WEBVTT
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So today, from today we will start the speed
control of induction motor, DC motor. So,
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before that we will study the working principle
of how the induction motor works and its steady
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state equivalent circuit and with the steady
state equivalent circuit, how we control the
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machine and for high dynamic performance applications,
what is the dynamic equivalent circuit model
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and from there, how we drive the machine for
high dynamic performance application; we will
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study.
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So, let us start the induction method; how
the working principle of induction motor,
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sinusoidal steady state and induction model,
we will start sinusoidal steady state induction.
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So, let us take the winding; how the winding
is distributed in slots to get a sinusoidal
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distribution? So far, we said the distribution
is sinusoidal. But with the distributed winding,
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the air gap flux distribution or MMF is a
stepped wave from as close as to a sine wave.
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Let us take the phase, let us say this is
one slot for A phase; A phase let us this
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way, so current is coming here and going like
this, this is for A phase. Let us take one
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time or all or let us say all the times are
embedded in one slot. So here, if you see
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here, the flux, the flux will be in this direction.
So, you have the flux distribution like this,
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here also it will be like this; so the current
movement is such that all the flux direction
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here will be like this. So, let say, this
is the starter with uniform air gap, the reluctance
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is uniform throughout. So the MMF, so let
us split this one and spread it; so if you
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see here, for the A phase, around 180 degree,
you will have this one like this. So, if you
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go like this, the MMF will be like this throughout
the air gap.
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We know that MMF is equal to n into i is equal
to phi into R. But this is a stepped wave;
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it is a rectangular waveform, the starter
sinusoidal distribution. So, in the slots,
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we have different conductors in adjacent slots
for phase A, they are short pitched also to
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get a nearly stepped waveform. So, let us
take to distributed winding like this; 3 slots,
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so we have one here, one here, let us take
this way, one here and one here.
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So, MMF for this one, we will use different
color. For this one will be, so same current,
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this conductors in this slots belong to A
phase and they are series connected and same
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current is flowing through. So, the MMF for
the conductors in the other slot will be like
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this. Then for this one, is are of equal amplitude;
I have slightly changed the amplitude to make
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it visible. So, if the resultant MMF if you
see here, you will get a stepped waveform
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here for the A phase with pre conductors like
this; here it will be like this, then here
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it will be like this, then you will go like
this, go like this so sorry here it will be
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like this, it will go like this. Now, so this
type of stepped waveform for A phase will
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be.
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When all the 3 phase are exited, then you
have B phase around 120 degree phase shifter
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from here. 120 means it will be somewhere
here it will be there, for the B phase. C
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phase again 120 degree means somewhere here
it will be there. So, if you see here, all
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the windings together and if you see the MMF
distribution in the air gap, you will have
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a nearly a sinusoidal a stepped waveform as
close as to a sinusoidal will be here. So,
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this winding has to take care of this to get
the nearly sinusoidal distribution of the
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MMF here.
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So, this we can approximate a sine wave. So,
as power electronics engineers, control engineers,
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we will not worry about the stepped distribution.
We will assume this is sinusoidal distribution.
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So this flux, sinusoidal distribution; so
we will be assuming the flux distributor,
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the sinusoidal distribution of the flux across
the air gap and if the machine is exited with
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sinusoidal currents, this peak value slowly
rotate with respect to a speed omega s, omega
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s is our singular speed or excitation frequency.
So, this peak keeps on moving. So, that is
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equal to say that or we have a flux, space
vector flux rotate smoothly, this way with
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omega s, the peak value rotates that means
this will move like this.
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Now, when it moves like this, what happens?
So, flux is sinusoidally distributed, so if
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you see here, let us go the next page, we
have a, now we will not show the stepped waveform,
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we will show it is sinusoidal distribution.
So this is the flux, air gap flux is sinusoidal
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distributed.
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Now, what happens? Let us take the starters,
starter it has conductors, conductors in various
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loads, these are the conductors in various
loads; starter, conductors. This starter conductor,
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there is a relative motion; the starter conductor
is having current. Now, there is current,
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there is current flow in starter conductors
and the flux is moving; there is a relative
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motion between a current carrying conductor
and the air gap flux. So, what will happen?
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Because of the relative motion between the
air gap flux, air gap flux is moving with
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a speed omega s, omega s is our excitation
frequency because of this relation relative
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motion, there will be induced voltage
induced voltage in starter conductors. So,
this conductor where the flux is maximum,
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you will have the maximum voltage. So, each
conductor will experience a bar magnitude
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voltage and that voltage is proportional to
the speed and also the flux, the flux at that
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point cutting the conductors. These are called
bar magnitude voltages, these are called induced
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voltage.
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So, let us take the machine is not loaded,
induction machine. So, if you see here, we
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are applying a sinusoidal voltage here, across
the machine and assuming starter drops are
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negligible. So, if you see here, then this
induced voltage Eb sinusoidal distribution,
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this is Vm will be exactly opposite to V.
That means we are sending a current, magnetizing
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current through the starter, no load. So,
let us see this is im magnetizing current;
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due to magnetizing current, flux is generated
and this flux, this im is generated due to
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the voltage V, these two are lagging by 90
degree, no load operation, no load, secondary.
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Due to this flux, there is an induced voltage
Vm; if you see here, according to notation,
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let the current is going like this im. So,
one is leaving this one, one is entering this
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terminal, so notation this bm induced voltage
is equal to minus d phi by dt. So, this will
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be Vm. So, this Vm and with the applied voltage
will be exactly opposite that is opposing
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each other; this is starter, under no load.
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Now, the flux is rotating but we have rotor
also we have conductors, the rotor slots we
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have rotor bars and short circuited at the
ends. That means we have short circuiter bar
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that means to applied rotor voltage is 0.
So, this short circuited rotor bars if you
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see here, this will act like a coil, these
are short circuiter and then these are short
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circuited at the end. So, these are like coils,
the various coils; see various coils will
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be there, all are short circuited, full periphery
you can have slots like this if you see spread.
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So, this rotor conductors; there is a flux
moving air gap and there is relation motion
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between the flux and the rotor conductors
also. Let us take the rotor conductors with
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a different color here. Let us see this is
or assuming the rotor also having the same
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number of times as the starter; so we have
rotor conductors here. But if you see, if
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the rotor is rotating with a speed, this is
rotating with omega s, rotor is rotating with
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omega r, let us say omega r these are rotor
conductors. Now, what will happen? Rotor will
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also get induced voltage, same bar magnitude
voltage will
the rotor also get. Now, rotor is short circuited,
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so each coil has an alternating voltage, this
way, bar magnitude voltage. So, this coil
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will also have a current flowing through the
coil and the coil means it has inductance
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and the resistance.
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So, that current will be lagging the voltage.
So, what you meant by lagging? Let us draw
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that one. See, let us draw that one, let us
draw the current; see this is the 180 degree,
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this is the 0 degree, this is the starter.
So, if you see here, the rotor bar magnitude
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voltage will be like this; this bar magnitude
voltage for the rotor also, this voltage will
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not be the same as the starter induced voltage.
Why? Rotor is moved with the speed omega r.
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So, relative speed will be that is omega s
minus omega r, this is called the omega slip.
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So correspondingly, this voltage that is starter
voltage and this voltage will not be equal
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but they are if you see here, they for the
same conductors, if you see here, the rotor
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also will get this type of bar magnitude voltage
and because of the rotor coil, you will get
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current in the starter rotor coils. That current
will be lagging this bar magnitude voltage.
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What do you mean by that one?
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Suppose, the conductor which is lying here
at this movement is having the maximum voltage;
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lagging, lagging is equal to the rotor leakage
inductance that is the phi will be tan inverse
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omega sl L divided by that rotor resistance
r, that phi. So what is meant by lagging rotor
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current? Rotor also will have this bar magnitude
current, sinusoidally distributed that is
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lagging the bar magnitude voltage by an angle
by phi. What is meant by that one?
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That means the conductors which is having
the maximum voltage now will have the maximum
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current after phi. So, that means the bar
magnitude current will distribution will be,
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for the conductors, it will be like this;
this is the phi. That means now this conductor
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is having the maximum current; after sometime,
this conductor will have the maximum current.
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So, current is lagging the coil. Now, what
happens? If you see the equivalent circuit,
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let us draw the equivalent circuit here itself.
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So, let us strike your phasor diagram now.
If you see here, the phasor diagram is like
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this; see we have, originally we have the
im which will generate the flux because of
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the flux we have the this im is generated
by V applied voltage, starter voltage for,
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this is the single phase equivalent circuit;
so im, these are 90 degree and this will im
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generate the induced voltage which is Vm,
these are quarterly conversion, Vm Vr, Vm
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and Vr are opposite.
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Now this Vm, this Vm is the voltages which
introduce the rotor also. So, this rotor will
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have a current which is lagging the Vm by
phi; this is phi, angle phi, this is the current,
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peak value of the current, peak value of the
current is here. Let say, this is Ir, peak
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value, so we will make the Ir here. Now, what
happens? There is a current distribution in
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the quarter circuits, the same way the current
distribution im in the starter. So, starter
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has produced a sinusoidal distribution of
the flux. Same way, rotor can also produce
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a sinusoidal distribution of the flux. Then
what happens?
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The original air gap flux will get disturbed.
So, what the starter will do? Flux cannot
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change instantaneously for steady state operation.
So, what happens? Starter will also to counter
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the effect, it will produce equivalent opposite
current ir. Now because of the rotor current,
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the starter current will be im parallel to
that addition we will do; this is the is,
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new is will be generated
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Initially, we assumed starter does not have
or starter drops negligible but starter has
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resistance on inductance. So, if you see here,
the equivalent circuit, we are applying V
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here; it has resistance, leakage inductance
and the magnetizing inductance. Now, this
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is the rotor current. Rotor will also have
induced voltage, so this induced voltage depending
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on the speed that is the slip, rotor also
will have an induced voltage here. If this
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is equal to Vm, this is equal to Vm dash that
is proportional to slip and you have resistance
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and inductance here. So, this drop will be
equal to the omega slip; this is the current
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ir, this is Vm dash. If Vm, if omega s is
equal or with the speed omega s, we will get
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omega m. Vm dash will Vm dash which approximately
we can find out, Vm dash will be equal to
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omega slip into Vm divided by omega s; this
way you will get it.
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Now, if the starter also has a leakage inductance,
leakage inductance; the starter voltage, applied
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voltage has to be for this distribution, not
be V. So, then our applied voltage V, this
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is Vm, we have starter drop along is that
is Rs, Rs leakage inductance ls, then perpendicular
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to that one, j omega s ls into is; that drop
will be there. So, depends on the is, this
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V may be here, this is may be our V. So, this
is the steady state equivalent circuit.
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Now, let say, this is for motoring operation;
now, let us say, for the same slip, motor
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is rotating in the opposite direction. For
the same slip, motor is rotating in the opposite
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direction or we can say let us I am slowly
reducing the speed, omega s, somehow if I
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can reduce the speed let us say, such that
the relative speed omega s omega r is greater
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than omega s and omega s minus omega r, absolute
value of the slip is the same.
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So, what happens? If you see here, the rotor
conductors, we will see the flux is rotating
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in the opposite direction because omega r
is greater than the omega s. So, if you stand
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on the rotor conductors when the omega slip
is omega s is less than omega r; the rotor
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conductors we will see that the flux is rotating,
flux is moving in the opposite direction,
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flux is moving in the opposite direction,
opposite direction. What is meant by this
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one? Let us go to the next page.
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See, we will again let us draw our bar magnitude
voltages here. See, this is the flux, this
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is the flux; from the starter, it will be
moving in this direction. Now the rotor conductors,
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rotor conductors, they are moving with the
speed omega r; this is omega s. Now, we will
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say omega r is greater than omega s and omega
r minus omega s is equal to omega slip is
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equal to same as our omega slip, same as before.
Now, the relative motion is in the opposite
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direction. So, what will happen? Rotor conductors,
we will see bar magnitude voltages in the
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opposite direction; when the slip is negative
that means slip omega s minus omega r is equal
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to minus omega slip, slip is negative.
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Now again, there is a rotor conductor that
is inductance and resistance, so it will take
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a lagging current. Here, what is the lagging
current means? Now, the relative motion between
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flux and the rotor is, it is now in this direction,
relative motion. So, the lagging means if
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you say this conductor has the peak value
now; according to the current, the current
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will be lagging now, the lagging direction
is in this direction; this is the rotor current.
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So, this is the rotor current, bar magnitude
current; so this is the phi now.
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So, as the relative motion is in this direction
if this voltage is having the maximum, the
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maximum current is in this direction. So,
it will come here after some period. But if
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you stand on the starter, see starter has,
it has provided the im now; again we will
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let us assume the starter drops are negligible.
So, V is here. Now our Vm, dash Vm dash is
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in this direction, this is the Vm dash for
omega slip negative, for minus omega slip.
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Now lagging current, as far the starter is
concerned, if you look from the starter, this
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will take a leading current, the rotor is
taking a leading current in this direction;
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this is the ir, now to counter this one, starter
will produce equivalent opposite current in
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this direction, ir.
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So, if you see, ir plus is, this is the is;
so new is, as far the starter is concerned
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more than 90 degree, this theta is greater
than 90 degree. So, slip is named, so import
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power is equal to Vi cos theta, theta more
than 90 degree, so is equal to power will
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be, theta is more than 90 degree means this
will become negative. So, power flow from
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the rotor side to the starter side. So, this
happens when during the slip is negative,
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this is the way what is happening inside.
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But as far the rotor is concerned, rotors
will always taking lagging current only but
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look from the starter, now the lagging direction
is changed because of the relative motion
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changed, the starter we will see, rotor is
taking a leading current and approximately
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it will equivalent opposite. So, net effect
current will be more than 90 degree. So, theta
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more than 90 degree, it will be minus cos
theta. So, it will become minus Vi cos theta,
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it will become, so power becomes negative;
this happens during regeneration.
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Now, we have approximately come to the steady
state equivalent circuit where the machine
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is excited with sinusoidal currents. Now,
how we have the speed control?
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So, assuming starter will have a voltage E,
starter and the rotor has same number of turns.
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Now, if relative motion between the starter
and that is omega r is equal to 0 let us take,
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rotor we have, this is the rotor Rr lr. Now,
omega r is equal to 0, then the applied voltage
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Vm and Vm; so instead of Vm, I will put the
excitation voltage now as E; so, that would
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make it simpler. Let us see, this is E, this
will also have E, omega r is equal to 0. That
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means this drop, the ir current; here also
the frequency is omega s.
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Now, omega r is not equal to 0 that is rotor
is running, omega r is equal to non-zero but
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still the starter will have the voltage E;
this is the starter. We will assume same number
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of turns for the starter as well as the rotor.
So, rotor will have a voltage. What is the
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voltage? It is proportional to the slip. What
is slip? Slip is equal to, so what is the
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voltage E dash? E dash is equal to E by omega
s into omega s minus omega r. So this turn,
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omega s minus omega r, that is a ratio; you
will say, the slip s. So, E dash is equal
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to SE you will get.
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So here, this leakage inductance lr, here
the frequency is equal to omega slip. Omega
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slip is equal to this is the omega s minus
omega r is equal to omega slip; this is the
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ratio s. Then you have the resistance r; omega
r is equal to not zero. Now see, starter also
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starter has E, rotor has SE. So, we can make
starter and the rotor same induced voltage
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by multiplying or dividing throughout by s.
If you see here, dividing throughout by s;
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this is also E, this is also E, then omega
slip divided by s, s is equal to omega s.
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So, the starter frequency also becomes omega
s lr but now the new resistance becomes Rr
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by s.
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So, when we do this one, power balance should
be there; power before and power after should
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be the same, Rr by s. Now, the resistance
has become Rr by s; so what is the power transferred?
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Let us see, the power transferred, the power
transferred from starter to a rotor to the
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air gap, transferred. Let us take for one
phase it is this is ir is equal to ir square
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into Rr by s is the power developed in the
rotor. For the three phase it is 3 that is
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the total power developed. This is the total
power in the rotor.
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See, we have divided by s to make the induced
voltage same as the rotor that is for convenient
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for analysis. But now that reflected a new
resistance Rr by s. But the rotors copper
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loss is the same. What is the rotor copper
loss? A rotor copper loss, loss is equal to
35:48.970 --> 35:55.970
3 into ir square into Rr, this is independent
of the frequency. So, total power generated
36:00.440 --> 36:07.020
is this one, copper loss is this one; so total
power minus a copper loss is the power which
36:07.020 --> 36:14.020
is responsible for the mechanical power that
is a rotation. So, P mechanical is equal to
36:17.790 --> 36:24.790
3 ir square into Rr by s minus Rr. So, what
is this one? Let us bring that to a simple
36:34.790 --> 36:36.070
form for analysis.
36:36.070 --> 36:43.070
So, P mechanical is equal to 3 into ir square
into Rr into 1 minus s by s, the previous
36:52.000 --> 36:56.990
equation if you simplify, you will get like
this. See previously also, we know this is
36:56.990 --> 37:03.990
equal to torque into omega rotor, omega mechanical
speed. So, what is omega mechanical? That
37:12.730 --> 37:19.730
is omega rotor that is omega mechanical is
the rotor of speed. So, omega mechanical is
37:24.140 --> 37:31.140
equal to omega synchronous minus omega slip
that is equal to, in s, omega s into 1 minus
37:39.150 --> 37:46.150
s it will be, slip where s is equal to omega
slip divided by omega s is equal to omega
37:52.310 --> 37:57.440
s minus omega r divided by omega s; this is
the excitation frequency, this is the rotor
37:57.440 --> 38:01.120
speed, this r is our mechanical speed.
38:01.120 --> 38:08.120
So, omega mechanical is equal to omega s minus
this one is equal to our omega r, this is
38:09.770 --> 38:16.770
equal to rotor speed. In all our analysis,
we assumed that rotor conductors have the
38:20.010 --> 38:26.070
same number of turns as the starter and the
distribution is also the same as starter and
38:26.070 --> 38:32.990
we have assumed that the distribution for
the full cycle is like this; this shows two
38:32.990 --> 38:39.990
pole, two pole, so omega r, the omega r is
equal to omega rotor speed. So, what will
38:42.220 --> 38:44.840
happen if number of poles is more? We will
come to that one later.
38:44.840 --> 38:50.350
So, T is equal to torque, develop T is equal
to why we want torque? We want the torque
38:50.350 --> 38:54.360
speed characteristics of the induction machine,
steady state for speed control that is some
38:54.360 --> 39:01.360
purpose. So, T is equal to P mechanical by
omega mechanical that is equal to I2 square
39:06.350 --> 39:13.350
Ir square Rr by s into 1 minus s divided by
omega synchronous into 1 minus s. This will
39:23.010 --> 39:30.010
be equal to Ir square into Rr by s into 1
by omega s is equal to T, torque developed;
39:48.100 --> 39:55.100
this is the torque developed. So, we have
related between the torque and the omega synchronous.
39:56.090 --> 40:03.090
Now, let us see, how we do the speed control
for the machine?
40:03.750 --> 40:10.750
So, we want a torque speed characteristics
of
40:21.950 --> 40:28.950
the induction machine
with the sinusoidal excitation that is also
called steady state excitation. So, the power
40:39.190 --> 40:46.190
transferred across the air gap, P is equal
to 3 Ir square into Rr by s. So, let us assume
40:54.560 --> 41:01.560
the equivalent circuit is starter drop. See,
now multiplying by s, starter and rotor, we
41:03.990 --> 41:10.990
have brought it to the same frequency. So,
this is ls Rs, this is lr, E is the same for
41:12.780 --> 41:19.780
both; then you have the Rr by s. See, this
is the applied voltage V.
41:22.470 --> 41:29.470
Now, let us assume applied voltage V is equal
to E that means the starter drops are negligible,
41:29.810 --> 41:36.810
first approximation. See, very high speeds
of operation, 50 hertz, 40 hertz, 30 hertz
41:37.830 --> 41:44.830
of operation, we can assume starter drops
of negligible; so, first approximation. Neglect
41:48.810 --> 41:55.810
the starter drops, so starter drops are negligible,
so applied voltage is equal to E. So, our
42:00.190 --> 42:07.190
magnetizing m is here, let us say this is
lm, then you have seen starter drops are negligible
42:14.090 --> 42:18.240
as far as the V and E are concerned. That
means what you mean by starter drops are negligible
42:18.240 --> 42:25.240
means? The applied voltage, V is equal to
approximately our E. But due to current, there
42:33.510 --> 42:38.720
is starter drop is there. So, this starter
drops, same current, so when we say these
42:38.720 --> 42:45.460
two are equal, the same current is, the current
flowing through Rr, s also going through starter.
42:45.460 --> 42:52.460
So, starter inductance, resistance we can
put it here, then lr, Rr by s.
42:59.790 --> 43:06.790
So, by putting this inductance here we are
assuming V is approximately equal to E. But
43:09.460 --> 43:16.460
still there is starter drop here. So, if you
see here, ir is what? See, we have to find
43:18.420 --> 43:25.420
out ir. If you take this one, this is im,
this is ir. So, to find out ir, assuming V
43:29.000 --> 43:35.840
is equal to this way, so ir will be equal
to this is good approximation is equal to
43:35.840 --> 43:42.840
V divided by from this equation, V divided
by root of Rs plus Rr by s whole square plus
43:54.900 --> 44:01.900
omega s into lr plus ls whole square; this
is our ir, this approximation we made V is
44:14.350 --> 44:21.350
equal to Em, so the same current passing the
rotor, it is coming through here also.
44:22.460 --> 44:29.460
So here, what we are doing is the im during
loading, whenever the Rr is your rotor current
44:32.080 --> 44:37.440
is there that means there is a loading. So,
the magnetizing current is magnetizing current
44:37.440 --> 44:44.410
is much less than the load current. So, the
drop due to the magnetizing at current at
44:44.410 --> 44:51.410
the starter is neglected. So here, what is
that? The drop due to im in the starter is
44:56.450 --> 45:03.450
neglected; that is an approximation we have
done.
45:07.570 --> 45:14.570
Now, this is ir; from this one, what is torque?
T is equal to 3 ir square that is this equation
45:18.570 --> 45:25.570
that is equal to 3 V square by Rs starter
plus Rr by s, this is slip, this notation
45:36.700 --> 45:43.700
subscribed only for the starter, this is the
slip into square plus let us say, this I will
45:45.440 --> 45:52.440
make it as X1 plus X2 square. X1 is the impedance
that is X1 is let us say starter ls omega,
45:57.200 --> 46:04.200
let us say X2 is ls omega or we can say s
X s plus Xr; that will be better, starter
46:07.800 --> 46:14.800
plus V, into Rr by s. So, this is the power,
power divided by what we require? Omega s,
46:22.430 --> 46:29.430
this is from the previous equation, omega
s. This is true if we have or number of poles
46:36.840 --> 46:43.840
is 2. Suppose the number of poles is more
than two than if P is more than 2, number
46:45.000 --> 46:52.000
of poles, P means poles, number of poles,
then omega mechanical, the mechanical speed
46:54.520 --> 46:57.480
will come down; so this one.
46:57.480 --> 47:04.480
So equivalently, omega s divided by P by 2
is the omega mechanical. So, here it will
47:06.560 --> 47:11.630
be that means divided by p by 2, it will go
to the numerator. So, with number of poles,
47:11.630 --> 47:16.860
this is the torque equation, steady state.
When the machine is excited with sinusoidal
47:16.860 --> 47:22.030
currents that is using, that is called the
steady state equivalent circuit. That means
47:22.030 --> 47:28.540
sinusoidal excitation means the frequency
is varying, the inductive drop is due to the
47:28.540 --> 47:34.260
omega, l omega, ir into drop. That is why
called sinusoidal excitation; so, this is
47:34.260 --> 47:39.530
the equation. Now, let us see how to find
out the torque speed characteristics here.
47:39.530 --> 47:46.530
So here, if you see here, the torque T is
equal to p by 2 3 V square divided by Rs plus
47:55.350 --> 48:02.350
Rr by s whole square plus Xs plus Xr whole
square into whole square into Rr by s, whole
48:19.200 --> 48:26.200
thing divided by 1 by omega s. See, if you
see here, this torque is this shows very sensitive
48:35.750 --> 48:42.750
to input voltage variation. That means this
shows torque is very sensitive, V square comes
48:51.650 --> 48:52.830
here.
48:52.830 --> 48:59.830
Now, let us say V is constant, let us take
s is small; s is small means what we can do?
49:08.400 --> 49:15.400
This we can neglect, s is small means this
will be very high and this we can remove,
49:15.610 --> 49:22.610
this also we can remove, then approximately
T is equal to 1 by omega s
V square by Rr into s that is in proportion
49:37.200 --> 49:43.080
and the proportionally constant into some
constant K that is because of the P by 2,
49:43.080 --> 49:44.540
3 all those things.
49:44.540 --> 49:51.540
So here, what it shows? For is s is small,
T is proportional to s. So, let us say torque
49:55.490 --> 50:02.490
speed characteristics for small slip; let
us say this is torque, this is slip, for small
50:11.030 --> 50:18.030
slip torque is proportional to this one. Now
s is large, s is large if you see here, then
50:29.710 --> 50:36.710
this is not the one we are going to remove;
s is large, then we can remove this one. So,
50:40.830 --> 50:47.720
this is a plus here, rotor resistance is this
also we can approximately we can remove this
50:47.720 --> 50:54.720
one. When s is large torque will be proportional
to 1 by omega s into V square.
50:57.270 --> 51:02.330
We can take a typical machine parameter and
we can check this one. Approximately, you
51:02.330 --> 51:07.840
take a 50 hertz operation, omega s is equal
to 50 hertz operation, V square divided by
51:07.840 --> 51:14.840
X1 Xs plus Xr whole square into Rr by s. So,
this shows torque is inversely proportional
51:25.450 --> 51:32.450
to s. That means T is inversely proportional
to s for high speed. So, that means it will
51:35.000 --> 51:39.800
come; inversely means it will come. So, in
between, it has to go through a peak and come
51:39.800 --> 51:46.800
for your continuous operation. So here, you
will get the X maximum. So, this is slip.
51:51.130 --> 51:58.130
Now, this is slip can be a maximum, s here,
s is equal to 0 here means s is equal to 1.
52:01.230 --> 52:08.230
s is equal to 1 means what? Omega r is equal
to 0 that means this is the starting speed,
52:12.700 --> 52:19.700
omega r is equal to 0, omega r is equal to
omega s. So, torque speed characteristics
52:20.380 --> 52:27.380
will be like this. So, now let us write this
one, torque speed characteristic.
52:32.460 --> 52:39.250
See, this will be torque speed characteristic
will be if you draw it, it will approximately
52:39.250 --> 52:46.250
that is what finally we want. We will write
torque and speed here that is the omega s.
52:50.450 --> 52:57.450
When you do it, see this is the same curve,
it will go like this. If you see here, this
53:04.690 --> 53:10.910
is the linear region, this is for one omega
s. Suppose if our omega is changed, omega
53:10.910 --> 53:17.910
s and this let us see omega s1; now omega
s2, then it be something like this or with
53:20.040 --> 53:25.280
various omega, it will go like this. So, that
shows this parallel curves you will get.
53:25.280 --> 53:32.280
Same like DC motor, DC is the stable operation.
See, if you represent it like this that is
53:34.560 --> 53:41.560
torque omega, we are this is the linear region,
this is the linear region and we will be operating
53:47.530 --> 53:51.900
in this linear region. So, torque speed characteristics
will be approximating something like this,
53:51.900 --> 53:58.900
it will go like this. This part we are using
it. So, how do you control the speed here,
54:00.270 --> 54:04.630
speed control for torque speeds characteristic
like this? And, we are restricting only to
54:04.630 --> 54:11.630
the linear region; this we will study, this
is for omega s1, omega s1, omega s2, omega
54:16.490 --> 54:23.490
s3, we will study in the next class.