Highlight text to annotate itX
We had commenced the simulation of a single machine connected to a voltage source through
a reactor. We had in fact just started that. The main point to be noted in that was the
behavior of the a V r. Now, a V r of course, is the automatic voltage regulator. The simulation
studies of interest are the step response of the a V r to a step change in the reference
voltage of the a V r. Also, we need to see in the simulation how with changing load conditions
whether the a V r is able to regulate the voltage at the terminal of a synchronous machine
connected to a voltage source or an infinite bus.
Of course, the important difference between the simulation we are carrying out now and
what we carried out few lectures back was that, the synchronous machine is not directly
connected to a stiff voltage source. But, it is connected via a very, a kind of toy
model of a transmission line. And the transmission line in fact, is a model just like a reactor.
Now, if you look at what we are trying to do; the synchronous machine was connected
via a transmission line to another stiff voltage source. This is the infinite bus, this is
a transmission line, this is a synchronous generator, this is the terminal voltage of
this machine was being regulated. So, we took the feedback of the voltage at this point
and fed it voltage regulator of this kind. Simple proportional kind of voltage regulator
and the excitation system was assumed to be a static one.
So, though the model of it with normalized quantity this simple a plants with a gain
of 1 and the limits of the static excitation system of course, were taken. This is fed
to the field winding and of course, the feedback here is the magnitude of the voltage which
is root of V d square plus V q square. We assume that there is no measurement delay
here. We could if we wished model this delay by a first order transfer function. But, we
will not do that right now. The simulation was is being carried out with the following
First, the synchronous machine is synchronized to this voltage source. It is in fact, it
is kind of bumpless synchronization. We have talked about synchronization of a synchronous
machine before. So, we will not talk of it again right now. Initially the voltage that
the power of course, initially is 0 at the time of synchronization and the open circuit
voltage is 1 per unit and the machine is running at practical at the rated speed. Now, in order
to create 1 per unit at the terminals of the machine, you need to have E f d is equal to
1. So, the output of the excitation system should be should give a field voltage which
results in the terminal voltage being at its rated value at rated speed.
So, E f d is one. Now, to create this E f d of course, E f d is within the limits because
the terminal voltage being 1 itself. So, 7 point 0 into 1 and minus 7 point itis within
the limit so it is not getting clip. And to create this E f d you have got an automatic
voltage regulator, V ref and this is V. V which is nothing but, root of V d square plus
V q square. Now, since this is equal to 1, this is one, the gain is K A. So it follows
that the value of this error here is 1.0 divided by K A. So, V ref in fact is nothing but,
V plus e f T by K A. So one thing you should notice this is the proportional controller
there is some steady small steady state error. In fact this steady state error is less if
K is large. So, V ref and V are not exactly equal if you
have got a proportional controller. You would need to a have a small steady state error
in order to get a nonzero E f d. So, this is something which you should keep in mind.
So V ref can be back calculated. So what is the V ref which results in V being equal to
1.0 under open circuit conditions? Well, it is equal to this. So, V ref is slightly greater
than what is required to that is what what V we require. So, if this is 1 this will be
E f d which is none 1 by say if K A is 2; 100 which is a typical gain per unit by per
unit then, your V ref is equal to 1 plus 1 upon 200 so that is equal to 1.0 0 5.
So, you see that V ref and V are almost the same provided K A is large which is indeed
the case. K is normally quite large. So, in our studies right now we will assume K is
to 200 T A, the time constant here is we will take it as 0.0 2. It is small value and once
we synchronize the machine there is no power flow. So, actually nothing nothing changes
because you have just had a bump less synchronization. The conditions on the machine do not change.
However if I increase the mechanical power out input for example, initially of course,
before synchronization the machine is simple rolling. So, if I increase the mechanical
power after synchronization, say I give a step change. It is not possible to give a
step change to the mechanical power. But, let us just take it as a mathematical idealization.
If I give a step change in the torque, electrical torque you will find at the power flow through
this increase and presumably it remain in synchronism.
So, the speed of this and this will be the same. These two machines or this this machine
and this voltage source is going to be same. That is, in spite of the change in power the
synchronous machine tends to remain in synchronism with this bus. But, it is important change.
As soon as some current starts flowing that is some power starts flowing, the voltage
here tends to drop. So, if you did not have a automatic voltage regulator what would happens
is this voltage would it would drop and E f d would remain the same. But, with a voltage
regulator, as soon V drops V ref is of course, the same right now. If V ref drops E f d increased.
So, this voltage here is regulated and E f d is not a constant.
So, an automatic voltage regulator saves as the trouble of trying to you know, manually
trying to change E f d very time we change the mechanical power. It does it automatically
by satisfying this regulation function. So, that is for first test signal we will give
to this machine. The second test signal will give is give a step change in V ref keeping
torque constant. So, what will do is increase the reference voltage of this a V r. That
is I am going to increase this voltage. Now, you know that if I increase this voltage will
the mechanical power of the electrical power be affected? No. What instead state will get
affected is, since this voltage is increased the reactive power output of the machine will
change. So, this q here will change. So, this is one important thing which we will see now
in our simulation. So, today’s lecture we continue this simulation of automatic voltage
regulator. In fact, it is system which contains an automatic voltage regulator.
So, what will do is first synchronize the machine it is a bump less, absolutely smooth
transfer smoothed interconnection. Thereafter that is of course, than time T is equal to
0. Thereafter, we increase the mechanical torque of the machine. We increase the mechanical
torque of the machine and thereafter we increase the reference voltage of the A V r to 1.0
5. Now, I will, since is the absolutely smooth synchronization we have assume that the rated
speed of the machine, the machine is running at exactly the speed which is equal to the
infinite bus and the line to line r m open circuited voltage is equal to the voltage
which appears right at the point whether circuit breaker as to be closed.
So, you can see that there is going to be a bump less transfer of the magnitude of the
voltage, is the phase angle also is 0 at a time of synchronization. So, what will have
is the bump less synchronization first at time T is equal to 0. You will not see any
transient because it is absolutely a smooth synchronization.
We will do the simulation right away. yeah So, I have already written the program. So,
this time I will not show you the program. It is a simulating the system using Euler
method. Euler method can be used only if I remove the stiffness out of the system remove
the stiffness in the system. So, I have neglected d psi stator transient
d psi d by d t d i d by d t and so on. So, this is something we have explained before
in our earlier class. So, what I will do is concentrate more on the results this time.
So, let us just see what happens to this speed. So, I will plot. yeah
So, what you see here is, a almost a bump less transfer here, a synchronization. Here
at this point I give a step change in the mechanical power. If you increase the mechanical
power in this system, the infinite, the machine does not loss synchronism in fact the step
change in torque is 0.5 per unit from 0 to 0.5 per unit. A machine remains in synchronism
in the sense that, it the speed of the synchronous machine although it deviates in transients;
there is this swing which you see, a low frequency swing it tends to come back to synchronism.
The speed tends to come back to synchronism at ten seconds. I have given again another
step in the voltage reference of the synchronous machine.
So, you see that in a both cases the machine tends to remain in synchronism. That is, the
mechanical the speed of the machine equals to the frequency of the infinite bus. Now,
you can also see various other parameters for example, delta.
If you look at delta of the machine this is of course, in radiant. The machine is synchronize
when delta is equal to 0 and speed is equal to 0. So, you have got this bump less synchronization
right in the beginning. And as soon as this give step change in torque, the phase angle
of the synchronous machine, the delta of the synchronous machine rather it changes. You
see this swing which is also observable in a speed, it is also observable here.
So, you see this swing this is one particular mode which is associated with a electro mechanical
variables; delta and omega especially. Also, when you change V ref, you find at delta actually
slightly decreases. Now why is that so? You are have not changed the mechanical torque
at this point. You have change delta to 1.0 5 from 1.0 0 5. So, if you make this step
change in V ref; delta reduces because E f d as changed. Once you change V ref, we are
effectively changing E f d also. That is why delta changes. You again see a swing and which
settles down to a steady state value. So the delta settles down to a steady state value.
So, if you actually plot the mechanical power or rather the electrical power output of the
generator which is e V i d plus E q i q we will find that d drive. yeah
So, what you see here is a of course, a mechanical power changes to 0.5. The mechanical sorry
the electrical power follows the mechanical power. You should remember, you have got an
infinite bus. The synchronous machine is connected to an infinite bus. So, if the speed remains
finally, the machine remains in synchronism and mechanical power become equals to the
electrical power. So you see that the mechanical, electrical power output P is equal to the
mechanical power. It eventually becomes 0.5. If I change the V ref of the machine, if I
change the V ref of the machine the reference voltage of the A V r of the machine; you find
that the output electrical power P does not change. did not suppressing This is not surprising.
After all, by changing the reference voltage of the machine you are not changing the mechanical
power. The mechanical power becomes equal to the electrical power in steady state if
the mechanical power does not change, the electrical power does not do so.
So, this is an interesting point here. What if I plot Q? Q is the reactive power output
of the machine okay.
If you look at the reactive power output of the machine; if I, if of course, the machine
is initially synchronized. There is no real or reactive power output at the point of synchronization.
But, if I am change the mechanical power; the electrical power also changes, E f d also
changes and you will find at the generator in fact supplies the bit of reactive power.
There is some reactive power output of the generator. If I change V ref on the other
hand, if I change the V ref, what you are doing is not changing much the real power
does not change. But, the reactive power changes quite substantially. So, by changing the reference
voltage of the AVR; we infect in fact changing the reactive power output of the generator.
and of course, the terminal voltage also will change under these circumstances.
So, let us just look at the terminal voltage of the machine. Remember that the first disturbance
or first step change is that of the synchronous mechanical torque at ten seconds you have
given step change in the reference voltage of the a V r from 1.0 0 5 to 1.0 5. So, let
us just plot the terminal voltage. I think it is called V g. yeah So, if you look at
the terminal voltage, well I plotted something wrong. I will just just movement a moment.
We will just look at the variable which is corresponding to it. It just split slipped
of my mind. It is Vgen. So you just see what vgen is. It is a terminal voltage of the synchronous
You see that the terminal voltage of the synchronous machine is initially one. When I increased,
if I increased the real power output of the machine by changing the mechanical power at
0.5 seconds; we notice that the voltage slightly drops. Now, why does the voltage drop? That
is the important point which you should pointed ponder upon. We will try to address it here.
What happens actually is that when you have got synchronous machine under open circuit
conditions, there is a certain field voltage which will produce 1 per unit at the terminals.
Once you load the machine, on the other hand that is you increase the mechanical power
in and therefore, the electrical power. What happens is that, the terminal voltage tends
to drop. Now, because the terminal voltage tends to drop, E f d is increased. So, if
E f d is increased, V ref is the same, so let us just do the, you know kind of Cos Cause
effect kind of analysis. Mechanical power increases, electrical power also tends to
increase because of that. If electrical power increases, there will be from current 0, there
is some current. So, current increases. Because current increases, you will find at the terminal
voltage tends to drop. The terminal voltage tending to drop will force this automatic
voltage regulator to increase E f d. So, if E f d is increased, it also means error has
changed. V ref we have not touched right now. We are just giving a step change in mechanical
power. So, what happens is effectively if the error as in steady state if E f d required
is more, we will have to drop more. This because you are using a propositional controller and
steady state error is nonzero. So, to get more E f d you will have the steady state
error will have to be more. So, V slightly decreases.
So, once you load the machine, since you got proportional type automatic voltage regulator,
you find that the voltage slightly decreases. Of course, if I give step change in V ref
itself; so if you look at what I am done here now. Now, E f d is certain value. Now, V ref
is increased. If V ref is increased of course, you will find that the error would have increased.
If error increased, E f d will increase. As a result of that, V will increase. But, remember
that V ref and V are not going to be exactly equal. That is why even though if I made V
ref is equal to point for 1.0 5 at ten seconds, you find at actually the voltage settles down
to around 1.0 4 2 5. So, this is the kind of interesting load look at the simulation.
It should correlate well with our steady state analysis. If you look at of course, this something
we have discussed even before. You have got to you should keep in I an eye on this swing.
Now, if you look at this swing which is there, which is seen in almost all the quantities,
it is a very prominent mode. You will find at it is oscillation of roughly 2 yards hertz,
slightly less than 2 yards hertz.
Now, will also look at E f d. We have not looked at E f d. Let see how E f d behaves.
So, look at E f d E f d. Once you give step change in V ref sorry mechanical power, E
f d as increased from one which is the value required to get 1 per unit at the terminals
of the generator under open circuit conditions or no load conditions. If I increase the mechanical
power, if I want to get a 1 per unit at the terminals, I will have to increase E f d and
that is what exactly the voltage regulator as done. It has increased E f d from around
1 to approximately 1.3 or 1.2. Now, if I give another step, if I give the step change in
V ref, you find at E f d again in increases. Of course, that is true. Because if I want
to get higher voltage, I will have to increase my E f d. Now, remember that once I, the moment
I give step you see that the change in E f d quite large. Why is that so?
That is because this is extremely high gain system. Remember I told you, in order to get
a very good response of the voltage regulator we do use very high gain in our automatic
voltage regulator. This is because in order to opposite offset the relatively slow response
of the field winding, what we do is we give a big push initially to the field voltage
so that to get the field winding moving in the sense that we change the field current.
So, that is why you provide for a very large, when you are designing excitation system and
a V r which provide for large gains and a fairly large range in which E f d can vary.
Normally E f d did not vary more than, under steady state condition more than 3 per unit
1 2 3 per unit is all what is required. But, normally the sealing is captivated more than
7 or 7 or around 7 time the terminal voltage. So, the sealing voltage is around 7 per unit
plus or minus 7 per unit. So of course, if I give sudden change like a step change in
V ref the field gets forced. But, of course, it cannot exceed 7. So, it is kind of gets
clipped at near about 7. So, what you see here is in the simulation is that the E f
d is try to increase to a very high value in order to get a very good response from
the sluggish field winding but, it is clipped at 7.
So you see this almost flat top here. You see this and of course, once the field winding
gets going there is no need to have such a large E f d. And E f d finally, settles down
to a more comfortable value which is roughly around 1.7. So, this is an interesting point.
So, this is the, showing the effective of the limit at the clipper which is there. You
can also have a situation if I give a step change in the reverse direction of course,
E f d will decrease. So, I will not do that simulation but, you can well imagine what
will happen and in fact you can try out this doing this particular simulation. It is very,
it is an interesting simulation. Now, let me tell you something which is very, very
interesting and important.
Suppose, I if you recall I have written down this program and disturbance is which I given
are at t greater than 5 seconds. I give a step change in mechanical power at t greater
than ten seconds, V ref is given a step. So what I will do is, I will not give step change
in V ref. I will keep it as it is. But, at ten seconds I will give an additional step
to T m. So, I will put mechanical power is 1 per unit which is at rated speed equal to
the rated power output of the machine. So, what I am done is I am trying to load the
Just one small point which I need to rewrite reiterate here is that, normally one cannot
give such step changes in mechanical torque. This is an, just an idealization. It just
is a kind of toy simulation in that sense. In later classes we shall see about modeling
of the prime mover system themselves and you will note that you cannot give such step changes
normally. So, let us just try to redo this simulation.
yeah In the mean time I will just close this graphics graphical window which is of the
previous simulation. So, if you look at the simulation now.
Now if you look at for example, just look at speed. There is an interesting problem
which seems to have reason risen now is that, the speed does not seems to settle down. In
fact the machine to off for all practical purposes seems to have loss lost synchronism.
You see that speed is increasing with time and if you plot for example, delta also, we
will of course, close this and redo this.
Yeah So you see delta, it does not seem to be reaching a steady state. It seems to be
growing with time. So, these oscillations are growing with time. So, why is that happening?
And in that is an interesting point which you should ponder on. I mean it is, it is
going on increasing with time. In fact, if I simulate for a longer time,
this is a simulation for 20 seconds. If I make it as 25 seconds simulation, let us see
what happens. Simulate for a slightly longer time and redo this. Close this.
The delta seems to have kind of, is not settling down. It in fact if you look at what is happening
is that it is increasing.
So, this looks as if this particular operating point or equilibrium point which it should
settle down to does not seem to be stable. It seems to be small signal unstable. So that
is an interesting thing. So, the feedback system which we’ve got seems to, seems to
be suggesting seems to be creating a situation for certain operating points which leads to
a oscillation which does not seem to be dyeing down.
So, if you, in fact, may be if I give a make vref also increase at the same time, let us
see what happens. Let us give you another simulation. So, what we saw in the previous
simulation was that, there is an oscillation in delta. But, it is kind of increasing with
time. That is not you know, usually we expected the delta decrease and you reach steady state.
So, by giving step changes extra we were expecting the thing to reach steady state. But, it appears
that there are situations in which steady state is not reached.
In fact, this seems to be even worst. Your angle seems to have increased even more. So,
or perhaps its moral more or less is the same but, you see that this is also unstable. So,
actually what we are effectively coming to a situation where the equilibrium point under
our study appears to be unstable. So, what we should expect is that, at this operating
point when do a linearize analysis of this system we should get Eigen values with positive
real parts. So, that is what we need to correlate in our analysis now. So, let me just again
point out what I want to say. We are having a situation, when we try to change the power
from 0.5 to 1; we see that the system is not settling down to the new equilibrium point.
Now, what it what this means is that, why is it so actually why should the behavior
depend on the equilibrium point? Well, one of the important point you should note is
that, this is a non-linear system. So, when we do a linearise analysis we will be linearlizing
the system around each operating point. So, when we do the linearise analysis around
different operating points our Eigen value is going to b,e are going to be different.
So, while step changes to 0.5, T m is equal to 0.5 or v ref is equal to 1.0 5 which you
did in the previous simulation were all stable. You see that in trying to come close to T
m is equal to 1.0, there seems to be a problem. So, the operating point to which we are trying
to get to, appears to be not stable not small signal stable itself. So, what we need to
do here is try to understand this instability of the swings. You see this low frequency
swing if you except to be stable and to die out is growing with time because of the feedback
feedback control system of the a V r. So, if we kind of try to a linearise analysis,
how do we go ahead? Now, linearise analysis is something we’ve done before. I will just
quickly retrace out the steps. So, what we have in our equations of a synchronous machine.
If you look at the equations of a synchronous machine, these are in fact non-linear. So,
we’ve done the linearization of such systems before. So, what we need to do is instead
of having a set of system x dot is equal to f of x; what we need to do really is do a
linearise analysis around an equilibrium point of this system.
So, if we’ve written down our equations in this form, we should now see we can do
linearise, we have done a simulation of this system. This is a full non-linear system.
We can try to understand the stability of the equilibrium point by doing a linearise
analysis in this form. Remember, that when doing a linearise analysis you have to take
we have to linearise it around an equilibrium point.
So, there is one important thing is, first thing is we have to linearise. Linearise in
the sense use tailor series expansion of the non-linear functions which appear here. That
is called linearization. And evaluate those you know after truncation of the higher order
terms. You evaluate this at the equilibrium point. So, second thing is you’ve to take
out the equilibrium points. So, linearise analysis would require first of all you to
take out the equilibrium point and the second part is you linearise around an equilibrium
point. Is that okay? So, this is what you need to do. How do you compute the equilibrium
point? So, in fact long ago, I mean, I am sure it is not too long ago, you must have
done the load flow analysis. A load flow analysis is in fact is a kind of an equilibrium analysis
of the system. But of course, that is a in some sense the starting point of computing
any equilibrium. In our case we’ve got a very simple system in fact computing the equilibrium
points is very, very easy.
So let us just go ahead and see how you can do it. Suppose, I want to find out this, the
equilibrium values of the states corresponding to this scenario. You have got this transmission
line which has got a plain reactants X. This is your infinite bus which we take as 1 angle
0. That is I have already defined what the infinite bus voltage is r V n V b n and V
c n in the previous class. Let us say the terminal voltage of the machine is 1 per unit.
it is near about 1 per unit and T m is equal to 1. If T m is equal to 1 and this is at
rated speed then it also means that electrical power output of the machine, we are of course,
neglecting the loses in the resistance of the synchronous generator they are really
very small so compared to this speed. So, we will assume that electrical power output
at the terminals of the machine is same as in per unit as T m because, the speed is equal
to the rated speed. We will assume that the frequency of the infinite bus is also equal
to the rated frequency. So, if P is the output of the machine, this is 1 per unit what is
the face angle of the terminal voltage of the synchronous machine? That can be easily
found out. You know that for a system with reactants x voltage magnitude 1 on both sides,
the phase angle of the voltage here theta is such that P e is equal to 1 per unit is
nothing but, voltage magnitudes at both ends divided by x.
This is of course, a multiplication sign and this is, it looks the same. So, we will just
make this larger here into sin of theta. So theta, you can calculate theta from this.
So, what, let me just remind you what we are doing. We are computing the equilibrium point
corresponding to this operating condition and thereafter doing a linearise analysis.
So, theta is sign inverse P e. This is at equilibrium divided by 1 into 1 into x. So,
this is theta. We also can compute the current flow through this. Once you’ve got theta
you can get the current. So, the current phasor can also be obtained. So once you’ve got
the current phasor. So, current is equal to V angle theta minus E. E angle 0 E is 1 per
unit e is the infinite bus voltage divided by j into x. So, this is effectively the current
I. Once you get the current I, we are effectively can get the, what the current looks like.
So, what we’ve effectively done is, if you actually look at, if you look at what we have
in another way what we’ve done effectively is, set the left hand side d by d t is equal
And thereof, therefore, we’ve got i d i q. If you look at the equations, which we
saw on the screen you can have a look at them again. 0 minus X and 0 is equal to minus of
V d V q minus E d E q. Now, this is effectively the same as saying i q. You can just write
this in very compact form i q plus i d is equal to V q plus j V d minus E q plus j E
d divided by j of x. You can just expand this and show that this is same as this. So, what
let me tell you what we know that we know what what we effectively get is this particularly
equation. That is why it appears like this.
So, the point here is that you’ve got 1 angle 0. These are the voltages of the infinite
bus E a n E b n and E c n.
So, what does it mean that e a n is equal to root 2 by 3 sin omega naught T and E b
n and E c n are of course, balance counter parts of it. Your terminal voltage is 1 angle
theta. What does it mean? It means V a n is equal to root 2 by 3 sin omega naught T plus
theta and V b n is of course, root 2 by 3 sin omega not T plus theta minus 2 pi by 3
and V c n is minus 4 pi by 3. So, the point is that if I know theta I can tell you what V a n, V
b n and V c n look like. As a result of it, we can get what V d and V q look like. That
is the most important thing which you should get out of this.
See if you know E a n, I told you that E d is equal to minus E times E is 1 in this particular
case sin delta and E q this is plus E Cos delta.
So, V d V q is therefore, if V is equal to 1 point 0 angle of theta, what it follows
is that V d is equal to and V q is equal to 1.0 yeah sin of delta minus delta and 1.0 Cos of theta
minus delta. So, we’ve got V d and V q and E q and E d. How did you get V d and V q?
We computed theta. Theta is obtained from the initial condition. So, in effect we’ve
obtained what V d and V q should be. If I know V d and V q, I can get i q plus j i d
or equivalently i d and i q separately from this. So, if I have got i d and i q V d V
q and E d E q, the next step is to try to back calculate what is the value of delta,
omega and all the other fluxes.
Now, delta omega is equal to omega naught in steady state. So, if we are talking of
an equilibrium point, omega will be equal to omega naught. So, we are taking out an
equilibrium, we are doing an equilibrium analysis, we are back calculating the equilibrium conditions.
Under equilibrium conditions psi d is equal to x d into i d plus E f d. How does one get
that? Well, what we have to do effectively is set d f dot d psi F by dot by d t is equal
to 0 and d psi H by d t is equal to 0. So, if you set this equal to 0 you will get psi
f and psi H in terms of psi d. After that, replace it in the algebraic equations which
relates psi d with i d. You will effectively get this particularly equation. Similarly,
in steady state this is not true in general. This is true only in steady state.
Please remember, psi q is equal to x q i q. Similarly, you will get psi q, so under steady
state conditions this is what you get. Now, we know that in steady state condition d psi
d by d t is equal to 0. So, what you’ll get is 0 is equal to omega into psi q under
equilibrium condition omega is equals to omega naught is equals to base value minus omega
b are a i d minus omega b into V d. Similarly, you will have sorry this should be minus omega
plus omega psi d. This is got by putting d psi by d t and d psi q by d t equals to 0
which is to under steady state conditions. This is minus I am sorry this should be minus
i d minus omega b V q. So what happens is, if you assumes is resistance
is very small of course, you have the value there is no harm. If you assume that it is
very small, what you will get is V d is equals to yeah, V d is equal to omega b and omega
the same. We will assume that the speed of the infinite bus is the same has the base
speed. In such a case V d is equal to minus i q and V q is equal to psi d. So, this is
one important thing which we get.
So, what we have is V d is equals to minus x q i q and what we have here is V q is equals
to x d i d plus E f d. So, if you have this particular equation;
what a simple thing we can do is, we have got V q plus j V d is equal to yeah E f d
plus is. We can write this j times x d yeah into i q plus j i d minus yeah. So, this
should be minus here. E f d minus j into x d into i q plus j i d. So, what I have done
is just added up this equation then written them in a particular compact fashion. So,
what we have here is, so what we have here is j and this is plus j x d x.
So, what we have is V q plus j V d under steady state is equal to E f d minus j x d i q plus
j i d minus j x q i q plus j x d i q. An alternative way of writing this is E f d minus j x q into
i q plus j i d. And instead of, so instead of x d you write it as x q and as a result
of which the additional term you will get is.
So, you can write it in this fashion also. The basic cracks crux of the matter is that,
if I know V q plus j V d, so I will just write this down again. We will have V q plus j V
d is equals to E f d minus j x q plus x d minus x q into i d. So, just yeah so what
we get from this is that, if I know this and I know this, so if I know this and I know
this, I can get this. But, remember I do not know delta. Remember I do not know delta,
so I cannot actually find out what V d and V q is even if I know what theta is. Similarly,
I cannot get E d and E q.
So what I do is very interesting thing, E f d I am multiplying by e raise to j delta.
So, I will get V q plus j V d into e raise to j delta. On both sides I just multiple
e raise to j delta. I multiplied by e raise to j delta. Now, the question is, do I know
this? The answer is yes because I know theta.
So, this you can just try to prove that if I know V d and V q, sorry if I know that V
d and V q have this form V q plus j V d will e into e raise to j delta will simple give
me what? 1.0 yeah and theta is something I know.
So, what we have essentially is that I know this, I also know this. So, if I know when
I know this I mean, I know this complete quantity here. Snd I know the complete quantity here.
As a result of which I know this complete quantity. Now, this is a real number. So if
I take out the magnitude of this this something I know and this is something I know, I can
take its magnitude an angle and equated to this.
As a result of which I will be able to get E f d and I will also be able to get what
delta is. So, that the thing which is important to notice V q plus j V d e raise to j delta
is nothing but, 1 angle that theta. Similarly, actually if you look at that infinite bus
voltage also, its very easy to show that this will be 1 angle 0 as per our initial situation
which we have talked of. Similarly, i q plus j i d into e raise to j delta is nothing but,
this minus this divided by j x. So, we know this complete term and this complete
term. We know the value of x q. As a result of which we can back cab back compute the
equilibrium value of delta so you you can directly get what the equilibrium value of
Once you got the equilibrium value of delta, the next step is get i q plus j i d. Remember
we know what i q plus j i d into e raise to j delta is. What is that equal to? 1 point
0 angle theta minus 1 angle 0 divided by j x. So, if I know delta I can actually find
out what i q and i d are. Once I find i d and i q, the next step is from i d you know
this value, the magnitude of this by equating these two compute E f d.
So we can compute E f d. Now, we got E f d initial value, delta. And once you know E
f d and delta in fact you can compute all the equilibrium conditions. So, what we have
is let me just recapsulate what we have done. It may sounded a bit complicated but, it isn’t
that as complicated as its sounds. We will just recapsulate recapitulate quickly.
If E a n is this, it means of course, that E d and E q are like this, which also means
E q plus j E d e raise to j delta is nothing but, 1.0 plus j 0.0. Similarly, V a n is this.
So what you get essentially is theta. Theta is something you get from the condition that,
1 per unit power is flowing through a reactance of X. The terminal voltage here is 1 point
angle 0 and terminal voltage here is angle theta. So, this theta is obtained that way.
So, once you got theta the next step is of course, getting the equilibrium values of
delta extra etc. Remember that, once have got theta you know this. You do not know what
delta is but, interestingly we q plus j V d into e raise to j delta is nothing but,
a function only of theta. So, if you know theta you know V q plus j V d into e raise
to j delta. So, this is how you get V q plus j V d into e raise to j delta. i q plus j
i d into e raise to j delta is nothing but, this so we get the values of i d and we can
get because of this, from this particular equation. Because we know this and this and
the value of this parameter x q, we get this value that is the magnitude and the angle
can be obtained. So, we know the magnitude of this is we equated to the magnitude of
the left hand side, the right hand side and left hand side magnitude equate. So, we know
what E f d plus x d plus x q into i d is. Thereafter we also know from the angle of
this, what delta is. Once we get this delta, we can compute, we know what this is i q plus
j i d into e raise to delta j delta we know. Once we know delta, we can find out what i
q and i d are so in that fashion. We can get the equilibrium values of all the states corresponding
to this situation, whether terminal voltage of this is 1, the terminal voltage of the
infinite bus is also one, the voltage of the infinite bus is also one. T e is one also
we can back calculate all the values of the states.
So that you get this particular situation once if computing, back calculated all the
values of the states. You can actually do a linearise analysis by plugging in the equilibrium
values wherever necessary in the linearized state space equations. Once you do that, one
can get the Eigen values. There is too it is little time for us to actually compute
the Eigen values. And I show you everything that are the important points corresponding
to these Eigen values. So, we will do that in the next class.