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Welcome to module number two, and in this
first lecture, I will discuss some of the
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introductory power of the equipment and their
stability constraints in the power system
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operation. The various issues that will be
discussed in this module is your capability
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and constraints of generators, exciters, governors
as well as the network element constraints
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that is basically your transmissions line
as well as the transformers. And also constraints
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related to the energy supply system, load
characteristics and the introduction to angle
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and voltage stability phenomena s, and what
are the various stability constraints will
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be discussed in detail.
This equability and stability constraints,
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your power system is operating with the various
equipments in the system. So, the constraints
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related to the equipments are also equally
important and here equipment constraints may
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be of two types.
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It may be thermal constraints or it may be
the dielectric constraints. So, all the equipments
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those are in the power system, they must operate
within the specified ratings; rating in terms
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of the voltage and the frequency. So, your
supply system should also have the standard
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rated voltage frequency, so that we can give
this supply to the various operators in the
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power system. In these equipments constraints
as I said, the thermal constraint is one of
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the very very important; that once your equipment
is operating with some electricity supply
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system. Means there some current will be flowing
in some of the conductors in those instruments.
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And once they have some currents in the various
part of the system, so there will be some
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losses as well and that may be your real and
reactive power losses. But we are concerned
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much about the real power losses. The real
power loss is nothing but we know it very
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well that is the current which is flowing
in the system that is I the square of this
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and the resistance of that element where the
current is flowing. So, this creates the loss,
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and if this loss is not dissipated outside,
then there will be constant increase in the
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temperature, and that may lead to your thermal
constraints, and there may be breakdown.
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This it may be heated; it may be burnt some
time. So, we have to see what is the equipment
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constraints in terms of thermal. In the dielectric
constraints as we know, if current is flowing
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at the certain voltage, then there is some
neutral or ground potential and the current
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which is flowing in that any part of the system
that will have some potential. So, there should
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be some insulation, so that there should not
be a short circuit. So, this insulation is
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providing from the current voltage that is
where the current is flowing this electricity
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current as well as your zero potential, ground
potential or you can say neutral potential.
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So, we have to provide some insulation and
that is called the dielectric.
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Now only there is different type of dielectric
we know; it may be your solid dielectric,
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may be your gaseous dielectric, may be your
liquid dielectric, but it depends upon system
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to system. Most of the operators that is the
utilization or you can say end users equipments;
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they use normally your solid dielectrics.
So, whenever there is some change in the voltage,
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there may be some overstress; your dielectric
may over state or, it may get ruptured. And
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there may be a short circuit; there is a fault.
So, the rating of equipments basically decided
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by thermal rating as well as the dielectric
rating; so, the operating as I said the equipment
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must operate within the specified ratings
in terms of voltage and it is in terms of
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your frequency.
Means there is your supply system is not providing
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the balance sinusoidal wave or current in
terms of current and voltages, what will happen?
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Then there may be more losses. So, your equipment
must sustain those losses and we have to do
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something. For that, we can take those losses;
that is the loss once it will be created in
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terms of feet i square r and that must be
taken out, so that we can maintain the temperature,
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we can reduce. We should not increase the
temperature that may lead to the damage of
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that equipment.
Another constraint in the whole system is
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your system constraints. One is related to
equipment, and we know there are various equipments
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in the power system. And if we will take the
combined considerations, so there are various
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constraints, system constraints are also arising.
Those constraints I can enumerate that is
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your stability constraints; means we must
operate our power system in stable fashion.
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Means whole synchronous operation means we
have to operate system stably without using
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the synchronous.
Our system must be reliable, so that we have
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to operate system in terms of the reliability
constraints as well, and another is your security
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constraints. I said the security constraints
that we have to operate our power system operating
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constraints that is here the two constraints
here your operating constraints, and another
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is called your load constraints. So, these
two constraints must be satisfied; also we
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call this operating constraints as in previous
lectures, I mentioned this is also called
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inequality constraints I and it is here your
equality constraints E.
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So, these two constraints must be satisfied
and operating constraints we know that the
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line loading must be less than its maximum
rated capacity. At the same time, the voltage
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limits must also well within its minimum and
the maximum value to operate the system in
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the reliable and the secure fashion, so that
we can achieve higher efficiency of the system.
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The stability constraints, there is various
type of stability.
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Normally, we call here either it is a voltage
instability. This is voltage instability;
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I will discuss in detail, and another is your
angle instability. So, we will discuss much
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in detail about these two instabilities in
the later lectures of this module. To go for
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all these analyses, it is very very important.
Let us review that what are the various relations
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in terms of power system; that is in terms
of voltage, current, power and impedance,
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and also how we are going to represent to
get the solutions.
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So, here the various relations that are normally
we know the supply system is our three phase
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balance supply system. And it can be either
in star connected or in delta connected. You
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know it very well that the star connected
here, this is your star connection, and this
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is your delta connection. From these, we can
get the three phase supply here, and this
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is also giving you a three phase supply system.
Now the quantities which are very much used
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those are your power, voltage, current and
the impedance. First, we will see the power
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that is the three phase power.
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It is nothing but the power p is equal to
under root three times multiplication of voltage.
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This VL is line to line; here this voltage
is your VL that is between two phases. So,
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it is called line to line VL. And the current
that is flowing in line here, it is your IL.
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So, the power which is under root times voltage
that is line to line; sometimes, it is also
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called phase to phase voltage multiplied by
your current that is line current, and that
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is multiplied by your power factor that is
power factor here cosine of the phi. And this
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phi is the angle between voltage and current
at any particular point.
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If you want to represent these quantities
means that is your power in terms of phasor
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quantity, we can write thrice that is three
times of the phase voltage as well as the
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phase current. What is the phase voltage here?
The voltage here that will be here VPs that
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is your line to your neutral position; so,
this is also sometime called VP is the phase
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to neutral voltage. So, the VP and the current
here again it depends upon what is the phase
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current which is flowing here that is your
IP. In this delta connection, you have the
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line current is this; that is your IL and
your VL that is the line to line is this voltage.
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But the phase voltage here, this voltage between
this it is your VP, and the current which
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is flowing here it is your IP that is your
phase current. So, whether it is a star connected
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or it is delta connected, you have the real
power equation. It will be the same whether
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you are representing in terms of line to line
current and voltages or in terms of phasor.
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So, here the real power in terms of line voltage
as well as the line current, we are getting
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under root three times voltage multiplied
by your line current and multiplied by that
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is the power factor cosine of the power factor
angle.
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If you are representing in this phasor phase
voltage and phase current, so it is thrice
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of here the phase voltage multiplied by phase
current multiplied by your power factor. What
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is happening here? You can see why it is thrice?
Because we have the three phases, one phase,
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second phase and third phase here. So, the
phase power here one, this and this all will
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be added together and this is the thrice.
We are assuming the phase power here this
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is balanced. So, the power in one phase will
be here in the same this phase second phase
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as well as in the third phase. Similarly,
here also the voltage here VP multiplied by
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IP for one phase, for here another phase,
here for this.
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So, if it is A phase; this is your B phase,
and this your C phase. So, the phase A power,
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phase B power, phase C power will be added
together, and that will give you real power.
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But now let us see this relation; that is
the voltage line to line voltage and the phase
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angle as well as the current relation for
these two configurations. So, line voltage
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here you can say this is between the two phases.
So, this VL will be under root three times
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of your phase voltage; why it is so? You can
see if we will draw the phasor diagram; that
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is here three phase.
This is say let us suppose your V a; this
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is your V b, because we know all these phasor
voltages, they are displayed by 120 degree.
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So, it is 120 degree and here also 120 degree,
here also 120 degree. So, if you will see
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that resultant voltage means I want to write
the voltage V ab; here this is your V ab means
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V a minus V b this voltage. So, this V ab
will be your line to line; here what is that?
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I am writing this is your VL. So, we can very
easily show that this magnitude this V ab
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will be means I can say this V ab will be
under root three times of your V a; that is
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this magnitude.
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So, we can say this line to line voltage will
be under root three times your phase voltage,
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because this is your phase voltage; this is
your phase voltage; this is phase voltage;
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this is your line voltage; this is line voltage;
this is also a line voltage. So, the line
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voltage is related with under root three times
of your phase voltage in the star connected.
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So, this is your star connected configuration
and this line current. Now you can see here
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the line current in this configuration which
is here line current that is equal to your
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phase current. So, whatever the current which
is flowing in the phase A is your line current
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here. So, that is written here in the star
connected this line current will be equal
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to your phase current.
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Let us see in the delta connected. Now you
can see the delta connected. What is happening
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here? This is your V p, and this is your V
L. Now you can see here this voltage and this
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voltage again it is VL. Similarly, you can
see this voltage, this phase voltage; this
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VP is equal to VL. So, I have written the
line voltage will be equal to your phase voltage.
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For the current, now it is reverse, because
here it is a line current; here it is your
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phase current. So, there are some current;
it is addition of the two currents which is
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flowing here; it is phase a, phase b. So,
similarly, if these are the currents, then
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this line current will be three times of your
phasor current.
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So, these are the relations which are very
widely used in the power system. We have seen
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that people get confused with this line voltage,
phase voltage and this and that. Normally,
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some instruments they ask that if the system
which is retained that it is 11 KV three phase
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system. So, this voltage whether it is a phase
voltage or line to line voltage. So, if nothing
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is mentioned. So, this voltage is always line
to line voltage. So, there should not be any
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confusion that the voltage which is mentioned
here, it is always line to line voltage rather
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than it is a phase neutral voltage if it is
a three phase supply system is there.
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In the single phase supply system, there is
no concept of line to line and the phase,
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because always it is your phase. We always
measure the single phase to neutral voltage.
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So, here this is your line to line voltage
unless until it is stated. If it is set this
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11 KV, it is a phase to neutral voltage or
phase voltage, then it will be phase voltage.
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So, this is basically the relation between
voltage, current and the power. Now in the
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power system if we will see, normally the
various equipments those are having with the
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different voltage rating, different power
ratings and especially we have the transformers
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and the generators.
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In the power system, the generators are of
different; they are generating at the different
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frequency not frequency. They generated at
the same frequency if it is an interconnected
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system, but it may be rated at the different
voltage. They may be generating different
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magnitude of power and the transformers those
are interconnecting the generating stations
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as well as the various transmission lines
and may have the different primary and secondary
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voltages.
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So, due to cost and technical reasons especially,
the different operating voltages represent
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the transformers, different rating of generators,
etcetera exist in the power system, and therefore,
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the calculation become very very difficult.
So, if you want to use the actual value calculation,
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so always you should be very much aware about
this the change in the voltage and also in
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terms of connection. Sometimes, a star connected
and delta connected that creates lot of confusion
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in calculating the voltage; whether you are
calculating the phasor voltage, whether you
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are calculating the line to line voltage.
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So, it is always convenient that we can use
the per unit system. In the per unit system,
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we will see the various advantages and another
you can see most of the machine data s,
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they are normally available in the per unit
system of it their own bases. Means machines
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and the systems data are generally available
in per unit values on its own base; therefore,
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it is essential to use the per unit system
of various physical quantities; what are that
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physical quantities? We have only four physical
quantities that is your power, voltage here
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and current as well as your impedance.
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These are the various four power system quantities,
and these four quantities must be represented
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into the per unit system, and the common base
that is also very important. No doubt all
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the elements are having the per unit system
on their own base, but we have to represent
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power system quantities; if you want to solve,
you want to use them for various calculation,
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etcetera, then you have to take the common
base for all these values, and then you have
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to convert it back. Suppose, our machine is
written here 0.2 per unit and its base is
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the different, then you have to change this
impedance at the different phase which is
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you can say system base.
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In the per unit system, the different voltage
levels disappears as I said using the per
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unit system the voltage level consume. Means
whether there is an 11 KV to 132 KV and so
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on, so forth, there will be no voltage variation.
So, it is disappeared and also the power network
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consisting of generators, transformers, transmission
lines and the load reduces to our system of
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simple impedances. Means we can remove; let
us suppose we want to represent a transformer.
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So, the transformer will be represented simply
by its impedance and then we calculate again
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this impedance in per unit.
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So, we can calculate the currents as well
as the voltages, and later, we can get it
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back to actual voltage and the current. The
per unit system as it is a dimensionless system,
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means it has no dimension, and it is represented
as per unit. Some people write here it is
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p dot u, some people write p u simply, so
it is up to you. So, it is per unit means
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p u we represent this. So, per unit is defined.
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The quantity in per unit that is p u is defined
at the ratio of actual quantity in any unit
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here divided by the base or reference value
of the quantity in the same unit. Why it is
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same unit? Suppose you want to write the per
unit for your voltage. So, the voltage here
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00:19:14,710 --> 00:19:25,640
p u will be the actual V actual divided by
V the voltage of the base. So, let us suppose
198
00:19:25,640 --> 00:19:31,200
actual it is in kilovolt. So, your base must
also be in the kilovolt; it is not in volt,
199
00:19:31,200 --> 00:19:35,950
so, that we will get the different value.
So, that is why here the quantity in per unit
200
00:19:35,950 --> 00:19:40,970
is defined at the ratio of actual quantity
in any unit. If you are using voltage, so
201
00:19:40,970 --> 00:19:45,200
it will be here it will be in the same voltage;
if you are using kilovolt, it should be in
202
00:19:45,200 --> 00:19:46,080
the same kilovolt.
203
00:19:46,080 --> 00:19:54,080
So, base or reference value of quantity in
the same unit that is the per unit definition.
204
00:19:54,080 --> 00:20:01,309
We know that a well chosen per unit can reduce
the computational effort, simplify evaluation,
205
00:20:01,309 --> 00:20:06,980
and facilitate the understanding of the system
characteristic; this is one of the advantages.
206
00:20:06,980 --> 00:20:14,130
The selection of base quantities is also very
important. Some of the base quantities are
207
00:20:14,130 --> 00:20:20,540
chosen independently and arbitrary, while
other follows the automatically depending
208
00:20:20,540 --> 00:20:25,960
upon the fundamental relationship between
the system variables. Because in the power
209
00:20:25,960 --> 00:20:29,910
system, I said there are various elements
including transformers, transmission lines,
210
00:20:29,910 --> 00:20:32,100
generators and they are of different rating.
211
00:20:32,100 --> 00:20:38,429
So, you have to start by taking one base for
one element, and then you have to follow with
212
00:20:38,429 --> 00:20:41,919
the various element which are keep on transforming
the voltage; therefore, the bases will be
213
00:20:41,919 --> 00:20:49,520
also changed. I will come to that point later.
So, out of four power system quantities that
214
00:20:49,520 --> 00:20:57,090
is the voltage, power, current and the impedance;
only two are independent. Means here we have
215
00:20:57,090 --> 00:21:05,549
the power, voltage, current and the impedance;
only the two are independents. So, we have
216
00:21:05,549 --> 00:21:10,730
to take the only two bases, and other two
bases, we calculate by knowing this one.
217
00:21:10,730 --> 00:21:18,169
So, what we did? The universal practice is
to use machine rating, power and basically,
218
00:21:18,169 --> 00:21:23,520
using the machine rating; that is your power
and the voltage at the base values. Means
219
00:21:23,520 --> 00:21:29,740
we are using the power; again here the real
power, or you are using complex power that
220
00:21:29,740 --> 00:21:36,020
is apparent power here not complex power.
And the voltage, they are taken as base quantities
221
00:21:36,020 --> 00:21:43,510
and other base values; means from these, we
can calculate the current base and your impedance
222
00:21:43,510 --> 00:21:50,600
base can be calculated. Now let us see as
we know in power system, mostly the power
223
00:21:50,600 --> 00:21:56,040
supply system is three phase, but in the distribution
areas, another sometimes in generation area
224
00:21:56,040 --> 00:22:00,200
also it is not conventional, then we can go
for the single phase.
225
00:22:00,200 --> 00:22:07,990
So, let us see what are the various base quantities
those are calculated with the help of knowing
226
00:22:07,990 --> 00:22:17,080
the two our power as well as our voltage base;
if your power base that is S I can say b;
227
00:22:17,080 --> 00:22:24,799
b denotes your base, it is unit is your V
a. It is sometimes also written as V a base;
228
00:22:24,799 --> 00:22:30,320
since, it is apparent power. So, V a base
if it is given to you and you know this voltage
229
00:22:30,320 --> 00:22:40,790
base; so we, know this relation that current
is always your S that is here divided by V.
230
00:22:40,790 --> 00:22:47,620
Here we are talking the magnitude. So, here
the I base will be your power base; sorry,
231
00:22:47,620 --> 00:22:50,330
this is b and this is your V a.
232
00:22:50,330 --> 00:22:57,179
So, knowing these two bases, we can calculate
the current base. So, that is written here
233
00:22:57,179 --> 00:23:04,110
you can here this is written. So, the current
base I b will be equal to the power base divided
234
00:23:04,110 --> 00:23:10,299
by your voltage base, and its units will be
ampere. Here if you are using volt as ampere
235
00:23:10,299 --> 00:23:16,150
volt ampere here you are using voltage; so,
it will be ampere. So, always you must be
236
00:23:16,150 --> 00:23:22,110
very careful what you are going to use; means
whether you are using kilo, then you have
237
00:23:22,110 --> 00:23:31,600
to change accordingly. So, your impedances
are now similarly we know this z will be sorry
238
00:23:31,600 --> 00:23:35,830
V upon I; we know it very well.
239
00:23:35,830 --> 00:23:43,820
Because impedance is ratio of voltage to impedance;
so, the z base similarly here voltage base
240
00:23:43,820 --> 00:23:51,330
divided by current base. So, here I have written
this. So, we can calculate the z base. We
241
00:23:51,330 --> 00:23:59,820
can also substitute the value of I base; from
here if you put this value, then we will get
242
00:23:59,820 --> 00:24:05,970
this relationship that is your voltage base
square divided by power base, and it will
243
00:24:05,970 --> 00:24:13,289
be in ohms. So, the z per unit suppose you
want to calculate in the power system, we
244
00:24:13,289 --> 00:24:19,140
want to calculate all the quantities especially
impedance per unit. Because now knowing the
245
00:24:19,140 --> 00:24:25,200
voltage base is known to you, actual value
you can simply you can divide. Means your
246
00:24:25,200 --> 00:24:33,610
voltage per unit here is actual voltage divided
by voltage base.
247
00:24:33,610 --> 00:24:42,760
But this z per unit is defined at actual impedance
z of any element; element in the power system
248
00:24:42,760 --> 00:24:48,340
is the broadly the three elements; that is
your branch includes your transformer and
249
00:24:48,340 --> 00:24:57,549
transmission line and it can be your generator
and some other devices like reactors, capacitors,
250
00:24:57,549 --> 00:25:03,640
etc. So, the actual impedance z per unit is
equal to your actual impedance divided by
251
00:25:03,640 --> 00:25:08,950
base impedance; this base impedance we calculated
from the previous equations. So, I can write
252
00:25:08,950 --> 00:25:17,720
here this z multiplied by the V V a divided
by voltage base square. Normally, in the power
253
00:25:17,720 --> 00:25:21,660
system, your voltage is here.
254
00:25:21,660 --> 00:25:26,870
It is given into the kilovolt; the unit is
not volt. It is in terms of kilo and your
255
00:25:26,870 --> 00:25:35,820
power that is your S; it is given in your
MVA. So, we know this is the M is the megavolt
256
00:25:35,820 --> 00:25:44,260
ampere, and it is a kilovolt ampere. So, we
can write here the Z multiplied by MVA base
257
00:25:44,260 --> 00:25:52,539
divided by KV base means the voltage base
in term of kilo and here in terms of mega,
258
00:25:52,539 --> 00:26:01,130
or if you are using the power base in terms
of here in KVA, it will be z multiplied by
259
00:26:01,130 --> 00:26:08,270
KVA divided by kilovolt; that is the base
in terms of kilo square multiplied by here
260
00:26:08,270 --> 00:26:09,110
1000.
261
00:26:09,110 --> 00:26:15,600
So, knowing all these base value, whether
we are knowing in the voltage or in kilovolt
262
00:26:15,600 --> 00:26:22,440
and power base in terms of actual V a or KVA
or MVA, we can calculate the z power unit
263
00:26:22,440 --> 00:26:27,770
for the single phase system. Let us see now
the three phase system. Three phase system
264
00:26:27,770 --> 00:26:32,780
is slightly different because single phase
system is very simple. In the three phase
265
00:26:32,780 --> 00:26:37,750
system, now we have to calculate this base
current.
266
00:26:37,750 --> 00:26:47,870
Here also we know this power S is nothing
but it is your under root 3 VL I L. So, this
267
00:26:47,870 --> 00:26:56,240
voltage this VL, we take this line to line
voltage as the base value. So, we can write
268
00:26:56,240 --> 00:27:04,700
this I, here base is your KVA divided by your
kilovolt of under root three; means here I
269
00:27:04,700 --> 00:27:13,600
will be your S over under root 3 V. So, this
I b will be here b power base divided by under
270
00:27:13,600 --> 00:27:18,809
root three times voltage base. So, this voltage
base is line to line; it is not phase. So,
271
00:27:18,809 --> 00:27:24,030
we take the line to line voltage at base,
then it will be your current base will be
272
00:27:24,030 --> 00:27:29,280
power. And here it is in kilo, so it will
be in kilo, because all these ten power three
273
00:27:29,280 --> 00:27:33,610
will be canceled, and we get the previous
here like this.
274
00:27:33,610 --> 00:27:39,070
But again most of the since we are operating
the high power that is in terms of megawatt,
275
00:27:39,070 --> 00:27:44,990
MVA, MVR and so on, so forth. So, the power
base is normally it is represented into the
276
00:27:44,990 --> 00:27:51,340
MVA. So, what we do here? If we are going
for here smaller unit, you have to here multiply
277
00:27:51,340 --> 00:27:57,380
by this smaller unit because here earlier
the kilo. Now we are going for mega. So, here
278
00:27:57,380 --> 00:28:01,690
already you have included ten power three,
so that should come out. So, that we can here
279
00:28:01,690 --> 00:28:03,169
unit is reduced.
280
00:28:03,169 --> 00:28:12,090
Means here let us suppose you are using 1000
KVA; what does it mean? It means it is equal
281
00:28:12,090 --> 00:28:20,120
to 1 MVA. So, the unit here it is reduced.
So, here we are writing only one. So, this
282
00:28:20,120 --> 00:28:26,250
thousand should come here. So, we are multiplying
by thousand that is the kilowatt and then
283
00:28:26,250 --> 00:28:34,429
this MVA divided by under root three KVA and
that is your MVA. Our base impedance z base
284
00:28:34,429 --> 00:28:40,059
can be written with this expression. Now this
is you can understand why we have written
285
00:28:40,059 --> 00:28:42,280
like this that can be explained.
286
00:28:42,280 --> 00:28:51,600
Here this z we know; this z is your V phase
divided by I phase, because the z is always
287
00:28:51,600 --> 00:28:57,110
there is no concept of z line to line. So,
z is always for let us suppose this is a transmission
288
00:28:57,110 --> 00:29:04,070
line; this is your inductance resistance.
So, this value is for phase value. So, this
289
00:29:04,070 --> 00:29:11,470
z is always z phase; there is no concept of
Z L it is no. So, this z phase will be equal
290
00:29:11,470 --> 00:29:19,340
to V phase divided by current phase. Since,
our base here V base was your V L and we know
291
00:29:19,340 --> 00:29:29,929
this VL is your V P under root three means
your V phase is nothing but V L over under
292
00:29:29,929 --> 00:29:36,950
root three. So, here I can write this Z will
be your V L under root three I phase.
293
00:29:36,950 --> 00:29:42,809
So, the current here it was your phasor in
the star connected if you are using this.
294
00:29:42,809 --> 00:29:48,740
So, in this case, this is equal to the phase
current will be equal to line current. So,
295
00:29:48,740 --> 00:29:57,539
using this, we can write here V L divided
by under root 3 I L. So, now already we have
296
00:29:57,539 --> 00:30:04,140
calculated means z base will be your V base;
that is line to line divided by under root
297
00:30:04,140 --> 00:30:13,419
three your I base. So, this expression is
used previously here if we will see; here
298
00:30:13,419 --> 00:30:19,020
I am using this. Here this thousand is multiplied
due to the units that I have used the KV;
299
00:30:19,020 --> 00:30:23,809
means KV if you are not using K, then this
thousand, this K will be canceled because
300
00:30:23,809 --> 00:30:25,200
this unit is reduced by k.
301
00:30:25,200 --> 00:30:32,480
So, it is nothing but your this value is your
V base divided by under root three I base
302
00:30:32,480 --> 00:30:38,450
or we can write this one. Again we know this
is our power system quantities. We do not
303
00:30:38,450 --> 00:30:44,210
want to represent the current anything in
the current value. We always try to represent
304
00:30:44,210 --> 00:30:49,289
in terms of the two bases; that is your voltage
base as well as the power base. So, this current
305
00:30:49,289 --> 00:30:54,080
base can be replaced from the previous equation,
and then if we reuse this, then we can go
306
00:30:54,080 --> 00:31:01,850
for we can get this here this voltage base
in the kilo square divided by the power base
307
00:31:01,850 --> 00:31:08,419
in the mega will be your z base, and this
unit will be your ohms.
308
00:31:08,419 --> 00:31:14,610
If you are using this power in terms of kilo,
then we can write 1000 will be going up, and
309
00:31:14,610 --> 00:31:20,070
then finally, we are going to get this one.
So, the z per unit will be your actual impedance
310
00:31:20,070 --> 00:31:27,620
z divided by as usual the base impedance,
and then we can write here z multiplied by
311
00:31:27,620 --> 00:31:34,350
here your VA divided by voltage base square.
And for the various different power and the
312
00:31:34,350 --> 00:31:40,230
voltage levels or you can say unit in terms
of kilo or voltage, we can represent in this
313
00:31:40,230 --> 00:31:48,549
either z multiplied by the MVA base divided
by the voltage base square in terms of kilovolt.
314
00:31:48,549 --> 00:31:56,429
Here we can write again the z base z per unit
will be your z that is actual z multiplied
315
00:31:56,429 --> 00:32:04,700
by your power wave in kilo divided by the
voltage square in terms of kilo multiplied
316
00:32:04,700 --> 00:32:14,559
by 1000. So, this is your z per unit. Now
it is very much usual to represent all the
317
00:32:14,559 --> 00:32:23,289
quantities on the common base. As I said even
though in the very beginning here these bases
318
00:32:23,289 --> 00:32:27,919
suppose your system your transformer or your
generator.
319
00:32:27,919 --> 00:32:38,010
Here this generator is rated at 11 KV, and
its power is your 50 MVA or you can say megawatt,
320
00:32:38,010 --> 00:32:44,600
whatever you say. Power normally here whether
it is megawatt or MVA, we take the same value.
321
00:32:44,600 --> 00:32:53,360
So, here and this x is given in the per unit
is let us suppose is 0.2. Now I want to write
322
00:32:53,360 --> 00:33:02,110
this x per unit on the different base; I want
to get here in the 11 kV< and let us suppose
323
00:33:02,110 --> 00:33:08,429
100 MVA base. So, we have to change this x
per unit, because this value is on this base
324
00:33:08,429 --> 00:33:16,279
and in this base. So, we have to change it
back to our base here this is called the new
325
00:33:16,279 --> 00:33:20,890
base, and this is called your old base.
326
00:33:20,890 --> 00:33:31,450
To know this here, what I can do? We know
this z per unit; whether it is old or new,
327
00:33:31,450 --> 00:33:37,919
it is equal to your various quantities in
terms of power and this one. And this value
328
00:33:37,919 --> 00:33:50,020
is nothing but here you can see this value
is your z MVA divided by KV square. So, it
329
00:33:50,020 --> 00:34:02,190
is your z that is your actual value multiplied
by here KV that was your here this is MVA
330
00:34:02,190 --> 00:34:12,539
divided by KV. So, it is your MVA base divided
your KV base square. So, if this is your for
331
00:34:12,539 --> 00:34:18,539
one base; let us suppose this per unit on
your base one. So, I can say it is one, here
332
00:34:18,539 --> 00:34:23,500
it is one; means here this is your old is
equal to your first base.
333
00:34:23,500 --> 00:34:30,540
Similarly, I can write this z per unit on
another base that is then it will be z p per
334
00:34:30,540 --> 00:34:36,070
unit at the base two quantities. Here the
actual will not change multiplied by your
335
00:34:36,070 --> 00:34:48,790
MVA base two divided by your KV base two square.
So, we want that z per unit, it is given you.
336
00:34:48,790 --> 00:34:54,159
These bases are also known to you; this is
given. I want to calculate this z per unit
337
00:34:54,159 --> 00:34:59,310
on the new base; this is the new base; this
is your new base. And then what I can do?
338
00:34:59,310 --> 00:35:10,540
This now your z per unit two will be we can
simply divide this equation here. So, I can
339
00:35:10,540 --> 00:35:15,369
write here as the simplified equation.
340
00:35:15,369 --> 00:35:23,109
This your z per unit two divided by your z
per unit one will be here now this actual
341
00:35:23,109 --> 00:35:27,490
quantity will vanish. So, in this, we do not
know the actual quantity; all though we can
342
00:35:27,490 --> 00:35:32,220
calculate it, but there is no need to this.
So, if you are dividing this z per unit here,
343
00:35:32,220 --> 00:35:41,890
what we again getting here MVA b 2 divided
by here MVA b 1; this quantity divided by
344
00:35:41,890 --> 00:35:49,770
this. Now multiplied by here this quantity
since you are dividing, it will be coming
345
00:35:49,770 --> 00:35:59,670
here two and this will be going up. So, we
are getting KV base 1 here KV base 2, and
346
00:35:59,670 --> 00:36:02,810
this is your square term.
347
00:36:02,810 --> 00:36:09,839
Now I can say this z per unit two will be
this multiplied by this unit. So, z per unit
348
00:36:09,839 --> 00:36:23,560
1 multiplied by your MVA base 2 divided by
MVA base 1 here multiplied by your this KV
349
00:36:23,560 --> 00:36:33,280
base 1 divided by your KV base 2 and this
is the square term. So, what we can say? At
350
00:36:33,280 --> 00:36:38,730
the new base, here this new base I represented
at 2; old base I have represented as 1. So,
351
00:36:38,730 --> 00:36:47,170
I can write here this value is the z per unit
at the new base and this z per unit I can
352
00:36:47,170 --> 00:36:56,650
say z per unit at the old base. So, this equation
can be again retained here. If we will see,
353
00:36:58,710 --> 00:37:05,280
this per unit impedance referred to the new
base; that is z per unit new will be equal
354
00:37:05,280 --> 00:37:15,060
to z per unit volt multiplied by your MVA
base new that is two to MVA base to old value.
355
00:37:15,060 --> 00:37:21,220
And multiplication here of the voltage base
of old divided by voltage base new, and then
356
00:37:21,220 --> 00:37:27,690
it is the square value. So, it is always easy,
and this formula is used very widely. Means
357
00:37:27,690 --> 00:37:32,940
whatever the per unit quantity you are knowing,
at the different base you have to calculate
358
00:37:32,940 --> 00:37:37,960
all the quantities of the power system on
the single base for your calculation purpose.
359
00:37:37,960 --> 00:37:42,349
So, this equation is very widely used. So,
you are knowing the new base, you are knowing
360
00:37:42,349 --> 00:37:47,190
the old bases; knowing the old per unit quantity,
you can calculate on the new per unit. So,
361
00:37:47,190 --> 00:37:55,940
this equation is very widely used. Now let
us see the various advantages we have explained
362
00:37:55,940 --> 00:37:56,570
this.
363
00:37:56,570 --> 00:38:05,030
The various advantages of using the per unit
is in large electric power systems, the capacities,
364
00:38:05,030 --> 00:38:10,310
rating of equipments are different and the
use of per unit quantity simplifies the calculation
365
00:38:10,310 --> 00:38:16,170
as mentioned earlier. So, it is basically
recap of the per unit representation advantages.
366
00:38:16,170 --> 00:38:22,849
The per unit representation of the impedance
of any equipments is more meaningful than
367
00:38:22,849 --> 00:38:27,990
its absolute value, why? Knowing the absolute
value, let us suppose the impedance of a transformer
368
00:38:27,990 --> 00:38:33,450
is known to you or machine it is known to
you two ohms. And if it is known in the 0.1
369
00:38:33,450 --> 00:38:39,650
per unit, why this absolute value is not so
meaningful than the per unit; that can be
370
00:38:39,650 --> 00:38:40,060
understood.
371
00:38:40,060 --> 00:38:47,589
By the per unit impedance of equipments of
the same general type based on their own rating
372
00:38:47,589 --> 00:38:53,520
fall in a narrow range regardless of the rating
of equipments, whereas their impedance in
373
00:38:53,520 --> 00:39:00,589
ohms vary greatly with the rating to rating.
So, using per unit system, the chances of
374
00:39:00,589 --> 00:39:07,099
making mistakes in phase and the line voltages,
single or three phase quantities are minimized
375
00:39:07,099 --> 00:39:12,859
as we saw the phase and the your line to line
voltages. So, the various advantages of per
376
00:39:12,859 --> 00:39:19,550
unit representation just as I mentioned using
per unit system, the chances of making calculations
377
00:39:19,550 --> 00:39:25,869
in terms of phase to line, line to line phase
and also three phase or single phase, it is
378
00:39:25,869 --> 00:39:28,080
greatly minimized.
379
00:39:28,080 --> 00:39:35,109
Already, I have shown that how this line to
line and per unit phase quantities are related,
380
00:39:35,109 --> 00:39:40,280
but in the per unit, this is per unit whether
you are line or base, it is a constant. So,
381
00:39:40,280 --> 00:39:43,970
there is no confusion in the phase; that is
always base value per unit quantities. So,
382
00:39:43,970 --> 00:39:50,390
it has no line to line or phase to phase calculation;
also in the single phase and the three phases,
383
00:39:50,390 --> 00:39:54,650
this calculation becomes very simple. In case
of transformers, this is the great advantage
384
00:39:54,650 --> 00:40:01,339
of per unit system. In case of transformer,
the per unit values of impedance voltage current
385
00:40:01,339 --> 00:40:06,830
refer to the primary side or secondary side
will be the same which further simplify the
386
00:40:06,830 --> 00:40:07,790
calculation.
387
00:40:07,790 --> 00:40:13,790
To understand this, here you can say this
is a transformer. This is your transformer
388
00:40:13,790 --> 00:40:22,010
primary winding; this is your secondary transformer
winding. Now the quantity which this side
389
00:40:22,010 --> 00:40:29,460
is going, it is called the v primary, I primary,
and here your v secondary and the current
390
00:40:29,460 --> 00:40:40,030
here your secondary current. If this transformer
is let us suppose is 11 KV to 132 KV, then
391
00:40:40,030 --> 00:40:44,310
the impedance which is referred this side
will be certainly different than the impedance
392
00:40:44,310 --> 00:40:50,400
referred to the side. If this transformer
is the z let us suppose z actual referred
393
00:40:50,400 --> 00:40:57,390
to the primary side that is here z p referred
to the primary side, it will be the z p impedance
394
00:40:57,390 --> 00:40:59,940
plus here your z s.
395
00:40:59,940 --> 00:41:07,530
And it will be the ratio of your N 1 or N
p over N s, and this is the square quantity.
396
00:41:07,530 --> 00:41:13,420
So, this quantity you have referred to the
primary. In actual quantity, it will be different
397
00:41:13,420 --> 00:41:20,900
than this impedance referred to the secondary
side. This will be your nothing but z s plus
398
00:41:20,900 --> 00:41:29,770
here it is your z p and here your N s over
N p square. So, these two impedances are different
399
00:41:29,770 --> 00:41:36,210
whether you are referring this side or that
side, but in per unit system whether you are
400
00:41:36,210 --> 00:41:43,730
referring this side or that side; that will
be same. Means your z per unit referred to
401
00:41:43,730 --> 00:41:50,849
the primary side in per unit will be equal
to your z secondary per unit referred to this side.
402
00:41:51,010 --> 00:41:56,220
So, whether it is referred to the secondary
side per unit or it is referred to this primary
403
00:41:56,220 --> 00:42:02,010
side, in the per unit, both will be the same.
So, it simplifies; means this transformer
404
00:42:02,010 --> 00:42:09,119
is simply represented by the impedance here
z and it is in per unit. So, now it is irrespective
405
00:42:09,119 --> 00:42:13,890
or referred to the primary side and the secondary
side. This is the great advantage of the per
406
00:42:13,890 --> 00:42:18,820
unit system but can be proved very easily.
And you will find all these proofs in the
407
00:42:18,820 --> 00:42:25,609
various books. Here normally what we do? The
base now, another important quantity here
408
00:42:25,609 --> 00:42:30,849
the base this primary side your here this
V p will be the base; however, in the secondary
409
00:42:30,849 --> 00:42:32,839
side, this voltage will be this base.
410
00:42:32,839 --> 00:42:40,070
For example, for this transformer, the voltage
base in the primary side, it is 11 KV. However,
411
00:42:40,070 --> 00:42:47,710
in the secondary side, I can say primary side,
and in the secondary side, it will be your
412
00:42:47,710 --> 00:42:58,750
v base. It will be 132. So, the base is also
going to be changed. So, in the power system,
413
00:42:58,750 --> 00:43:05,089
here suppose we have a different transformer
of different ratings in terms of power as
414
00:43:05,089 --> 00:43:09,440
well as the voltage rating, generators are
also of the different voltage and the power
415
00:43:09,440 --> 00:43:14,050
ratings. So, all they are interconnected as
we saw the various advantages of interconnection.
416
00:43:14,050 --> 00:43:17,640
So, we have the interconnected complex power
system.
417
00:43:17,640 --> 00:43:25,570
Then we have to the power base will be constant
throughout the system. Let us go for the various
418
00:43:25,570 --> 00:43:33,020
advantages, then I come back to that one.
Here this other advantage is the power and
419
00:43:33,020 --> 00:43:39,730
the voltage equations are simplified as a
factor of and the three are eliminated in
420
00:43:39,730 --> 00:43:45,140
the per unit system. Means here the three
concept is basically eliminated means every
421
00:43:45,140 --> 00:43:52,980
time we can write in the per unit in the power
your this voltage base as well as your current
422
00:43:52,980 --> 00:43:57,890
base that will give you the power base. So,
this three phase, etcetera, it is eliminated,
423
00:43:57,890 --> 00:44:03,660
and it is very easy to calculate all the power
system quantities in terms of your steady
424
00:44:03,660 --> 00:44:07,980
state as well as your dynamic quantities that
can be calculated.
425
00:44:07,980 --> 00:44:12,869
And also all loss are equally valid in the
per unit system; that is very important. So,
426
00:44:12,869 --> 00:44:17,710
that you should not worry that whether Kirchhoff s
current law and voltage law are valid in per
427
00:44:17,710 --> 00:44:25,130
unit system equally valid in that one. To
understand this, let us see a power system
428
00:44:25,130 --> 00:44:32,630
here consist of transformers, generators and
other systems. This is a system which I am
429
00:44:32,630 --> 00:44:35,320
representing; let us suppose we have a generator
430
00:44:35,320 --> 00:44:40,630
This is the connected with the transformer;
here transformer I can represent like this.
431
00:44:40,630 --> 00:44:46,099
Here we have another bus; we have a transmission
line, and we have the various load as well
432
00:44:46,099 --> 00:44:54,070
as we can have the motor loads. So, this is
your generator; this is your motor load, and
433
00:44:54,070 --> 00:45:02,869
here this is any other static load. Now this
transformer actually this generator is a three
434
00:45:02,869 --> 00:45:07,820
phase; transformer is also a three phase.
Let us suppose this connection will be always
435
00:45:07,820 --> 00:45:14,849
not always it is a star and delta and this
may be your grounded as well. Now the rating
436
00:45:14,849 --> 00:45:22,030
of this is let us suppose is 11 KV, and the
power which is generating it is a 200 megawatt.
437
00:45:22,030 --> 00:45:31,940
Basically, for the 200 megawatt units in our
India, it is 16 KV normally. In India, we
438
00:45:31,940 --> 00:45:38,609
have the different rating generators. So,
the generators which are rated at the 500
439
00:45:38,609 --> 00:45:46,190
megawatt, they are basically operating at
21 KV. Those are 200 and 210 megawatt; they
440
00:45:46,190 --> 00:45:56,810
operate 15.75 kilowatt and others like 50,
100, 110, they operate 11 KV. So, let us suppose
441
00:45:56,810 --> 00:46:04,540
this is a 200 megawatt generator, and it is
rated at 16 KV this per unit. Now this transformer
442
00:46:04,540 --> 00:46:14,180
is we have here 16 KV to 400 KV and then here
we have another basically the transformer
443
00:46:14,180 --> 00:46:17,390
we have to use, because we cannot use the
motor load at the 400 KV.
444
00:46:17,390 --> 00:46:27,700
So, it is another transformer I can say it
is your delta delta and it is 400 by 16 KV.
445
00:46:27,700 --> 00:46:38,829
So, now we have to take the common base for
this complete system, and we have to start
446
00:46:38,829 --> 00:46:45,900
from one end and especially we start from
the generating end. So, our voltage base if
447
00:46:45,900 --> 00:46:53,099
we start from this generator, let us suppose
it is 16 KV. So, this is your V b; I said
448
00:46:53,099 --> 00:46:59,230
it is a three phase generator, and it sets
in 16 KV voltage means it is line to line
449
00:46:59,230 --> 00:47:05,660
and that will be the base value. Now the power
here it is even though shown at 200 megawatt,
450
00:47:05,660 --> 00:47:13,329
but normally, we take the power base in normal
calculation it is 100 MVA.
451
00:47:13,329 --> 00:47:22,480
So, this power base will continue, and it
will be same throughout the system; however,
452
00:47:22,480 --> 00:47:28,040
the voltage base will keep on changing. Now
you can see at here at the primary side of
453
00:47:28,040 --> 00:47:34,390
this transformer, the base will be your 16
kV; however, this side since we are transforming
454
00:47:34,390 --> 00:47:43,990
with this transformer, the voltage base at
here, it will be 400 KV. So, the voltage base
455
00:47:43,990 --> 00:47:49,339
of this transmission line will be this is
your voltage base and the power base will
456
00:47:49,339 --> 00:47:55,099
be your 100 MVA. Now again if you are going
for again this transformer is used means again
457
00:47:55,099 --> 00:48:00,339
we are using 400 to 16 KV; means we are stepping
down.
458
00:48:00,339 --> 00:48:06,000
So, the voltage base at this bus, again it
will become the 16 kilovolt and the power
459
00:48:06,000 --> 00:48:12,920
base as I said it will be continued throughout
the system. So, what happens? Then you have
460
00:48:12,920 --> 00:48:18,200
to calculate, because this transformer is
one element; this is one element; this is
461
00:48:18,200 --> 00:48:22,040
another element; this is another element,
and this is also an element. So, we have to
462
00:48:22,040 --> 00:48:25,700
convert all the quantities and the single
phase.
463
00:48:25,700 --> 00:48:31,410
So, for example, if here we have given the
per unit impedance of let us suppose this
464
00:48:31,410 --> 00:48:37,190
generator is given 0.2 per unit; for this
transformer, it is written let us suppose
465
00:48:37,190 --> 00:48:44,990
0.5 per unit, and again, here this is the
voltage rating. And all that as I said all
466
00:48:44,990 --> 00:48:49,290
the elements, they will be rated with its
maximum power that it can handle. And again
467
00:48:49,290 --> 00:48:54,319
it depends upon the constraints; how much
power it can handle without excessively heating,
468
00:48:54,319 --> 00:49:00,480
and it should not burn means continuous rating
of the operators. So, let us suppose it has
469
00:49:00,480 --> 00:49:04,630
your 300 MVA rating, and this line is there.
470
00:49:04,630 --> 00:49:11,730
Let us suppose impedance is 50 ohm, and it
is let us suppose only x is there, and then
471
00:49:11,730 --> 00:49:16,050
we have another transformer of the same rating,
and then we have the motor load. Now what
472
00:49:16,050 --> 00:49:23,569
we have to do? We have to basically calculate
the impedance of this on this base, because
473
00:49:23,569 --> 00:49:31,700
this value is given on its base. This value
is given on this base and in this base. So,
474
00:49:31,700 --> 00:49:38,930
we have to change it; we have to change in
this this value base, and so here we can as
475
00:49:38,930 --> 00:49:47,290
I said, I should z per unit at the new base
will be equal to here z per unit one here
476
00:49:47,290 --> 00:50:00,220
multiplied by MVA new base over MVA 1, and
here multiplied by your KV 1 divided by KV
477
00:50:00,220 --> 00:50:02,619
2 of square.
478
00:50:02,619 --> 00:50:10,780
So, we have to see the voltage is same, so
there is no use of this and here 200 means
479
00:50:10,780 --> 00:50:17,700
now it will be your z per unit volt is two
multiplied by MVA two is now your 100. New
480
00:50:17,700 --> 00:50:24,410
one is 100, and the old on which it is given
it is 200. So, it is 200. So, now we are getting
481
00:50:24,410 --> 00:50:34,200
this one per unit. So, for the calculation
on this base, now this impedance of this generator
482
00:50:34,200 --> 00:50:42,530
will be taken at here z per unit of this generator
is this; here resistance is neglected basically.
483
00:50:42,530 --> 00:50:48,609
Similarly, we have to go for this one and
then for the other element. Here for the transmission
484
00:50:48,609 --> 00:50:54,450
line, we have not written in the per unit,
because the transmission line the voltage
485
00:50:54,450 --> 00:50:56,069
rating is very difficult to define.
486
00:50:56,069 --> 00:51:00,900
It is very easy to calculate r real and see
as you know the elements parameters of that
487
00:51:00,900 --> 00:51:05,710
transmission line, we will see later also.
And then, we have to calculate what will be
488
00:51:05,710 --> 00:51:13,609
the base. For this line, your power base will
be again it is your 100 MVA; however, the
489
00:51:13,609 --> 00:51:21,210
voltage base will be your S will be your 400
KV. So, you have to calculate the z base,
490
00:51:21,210 --> 00:51:28,420
and then we can calculate the z per unit of
this line will be z upon z base. And then
491
00:51:28,420 --> 00:51:33,990
whole this system will be represented by the
impedance diagram, and that impedance diagram
492
00:51:33,990 --> 00:51:42,020
of this will be here this is your only generator
is represented by its impedance, then this
493
00:51:42,020 --> 00:51:43,069
is this bus.
494
00:51:43,069 --> 00:51:49,380
And since, it is the voltage source, here
it is generating, and here there will be some
495
00:51:49,380 --> 00:51:54,930
ground; this is the star interconnected generator.
Now for the transformer, transformer here
496
00:51:54,930 --> 00:52:00,890
we are simply representing as the impedance;
this is another bus. So, this bus corresponds
497
00:52:00,890 --> 00:52:07,550
to this; this bus corresponds to this, and
then we have this impedance of the transmission
498
00:52:07,550 --> 00:52:17,119
line, and then here we are having the impedance
of again another transformer here. And then
499
00:52:17,119 --> 00:52:23,060
we have again at this bus, we have a load,
and we have a motor, and this motor is again
500
00:52:23,060 --> 00:52:24,480
motor, then it is a grounded.
501
00:52:24,480 --> 00:52:30,060
So, basically this is the impedance diagram,
and based on that, now we can calculate this
502
00:52:30,060 --> 00:52:33,950
current knowing this voltage here, and we
can calculate other quantity of the power
503
00:52:33,950 --> 00:52:39,690
system. So, this is whole duty of the per
unit system; that is why I have reviewed,
504
00:52:39,690 --> 00:52:44,819
because all our calculations later on, we
will go in the per unit calculation. Whether
505
00:52:44,819 --> 00:52:49,230
we are calculating our AGC that is the automatic
load frequency control or automatic generation
506
00:52:49,230 --> 00:52:54,560
control; in that, I will be using the per
unit system, also in the stability we will
507
00:52:54,560 --> 00:52:57,130
see the various here in the per unit system.
508
00:52:57,130 --> 00:53:02,740
So, this is basically total about your per
unit system representation. Now let us come
509
00:53:02,740 --> 00:53:09,630
to the various equipments of the power system
and those are here the first one is that is
510
00:53:09,630 --> 00:53:18,380
called the generators. Generators are the
source of a power system means they supply
511
00:53:18,380 --> 00:53:25,829
power to the grid to the system, and that
power must carried by the transmission lines,
512
00:53:25,829 --> 00:53:29,490
transformers, distribution lines, and finally,
it is reaching to the various customers.
513
00:53:29,490 --> 00:53:37,829
So, for the electric power generation, normally
it is the three phase synchronous generators
514
00:53:37,829 --> 00:53:45,130
driven by either steam turbines, hydro turbines,
or it is a gas turbine known as the alternator
515
00:53:45,130 --> 00:53:50,690
are used. Normally, these generators are also
called the alternators. Why it is actually
516
00:53:50,690 --> 00:53:55,530
this is a very old term; alternator means
they provide the alternating voltage that
517
00:53:55,530 --> 00:54:02,310
is here this voltage normally they are providing.
So, it is alternators, and since, there is
518
00:54:02,310 --> 00:54:06,880
alternating current and you know they provide
the ac current; this is alternating current.
519
00:54:06,880 --> 00:54:15,530
So, they are called the alternators as well.
In other cases as I said, it is a synchronous
520
00:54:15,530 --> 00:54:16,089
generator.
521
00:54:16,089 --> 00:54:23,220
We have the two types of generators, and broadly
in the ac, we have two type of machines. We
522
00:54:23,220 --> 00:54:32,710
can say the synchronous, another is your induction.
Induction generators are having inherent problem,
523
00:54:32,710 --> 00:54:37,819
and therefore, they are not widely used. They
are only used for some specific purpose like
524
00:54:37,819 --> 00:54:44,970
a wind turbine where there is a variable speed
is there and other related, but the synchronous
525
00:54:44,970 --> 00:54:48,790
alternators for all the other conventional
power system; that is your stream, hydro and
526
00:54:48,790 --> 00:54:57,020
the gas they are used. So, it is very very
important to know about their limitations,
527
00:54:57,020 --> 00:55:03,670
their constraints, so that we can see and
we can feel about the power system constraints.
528
00:55:03,670 --> 00:55:09,609
So, the equipments constraints in terms of
generators that is your alternator, and the
529
00:55:09,609 --> 00:55:13,859
alternator I am talking about only the conventional
power generators; that is whether it is a
530
00:55:13,859 --> 00:55:19,790
cool base that is steam, hydro or it is a
gas turbines. Basically, we have a four type
531
00:55:19,790 --> 00:55:26,000
of conventional; that is your steam, hydro,
gas, diesel as well as the nuclear power station.
532
00:55:26,000 --> 00:55:30,930
So, we use the synchronous generator. So,
we will see the limitations, and to know the
533
00:55:30,930 --> 00:55:37,170
limitations and the capabilities of the generators,
we must see the various types of generators
534
00:55:37,170 --> 00:55:42,140
and how their phasor and other relations are
derived, so that we can go for their capabilities
535
00:55:42,140 --> 00:55:42,640
limits.
536
00:55:42,640 --> 00:55:50,500
So, in this lecture, now I can recap. We saw
various constraints that is I briefly said
537
00:55:50,500 --> 00:55:54,440
that it is related to the system constraints
or related to the equipments constraints.
538
00:55:54,440 --> 00:56:00,380
They are very very important, and then, we
saw the per unit representation, their advantages,
539
00:56:00,380 --> 00:56:07,410
and those advantages are very very useful
for the power system analysis. And also in
540
00:56:07,410 --> 00:56:13,290
this power system operation and control, we will use those quantities in the further lectures Thank you.