Welcome. So, we were discussing about the
equivalent circuit of an induction motor.
See initially, we just told about how torque
is produced in a motor. And try to understand
it, physically what is happening that is the
rotor field, stator field this that. And the
interaction of the stator result in field
and rotor field gives you the torque.
Now, we have gone further, and we try to develop
the equivalent circuit of a 3-phase induction
motor per phase basis. And from that looking
at the power flow that takes place in the
machine starting from supply source to the
mechanical power developed by the motor.
So, I we develop this equivalent circuit,
and told you that this is the equivalent circuit
which is a stator resistance stator leakage
reactance per phase. Now, it is similar to
that of a transformer, and then you have this
magnetizing branch which is j x m. I mean
all reactance's are j that is there.
And then on this side, the reflected rotor
parameters came out to be x 2 dashed, and
this is r 2 dashed by s ok. And slip depends
upon the operating point of the induction
motors. And slip is a measure of speed as
well, because you know actual rotor speed,
all speed are mechanical here is 1 minus s
into n s. So, so and n s is this synchronous
speed.
So, now this is the per phase applied voltage
V 1 this is the thing. Then what I did between
these two points, we found out the thevenins
equivalent. And this circuit was redrawn as
you know r thevenins x thevenins m series,
and then this remains V 1. And then x m goes,
and you are having this equivalent circuit
ok. And this current is suppose I 2 dashed.
When we just considered the power balance,
after we have got the equivalent circuit the
rest of the work is somewhat routine network
type analysis, no field is coming here that
is one good thing.
So, the input power to this circuit is and
I drew this this I will not waste time. Last
time we told you that this equivalent circuit
means, you are giving some input power drawn
from the supply out of which a portion will
be lost in the stator coils, as stator copper
loss 3 I 2 square r thevenins that is value
will be I mean I 2 dashed squared r thevenins.
Then rest of the power is as you crossing
the air gap, and entering into the rotor that
is why, it is called air gap power air gap
power for our own understanding ok, we will
interpret it like this. Total input power
minus this stator copper loss or stator loss
whatever is there. There will be cold loss
also that we will take care later P a g. And
then this power after entering into the rotor
circuit, there will be a power loss in the
rotor resistance, which mind you is 3 I 2
dew dashed square r 2, which is same as 3
I 2 square r 2, because we have referred everything
to there ok.
Now, recall that this equivalent circuit,
the power involved in the circuit is the air
gap power, because this is the power power
real power enters. So, P a g P a g is 3 I
2 dash squared r 2 dashed by s. And obviously,
rotor copper loss is equal to 3 I 2 dash squared
r 2 dashed. So, it looks like rotor copper
loss is nothing but s in to P a g.
Now, so from P a g, if you subtract this rotor
copper loss if you subtract it, then you get
P gross mechanical which will be equal to
P a g minus P rotor copper loss. If you subtract,
this becomes 1 minus s into P a g. And this
is shown in this one P a g comes, then you
subtract rotor copper loss, then you get P
gross mechanical power whatever mechanical
power, it is called gross because a portion
of it will be lost to supply the mechanical
losses in friction etcetera.
This is not the net mechanical power output,
but gross mechanical power output. Because,
this gross mechanical power output only decides
what is the total opposing torque developed
by the machine, put on the shaft of the machine
including the actual mechanical rotor and
the frictional power. Anyway, so this is the
thing.
Now, our target is to find out, what will
be the torque developed by the machine. So,
this we will do in this way, I have got the
P gross mechanical power from which I will
find out, what is the mechanical torque present
on the shaft of the machine. And we will assume
that the machine is running with a with a
fixed value of slip s that means, at a fixed
RPM.
Therefore, the if I can estimate the mechanical
torque, then the electric torque developed
by the machine must be same. So, so remember
this visual picture any motor, this is the
electromagnetic torque, this is the direction
of rotation n r. And the gross opposing torque
T mechanical gross, this two must be same,
so that it is running with a particular RPM
n r or with a particular slip. Because, every
n r has got a if you map it to slip, it has
got a unique value of slip, so that is the
thing.
Therefore, from this I will say and then I
told one very important relationship that
is the this must be remembered that is P a
g is to P rotor copper loss is to gross mechanical
is in the ratio 1 is to s is to 1 minus s
that we have found out. So, this is the thing
P gross mechanical.
Therefore, total opposing torque opposing
torque, which is sum of the frictional torque
to be overcome fictional torque plus the actual
net mechanical net mechanical torque, which
is doing useful mechanical work for me that
is the thing. Therefore, this must be equal
to this must be equal to 1 minus s into P
a g that is P gross mechanical divided by
2 pi n r, where n r is the actual rotor speed
actual mechanical I will repeatedly not write,
but any way actual mechanical speed n r in
rps if you want to get torque. So, this is
the thing.
Now, what I will do this for n r, I will substitute
this thing here. So, it is like this then
1 minus s into P a g divided by 2 pi 1 minus
s into n s. And then you see 1 minus s goes,
and gross mechanical power developed will
be equal to 2 pi n s. And this must be equal
to the electromagnetic torque developed, because
machine is running at a constant speed or
constant value of slip.
So, we are trying to get an expression for
electromagnetic torque not in terms of delta
or b r or b s, but in terms of equivalent
circuit parameters that is the idea here.
Because, this equivalent circuit parameters
if these are known to me, then I will be able
to calculate torque simply by using the equivalent
circuit parameter. So, this is the also equal
to electromagnetic torque developed by the
machine at slip s that is the thing.
So, this is the expression of electromagnetic
torque of which you must note this point that
this n s is constant n s sorry n s is equal
to 2 f by p is constant. So, long the supply
frequency is constant, and machine is given
so p is also constant. Therefore, the electromagnetic
torque developed by the machine, if you divide
by 2 pi n s the air gap power, then its unit
will be in newton newton so much newton. If
this is in watt, and this is in rps.
Now, note that to calculate the electromagnetic
torque, therefore if you have estimated the
air gap power that itself is a measure of
electromagnetic torque developed by the machine
why, because 2 pi n s is constant. So, if
you have calculated P a g, electromagnetic
torque is as such calculated I have to simply
divided by a constant number 2 pi n s. So,
sometimes P a g is considered to be the torque
developed and P a g alone its unit is watt,
so it is called a a given a name synchronous
watt.
So, P a g is also T e expressed in synchronous
watt, this is a term synchronous watt. So,
sometimes torque is expressed in sometimes
torque is expressed in newton, then so so
torque T e is equal to P a g you can write
absolutely correct, but write synchronous
watt its unities or you write P a g by 2 pi
n s. And this will be in newton meter that
is the thing should just recall.
Now, after I have learned that in torque in
newton meter is this, let us further simplify
this expression that is T e P a g P a g is
nothing but 3 I 2 dash squared r 2 dashed
by s by 2 pi n s. Now, I want to express this
expression of the electromagnetic torque developed
by the machine in terms of equivalent circuit
parameter. And equivalent circuit is this
which we will consider the thevenins equivalent
this circuit.
So, so this I 2 dashed is nothing but V thevenins
divided by square root of r thevenins plus
r 2 dashed by s whole squared plus x 1 plus
x 2 dashed x thevenins plus x 2 dashed whole
square. This is so I have to put this value
here. And if you do that, you will get 3 V
thevenins squared by 2 pi n s square it. And
in the numerator you have r 2 dashed by s
divided by in the denominator it will be a
these are all algebraic manipulation r 2 dashed
by s whole squared plus x thevenins plus x
2 dashed whole square. So, this is the expression
of the torque in terms of equivalent circuit
parameter.
So, if somebody says machine is running, steadily
with a slip s equal to 0.05. And suppose the
all the parameter values of unknown, and I
will just those that value of slip, and tell
ok. If the machine has to run at this slip
as a motor, then the electromagnetic torque
will be so much of newton meter. So, by putting
different values of slip, I will be able to
calculate, how much torque the machine is
capable of developing that is I can make a
table here. I will write slip, and value of
the slip range is known, and therefore by
putting different values of slip slip varies
from 0 to 1.
So, if you can easily see, if you flew put
a s equal to 0, this torque will be approaching
0, because r 2 dashed by s is there this will
be if you if a s tends to 0, torque will be
0 tend to 0. Because, there is a square term
here r 2 dashed by whole square, this will
dominate others are negligible, and on the
top also there is r 2 dashed by s. So, ultimately
the denominator becomes infinitely large,
I mean just telling mathematically.
So, torque will be 0, physically I know why
the torque will be 0, because its slip equal
to 0 means n r equal to n s, and machine cannot
develop any torque. Similarly, I will put
say s equal to 0.0,1 and I will get some number
I will put it here like that 0.02 and so on,
and go up to s equal to 1. So, all these entries,
I can make. Provided I know the equivalent
circuit parameters, suppose these are given
to me.
So, I will be able to calculate make a table,
where y axis will be torque and slip is vary
from 0 to 1. So, this I can make a table like
this. And after that what I can do is this,
I can plot this for a given machine, and suppose
this is slip equal to 0, and this is slip
equal to 1, this is this axis is slip. And
when you are potting this slip this way, I
know this is speed. One and the same way either
you tell the speed, and this is n r equal
to 0, and this is n r equal to n s you must
keep this picture in mind.
Now, if you sketch this tabulated value of
torque, again slip which I got from this expression.
The typical curve will be somewhat like this.
The slip varies between 0 to 1 for motoring
mode, and the characteristics typical characteristics
will be like this understood this is the shape
this curve gives you. And you will note that
once this torque slip characteristics, this
is called torque slip characteristics torque
versus slip characteristics.
And if you get this torque, you will note
that there is a slip at which the the torque
increases as slip value increases, then it
attains some maximum value T max at some slip
say s equal to alpha. At s equal to alpha
there is a maximum value of the torque, and
then torque once again decreases, and this
is a typical torque slip characteristics typical
characteristics.
And also for a well designed induction motor
the the value of the slip at which the machine
is capable to develop maximum torque is valued
typically will be about if this is 0 to 1,
this will be one-tenth of this about it is
no fixed number, I am telling you the the
expected value of the slip for which the torque
attains maximum is about alpha value will
be about 0.1.
So, this of course is one-tenth of this one,
indicating that the curve go to to maximum
at a very small value of slip, then it decreases,
and this comes here. And this is the value
of the obviously, starting term T starting
of the induction motor, because s equal to
1.
As we switch on the supply to the stator s
equal to 1, machine is capable of developing
this much of starting torque. Obviously, frictional
torque if it is opposing torque is less that
this, then machine will accelerate. If the
opposing torque present is this much, oh machine
will not start, because starting torque will
never be able to match this opposing torque,
and it will remain stationary blocked condition.
Although, the induction motor has got a starting
torque definite starting torque T starting,
but it must overcome whatever opposing frictional
torque is there, so that it accelerates ok.
So, this is the typical torque slip characteristics.
Then the next question comes in ok in terms
of equivalent circuit parameters can we tell
what will be the value of T max, and at what
slip that T max occurs. So, so so at at what
slip T e max occurs, and what is the value
of T e max can we derive some formula, what
is the value of x T e max.
At what slip that slip I have identified as
a alpha. So, to do this, what should I do
so we write down the general expression for
T e that is T e is equal to 3 V thevenins
square, it is to be written actually rewritten
it 2 pi n s r 2 dashed by s divided by this
is the general expression of the torque r
thevenins plus r 2 dashed by s whole square
plus x thevenins plus x 2 dashed whole square.
This is the general expression of the torque
of which is the equivalent circuit parameters
are constant V thevenins is depends on primarily
applied voltage, and small r 1 x 1. So, all
things are constant except s. And I want to
find out, at what slip T e max occurs. So,
it comes to our mind that then what we have
to do is this, we have to calculate d T e
d s, you differentiate this equated to 0.
And from that and try to get try to get the
value of alpha, alpha is the slip at which
maximum torque occurs, but it looks like it
will be slightly involved task.
So, we will not calculate in that way. Rather
we will try to get that expression in a much
elegant way, because I know the equivalent
circuit of this induction motor looks like
this this is r thevenins, this is x thevenins,
and this is your magnetizing inductance j
x m, and this is x 2 dashed, and this is r
2 dashed by s. And sometimes people put some
variable resistance sign to indicate that
machine may operate at different slip, so
it is only this parameter which goes on changing.
As you change the degree of loading of the
machine, and this is your per phase applied
voltage.
So, with this equivalent circuit, the value
of the slip that is s equal to alpha, we are
looking for which will give you the maximum
torque developed can be very nicely obtained,
and this we will continue in the next class.
Thank you.
