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Welcome to lesson 7 in Power System Analysis.
In this lesson, we will talk about
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Transmission Line Modeling.
.
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In this first we will talk about distributed
versus lumped parameter models. Then, we
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will talk about short line model, medium line
model and long line model. We will also
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discuss voltage regulation for transmission
lines. Well, as we have seen in our previous
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lessons, when we discussed the calculation
for transmission line parameters. We found
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that transmission line have in general three
parameters, the line resistance, the line
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inductance and a capacitance of the line to
neutral or ground. Now, as we had seen in
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the
previous lessons, these parameters are distributed
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all along the line. That is if we see a
line it will be something like this, where
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I have taken a small section of a line.
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..
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For this line of a length delta x, we have
the transmission line impedance. Or the series
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impedance of the transmission line, which
is given by z into delta x. It consists of
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the
resistance of the line and the inductance
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of the line. These will be per unit length
or per
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meter multiplied by delta x will give us the
value of series impedance, z delta x for the
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delta x length of the line.
Similarly, we will have a capacitance to ground
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for the line which will be y; and for delta
x of the line this shunt admittance is going
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to be y into delta x. There is capacitance
will
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provide us an admittance j omega c, which
will be equal to y. And for that will be per
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unit length of the line multiplying it by
delta x the length of the line we get y delta
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x.
Here I had also shown a conductance part for
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the line normally for 60 Hertz line or 50
Hertz line the power frequency line.
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This conductance is negligible and most of
the time we neglect this conductance. So,
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we
have resistance and inductance per unit length
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of the line. This will form the series
impedance per unit length of the line. And
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we have the capacitance per unit length of
the
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line, which forms the shunt admittance of
the line. Now, when we talk about modeling
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of
transmission lines, what we have to see is.
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If we use distributed parameter model of the
line, then the modeling of the transmission
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line becomes very complex. And it will result
into differential equations.
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..
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What we will see is that for sinusoidal voltage
and current waves on overhead line. The
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wavelength of the line is given by this lambda;
wavelength of the line is given by C by f,
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where, C is the velocity of light or the velocity
of propagation of the waves, current or
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voltage waves on the transmission line. For
overhead transmission line, this will be very
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nearly equal to the velocity of light.
Therefore, wavelength lambda is equal to C
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by f, where C is 3 into 10 to the power 8
meters per second. If we substitute this value
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of C and for frequency of 50 hertz, we get
lambda is equal to 6000 kilometers. That is
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the wavelength of a transmission line at 50
Hertz is of the order of 6000 kilometers.
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Since most of the transmission lines that
we
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have, are only a of a few 100 kilometer length.
Therefore, in most of the cases, we can neglect
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the distributed parameter model of the
line. That is the distributed effect can be
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neglected and it is sufficient to model the
line as
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lumped parameter model. That is if line length
is less than about 250 kilometers, then
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one can neglect the distributed effect of
the line parameters. And it is sufficient
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to
consider lumped parameter model for transmission
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lines. Now, once we have used a
lumped parameter model for the transmission
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line. Then, we can consider this
transmission line as a simple two-port network.
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..
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Where we have an input port, where we have
the voltage V S and I S. That is the sending
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end port, V S is the sending end voltage,
I S is the sending end current. And V R is
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the
receiving end voltage and I R is receiving
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end current. So, transmission line can be
visualized as a two-port network. For this
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two-port network, we can write down the
relationship between the sending and voltage
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and current. In terms of receiving and
voltage and current as V S is equal to A times
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V R plus B times I R volts.
And I S the sending and current is equals
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to C times V R, that is receiving an voltage.
Plus D times I R the receiving in a current.
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Of course the unit will be amplix. Now, this
is a general relationship for two-port network.
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It is valid for any two passive bilateral
linear two-port network. And one of the properties
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of this from network theory, we get
one of the properties for this network as
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A D minus B C is equal to 1. That is if we
are
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putting this relationship in a matrix form,
we get V S, I S is equal to ABCD into V R
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I R.
This is the general transmission line ABCD
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parameter model for any transmission line
with the relationship that A D minus B C is
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equal to 1.
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..
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Now, transmission lines which are less than
80 kilometers long or about 50 miles long,
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we neglect the effect of capacitance. Because,
the capacitive effect or the charging
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current is not very large and it is effect
can be neglected. Also these lines are generally
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medium voltage and low voltage lines, because
these are very short lines. And therefore,
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also the charging current, because the voltage
being small will also be much smaller.
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So, generally for line lengths less than 80
kilometer, which are generally medium voltage
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or low voltage lines, sometimes even high
voltage lines can be there. The capacitance
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for
these lines can be neglected, in that case
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the lumped parameter model for the
transmission line, will consist of only the
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resistance and inductance. That is the series
impedance of the line, the series resistance
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of the line and the series inductance of the
line.
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Here, the series impedance of the line Z is
equal to z into l, small z into l, where small
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z
is the per unit length series impedance. That
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is R plus j omega l per unit length, that
is
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resistance per unit length and inductance
per unit length, as we have seen we can
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calculate it for any line. So, Z is equal
to z into l, this is equal to R plus j omega
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l into l,
where R is the resistance per unit length
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of the line and j omega l is the reactance
per unit
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length of the line.
So, from this model we have V S the sending
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end voltage, I S the sending end current
only series impedance, that is here R and
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this j omega l or X. And I R is the receiving
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.end current and V R is the receiving end
voltage. From this circuit it is very clear
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I S and
IR will be equal. Therefore, we can write
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down the circuit equation for this transmission
line model as V S is equal to V R plus Z into
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I R.
.
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V S is going to be equal to this voltage plus
the drop in this transmission line impedance.
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Series impedance of the transmission line,
this will be equal to Z into I R. So, we have
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V
S is equal to V R plus Z into I R. And as
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we have all ready seen I S will be equal to
I R,
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therefore writing it in terms of matrix equation
for ABCD parameters. We have V S I S
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is equal to V R plus Z into I R and I S is
equal to 0 V R into plus 1 into I R. That
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is
ABCD parameters from here, if you see are
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A is equal to D is equal to 1 per unit.
And B is equal to Z ohms and C is equal to
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0 Siemens, that is the unit of A and D is
basically dimensionless, it will be in terms
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of per unit. Whereas, the unit of B is the
unit
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of impedance or ohms and unit of C is the
unit of admittance or Siemens. So, from this
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relationship for a short line, we have A and
D is equal to 1 B is equal to Z and C is equal
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to 0.
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00:12:35,590 --> 00:12:36,590
..
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The Phasor diagram for short line can be developed,
we can start with the voltage at the
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receiving end at full load. So, V R f l as
the reference voltage and we have a current
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I R
at full load, which is in this direction,
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that is lagging the voltage by some angle
theta.
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Now, with this we can find out using the circuit
model, voltage at the sending end, which
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will be equal to V R f l plus R into I R f
l.
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That is resistance, the drop in the resistance
due to the current, receiving end current
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at
full load. Plus the drop in the reactance,
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that is j X into I R f l, so the drop in the
reactance. So, when these drops are added,
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we get the voltage V S. That is what we see
is
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for lagging power factor load, where I R f
l is lagging V R f l by an angle theta. We
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get
V S which will generally be larger than V
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R, that is there is going to be a positive
drop of
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voltage for the system.
Similarly, if the current is of leading power
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factor, that is again choosing V R full load
as
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a reference voltage. And since, the current
I R full load is a leading power factor load.
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So, it leads this voltage by some angle theta.
Then, again by adding the voltage drops, we
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will get R times I R f l, which is the drop
in the resistance plus j X times I R f l,
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which is
the drop in the reactance. Or the inductive
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reactance this will give us the value V S,
and
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this V S as we see will be many times smaller
than V R.
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That is in case of leading power factor load,
the sending end voltage magnitude can be
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lowered than the receiving end voltage magnitude.
Of course, in case of short lines the V
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.S is also equal to V R at no load. Because,
if there is no current flowing in this, if
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there is
no current flowing in this ((Refer Time: 15:29)),
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then the voltage V R and V S will be
equal. So, V R at no load is equal to V S,
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that is what we have shown here. That is V
S is
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equal to V R at no load V R n l, this we required
for calculating regulation as we will see
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later.
Now, for line lengths or lines having a length
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between 80 to 250 kilometers, we call
these lines as medium Lang lines. These lines
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are generally high voltage or extra high
voltage lines. And for these lines, we cannot
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neglect the shunt capacitance or the
capacitive admittance of the line. The shunt
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admittance needs to be taken care of.
.
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What we do here is, again we are use a lumped
parameter model where the complete
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series impedance that it is total resistance
of the line, is lumped as one resistance.
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And
the total inductance of the line is again
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lumped together. So, we get this R plus j
omega l
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as the series impedance of the line. So, Z
the total series impedance is equal to z into
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l,
where small z is the series impedance of the
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line per unit length.
Now, the total shunt admittance of the line
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will be Y, which will be equal to small y
the
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shunt admittance of the line per unit length
into l. Now, what we do is normally, we
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divide this total shunt admittance into two
parts and put them at the two ends. So, we
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have shunt admittance Y by 2, put at the sending
end and another Y by 2 put at the
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receiving end.
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.So, this model it looks like a pi model pi
and therefore, we call the model as a nominal
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00:17:48,669 --> 00:17:52,930
pi
circuit for medium length line. Now, again
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here we have this as a two-port network with
V S, I S the sending end voltage end current
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and V R, I R the receiving end voltage end
current.
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.
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And we can use the ABCD parameter for this
as V S, from here we can see is equal to V
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00:18:14,830 --> 00:18:26,020
R plus Z into I R plus V R into Y by 2. Now,
if you look at this circuit, what we are
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00:18:26,020 --> 00:18:36,770
finding is, V S this voltage will be equal
to the drop in this. Now, what will be the
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00:18:36,770 --> 00:18:42,280
drop
in this, this drop will be equal to how much
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current is flowing in this part IR plus VR
into Y by 2.
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So, I R plus V R into Y by 2 is the current
this multiplied by Z will give us the drop
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in
this. So, that is what we are writing V R
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is the receiving end voltage plus the drop,
drop
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00:19:03,800 --> 00:19:15,890
is Z into I R plus V R Y by 2. So, this is
the current which is flowing in the charging
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00:19:15,890 --> 00:19:23,420
admittance of the line placed at the receiving
end. So, V S is equal to V R plus Z into I
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00:19:23,420 --> 00:19:30,770
R
V R Y by 2. If we rearrange this by combining
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00:19:30,770 --> 00:19:39,800
all the terms of V R together, we get this
as 1 plus Y Z by 2 into V R plus Z into I
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R. Similarly, we can write down the
relationship for the sending end current.
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00:19:48,150 --> 00:19:53,799
Now, sending end current sending end current
I
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00:19:53,799 --> 00:20:07,760
S is going to be equal to this I R plus the
current here. That is V R into Y by 2 plus
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00:20:07,760 --> 00:20:12,270
the
current here which is V S into Y by 2.
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00:20:12,270 --> 00:20:13,270
..
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00:20:13,270 --> 00:20:23,179
So, we are writing I S is equal to I R plus
V R into Y by 2 plus V S into Y by 2. Now,
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00:20:23,179 --> 00:20:32,040
again rearranging this by combining terms,
we write this as I R plus V R into Y by 2
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plus
V S. We are writing the relationship for V
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S from the previous equation here, V S is
equal to 1 plus Y Z by 2 into V R plus Z I
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R. So, substituting this value for V S, we
are
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00:20:47,780 --> 00:20:52,180
putting it here, so V S into Y by 2 we write
here.
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So, this is 1 plus Y Z by 2 V R plus Z I R
for V S into Y by 2. Now, again rearranging
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00:20:59,470 --> 00:21:04,570
it,
we will get this as Y into 1 plus Y Z by 4
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00:21:04,570 --> 00:21:10,750
into V R plus 1 plus Y Z by 2 into I R. This
is
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00:21:10,750 --> 00:21:19,480
the relationship for the current I S, in terms
of V R and I R. So, now using these
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00:21:19,480 --> 00:21:22,520
relationships, we can write down this.
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00:21:22,520 --> 00:21:23,520
..
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In terms of ABCD parameters as V S is equal
to 1 plus Y Z by 2 V R plus Z I R, I S is
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equal to 1 Y into 1 plus Y Z by 4 VR plus
1 plus Y Z by 2 I R, which shows that A is
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equal to D, that is A is equal to D is equal
to 1 plus Y Z by 2 this will be in pi unit.
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00:21:53,360 --> 00:21:58,570
B will
be equal to Z ohms and C will be equal to
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00:21:58,570 --> 00:22:08,840
Y into 1 plus Y Z by 4 Siemens. So, for a
medium length line, when we are using a nominal
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00:22:08,840 --> 00:22:14,299
pi equivalent for this medium length
line, we have the ABCD parameters.
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Given as A is equal to D is equal to 1 plus
Y Z by 2, B is equal to Z ohms and C is equal
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to Y into 1 plus Y Z by 4 Siemens. Now, we
will talk about voltage regulation. Now,
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voltage regulation, what we mean by voltage
regulation is basically in when there is no
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load on the line, the voltage on the line
will be somewhat higher. When we load the
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line,
generally the loads will be of a lagging power
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factor loads.
So, when we load these lines, then the voltage
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at the receiving end drops. And voltage
regulation is basically telling us how much
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00:23:03,470 --> 00:23:10,440
this voltage drop is, in terms of the rated
voltage or the sending end voltage. Therefore,
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we define this voltage regulation as a
percentage voltage regulation.
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00:23:15,500 --> 00:23:16,500
..
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00:23:16,500 --> 00:23:23,340
So, percent voltage regulation is defined
as V R no load minus V R full load divided
204
00:23:23,340 --> 00:23:28,790
by
V R full load into 100 percent. So, this is
205
00:23:28,790 --> 00:23:31,700
what we define, basically what we are trying
to
206
00:23:31,700 --> 00:23:41,070
say here is that suppose we have a line which
is loaded to it is full load value. And if
207
00:23:41,070 --> 00:23:43,970
the
load is suddenly thrown off what will be the
208
00:23:43,970 --> 00:23:50,310
voltage at the receiving end. How much rise
of voltage is there at the receiving end.
209
00:23:50,310 --> 00:23:58,640
So, this V R n l as showing what will be the
voltage of the receiving end? When the load
210
00:23:58,640 --> 00:24:08,929
is thrown off and V R f l is the voltage when
the full load is there on the transmission
211
00:24:08,929 --> 00:24:16,090
line. So, this percentage voltage regulation
gives us a very good parameter. Because,
212
00:24:16,090 --> 00:24:22,660
when we are designing a transmission line,
we cannot allow more than about 8 to 10
213
00:24:22,660 --> 00:24:26,600
percent voltage drop.
So, voltage regulation will tell us what is
214
00:24:26,600 --> 00:24:31,600
going to be the voltage drop, when the
transmission line is fully loaded, because,
215
00:24:31,600 --> 00:24:35,610
if the voltage drop is more, then we need
some
216
00:24:35,610 --> 00:24:43,910
compensation. We will also we had seen that
when the load is capacitive in nature or of
217
00:24:43,910 --> 00:24:51,530
leading power factor, then the receiving end
voltage can go higher than the sending end
218
00:24:51,530 --> 00:24:57,390
voltage. And this also needs to be checked
for the transmission line.
219
00:24:57,390 --> 00:25:03,419
Because, if the receiving end voltage becomes
very high, then it may endanger the
220
00:25:03,419 --> 00:25:12,049
transmission line insulators, because the
design voltage will be based on the nominal
221
00:25:12,049 --> 00:25:19,020
value or the rated value for the transmission
line. Therefore, if you see here for the short
222
00:25:19,020 --> 00:25:27,620
line, when we want to calculate the voltage
regulation we need to calculate the voltage
223
00:25:27,620 --> 00:25:28,620
at
224
00:25:28,620 --> 00:25:34,809
.the receiving end at no load. And as we have
seen earlier the voltage at the receiving
225
00:25:34,809 --> 00:25:41,400
end
at no load, for the short the line is nothing
226
00:25:41,400 --> 00:25:50,350
but, equal to V S.
That is when I R is equal to 0, that is I
227
00:25:50,350 --> 00:25:54,740
R and I S both are equal to 0, because I R
there is
228
00:25:54,740 --> 00:26:02,870
only I S and I R are equal. So, when the load
current is 0, that is the system is at no
229
00:26:02,870 --> 00:26:07,880
load,
then this drop is also equal to 0, so V R
230
00:26:07,880 --> 00:26:16,850
is equal to V S. For a medium length line,
that is
231
00:26:16,850 --> 00:26:30,350
the pi model that we have used, we have seen
that V S is equal to A V R plus Z into I R.
232
00:26:30,350 --> 00:26:39,950
Now, if we see this then when the I R is 0,
that is at no load we have V S is equal to
233
00:26:39,950 --> 00:26:45,520
A V
R plus Z into 0. So, V S will be equal to
234
00:26:45,520 --> 00:26:52,750
A V R only.
And therefore, we have the no load voltage
235
00:26:52,750 --> 00:27:01,450
at the receiving end, that is V R at no load
will be equal to A into V S. And therefore,
236
00:27:01,450 --> 00:27:16,809
we can calculate V R at no load from this
relationship. Now, we will take up the ABCD
237
00:27:16,809 --> 00:27:23,440
matrix for various kinds of line
configurations. That is various kinds of network
238
00:27:23,440 --> 00:27:27,370
models or parameters that we use.
.
239
00:27:27,370 --> 00:27:33,929
So, if we have a simple series impedance which
is the case for a short line. Then we have
240
00:27:33,929 --> 00:27:41,049
the ABCD parameters as A is equal to 1, B
is equal to Z, C is equal to 0 and D is equal
241
00:27:41,049 --> 00:27:45,260
to
1. If we have a simple capacitance, this is
242
00:27:45,260 --> 00:27:54,100
what we many times use when we have to
compensate for the reactive power drawn by
243
00:27:54,100 --> 00:28:00,870
the load. So, many times when the load
which is highly inductive, draws a large amount
244
00:28:00,870 --> 00:28:06,510
of reactive power. We need to
compensate this reactive power by putting
245
00:28:06,510 --> 00:28:09,049
capacitance at the receiving end.
246
00:28:09,049 --> 00:28:18,010
.The shunt capacitance at the receiving ends,
will then provide some reactive power. And
247
00:28:18,010 --> 00:28:24,880
this will compensate for the large amount
of reactive power drawn by the load or
248
00:28:24,880 --> 00:28:33,190
consumed by the load. So, capacitance or the
capacitor will generate reactive power at
249
00:28:33,190 --> 00:28:37,990
the receiving end itself. And compensate for
the large amount of reactive power drawn
250
00:28:37,990 --> 00:28:46,570
by the load. For simple shunt capacitance
the ABCD parameters will be A is equal to
251
00:28:46,570 --> 00:28:50,380
1,
B is equal to 0, C is Y and D is equal to
252
00:28:50,380 --> 00:28:53,400
Y.
.
253
00:28:53,400 --> 00:29:03,740
Similarly, instead of using a pi equivalent
model, if you use a T equivalent model, where
254
00:29:03,740 --> 00:29:12,700
what we do is we divide the total series impedance
into two equal parts. And lump the
255
00:29:12,700 --> 00:29:19,260
total capacitance as one capacitance and put
it at the centre of the line. Then in this
256
00:29:19,260 --> 00:29:28,270
model we have Z by 2 and Z by 2 here, and
total Y is here, in this kind of a model which
257
00:29:28,270 --> 00:29:34,760
is a T equivalent circuit for the transmission
line. The ABCD parameters will be given
258
00:29:34,760 --> 00:29:42,530
by A is equal to 1 plus Y Z 1, where this
is Z 1 and this is Z 2.
259
00:29:42,530 --> 00:29:54,940
So, 1 plus Y Z 1 and B will be equal to Z
1 plus Z 2 plus Y Z 1 into Z 2. So, and C
260
00:29:54,940 --> 00:29:59,520
will
be equal to Y and D will be equal to 1 plus
261
00:29:59,520 --> 00:30:04,809
Y Z 2. Again for this if you see A D minus
B
262
00:30:04,809 --> 00:30:16,520
C will be equal to 1, for the pi equivalent
circuit, we have all ready seen. Now, here
263
00:30:16,520 --> 00:30:20,230
for
this pi circuit we have Y 1 and Y 2 which
264
00:30:20,230 --> 00:30:23,059
may not be equal, whereas for the nominal
pi
265
00:30:23,059 --> 00:30:30,169
circuit we had both Y 1 Y 2 where equal.
That is the total shunt capacitance was divided
266
00:30:30,169 --> 00:30:38,240
equally and placed half of it was placed at
the sending end and half was placed at the
267
00:30:38,240 --> 00:30:41,870
receiving end. Now, for this general pi circuit,
268
00:30:41,870 --> 00:30:52,110
.if we calculate the ABCD parameters, we will
get A is equal to Y 1 plus Y 2 Z, B will be
269
00:30:52,110 --> 00:30:59,640
equal to Z, C will be equal to Y 1 plus Y
2 plus Y 1 Y 2 into Z and D will be equal
270
00:30:59,640 --> 00:31:04,520
to 1
plus Y 1 Z. Again here if you see A D minus
271
00:31:04,520 --> 00:31:14,950
B C will be equal to 1.
Now, as I said many times we use compensation
272
00:31:14,950 --> 00:31:22,669
like adding a capacitance at the sending
end. And then the line model will consist
273
00:31:22,669 --> 00:31:25,550
of two parts, one will be the transmission
line
274
00:31:25,550 --> 00:31:29,860
model, another will be for the compensating
equipment.
275
00:31:29,860 --> 00:31:30,860
.
276
00:31:30,860 --> 00:31:40,200
And in that case we have two cascade ABCD
systems. And for this complete system also
277
00:31:40,200 --> 00:31:50,669
we can find out the ABCD parameters, which
will be nothing but, this A B, A 1 B 1 C 1
278
00:31:50,669 --> 00:31:59,350
D 1 multiplied by A 2 B 2 C 2 D 2 will give
the composite ABCD of this. That is A for
279
00:31:59,350 --> 00:32:06,549
the composite system or cascaded network will
be A 1 A 2 plus B 1 C 2. B will be equal
280
00:32:06,549 --> 00:32:15,120
to A 1 B 2 plus B 1 D 2, C will be equal to
C 1 A 2 plus D 1 C 2 and D will be equal to
281
00:32:15,120 --> 00:32:22,610
C
1 B 2 plus D 1 D 2.
282
00:32:22,610 --> 00:32:30,840
So, in this way we had seen how we can model
the transmission line, for short lengths
283
00:32:30,840 --> 00:32:37,870
and medium lengths. But, when transmission
line lengths are very long, which is the case
284
00:32:37,870 --> 00:32:44,850
in extra high voltage lines. Or some of the
lines which are connecting the remote power
285
00:32:44,850 --> 00:32:50,559
generating sources, such as hydel plants to
the load centered. Then in that case it is
286
00:32:50,559 --> 00:32:53,990
no
longer possible to use a lumped parameter
287
00:32:53,990 --> 00:32:54,990
model.
288
00:32:54,990 --> 00:33:00,909
.And we will have to use the distributed parameter
model, or the more accurate model
289
00:33:00,909 --> 00:33:09,450
otherwise. The errors which will come, because
of using the lumped parameter model
290
00:33:09,450 --> 00:33:15,780
may not justify the simplification that we
have used.
291
00:33:15,780 --> 00:33:16,780
.
292
00:33:16,780 --> 00:33:25,809
So, here again we come back to the distributed
parameter model. So, for long line as I
293
00:33:25,809 --> 00:33:32,760
said, we will use this distributed parameter
model. We have taken a section of line which
294
00:33:32,760 --> 00:33:42,870
is delta x in length a small length delta
x. The series impedance of the line is z into
295
00:33:42,870 --> 00:33:46,150
delta
x, where z is the series impedance of the
296
00:33:46,150 --> 00:33:53,190
line the resistance plus the inductive reactance
of the line R plus j omega l per unit length.
297
00:33:53,190 --> 00:34:01,190
So, that is represented here as z into delta
x.
298
00:34:01,190 --> 00:34:11,409
And the total shunt admittance is y into delta
x, for this delta x part of the line where
299
00:34:11,409 --> 00:34:19,700
small y is the shunt admittance of the line
per unit length. As I said for the transmission
300
00:34:19,700 --> 00:34:26,520
line at power frequency 50 Hertz or 60 Hertz,
this part the conductance is negligible and
301
00:34:26,520 --> 00:34:38,500
this is generally neglected and only capacitance
is there. Now, we have taken the
302
00:34:38,500 --> 00:34:43,230
parameter x or the distance parameter from
the receiving end side.
303
00:34:43,230 --> 00:34:49,659
So, the receiving end voltage or the voltage
at a point at a distance x from the receiving
304
00:34:49,659 --> 00:35:00,150
end we write this voltage as V at distance
V x. And the current at this point as I at
305
00:35:00,150 --> 00:35:02,849
a
distance x form the receiving end that is
306
00:35:02,849 --> 00:35:10,561
I x. Similarly, the voltage of the line delta
x
307
00:35:10,561 --> 00:35:18,980
meter in this positive direction, that is
from receiving end towards the sending end
308
00:35:18,980 --> 00:35:22,890
will
be given by V into x plus delta x. That is
309
00:35:22,890 --> 00:35:26,060
voltage at x plus delta x from the receiving
end;
310
00:35:26,060 --> 00:35:34,460
current will be given by I at x plus delta
x, that is at current at x plus delta x distance
311
00:35:34,460 --> 00:35:46,310
.from the receiving end. Now, with this these
values now we will write the Kirchhoff's
312
00:35:46,310 --> 00:35:48,380
voltage law.
.
313
00:35:48,380 --> 00:35:58,619
Here, we have V plus V at x plus delta x will
be equal to V x. That is the voltage at
314
00:35:58,619 --> 00:36:07,470
distance x plus the voltage drop, that is
the voltage drop here will be I x into z delta
315
00:36:07,470 --> 00:36:12,820
x.
So, the voltage at this end, that is V at
316
00:36:12,820 --> 00:36:16,750
x plus delta x is going to be equal to V at
x plus
317
00:36:16,750 --> 00:36:24,960
the voltage drop, which is I at x into z delta
x. That is what we have written here, V at
318
00:36:24,960 --> 00:36:28,619
x
plus delta x is equal to V at x plus z into
319
00:36:28,619 --> 00:36:38,770
delta x into I x I at x.
Now, here as I said earlier z is the series
320
00:36:38,770 --> 00:36:41,840
impedance per unit length, which will be R
plus
321
00:36:41,840 --> 00:36:50,040
j omega L ohms per meter. And y is G plus
j omega C Siemens per meter, where G for
322
00:36:50,040 --> 00:37:01,140
power frequency or transmission lines we have
is almost 0 or negligible. Now, from this
323
00:37:01,140 --> 00:37:09,890
relationship of voltage at distance x plus
delta x, in terms of voltage at distance x,
324
00:37:09,890 --> 00:37:13,849
if you
rearrange, this we will get as V at x plus
325
00:37:13,849 --> 00:37:20,960
delta x minus V at x divided by delta x.
This delta x term we have taken on this side
326
00:37:20,960 --> 00:37:23,690
and this V x, we have taken on this side,
so
327
00:37:23,690 --> 00:37:30,320
this becomes negative. And this delta x comes
in the denominator. So, V at x plus delta
328
00:37:30,320 --> 00:37:33,660
x
minus V at x divided by delta x is equal to
329
00:37:33,660 --> 00:37:44,310
z into I at x. Now, if we take a limit with
delta
330
00:37:44,310 --> 00:37:55,020
x tending to 0, then we can write this as
d V x by d x is equal to z I x. That is derivative
331
00:37:55,020 --> 00:38:03,060
of voltage at x will be equal to z into I
at x.
332
00:38:03,060 --> 00:38:04,060
..
333
00:38:04,060 --> 00:38:13,140
Similarly, writing the Kirchhoff's current
law. We will have I at x plus delta x will
334
00:38:13,140 --> 00:38:19,390
be
equal to I at x plus y into delta x into V
335
00:38:19,390 --> 00:38:26,730
at x plus delta x. That is if we look at this
((Refer
336
00:38:26,730 --> 00:38:39,020
Time: 38:27)) I at x plus delta x is going
to be equal to I at x. Plus the current which
337
00:38:39,020 --> 00:38:43,089
is
flowing in this admittance which will be equal
338
00:38:43,089 --> 00:38:51,830
to y into delta x multiplied by voltage at
this point which is V at x plus delta x.
339
00:38:51,830 --> 00:39:02,200
That is why we have got I at x plus delta
x is equal to I x, the current at x plus y
340
00:39:02,200 --> 00:39:06,240
into
delta x the admittance multiplied by V at
341
00:39:06,240 --> 00:39:09,540
x plus delta x. Again rearranging by taking
this
342
00:39:09,540 --> 00:39:16,580
I x on this side and dividing by delta x,
we have I at x plus delta x minus I at x divided
343
00:39:16,580 --> 00:39:24,450
by delta x is equal to y into V at x plus
delta x. Now, taking limit delta x tending
344
00:39:24,450 --> 00:39:30,710
to 0, we
can write this as equal to d I x by diagnosis,
345
00:39:30,710 --> 00:39:40,960
that is first derivative of current at x with
respect to x is equal to y into V at x.
346
00:39:40,960 --> 00:39:52,060
Now, what we have is if we see our previous
equation for V. If we differentiate it again
347
00:39:52,060 --> 00:40:02,690
with respect to x, then what we will get is
d 2 V x by d x 2 is equal to z d I x by d
348
00:40:02,690 --> 00:40:10,920
x, d 2
V x by d x 2 is equal to z d I x by d x. Now,
349
00:40:10,920 --> 00:40:18,000
substituting the value of d I x by d x from
this point or this equation, we get this is
350
00:40:18,000 --> 00:40:27,470
equal to z into y into V x. We can write this
as d
351
00:40:27,470 --> 00:40:37,630
2 V x by d x 2 minus z y V x is equal to 0.
That is what we have done is taken this term
352
00:40:37,630 --> 00:40:47,140
on this side. So, d 2 V x by d x 2 minus z
y V x is equal to 0.
353
00:40:47,140 --> 00:40:48,140
..
354
00:40:48,140 --> 00:40:56,000
Now, this a second order homogeneous equation.
And we know the solution for this
355
00:40:56,000 --> 00:41:03,650
equation can be obtained in this form, where
V x will be given by A 1 e to the power
356
00:41:03,650 --> 00:41:10,589
gamma x plus A 2 e to the power minus gamma
x volts. So, this will be a general
357
00:41:10,589 --> 00:41:20,230
solution for this second order differential
equation, where gamma is equal to root z y
358
00:41:20,230 --> 00:41:26,980
and
the unit for this will be per meter.
359
00:41:26,980 --> 00:41:35,740
Now, again differentiating this equation,
we can write d V x by d x differentiating
360
00:41:35,740 --> 00:41:38,530
this
equation with respect to x. We have got d
361
00:41:38,530 --> 00:41:45,560
V x by d x is equal to gamma A 1 e to the
power gamma x minus gamma A 2 e to the power
362
00:41:45,560 --> 00:41:53,160
minus gamma x. And this will be
equal to z into I x, because d V x by d x
363
00:41:53,160 --> 00:42:00,950
is z into I x. Therefore, we can write that
I x is
364
00:42:00,950 --> 00:42:10,119
equal to the this term divided by z.
365
00:42:10,119 --> 00:42:11,119
..
366
00:42:11,119 --> 00:42:19,500
So, I x is equal to A 1 e to the power gamma
x minus A 2 e to the power minus gamma x
367
00:42:19,500 --> 00:42:26,010
divided by z. And the gamma which was here
multiplied with A 1 and A 2 that we have
368
00:42:26,010 --> 00:42:33,930
taken on in the denominator, so this is z
divided by gamma. Therefore, we can write
369
00:42:33,930 --> 00:42:39,099
I at
x is equal to A 1 e to the power gamma x minus
370
00:42:39,099 --> 00:42:47,829
A 2 e to the power minus gamma x
divided by Z c. Now, this term z by gamma
371
00:42:47,829 --> 00:42:51,430
we are writing this as Z c, Z c is called
a
372
00:42:51,430 --> 00:42:55,670
characteristic impedance of the transmission
line.
373
00:42:55,670 --> 00:43:03,740
Now, if you see Z c as we have put here is
z by gamma. This is equal to z divided by
374
00:43:03,740 --> 00:43:11,520
gamma is square root of z y and therefore,
this is equal to square root of z by y. That
375
00:43:11,520 --> 00:43:16,800
is
square root of series impedance by shunt admittance.
376
00:43:16,800 --> 00:43:25,560
And the unit for this is ohms and
therefore, this Z c is termed as the characteristic
377
00:43:25,560 --> 00:43:37,170
impedance of a transmission line. Now,
we need to find out the constants of integration
378
00:43:37,170 --> 00:43:47,270
that is this A 1 and A 2. So, for this what
we do is, we use the boundary conditions.
379
00:43:47,270 --> 00:43:48,270
..
380
00:43:48,270 --> 00:43:57,650
So, here using the boundary, that is at x
is equal to 0, V x is equal to V R. So, at
381
00:43:57,650 --> 00:44:03,800
0 V at 0
is equal to V R and I at 0 is equal to I R,
382
00:44:03,800 --> 00:44:08,800
therefore V R will be equal to A 1 plus A
2.
383
00:44:08,800 --> 00:44:18,590
That is if we substitute in this relationship
x is equal to 0, then we get V 0 which is
384
00:44:18,590 --> 00:44:22,430
equal
to V R is equal to A 1 plus A 2 as these terms
385
00:44:22,430 --> 00:44:35,320
will become 1. So, V R is equal to A 1
plus A 2 and again substituting it in this
386
00:44:35,320 --> 00:44:40,300
((Refer Time: 44:36)) I x here x is equal
to 0 in
387
00:44:40,300 --> 00:44:50,470
these we will get I x is equal to A 1 minus
A 2 by Z c, A 1 minus I R is equal to A 1
388
00:44:50,470 --> 00:45:01,710
minus A 2 by Z c. From these two relationships
A 1 plus A 2 is equal to V R and A 1
389
00:45:01,710 --> 00:45:09,330
minus A 2 by Z c is equal to I R, we can calculate
A 1 and A 2.
390
00:45:09,330 --> 00:45:10,330
.
391
00:45:10,330 --> 00:45:17,900
.And A 1 comes out to be equal to VR plus
Z c into I R by 2 and A 2 will come out to
392
00:45:17,900 --> 00:45:24,770
be
equal to V R minus Z c into I R by 2. Now,
393
00:45:24,770 --> 00:45:27,100
substituting the values of this A 1 and A
2, in
394
00:45:27,100 --> 00:45:34,360
the relationship for V x, we can get V x is
equal to V R plus Z c into I R by 2. That
395
00:45:34,360 --> 00:45:39,980
is A
1 into e to the power gamma x plus V R minus
396
00:45:39,980 --> 00:45:48,250
Z c into I R by 2, this is A 2 into e to the
power minus gamma x.
397
00:45:48,250 --> 00:45:49,250
.
398
00:45:49,250 --> 00:45:55,700
And similarly putting it for the value of
A 1 A 2 for I x relationship, we get I x is
399
00:45:55,700 --> 00:46:01,250
equal to
V R plus Z c into I R by 2 into Z c e to the
400
00:46:01,250 --> 00:46:09,250
power gamma x minus V R minus Z c I R
divided by 2 into Z c into e to the power
401
00:46:09,250 --> 00:46:18,440
minus gamma x. Now, if we again rearrange
the
402
00:46:18,440 --> 00:46:27,140
equations for V x and I x, that is in terms
of V R and I R, then combining all the terms
403
00:46:27,140 --> 00:46:31,470
of
V R together and terms of I R together. Then
404
00:46:31,470 --> 00:46:37,820
we can write from this equation take all the
terms which are having V R together and take
405
00:46:37,820 --> 00:46:40,099
all the terms which are having I R
together.
406
00:46:40,099 --> 00:46:46,070
Then, we can write V x is equal to e to the
power gamma x plus e to the power minus
407
00:46:46,070 --> 00:46:53,089
gamma x divided by 2 into V R plus Z c e to
the power gamma x minus e to the power
408
00:46:53,089 --> 00:47:02,589
minus gamma x by 2 into I R. And I x will
be equal to 1 by Z c into e to the power
409
00:47:02,589 --> 00:47:08,750
gamma x minus e to the power minus gamma x
by 2 V R plus e to the power gamma x
410
00:47:08,750 --> 00:47:15,329
plus e to the power minus gamma x by 2 into
I R, this is the relationship that we will
411
00:47:15,329 --> 00:47:16,329
get.
412
00:47:16,329 --> 00:47:24,500
.Now, here we have got a relationship of V
at a distance x, in terms of receiving end
413
00:47:24,500 --> 00:47:30,960
voltage and current. Similarly, current at
any distance x, in terms of receiving end
414
00:47:30,960 --> 00:47:35,960
voltage and current.
.
415
00:47:35,960 --> 00:47:47,220
Now, from this we can ((Refer Time: 47:46))
this term e to the power gamma x plus e to
416
00:47:47,220 --> 00:47:56,810
the power minus gamma x by 2 is equal to cos
hyperbolic gamma x. And e to the power
417
00:47:56,810 --> 00:48:02,191
gamma x minus e to the power gamma x by 2
will be equal to sin hyperbolic gamma x.
418
00:48:02,191 --> 00:48:13,270
Therefore, we have this cos hyperbolic gamma
x into V R plus Z c into sin hyperbolic
419
00:48:13,270 --> 00:48:21,250
gamma x into I R, that is what we get here.
That is voltage at any point in the transmission
420
00:48:21,250 --> 00:48:29,940
line can be found, in terms of the
hyperbolic functions and the receiving end
421
00:48:29,940 --> 00:48:34,300
voltage and currents. Similarly, the current
at
422
00:48:34,300 --> 00:48:39,410
any point in the transmission line any distance
x from the receiving end, can be found
423
00:48:39,410 --> 00:48:47,320
out. Similarly, using the relationship I x
is equal to 1 by Z c sin hyperbolic gamma
424
00:48:47,320 --> 00:48:52,050
x into
V R plus cos hyperbolic gamma x into I R.
425
00:48:52,050 --> 00:49:00,520
Now, if you look at this, this is giving us
a relationship of voltage and current, at
426
00:49:00,520 --> 00:49:03,760
any
point x, in terms of receiving end voltage
427
00:49:03,760 --> 00:49:06,200
and current. Therefore, we can write this
in
428
00:49:06,200 --> 00:49:17,280
terms of ABCD parameters as V x I x is equal
to A x B x C x D x V R I R, where
429
00:49:17,280 --> 00:49:24,730
ABCD, A is equal to cos hyperbolic gamma x,
B is equal to Z c sin hyperbolic gamma x,
430
00:49:24,730 --> 00:49:30,540
C is equal to 1 by Z c sin hyperbolic gamma
x. And D is equal to cos hyperbolic gamma
431
00:49:30,540 --> 00:49:36,070
x, again you will see A D minus B C is equal
to 1.
432
00:49:36,070 --> 00:49:44,300
.Now, since most of the time we are interested
only in the terminal conditions. That is
433
00:49:44,300 --> 00:49:49,619
conditions at the sending end and the receiving
end, not anywhere in between the lines.
434
00:49:49,619 --> 00:49:57,710
So, we need to just substitute l instead of
x and then we will get the sending end voltage
435
00:49:57,710 --> 00:49:58,960
and sending end current.
.
436
00:49:58,960 --> 00:50:07,520
So, this relationship can also be put in the
form of sending end and receiving end voltage
437
00:50:07,520 --> 00:50:12,090
end current, where x is replaced by l.
.
438
00:50:12,090 --> 00:50:19,520
.That is at sending end x is equal to l, therefore
we can write this relationship as V S I S
439
00:50:19,520 --> 00:50:25,060
is
equal to ABCD V R I R. And where A is equal
440
00:50:25,060 --> 00:50:33,200
to D is cos hyperbolic gamma l per unit,
B is equal to Z c sin hyperbolic gamma l,
441
00:50:33,200 --> 00:50:37,820
C is equal to 1 by Z c sin hyperbolic gamma
l
442
00:50:37,820 --> 00:50:47,339
in Siemens. So, again we have got the ABCD
parameters for the transmission line and
443
00:50:47,339 --> 00:50:55,480
with for a long transmission line, the only
thing here, what we are seeing is we have
444
00:50:55,480 --> 00:50:59,030
to
use hyperbolic trigonometric functions.
445
00:50:59,030 --> 00:51:06,780
That is ABCD parameters are in terms of hyperbolic
trigonometric functions, not just Y
446
00:51:06,780 --> 00:51:18,609
and Z as we had for lumped parameter models.
So, with this we stop today and we will
447
00:51:18,609 --> 00:51:27,890
be continuing this line long transmission
line model. In the next class, where we will
448
00:51:27,890 --> 00:51:34,700
talk
about the velocity of propagation the speed,
449
00:51:34,700 --> 00:51:37,650
the phase angles and another things. Also
we
450
00:51:37,650 --> 00:51:42,490
will talk about regulation and efficiency
of transmission lines.
451
00:51:42,490 --> 00:51:43,490
Thank you.
Preview of Next Lecture
452
00:51:43,490 --> 00:52:02,450
Lecture - 08
Transmission Line Modeling Long Line (Contd.)
453
00:52:02,450 --> 00:52:08,900
Welcome to lesson 8 on Power System Analysis.
This lesson is a continuation on
454
00:52:08,900 --> 00:52:15,120
Transmission Line Modeling, specially Modeling
of the Long Line. Now, if you
455
00:52:15,120 --> 00:52:21,680
remember in the previous lesson that is lesson
7, we talked about distributed parameter
456
00:52:21,680 --> 00:52:29,549
model for long transmission lines. That is
lines which are longer than 250 kilometers.
457
00:52:29,549 --> 00:52:30,549
..
458
00:52:30,549 --> 00:52:42,510
For these lines we said that the line voltage
V x, at a distance x from the receiving end
459
00:52:42,510 --> 00:52:49,000
is
given by cos hyperbolic gamma x into V R plus
460
00:52:49,000 --> 00:52:59,580
Z c sin hyperbolic gamma x into I R,
where gamma is called the propagation constant,
461
00:52:59,580 --> 00:53:08,640
x is the distance of the point from the
receiving end, V R and I R are the voltages
462
00:53:08,640 --> 00:53:14,670
and currents at the receiving end. Similarly,
the current at a distance x from the receiving
463
00:53:14,670 --> 00:53:23,220
end, I x is equal to 1 by Z C sin hyperbolic
gamma x into V R plus cos hyperbolic gamma
464
00:53:23,220 --> 00:53:30,359
x into I R, where Z c is the characteristic
impedance of the transmission line.
465
00:53:30,359 --> 00:53:39,310
And it is given by square root of Z by Y,
where Z is the series impedance of the
466
00:53:39,310 --> 00:53:46,790
transmission line per unit length. And Y is
the shunt admittance of the transmission line
467
00:53:46,790 --> 00:53:57,920
per unit length. As we see these models these
equations involve hyperbolic functions.
468
00:53:57,920 --> 00:54:03,589
Now, since we have been writing all these
transmission line equations, in terms of
469
00:54:03,589 --> 00:54:11,250
ABCD parameters, we can write this equation
also in those terms. So, in matrix form we
470
00:54:11,250 --> 00:54:27,670
can use this as V x, I x is equal to A x B
x C x D x and V R and I R this form, where
471
00:54:27,670 --> 00:54:32,430
this
relationship is in terms of voltage and current
472
00:54:32,430 --> 00:54:37,859
at any distance x from the receiving end.
473
00:54:37,859 --> 00:54:38,859
..
474
00:54:38,859 --> 00:54:50,430
Now, here if you see A x is equal to D x and
that is equal to cos hyperbolic gamma x in
475
00:54:50,430 --> 00:55:01,400
per unit. That is if you see this relationship
this is your A x and this is your B x, this
476
00:55:01,400 --> 00:55:07,859
is C
x and this is D x. So, A x is equal to D x
477
00:55:07,859 --> 00:55:11,990
is equal to cos hyperbolic gamma x B x is
equal
478
00:55:11,990 --> 00:55:22,500
to Z c sin hyperbolic gamma x, V x is equal
to Z c sin hyperbolic gamma x. C x is equal
479
00:55:22,500 --> 00:55:31,000
to 1 by Z c sin hyperbolic gamma x and D x
as we have all ready seen is equal to A x.
480
00:55:31,000 --> 00:55:40,849
So, this is the model, where A and D are basically
dimensionless, B has a dimension of
481
00:55:40,849 --> 00:55:49,770
impedance and C has a dimension of Siemens
that is admittance. Now, normally we are
482
00:55:49,770 --> 00:55:55,390
interested only in the terminal conditions.
That is the sending end voltages end currents
483
00:55:55,390 --> 00:56:00,569
and the receiving end voltages end current.
Rather than, the voltage end current at any
484
00:56:00,569 --> 00:56:06,520
intermediate point on the transmission line.
Therefore, we can find out the voltage end
485
00:56:06,520 --> 00:56:11,600
currents at the sending end, in terms of voltage
end current at the receiving end by
486
00:56:11,600 --> 00:56:17,130
substituting x is equal to l, where l is total
line length.
487
00:56:17,130 --> 00:56:18,130
..
488
00:56:18,130 --> 00:56:25,109
See this relationship, the voltage that is
the power flowing is proportional to square
489
00:56:25,109 --> 00:56:29,150
of
the voltage. So, if you are doubling the voltage,
490
00:56:29,150 --> 00:56:43,710
you are able to transmit 4 times the
power and so on. Now, I will show you the
491
00:56:43,710 --> 00:56:48,039
characteristics of this long lose less
transmission line.
492
00:56:48,039 --> 00:56:49,039
.
493
00:56:49,039 --> 00:56:58,140
When we have a surge impedance loading. The
voltage across the line all along from the
494
00:56:58,140 --> 00:57:10,349
sending end to receiving end is going to be
same. If the line is unloaded, then the sending
495
00:57:10,349 --> 00:57:15,619
end voltage.
496
00:57:15,619 --> 00:57:16,619
..
497
00:57:16,619 --> 00:57:27,890
V S or V no load at any distance x which we
can write as distance l is equal to cos beta
498
00:57:27,890 --> 00:57:31,160
x
or cos beta l, for sending end voltage into
499
00:57:31,160 --> 00:57:40,530
V RNL. Now, in this case we find that the
sending end voltage is going to be less than
500
00:57:40,530 --> 00:57:46,640
the receiving end voltage. And if we keep
the sending end voltage as 1 per unit, then
501
00:57:46,640 --> 00:57:52,670
what we find at receiving end voltage is going
to be much higher.
502
00:57:52,670 --> 00:59:50,900
Thank you and in the next class, we will take
up some problems on transmission lines.
503
00:59:50,900 --> 00:59:50,900
.