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Welcome to lesson 10, on Power System Analysis.
In this lesson, we will be continuing
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from what we did in lesson 9. That is Transmission
Line Steady State Operation. In
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lesson 9, we talked about the power flow equations
on transmission line, under steady
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state operating conditions. We also discussed
how we can compensate for the reactive
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power flow on the line. And thereby, maintain
the voltage profile on the line.
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We discussed there static compensation equipment.
We also discussed about the rotating
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compensation equipment. And this lesson, we
will talk about the transformer taps and the
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combination of these equipments. We will also
discussed, something about the regulating
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transformers, how we can control the voltage
and power flow on the line, using these
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regulating transformers. And after this, we
will try to solve some problems, which will
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clarify most of the ideas, that we discussed
in lesson 9. And what we will be discussing
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in lesson 10, today.
.
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.So, we will start with voltage control of
a line using tap changing transformers. Here,
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we
have this transmission line, given by it is
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simple short line model of R plus j x. We
have
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neglected the shunt capacitance for the line
in this model. And we have step up
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transformer, at the sending end. And we are
assuming that we have a generator on this
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side and we have a step up transformer here.
On this transformer, we have a number of taps
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on the high voltage side. There by,
changing the taps we change the turn’s ratio
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and therefore, the output voltage of the
transformer or the voltage of the transformer
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on the high voltage side. Then, on the
receiving end, we again have step down transformer,
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where we have again taps on the
high voltage side. And by changing the tap
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on this side, we can get the change in voltage
at a load.
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Therefore, for any given at the voltage at
the supply end, we can by managing the taps
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on
the two sides of the transmission line. That
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is on the two transformers on the two sides
of
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the transmission line, we can control the
voltage to the load. Now, these taps are what
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we
call as off nominal tap. That is, if it is
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placed in a position, where it is the nominal
turns
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ratio of the transformer.
Suppose, this is a 11 kV to 220 kV transformer.
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Then, when we are putting 11 kV on this
side and the low volt this side and if we
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put the tap at the nominal position, then
we will
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get 220 kV on the high voltage side. And if
we place the tap on the upper side of the
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nominal tap or on the lower side of the nominal
tap, we will get different voltage. If you
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put it on the upper side, we will get higher
voltages, if we put it on the lower side,
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we
will get lower voltages.
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Normally, the transformers may have taps,
which can be in terms of variation of plus
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minus 5 to 10 percent, with variation at each
tap of the order of 2 percent or 2.5 percent
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or so. Similar thing, we have on the low voltage
side. Here, also we can place the taps at
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the nominal position and then we will get
the nominal voltage ratios. If we place the
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tap
above or below, then we will get different
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voltage is like, if we put it above the nominal
position.
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Then, the load voltage is going to be somewhat
lower than, what it will be, if it is place
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at the nominal position. Because, high voltage
side number of turns will increase, where
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as the low voltage side the number of turns
being same, the ratio will get reduced. So,
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.similarly if we put it on the lower side
of the nominal, we will get somewhat higher
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voltage on the low voltage side.
Now, normally these taps are always placed
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on the high voltage winding. One reason is,
the high voltage winding will be carrying
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much less current. And therefore, when we
are
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changing this tap especially, if we have on
load tap changing transformer, that is OLTC.
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Then, the current will be much lower here.
Also it has been found, that if we place the
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taps on the high voltage side, we get a much
better voltage control, then if we place it
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on
the low voltage side.
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So, here, I am showing in this diagram, that
we have a voltage t s into V 1, at the sending
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end, where t s is the off nominal tap ratio.
It can be in terms of say, if it is at 100,
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10
percent, 10 percent above the normal, then
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the value of t s is 1.1. If it is say at 5
percent
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below the nominal, then the value of t s will
be 0.95. Similarly, on the receiving end side,
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we have the off nominal tap ratio as t r.
And therefore, the voltage here will be t
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r into V
2.
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.
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Now, writing the circuit equation for this.
We have t s into V 1.
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..
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That is, if we go here t s into V 1 will be
equal to t r into V 2 plus the drop, which
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takes
place here. If the current flowing in this
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line is I, then it will be I into Z, Z is
equal to R
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plus j x.
.
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Therefore, we have t s into V 1 is equal to
t r, V 2 plus I into Z. Now, this will be
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approximately equal to t r into V 2 plus I
into Z is I into R plus j x, which we can
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write
as approximately equal to I into R cos theta
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plus I into X sin theta. Now, we also know
that, I cos theta will be equal to P by t
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r into V 2, because t r into V 2 is the voltage.
So,
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.V I cos theta is equal to the real power
P, therefore I cos theta is equal to P by
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t r into V
2.
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And similarly, we have the reactive power
Q is equal to t r into V 2 into I sin theta.
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This
is the voltage into I sin theta is equal to
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Q. So, I sin theta is equal to Q by t r into
V 2.
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Therefore, substituting these we will get
t s into V 1 is equal to t r, V 2 plus this
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is R into
I cos theta. So, R into I cos theta we are
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replacing by P by t r, V 2. So, R into P by
t r, V
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2 plus I X into I sin theta. So, X into Q
by t r, V 2. This is I sin theta, we have
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substituted
here. Normally, what we do is, we want to
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keep t s into t r as equal to 1. Because,
this is
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what will make the voltages proper on both
sides, so the product t s, t r is generally
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kept
as 1.
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.
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Then, if we do that, then we can write t r
is equal to 1 by t s. And therefore, substituting
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that, we get t s is equal to 1 by V 1. That
is V 1 t s is equal to V 2 by t.
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..
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Here t r into V 2. So, t r into V 2, I am
writing t r as 1 by t s.
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.
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So, this becomes V 2 by t s plus t s into
R P plus X Q by V 2, which if we multiply
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the
both sides by t s. Then, we get t s squared
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is equal to V 2 by V 1, because this t s will
cancel out plus t s squared, because it is
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multiplied by t s. So, t s squared R P plus
X Q
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divided by V 2, I am sorry, this should be
V 1, V 2. And therefore, if we take this taking
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on this side, then we will get t s squared
one minus R P plus X Q divide by V 1, V 2.
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.Because, this here is V 1, V 2, so 1 minus
R P plus X Q, V 1 divided by V 1, V 2 is equal
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to this term V 2 by V 1.
.
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So, from this, we can find out the value of
t s for a complete compensation. That is,
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when
V 1 is equal to V 2. Mind it, we are writing
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all these values V 1, V 2 as per unit values.
So, if we want both of them to be 1 per unit,
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then we can write this as V 1 is equal to
V 2
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or both side voltages, becoming equal.
.
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.In this case here and here, both the voltages
on the transmission line, two ends being
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equal.
.
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That is the line is fully compensated.
.
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Then, V 1 is equal to V 2 and t s can be found
out from this relationship.
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..
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Form here, because V 2 by V 1 will be equal
to 1 and solving this, we will get the value
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of t s.
.
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Once, we get t s we know t r is equal to 1
by t s. So, we can also solve for t r. In
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fact, for
any ratio of V 1, V 2 we solve this for t
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s and then t r.
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..
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Now, sometimes, we would like to use a combination
of the transformer taps as well as
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the reactive power injection devices. Such
as, static power systems or static
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compensation, any static compensation device,
like may be stat com, another devices.
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So, here I am showing this, there is between
these two buses voltage V 1 is at a sending
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end and voltage V 2 as at the receiving end.
We have a transmission line and may be a
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three winding transformer, which is connected
near the receiving end.
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And the tertiary of the transformer, we connect
the static var system or static
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compensation device. The primary side we have
this transmission line, connected to the
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sending end. And secondary side, we have the
load connected. That is, it is the receiving
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end bus.
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00:13:32,589 --> 00:13:33,589
..
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The whole diagram an electrical circuit can
be represented by this, where, we have V 1
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the sending end, the current I is following
through this. X 1 is the reactance of the
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transmission line, we have neglected the resistance.
So, it is the reactance of the
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transmission line, plus the reactance of the
primary side of the transformer winding. X
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2
is the reactance of the secondary side of
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the transformer winding. And X 3 is the
reactance of the tertiary winding.
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And we have this as the transformation ratio
the transformer is shown here. Here, we
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have the transformation ratio. And we have
a variable tap. That is T 12 is showing a
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variable tap between primary and secondary.
So, by changing this T 12, we are basically
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changing the tap on the secondary side of
this. Similarly, we have here on the tertiary;
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we have connected the static var system or
the stat com, which is a static compensation
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device.
Normally, all these compensating devices,
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we connect to the tertiary. Also, it is always
connected to the low voltage side. For simple
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reason, because developing or making
these devices with low voltage is much cheaper
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compare to making these devices for
very high voltage system. So, therefore, we
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normally connect these devices at the low
voltage end.
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So, this is connected on the tertiary. Now,
if you see if P plus j Q is the power, which
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is
coming from the sending end. We will have
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at the receiving end, the power P R plus j
Q
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.R. And if we want that the voltage of the
system to be maintained, then what we need
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is
to transfer less amount of reactive power
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from here.
.
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Less amount of reactive power from here, because
we know that, if we transfer less
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amount of reactive power, the voltage at this
point will be much higher or much nearer to
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the voltage here. And therefore, by trying
to inject the amount of reactive power, through
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this compensating device, we can reduce or
make this reactive power flow on the line
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and in the primary of the transformer to very
low value or even 0. Sometimes, we can
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even make it leading and thereby, we can control
the voltage here. The other option is by
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changing that transformer taps; we can again
regulate the voltage at V 2. So, a
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combination of these two, will give me a much
larger variation over a larger load
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variation in the system.
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00:16:58,790 --> 00:16:59,790
..
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00:16:59,790 --> 00:17:11,120
Now, many times we use regulating transformers.
Basically, these transformers are used
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for injecting voltage into the transmission
system. If we inject the voltage in the same
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phase as the voltage on the transmission system,
then what we will be doing is, either we
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can add the voltage. And therefore, boost
the voltage or if we want we can inject the
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voltage in opposite phase. Then, we will be
reducing the voltage or bucking the voltage.
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So, we can have a buck transformer or a boost
transformer. Normally, in power system,
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we will be using boost transformers, most
of the time, because with the increase in
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00:17:51,300 --> 00:17:54,570
load,
it is always the voltage which drops down.
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So, we try to inject voltage in phase with
the
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voltage on the transmission line. By that
way, we can control or the magnitude of the
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voltage.
If we inject the voltage, in not in phase,
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with the transmission line voltage, but in
quadrature or at some different angle, then
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we can be making a phase shift of the voltage
as well. Because, the final voltage which
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is coming at the receiving end will be the
some
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of these two voltages, the sending end voltage
plus the voltage, which we are adding by
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this transformer. And since to these two are
phases and they are not in same phase. We
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00:18:45,630 --> 00:18:51,270
will get the resultant voltage having some
phase difference, from the voltage at the
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sending end or at the primary side of these
transformers.
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00:18:54,930 --> 00:18:55,930
..
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00:18:55,930 --> 00:19:06,220
So, here, this is voltage magnitude controller
or regulating transformer, which controls
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00:19:06,220 --> 00:19:15,220
the voltage magnitude only. If you look at
this, here we have transformer winding, which
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00:19:15,220 --> 00:19:20,419
is a three phase winding, connected to the
three phase transformer. And we have taps
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00:19:20,419 --> 00:19:25,250
on
this winding. Now, based on all these three
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00:19:25,250 --> 00:19:30,070
taps on the three phases are all ganged. So,
they will move together.
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So, what we get is, this voltage, whatever
voltage we want here, between these two
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points, will get injected here. So, this voltage
can be added to the transmission line
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00:19:47,309 --> 00:19:53,669
voltage in the same phase here. This voltage
is in the same phase as this voltage. And
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00:19:53,669 --> 00:20:00,280
therefore, the voltage here is going to be
the voltage at this point plus this voltage
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00:20:00,280 --> 00:20:01,280
that we
have added.
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Because, it is in phase, the same thing happens
in all the three phases. And therefore, the
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00:20:06,799 --> 00:20:14,570
secondary voltage gets increased by the amount
of the voltage that we have injected.
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This is the way, how we use the regulating
transformer for increasing the magnitude of
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00:20:23,010 --> 00:20:28,120
the voltage at the secondary of this regulating
transformer.
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00:20:28,120 --> 00:20:29,120
..
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Now, the same thing, if we do by injecting
voltage, which is not in phase, but in
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00:20:38,679 --> 00:20:45,110
quadrature. If you look at this transformer,
we have a transformer here, like this, which
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00:20:45,110 --> 00:20:50,730
is
connected to the primary three phases a n,
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00:20:50,730 --> 00:20:54,550
b n and c n are connected at these points.
We
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have another winding, which has half on this
side and half on this side. That is the mid
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00:21:01,560 --> 00:21:08,140
points is connected to this point a and this
winding is having the phase of b c.
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00:21:08,140 --> 00:21:18,820
Similarly, we have a winding here, which is
connected at the midpoint to b and this has
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a
phase which is same as a c. Similarly, we
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00:21:24,410 --> 00:21:33,940
have another winding, which is connected at
the midpoint to phase c and this has a phase
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00:21:33,940 --> 00:21:41,260
which is same as a b. In this way, what we
do is, if you look at the phase a voltage,
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phase a voltage will be like this. Phase a
voltage
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will be like this, phase b voltage will be
like this and phase c voltage, will be like
205
00:21:54,549 --> 00:21:57,970
this.
Therefore, we have seen, we are adding or
206
00:21:57,970 --> 00:22:04,890
subtracting the voltage or adding any voltage,
which is in quadrature, which phase a in this
207
00:22:04,890 --> 00:22:12,460
side or in this side. So, if we see, in this
present position, the three taps are place
208
00:22:12,460 --> 00:22:17,030
like this. Then, we have a voltage V a on
which,
209
00:22:17,030 --> 00:22:25,230
we have this much voltage added, that is delta
Van is added here. Similarly, we have
210
00:22:25,230 --> 00:22:34,639
voltage b plus delta b n is added here. We
have voltage c plus delta c n is added here.
211
00:22:34,639 --> 00:22:35,639
..
212
00:22:35,639 --> 00:22:41,260
And if we see the phasor, it comes out to
be V a is the voltage plus delta V a n is
213
00:22:41,260 --> 00:22:45,990
added
here. V b is the voltage plus delta V b n
214
00:22:45,990 --> 00:22:49,620
is added here. V c is the voltage and delta
c n is
215
00:22:49,620 --> 00:22:59,101
added here. So, now the new voltage a dash,
b dash and c dash will be these points. And
216
00:22:59,101 --> 00:23:06,399
we see that, though there may not be large
change in the magnitude, the angle of these
217
00:23:06,399 --> 00:23:15,190
phases has change. That is a dash is now leading
a by this angle theta.
218
00:23:15,190 --> 00:23:23,700
Similarly, c is leading this c by this angle
theta and b is leading this b, b dash is leading
219
00:23:23,700 --> 00:23:29,630
b
by this angle theta. So, this is, what we
220
00:23:29,630 --> 00:23:36,159
call as a phase regulating transformer or
phase
221
00:23:36,159 --> 00:23:43,220
shifting transformer, this is how it works.
And we can change the phase angle of the
222
00:23:43,220 --> 00:23:50,179
voltage and what is the consequence of changing
this phase angle. As we have seen
223
00:23:50,179 --> 00:23:59,789
earlier, the real power P is given by V s,
V r, by X into sin delta.
224
00:23:59,789 --> 00:24:07,039
Since, V s, V r are regulated to be very near
to 1 per unit and X of the transmission line
225
00:24:07,039 --> 00:24:13,130
is a constant, therefore to change the real
power of flow on the transmission. The only
226
00:24:13,130 --> 00:24:20,880
thing that we can do is, change this angle
power angle delta, which is the angle by which
227
00:24:20,880 --> 00:24:30,889
V s leads V r. And therefore, if you see here
by changing this angle, we can change the
228
00:24:30,889 --> 00:24:36,130
power flow.
So, by changing the angle of at V r or at
229
00:24:36,130 --> 00:24:42,149
V s, we can change the power flow on a
transmission line and phase shifting transformers
230
00:24:42,149 --> 00:24:48,990
are used for this purpose. Of course,
nowadays with power electronic devices coming
231
00:24:48,990 --> 00:24:57,860
into picture, we have new kinds of
controllers, which we use for power system.
232
00:24:57,860 --> 00:24:59,860
Transmission line, voltage and power
233
00:24:59,860 --> 00:25:07,539
.control. These devices are called flexible
ac transmission system devices or facts
234
00:25:07,539 --> 00:25:17,409
devices, because they allow you the flexibility
of controlling the power flow and the
235
00:25:17,409 --> 00:25:22,850
voltage. And thereby, the reactive powers
flow on the transmission line. So, both real
236
00:25:22,850 --> 00:25:29,110
and reactive power flow in a transmission
line can be controlled using these facts
237
00:25:29,110 --> 00:25:42,429
devices. Now, we will take up some simple
examples, on transmission line operation to
238
00:25:42,429 --> 00:25:50,220
understand the concepts. That we learnt in
lesson 9 and 10.
239
00:25:50,220 --> 00:25:51,220
.
240
00:25:51,220 --> 00:26:02,420
So, first example is a 50 hertz, 132 kV, 138
kV 3 phase transmission line is 200
241
00:26:02,420 --> 00:26:10,110
kilometer long. The distributed line parameters
are R is equal to 0.1 ohm per kilometer,
242
00:26:10,110 --> 00:26:17,510
L that is the inductance is equal to 1.2 milli
Henry per kilometer. The capacitance c is
243
00:26:17,510 --> 00:26:26,820
0.01 micro farad per kilometer the conductance
is 0. The transmission line delivers 40
244
00:26:26,820 --> 00:26:38,909
Mega Watt at 132 kV with 0.95 power factor
lagging. Find the sending end voltage and
245
00:26:38,909 --> 00:26:47,720
current and also the transmission line efficiency.
So, what we are given is the transmission
246
00:26:47,720 --> 00:26:50,450
line parameters, it is resistance set in,
it is
247
00:26:50,450 --> 00:26:58,170
inductance and capacitance per kilometer length.
And we have the line length given; we
248
00:26:58,170 --> 00:27:06,980
also know that the transmission line has to
deliver 40 Mega Watt at 132 kV. Means, the
249
00:27:06,980 --> 00:27:15,580
receiving end voltage is 132 kV. And the receiving
end power is 40 Mega Watt at 0.95
250
00:27:15,580 --> 00:27:23,980
power factor lagging. We have to find the
sending end voltage, that is V s and current
251
00:27:23,980 --> 00:27:31,320
that is I s and also the transmission line
efficiency. So, let us see how we work it
252
00:27:31,320 --> 00:27:32,940
out.
253
00:27:32,940 --> 00:27:33,940
..
254
00:27:33,940 --> 00:27:47,080
Now, for the given values of R L and C, we
have omega. That is the equal to twice pi
255
00:27:47,080 --> 00:27:55,590
into f f is 50 hertz, therefore, we can calculate
the series impedance per unit length or per
256
00:27:55,590 --> 00:28:05,409
kilometer as 0.1, which is the resistance
plus j, 0.377. That is twice pi f into L.
257
00:28:05,409 --> 00:28:09,929
That is
1.2 milli henry, per kilometers. So, this
258
00:28:09,929 --> 00:28:14,450
comes out to be 0.1 plus j 0.377, which is
equal
259
00:28:14,450 --> 00:28:22,570
to 0.39 with an angle of 75.14 in ohms, per
kilometer.
260
00:28:22,570 --> 00:28:32,799
Similarly, the shunt admittance can be computed
as j omega c, which comes out to be
261
00:28:32,799 --> 00:28:43,450
twice pi into 50 in to 0.1 microfarad per
kilometer. So, when we do that multiplication,
262
00:28:43,450 --> 00:28:53,380
this is j omega into c and then this comes
out to be j 3.14 into 10 to power minus 6.
263
00:28:53,380 --> 00:28:57,250
This
is equal to 3.14 into 10 to power minus 6,
264
00:28:57,250 --> 00:29:00,950
with an angle of 90 degrees; the unit will
be
265
00:29:00,950 --> 00:29:06,820
mho per kilometer.
Now, using this z and y, we can calculate
266
00:29:06,820 --> 00:29:16,130
the characteristic impedance Z c. Z c is given
by square root of z by y. So, substituting
267
00:29:16,130 --> 00:29:21,610
a value of z and y, we get Z c is equal to
352.42
268
00:29:21,610 --> 00:29:29,960
with an angle of minus 7.43, this will have
unit is of ohm, because the length per
269
00:29:29,960 --> 00:29:36,909
kilometer, will cancel out and this, will
have the unit ohm. Similarly, we can calculate
270
00:29:36,909 --> 00:29:45,100
gamma l, this is l is 200 and gamma is given
by root z into y.
271
00:29:45,100 --> 00:29:53,690
So, substituting the values, we get this as
equal to 0.2213 with an angle of 82.75 degrees,
272
00:29:53,690 --> 00:30:05,380
which can be written in rectangular coordinates
as 0.0286 plus j 0.2194. So, now that we
273
00:30:05,380 --> 00:30:14,009
have calculated gamma l and Z c, we can find
out also the values of sin hyperbolic
274
00:30:14,009 --> 00:30:15,940
gamma l and cos.
275
00:30:15,940 --> 00:30:16,940
..
276
00:30:16,940 --> 00:30:21,120
Hyperbolic gamma l sign hyperbolic gamma l
is equal to e to the power gamma l minus e
277
00:30:21,120 --> 00:30:26,240
to the power of minus gamma l divided by 2,
which is after substituting the value of
278
00:30:26,240 --> 00:30:35,039
gamma l will come out to be 0.2195 with an
angle 82.67. Similarly, cos hyperbolic
279
00:30:35,039 --> 00:30:40,399
gamma l will be equal to e to the power gamma
l plus e to the power minus gamma l.
280
00:30:40,399 --> 00:30:46,779
This whole divided by 2, which will come out
to be on substituting the values of gamma
281
00:30:46,779 --> 00:30:52,450
l. This comes out to be 0.975 with an angle
0 .37 degrees.
282
00:30:52,450 --> 00:30:53,450
.
283
00:30:53,450 --> 00:30:59,730
.So, knowing sign hyperbolic gamma l cos hyperbolic
gamma l and Z c. We can calculate
284
00:30:59,730 --> 00:31:10,990
a, b, c, d parameters for the transmission
line. Now, the value of power and voltage
285
00:31:10,990 --> 00:31:14,740
is
specified in the problem refers to three phase
286
00:31:14,740 --> 00:31:22,110
and line to line quantities. That is the
values, that we are given in the problem is
287
00:31:22,110 --> 00:31:28,530
in terms of 132 kV, which is a line to line
voltage and 40 Mega Watt, which is a three
288
00:31:28,530 --> 00:31:33,101
phase power.
Therefore, the V 2 or the receiving and voltage
289
00:31:33,101 --> 00:31:39,539
magnitude is equal to 132 by root 3. We
are talking of per phase value or phase to
290
00:31:39,539 --> 00:31:43,470
neutral value. So, it is 132 divide by root
3,
291
00:31:43,470 --> 00:31:49,649
which come out to be 76.2 kV, because we would
like to work on a single phase bases.
292
00:31:49,649 --> 00:31:56,840
Therefore, we will convert all the line to
line voltages into line to neutral voltages
293
00:31:56,840 --> 00:32:01,299
and
three phase power into single phase power.
294
00:32:01,299 --> 00:32:05,659
That is power per phase. Also we will use
V
295
00:32:05,659 --> 00:32:12,090
2, the receiving end voltage as a reference.
Therefore, the angle of V 2, we will take
296
00:32:12,090 --> 00:32:16,779
as 0
degree. Therefore, the phasor V 2 is equal
297
00:32:16,779 --> 00:32:24,820
to 76.2 angle 0 degree in Kilo Volts.
.
298
00:32:24,820 --> 00:32:29,799
And the per phase power supplied to the load,
will be equal to 40, is the three phase
299
00:32:29,799 --> 00:32:37,090
powers. So, 40 divided by 3; that is 13.33
Mega Watt. Given the value of power of factor
300
00:32:37,090 --> 00:32:45,600
which is 0.95, we can find out the current
at the receiving end I 2. Now, we know P is
301
00:32:45,600 --> 00:32:54,630
equal to V I cos 5. So, the receiving end
power P load will be equal to this is cos
302
00:32:54,630 --> 00:33:01,529
5, 0.95
into V into I. And therefore, substituting
303
00:33:01,529 --> 00:33:06,269
the value of P load and V 2 we get I 2, as
equal
304
00:33:06,269 --> 00:33:08,830
to 1 184.1 amperes.
305
00:33:08,830 --> 00:33:16,980
.We also know that, the phase angle between
V 2 and I 2, is given by cos inverse 0.95,
306
00:33:16,980 --> 00:33:25,970
which is the power factor angle. So, also
since I 2 lags V 2 by cos invar 0.95, which
307
00:33:25,970 --> 00:33:29,760
is
18.195 degrees. Therefore, we can write the
308
00:33:29,760 --> 00:33:34,529
phasor I 2, as I 2 is equal to 184.1 with
an
309
00:33:34,529 --> 00:33:42,929
angle of minus 18.195. Because, I 2 is going
to lags the voltage V 2, because the power
310
00:33:42,929 --> 00:33:46,559
factor is 0.95 lagging.
.
311
00:33:46,559 --> 00:33:54,740
Therefore, now we can calculate the sending
end voltage V 1, which will be given by a
312
00:33:54,740 --> 00:34:02,240
V
2 plus b, I 2. And a is cos hyperbolic gamma
313
00:34:02,240 --> 00:34:09,011
l and V is Z c sin hyperbolic gamma l.
Therefore, V 1 is equal to V 2 cos hyperbolic
314
00:34:09,011 --> 00:34:15,950
gamma l plus I 2 into Z c sin hyperbolic
gamma l. Now, if we substitute the value of
315
00:34:15,950 --> 00:34:22,410
cos hyperbolic gamma l, sin hyperbolic
gamma l and Z c. Then, we will get the value
316
00:34:22,410 --> 00:34:35,090
of V 1 as equal to 82.96 with an angle, 8.6
degrees in Kilo Volts. So, V 1 is 82.96 Kilo
317
00:34:35,090 --> 00:34:42,810
Volts with an angle of 8.6 degrees.
Now, this is a line to neutral voltage at
318
00:34:42,810 --> 00:34:47,159
the sending end. Similarly, we can calculate
the
319
00:34:47,159 --> 00:35:00,760
sending end current I 1, which will be equal
to c, sorry which will be equal to C V 2 plus
320
00:35:00,760 --> 00:35:10,110
D I 2. D is equal to A, which is equal to
cos hyperbolic gamma l and C is sin hyperbolic
321
00:35:10,110 --> 00:35:17,770
gamma l by Z c. Therefore, I 1 is equal to
I 2 cos hyperbolic gamma l plus V 2 by Z c
322
00:35:17,770 --> 00:35:21,160
sin
hyperbolic gamma l. Substituting, the value
323
00:35:21,160 --> 00:35:28,820
of V 2, Z c sin hyperbolic gamma l, cos
hyperbolic gamma l and I 2, we will get I
324
00:35:28,820 --> 00:35:38,910
1 as equal to 179.46 with an angle of 17.79
degrees.
325
00:35:38,910 --> 00:35:39,910
..
326
00:35:39,910 --> 00:35:47,720
Now, we can calculate the efficiency of the
transmission, so per phase, input power will
327
00:35:47,720 --> 00:35:52,790
be given by real part of V 1, I 1 conjugate.
.
328
00:35:52,790 --> 00:36:00,880
If we do this multiplication, because we know,
V 1 is this and I 1 is this. We will take
329
00:36:00,880 --> 00:36:03,460
a
conjugate, which means, we will put a negative
330
00:36:03,460 --> 00:36:05,240
sign to this angle.
331
00:36:05,240 --> 00:36:06,240
..
332
00:36:06,240 --> 00:36:10,930
And then we multiply these two, then we are
going to get and we take only the real part
333
00:36:10,930 --> 00:36:19,400
of that, then we get this as equal to 14.69
Mega Watt. Now, per phase power, which is
334
00:36:19,400 --> 00:36:27,130
received, is 40 by 3, which is 13.33. Therefore,
efficiency of transmission is output
335
00:36:27,130 --> 00:36:36,500
divided by input, which comes out to be 13.33
divided by 14.69. That is 0.907 or the
336
00:36:36,500 --> 00:36:46,470
efficiency is 90.7 percent. So, this is one
simple problem, that we solved.
337
00:36:46,470 --> 00:36:47,470
.
338
00:36:47,470 --> 00:36:56,350
Now, we will take up another problem. Now,
this problem states a three phase 132 kV
339
00:36:56,350 --> 00:37:05,010
overhead line delivers 60 MVA. Mind it, this
is 60 MVA not Mega Watt, 60 MVA at
340
00:37:05,010 --> 00:37:13,500
.132 kV and power factor 0.8 lagging at it
is receiving end. That is V r or V 2 is given
341
00:37:13,500 --> 00:37:22,350
as
132 kV and the P S R is the MVA value is given
342
00:37:22,350 --> 00:37:31,870
as 60, MVA at 0.8 power factor
lagging. The constants of the line are given
343
00:37:31,870 --> 00:37:39,810
as A is equal to 0.98 with an angle 3 degrees
and B is equal to 100 with an angle 75 degrees
344
00:37:39,810 --> 00:37:47,140
ohms per phase. Now, we need to find for
this system, sending end voltage and power
345
00:37:47,140 --> 00:37:56,920
angle, B sending and active and reactive
power, C line losses and vars absorbed by
346
00:37:56,920 --> 00:38:00,900
the line.
.
347
00:38:00,900 --> 00:38:07,170
Also we would like to find out the capacity
of static compensation equipment at the
348
00:38:07,170 --> 00:38:13,610
receiving end to reduce the sending end voltage
to 145 kV for the same load conditions.
349
00:38:13,610 --> 00:38:14,610
.
350
00:38:14,610 --> 00:38:24,060
.That is whatever voltage, we have found out
in A will be somewhat higher. And if you
351
00:38:24,060 --> 00:38:31,200
want to reduce this voltage by keeping the
receiving end voltages kept at 132. And now,
352
00:38:31,200 --> 00:38:39,120
we want to keep the sending end voltage also
at 145 only. That is, we do not want the
353
00:38:39,120 --> 00:38:47,230
voltage at the sending end two go beyond 145
kV. And therefore, we will require some
354
00:38:47,230 --> 00:38:51,280
compensation, reactive power compensation
at the receiving end. So, we need to find
355
00:38:51,280 --> 00:38:58,560
out, what is the rating of the compensating
equipment for such a condition.
356
00:38:58,560 --> 00:38:59,560
.
357
00:38:59,560 --> 00:39:03,770
Also we need to find, the unity power factor
load, which can be supplied at the receiving
358
00:39:03,770 --> 00:39:10,500
end with 132 kV as the line voltage at both
the ends. Now, if we want to keep the voltage
359
00:39:10,500 --> 00:39:17,119
at both the ends at 132 kV, then what is the
amount of unity power factor load?
360
00:39:17,119 --> 00:39:18,119
..
361
00:39:18,119 --> 00:39:24,310
That is purely resistive load, which can be
supplied by this transmission line, when we
362
00:39:24,310 --> 00:39:29,320
are working with three phase system, especially
when we are calculating voltages and
363
00:39:29,320 --> 00:39:35,830
current. It is always better to work with
the phase to neutral voltages and the currents
364
00:39:35,830 --> 00:39:44,810
flowing in each phase. So, what we do is,
first we calculate the phase voltage, now
365
00:39:44,810 --> 00:39:46,940
for
the receiving end, we have been given the
366
00:39:46,940 --> 00:39:54,200
voltage as 132 kV line to line.
So, we divided by root 3 to get the phase
367
00:39:54,200 --> 00:40:06,660
voltage, which comes out to be 76210, volts.
That is 76.21 kV and we are choosing this
368
00:40:06,660 --> 00:40:14,160
receiving end voltage as a reference voltage.
So, we fix it is angle as 0 degree. Now, next,
369
00:40:14,160 --> 00:40:19,570
what we have to do is, we need to find out
the current, at the receiving end. We know
370
00:40:19,570 --> 00:40:22,520
that the power at the receiving end is given
as
371
00:40:22,520 --> 00:40:29,660
60 and the a, at a power factor of 0.8 power
lagging.
372
00:40:29,660 --> 00:40:37,400
So, what we do is 60 into 10 to power 6, gives
us the 60 MVA with divided by the phase
373
00:40:37,400 --> 00:40:47,000
voltage. So, root 3 into, sorry, root 3 into
the line voltage. So, this is 3 into 0.5,
374
00:40:47,000 --> 00:40:51,230
3 to the
power 0.5. That is root 3 into the line voltage,
375
00:40:51,230 --> 00:40:59,300
which is 132 kV. And the angle is given as
cos inverse 0.8 with a minus sign to take
376
00:40:59,300 --> 00:41:03,980
care of the lagging power factor. When, we
do
377
00:41:03,980 --> 00:41:13,410
this mathematics, we will get the receiving
end current as 262.43 amperes, with an angle
378
00:41:13,410 --> 00:41:20,760
of minus 36.87 degrees.
Next is, we need to calculate the sending
379
00:41:20,760 --> 00:41:24,900
end voltage V s. We know is equal to A into
V
380
00:41:24,900 --> 00:41:32,310
r plus B into I r. Since, we have already
calculated V r as shown here and I r as shown
381
00:41:32,310 --> 00:41:37,620
here. So, we substitute these values. So,
A is given as 0.98 angle 3 degrees, V r is
382
00:41:37,620 --> 00:41:39,910
76210
383
00:41:39,910 --> 00:41:50,660
.volts plus B is given as 100 angle 75 degrees.
And I r is given as 262.43 with an angle of
384
00:41:50,660 --> 00:41:55,930
minus 36.87 degrees.
Now, here we have not taken any angle here,
385
00:41:55,930 --> 00:42:02,870
because the angle of V r is 0 degree, which
is the reference angle, with respect to which
386
00:42:02,870 --> 00:42:11,470
we are calculating all angles. So, when we
do this calculation, we find this as 74685.8,
387
00:42:11,470 --> 00:42:24,300
with an angle of 3 degrees plus 26243, with
an angle of 38.13 degrees. And finally, when
388
00:42:24,300 --> 00:42:36,010
we calculate this, we get this, out as 74.59
plus j 3.909, which we are this polar coordinate
389
00:42:36,010 --> 00:42:38,990
value is converted into rectangular
coordinate values.
390
00:42:38,990 --> 00:42:46,020
And this polar coordinate values is put in
to rectangular coordinate values, then it
391
00:42:46,020 --> 00:42:51,680
comes
out to be 20.64 plus j 16.20 and since these
392
00:42:51,680 --> 00:42:59,990
are in 1000’s. So, we are multiplying this
into 10 to the power 3. So, here instead of
393
00:42:59,990 --> 00:43:03,240
writing this 1000’s, we have taken this
1000’s
394
00:43:03,240 --> 00:43:04,240
here.
.
395
00:43:04,240 --> 00:43:12,231
So, when we do this calculation, this comes
out to be 95.23 plus j 20.11 into10 to the
396
00:43:12,231 --> 00:43:21,060
power 3. That is equal to 97.33 into 10 to
the power 3, volts with an angle of 11.92,
397
00:43:21,060 --> 00:43:29,900
which tells us that the sending end voltage
phase to neutral voltage is 97.33 kV. And
398
00:43:29,900 --> 00:43:34,570
the
power angle is 11.92 volts, because this is
399
00:43:34,570 --> 00:43:36,911
the angle of the sending end with respect
to
400
00:43:36,911 --> 00:43:37,911
the reference.
401
00:43:37,911 --> 00:43:42,600
.If you want to find out the line to line
voltage at the sending end, then we multiply
402
00:43:42,600 --> 00:43:46,180
this
by root 3. So, when we do this, we get this
403
00:43:46,180 --> 00:43:54,060
as 168.58 kV and the power angle as we have
seen is equal to 11.92 degrees.
404
00:43:54,060 --> 00:43:55,060
.
405
00:43:55,060 --> 00:44:02,800
Now, for the b part, where we need to find
out the sending end real and reactive power.
406
00:44:02,800 --> 00:44:09,560
We have three phase power in Mega Watt given
as S s is equal to A V s squared divided
407
00:44:09,560 --> 00:44:17,930
by B, with an angle of beta minus alpha for
division by B. We are writing B inverse, that
408
00:44:17,930 --> 00:44:28,420
is 1 by B minus V r, V s divided by B with
an angle beta plus delta. Now, here, mind
409
00:44:28,420 --> 00:44:31,080
it,
when we are writing this as three phase power,
410
00:44:31,080 --> 00:44:35,740
then we are writing V s, V r all as line to
line voltages.
411
00:44:35,740 --> 00:44:48,730
Therefore, substituting the values we have
A is 0.98, V s is 168.58. So, into 168.58
412
00:44:48,730 --> 00:45:04,300
square divided by B which is 100 at an angle
of beta is 75 and alpha is 3 degrees. Minus
413
00:45:04,300 --> 00:45:17,990
V r into V s, that is 132 into 168.58 divided
by 100, which is B in here. So, with an angle
414
00:45:17,990 --> 00:45:24,910
of beta plus deltas, that is 75 degrees is
beta plus delta is 11.92, as we have calculated
415
00:45:24,910 --> 00:45:26,990
in
the previous part.
416
00:45:26,990 --> 00:45:36,320
So, putting all these values and solving,
we have got this S s is equal to 278.49 at
417
00:45:36,320 --> 00:45:40,851
an
angles of 72 degrees minus 222.53 at an angle
418
00:45:40,851 --> 00:45:51,990
of 86.92. We do this calculation for
calculating the real part, what we need to
419
00:45:51,990 --> 00:45:57,620
is, take the cos of these angles. And for
reactive power, we take the sin of these angles.
420
00:45:57,620 --> 00:46:05,910
So, P s is equal to 278.49 cos 72 minus
421
00:46:05,910 --> 00:46:14,970
.222.53 cos 86.92, which comes out to be 86.06
minus 11.96. That is the sending end real
422
00:46:14,970 --> 00:46:22,020
power is 74.10 Mega Watt.
Now, since we have used these kilovolt values
423
00:46:22,020 --> 00:46:27,710
here, the power will be in Mega Watt,
because multiplication of 10 to the power
424
00:46:27,710 --> 00:46:31,250
3 and 10 to the power 3, will give me 10 to
the
425
00:46:31,250 --> 00:46:38,030
power 6. So, the power here is directly in
Mega Watt. When we are using the voltages
426
00:46:38,030 --> 00:46:41,050
as
Kilo Volts and if we are using line to line
427
00:46:41,050 --> 00:46:48,160
voltages, then the power is a three phase
power. So, this is a three phase power sending
428
00:46:48,160 --> 00:46:53,070
end power is equal to 74.10 Mega Watt.
.
429
00:46:53,070 --> 00:47:00,830
Similarly, we can calculate the sending end
reactive power Q s, which will be 278.49
430
00:47:00,830 --> 00:47:10,360
into sin of 72 degrees minus 222.53 into sin
of 86.92 degrees. That is, we have to take
431
00:47:10,360 --> 00:47:18,450
the sin of the angles, what we had got for
S s. Therefore, this comes out to be 264.89
432
00:47:18,450 --> 00:47:28,260
minus 222.21 which is equal to 42.65 Mega
Vars lagging. So, this way, we have
433
00:47:28,260 --> 00:47:33,530
calculated the sending end real and reactive
power.
434
00:47:33,530 --> 00:47:41,940
Now, in part c, we need to calculate the line
loses and the Mega Var absorbed by the
435
00:47:41,940 --> 00:47:47,530
line. So, line loses will be nothing but,
whatever sending end power is there and
436
00:47:47,530 --> 00:47:52,350
whatever power is received. If we take the
difference of the two, that much amount of
437
00:47:52,350 --> 00:47:59,460
real power is lost in the line or lost in
the resistance of the line or dissipated by
438
00:47:59,460 --> 00:48:01,270
the
resistance of the line.
439
00:48:01,270 --> 00:48:09,250
.Therefore, line loses is equal to P s minus
P r, which is equal to 74.10 minus the
440
00:48:09,250 --> 00:48:19,710
receiving end real power will be 60 into 0.8.
So, MVA into cos 5, so 60 into 0.8, that is
441
00:48:19,710 --> 00:48:28,420
48. So, therefore, line loses comes out to
be 26.10 Mega Watt, which is quite a large
442
00:48:28,420 --> 00:48:36,870
amount us loss, almost one-third power is
getting lost in the lag. Similarly, MVar
443
00:48:36,870 --> 00:48:41,530
absorbed by the line, will be whatever is
then we are at the sending end, whatever is
444
00:48:41,530 --> 00:48:45,700
being sent on the line and whatever is being
received at the receiving end.
445
00:48:45,700 --> 00:48:55,491
So, Q s minus Q r, which is coming out to
be 42.65 minus 60 into 0.6, that is sin of
446
00:48:55,491 --> 00:49:01,180
the
power factor angle. So, this is 60 into 0.6,
447
00:49:01,180 --> 00:49:05,480
which comes out to be 6.65, MVar, that is
this
448
00:49:05,480 --> 00:49:11,400
much 6.65 and MVar is being absorbed by the
line itself.
449
00:49:11,400 --> 00:49:12,400
.
450
00:49:12,400 --> 00:49:21,410
Next, we say that, we want to keep the sending
end voltage at 145 kV and the receiving
451
00:49:21,410 --> 00:49:28,541
end voltage is kept at 132 kV, then what is
the amount of compensation, that is required.
452
00:49:28,541 --> 00:49:36,860
Now, the since we are transmitting 60 MVA
at 0.8 power factor lagging, which comes
453
00:49:36,860 --> 00:49:45,140
out be 48 Mega Watt. Therefore, for this P
r is equal to 48 Vs is equal to 145 kV, V
454
00:49:45,140 --> 00:49:48,230
r is
equal to 132 kV.
455
00:49:48,230 --> 00:49:54,160
Since, this is in Mega Watt and these are
in K v. So, we need not write those 10 to
456
00:49:54,160 --> 00:49:58,010
power
3 and 10 to power 6. So, P r is equal to V
457
00:49:58,010 --> 00:50:05,470
s, V r divide by B into cos beta minus delta
minus A into V r square by B into cos beta
458
00:50:05,470 --> 00:50:15,290
minus alpha. Now, here we know P r value as
48. So, we substituting all the values we
459
00:50:15,290 --> 00:50:20,660
will get 48 is equal to 145 into 132 V s into
V r
460
00:50:20,660 --> 00:50:24,390
divided by 100 into cos of beta minus delta.
461
00:50:24,390 --> 00:50:32,320
.Now, here we do not know, we know beta, but
we do not know delta, for this particular
462
00:50:32,320 --> 00:50:39,660
operating condition. Therefore, I have not
substituted any value here and for these terms
463
00:50:39,660 --> 00:50:49,500
A is 0.98 V r is 132. So, A V r square divided
by b into cos beta minus alpha, that is 75
464
00:50:49,500 --> 00:50:58,010
minus 3. Now, from this, again if we solve
this, we will get cos beta minus delta is
465
00:50:58,010 --> 00:51:05,710
equal
to 0.5275 or angle beta minus delta is equal
466
00:51:05,710 --> 00:51:15,930
to 58.16 degrees. Now, we know beta as 75.
So, we can calculate delta very easily, but
467
00:51:15,930 --> 00:51:20,583
we do not need the delta value as such, what
we need is angle beta minus delta.
468
00:51:20,583 --> 00:51:21,583
.
469
00:51:21,583 --> 00:51:29,400
So, we will use this. Now, we know Q r is
equal to V s, V r divided by B into sin beta
470
00:51:29,400 --> 00:51:37,210
minus delta minus A V r square divided by
B into sin beta minus alpha. So, again
471
00:51:37,210 --> 00:51:46,490
substituting the values, we will get Q r is
equal to 145 into 132 divided by 100 into
472
00:51:46,490 --> 00:51:51,980
sin
58.16. This is beta minus delta into A 0.98
473
00:51:51,980 --> 00:52:01,200
into V r square 132 squared divided by 100
into sin 72 degrees. So, after calculating
474
00:52:01,200 --> 00:52:07,050
this comes out to 162.60 minus 162.40, which
is
475
00:52:07,050 --> 00:52:16,220
equal to 0.2 Mega Var.
Thus for a V s of 145 kV V r and V r of 132
476
00:52:16,220 --> 00:52:24,380
kV with P r 48 Mega Watt a lagging, Mega
Var of 0.2 will be supplied from the line
477
00:52:24,380 --> 00:52:32,950
along with the real power of 48 Mega Watt.
Now, since the load requires 60 into 0.6,
478
00:52:32,950 --> 00:52:39,950
that is 36 Mega Vars lagging, the static
compensation equipment must supply the rest,
479
00:52:39,950 --> 00:52:43,380
because the line will supply only 0.2
Mega Var.
480
00:52:43,380 --> 00:52:44,380
..
481
00:52:44,380 --> 00:52:51,540
So, the static equipment must deliver 36,
minus 0.2, that is 35.8 Mega Var lagging or
482
00:52:51,540 --> 00:52:56,250
it
we can say it must absorb 35.8 Mega Var leading.
483
00:52:56,250 --> 00:53:02,670
That means, the static compensation
there has to be capacitive to provide this
484
00:53:02,670 --> 00:53:12,750
much Mega Var. The capacity of static
capacitor is therefore, 35.8 Mega Var. The
485
00:53:12,750 --> 00:53:21,820
third part was if we keep both V s and V r
equal to 132 kV and the load is unity power
486
00:53:21,820 --> 00:53:28,620
factor, that is Q r is equal to 0, then what
will be the power, that we can supply. So,
487
00:53:28,620 --> 00:53:33,110
using this identity Q r is equal to 0, we
will
488
00:53:33,110 --> 00:53:40,100
write Q r relationship as Q r is equal to
V s, V r by B into sin beta minus delta, minus
489
00:53:40,100 --> 00:53:43,960
A
V r square by B the into sin beta minus alpha.
490
00:53:43,960 --> 00:53:51,860
And substituting all the values, here we
can calculate the angle beta minus delta,
491
00:53:51,860 --> 00:53:57,030
which comes out to be 68.75 degrees.
492
00:53:57,030 --> 00:53:58,030
..
493
00:53:58,030 --> 00:54:01,800
Now, since we have calculated beta minus delta,
we can write down the equation for P r,
494
00:54:01,800 --> 00:54:10,100
V s, V r by B into cos beta minus delta minus
A V r square by B into cos beta minus
495
00:54:10,100 --> 00:54:15,760
alpha substituting all the values, we will
get this as 10.36 Mega Watt. That is, if we
496
00:54:15,760 --> 00:54:18,580
want
to keep the voltage at both ends equal to
497
00:54:18,580 --> 00:54:27,880
132 kV and we want to supply unity power
factor load. Then the load that we can supply
498
00:54:27,880 --> 00:54:32,400
will be only 10.36 Mega Watt.
.
499
00:54:32,400 --> 00:54:42,550
Now, let us take another example, a 220 kV
line has tap changing transformers at both
500
00:54:42,550 --> 00:54:49,890
ends. The transformers at the sending end
has nominal ratio of 11 to 220 kV and at the
501
00:54:49,890 --> 00:54:56,200
.receiving end 220 to 11 kV. So, step up transformer
at the sending end step down
502
00:54:56,200 --> 00:55:02,460
transformer at the receiving end the line
impedance is ten plus j 40, again we are using
503
00:55:02,460 --> 00:55:06,830
a
short line model. So, 10 plus j 40 ohms and
504
00:55:06,830 --> 00:55:13,070
the load at the receiving end is 100 Mega
Watt at 0.9 power factor lagging. Assuming
505
00:55:13,070 --> 00:55:15,650
that the product of off nominal tap setting
is
506
00:55:15,650 --> 00:55:20,550
1, that is T s into t r is equal to 1, as
we said earlier, find the tap settings to
507
00:55:20,550 --> 00:55:23,980
give 11 kV at
load bus.
508
00:55:23,980 --> 00:55:24,980
.
509
00:55:24,980 --> 00:55:30,930
So, this is the example, for what we take
for tap changing transformer in today’s
510
00:55:30,930 --> 00:55:41,500
class.
So, now S is equal to 100 MVA, let us 100
511
00:55:41,500 --> 00:55:46,950
Mega Watt is being transmitted at 0.9 power
factor lagging. So, S per phase will be 100
512
00:55:46,950 --> 00:55:55,020
into 10 to power 6, this is Mega Watt divided
by 3. So, that is Mega Watt per phase divided
513
00:55:55,020 --> 00:56:02,210
by 0.9 will give me because 0.9 is the
power factor, this will give me the MVA.
514
00:56:02,210 --> 00:56:08,350
So, this is Mega Watt divided by the power
factor gives me the MVA. So, this comes out
515
00:56:08,350 --> 00:56:20,140
to be 37.03 into 10 power 6 volt ampere 37.06
MVA per phase and P is known as
516
00:56:20,140 --> 00:56:28,300
hundred Mega Watt. So, per phase is 33.33
Mega Watt. Therefore, we can calculate what
517
00:56:28,300 --> 00:56:34,110
is the value of Q, Q will be S square minus
P square root of that. So, this comes out
518
00:56:34,110 --> 00:56:40,650
to be
16.16 Mega Var per phase.
519
00:56:40,650 --> 00:56:41,650
..
520
00:56:41,650 --> 00:56:49,840
Now, we have V 1 is equal to V 2 is equal
to 220 kV and since, we want to work out on
521
00:56:49,840 --> 00:56:52,200
a
per phase basis. So, we divide it by root
522
00:56:52,200 --> 00:56:56,450
3. So, we will get the line to neutral voltage.
So,
523
00:56:56,450 --> 00:57:04,080
220 into 10 to power 3, divide by root 3 volts.
This is coming out to 127.02 Kilo Volts.
524
00:57:04,080 --> 00:57:05,080
.
525
00:57:05,080 --> 00:57:11,580
Now, R is given as 10 X is given as 40, therefore
substituting this on the relationship for
526
00:57:11,580 --> 00:57:18,780
the transformer. That we had derived today,
1 by t s squared is equal to 1 minus R into
527
00:57:18,780 --> 00:57:22,630
P
plus X into Q divided by V 1, V 2. So, substituting
528
00:57:22,630 --> 00:57:29,240
all these values of R, P, X, Q, V 1, V
529
00:57:29,240 --> 00:57:37,750
.2. We get t s is equal to 1.0317. And therefore,
t r is equal to 1 by t s, which will be
530
00:57:37,750 --> 00:57:43,651
0.9692.
So, if we keep this of nominal taps t s as
531
00:57:43,651 --> 00:57:56,070
1.0317 and t r is equal to 0.9692 both end
voltages will be equal to 220 kV line to line.
532
00:57:56,070 --> 00:58:07,210
So, with these three examples, we will end
today’s lesson. And in the next lesson,
533
00:58:07,210 --> 00:58:11,530
we will review whatever we have learnt about
the
534
00:58:11,530 --> 00:59:23,670
transmission line.
So, thank you very much.
535
00:59:23,670 --> 00:59:23,670
.