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Welcome to lesson 8 on Power System Analysis.
This lesson is a continuation on
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Transmission Line Modeling, especially modeling
of a long line. Now, if you remember
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in the previous lesson, that is lesson 7,
we talked about distributed parameter model
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for
long transmission lines. That is lines which
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are longer than 250 kilometers. For these
line
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we said that the line voltage V x.
.
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And at distance x from the receiving end is
given by cos hyperbolic gamma x into V R
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plus Z c sin hyperbolic gamma x into I R,
where gamma is called the propagation
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constant, x is the distance of the point from
the receiving end. V R and I R are the
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voltages and currents at the receiving end.
Similarly, the current at a distance x from
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the
receiving end I x is equal to 1 by Z c sin
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hyperbolic gamma x into V R plus cos
hyperbolic gamma x into I R.
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Where, Z c is the characteristic impedance
of the transmission line. And it is given
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by
square root of Z pi y, where Z is the series
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impedance of the transmission line, per unit
length and y is the shunt admittance of the
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transmission line per unit length. As we see
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.these models, these equations involved hyperbolic
functions. Now, since we have been
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writing all these transmission line equations,
in terms of ABCD parameters. We can
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write this equation also in those terms. So,
in matrix form, we can use this as V x I x
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is
equal to A x B x C x D x and V R I R. These
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form where this relationship is in terms of
voltage and current at any distance x from
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the receiving end.
.
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Now, here if you see A x is equal to D x and
that is equal to cos hyperbolic gamma x, in
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per unit. That is if you see this relationship
this is your A x and this is your B x, this
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is C
x and this is D x. So, A x is equal to D x
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is equal to cos hyperbolic gamma x, V x is
equal to Z c sin hyperbolic gamma x, V x is
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equal to Z c sin hyperbolic gamma x. C x is
equal to 1 by Z c sin hyperbolic gamma x and
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D x as we have already seen is equal to A
x.
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So, this is the model where A and D are basically
dimensionless, B has a dimension of
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impedance and C has a dimension of Siemens,
that is admittance. Now, normally we are
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interested only in the terminal conditions.
That is the sending end voltages and currents,
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and the receiving end voltages and currents
rather than the voltage and current at any
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intermediate point on the transmission line.
Therefore, we can find out the voltage and
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currents at the sending end, in terms of voltage
and current at the receiving end by
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substituting x is equal to l, where l is the
total line length.
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.In that case we will get the equation V S
I S is equal to ABCD. Now, we are not writing
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x because, we are now dealing with the terminal
conditions only their fore we will have
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A, B, C and D parameter into V R I R. Here,
A is equal to D is equal to cos hyperbolic
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gamma l, instead of x we are substitute l.
This is the dimensional less quantity, it
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will be
in per unit if voltage and currents are all
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in per unit. B is equal to Z c sin hyperbolic
gamma l, C is equal to 1 by Z c sin hyperbolic
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gamma l. This B has a dimension of
impedance, C has a dimension of admittance.
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Now, here the term gamma is what we call,
propagation constant of the line.
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.
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And gamma in general will a complex quantity.
That is gamma is equal to alpha plus j
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beta, it is dimension will be per meter. In
fact, if you see gamma l, the dimension of
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gamma l will be dimensions. So, gamma is equal
to alpha plus j beta per meter. Now, if
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we write e to the power gamma l, this will
be equal to e to the power alpha l plus j
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beta l.
And this will be equal to e to the power alpha
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l into e to the power j beta l, this we can
write as e to the power alpha l angle of beta
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l.
Now, here alpha if you see is called the attenuation
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constant of the line, and beta gives
the phase angle. So, it is called the phase
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constant of the line. So, propagation constant
has two terms alpha, the attenuation constant
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and beta the phase constant. Now, if we
write cos hyperbolic gamma l, this will be
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equal to e to the power gamma l plus e to
the
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.power minus gamma l by 2. This is equal to
half e to the power alpha l angle beta l plus
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e
to the power minus alpha l angle minus beta
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l.
Similarly, sin hyperbolic gamma l is equal
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to e to the power gamma l minus e to the
power minus gamma l divided by 2. This is
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equal to half into e to the power alpha l
angle
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beta l minus e to the power minus alpha l
angle beta l. So, in this way, if we know
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the
alpha and beta, we can calculate the terms
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sin hyperbolic gamma l and cos hyperbolic
gamma l. And therefore, we can calculate V
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S I S, in terms of V R I R.
.
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Or we can use the other identity, that is
cos hyperbolic alpha l plus j beta l. That
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is cos
hyperbolic gamma l is equal to cos hyperbolic
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alpha l into cos beta l plus j sin hyperbolic
alpha l into sign beta l. Or sin hyperbolic
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gamma l, that is sin hyperbolic alpha l plus
j
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beta l is equal to sin hyperbolic alpha l
into cos beta l plus j cos hyperbolic alpha
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l into
sin beta l. So, knowing the Z and Y, we can
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calculate the propagation constant gamma
and therefore, alpha and beta.
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And we can also calculate the characteristic
impedance Z c. So, knowing these values,
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we can evaluate the sending end voltage and
current, in terms of receiving in voltage
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and
current. That is using these relationship,
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we can evaluate the terminal condition of
the
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line. Now, in most of the analysis that we
do, especially using computers most of the
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analysis that we do, we need circuit parameter
model rather than the ABCD model of the
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transmission line.
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.As we have said earlier we had created a
pi model or a nominal pi model for a medium
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length line. Similarly, we can find out or
we can represent a long line, in terms of
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a
equivalent pi model. The difference between
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nominal pi and equivalent pi is only that,
instead of using Z, which is the series impedance
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of the line. And Y the shunt impedance
of the line, we have to use here Z dash which
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is a not equal to Z, but modified value of
Z.
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And Y dash which is again a modified value
of Y. So, some modification into on series
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impedance and shunt admittance of the transmission
line, will have to be used here in the
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long line model. Now, this is the pi model,
where we have V S, the sending end voltage
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Is the sending end current V R the receiving
end voltage I R the receiving end current.
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Z
dash is the series impedance of the line and
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the total shunt admittance, Y dash is divided
into 2 parts Y dash by 2 at the sending end
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and Y dash by 2 at the receiving end. This
is
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the pi equivalent model for the long line.
Now, we need to find out these values of Z
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dash and Y dash for making this model complete.
.
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Now, we know that for the pi model A is equal
to D is equal to 1 plus Y dash Z dash by
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2. This is the same thing that we had done
in lesson 7, where we had found the ABCD
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parameters. And transmission line nominal
pi model parameters. There we had seen A is
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equal to D is equal to 1 plus Y dash Z dash
by 2 in per unit. This is the dimension less
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quantity and B is equal to Z dash, this is
ohms, that is a impedance units. And C is
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equal
to Y dash into 1 plus Y dash Z dash by 4 Siemens,
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that is the unit for admittance.
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.Now, using this here we have Z dash is equal
to Z c sin hyperbolic gamma l, which we
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can write as Z, which is the series impedance
of the line. Total series impedance of the
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line into F 1 or this is equal to Z into sin
hyperbolic gamma l by gamma l. Similarly,
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Y
dash by 2 will be equal to tan hyperbolic
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gamma l by 2 divided by Z c. This is equal
to Y
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by 2 into F 2, this is equal to Y by 2 into
tan hyperbolic gamma l by 2 divided by gamma
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l by 2. We will see how we get these relationships.
.
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Z dash is equal to Z c sin hyperbolic gamma
l. This is equal to square root of Z by Y,
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that is Z c is Z by Y, square root of Z by
Y into sin hyperbolic gamma l. Therefore,
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Z
dash is equal to Z l into square root of Z
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by Y sin hyperbolic gamma l by Z into l, this
is
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equal to Z into l sin hyperbolic gamma l by
square root of Z y l.
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..
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So, this is sin hyperbolic gamma l, Z l into
sin hyperbolic gamma l divided by square
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root of Z y into l, this is equal to Z into
F 1, where F 1 is sin hyperbolic gamma l divided
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by gamma l in per unit. And 1 plus Y dash
Z dash by 2 that is A is equal to cos
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hyperbolic gamma l.
.
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Therefore, 1 plus Y dash Z dash by 2, it is
the same equation or Y dash by 2 is equal
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to
cos hyperbolic gamma l minus 1 divided by
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Z c sin hyperbolic gamma l. This is equal
to
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tan hyperbolic gamma l by 2 divided by Z c,
this is equal to tan hyperbolic gamma l by
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2
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.divided by square root of z by y, that is
Z c. Therefore Y dash by 2 is equal to y into
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l by
2 into tan hyperbolic gamma l by 2 this term.
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Now, here we have multiplied it by y l by
2, so in the numerator and therefore, we have
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divided in the denominator.
So, square root of z by y into y l by 2, therefore
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we will get this equal to y l by 2 into tan
hyperbolic gamma l by 2 divided by square
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root of z into y into l by 2, which is equal
to
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Y by 2 in to y l will give the Y, capital
Y that is total shunt admittance of the line.
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So,
this is Y by 2 into F 2 Siemens, where F 2
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is equal to tan hyperbolic gamma l by 2 and
root Y Z is nothing but, gamma, so this is
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gamma l by 2 in per unit.
Now, these relationships give us the model
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for finding out the voltage and current at
any
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point on the line or the terminal conditions
of the line. That is sending end voltage and
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current of the line. These are the accurate
ABCD parameters of the transmission line or
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the accurate parameter for the pi equivalent
circuit of the line. Now, most of the time
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since, we see this relationship that we have
seen earlier, these relationships are involving
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hyperbolic functions.
So, doing computation with these is going
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to be some more tedious. Also many times
what we need to do is try to get some approximate
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idea, of what is happening in the
transmission line. This is very important,
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especially when we are designing a
transmission line, because we will have to
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work out large number of options. So, for
this
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purpose what we do is, we try to introduce
the concept of a loss less line.
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That is a line where the series impedance
consists of only the inductance part, or the
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reactance of the line and the resistance is
negligible. Similarly, the shunt admittance
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consists of only the capacitance and the shunt
conductance is neglected. One of the
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advantages of this kind of a modeling is also
that, it gives us a very good idea about
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some of the important concepts of power flow,
on the transmission line another aspects.
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..
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Now, in case of the loss less line, we have
the series impedance per meter length, is
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equal to j omega L, where L is the inductance
of the line per meter length. And shunt
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admittance y per meter length will be equal
to j omega C, where C is the shunt
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capacitance of the line per meter length.
So, this is y is in terms of Siemens per meter,
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z
is in terms of ohms per meter.
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Now, here if we take the characteristic impedance
of this line. Then it will be z by y,
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since this line is loss less, we call this
characteristic impedance for a loss less line
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as the
surge impedance of a line. So, for this line,
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the surge impedance Z s is equal to square
root of z by y which is equal to square root
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00:18:24,790 --> 00:18:31,940
of j omega L by C j omega C, which is equal
to square root of L by C ohms.
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Now, if you look at this Z s, you find that
this quantity Z s is a pure real number. That
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is
Z is or surge impedance is pure resistive
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in nature, it is not a complex number. As
we
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had seen earlier, Z C is a complex number.
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00:19:00,910 --> 00:19:01,910
..
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Similarly, the propagation constant gamma
is equal to square root of z y, which will
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be
equal to square root of j omega L into j omega
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C. This comes out to be j omega square
root of root L C, which we write as j beta.
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Because, we find that this term, now is purely
imaginary. And therefore, we have only imaginary
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term for the propagation constant, or
it is in terms of only the phase constant.
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The attenuation constant is 0 which is natural
because there are no losses in this line.
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So, gamma is equal to j beta per meter. And
if
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you are talking in terms of the whole length,
then gamma l will be dimensionless. Here,
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beta is equal to omega into root L C per meter,
which is the phase constant. Now, we will
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try to develop the ABCD parameters for this
loss less transmission line.
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00:20:21,850 --> 00:20:22,850
..
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Now, again we had seen earlier A x is equal
to D x is equal to cos hyperbolic gamma x.
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Now, gamma is equal to j beta, therefore this
is equal to cos hyperbolic j beta x, which
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will be equal to e to the power j beta x plus
e to the power minus j beta x divided by 2,
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which is equal to cos beta x in per unit,
this is a dimensionless quantity. Now, here
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we
see that A and D are in terms of cos beta
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x, not hyperbolic functions. Similarly, sin
hyperbolic gamma x is equal to sin hyperbolic
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j beta x, which is equal to e to the power
j
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beta x minus e to the power minus j beta x
divided by 2, which again comes out to be
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equal to j sin beta x. Again this is a pure
trigonometric function, it is not a hyperbolic
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00:21:20,310 --> 00:21:23,070
function.
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00:21:23,070 --> 00:21:24,070
..
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Therefore, B x will be equal to Z c sin hyperbolic
gamma x, this is equal to j times Z c
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sin beta x. This is equal to j times square
root of L by C Z c is root over L by C into
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00:21:40,740 --> 00:21:47,330
sin
beta x. Now, again this, if we see is a purely
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00:21:47,330 --> 00:21:55,560
imaginary quantity or the quantity for
reactance, this has a unit of ohm. Similarly,
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00:21:55,560 --> 00:21:58,590
C x will be equal to sin hyperbolic gamma
x
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00:21:58,590 --> 00:22:09,000
by Z c, that is j sin beta x divided by root
L by C. So, this is again an imaginary quantity,
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which is very similar to capacitance. So,
which Z dash will be equal to j Z c sin beta
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l,
which we can write as j X dash.
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.
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00:22:30,520 --> 00:22:41,330
.Similarly, Y dash by 2, we can write as equal
to Y by 2 sin hyperbolic gamma l by 2 was
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there, so this is j beta l by 2 divided by
gamma l by 2 was their earlier. Now, gamma
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is
equal to j beta only, so it is j beta l by
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00:22:52,780 --> 00:22:58,220
2, this is equal to Y by 2 sin hyperbolic
j beta l by
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00:22:58,220 --> 00:23:08,100
2 divided by j beta l by 2 into cos hyperbolic
j beta l by 2. That is tan hyperbolic j beta
201
00:23:08,100 --> 00:23:11,860
l
by 2, we have put it as sin hyperbolic j beta
202
00:23:11,860 --> 00:23:16,360
l by 2 divided by cos hyperbolic j beta l
by 2.
203
00:23:16,360 --> 00:23:25,040
This is equal to j omega C into l, this is
for Y total line capacitance divided by 2,
204
00:23:25,040 --> 00:23:30,180
so Y by
2 into j sin beta l by 2. Because, sin hyperbolic
205
00:23:30,180 --> 00:23:38,070
j beta l by 2 we have seen is equal to j sin
beta l by 2 divided by j beta l by 2 this
206
00:23:38,070 --> 00:23:41,750
term, into cos hyperbolic j beta l by 2. We
have
207
00:23:41,750 --> 00:23:48,370
seen is equal to cos beta l by 2. So, this
terms comes out to be j omega C into l by
208
00:23:48,370 --> 00:23:54,790
2 into
tan beta l by 2 divided by beta l by 2.
209
00:23:54,790 --> 00:24:05,630
This we can write as j omega C dash l by 2,
where C dash is now the modified
210
00:24:05,630 --> 00:24:15,260
capacitance of the transmission line or j
omega C dash l is the modified admittance
211
00:24:15,260 --> 00:24:22,750
of the
line. And that C dash is equal to C times
212
00:24:22,750 --> 00:24:31,900
tan beta l by 2 divided by beta l by 2. So,
for a
213
00:24:31,900 --> 00:24:39,401
loss less line, we see that we reduced the
relationship from hyperbolic functions, for
214
00:24:39,401 --> 00:24:45,260
a
normal loss long line to a pure trigonometric
215
00:24:45,260 --> 00:24:53,280
function relationship for a lossless line.
This reduces our computational burden, considerably
216
00:24:53,280 --> 00:24:58,920
and one can do hand calculations
with this kind of relationships.
217
00:24:58,920 --> 00:24:59,920
.
218
00:24:59,920 --> 00:25:08,520
Therefore, the pi equivalent circuit for a
loss less line will look like this, where
219
00:25:08,520 --> 00:25:13,610
V S is
equal to V S angle delta, I S is the current
220
00:25:13,610 --> 00:25:18,350
flowing Y dash by 2 is the half of the total
221
00:25:18,350 --> 00:25:24,010
.modified admittance of the line. Z dash is
the total series impedance of the line, that
222
00:25:24,010 --> 00:25:26,700
is
the modified series impedance of the line.
223
00:25:26,700 --> 00:25:34,640
Y dash by 2 the other half of the total shunt
modified, total shunt admittance of the line
224
00:25:34,640 --> 00:25:38,820
is placed at the receiving end, I R is the
receiving end.
225
00:25:38,820 --> 00:25:44,750
Current and V R is equal to V R angle 0 degrees
which is the receiving end voltage, we
226
00:25:44,750 --> 00:25:50,920
have chosen the receiving end voltage as a
reference voltage, where we have as already
227
00:25:50,920 --> 00:25:59,630
seen Z dash is equal to j omega L into l is
the length of the line, the sin beta l divided
228
00:25:59,630 --> 00:26:04,410
by
beta l which is a pure imaginary quantity,
229
00:26:04,410 --> 00:26:13,230
or the reactance j X dash. And similarly Y
dash by 2 can be written as j omega C dash
230
00:26:13,230 --> 00:26:26,120
l by 2 Siemens, which is purely capacitance.
Now, for lossless lines, we can also introduce
231
00:26:26,120 --> 00:26:36,030
the concept of wave length, that is the
length of the voltage and current wave on
232
00:26:36,030 --> 00:26:42,370
the line as it moves by 360 degrees. That
is a
233
00:26:42,370 --> 00:26:50,700
total change in face of voltage and current
wave, on the transmission line over a 360
234
00:26:50,700 --> 00:26:52,700
degree displacement.
.
235
00:26:52,700 --> 00:27:00,980
Now, if lambda is the wavelength of the line,
then lambda will be equal to 2 pi, which is
236
00:27:00,980 --> 00:27:09,230
360 degrees divided by beta the phase constant
of the line. Therefore, we can write
237
00:27:09,230 --> 00:27:17,630
lambda the wavelength of the line is equal
to 2 pi by omega root L C, beta is omega root
238
00:27:17,630 --> 00:27:25,730
L C which is equal to 1 by f root L C, because
omega is twice pi f. So, if we write here
239
00:27:25,730 --> 00:27:33,440
twice pi f twice pi will cancel out, this
is 1 by f root L C which comes out to be about
240
00:27:33,440 --> 00:27:37,740
6000 kilometer for 50 Hertz line.
241
00:27:37,740 --> 00:27:46,070
.Now, this will be approximately 6000 kilometer
of the line. Because, if you see the
242
00:27:46,070 --> 00:27:55,130
relationship for L and C. And since, we can
take R dash the effective radius in
243
00:27:55,130 --> 00:28:01,500
calculating the inductance to be all most
same as that of R. Because, D is much larger
244
00:28:01,500 --> 00:28:08,270
compare to R. Therefore, in that case if you
substitute the relationship for L and C here,
245
00:28:08,270 --> 00:28:11,960
then we will find that this comes out to be
this much.
246
00:28:11,960 --> 00:28:19,190
Similarly, we will get f lambda, that is if
we multiply it here 1 by root L C, which is
247
00:28:19,190 --> 00:28:22,900
the
velocity of propagation of voltage and current
248
00:28:22,900 --> 00:28:31,320
wave on a lossless line. And this for a
lossless line comes out to be same as that
249
00:28:31,320 --> 00:28:40,720
of speed of light. So, the velocity of
propagation of voltage and current wave on
250
00:28:40,720 --> 00:28:50,400
a lossless long transmission line is the same
as that of light. This relationship, again
251
00:28:50,400 --> 00:28:55,470
you can see by simply substituting the
relationship for L and C.
252
00:28:55,470 --> 00:29:03,550
Here, this will then come out to be 1 by mu
0 into root over mu 0 into epsilon 0, which
253
00:29:03,550 --> 00:29:11,140
comes out to be 3 into 10 to the power 8 meters
per second, which will be same as the
254
00:29:11,140 --> 00:29:20,050
velocity of the light. Now, for lossless line,
we can introduce one more concept which is
255
00:29:20,050 --> 00:29:26,500
very important, especially when we are designing
the transmission lines or transmission
256
00:29:26,500 --> 00:29:33,070
systems. This concept is of surge impedance
loading.
257
00:29:33,070 --> 00:29:34,070
.
258
00:29:34,070 --> 00:29:41,420
As we had seen earlier V X is equal to A X
into V R plus B X into I R, where A X is cos
259
00:29:41,420 --> 00:29:52,340
beta x in case of lossless line. So, and V
X is equal to Z c sin beta x, therefore V
260
00:29:52,340 --> 00:29:55,900
X is
equal to cos beta x into V R plus j Z c sin
261
00:29:55,900 --> 00:29:59,990
beta x into I R. And similarly I X will be
equal
262
00:29:59,990 --> 00:30:08,380
.to C X V R plus D X into I R, C X is equal
to j sin beta x by Z c and D X is equal to
263
00:30:08,380 --> 00:30:10,940
A X
which is equal to cos beta x. Therefore I
264
00:30:10,940 --> 00:30:14,470
X is equal to j sin beta x by Z c into V R
plus
265
00:30:14,470 --> 00:30:18,350
cos beta x into I R.
.
266
00:30:18,350 --> 00:30:27,960
Now, if we terminate this transmission line
into an impedance Z c. Now, here this figure
267
00:30:27,960 --> 00:30:36,340
is showing you the transmission line, or a
single phase transmission line, we can consider
268
00:30:36,340 --> 00:30:48,110
it also as a single phase to neutral for this
transmission line. Now, if we look at this
269
00:30:48,110 --> 00:30:55,700
transmission line and terminate it into Z
c. That is if we load this line with an impedance
270
00:30:55,700 --> 00:31:00,640
equal to the characteristic impedance or the
surge impedance for a lossless line, that
271
00:31:00,640 --> 00:31:02,471
is Z
c is equal to root L by C.
272
00:31:02,471 --> 00:31:03,471
..
273
00:31:03,471 --> 00:31:14,740
Then we get V X is equal to cos beta x into
V R plus j Z c sin beta x into I R. Now, I
274
00:31:14,740 --> 00:31:22,210
R
will be equal to, if you look at this circuit,
275
00:31:22,210 --> 00:31:25,440
then I R will be equal to V R, the voltage
at
276
00:31:25,440 --> 00:31:34,291
this point divided by Z c, the impedance connected
here. So, we replace this I R by V R
277
00:31:34,291 --> 00:31:42,970
by Z c, therefore we will get this as cos
beta x plus j sin beta x into V R. This Z
278
00:31:42,970 --> 00:31:47,560
c, this Z
c cancel out, so we get this as equal to e
279
00:31:47,560 --> 00:31:56,780
to the power j beta x into V R volts.
Now, here if we take the magnitude of this,
280
00:31:56,780 --> 00:32:03,460
then what we get is that V X is equal to,
magnitude of V X is equal to magnitude of
281
00:32:03,460 --> 00:32:13,850
V R. That is the voltage at any point on the
line is constant, that is it is equal to the
282
00:32:13,850 --> 00:32:17,490
receiving end voltage. So, from sending end
to
283
00:32:17,490 --> 00:32:23,270
receiving end at all point the voltage is
same, that is the magnitude of the voltage
284
00:32:23,270 --> 00:32:30,500
is
same. This is one great advantage in using
285
00:32:30,500 --> 00:32:34,090
the loading on the line, which is equal to
the
286
00:32:34,090 --> 00:32:43,730
surge impedance. And this is a particularly
important characteristic of lossless line.
287
00:32:43,730 --> 00:32:48,060
That
is if you terminate this line into surge impedance,
288
00:32:48,060 --> 00:32:54,440
then the voltage all along the length of
the line is going to remain constant.
289
00:32:54,440 --> 00:32:55,440
..
290
00:32:55,440 --> 00:33:01,480
Similarly, we can find out the current at
any point X from the receiving end. So, I
291
00:33:01,480 --> 00:33:05,040
X is
equal to j sin beta x by Z c into V R plus
292
00:33:05,040 --> 00:33:09,771
cos beta x into V R by Z c, that is I R V
R
293
00:33:09,771 --> 00:33:19,030
replacing by V R by Z c. So, this comes out
to be cos beta x plus j sin beta x into V
294
00:33:19,030 --> 00:33:24,510
R by
Z c, which again is equal to e to the power
295
00:33:24,510 --> 00:33:32,780
j beta x into V R by Z c. Again the magnitude
if you take is going to remain constant all
296
00:33:32,780 --> 00:33:39,630
along the line. Now, if we talk in terms of
the
297
00:33:39,630 --> 00:33:47,781
power, which is flowing in the line when we
terminate this lossless transmission line
298
00:33:47,781 --> 00:33:51,190
through a surge impedance.
.
299
00:33:51,190 --> 00:33:57,730
.Then we have S x the power the complex power
flowing, at any point in the line at a
300
00:33:57,730 --> 00:34:06,150
distance x from the receiving end is equal
to P x plus j Q x, where P x is the real part
301
00:34:06,150 --> 00:34:09,929
or
the real power flowing at a distance x from
302
00:34:09,929 --> 00:34:16,359
the receiving end. And Q x is the reactive
power which is flowing at a distance x from
303
00:34:16,359 --> 00:34:21,730
the receiving end. This is equal to V x into
I
304
00:34:21,730 --> 00:34:28,649
x conjugate.
Now, if we do this then we write for V x as
305
00:34:28,649 --> 00:34:32,839
e to the power j beta x into V R and for I
x,
306
00:34:32,839 --> 00:34:38,929
we write e to the power j beta x into V R
by Z c, we take the conjugate of that. Then,
307
00:34:38,929 --> 00:34:43,179
if
we do this calculation e to the power j beta
308
00:34:43,179 --> 00:34:50,529
x will cancel out. And we get V R square,
that is the mod of V R square and divided
309
00:34:50,529 --> 00:34:55,289
by Z c, that is the real power flow along
the
310
00:34:55,289 --> 00:35:02,640
line is constant. That is all along the line
the real power flow is constant. This is natural,
311
00:35:02,640 --> 00:35:07,509
because there is no losses which are taking
place in the line.
312
00:35:07,509 --> 00:35:13,029
So, whatever power is being sent from the
sending end, we are receiving the same power
313
00:35:13,029 --> 00:35:19,770
at the receiving end. And real power all along
the line is constant. And also a very
314
00:35:19,770 --> 00:35:25,920
important characteristic of this lossless
transmission line being terminated into it
315
00:35:25,920 --> 00:35:28,859
is
characteristic impedance, that reactive power
316
00:35:28,859 --> 00:35:38,340
flow is 0. That is whatever reactive power
losses which take place in the line, because
317
00:35:38,340 --> 00:35:48,329
of the series reactance of the line is being
produced by the shunt capacitance of the line.
318
00:35:48,329 --> 00:35:53,630
That is whatever is the reactive power generated
by the shunt capacitance of the line, that
319
00:35:53,630 --> 00:36:01,099
is being consumed by the series impedance
of the line. And therefore, the reactive power
320
00:36:01,099 --> 00:36:14,910
flow on along the line is 0. So, the surge
impedance and surge impedance loading gives
321
00:36:14,910 --> 00:36:23,970
us very important characteristic. That is
what we find from this is that, if we load
322
00:36:23,970 --> 00:36:27,980
the
transmission line at it surge impedance. Or
323
00:36:27,980 --> 00:36:34,710
the surge impedance loading of the line will
provide us a constant voltage all along the
324
00:36:34,710 --> 00:36:38,520
line.
That is both V S and V R will be same, this
325
00:36:38,520 --> 00:36:45,750
is a very great characteristic. That is the
regulation of the transmission line is going
326
00:36:45,750 --> 00:36:52,770
to be 0 percent there is no voltage drop which
takes place. Similarly, we do not have to
327
00:36:52,770 --> 00:36:59,170
provide any reactive power support for the
reactive power consumed by the series impedance
328
00:36:59,170 --> 00:37:04,089
of the line. Because, reactive power
consumed by the series impedance of the line
329
00:37:04,089 --> 00:37:10,420
is being provided by the shunt impedance
of the line itself.
330
00:37:10,420 --> 00:37:22,140
.And there is not reactive power flow, so
we can transmit the full capacity of the line
331
00:37:22,140 --> 00:37:29,710
in
this case. Now, this surge impedance value
332
00:37:29,710 --> 00:37:35,069
for different voltage levels of the line is
given
333
00:37:35,069 --> 00:37:36,170
in this table.
.
334
00:37:36,170 --> 00:37:42,279
These are typical values, these may not be
the exact values, but their approximately
335
00:37:42,279 --> 00:37:46,609
the
values which we have. So, for a rated 230
336
00:37:46,609 --> 00:37:51,490
KV line to line voltage three phase system,
Z c
337
00:37:51,490 --> 00:38:00,410
is about 380 ohms. And the surge impedance
loading in that case is going to be above
338
00:38:00,410 --> 00:38:10,500
140 mega watt. Similarly, same for 345 KV,
it is about 285 ohms and the surge
339
00:38:10,500 --> 00:38:19,140
impedance loading is four 120 mega watts.
For 500 KV this Z c, that is the characteristic
340
00:38:19,140 --> 00:38:26,319
impedance or surge impedance is equal to
250 ohms, and the surge impedance loading
341
00:38:26,319 --> 00:38:37,550
of the line is 1000 mega watt. And if we go
to 765 KV, then the value of Z c is again
342
00:38:37,550 --> 00:38:48,490
approximately around 257 ohms. This can vary
from 250 to 260 ohms and the surge impedance
343
00:38:48,490 --> 00:39:04,400
loading will be about 2280 or 2300 mega
watt. Now, what we see from here is if we
344
00:39:04,400 --> 00:39:10,730
go for higher and higher voltages, our surge
impedance loading keeps on increasing.
345
00:39:10,730 --> 00:39:17,940
And that is why when we need to transmit more
and more power, we go for higher and
346
00:39:17,940 --> 00:39:24,599
higher voltages. Because, if you see this
relationship the voltage, that is the power
347
00:39:24,599 --> 00:39:31,941
flowing is proportional to square of the voltage.
So, if you are doubling the voltage, you
348
00:39:31,941 --> 00:39:46,900
are able to transmit 4 times the power and
so on. Now, I will show you the characteristics
349
00:39:46,900 --> 00:39:50,839
of this long lossless transmission line.
350
00:39:50,839 --> 00:39:51,839
..
351
00:39:51,839 --> 00:40:00,940
When we have a surge impedance loading, the
voltage across the line all along from the
352
00:40:00,940 --> 00:40:13,060
sending end to receiving end is going to be
same. If the line is unloaded, then the sending
353
00:40:13,060 --> 00:40:28,890
end voltage V S or V no load at any distance
x, which we can write as at distance L is
354
00:40:28,890 --> 00:40:38,160
equal to cos beta x or cos beta L for sending
end voltage into V RNL. Now, in this case
355
00:40:38,160 --> 00:40:46,200
we find that the sending end voltage is going
to be less than the receiving end voltage
356
00:40:46,200 --> 00:40:51,009
And if we keep the sending end voltage as
1 per unit, then what we find at receiving
357
00:40:51,009 --> 00:40:59,170
end
voltage is going to be much higher. Similarly,
358
00:40:59,170 --> 00:41:06,010
if we do a short circuit of the transmission
line at the receiving end, then the receiving
359
00:41:06,010 --> 00:41:16,619
end voltage is going to be 0 and the sending
end voltage being 1 per unit. We will get
360
00:41:16,619 --> 00:41:20,460
the short circuit voltage at any point x is
equal
361
00:41:20,460 --> 00:41:31,869
to Z c sin beta x into I R short circuit.
So, in this case again the voltage, that we
362
00:41:31,869 --> 00:41:36,400
get will be following this profile coming
down
363
00:41:36,400 --> 00:41:44,609
to 0 at the receiving end, where the short
circuit has occurred and it will be V s at
364
00:41:44,609 --> 00:41:51,390
the
sending end. When the line is loaded at some
365
00:41:51,390 --> 00:42:01,240
other loading, then the surge impedance
loading, generally lines which are not very
366
00:42:01,240 --> 00:42:09,060
long can be loaded more at loads, which are
more than the surge impedance loading. That
367
00:42:09,060 --> 00:42:15,460
is full load is in general for not very long
lines will be larger than the surge impedance
368
00:42:15,460 --> 00:42:19,260
loading.
And therefore, the voltage for that case is
369
00:42:19,260 --> 00:42:25,609
going to be somewhere, in between these two
conditions which is the short circuit and
370
00:42:25,609 --> 00:42:28,640
the surge impedance loading. There is going
to
371
00:42:28,640 --> 00:42:36,130
be some voltage drop and this is what we will
find there is going to be some kind of a
372
00:42:36,130 --> 00:42:46,579
.voltage regulation for this system. So, again
from this we find one thing, that if we load
373
00:42:46,579 --> 00:42:52,930
the line to surge impedance, then the voltage
remains constant. And that is why when we
374
00:42:52,930 --> 00:42:58,809
are designing the transmission line, we try
to work on the bases of surge impedance
375
00:42:58,809 --> 00:43:02,819
loading.
Because that is the ideal condition, that
376
00:43:02,819 --> 00:43:10,690
we would like to have, that is voltage remains
constant all along the transmission line.
377
00:43:10,690 --> 00:43:17,910
Now, for these lossless lines, there is another
very important concept that we would like
378
00:43:17,910 --> 00:43:25,619
to discuss. And that is steady state stability
limit for a transmission line.
379
00:43:25,619 --> 00:43:26,619
.
380
00:43:26,619 --> 00:43:41,670
Now, if you look at the pi equivalent circuit,
then we can calculate this
381
00:43:41,670 --> 00:43:49,150
I R will be equal
to V S minus V R divided by this Z, which
382
00:43:49,150 --> 00:43:53,019
will be the current flowing in this, minus
the
383
00:43:53,019 --> 00:44:04,960
current flowing here which will be V R into
Y dash by 2. So, if you look at that, then
384
00:44:04,960 --> 00:44:08,710
we
have I R is equal to V S minus V R by Z dash
385
00:44:08,710 --> 00:44:16,451
minus Y dash by 2 into V R. Now,
substituting the value of V S and V R, V S
386
00:44:16,451 --> 00:44:24,230
we said is the sending end voltage with a
magnitude V S and angle delta.
387
00:44:24,230 --> 00:44:32,809
And V R is the magnitude of V R at an angle
0 degree, because that is chosen as
388
00:44:32,809 --> 00:44:39,640
reference. So, and Z dash in this case is
only j X dash, because we are considering
389
00:44:39,640 --> 00:44:43,619
an
lossless line. So, V S minus V R by Z dash,
390
00:44:43,619 --> 00:44:51,410
can be written as V S e j delta minus V R
divided by j X dash and Y dash by 2 can be
391
00:44:51,410 --> 00:45:04,009
written as j omega C dash l by 2 into V R.
Now, the receiving end power as we had talked
392
00:45:04,009 --> 00:45:08,200
earlier will be given by this S R is equal
393
00:45:08,200 --> 00:45:17,880
.to V R into I R conjugate, so V R into I
R conjugate is this terms. So, we take the
394
00:45:17,880 --> 00:45:27,690
conjugate of that. So, V S e j delta minus
V R by j X dash conjugate and the conjugate
395
00:45:27,690 --> 00:45:30,830
of
this will be negative of this. So, plus j
396
00:45:30,830 --> 00:45:37,849
omega C dash l divided by 2 into V R square,
because with this V R gets multiplied on this
397
00:45:37,849 --> 00:45:41,240
side also.
.
398
00:45:41,240 --> 00:45:53,119
So, this is equal to V R into V S e to the
power minus j delta minus V R minus j X dash,
399
00:45:53,119 --> 00:45:58,200
because we are taken the conjugates. So, this
minus of angle comes and minus j X dash
400
00:45:58,200 --> 00:46:07,330
for that is shifting of the angle will come.
So, this minus j X dash plus j omega C l
401
00:46:07,330 --> 00:46:17,490
divided by 2 into V R square. Now, if we expand
this, then we will get this as j V R into
402
00:46:17,490 --> 00:46:29,049
V S cos delta plus V R into V S sin delta
minus j V R square by X dash plus j omega
403
00:46:29,049 --> 00:46:32,339
C l
by 2 into V R square.
404
00:46:32,339 --> 00:46:40,550
There is a slight mistake here ((Refer Time:
46:36)), this should be C dash and this
405
00:46:40,550 --> 00:46:47,119
should be C dash. Now, if we look at the real
power only, that is we take the real part
406
00:46:47,119 --> 00:46:49,980
of
this when P is equal to P S is equal to P
407
00:46:49,980 --> 00:46:55,170
R, which is real part of the complex power
S R.
408
00:46:55,170 --> 00:47:04,049
Then, this will be equal to V S V R sin delta
by X dash all other terms are having j, that
409
00:47:04,049 --> 00:47:10,069
is their having imaginary, their imaginary
components. That is their terms for the
410
00:47:10,069 --> 00:47:15,160
reactive power.
So, P the real power at the sending end is
411
00:47:15,160 --> 00:47:22,210
equal to the real power at the receiving end.
And is given by V S into V R by X dash into
412
00:47:22,210 --> 00:47:28,549
sin delta, this will be watts or mega watts
depending on whether we are using volts or
413
00:47:28,549 --> 00:47:32,720
kilo volts. Therefore, P max is equal to V
S
414
00:47:32,720 --> 00:47:39,019
.V R by X dash, that is the maximum power
that can be transmitted over this transmission
415
00:47:39,019 --> 00:47:47,640
line will be equal to V S V R by X dash. Because,
this terms comes, this will be
416
00:47:47,640 --> 00:47:56,869
maximum when delta is 90 degree, that is sin
delta is equal to 1.
417
00:47:56,869 --> 00:47:57,869
.
418
00:47:57,869 --> 00:48:04,859
The relationship is plotted here and it shows
how with change in angle delta the real
419
00:48:04,859 --> 00:48:14,420
power varies. Now, this is a very, very important
relationship, this tells us one very
420
00:48:14,420 --> 00:48:20,890
important thing, that the maximum power transmitted
over the transmission line depends
421
00:48:20,890 --> 00:48:26,530
on the square of the voltage, because normally
V S and V R are will be very close to 1
422
00:48:26,530 --> 00:48:34,069
per unit. So, we will be talking in terms
of square of the voltage only, the rated voltage
423
00:48:34,069 --> 00:48:35,780
of
the transmission line.
424
00:48:35,780 --> 00:48:42,469
So, P max is proportional to the square of
the voltage. So, when we double voltage,
425
00:48:42,469 --> 00:48:50,079
system voltage we are able to transmit 4 times
the power. Also the maximum power
426
00:48:50,079 --> 00:48:55,130
which can be transmitted over the line is
inversely proportional to the reactance of
427
00:48:55,130 --> 00:48:59,161
the
line. Now, this is again very important. And
428
00:48:59,161 --> 00:49:06,380
we have seen earlier that, when we use
bundle conductors, then the inductance of
429
00:49:06,380 --> 00:49:14,480
the line gets reduced, because the effective
radius of the line gets increase considerably.
430
00:49:14,480 --> 00:49:21,400
And that produces this reactance and
therefore, use of bundle conductor will allow
431
00:49:21,400 --> 00:49:28,509
you to transmit more power on the
transmission line. There is another very important
432
00:49:28,509 --> 00:49:36,810
aspect is that, suppose we have a
synchronous machine which is connected to
433
00:49:36,810 --> 00:49:42,819
the system by means of a transmission line.
434
00:49:42,819 --> 00:49:54,599
.If this synchronous machine is supplying
certain amount of power to this system, let
435
00:49:54,599 --> 00:49:57,950
us
say it is working at this point somewhere.
436
00:49:57,950 --> 00:50:10,349
So, the delta angle for this will be there,
the power which is being transmitted. Suppose,
437
00:50:10,349 --> 00:50:35,890
here the power which is being transmitted
here, and the angle power angle or the voltage
438
00:50:35,890 --> 00:50:46,430
angle at the machine is delta 0. Now, if we
slowly start increasing the input to the
439
00:50:46,430 --> 00:50:54,619
machine, what will happen is the machine output
is remaining same, then input has been
440
00:50:54,619 --> 00:50:59,220
increase. So, it will experience some acceleration,
since it is synchronized it cannot
441
00:50:59,220 --> 00:51:02,260
accelerate as such as speed will not change
much.
442
00:51:02,260 --> 00:51:07,869
But, what will happen is the delta angle will
start increasing. If we keep doing this, what
443
00:51:07,869 --> 00:51:14,180
will happen is delta angle will keep increasing
and as delta angle increases, the electrical
444
00:51:14,180 --> 00:51:22,319
power output also increases. So, the mechanical
power and electrical power are
445
00:51:22,319 --> 00:51:27,980
becoming equal and the machine will be operating
in stable region. Suppose, we keep
446
00:51:27,980 --> 00:51:34,460
doing it very slowly, we keep on moving on
this, when we reach this point, when delta
447
00:51:34,460 --> 00:51:36,470
is
equal to 90 degrees.
448
00:51:36,470 --> 00:51:43,080
If we now increase the mechanical input to
the synchronous machine, what is going to
449
00:51:43,080 --> 00:51:51,210
happen is delta angle will increase, but electrical
power output is going to get reduced,
450
00:51:51,210 --> 00:51:58,200
which means that, there is going to be some
accelerating power available mechanical
451
00:51:58,200 --> 00:52:03,960
input is more electrical output is less. So,
delta angle will keep increasing, as delta
452
00:52:03,960 --> 00:52:10,150
angle
increases the accelerating power also increases.
453
00:52:10,150 --> 00:52:15,910
And which will further increase this
delta angle and machine will become unstable.
454
00:52:15,910 --> 00:52:22,450
That is what we see is, if we gradually increase
the input to the machine, the machine
455
00:52:22,450 --> 00:52:30,500
will adjust to that. And will increase it
is electrical power output till delta is equal
456
00:52:30,500 --> 00:52:35,700
to 90
degree. Beyond that, it will become unstable,
457
00:52:35,700 --> 00:52:42,940
because the mechanical input increase is
not going to bad accompanied by similar amount
458
00:52:42,940 --> 00:52:51,279
of increase in electrical output. And
that is why this P max at delta is equal to
459
00:52:51,279 --> 00:52:57,210
90 degree, which is given by V S into V R
by
460
00:52:57,210 --> 00:53:19,769
X dash is also called as steady state stability
limit.
461
00:53:19,769 --> 00:53:20,769
..
462
00:53:20,769 --> 00:53:50,359
Now, we can write this P of the real power
for a lossless line, P is equal to V S V R
463
00:53:50,359 --> 00:53:54,809
into
sin delta by Z c sin beta l, which is equal
464
00:53:54,809 --> 00:54:04,210
to V S V R by Z c into sin delta by sin twice
pi
465
00:54:04,210 --> 00:54:13,279
l by lambda; which, that is we are trying
to write the real power in terms of surge
466
00:54:13,279 --> 00:54:22,119
impedance loading. So, P is equal to V S by
V rated, we want to put everything in terms
467
00:54:22,119 --> 00:54:30,480
of per unit system. Therefore, V S by V rated
into V R by V rated into V square rated by
468
00:54:30,480 --> 00:54:37,809
Z c, that is this V rated V rated we will
get canceled here. So, V square rated by Z
469
00:54:37,809 --> 00:54:46,559
c into
sin delta divided by sin twice pi l by lambda,
470
00:54:46,559 --> 00:54:53,970
which is equal to V S per unit into V R per
unit into this term is surge impedance loading.
471
00:54:53,970 --> 00:54:58,180
So, into SIL into sin delta by sin twice pi
l
472
00:54:58,180 --> 00:55:02,339
by lambda, in terms of watts or mega watts.
473
00:55:02,339 --> 00:55:03,339
..
474
00:55:03,339 --> 00:55:10,440
Or P max is equal to V S per unit into V R
per unit into surge impedance loading divided
475
00:55:10,440 --> 00:55:18,809
by sin twice pi l by lambda, where lambda
is the wavelength of the line and l is the
476
00:55:18,809 --> 00:55:32,430
length of the line. That is we can see that,
the transmission line
477
00:55:32,430 --> 00:55:42,440
value of P max is going
to depend on V S V R. And it is going to,
478
00:55:42,440 --> 00:55:46,069
that is if you increase V S and V R, then
your
479
00:55:46,069 --> 00:55:53,089
maximum power transfer will increase. And
if you increase l your maximum power
480
00:55:53,089 --> 00:55:55,519
which can be transmitted will reduce.
.
481
00:55:55,519 --> 00:56:02,859
.And if we plot this curve, we will get this
curve like this, this is the theoretical one,
482
00:56:02,859 --> 00:56:10,569
where as the practical one will be much lower,
because we normally work at an angle of
483
00:56:10,569 --> 00:56:18,150
delta around 30, 35 degrees only not beyond
that. Normally between 20 to 30 degrees.
484
00:56:18,150 --> 00:56:26,460
And here we see that for short lines, that
is lines up to 80 kilo meters also, it is
485
00:56:26,460 --> 00:56:30,420
the
thermal limit which is the limiting factor
486
00:56:30,420 --> 00:56:37,480
for the transmission line load ability.
And this can be as high as 3 times as higher,
487
00:56:37,480 --> 00:56:45,859
but as we go beyond this 80 kilo meter
length. Then the line load ability keeps on
488
00:56:45,859 --> 00:56:52,980
increasing with the practical line
characteristics with delta around 30 degrees.
489
00:56:52,980 --> 00:57:01,859
We will get this kind of a characteristics
which shows that beyond 500 kilo meters we
490
00:57:01,859 --> 00:57:09,509
have to load the line at a limit, which is
lower than surge impedance loading. So, we
491
00:57:09,509 --> 00:57:19,690
see that if the line length increases our
power transfer capability also reduces. This
492
00:57:19,690 --> 00:57:27,680
is the typical surge impedance loading at
different voltages. And this is the thermal
493
00:57:27,680 --> 00:57:33,839
rating of the line. So, we see that for the
lines,
494
00:57:33,839 --> 00:57:42,650
the thermal rating for the most of the cases
is much higher than the surge impedance
495
00:57:42,650 --> 00:57:46,599
loading.
.
496
00:57:46,599 --> 00:58:02,910
The same concept we can derive for the lossy
lines, where instead of using the lossless
497
00:58:02,910 --> 00:58:10,249
line that is gamma being equal to beta, we
will use the gamma as a complex number.
498
00:58:10,249 --> 00:58:15,470
Then, we can write this A is equal the cos
hyperbolic gamma l, which is equal to A angle
499
00:58:15,470 --> 00:58:22,749
theta A, B is equal to Z dash is equal to
Z dash theta Z. Then, we can write I R in
500
00:58:22,749 --> 00:58:23,749
the
501
00:58:23,749 --> 00:58:29,880
.same way as V S minus A V R divided by B.
And substituting these values we will get
502
00:58:29,880 --> 00:58:34,950
S
R, that is equal to V R I R conjugate which
503
00:58:34,950 --> 00:58:38,849
is equal to V R into this I R conjugate.
.
504
00:58:38,849 --> 00:58:47,549
This terms, this will come out to be V R V
S Z dash e to the power j theta Z minus delta
505
00:58:47,549 --> 00:58:56,410
minus A V R square by Z dash e to the power
j theta Z minus theta A. And P R which is
506
00:58:56,410 --> 00:59:03,150
the real part of this complex power at the
receiving end will be equal to V R V S by
507
00:59:03,150 --> 00:59:06,990
Z
dash into cos theta Z minus delta minus A
508
00:59:06,990 --> 00:59:15,410
V R square by Z dash cos theta Z minus theta
A. That is what we are seeing is this is the
509
00:59:15,410 --> 00:59:23,769
term, which is getting subtracted from the
earlier term that we had for a lossless line.
510
00:59:23,769 --> 00:59:31,660
That is for a lossy line, the real power is
transfer is going to be less. And the maximum
511
00:59:31,660 --> 00:59:39,849
power that we can get, for a lossy
transmission line will be when theta Z is
512
00:59:39,849 --> 00:59:49,710
equal to delta.
Thank you. And in the next class, we will
513
00:59:49,710 --> 00:59:52,829
take up some problems on transmission lines.
514
00:59:52,829 --> 00:59:52,829
.