So, increasing effective conductor
radius will result in decreasing line
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inductance. And that is why, as I said earlier,
sometime, we use expanded ACSR
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conductors for transmission lines, specially
in extra high voltage transmission lines.
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.
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.Now, we will start our lesson 4, which will
be on inductance calculations for three phase
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transmission lines. Well, in this lesson,
we will discuss inductance calculation for
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three
phase transmission lines with equilateral
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spacing, then with transposition. Again, with,
when we are using bundled conductor lines
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and finally, lines with ground return. Well,
first, we will take the calculation of inductance
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for a 3 phase line with equilateral spacing
or equal spacing.
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.
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Now, here in this diagram, you can see a,
b, c are the three conductors, which are placed
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at the vertices of an equilateral triangle.
That is the distance, among all three conductors
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are same. That is distance D. Now, if we calculate
the flux linkage of conductor a, due to
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currents flowing in conductor a, b and c.
Then, it will be equal to 2 into 10 to power
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minus 7 into I a log n 1 by r dash, where
r dash is the effective radius of conductor
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a.
As we have seen, the effective radius of conductor
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a, is generally lower. It is about
0.7788 times the radius of the conductor,
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plus I b into log n 1 by D. That is the distance
of conductor b from conductor a plus I c,
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the current flowing in conductor c into log
n 1
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by D, again the distance between conductor
c and conductor a. This is equal to 2 into
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10
to power minus 7, I a log n 1 by r dash plus
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I b plus I c log n 1 by D. That is, we have
combined these two terms, we have got this.
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..
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Now, we know that, for a 3 phase three conductor
line I a plus I b plus I c is equal to 0.
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Therefore, I b plus I c is equal to minus
I a. Therefore, we can write lambda a is equal
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to
2 into 10 to the power minus 7, I a log n
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1 by r dash minus I a log n 1 by D, which
results
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into 2 into 10 to power minus 7, I a log n
D by r dash Weber turn per meter.
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Now, once we have got the flux linkage. We
can calculate the inductance very easily,
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L a
inductance of conductor a is equal to lambda
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a by I a, which comes out to be 2 into 10
to
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power minus 7 log n D by r dash Henry’s
per meter. Since, all the three conductors
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are
equally space. Therefore, inductance of phase
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b and phase c conductors will also be
same.
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..
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Now, we take the inductance calculation for
a 3 phase transpose line. As we have said
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earlier, because of physical limitations,
it is not always possible to have an equilateral
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configuration for the conductors. So, by transposition,
what we do is, we try to make the
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flux linkage of all the three phase conductance’s
more or less same. Now, what we do
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here is, if we have transmission line, say
from this point to this point is three phase
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line.
What we do is, we change the position of the
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three phase conductance over one-third of
length of the line. Like, if we see here for
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first one-third length of the line. The position
of conductor of phase a is in position 1,
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phase b is in position 2, phase c is in position
3.
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Now, for the next one-third length, what we
do is, the phase a conductor goes to position
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2, phase b conductor goes to position 3 and
phase c conductor goes to position 1.
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And similarly, for the next one-third length
of the line, that is the last one-third length
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of
the line. The phase a conductor, now moves
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to position 3, phase b conductor moves to
position 1 and phase c conductor move to position
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2. In this way, the each phase
conductor has gone through all the three positions
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for one-third length of the line.
And therefore, the total flux linkage for
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each of the phase conductors will be almost
same. And this, because of this, the inductance
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will also be on the average being same
for the three phase conductors. This makes
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the transmission line more balanced.
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..
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So, now, let us find out the flux linkage
of phase a conductor in first one-third position
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of
the conductor. This is, we write as lambda
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a 1, this is equal to 2 into 10 to power minus
7
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into I a log n 1 by D s. Here, I am writing
instead of r dash, because sometimes, we use,
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instead of one conductor more conductors.
So, we are taking about the self distance,
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which is r dash for a single solid conductor
line.
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So, it I a log n 1 by D s plus I b log n 1
by D 12, because in this first one-third position
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current I b is following in the conductor
2. And here, we have the distance D 12.
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Similarly, this plus I c log n 1 by D 31,
which is again I c is flowing in this conductor
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and the distance between them is D 31. Similarly,
for the phase a conductor in the second
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one-third length of the line will be 2 into
10 to the power minus 7 into I a log n 1 by
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D s.
That is it is own distance plus I b log n
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1 by D 23.
If we see here, in this I b’s, now in this
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position I a’s is in this position the distance
between this two is D 23 plus I c log n D
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12. Again, if we see I c is flowing in this
and I
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a is flowing in this. So, distance between
them is D 12. Similarly, for the last one-third
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position portion of the line, we have a flux
linkage with phase conductor equal to 2 into
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10 to power minus 7 into I a 1 by D s plus
I b log n by D 31. Because, now I b is
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occupying position 1 and I a is occupying
position 3 plus I c log n 1 by D 23. As we
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see
here, I c now is occupying this position and
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I a is occupying this position, this is D
23.
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..
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Now, the total flux linkage of the conductor
over the whole length, can be given by sum
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of these multiplied by their lengths. And
if you want to take average per meter length,
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the flux linkage will be lambda a 1 into l
by 3 plus lambda a 2 into l by 3 plus lambda
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a 3
into l by 3 divide by l. Or the average flux
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linkage of phase a conductor, which occupies
all the three position is going to be the
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average of lambda a 1 plus a 2 plus lambda
a 3,
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divided by 3. That is the average of the lambda
in all the three positions.
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This, when we substitute the values of lambda
a 1 lambda a 2 and lambda a 3, comes out
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to be 2 into 10 to power minus 7, divided
by 3 into 3. I a log n 1 by D s plus I b log
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1 by
D 12 into D 23 into D 31 plus I c log n 1
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by D 12 into D 23 into D 31, which will finally,
turn out to be, because we can now combine
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these two terms I b and I c terms. Because,
the denominator, the two terms log n terms
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are same. Therefore, this we can write as
lambda a is equal to 2 into 10 power minus
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7 divided by 3 into 3 I a log n 1 by D s.
Now, I b plus I c is equal to minus I a. So,
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we can write this as minus I a log n 1 by
D 12
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plus D 23 plus D 31. Now, we can combine these
two terms and then we will get 2 into
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10 to power minus 7, I a log n cube root of
D 12 into D 23 into D 31, divided by D s,
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Weber turn per meter. So, the average flux
linkage per meter, for a conductor a, comes
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out to be this much. And as we have seen,
since all the three phase conductors, occupy
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all the three phases. So, the average flux
linkage is for phase b and phase c conductors
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will also be same.
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..
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Now, once, we have got the flux linkage, average
flux linkage, we can get the
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inductance, average inductances per meter.
Length of the line as lambda a by I a, which
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turns out to be 2 into 10 to power minus 7,
log n cube root of D 12 into D 23 into D 31
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divided by D s. Where, this term cube root
of D 12 into D 23 into D 31 is called the
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equivalent distance of the three phase system.
That is the equivalent distance between the
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three conductances; this is also seen as the
GMD of the three phase conductances. Therefore,
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L a is equal to or the inductance of
phase a is equal to 2 into 10 power minus
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log n D e q by D s, where D e q is the GMD
of
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the 3. Now, as we have said earlier, that
in order to reduce the inductances of the
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transmission system. We can do this by increasing
the resistance, increasing the radius of
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the conductor. That is the effective radius
of the conductor, should be increased. And
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this
is can be d1 by means of bundle conductances.
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In fact, bundle conductances are used for
two purposes.
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..
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One is it reduces the electric field strength
on conductor surface. And therefore, it
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reduces corona loses and radio interference
and audible corona loses all the aspects
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which is associated with corona get reduced.
Now, how does this happen. Now, if you
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see, if we use the conductor with the radius
r, then it is volume will be phi r square.
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Now,
if we take two conductors with the same, sorry,
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it is surface area is phi r square.
If we take the two conductors with the same
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cross sectional area, then it is radius is
not
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going to be r by 2, it is radius is going
to be r by root 2. Now, if we have this, then
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the
surface area of one meter length of the conductor.
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In case, we are using two conductors
instead of 1, will be now, phi into r by root
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2 multiplied by 2. So, it will be root into
r,
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whereas in the other case, it will be phi
r only.
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So, here, what we see the surface area of
the conductor increases considerably and
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because of this the electric stress reduces.
Another effect, that we get, when we bundle
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the conductors is that, it increases the effective
radius or the GMR of the self distance of
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the conductor considerably. And this reduces
the inductance and this is one of the
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reasons, why we use this bundle conductors.
These bundled conductors are used by having
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more than 1 conductor, which are
supported by a conducting frame at regular
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intervals, along the transmission line. Here,
you see a bundle conductor with two conductors
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here. This is a bundle conductor with
three conductors and bundle conductor with
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four conductors. In fact, for very high
voltage line, sometimes we use more than four
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conductors also.
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.In India, normally we are using two conductors
or four conductors for 400 kV line. And
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two conductors sometimes are used for 220
kV line.
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.
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Now, if you look at the effective radius of
the conductor. When, we are using two
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conductors, which are placed at a distance
d, normally this distance d is approximately
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10 times the diameter of the conductor. So,
here, if we look at this two conductor
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consideration a bundle with two conductors,
then D s will be equal to, we are taking all
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the distances. The distance, it is own distance
from itself will be r dash.
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Then, it is distance from this conductor will
be d, then distance of this conductance with
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itself will be r dash and distance of this
conductor with this will be d. So, there are
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four
distances involved and it will be 4th root
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of r dash into d whole square. That is r dash
square into d square, which is equal to square
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root of r dash d, which is certainly much
higher than r dash, because d is much larger
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than r dash.
Same thing, for three conductors we can get,
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line distances. That is, three distances for
each conductor, one is self distance and two
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for the other two conductors and same thing,
for each one of the three conductors. Therefore,
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we get self distance D s as 9th root of r
dash into d into d whole cube. This is equal
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to cube root of r dash into d square.
Similarly, when we are using four conductors,
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we are got 16 distances, one for itself and
three more for the other three conductors
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distance, from this conductor to other three
conductors.
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.Therefore, we get the self distance D s is
equal 16th root of r dash into d into d into
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d
root 2. This is the distance between this
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conductor and this conductor. Same thing will
happen for this and this conductor and so
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on. So, in 2 to the power to 4, which is comes
out to be 1.091 into 4th root of r dash d
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cube. And as we have seen, earlier an inductance,
we can get for these bundle conductances as
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l is equal to 2 into 10 to the power minus
7,
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log n D e q by D s.
Where, D s, is what, is the GMR for the bundle
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conductance. Since, this has increase
considerable as compared to single solid conductor,
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having the cross sectional area.
Therefore, the inductance, get reduced considerably.
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.
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Now, we will take up three phase line with
earth return. Now, this is a normal situation.
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Those three conductors may be horizontally
placed and we have at the top of the
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transmission tower earth wires or the ground
wires. These ground wires as I have said
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earlier in lessons 2. That these ground wires
are used to protect the phase conductors
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from direct lightening stroke.
Now, these ground conductors are normally
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conducted or connected to the tower and at
each tower footing, they are grounded. So,
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earth conductors are connected in parallel
and
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grounded at regular intervals along the transmission
line. If they are not grounded at each
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00:22:45,620 --> 00:22:56,100
tower, may be, they will be grounded at every
alternate tower or so. So, how do, we find
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out the inductances for such a system?
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.Normally, when the system is working as a
balance system, they would not be any return
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current flowing. Because, the sum of the three
currents will be 0 all the time, but in case
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of unbalance current flowing in the system.
We have some currents flowing through
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these return conductors, which will again
go to the ground and will flow through the
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earth.
.
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So, if phase currents are not balanced, return
currents will flow in earth conductors and
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ground. Because, since these earth conductors
are connected to the tower and again,
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there are grounded by the footing, ground
tower footing. So, the current flowing in
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these
earth conductors will go into the ground.
202
00:23:56,980 --> 00:24:02,570
And will this disperse and flow in the ground
also.
203
00:24:02,570 --> 00:24:15,720
Now, it was J.R. Carson in 1923, who effectively
or who proposed, how to take care of
204
00:24:15,720 --> 00:24:23,679
these ground currents in the system, in the
transmission system. So, which was later
205
00:24:23,679 --> 00:24:30,090
modified, he was the first person to model
this ground currents into a system. What he
206
00:24:30,090 --> 00:24:38,410
did was, he said that, earth return conductors
are basically or the earth current can be
207
00:24:38,410 --> 00:24:47,670
basically, represented by earth return conductors,
which are having the same GMR as the
208
00:24:47,670 --> 00:24:49,510
overhead conductors.
209
00:24:49,510 --> 00:24:50,510
..
210
00:24:50,510 --> 00:24:59,500
And are located directly below the overhead
conductors, at a distance D e, as shown in
211
00:24:59,500 --> 00:25:08,540
this figure, that is here we have three phase
conductors a, b and c. And we have two earth
212
00:25:08,540 --> 00:25:17,140
conductors are the ground conductors, which
are n 1 and n 2. Now, what we he proposed
213
00:25:17,140 --> 00:25:24,800
as for such a system. He said that, the ground
current flowing can be represented by
214
00:25:24,800 --> 00:25:33,040
conductors a dash, which is directly below
a into the ground at a distance D, from that
215
00:25:33,040 --> 00:25:38,730
conductor.
Same thing for n 1, we have a conductor n
216
00:25:38,730 --> 00:25:46,220
1 dash, which is in the ground or which is
directly below this in the ground at a distance
217
00:25:46,220 --> 00:25:54,310
D and so on. Where, he said D e, the
distance at which these conductors are from
218
00:25:54,310 --> 00:26:02,090
the overhead conductors is proportional to
rho, the earth resistivity. And inversely
219
00:26:02,090 --> 00:26:06,910
proportional to frequency of the system or
the
220
00:26:06,910 --> 00:26:12,550
frequency of the current, which is flowing
in these conductors.
221
00:26:12,550 --> 00:26:13,550
..
222
00:26:13,550 --> 00:26:25,790
So, now we can see that, D for any conductor
k, we have the ground conductor D k k
223
00:26:25,790 --> 00:26:34,001
dash. The distance of it is own, that is same
as D k k, which is the GMR of that
224
00:26:34,001 --> 00:26:45,380
conductor. And the distance D k to k dash
for any conductor k to k dash is D e. And
225
00:26:45,380 --> 00:26:50,530
the
D k k dash, which is the distance of this
226
00:26:50,530 --> 00:26:54,240
conductor to itself is same, as the GMR of
this
227
00:26:54,240 --> 00:27:10,510
conductor. And he said that, D e is equal
to 658.5 into square root of rho by f.
228
00:27:10,510 --> 00:27:19,890
Now, this is an empirical formula, which was
found after lots of experimentation. Where,
229
00:27:19,890 --> 00:27:28,559
rho is as resistivity and if one does not
know the earth resistivity properly. Then,
230
00:27:28,559 --> 00:27:33,580
one can
always choose 100 ohm meter as the earth resistivity,
231
00:27:33,580 --> 00:27:47,380
which is the resistivity for that
earth. He also found out, that the resistance
232
00:27:47,380 --> 00:27:55,070
of these image conductors are the
conductors, which are representing the ground
233
00:27:55,070 --> 00:28:04,410
current is given by R k dash is equal to
9.869 into 10 to power minus 7 f, where f
234
00:28:04,410 --> 00:28:15,000
is the frequency per ohms per meter.
So, these D e and R k dash are empirical values
235
00:28:15,000 --> 00:28:23,540
and these have been found out after lots
of experimentation. Now, for this system,
236
00:28:23,540 --> 00:28:34,980
that we have, we have five conductors here
and five conductors as images, below these
237
00:28:34,980 --> 00:28:42,280
conductors at a distance D e, from the
overhead conductors. Now, this makes a total
238
00:28:42,280 --> 00:28:49,679
of 10 conductors. Now, some of the
currents in all the 10 conductors will have
239
00:28:49,679 --> 00:28:57,130
to be equal to 0.
So, in this system, sum of the current I k
240
00:28:57,130 --> 00:29:03,540
is equal to 0, for k is equal to 1 to 2 times
into
241
00:29:03,540 --> 00:29:08,640
three phase conductors plus 2 neutral conductors.
That is sum of all the 10 conductor
242
00:29:08,640 --> 00:29:16,090
currents is equal to 0. Now, for such a system,
we can find out the flux linkage, for any
243
00:29:16,090 --> 00:29:25,040
conductor very easily. And lambda k, in this
case is given by 2 into 10 power minus 7
244
00:29:25,040 --> 00:29:32,250
.sigma m is equal to 1, 2, 3 plus 2. That
is three phase conductors plus 2 ground
245
00:29:32,250 --> 00:29:42,010
conductors, I m log n D km dash.
That is distance from the conductor to the
246
00:29:42,010 --> 00:29:53,799
conductors, which are the image conductors
divided by D k m, the distance between 2 overhead
247
00:29:53,799 --> 00:30:00,580
conductors. And these include
ground conductor as we are seeing. So, lambda
248
00:30:00,580 --> 00:30:08,559
k is 2 into 10 to power minus 7, sigma m
is equal to 1 to 5, I m log n D k m dash by
249
00:30:08,559 --> 00:30:12,380
D k m Weber turn per meter.
.
250
00:30:12,380 --> 00:30:21,180
Since, lambda is equal to L I, where L is
the inductance and I is the current. Now,
251
00:30:21,180 --> 00:30:24,280
in this
case, since there are five conductors, we
252
00:30:24,280 --> 00:30:28,799
have lambda for all the five conductors. That
is
253
00:30:28,799 --> 00:30:38,050
lambda a, lambda b, lambda c and lambda n
1 lambda n 2. So, lambda is a 3 plus 2
254
00:30:38,050 --> 00:30:45,299
vector, I again, since the current is following
in all these five conductors. So, I is a 3
255
00:30:45,299 --> 00:30:47,210
plus
2 vector.
256
00:30:47,210 --> 00:30:55,860
Now, L which is the inductance matrix is also
a 3 plus 2 into 3 plus 2 matrix. Whose
257
00:30:55,860 --> 00:31:05,100
elements L k m is equal to 2 into 10 to power
minus 7 log n D k m dash by D k m. This
258
00:31:05,100 --> 00:31:13,160
comes out from the previous relation of lambda
k. So, lambda k, we know. So, if we sum
259
00:31:13,160 --> 00:31:23,679
this up and then we find out the inductance,
sorry, the conductors, then L k m will come
260
00:31:23,679 --> 00:31:25,899
out to be this much.
261
00:31:25,899 --> 00:31:26,899
..
262
00:31:26,899 --> 00:31:35,330
And therefore, we can write this as matrix
equation, where we are saying E A a is the
263
00:31:35,330 --> 00:31:41,049
voltage drop in phase A conductor. E B b is
the voltage drop in phase B conductor. E C
264
00:31:41,049 --> 00:31:45,470
c
is voltage in phase C conductor. For the neutral
265
00:31:45,470 --> 00:31:48,890
conductors is there are grounded, there
potential is 0.
266
00:31:48,890 --> 00:31:56,290
So, voltage drops are going to be 0, for the
two ground conductors, this is equal to R
267
00:31:56,290 --> 00:32:05,740
plus
j Omega L, which will be again a 5 by 5 matrix
268
00:32:05,740 --> 00:32:14,450
into I a current in phase a. I b current in
phase b. I c current in phase c, I n 1 current
269
00:32:14,450 --> 00:32:22,990
in neutral 1 and I n 2 current in neutral
2 or
270
00:32:22,990 --> 00:32:25,340
ground 2.
.
271
00:32:25,340 --> 00:32:37,480
.Here, the matrix R is a 5 by 5 matrix and
this is represented by for R A a, what we
272
00:32:37,480 --> 00:32:43,390
have
is, R a plus R k dash. That is this is for
273
00:32:43,390 --> 00:32:46,930
R a, the resistances for conductor A. Now,
R k
274
00:32:46,930 --> 00:32:57,919
dash is the resistances for the image conductor,
which is below A. Then, we have R k
275
00:32:57,919 --> 00:33:13,640
dash for the other 4. Similarly, for R will
be R 12 will be R k dash R 2 with 2 will be
276
00:33:13,640 --> 00:33:18,310
R b
plus R k dash R 23 will R k dash and so on.
277
00:33:18,310 --> 00:33:26,470
So, this is a 5 by 5 matrix, which is a
resistance in ohms per meter for this five
278
00:33:26,470 --> 00:33:27,470
conductor system.
.
279
00:33:27,470 --> 00:33:36,260
Now, if you write the complete system equation
for the voltages across the line
280
00:33:36,260 --> 00:33:43,220
conductors. Then, we will have the voltage
as E A a, E B b, E C c. That is the voltage
281
00:33:43,220 --> 00:33:49,160
across the two points of the phase A conductor,
phase B conductor and phase C
282
00:33:49,160 --> 00:33:53,600
conductors. Similarly, for the neutral conductors,
since the neutral conductors are
283
00:33:53,600 --> 00:34:01,960
grounded, this voltage will be 0.
This voltage is going to be equal to the impedance,
284
00:34:01,960 --> 00:34:07,970
multiplied by the current which gives
the voltage drop across the two points. Now,
285
00:34:07,970 --> 00:34:18,549
here the impedances will be Z 11, Z 12, Z
13, Z 14, Z 15. All these impedances, this
286
00:34:18,549 --> 00:34:22,589
will give a 5 by 5 matrix. And the currents
will
287
00:34:22,589 --> 00:34:30,659
be the, three phase currents I a, I b, I c,
I n 1, the current in neutral conductor 1
288
00:34:30,659 --> 00:34:32,610
and I n 2
in the conductor 2.
289
00:34:32,610 --> 00:34:41,079
So, this for the five conductor system, we
have five set of equations and this impedance
290
00:34:41,079 --> 00:34:49,510
matrix is a 5 by 5 matrix, which we can divided
into four different sub matrices. As
291
00:34:49,510 --> 00:35:02,539
shown here, Z A which is 3 by 3 matrix, which
indicates the Z 11, Z 12 Z 21, Z 22, Z 23,
292
00:35:02,539 --> 00:35:09,660
.Z 31, Z 32,, Z 33. These impedances are relating
the voltages across the phase
293
00:35:09,660 --> 00:35:13,799
conductors with the currents following in
the phase conductor.
294
00:35:13,799 --> 00:35:19,539
Similarly, the voltages across the phase conductors,
due to currents following in the
295
00:35:19,539 --> 00:35:27,250
neutrals, can be given by the impedances as
shown in the Z B. That is Z 14, Z 15, Z 24,
296
00:35:27,250 --> 00:35:35,940
Z 34, Z 35. And Z C indicates the voltage
across the neutral conductors, due to currents
297
00:35:35,940 --> 00:35:44,579
in the phase conductors. These impedances
are Z 41, Z 42, Z 43, Z 51, Z 52, Z 53.
298
00:35:44,579 --> 00:35:49,920
Similarly, for the voltage n th across the
neutral conductors, due to currents in the
299
00:35:49,920 --> 00:35:53,050
neutral
conductors will be given by this matrix Z
300
00:35:53,050 --> 00:36:01,059
D, which is Z 44, Z 45, Z 54, Z 55. So, we
have three sub matrices Z A, which is a 3
301
00:36:01,059 --> 00:36:09,650
by 3, Z B which is 3 by 2 matrixes, Z C, which
is 2 by 3 matrixes and Z D, which is a 2 by
302
00:36:09,650 --> 00:36:15,451
2 matrix. This is the complete system
equation that we have, where the impedance
303
00:36:15,451 --> 00:36:19,729
terms are given as for the diagonal
elements.
304
00:36:19,729 --> 00:36:20,729
.
305
00:36:20,729 --> 00:36:27,789
That is Z k k, which means Z 11, Z 22 and
so on are given by R k plus R k dash plus
306
00:36:27,789 --> 00:36:32,930
j
omega 2 into 10 to power minus 7, log n D
307
00:36:32,930 --> 00:36:41,050
k k dash by D k k ohms per meters. Whereas,
these are the distances, as we have seen in
308
00:36:41,050 --> 00:36:43,920
the system in the system diagram earlier.
Off
309
00:36:43,920 --> 00:36:51,650
diagonal elements that is Z k m, which is
basically Z 12, Z 13 or Z 54, all these off
310
00:36:51,650 --> 00:36:55,660
diagonal elements.
That is, when k is not equal to m is given
311
00:36:55,660 --> 00:37:04,619
by R k dash plus j omega into 2 into 10 to
power minus 7 log n D k m dash by D k m ohms
312
00:37:04,619 --> 00:37:08,519
per meter. So, we can compute all these
313
00:37:08,519 --> 00:37:15,140
.diagonal and off diagonal elements, that
is the elements of this imperial matrix using
314
00:37:15,140 --> 00:37:16,770
these relationships.
.
315
00:37:16,770 --> 00:37:24,619
Now, what we can do, since we have already
divided this system into four sub matrices.
316
00:37:24,619 --> 00:37:31,640
Now, we can write this whole system of equation
in a short form like E p, where E p is
317
00:37:31,640 --> 00:37:37,640
the voltage drop across the face conductors,
0 is the voltage drop across the neutral
318
00:37:37,640 --> 00:37:45,609
conductor, this is equal to Z A, Z B, Z C,
Z D into I p. Where I p is the current flowing
319
00:37:45,609 --> 00:37:51,550
through the face conductors and I n is the
vector of current flowing through to neutral
320
00:37:51,550 --> 00:37:57,289
conductors.
As shown here, E p is equal to E A a, E B
321
00:37:57,289 --> 00:38:04,920
b and E C c are three phase conductor voltage
drops, I p is equal to the phase currents
322
00:38:04,920 --> 00:38:15,380
I a, I b, I c. And I n is the current following
through the two neutrals I n 1, I n 2. Now,
323
00:38:15,380 --> 00:38:22,369
from this set of equations, we can write this
into two separates sets of equation as E p
324
00:38:22,369 --> 00:38:26,440
is equal to Z A, I P plus Z B I n. E p is
equal to
325
00:38:26,440 --> 00:38:30,099
Z A, I P plus Z B into I n.
326
00:38:30,099 --> 00:38:31,099
..
327
00:38:31,099 --> 00:38:38,670
Similarly, 0 is equal to Z C into I P plus
Z D into I n. 0 is equal to Z C into I P plus
328
00:38:38,670 --> 00:38:42,589
Z D
into I n. Now, from this second equation,
329
00:38:42,589 --> 00:38:50,869
we can take this term. This term on the other
side then we have I n is equal to minus Z
330
00:38:50,869 --> 00:38:56,589
D inverse Z C into I P. That is I n is here
and
331
00:38:56,589 --> 00:39:01,230
we have taken this on this side. So, it is
minus Z C, I P. Now, we pre multiply both
332
00:39:01,230 --> 00:39:03,069
sides
by Z D inverse.
333
00:39:03,069 --> 00:39:11,519
So, we will get I n, I n is equal to minus
Z D inverse Z C into I P. And therefore, putting
334
00:39:11,519 --> 00:39:20,809
for this I n here in this first equation,
we will get E P is equal to Z A into I P minus
335
00:39:20,809 --> 00:39:26,599
because this minus term is coming Z B into
I n. So, I n is given by this relationship,
336
00:39:26,599 --> 00:39:34,440
so
minus Z B into Z C into Z D, sorry, Z B into
337
00:39:34,440 --> 00:39:41,070
Z D inverse into Z C, I P. So, we have
substituted for I n from here and then we
338
00:39:41,070 --> 00:39:46,520
are writing this expression like this.
Now, we can write this whole as E P is equal
339
00:39:46,520 --> 00:39:55,339
to Z P into I P, where ZP is this matrix.
Here, this will be a 3 by 3 matrix Z p is
340
00:39:55,339 --> 00:40:04,170
equal to Z A minus Z B, Z D inverse Z C. So,
this is what we will get, that is we have
341
00:40:04,170 --> 00:40:09,460
eliminated in this equation, the current I
n and
342
00:40:09,460 --> 00:40:16,160
we are writing the all the equations in terms
of the phase variable. Only thing is, this
343
00:40:16,160 --> 00:40:21,819
is
taken care by using Z D inverse and Z C from
344
00:40:21,819 --> 00:40:24,260
this expression.
345
00:40:24,260 --> 00:40:25,260
..
346
00:40:25,260 --> 00:40:37,319
Therefore, we will get now Z p, which will
be a 3 by 3 matrix as Z a a, Z a b, Z a c,
347
00:40:37,319 --> 00:40:42,710
Z a b
or Z b a will be same Z b b, Z b c and Z a
348
00:40:42,710 --> 00:40:48,839
c, Z b c, Z c c. So, this is now a 3 by 3
matrix,
349
00:40:48,839 --> 00:41:01,119
from where we have eliminated the currents
in the neutral conductors or the ground
350
00:41:01,119 --> 00:41:10,609
conductors. So, here we are getting a relationship,
only for the phase conductors, E p is
351
00:41:10,609 --> 00:41:16,880
equal to Z p into I p.
So, the current following in the ground conductors
352
00:41:16,880 --> 00:41:25,960
or the ground is now eliminated. And
we can now get a relationship, only for the
353
00:41:25,960 --> 00:41:36,690
conductor’s currents. In the phase conductors
relating the voltage drop in the phase conductors.
354
00:41:36,690 --> 00:41:48,190
Now, for a fully transpose line, what
we will have is since these phase conductors
355
00:41:48,190 --> 00:41:51,890
will be occupying all the three positions
for
356
00:41:51,890 --> 00:42:00,410
one-third length of the line. Therefore, we
will get the average values for all the three
357
00:42:00,410 --> 00:42:08,099
flux linkages and the inductances. And therefore,
we will have the same impedance for
358
00:42:08,099 --> 00:42:09,259
all the three phased.
359
00:42:09,259 --> 00:42:10,259
..
360
00:42:10,259 --> 00:42:17,039
So, self impedance Z a a, will be same as
Z b b, will be same as Z c c. That is all
361
00:42:17,039 --> 00:42:21,099
the
three phase will have the same self impedances.
362
00:42:21,099 --> 00:42:28,829
And the mutual impedances will also be
equal for all the three phases. So, Z a b
363
00:42:28,829 --> 00:42:32,510
and Z a c will be equal which will be also
equal
364
00:42:32,510 --> 00:42:40,499
to Z b c and Z c a.
So, we get finally, for a fully transpose
365
00:42:40,499 --> 00:42:45,529
line, we get the values or thus impedance
series
366
00:42:45,529 --> 00:42:53,000
impedance for the three phase system with
ground return. As Z p is equal to Z a a, Z
367
00:42:53,000 --> 00:43:00,030
a b,
Z a b, Z a b, Z a a, Z a b, Z a b, Z a b,
368
00:43:00,030 --> 00:43:09,300
Z a a. This matrix will give me the impedance,
series impedance of the transmission line,
369
00:43:09,300 --> 00:43:13,030
with ground return.
.
370
00:43:13,030 --> 00:43:23,950
.So, Z a a is equal to as I said the average
value and Z a b will be also equal to the
371
00:43:23,950 --> 00:43:29,990
average value the mutual and the self, will
be given by these two relationship.
372
00:43:29,990 --> 00:43:30,990
.
373
00:43:30,990 --> 00:43:39,180
Now, let us take an example, for finding out
the inductance of a three phase transmission
374
00:43:39,180 --> 00:43:45,390
system. The example, that we are taking is
for the conductor consideration of a
375
00:43:45,390 --> 00:43:51,069
completely transpose three phase over a transmission
line, with bundle conductor is
376
00:43:51,069 --> 00:43:58,900
shown. All the conductors have a radius of
0.74 centimeter with is 30 centimeter bundle
377
00:43:58,900 --> 00:44:00,260
spacing.
.
378
00:44:00,260 --> 00:44:08,200
.That is, we have a 30 centimeter bundle spacing
for the conductors and each conductor
379
00:44:08,200 --> 00:44:17,799
as a radius of 0.74 centimeter. The distance
between the two phase conductors is 6
380
00:44:17,799 --> 00:44:29,359
meters. That is from center to center. And
this is a horizontally spaced configuration.
381
00:44:29,359 --> 00:44:34,229
The
line is fully transposed. For this line, determine
382
00:44:34,229 --> 00:44:41,309
the inductance per phase in milli Henry
per kilometer and in milli Henry per meter.
383
00:44:41,309 --> 00:44:44,099
Find the inductive line reactance per phase
in
384
00:44:44,099 --> 00:44:49,000
ohms per meter at 50 hertz.
So, this is the question, that we will have
385
00:44:49,000 --> 00:44:53,450
to find out the inductances per phase of the
line
386
00:44:53,450 --> 00:45:01,829
in milli Henry per kilometer or a milli Henry
per meter. We will also need to find out the
387
00:45:01,829 --> 00:45:08,049
inductive line reactance per phase in ohms
per meter at 50 hertz.
388
00:45:08,049 --> 00:45:09,049
.
389
00:45:09,049 --> 00:45:21,130
So, for this system, what we have to do is,
we have to find out the equivalent distances.
390
00:45:21,130 --> 00:45:30,200
So, for the given configuration as shown is
figure, we have the mutual GMD between the
391
00:45:30,200 --> 00:45:58,009
different phases given by D a b is equal to
r 13, r 14, r 23, r 24. That is distance from
392
00:45:58,009 --> 00:46:03,190
here to this conductor, distance from this
to this conductor, distance from this to this
393
00:46:03,190 --> 00:46:09,080
conductor, distance from this to this conductor.
And 4th root of that, because again if you
394
00:46:09,080 --> 00:46:12,259
take the distances from this to this and this
to
395
00:46:12,259 --> 00:46:19,859
this, this to this and this to this, that
will also be same. So, it will 4th root of
396
00:46:19,859 --> 00:46:23,690
D a b is
equal is the 4th root of distance between
397
00:46:23,690 --> 00:46:33,040
13, 14, 23, 24. So, this is equal to 6 into
6 .3
398
00:46:33,040 --> 00:46:53,109
into 5.7 into 6. That is 13 is 6, 23 is 5.7,
14 is 6 plus 30 centimeters. So, 6.3 and 24
399
00:46:53,109 --> 00:46:58,829
is 6.
So, we take the 4th root, it comes out to
400
00:46:58,829 --> 00:47:04,849
be 5.9962, it is same as almost 6 meters.
401
00:47:04,849 --> 00:47:14,460
.Similarly, for between B and C, if we do
this, we will get that same distance or the
402
00:47:14,460 --> 00:47:22,650
same
GMD. But, for D c a, if we look at this between
403
00:47:22,650 --> 00:47:31,499
C and A, then we have the distance as A
1 to 5, 1 to 6, 2 to 5 and 2 to 6.
404
00:47:31,499 --> 00:47:32,499
.
405
00:47:32,499 --> 00:47:44,799
Therefore, D c a is equal to r 15 into r 16
into r 25 into 26, the 4th root of this. That
406
00:47:44,799 --> 00:47:50,430
is
equal to 12 into 12.3 into 11.7 into 12, 4th
407
00:47:50,430 --> 00:47:54,890
root of that is 11.9981, which is very near
to
408
00:47:54,890 --> 00:48:04,249
12 meter. Now, the equilateral spacing between
the phases is given by the D e q defined
409
00:48:04,249 --> 00:48:11,490
as D e q is equal to D a b into D b c into
D c a and cube root of that. So, if we find
410
00:48:11,490 --> 00:48:16,140
out
the equilateral spacing. The D e q, that comes
411
00:48:16,140 --> 00:48:26,029
out to be 5.9962 into 5.9962 into 11.9981
cube root of that, that comes out to be 7.5559
412
00:48:26,029 --> 00:48:28,239
meter.
413
00:48:28,239 --> 00:48:29,239
..
414
00:48:29,239 --> 00:48:36,019
Now, the self GMD of all the three phases
point to symmetry are equal. That is, the
415
00:48:36,019 --> 00:48:39,700
self
distance each conductor has the same radius
416
00:48:39,700 --> 00:48:46,200
and the distance between the two conductors
of a bundle is same. That is 30 centimeter.
417
00:48:46,200 --> 00:48:55,289
So, therefore, the D s for each phase
conductor is going to be equal to r dash.
418
00:48:55,289 --> 00:49:01,309
That is 0.7788 into r for the conductor into
30
419
00:49:01,309 --> 00:49:10,650
centimeter, which is the distance between
the two conductors of the same bundle.
420
00:49:10,650 --> 00:49:17,999
The square root of this comes out to be 4.1580
centimeter. In fact, we could have taken
421
00:49:17,999 --> 00:49:24,650
in all the four distances as we have seen
earlier D s is r dash into D square root of
422
00:49:24,650 --> 00:49:29,789
that.
So, this is equal to 4.1580 centimeter. Now,
423
00:49:29,789 --> 00:49:43,200
you see this is much larger than 2 times of
0.7788 into r which is 0.74 centimeter. Therefore,
424
00:49:43,200 --> 00:49:48,209
bundling has increased the effective
radius considerably.
425
00:49:48,209 --> 00:49:54,309
Therefore, now we can find out inductance
per phase for the given system as L is equal
426
00:49:54,309 --> 00:50:00,500
to 2 into 10 to power minus 7 into log n.
D e q by Ds Henry’s per meter per phase,
427
00:50:00,500 --> 00:50:06,339
which
is equal to 1.04049 into 10 to power 6 Henry’s
428
00:50:06,339 --> 00:50:15,530
per meter, per phase, which can be
written as 1.04049 into 10 power minus 3 milli
429
00:50:15,530 --> 00:50:33,759
Henry per meter, per phase or 1.04049
milli Henry per kilometer, per phase.
430
00:50:33,759 --> 00:50:34,759
..
431
00:50:34,759 --> 00:50:42,130
The inductive line reactance per phase at
50 hertz is given by x is equal to 2 phi f
432
00:50:42,130 --> 00:50:50,710
into the
inductance. So, this is 2 phi 50 into 1.04049
433
00:50:50,710 --> 00:50:59,180
into 10 to power minus 6 ohms per meter
per phase. This comes out to be 3.270 into
434
00:50:59,180 --> 00:51:07,249
10 to power minus 4 ohms per meter per
phase. So, this is, how we can calculate the
435
00:51:07,249 --> 00:51:15,309
inductance and the inductive reactance for
any given three phase transmission system.
436
00:51:15,309 --> 00:51:16,309
.
437
00:51:16,309 --> 00:51:25,089
Now, before we finish I would like you to
answer the following questions. First is,
438
00:51:25,089 --> 00:51:33,170
why
bundled conductors are used in EHV lines.
439
00:51:33,170 --> 00:51:41,589
Second question is, what is transposition.
And third question is, how the effect of earth
440
00:51:41,589 --> 00:51:45,170
return current is taken into account in
441
00:51:45,170 --> 00:51:54,059
.inductance calculation for a 3 phase line
with ground return system. Specially, when
442
00:51:54,059 --> 00:52:02,579
this
system is carrying unbalanced current, so
443
00:52:02,579 --> 00:52:07,930
with this we finish these lessons.
Thank you very much.
444
00:52:07,930 --> 00:52:16,680
We will meet again for lesson 5. In which,
we will talk about calculating the capacitance
445
00:52:16,680 --> 00:52:22,609
of the transmission line.
Thank you.
446
00:52:22,609 --> 00:52:32,430
Preview of next lecture
Lecture - 05
447
00:52:32,430 --> 00:52:41,180
Transmission Line Capacitance
Welcome to lesson 5, on power system analysis
448
00:52:41,180 --> 00:52:50,401
course. In this course, we will talk about
the transmission line capacitance, before
449
00:52:50,401 --> 00:52:54,780
we get in to the calculation of transmission
line
450
00:52:54,780 --> 00:52:55,780
capacitance.
.
451
00:52:55,780 --> 00:53:04,401
I would like to answer those questions. That
I ask in lesson 4. First question was, why
452
00:53:04,401 --> 00:53:12,150
bundle conductors are used in EHV lines? Well,
the answer to this question is bundling
453
00:53:12,150 --> 00:53:18,030
of conductors. That is, instead of using one
single conductor, use of a number of
454
00:53:18,030 --> 00:53:27,829
conductors connected by conducting frames,
reduces electric fields strength on conductor
455
00:53:27,829 --> 00:53:35,470
surface, which in effect reduces the corona
losses, which result in power loss.
456
00:53:35,470 --> 00:53:49,130
.As well as radio interference and audible
noise in the system, bundling also increases
457
00:53:49,130 --> 00:53:53,069
the
effective radius of the conductor. And there
458
00:53:53,069 --> 00:54:00,569
by, reduces the inductance of the
transmission line. This in effect will improve
459
00:54:00,569 --> 00:54:10,390
the regulation of the transmission line. As
seen from here, the effective radius for a
460
00:54:10,390 --> 00:54:21,119
3 conductor bundle, which are spaced at a
distance d. From the center d, from each other
461
00:54:21,119 --> 00:54:32,130
with a radius r is given by 9th root of r
dash into d into d whole cube, which is equal
462
00:54:32,130 --> 00:54:44,640
to cube root of r dash d square. This is
much larger than r dash, which is used, when
463
00:54:44,640 --> 00:54:55,759
a single conductor is used. And therefore,
bundling helps in reducing the inductance
464
00:54:55,759 --> 00:54:59,099
as it increases the effective radius.
.
465
00:54:59,099 --> 00:55:05,390
Therefore, we will take the case of a three
phase system. We will start with the three
466
00:55:05,390 --> 00:55:11,369
phase line with equilateral spacing. Because,
as we have seen for inductance calculation,
467
00:55:11,369 --> 00:55:19,150
we can always covert, if the line is transpose.
We can always convert any system into the
468
00:55:19,150 --> 00:55:28,029
equivalent three phase equilateral spacing
of the conductors by finding out the equivalent
469
00:55:28,029 --> 00:55:37,960
distance D e q. So, here we have equilateral
spacing conductor a, b and c, each with a
470
00:55:37,960 --> 00:55:44,630
distance d from each other. We also assume
that this system consist of only three
471
00:55:44,630 --> 00:55:53,729
conductors. So, the sum of the total charges
will be equal to 0. That is q a plus q b plus
472
00:55:53,729 --> 00:55:57,199
q
c is equal 0.
473
00:55:57,199 --> 00:55:58,199
..
474
00:55:58,199 --> 00:56:08,640
So, for solving this, we will take case of
a fully transpose three phase line. So, we
475
00:56:08,640 --> 00:56:12,400
have
the fully transpose three phase line. The
476
00:56:12,400 --> 00:56:15,949
voltage is given by the relationship V a b
is
477
00:56:15,949 --> 00:56:24,230
equal to 1 by twice phi Epsilon, I am sorry,
this is Epsilon into q a into log n D e q
478
00:56:24,230 --> 00:56:29,999
by r
plus q b into log n r by D e q. Where D e
479
00:56:29,999 --> 00:56:35,349
q is equal to D 12 into D 23 into D 31 into
the
480
00:56:35,349 --> 00:56:42,079
power 1 by 3. That is cube root of D 12, D
23 into D 31.
481
00:56:42,079 --> 00:56:49,799
Now, substituting these values, we get this
as equal to 12 meters into 12 meters into
482
00:56:49,799 --> 00:56:54,229
24
meters. That is the distance between the conductor
483
00:56:54,229 --> 00:57:04,109
a and c, the cube root of this, will
give us equal to 15.119 meter. Similarly,
484
00:57:04,109 --> 00:57:08,109
we can write the relationship for V a c and
V a
485
00:57:08,109 --> 00:57:16,989
c will be equal to 1 by twice pi Epsilon into
q a into log n D e q by r plus q c into log
486
00:57:16,989 --> 00:57:24,199
n r
by D e q. Calculate these quantities for a
487
00:57:24,199 --> 00:57:25,900
transmission system.
488
00:57:25,900 --> 00:57:26,900
..
489
00:57:26,900 --> 00:57:33,579
So, next, this example, we have a three phase
400 kV 50 hertz 350 kilometer overhead
490
00:57:33,579 --> 00:57:41,119
transmission line. That has a flat horizontal
spacing, which three identical conductors.
491
00:57:41,119 --> 00:57:48,640
That is, we have three identical conductors,
places in a plat horizontal spacing. The
492
00:57:48,640 --> 00:57:54,539
conductors have an outside diameter of 3.28
centimeter. And that is the diameter of the
493
00:57:54,539 --> 00:57:59,680
conductor is 3.28 centimeter. And the distance
between the adjacent conductors is and
494
00:57:59,680 --> 00:58:04,440
the distance between the adjacent conductors
is 12 meter. Now, for this system determine
495
00:58:04,440 --> 00:58:10,920
the capacitive reactance to neutral in ohms
per meter per phase and the capacitive
496
00:58:10,920 --> 00:59:43,849
reactance, for the line in ohms per phase.
Thank you.
497
00:59:43,849 --> 00:59:43,849
.