So, welcome to synchronous machine lecture. So, in my last lecture. I was telling you that how to draw, phasor
diagram is an important thing for synchronous machine and computation will be very simple,
if you have a scientific calculator. What all you have to connect fora given operating
point, if delta is given, delta is not given? If ra is neglected straight away go there. In fact, from the phasor diagram also using
properties of triangle, under this condition this angle 90 degree several problems can
be very quickly solved. I will try to devote one or two sessions for
problem solving later, but we have now a fare idea. Once again I am telling in case of generator
Ef will be always above V; no matter whether it is lagging and leading power factor, what
will change is either over excitation or under excitation. Similarly, for motor mode and the expression
of power with this condition that ra is neglected becomes your excitation voltage very easy
to remember, product of these and this divided by the reactants and sin delta that’s all,
ok. So, we have done. Now, in this lecture after knowing these things,
I will introduce to you the, what will be the effect of variation of excitation of a
synchronous machine. What I mean to say is this; see first try
to understand the operation. What I will do that, suppose the effect of,
this is very important topic, effect of variation of excitation. That means, excitation means what? DC field current, variation of excitation
or DC field current that we will investigate. So, first we will do for generator motor,
then I will do for motor mode. Suppose generator motor and in this explanation
what I will assume ra is pretty small. So, that I will come to know very quickly
what will be the effect of course, with ra also it can be taken into account that I have
included in my note, at this stage you need not go to that ok. First, let us see this can be very nicely
explained. Now, effect of variation of excitation in
generator mode what does this mean? This is the generator; let me first draw this
is the thing and here is the field winding. What I am trying to tell here is some If DC
current and here is some voltage and it is connected to the bus A phase, B phase and
C phase; how to synchronize those things we have discussed. So, the synchronous and it is operating as
a generator. How it will operate as a generator, because
it this rotator is driven by a prime mover, ok. And what is a prime mover? Prime mover could be a steam engine, a water
turbine, a steam turbine or a diesel engine this sort of thing. Now, that it will operate as a generator;
see after you have synchronized, let me spend few sentences. After you have synchronized the machine, this
synchronous machine is said to be floating on the bus, it is neither operating as a generator
or as a motor. If you want to operate it as a generator means
that this generator will fit power or current into the bus, this is bus. What you have to do is, you have to increase
the steam input if it is a steam turbine, then only real power, the rotor will then
become in advance with the resultant field, that is this diagram. In steady state everything will be rotating
so, that this rotor field will be behind the resultant field, rotor field should be in
advance so resultant field. So, if this is rotor field Bf or PHIf; you
are in case of generator mode of operation, your rotor field that is PHIf will be PHIresultant,
that we have seen and in case of generator everything is rotating in this direction omega
torque will be in the opposite direction. Anyway, so input increase the steam input
to the bus then it will feed power you have to burn more coal then, ok, and it is generator
mode ra equal to 0. So, suppose for a given steam input, the real
power input from the mechanical side is constant or I will assume that in that case the output
power delivered to the bus will be also constant. So, per phase circuit diagram is this and
I will assume only j xs is present and it delivers a current Ia. One may wonder where is the load if it is
supplying lagging power factor you do not own the, in the infinite bus all nodes are
connected all big generators are feeding, the bus a common bus nodes are already connected
nothing that it is not there. So, your this small generator in the laboratory
if you have synchronized and increase the prime mover input it will also contribute
to this total demand of the load which is connected to the bus by a small amount, some
kilowatt say if it is a kilowatt rating machine. So, load is already connected there in the
bus and this voltage is V is that clear? Now suppose for a constant output power for
constant generator mode output power output real power constant output real power. I can say this output of the generator will
be 3 V Ia costheta. Where theta is the power factor angle of the
motor because angle between V and Ia, this is Ia, it is delivering. So, this will be the thing if the output power
remains constant. I look at this expression P out V, V is the
bus voltage, it cannot be changed in any case, bus voltage is bus voltage, fixed it is. So, many generators are connected infinite
sort of bus. So, V remains constant, also I will say that
this P output it is equal to 3 V Ia cos theta, no doubt, from the terminal condition and
also I know this power is 3 Ef V by Xs sin delta and this two will be same, is not? There is no real power drop here that is,
if you calculate Ef Ia*; it will give you what is the total power output from looking
from the excitation, I mean excitation, I mean emf induced in the generator side, Do
you think this power Ef Ia* in this case Ef Ia* must be equal to V Ia*. Why not, because there is no power loss it
is a good exercise you can easily show this. But, none the less this is true. Now, what I am telling this Ef you recall,
this Ef is equal to root 2 pi f flux per pole Kw into N phase of the stator phase. Now, what I am telling that I will vary this
field current If by connecting at rheostat here, I will simply play with this without
changing the prime mover input or output power. How much will be the prime mover input, needed
power delivered, same as this whatever you are delivering here that must come from the
mechanical side it is doing like that. But, the one thing is very clear if vary If
, PHIf is directly proportional to If, therefore, as you vary a field current of the alternator,
induced emf in the alternator will change, Ef is bound to change, that is the magnitude
of these will change if If is varied, why not? But, what I am saying is this by doing this
you cannot change the magnitude of V, terminal voltage is fixed. So, something else then has to change we are
trying to investigate what is that thing how things will now look like. Therefore, if I am not changing this team
input to this prime mover output power delivered in kilowatt will remain same and for an excitation
present, these two must be same. So, let me draw the phasor diagram this is
quite interesting, you just see, ok. Suppose this is your V. I will once again
redraw, let me spoil this pen, see this is Va. Suppose at some excitation
this is your Ia1 and machine is delivering lagging power factor load theta1, this is
general operating point I have told Ia, theta. Suppose, it is doing like this. Then, if you are not from this equation, I
come to the conclusion that Ia cos theta has to remain constant, because of the fact V
is said constant, ok. Therefore, as I play with the field current
perhaps magnitude of Ia will change, perhaps theta will change, I am not sure that this
stage. But, what I will demand is that since output
power is constant no matter how you are playing with your field current, Ia cos theta this
product has to remain constant that is the important point. In other words, what is the Ia cos theta? The projection of this on this V axis. I do not know how Ia and theta is going to
change, but I will say that the tip of the current phasor Ia must lie on this vertical
line. So, that Ia cos theta is constant; I will
come to this. So, this is one thing second thing is at this
operating point with some field current If1. I now complete the phasor diagram. So, where is your Ef, suppose I want to find
out. Ef will be Ia1 plus j Ia x1, is not? Synchronous reactors and this voltage excitation
voltage I say this is Ef1 at some field current If1. I found that the machine is operating as a
generator, and supplying a lagging power factor load, delivering a current of Ia1 at a power
factor angle of theta1 at that time this is the thing V plus Ia jxs is this one, and this
angle is the delta1, at this load angle sometimes delta is called, at this load angle the synchronous
machine is operating as a generator why generator Ef is about V and it is delivering current
Ia, understood? Now, this power I am saying this is, this
remains constant this cannot change. If it cannot change V is constant, I told
Ia cos theta has to remain constant that is why the tip of current phasor armature current
phasor must lie on this vertical line, then come to this equation. This power constant, what it implies here
in this equation? it implies that V is already constant bus voltage, xs is constant. So, I will say, under this condition power
is constant Ef sin delta also as to be constant. Now what is Ef sin delta? Ef sin delta is this length, this vertical
length. This is your Ef sin delta in general, Ef1
sin delta1 in this case and I am telling you play with your field current, this thing must
remain constant, understood. Now, suppose I say now the most important
point. So, this is for a general operating point
generator is delivering a certain armature current Ia1 at a power factor angle theta1
at that time this is Ef1 and I am demanding that this power remains constant then I will
conclude that Ef sin delta will remain constant, that is this vertical length will remain constant,
and Ia another way of calculating power V Ia cos theta, this to must be same V constant
so Ia cos theta is constant means this projection of the tip of the current, this will be constant. So, draw a vertical line here. Now, see suppose I have increased the field
current it is operating nicely at this operating point 1. What is the operating point 1, armature current
Ia1 power factor angle theta1, load angle delta1 excitation voltage Ef1, terminal voltage
V in other operating point this is constant V is constant bus voltage. Now, suppose I have increased the excitation
what will be the consequence of that, if you change the excitation; Ef has to change. If you increase the field current; Ef will
become bigger because Ef depends on root 2 pi f PHIf Kw Nphase. PHIf is directly proportional to If, therefore,
excitation voltage Ef must be increased. So, this length must be increased, it will
be bigger, it will become bigger, now the interesting point is it will be bigger no
doubt, but since the power is constant the tip of this current phasor cannot, but lie
here. Suppose the field a current you have increased
excitation voltage has become more then take that length cut this line because Ef2 sin
delta2 has to be equal to Ef1 sin delta1, because Ef sin delta this thing has to remain
constant, Ef into sindelta. So, what do you do you have increased the
excitation. So, take this to Ef2 and then I know this
is by delta2 because the angle between V and Ef2 is there and on the top of this Ef2 will
be equal to V + jIa2xs, V plus Ia j xs is your Ef. So, Ef2 will be here, then I must say V this
length, follow me carefully, this length must be your j Ia2 xs, it has to be V plus jIa2
xs is your Ef2, xs is constant j Ia2 xs I know. Therefore, I will be able to fix up where
is my Ia2 where it will be this is jIa2 x2. Therefore, Ia2 must be perpendicular to, this
must be the direction of Ia2, but the tip of Ia2 must lie on this green line. So, this will be your Ia2 and this will be
your new power factor angle theta2. So, if you have increased the excitation of
a synchronous generator at under constant power, condition earlier it was operating
with Ef1 delta1 etc, your new current will now become if If is increased your new operating
point will be Ia2, power factor angle theta2, new load angle delta2, and new induced voltage
Ef2. So, how did I get the new operating point,
we have increase the field excitation. So, Ef is bound to rise. If Ef is bound to rise, I know Ef sin delta
has to remain constant no matter what is the excitation. So, Ef sin delta, tip of Ef must lie on this
horizontal red line. So, depending upon by what percentage you
have increase the field current this length will increase that cut it, it will be Ef2
and this is Ef2 V is constant. So, this must be this length dotted line red
must be your jI2 x2 if the jI2 xs is known then Ia2 must be lagging this a voltage across
that xs is this length therefore, the current must lag this length by ninety degree Ia2
and tip of Ia2 must lie on this green line. So, this will be the new thing. Therefore, what you will see this if you have
connected an ammeter here as you increase the excitation armature current will rise. It will become more understood, ok. So, I will now neatly I will draw this once
again. So, that you do not miss any point. The point is Ia cos theta has to remain constant. Ef sin delta has to remain constant as you
vary field current of the synchronous machine synchronous machine. I have taken generator case, so in the generator
case it will be then like this, suppose initial operating point is V and it is delivering
a current of Ia1 at a loading power factor of theta1 then Ef1 will be V plus jIax. So, this angle is 90 degree, j Ia1 xs and
you get your Ef1 and this angle is delta1 initial operating point. Suppose, I say I have reduced the excitation,
what is going to happen? If If is reduced, Ef reduces, that is length
of Ef will reduce, but I know Ef sin delta is constant. So, draw a line like this. So, tip of Ef must lie on this line. So, if you have reduced If your Ef2 will be
suppose here and your new delta will be this. Now, if this is your Ef2 then V plus Ef2 j
Ia2 xs this will be j, this point is crucial this will be your j Ia2 xs, that is the, this
is the thing you should always, j xs and this is V, and this is this is the direction of
Ia2. So, I know the voltage across the inductance
that gets fixed and I will claim then Ia2 should be here 90 degree lagging Ia2. This is the voltage across the inductance
90 degree lagging Ia2. Now the thing is direction of Ia2 I have got,
it is like this, is not? So, Ia2 must be here, I mean parallel to this
Ia2, I mean something like this. But tip of Ia2 has to be on this line. So, calculate j I2 x2, then and decide what
is the deduction of Ia2 draw it and this is Ia cos theta constant line. So, it will be here. Therefore if you have reduced the field current
of this generator this is suppose the field current If and I have reduced it. So, new excitation voltage will be Ef2 new
armature current will be Ia2 new power factor angle will be this and it will be leading. And when you go leading you see the length
of Ef has become less than length of V indicating that it is under excitation. So, if you have under excited the machine
for a synchronous generator operation, power factor may become leading. In fact, you can find out an excitation such
that the generator will deliver the same amount of power at unity power factor even if you
can find out such a excitation, got the point? So, I will just draw that case and then I
will stop here, I will do this for motor as well. Therefore this is the thing suppose this is
V, very quickly, V and this is the initial operating point Ia1 theta1 and V plus Ia1
x1 is your Ef1, this angle is 90 degree. This is Ef1 and I want to see that the, and
this is Ef sin delta business, this is delta1 and this is theta1 it is doing fine. Now, if I can select, if you it is lagging
power factor, reduce the excitation speed such that here is your Ef2, I reduce the excitation
sorry this is Ef1 deduce the excitation such that it becomes perpendicular V to this one
then where is your Ia1 this becomes j Ia2 xs and this will be the direction of I2 in
phase with V and this will be Ia2. So, by simply by concluding the field current
I will be able to operate the generator either at lagging power factor or leading or at unity
power factor. We will continue with this.
