Welcome to this Synchronous Machine lectures
and we were discussing generator mode of operation of cylindrical type synchronous machines,
cylindrical rotor synchronous machine where the air gap is uniform air gap is uniform
and synchronous machine equivalent circuit is much simpler than the case of induction
machine because of the fact there is no magnetising branch present here. So, the equivalent circuit will be something
like this plus minus Ef and then ra plus jxs synchronous impedance which is often called
Zs and I was writing it as some magnitude Zs angle beta, beta will be closed to 90 degree
because the value of xs is much higher than ra and this is the thing and suppose the terminal
voltage. This is per phase all the voltages and currents are here. And, therefore, in
case of generator mode of operation suppose the generator delivers a current of Ia that
is the per phase current, then it is true that Ef will be equal to V plus Ia into zs
bar. And, based on that we did last time very quickly,
I will tell this is the terminal voltage, this is generally taken on reference V 0 degree
and then suppose it is supplying a lagging power factor load Ia. So, this is the power
factor angle of the generator at which it is operating to the external world and this
plus Ia ra, Ia ra plus jIaxs, jIaxs gives you your Ef and in case of generator mode
of operation, Ef will be always above V and this gives you delta. Ef the length of Ef
is a measure of the field excitation and length of V is a measure of approximately the resultant
field, ok. Therefore, this generator supplying a lagging
power factor load is over excited. So, it is lagging power factor, it is over excited
we know that, over excited. And then I told you if you want to find out the expression
of power and torque so, power delivered total power delivered by the machine should be calculated
like this S is equal to this is the current delivered to the load. So, it should be Ef
this voltage I mean I will do like this, V into Ia*, this is the visual thing we will
do so that we get the expression of real and reactive power simultaneously.
And, since V is Vangle0, we know that Ef can be written as Ef angledelta where delta is
a positive number and then what I did, I just wrote Ia; Ia is this way. So, Ia should be
equal to Ef angledelta minus V angle0 degree divided by this Zs anglebeta, this is Ia.
So, Ia* is needed. So, Ia* will be simply Ef, all angle should be made negative minus
V angle0 divided by Zs minus beta which will be equal to Ef by Zs and this will be minus
delta plus beta. No, this will be beta minus delta, beta minus delta, this first term and
this will be minus V by Zs angle beta. So, this we substitute it here. I will not
repeat. So, I have substituted this here. Then, total power will be 3 times this and
we get an expression of the complex power delivered to the load. If both the terms become
positive; that means, it is delivering really kilowatt which in case of generator, it has
to be always positive the real part and imaginary part, if it is positive it means it is delivering
a lagging power factor load. And, then I told you that this Zs which is
equal to ra plus jxs, xs is much higher than ra. So, what people do? Often people use this
approximation that this is jxs which is equal to xs angle90 degree. And, if you do that
the real power output of the generator is a very famous equation and that is equal to
magnitude of Ef magnitude of V divided by xs into sin delta.
Since synchronous machine runs at constant speed, this itself will be the expression
of the torque if you want to know it in Newton metre divided by 2 pi into speed of the machine
that is ns mechanical speed, you will get the expression of the torque and P delta curve
will be sinusoidal which at distinct magnitude Pmax and P value of Pmax is proportional to
terminal voltage into Ef by xs, this is delta. Maximum power or torque it can develop is
90 degree, this side is power ok. So, we did this. Now, today what I will first; so, we are still
dealing with cylindrical rotor synchronous machine, but what will be the expression for
power and torque for a synchronous motor operation synchronous motor operation, ok. We will first
draw the phasor diagram and try to find out the expression of the torque.
In case of synchronous motor operation, I will draw like this, this is this, the impedance
is Zs as usual and here it is bus voltage V which is angle 0 degree volts and since
it is operating as a motor I will show that it is drawing power from the source. So, Ia
I will assume inconsistent with the supply voltage which is supplying the motor. So,
this will be the thing and this voltage will be your Ef, voltage between these two points.
Now, so, let the motor draws lagging power factor load, lagging power factor load, lagging
power factor current. It is load is mechanical load on the shaft and it draws at some steady
operating point it draws like that. Then, I will say that this Ef should be equal to
V minus Ia zs; this will be the magnitude of the rms voltage.
So, how to draw it? So, the drawing of this phasor diagram will be Ia ra minus jIa xs
ok. So, we first draw; please carefully follow me what I am doing this is V. Then I have
told it is drawing a lagging power factor current. So, this is my Ia ok, then from V
I have to subtract this ra and xs draw first I subtract. So, minus Ia ra, it will be like
this is minus Ia ra and then jIaxs is like this. So, minus jIaxs will be 90 degree to
this and this length is much higher compared to this length although it is.
So, you will get your excitation voltage Ef and this angle is theta, the operating power
factor of the machine angle between V and Ia is your theta. So, this will be Ef. One
thing I can say; say a motor draws lagging power factor current length of Ef is nothing,
but length of excitation EMF means the rotor field this length denotes. V denotes what?
The resultant field. Therefore, you can easily see resultant field is more than Ef and it
will act as a motor. So, in motor mode of operation, it is essential
to draw the phasor diagram in this way at least this is the convention I will follow.
In case of generator when you draw V, it delivers power to the outside world, in case of motor
it draws power from the source. So, it should be consistent with the supply. So, Ia I have
drawn like that. Now, in the same way as we have done in case
of generator we can suppose I want to know where is my PHIf, where is my PHIr and where
is my PHIa. Obviously, this will be the direction of armature flux or armature mmf along the
maximum value direction, it will be pointed this is PHIa. Now what is Ef? This point you
listen carefully unlike generator. Now, Ef is the generated voltage inside the
machine and it’s polarity will be like this and it will be inconsistent. If this is EMF
is allowed to act alone, it would have driven current in this direction. Therefore, to take
that into account your actual Ef is actually here, minus Ef you have to take if you want
to get information about PHIf correctly. And, then I know this is PHIf this will be your
PHIf, 90 degree lagging the induced voltage and then PHIa plus PHIa will give you the
resultant on net EMF PHIr. This r is not rotor resultant.
So, here also I find that PHIresultant is more than PHIf. Therefore, it is under excited.
Of course, while solving for problem, you need not it is not necessary to draw Ef PHIf
PHIf PHIr all the time, but only one thing I will tell that this is all phasors are rotating
in the anticlockwise direction omega like this. In this case, this is the rotor field
PHIf this is the net field. Therefore, the torque will be in this direction electromagnetic
torque developed by the machine. And, we know it should be because in case
of motor mod, if this is the direction of rotation electromagnetic torque must act in
this direction and it is therefore, consistent. In case of generator, I showed you that PHIf
was behind PHIresultant. So, the electromagnetic torque was in opposite direction as that of
the direction of rotation and we know that therefore, this will be this thing. So, in
this case once again, I can find out what is the total power drawn by the motor. Complex power drawn by the motor drawn by
the motor will be how much? This is the, I will draw it here once again. This is your
Zs anglebeta and this is your here you do not fumble ok. This is terminal voltage and
it is drawing a current like this and also you note that in case of motor mod Ef will
be always below V. In case of generator mode, Ef will be always above V, no matter whether
it is lagging or leading. So, Ef will be always above V in case of generator and Ef will be
below the terminal voltage when it is motor mode.
Therefore, I will assume this voltage Ef to be Ef angle-delta, this is the thing and this
is the armature current. Got the point? Therefore, I will say that power drawn should be once
again power delivered by the source complex power drawn by the motor is equal to complex
power. Anyway it need not be written complex power delivered by the source it will be equal
to V Ia*. This will be the thing, okay. Now, once again I have to calculate Ia. Ia*
I have to calculate. So, first I calculate Ia. This time Ia is Vangle0 degree minus Ef
angle- delta by Zsanglebeta and this will be equal to V by Zs angle- beta minus Ef by
Zs and angle-(delta+beta); this will be the thing. So, minus delta plus beta over I have
written. So, it is equal to like this; if we this bar confused as you it is like this.
Therefore, this will be the thing. Therefore, I need Ia* in this one; so, I will
calculate it here. So, this is V by Zs angle beta minus Ef by Zs and this angle will be
delta +beta. Therefore, your S will be, complex power delivered will be V angle0 degree into
this, that will be V square by Zs anglebeta minus V Ef by Zs angledelta+beta.
Therefore, real power drawn by the machine, P part will be V square by Zs into cos beta
minus V Ef by Zs cos (delta+beta). This is the expression for the real part that
is the how much kilowatt it is drawing and that is the work which will it will do on
the motor. Of course, after subtracting the rotor copper loss; not rotor copper loss,
stator armature loss. This is ra and Zs. Total power drawn is real power drawn is this of
which a portion will be lost in this one Ia2 ra, and the remaining portion it will do the
mechanical work. That way when you do the numerical, you will come to know.
So, this is similarly I can find out Q which I am not doing, the imaginary part how much
reactive power it is drawing, ok. So, this is the thing. Then what I told? These expressions
you can memorize; if you have lot of memory, otherwise it is not necessary. What I will
do with the numbers I will calculate straight away with a scientific calculator, you get
quickly this results, no doubt about it. But, one result is important which is a special
case that is since once again xs is much higher than ra. Therefore Zs can be written approximately
as jxs only and which is nothing, but xs angle90 degree. So, in this case if for this case
if ra is neglected, then the power drawn will be V2 by Zs, magnitude will be only xs. So,
V2 by xs and cos 90; this fellow goes and minus V Ef by xs magnitude of Zs into cosine
(delta+90) degree and cos 90+theta is minus. So, this will become Ef V by xs into sin delta.
Got the point? Therefore, the expression of the power if
ra is neglected which is often the case, power system engineers will neglect, do not bother
about ra. They will simply neglect ra represent the power system synchronous machine in terms
of xs and alone with the understanding that armature resistance is much smaller compared
to xs synchronous reactance of the machine. And you will get this, same expression as
that of generator and motor; the question is where it is.
Therefore, to conclude this so, this a lot of problems you can solve by whether ra is
given or not, whether ra can be neglected or not; you can solve a lot of problems in
this method. So, the expression of the real power which is most important Ef V by xs sin
delta is same for motor and generator mode. So, nothing to worry, but only thing you should
be clearly knowing whether the machine is running as motor or generator because the
direction of armature current, you have to them assume accordingly and calculate; whether
it is Ef minus V by Zs in case of generator it was like that and delta was positive and
in case of motor current is being drawn from the supply. So, that is the case. Now, what I will do? I will once again because
this is often done. If ra is neglected; neglected compared to whom? Compared to xs. Neglected
compared to xs, then your Zs becomes just xsangle90 degree, is not?
Under this condition I will just draw the phasor diagram very quickly. It is a worthwhile
exercise to pursue. See, suppose I will draw it here like this, this side I will draw generator
mode and this side I will draw motor mode motor mode. So, in generator mode what is
the thing and draw this circuit diagram that is only jxs, nothing else and here is it is
V and here is it is your excitation voltage. And, it is generator mode, current is delivered
to the bus or to the load at a terminal voltage V. So, it must be, it is direction must be
properly drawn, it is this. Then, I know that Ef is equal to V plus j
Ia xs, is not? Now, if it is lagging power factor, I will
draw it like this. Terminal voltage, begin with terminal voltage. Suppose lagging power
factor Ia, then there is no Ia ra now. So, Ia jxs will be perpendicular to this, j Ia
xs, and it will be your Ef and this angle is delta, over and so, this is the thing.
If suppose, the generator is supplying leading power factor load, then what should I do?
V is here, suppose the generator is supplying a leading power factor load Ia. Whether leading
or lagging this equation is true. So, V plus j Ia xs I have to draw. Here is your Ia direction.
So, draw a line which will be perpendicular to this Ia line and this will be your j Ia
xs and this will be your excitation voltage and this will be angle delta.
So, this is leading power factor load, leading power factor load. So, in case of leading
power factor load, lagging or leading no matter Ef will be always above V for generator mode.
Only thing in case of leading power factor load length of Ef is less than V, length of
Ef is a measure of rotor field and rotor excitation field that one and V this length is approximately
equal to Er resultant field. Therefore, this will be measure of the resultant
field. So, what should I say leading power factor generator is it over excited or under
excited? Under excited because length of Ef is less. So, under excited and this is Ef
length of Ef is greater than V. So, it is over excited. I am not going in this phasor
diagram to the level of PHIf and PHIr because length of Ef I know, it is a measure of PHIf;
length of V is a measure of resultant field. If I am asked I will be able to draw, but
for problem solve solving from electrical sides it is not necessary what I am telling.
So, this is the thing. In case of motor mode, what is the expression
of real power drawn by the machine? It will be simply Ef V by xs sin delta that is all
I am not writing the expression of Q, that can be easily found out. In case of motor
mode, so, first you draw this Ef, here is your Zs; Zs means only jxs reactance and here
is your supply voltage V and it draws a current of Ia. This direction of current assumption
is important, you can just compare. Then what I will do? I will once again start
from the V terminal voltage, suppose the motor is drawing a lagging power factor, so this
was lag, this was lead leading power factor. So, suppose lag power factor. So, motor is
drawing a current of Ia like this and this angle is the power factor angle at which it
is drawing and there is. So, from V, Ef is V minus Ia jxs. So, V minus Ia jxs. So, j
Ia xs is this way; so, negative of that. So, it will be what you should do? You should
draw a line perpendicular to this and this length will give you minus j Ia xs and this
fellow will be your Ef and this angle will be delta. Ef will be below V, everything will
be in place; if this convention is followed. Is it under excited or over excited? Length
of Ef is less than less than length of V. So, it is under excited; just opposite to
that of a synchronous generator if it is drawing lagging power factor current, it will be under
excited. And, what will be the leading power factor
case? V; suppose this is Ia, this is theta. So, V minus j Ia xs will be your Ef it is
written their this is the equation. So, V, so, j Ia xs is like this. So, negative of
that. So, it will go like this such that this angle is 90 degree and this will be your excitation
voltage and this is delta. So, if this is the reference excitation Ef is greater than
V it is over excited. So, we will continue with this in the next class.
Thank you.
