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Welcome to lesson 9, in Power System Analysis.
In this lesson, we will discuss about
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Transmission Line Steady State Operation.
Well, when we talk about the steady state
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operation on transmission line. What we really
mean is, how the line is going to perform,
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when we want to transmit certain amount of
power through it. So, here what we will do
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is, we will discuss and derive the relationship
between the voltage currents and the
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power, which is transmitted over the transmission
line.
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In the earlier lessons, we had derived the
models for the transmission line. Mostly,
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we
had used the A, B, C, D parameter model and
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the pi equivalent circuit module. Here, we
will try to take help of the A, B, C, D model
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and develop the power transfer relationship
on the transmission line, using these A, B,
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C, D parameters. First, we will consider two
bus power systems, the simplest one, a generating
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station supplying power to a load,
through a transmission line.
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.
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The system as shown in this figure has this
generating station, which is connected to
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a
bus power the voltage of this bus. Bus power
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is V s, because we call this bus as the
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.sending end bus. The current flowing from
the generator into this bus is I s. And then
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this
bus sending end bus is connected to the receiving
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end bus at which the loads are
connected, this by means of a transmission
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line.
The transmission line as we are said earlier
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is characterized by the A, B, C, D
parameters. So, for this system V r is the
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voltage at the receiving end. And normally,
what we assume is that, we take this receiving
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end voltage phase angle as 0 degrees,
whereas the sending end voltage phase angle
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as angle delta. The sending end power, the
complex power at the sending end S s is equal
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to P s plus j Q s. Where, P s is the real
power and Q s is the reactive power at the
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sending end. Similarly, S r is the complex
power at the receiving end, where P r is the
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real power at the receiving end and Q r is
the
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reactive power at the receiving end.
.
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Now, we will try to develop the power flow
relationship for this system. We have
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already develop the relationship using the
A, B, C, D. The parameters for the
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transmission line as V s, I s is equal to
A, B, C, D, V R, I R. That is the resending
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end
voltage and current, in terms of receiving
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end voltage and current, with A, B, C, D
parameters for the transmission line.
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If we solve these equations, then in terms
of the sending end and receiving end currents.
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Then, we will get the receiving end current
I r is equal to 1 by B, V s minus A by B,
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V r
and I s is equal to D by B, V s minus 1 by
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B, V r. We have already seen, that A is equal
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.to D. Therefore, we will replace this D by
A and we will have I s is equal to A by B
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into
V s minus 1 by B into V r. So, once we have
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found out these currents, the receiving end
and sending end currents, we can write down
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the relationship for power.
.
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Now, here, as I said earlier, the receiving
end voltage V r, we are assuming it to have
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a
magnitude V r and the angle 0. That is, we
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are making this receiving end voltage as a
reference phase. And V s is equal to magnitude
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of V s and an angle of delta, which is the
angle by which the sending end voltage leads
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the receiving end voltage. Also, we have D
is equal to A. And A, since we know is a complex
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number.
We are writing this A as a magnitude of A
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with a phase angle alpha. And B, as we know
is comes out to be equal to Z for a pi equivalent
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circuit. So, this is also a complex
number and we writing B as magnitude of B
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with an angle beta. So, once we substitute
these values for I s and I r. then we have
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I r is equal to V s, magnitude of V s by
magnitude of B, with an angle delta minus
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beta. And minus magnitude of A into
magnitude of V r divided by magnitude of B,
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with an angle alpha minus beta.
We are just substituting the phasar values
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for A, B, C, D and V s, V r in this relationship.
Similarly, we can get the value of I s as
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A V s by B, all these magnitudes with an angle
alpha plus delta minus beta minus V r by B
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with an angle of minus beta. Now, you try
to
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find out the conjugates for these currents.
Because, we know that the complex power s
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is
given by V I conjugate.
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.So, we will get the conjugates of these currents
as I r conjugate will be equal to V s by B
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with an angle beta minus delta. This angle,
what we had earlier is now replaced by the
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negative of it, when we take the conjugate.
So, instead of delta minus beta, we have now
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beta minus delta here. Similarly, this other
term is minus A V r by B and the angle
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instead of alpha minus beta as we had for
I r. For I r conjugate, this angle becomes
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beta
minus alpha.
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Similarly, we find out the conjugate for the
sending end current I s. This is equal to
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A V
s by B, angle beta minus alpha minus delta.
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Again, we take the negative of the angle that
we had for I s minus V r by B, with an angle
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beta.
.
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Because, earlier we had here, we had I s here,
we had an angle of minus beta. Therefore,
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I s conjugate is given by this relationship
A V s by B, angle beta minus alpha minus delta
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minus magnitude of V r by magnitude of B angle
beta. So, once we have got these
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conjugate of the currents at the receiving
end and sending end. We can write down the
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power relations now.
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..
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The power, which is received at the receiving
end or received by the load S r is equal to
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P r plus j Q r, where P r is the real power
and Q r is the reactive power and this is
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equal
to V r into I r conjugate. And since, we have
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already calculated the value of I r conjugate
for getting the value of S r. We need to multiply
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that relationship by V r. Since, we know
V r is equal to magnitude of V r an angle
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0.
So, we multiply that relationship by V r with
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an angle 0. And we get this as V r angle 0
multiplied to V s by B angle beta minus delta
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minus A V r by B angle beta minus alpha.
This can be written as V s into V r, magnitude
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of V s into V r by magnitude of B, with an
angle beta minus delta. Minus magnitude of
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A into magnitude of V r square divided by
B
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with an angle beta minus alpha.
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..
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Similarly, the power at from the sending end,
that is the power, which is sent by the
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generator is S s. And that is equal to P s
plus j Q s, the real power of P s, from the
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generator and the reactive power Q s from
the generator side. This is equal to V s into
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I s
conjugate. So, again, what we had already
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calculated this I s conjugate. We multiply
it by
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V s and V s; we know is magnitude V s with
an angle delta.
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So, me multiply this, then we get V s angle
delta, multiplied by I s, which is equal to
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A,
magnitude of A into magnitude of V s divided
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by magnitude of V, with an angle beta
minus alpha minus delta. Minus of magnitude
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of V r divided by magnitude of B, with an
angle beta, which when we simplify this whole
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thing, turns out to be magnitude of A into
magnitude of V s square, divided by magnitude
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of B with an angle beta minus alpha.
Because this delta which is plus delta, will
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cancel out with this minus delta, minus V
r
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magnitude of V r into magnitude of V s divided
by magnitude of B into angle beta plus
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alpha.
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..
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Now, if we take out the real and reactive
parts, from this complex powers S r and S
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s.
That is we take out the real and reactive
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real and imaginary part of these equations,
these
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complex equations. Then, we will get the real
power and reactive power. So, P r will
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come out to be V s into V r by B cos of beta
minus delta minus A into V r square by B
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into cos of beta minus alpha. And Q r, the
reactive power at the receiving end will be
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equal to V s into V r by B into sin beta minus
delta minus A into V r square by B sin beta
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minus alpha. Similarly, the real power at
the sending end will become equal to P s.
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And
P s is equal to A V s square by B into cos
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beta minus alpha minus V r into V s by B,
cos
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beta plus delta.
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..
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And Q s the reactive power at the sending
end will be given by A into V s square by
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B,
sin beta minus alpha minus V r into V s by
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B into sin beta plus delta. Now, these
quantities P r, Q r; P s, Q s are giving us
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the relationship for real and reactive power,
supplied at the sending end and received at
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the receiving end. Now, from this equation
Pr
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equation, we can see that, for a given system
voltage level V s and V r will be very near
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to the system voltage. And they do not change
much.
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That is, if you have a 400 kV system or a
220 kV system. These values will be very near
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to 400 kV or 220 kV. Whereas, this beta is
basically the angle of the impedance of the
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transmission line, that is, if we are using
a pi equivalent circuit or these parameters
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beta,
alpha, all these are transmission line parameters.
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And once the transmission line is
already there and then these parameters are
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fixed.
So, we see for any given transmission line,
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with the system voltage. For a given system
voltage, we have only one parameter, this
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parameter delta which can be varied, which
is
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the power angle or the voltage phase angle
at the sending end. And we can get a
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maximum power, which can be transmitted or
which can be received by the load on this
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transmission line for a given system voltage
level.
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That will be equal to P r max. This is equal
to V s, V r by B and when, we make this beta
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equal to delta. Then, this cos beta minus
delta is equal to 1 and this will be the maximum
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values. And that is what we get here Vs, V
r by B minus A V r square by B into cos beta
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.minus alpha. And corresponding to this maximum
power, that will be flowing; we have
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the reactive power at the receiving end given
by minus A V r square by B into sin beta
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minus alpha.
Now, what we see from here is, that if the
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real power is maximum. Then, the
corresponding reactive power received at the
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load end has to be negative. Means, the
power has to be transmitted for a leading
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power factor load. That is the load power
has to
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be leading for this condition.
.
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Now, in order to understand these phenomena
of what actually is happening on the
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transmission line. And get a better concept
of this power flow and how it gets affected.
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We would like to make to our system much more
simpler. So, let us first make this line
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as a short line. And for short line, we know
A is equal to D is equal to 1 angle 0. And
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B
is equal to Z angle theta. Z is the series
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impedance of the line and theta is the angle
of the
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impedance.
Then, substituting these values, we get Pr
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is equal to V s, V r by B, B is now Z. So,
we
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have got V s, V r by Z cos of beta minus delta,
beta is now theta. So, theta minus delta
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minus V r square, because A is now 1, V r
square by Z into cos beta was the term. So,
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here it is cos theta. Similarly, we can write
Q r, Q r is equal to V s, V r by Z sin theta
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minus delta minus V r square by Z sin theta.
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.For the sending end powers, we can write
the similar relations for the short line.
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P s is
equal to V s square by Z cos theta minus V
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s, V r by Z cos theta plus delta.
.
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And Q s the reactive power at the sending
end will be equal to Vs square by Z sin theta
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minus V s, V r by Z sin theta plus delta.
Now, in most cases for the transmission line,
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the
resistance is much smaller compared to it
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is reactance. And therefore, we can say that
magnitude of Z is very much equal to the magnitude
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of x, the reactants. Because, we
know Z magnitude of Z is equal to the square
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root of r square plus x square.
Since, r is much smaller, most of the transmission,
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like overhead transmission line. The
value of r will be either one-third or even
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less than that of the reactance of the line.
And
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therefore, we can say the magnitude of Z is
approximately equal to the magnitude of x.
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And we also can say that the angle theta in
this case will be very close to 90 degree.
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That
is, we are saying, that the line is purely
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reactive in this case, because resistance
is much
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smaller and can be neglected. Therefore, substituting
these values, that is Z is equal to x
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and theta is equal to 90 degrees. We get P
r is equal to V s, V r by x into sin delta.
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This is
a very important relationship for power flow
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on the transmission line.
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00:20:02,849 --> 00:20:03,849
..
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Similarly, we will get Q r is equal to V s,
V r by x into cos delta minus V r square by
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x.
Now, for power system operation, normally
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the angle delta is small. And therefore, we
can say that cos delta is nearly equal to
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1. Therefore, if we substitute this, then
we have
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Q r is equal to V s, V r by x minus V r square
by x, which can be simplified into V r by
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x
into V s minus V r.
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Now, these this relationship for Q r and this
relationship for P r. The real power received
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at the lower end and the reactive power at
the lower end, as these are very important
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relation, especially for understanding what
is happening. From these relationships, we
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can draw following conclusions.
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..
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That is for fix values of V s, V r and x.
That is, for a given transmission line and
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given
system voltage. The real power depends on
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angle delta, the phase angle by which V s
leads V r. This angle delta is called power
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angle. This is mainly, because this is the
angle
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delta, which is determinant, how much power
is going to flow on the line. Therefore, we
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call this as power angle.
When, delta is equal to 90 degrees; P is maximum.
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That is the maximum power. That can
be transmitted through this system can be
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when this delta is 90 degree. Then, sin delta
is
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equal to 1 and then the natural power that
can be transmitted is V s, V r by x. So, when
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delta is equal to 90 degree; P is maximum,
that is maximum power. That can be
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transmitted. For system stability considerations,
delta is generally kept much smaller,
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that 90 degree. It is normally work, in the
range of 20 to 30 degrees only. Because, any
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certain large disturbances can make system
unstable, if we work very near to 90 degrees.
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..
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The other inference, that we get from this
relationship is, that power can be transferred
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over a line, even when V s magnitude of the
sending end voltage is less than or equal
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to
the receiving end voltage magnitude. Normally,
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00:23:15,059 --> 00:23:24,809
we always assume that the voltage at the
sending end must be greater for power to be
204
00:23:24,809 --> 00:23:28,190
transferred. This is the case for a DC
system.
205
00:23:28,190 --> 00:23:32,870
If a power has to be transferred from one
end to other end, than the sending end voltage
206
00:23:32,870 --> 00:23:38,710
must be larger than the receiving end voltage.
But, in case of AC transmission system,
207
00:23:38,710 --> 00:23:45,710
this is not necessary. The value of the sending
end voltage magnitude, can be lower than
208
00:23:45,710 --> 00:23:52,240
the receiving end voltage magnitude, still
real power can be transmitted over this line.
209
00:23:52,240 --> 00:23:55,929
It
is the phase difference, delta between V s
210
00:23:55,929 --> 00:24:00,530
and V r, which determines how much power,
is
211
00:24:00,530 --> 00:24:05,929
going to flow.
If delta is positive, that is V s leads V
212
00:24:05,929 --> 00:24:10,260
r, then the power will be flowing from the
sending
213
00:24:10,260 --> 00:24:16,779
end to the receiving end. If delta is negative,
that is if V s lacks V r, then the power will
214
00:24:16,779 --> 00:24:24,580
be flowing from V r side to V s. So, it is
this phasar angle, which determines the
215
00:24:24,580 --> 00:24:31,379
direction of power flow and not the voltage
magnitude. This is a very, very important
216
00:24:31,379 --> 00:24:37,740
concept in AC transmission.
Now, most of the power systems are operated
217
00:24:37,740 --> 00:24:45,200
with almost the same voltage magnitude.
That is, what we do is, in order to keep out
218
00:24:45,200 --> 00:24:52,159
system properly. We keep the system voltage
at most of the busses in the system, very
219
00:24:52,159 --> 00:24:57,409
nearly equal to nominal system voltage. So,
say
220
00:24:57,409 --> 00:25:03,169
.for a 220 kV system or a 400 kV system, we
would be keeping the voltage magnitudes at
221
00:25:03,169 --> 00:25:10,509
most of busses very near to 220 or 400 kV.
Because, this provides us, a much better
222
00:25:10,509 --> 00:25:24,129
operating condition, for the system.
That is power systems are operated with almost
223
00:25:24,129 --> 00:25:29,860
the same voltage magnitude. That is one
power unit at important busses by using methods
224
00:25:29,860 --> 00:25:36,929
of voltage control. This is, what we will
talk about later in this lesson.
225
00:25:36,929 --> 00:25:37,929
.
226
00:25:37,929 --> 00:25:44,080
The third inference, that we can derive from
this relationship is the maximum real power
227
00:25:44,080 --> 00:25:50,590
transferred over a line increases with increasing
V s and V r. That is the sending end and
228
00:25:50,590 --> 00:25:56,650
there receiving end voltage. An increase of
100 percent in V r and Vs, that is, if we
229
00:25:56,650 --> 00:26:05,029
double the voltage level of the transmission
line. Then, the power transfer increases to
230
00:26:05,029 --> 00:26:12,429
400 percent. That is, if the line initially
was capable of carrying 100 MVA.
231
00:26:12,429 --> 00:26:20,200
Now, if we double the voltage, it will be
able to carry 400, sorry, if it was able to
232
00:26:20,200 --> 00:26:24,070
carry
100 MVA. Now, if we double the voltage, it
233
00:26:24,070 --> 00:26:38,230
will be able to carry 400 MVA. And this is
one of the main reasons, why we keep going
234
00:26:38,230 --> 00:26:50,749
for higher and higher voltage levels. When,
we want to transfer more and more power.
235
00:26:50,749 --> 00:26:51,749
..
236
00:26:51,749 --> 00:26:56,740
Another important inference, that we can derive
from this expression is the maximum
237
00:26:56,740 --> 00:27:04,809
real power depends on reactance x, which is
directly proportional to the line inductance.
238
00:27:04,809 --> 00:27:14,399
Now, as we have seen, P max is V s, V r by
x. Therefore, if we decrease the value of
239
00:27:14,399 --> 00:27:20,990
x, P
max will increase. A decrease in inductance,
240
00:27:20,990 --> 00:27:26,299
increases the line capacity or the power
transferred capability of the transmission
241
00:27:26,299 --> 00:27:30,889
system.
Now, which means, we need to decrease the
242
00:27:30,889 --> 00:27:37,830
inductance. As we had seen earlier, we can
decrease the line inductance by using bundle
243
00:27:37,830 --> 00:27:48,059
conductors, because bundle conductors will
increase the effective radius of the conductor.
244
00:27:48,059 --> 00:27:54,600
And therefore, reduce the inductance of
the transmission line. Reducing the inductance
245
00:27:54,600 --> 00:28:02,249
of the transmission line, will reduce the
reactance. And therefore, it will increase
246
00:28:02,249 --> 00:28:09,279
the power transfer capability or the maximum
power that can be transmitted on that line.
247
00:28:09,279 --> 00:28:10,279
..
248
00:28:10,279 --> 00:28:22,460
Another method of reducing the reactance is
by connecting a capacitor in series, with
249
00:28:22,460 --> 00:28:25,730
the
line. Now, if we connect a capacitor in series
250
00:28:25,730 --> 00:28:33,630
with the line, then the total effective
reactance of the line will get reduced. Because,
251
00:28:33,630 --> 00:28:43,450
the reactance of the capacitor will be
negative and will be reducing the effective
252
00:28:43,450 --> 00:28:51,789
reactance of the total, the effective series
reactance of the transmission line.
253
00:28:51,789 --> 00:28:57,269
This method of connecting a capacitor in series
with the line is called a series
254
00:28:57,269 --> 00:29:03,950
compensation of the transmission line. And
this is generally done, in order to increase
255
00:29:03,950 --> 00:29:11,239
the power transfer capacity of the transmission
line. Because, as we had seen in the
256
00:29:11,239 --> 00:29:20,009
earlier lesson, most of the time, we run our
system at much below the thermal limits.
257
00:29:20,009 --> 00:29:28,389
Mainly, because the stability limit is are
the major problem and we have ensure that,
258
00:29:28,389 --> 00:29:30,659
we
run our system properly.
259
00:29:30,659 --> 00:29:37,499
Even, when there is sudden large disturbance,
which takes place in the system, like short
260
00:29:37,499 --> 00:29:45,460
circuits and tripping of lines or generators,
even, when these take place. Still, the system
261
00:29:45,460 --> 00:29:54,059
should run synchronously. And for this, we
need that delta angle should be much lower
262
00:29:54,059 --> 00:30:00,909
than 90 degrees. That is in the range of around
20 to 30 degrees. And that is why, unless
263
00:30:00,909 --> 00:30:09,259
we reduce the reactance of the transmission
system. We cannot transfer large amount of
264
00:30:09,259 --> 00:30:20,649
power through that transmission system.
Now, as I said we can reduce by using series
265
00:30:20,649 --> 00:30:25,120
capacitances. That is connecting a
capacitance in series with the line. The series
266
00:30:25,120 --> 00:30:27,970
capacitors are usually installed at the
267
00:30:27,970 --> 00:30:34,600
.middle of the line. This is generally done,
for one simple reason, that if we install
268
00:30:34,600 --> 00:30:38,460
them at
either the sending end or at the receiving
269
00:30:38,460 --> 00:30:46,400
end. Then, when the loads are low, the voltage
values at these ends can go very high. Because,
270
00:30:46,400 --> 00:30:54,750
of the large amount of compensation,
which is available at those ends. Whereas,
271
00:30:54,750 --> 00:30:58,400
if we put them in the middle of the line.
The
272
00:30:58,400 --> 00:31:06,649
voltage profile generally will be much better.
.
273
00:31:06,649 --> 00:31:14,710
The from the reactive power relationship,
we derived the following inferences. The
274
00:31:14,710 --> 00:31:20,750
reactive power transferred over a line is
directly proportional to V s minus V r. That
275
00:31:20,750 --> 00:31:25,289
is, if
we go through this relationship Q r is equal
276
00:31:25,289 --> 00:31:31,000
to V r by x into V s minus V r. That is Q
r is
277
00:31:31,000 --> 00:31:36,120
directly proportional to the difference of
the voltage magnitude V s minus V r.
278
00:31:36,120 --> 00:31:44,399
Therefore, we see that the regulation, that
is the voltage difference. That comes, when
279
00:31:44,399 --> 00:31:50,009
the line is loaded is very important. And
that governs the reactive power flow on the
280
00:31:50,009 --> 00:31:58,690
transmission line. So, reactive power transferred
over a line is directly proportional to V
281
00:31:58,690 --> 00:32:05,190
s minus V r magnitude of Vs minus magnitude
of V r. That is the voltage drop along the
282
00:32:05,190 --> 00:32:09,529
line and is independent of power angle.
283
00:32:09,529 --> 00:32:10,529
..
284
00:32:10,529 --> 00:32:24,419
That is in this relationship, we do not have
any power angle delta coming into picture.
285
00:32:24,419 --> 00:32:31,710
This means, the voltage drop on the line is
due to the transfer of reactive power over
286
00:32:31,710 --> 00:32:36,030
the
line. That is, it is because of the reactive
287
00:32:36,030 --> 00:32:39,129
power, which is being transferred over the
line,
288
00:32:39,129 --> 00:32:45,580
that there is a large drop in the voltage.
To maintain a good voltage profile, reactive
289
00:32:45,580 --> 00:32:52,679
power control is therefore, necessary.
That is, if we want that the voltage drop
290
00:32:52,679 --> 00:33:00,259
is not large. Then, we have to see that, the
reactive power transferred, over the line
291
00:33:00,259 --> 00:33:05,690
is reduced as much as possible. This is done
by
292
00:33:05,690 --> 00:33:15,290
means of, what we call reactive power compensation
of transmission lines. So, now, we
293
00:33:15,290 --> 00:33:24,009
will look into, how we control the voltage.
And so the reactive power flows on the
294
00:33:24,009 --> 00:33:26,090
transmission line.
295
00:33:26,090 --> 00:33:27,090
..
296
00:33:27,090 --> 00:33:33,129
Reactive power compensation equipment has
following effects. Basically, why we need
297
00:33:33,129 --> 00:33:39,029
this compensation, we need this compensation
mainly, to see that the voltage profile of
298
00:33:39,029 --> 00:33:46,059
the line is maintained, as near the nominal
value as possible. Now, the question comes
299
00:33:46,059 --> 00:33:53,820
why it is so important. Well, if we go at
higher and higher voltages, what we have is,
300
00:33:53,820 --> 00:34:00,419
we
need to insulate these lines from ground.
301
00:34:00,419 --> 00:34:03,830
As well as the insulation, between line to
line,
302
00:34:03,830 --> 00:34:12,379
which means, that we need to use insulators.
And these insulators will be expensive, if
303
00:34:12,379 --> 00:34:20,830
we need to provide large amount of safety
factor. That is why, at extra high voltages
304
00:34:20,830 --> 00:34:29,379
of 400 kV and above the safety factors, may
not be very large. In order to make the transmission
305
00:34:29,379 --> 00:34:36,570
system economic and for this
purpose, we say, that we work these lines
306
00:34:36,570 --> 00:34:40,580
or these transmission systems at a level of
only
307
00:34:40,580 --> 00:34:47,109
around plus minus 5 percent.
That is the voltage, which is allowed to change
308
00:34:47,109 --> 00:34:54,919
is up to a level of plus minus 5 percent,
from the nominal value, which means, that
309
00:34:54,919 --> 00:35:00,680
the maximum voltage drop, that can be
allowed on a transmission system is only of
310
00:35:00,680 --> 00:35:05,200
an order of 10 percent. And therefore, it
is
311
00:35:05,200 --> 00:35:14,370
necessary, that we use some control equipment
to keep this voltage within these limits.
312
00:35:14,370 --> 00:35:20,260
Where, the voltage drop is not allowed to
go beyond this. And if it goes some
313
00:35:20,260 --> 00:35:27,080
compensation equipment has to be there to
boost it up. And keep the voltage drop within
314
00:35:27,080 --> 00:35:29,450
the permissible limits.
315
00:35:29,450 --> 00:35:36,940
.Now, the reactive power compensation equipment
has the following effects. First is
316
00:35:36,940 --> 00:35:42,180
reduction in current. When, we are reducing
the reactive power flow in the transmission
317
00:35:42,180 --> 00:35:51,320
line, what we basically mean is the current
component of the reactive power flow is also
318
00:35:51,320 --> 00:35:58,130
getting reduced. Because, we known the complex
power is P plus j cube and if V and this
319
00:35:58,130 --> 00:36:06,530
is equal to V into I conjugate. And therefore,
if we reduce Q, then we are reducing S and
320
00:36:06,530 --> 00:36:15,750
we are there by also reducing I. Because,
V is very near to the nominal voltage.
321
00:36:15,750 --> 00:36:21,720
Therefore, if we reduce the reactive power
flow on the line we are also reducing the
322
00:36:21,720 --> 00:36:28,780
current. We are also able to maintain the
voltage profile within limits. If we reduce
323
00:36:28,780 --> 00:36:33,430
the
reactive power as we said, if the Q r is given
324
00:36:33,430 --> 00:36:39,670
by V r by x into V s minus V r and this V
s
325
00:36:39,670 --> 00:36:49,810
minus V r will be reduced, if Q r is reduced.
Since, the current flowing gets reduced,
326
00:36:49,810 --> 00:36:55,490
therefore the losses I square r losses on
the transmission line will also get reduced.
327
00:36:55,490 --> 00:37:00,180
This
is another advantage in having reactive power
328
00:37:00,180 --> 00:37:03,820
compensation on the system.
.
329
00:37:03,820 --> 00:37:11,400
Of course, this means that, the there is going
to be reduction in investment in the system
330
00:37:11,400 --> 00:37:17,280
per Kilo Watt of load supply. For this simple
reason, that if we are reducing the reactive
331
00:37:17,280 --> 00:37:22,580
power, we are reducing the current. And therefore,
the transmission line, that we are
332
00:37:22,580 --> 00:37:30,790
using, can be of a smaller diameter as such,
which will reduce our investment. Also, all
333
00:37:30,790 --> 00:37:37,940
the equipment that we have is rated in terms
of their k V A or MVA capacity. That is the
334
00:37:37,940 --> 00:37:44,860
volt ampere capacity, not the Kilo Watt or
Mega Watt or watt capacity.
335
00:37:44,860 --> 00:37:53,970
.This is because, all the equipment has current
rating associated with it, which is basically
336
00:37:53,970 --> 00:38:00,030
dependent on the, I square r losses, that
can be dissipated from that equipment. So,
337
00:38:00,030 --> 00:38:04,470
that
the temperature of the windings and other
338
00:38:04,470 --> 00:38:07,330
parts of the equipment is maintained within
a
339
00:38:07,330 --> 00:38:15,510
give limit. Otherwise, the insulation of the
equipment, well deteriorate very fast and
340
00:38:15,510 --> 00:38:20,630
the
machine will go out of order, because it will
341
00:38:20,630 --> 00:38:30,100
lead to short circuits.
So, by using reactive power compensation,
342
00:38:30,100 --> 00:38:35,630
we are reducing the reactive power
requirement of the system. And therefore,
343
00:38:35,630 --> 00:38:44,500
we are also reducing it is k V A requirement
or we are reducing the investment in system
344
00:38:44,500 --> 00:38:55,120
per Kilo Watt of the load supply. Now,
decrease k V A loading of generators are lines,
345
00:38:55,120 --> 00:39:04,780
happens because of this. And this
decrease helps us, because this decreasing
346
00:39:04,780 --> 00:39:11,970
k V A loading relieves the overload or
condition or releases capacity.
347
00:39:11,970 --> 00:39:20,070
Suppose, I am running a system to it is limit,
say I have a 50 MVA generator and I am
348
00:39:20,070 --> 00:39:29,520
running this generator at 50 MVA at 0.8 power
factor. Now; that means, the load is
349
00:39:29,520 --> 00:39:41,740
going to be only 40 Kilo Watt and the MVA
is 50. Now, suppose I increase the power
350
00:39:41,740 --> 00:39:51,050
factor 2.9, then what happens? Keeping the
generator capacity or still running this
351
00:39:51,050 --> 00:40:02,110
generator at 50 MVA. Now, I will be able to
supply 45 Mega Watt. That means, this 5
352
00:40:02,110 --> 00:40:09,130
Mega Watt capacity is now released. So, I
can supply more load from the same
353
00:40:09,130 --> 00:40:19,400
generator. Therefore, the reactive power compensation
helps in relieving extra capacity
354
00:40:19,400 --> 00:40:29,440
or relieving overload on the equipment. And
that is generators transmission lines,
355
00:40:29,440 --> 00:40:31,960
etcetera.
356
00:40:31,960 --> 00:40:32,960
..
357
00:40:32,960 --> 00:40:39,930
It also helps, as I said if we increase the
power factor. It helps the generator also,
358
00:40:39,930 --> 00:40:44,020
because
it will have to generate, less amount of reactive.
359
00:40:44,020 --> 00:40:51,880
Power, which means, that the excitation
system, will be working with lesser current.
360
00:40:51,880 --> 00:41:03,340
We do not have to over excite, the
generators. Of course, for the consumers the
361
00:41:03,340 --> 00:41:06,910
advantage is, that is going to be reduction
in
362
00:41:06,910 --> 00:41:12,080
their k V A demand charge.
Because, large consumers are charged, not
363
00:41:12,080 --> 00:41:20,940
only on the basis of the energy they consume.
But, also on what is their maximum k V A demand.
364
00:41:20,940 --> 00:41:27,570
Because, basically the utility, which
is supplying to these consumers, have to build
365
00:41:27,570 --> 00:41:33,180
the transmission system, the transformers
have to be installed. All this have to be
366
00:41:33,180 --> 00:41:36,730
done on the basis of, what is going to be
their
367
00:41:36,730 --> 00:41:41,760
maximum k V A.
So, if we have improved the reactive power
368
00:41:41,760 --> 00:41:49,760
requirement. Then, what we have done is,
we have reduced the k V A requirement. And
369
00:41:49,760 --> 00:41:58,480
therefore, the k V A demand charge, which
is charged by the utility will also get reduced.
370
00:41:58,480 --> 00:42:06,750
Now, how do, we provide this
compensation? Now, there are different ways
371
00:42:06,750 --> 00:42:11,900
in which we can provide this
compensation. Like, static compensation, we
372
00:42:11,900 --> 00:42:19,220
have rotating compensation and also, we
can use transformers for compensation or control
373
00:42:19,220 --> 00:42:21,930
of voltage on the transmission line.
374
00:42:21,930 --> 00:42:22,930
..
375
00:42:22,930 --> 00:42:28,900
So, let us take first the static compensation.
Well, static compensation basically consists
376
00:42:28,900 --> 00:42:37,730
of two ways in which we can do it. One is,
the shunt compensation and the other is series
377
00:42:37,730 --> 00:42:45,550
compensation. Normally, series compensation
is done, mainly for increasing the
378
00:42:45,550 --> 00:42:54,820
transmission capacity of the line. Whereas,
the shunt compensation is used, mostly for
379
00:42:54,820 --> 00:43:02,480
voltage control, because, what it does is,
it supplies the reactive power to the load.
380
00:43:02,480 --> 00:43:08,850
And the reactive power, which is demanded
by the load, is no longer flowing from the
381
00:43:08,850 --> 00:43:14,750
sending end to the receiving end, because
this is now being supplied locally at the
382
00:43:14,750 --> 00:43:22,650
receiving end by the compensating equipment.
And there by, releasing capacity on the
383
00:43:22,650 --> 00:43:32,150
transmission system, now what this shunt compensating
equipment. Consist of normally,
384
00:43:32,150 --> 00:43:39,280
at the receiving end of the transformer, which
will be a three winding transformer. The
385
00:43:39,280 --> 00:43:43,000
tertiary winding is normally a delta connected
winding.
386
00:43:43,000 --> 00:43:53,600
So, suppose we have a 400 kV to 220 kV transformer
and it is tertiary is a 11 kV. Now,
387
00:43:53,600 --> 00:44:01,810
the compensating equipment is normally connected
at the low voltage or the tertiary
388
00:44:01,810 --> 00:44:09,160
winding, which is at low voltage. And this
equipment consists of basically some
389
00:44:09,160 --> 00:44:16,090
capacitors, which are permanently connected.
Some capacitors, which are switchable and
390
00:44:16,090 --> 00:44:23,760
can be connected or disconnected depending
on the demand. And variable reactor or a
391
00:44:23,760 --> 00:44:35,360
saturable reactor, the current of which can
be controlled by means of thyristor or GTO.
392
00:44:35,360 --> 00:44:43,090
So, this is a controllable reactor here. The
current following through this reactor can
393
00:44:43,090 --> 00:44:46,890
be
controlled. And therefore, this reactor becomes
394
00:44:46,890 --> 00:44:50,900
a variable reactor. And therefore, using
395
00:44:50,900 --> 00:45:01,060
.equipment with this capacitors and variable
reactor, we can give any value of
396
00:45:01,060 --> 00:45:03,080
compensation. That we want.
.
397
00:45:03,080 --> 00:45:11,190
Basically, if you see, it is how much the
capacitance, that we are providing, that will
398
00:45:11,190 --> 00:45:20,050
decide, what how much compensation we are
doing. Normally, when the load is high or
399
00:45:20,050 --> 00:45:27,500
during the peak load period, the loads being
mostly inductive in nature. It will be the
400
00:45:27,500 --> 00:45:34,170
lagging reactive power, which will be demanded
by the load. So, the loads will be
401
00:45:34,170 --> 00:45:40,480
absorbing this lagging reactive power.
And therefore, the compensating equipment
402
00:45:40,480 --> 00:45:46,660
will have to generate this lagging reactive
power to supply it to the load. So, that the
403
00:45:46,660 --> 00:45:53,980
transmission system is relieved of supplying
this reactive power. Now, how much is that,
404
00:45:53,980 --> 00:46:02,660
that can be found very easily. We have the
current flowing in the capacitive, shunt capacitors
405
00:46:02,660 --> 00:46:10,510
given by V s by root 3 into x c, where
x c is the capacitive or the reactance of
406
00:46:10,510 --> 00:46:21,350
the capacitors and V s is the voltage, which
is
407
00:46:21,350 --> 00:46:28,580
line to line voltage or the system voltage.
Now, for a three phase system, if we assume
408
00:46:28,580 --> 00:46:36,560
these capacitors to be connected in star,
then the for the three phase system, we will
409
00:46:36,560 --> 00:46:43,350
have the total three phase reactive power.
Supplied by these capacitors, will be equal
410
00:46:43,350 --> 00:46:49,710
to three times V by root 3 x c into this,
is I c
411
00:46:49,710 --> 00:46:57,010
into V. That is per phase voltage or phase
to neutral voltage V by root 3. This comes
412
00:46:57,010 --> 00:47:00,380
out
to be equal to V square by x c.
413
00:47:00,380 --> 00:47:13,400
.Similarly, for this reactor will be absorbing
the reactive power and that will be given
414
00:47:13,400 --> 00:47:18,230
by,
again this relationship Q L is equal to minus
415
00:47:18,230 --> 00:47:26,540
V square by x L. If we are writing this
voltage in Kilo Volts, then these values will
416
00:47:26,540 --> 00:47:37,510
be in terms of Mega Vars. Now, when the
load is low or we have very light load condition.
417
00:47:37,510 --> 00:47:42,840
In those conditions, what happens is,
these high voltage transmission line has large
418
00:47:42,840 --> 00:47:49,710
amount of charging capacitance.
So, there is large amount of charging current,
419
00:47:49,710 --> 00:47:56,350
which flows through them, which produces
large amount of reactive power. And since
420
00:47:56,350 --> 00:48:03,810
the load is low, there is not enough absorption
of this reactive power. Because, of this,
421
00:48:03,810 --> 00:48:06,960
what happens is the receiving end voltage
goes
422
00:48:06,960 --> 00:48:15,700
up right. Sometimes, this voltage can go much
higher than that of the sending end. This
423
00:48:15,700 --> 00:48:23,400
again is problematic, because we have said
earlier, that we work within a voltage range
424
00:48:23,400 --> 00:48:28,720
of plus minus 5 percent.
And if this voltage goes beyond that, then
425
00:48:28,720 --> 00:48:32,710
there is a chance of insulation failure for
the
426
00:48:32,710 --> 00:48:39,330
insulators on the transmission line. Because,
the voltage will go beyond their rated
427
00:48:39,330 --> 00:48:47,380
capacity, this cannot be allowed. And therefore,
we under light load conditions, we have
428
00:48:47,380 --> 00:48:57,980
to absorb the reactive power and that is done
by this reactor. So, we have both absorption
429
00:48:57,980 --> 00:49:03,680
of reactive powers, when we want and generation
of reactive power. Most of the time,
430
00:49:03,680 --> 00:49:10,740
when the load is low high, we need this part
more and this part very less. Whereas, when
431
00:49:10,740 --> 00:49:17,280
the load is very low and the voltage is going
higher, then the current through this can
432
00:49:17,280 --> 00:49:21,220
be
increased and these can be switched off. And
433
00:49:21,220 --> 00:49:24,380
again, the voltage can be maintained.
.
434
00:49:24,380 --> 00:49:37,060
.Now, instead of using static compensators,
earlier people where using rotating
435
00:49:37,060 --> 00:49:43,530
compensators. These rotating compensators
are nothing but synchronous motors, which
436
00:49:43,530 --> 00:49:51,240
are not mechanically loaded. That is, they
run at low load. So, they are designed as
437
00:49:51,240 --> 00:49:58,530
synchronous motors, running at low load. It
is only the excitation, that we vary and
438
00:49:58,530 --> 00:50:06,040
varying this excitation, we can make these
machines either absorb reactive power or
439
00:50:06,040 --> 00:50:11,250
generate reactive power.
So, a synchronous compensator, can supply
440
00:50:11,250 --> 00:50:15,520
lagging vars up to it is rating. That is,
up to it
441
00:50:15,520 --> 00:50:24,070
is design capacity. So, by over exciting this,
synchronous machine, we can supply large
442
00:50:24,070 --> 00:50:34,510
amount of lagging vars. And by making this
machine, run under excited, we can absorb
443
00:50:34,510 --> 00:50:42,870
the lagging vars. That is, generally these
are rated to absorb up to 50 percent of it
444
00:50:42,870 --> 00:50:46,820
is rated
capacity, in terms of Mega Watts.
445
00:50:46,820 --> 00:50:55,841
So, thus a synchronous compensator of a certain
rating is equivalent to a static capacitor
446
00:50:55,841 --> 00:51:04,150
of that rating. That is, because it can supply
lagging vars with that capacity, which a
447
00:51:04,150 --> 00:51:11,730
capacitor also can do. Therefore, we can say
it is equivalent to a capacitor or a static
448
00:51:11,730 --> 00:51:18,820
capacitor of that rating. And since, it can
absorb up to half of that rating, it is equivalent
449
00:51:18,820 --> 00:51:27,530
to that of inductor of half it is rating.
.
450
00:51:27,530 --> 00:51:43,780
The advantage of synchronous compensator is,
that we can very smoothly vary the
451
00:51:43,780 --> 00:51:55,950
reactive power absorption or generation by
this machine. Another advantage of the
452
00:51:55,950 --> 00:52:04,020
synchronous compensator is that, it can be
overloaded for short periods. Whereas, this
453
00:52:04,020 --> 00:52:05,020
is
454
00:52:05,020 --> 00:52:12,260
.not possible with static capacitors, because
if you try to overload them, they will
455
00:52:12,260 --> 00:52:19,970
normally puncture, whereas these synchronous
machines have some amount of overload
456
00:52:19,970 --> 00:52:31,540
capacity for short periods.
Another major advantage for the synchronous
457
00:52:31,540 --> 00:52:37,780
compensators is that a drop in voltage
causes the synchronous compensator to supply
458
00:52:37,780 --> 00:52:44,920
greater vars. Because, the difference
between generated EMF and the terminal voltage
459
00:52:44,920 --> 00:52:51,610
will increase, which means the
excitation will have to be raised, which helps
460
00:52:51,610 --> 00:53:02,400
in arresting the voltage dip. So, it can
supply more var, when the load demands more.
461
00:53:02,400 --> 00:53:09,630
Even, if the voltage drops and if the
voltage drops by increasing the excitation,
462
00:53:09,630 --> 00:53:17,381
we can raise the voltage of the terminal.
Whereas, in case of static compensation, if
463
00:53:17,381 --> 00:53:23,850
we are using static capacitor, if the voltage
drops. Then, what happens is the reactive
464
00:53:23,850 --> 00:53:32,190
power generated by these capacitors will also
drop. If you can see this relationship, this
465
00:53:32,190 --> 00:53:40,590
Q c is proportional to or is equal to V square
by x c. And if this voltage drops, then Q
466
00:53:40,590 --> 00:53:47,880
c generated will also drop, which simply means,
that when we need more reactive power. That
467
00:53:47,880 --> 00:53:50,540
is, when the voltage drops, we need more
reactive power.
468
00:53:50,540 --> 00:53:56,640
The voltage drop occurs mainly, because the
load is drawing more reactive power, at that
469
00:53:56,640 --> 00:54:04,770
time, it is unable to supply larger reactive
power. In fact, it reduces it reactive power
470
00:54:04,770 --> 00:54:11,200
supply, which will further dip the voltage
unless we have other sources of reactive
471
00:54:11,200 --> 00:54:14,500
power, around.
.
472
00:54:14,500 --> 00:54:21,100
.So, in case of static capacitors a drop in
voltage causes a reduction in var delivered
473
00:54:21,100 --> 00:54:26,500
to the
system. That is, it is unable to do or the
474
00:54:26,500 --> 00:54:31,911
function for which it is basically used, that
is, it
475
00:54:31,911 --> 00:54:44,320
can help. But, if the voltage goes down the
help also reduces. One another advantage of
476
00:54:44,320 --> 00:54:50,060
synchronous compensator is that, it is a rotating
machine, which means it has large
477
00:54:50,060 --> 00:54:56,290
amount of kinetic energy stored in it. Because,
of the inertia of the rotating path and
478
00:54:56,290 --> 00:55:03,690
therefore, it helps in making the system more
robust and more stable.
479
00:55:03,690 --> 00:55:04,690
.
480
00:55:04,690 --> 00:55:09,810
So, inertia of a synchronous compensator improves
system stability and reduces the
481
00:55:09,810 --> 00:55:15,550
effect of sudden change in load. Because,
any sudden jerk is taken up by the larger
482
00:55:15,550 --> 00:55:22,501
inertia and the system response is not that
great. Static capacitor is installations,
483
00:55:22,501 --> 00:55:26,290
can be
distributed in the system. However, small
484
00:55:26,290 --> 00:55:30,090
size synchronous compensators are very
uneconomical.
485
00:55:30,090 --> 00:55:37,440
The main reason of not using synchronous compensators
nowadays is, synchronous
486
00:55:37,440 --> 00:55:43,370
compensators are comparatively much more expensive
as well as they need maintains.
487
00:55:43,370 --> 00:55:52,280
Whereas, static compensators can be installed
at various locations, almost any substation,
488
00:55:52,280 --> 00:55:59,300
we can install them and they do not need any
maintenance or personal to look after them,
489
00:55:59,300 --> 00:56:06,830
because they are static equipment. Therefore,
the static capacitors can be located near
490
00:56:06,830 --> 00:56:09,180
the
loads and are more effective.
491
00:56:09,180 --> 00:56:16,290
We can use small static compensators, as per
the requirements and put them very near to
492
00:56:16,290 --> 00:56:21,320
the loads substations or at the load substations
depending on the load requirement.
493
00:56:21,320 --> 00:56:31,000
.Especially, with the use of power electronic
devices, this has become now very easy to
494
00:56:31,000 --> 00:56:39,680
do. And therefore, this is what we are putting
more and more in the system nowadays,
495
00:56:39,680 --> 00:56:51,650
with this we will stop here. We will talk
about some other compensating equipment in
496
00:56:51,650 --> 00:56:57,570
the next class. And we will work out some
problems on transmission lines, which will
497
00:56:57,570 --> 00:57:10,620
make things, much more clearer.
Thank you.
498
00:57:10,620 --> 00:57:11,620
.
499
00:57:11,620 --> 00:57:12,620
.
500
00:57:12,620 --> 00:57:21,200
.Welcome to lesson 10 on power system analysis.
In this lesson, we will be continuing
501
00:57:21,200 --> 00:57:27,620
from what we did in lesson 9. That is Transmission
Line Steady State Operation. In
502
00:57:27,620 --> 00:57:34,860
lesson 9, we talked about the power flow equations
on transmission line and the steady
503
00:57:34,860 --> 00:57:44,430
state operating conditions. We also discussed
how we can compensate for the reactive
504
00:57:44,430 --> 00:57:50,410
power flow on the line and thereby, maintain
the voltage profile on the line.
505
00:57:50,410 --> 00:57:51,410
.
506
00:57:51,410 --> 00:58:00,950
Now, we will take up another problem. Now,
this problem states a three phase 132 k V
507
00:58:00,950 --> 00:58:08,700
overhead line, delivers 60 MVA. Mind it, this
is 60 MVA not Mega Watt 60 MVA at
508
00:58:08,700 --> 00:58:17,260
132 k V and power factor 0.8 lagging at it
is receiving end. That is, V r or V 2 is given
509
00:58:17,260 --> 00:58:25,800
as
132 k V and the P S r is the MVA value is
510
00:58:25,800 --> 00:58:36,900
given as 60 MVA at 0.8 power factor lagging.
The constants of the line are given as A is
511
00:58:36,900 --> 00:58:40,500
equal to 0.98 with an angle 3 degrees and
B is
512
00:58:40,500 --> 00:58:45,230
equal to 100 with an angle 75 degrees ohms
per phase.
513
00:58:45,230 --> 00:58:53,710
Now, we need to find for this system sending
end voltage and power angle. B sending
514
00:58:53,710 --> 00:59:04,750
end active and reactive power, C line losses
and vars absorbed by the line. We will end
515
00:59:04,750 --> 00:59:12,790
today’s lessons and in the next lesson,
we will review. Whatever, we have learnt about
516
00:59:12,790 --> 00:59:15,600
the transmission line.
So, thank you very much.
517
00:59:15,600 --> 00:59:15,600
.