lec7.pdf

Power System Dynamics

Prof. M. L. Kothari

Department of Electrical Engineering

Indian Institute of Technology, Delhi

Lecture - 07

Transient Stability Analysis of a Multi Machine System

(Refer Slide Time: 00:55)

Friends, today we will study the transient stability analysis of a multi machine system

using a classical approach.


We shall address to some of the important issues related to the transient stability analysis

of a multi machine system, I shall discuss the node elimination by matrix algebra,

classical model of the system using that model we will develop the power output equations of generators in the multi machine system, then we will write down the swing

equations in a multi machine system and then for transient stability analysis whatsoever

preliminary computations are required to be made then we will discuss those.



One of the requirements while pursuing the multi machine stability problem is the

elimination of some nodes in the system where there is no current injection that is if you

have a system right and at some nodes in the system if there is no injection then these

nodes can be eliminated okay.

Now to understand how these nodes can be eliminated we start with the general system

and the mathematical model of the system is expressed as I is equal to Y bus into V

where I is the current vector and V is the voltage vector that is I are the currents which

are injected at the buses and V are the bus voltages okay, Y bus is a matrix which is a

square matrix okay and this is symmetric matrix you all know about it.

Now to understand how the nodes can be eliminated what we will do is that we will

partition this equation that is the current vector I can be partitioned into two sub vectors

similarly, the voltage vector will be appropriately partition and then Y bus matrix is also

partitioned, after partitioning this matrices we can write down IA IX this is the I vector and

IA is a vector, IX is also a vector.


It can be partitioned as the sub matrix Y matrix is partitioned as K, L, L transpose M,

VA, VX that is we are when we are doing the partitioning right of the system then when I

partition this current vector then the Y bus matrix is also appropriately partitioned and

bus voltage that is also appropriately partitioned. After this partitioning we can write

down the equations, the two basic equation in the form, IA equal to K times VA plus L

times VX and IX can be written as L transpose VA plus M times VX that is this is simply

obtained by multiplying the uh this two matrices on the right hand side of this equation

okay therefore, I get IA is equal to K times VA plus L times VX, IX is equal to L transpose

VA plus M times VX.


Now in case in case this IX is a null vector in the sense all elements are 0, okay in that

case what we can do is that we can write down VX in terms of VA and other elements of

this equation, what we do is that you subtract both the sides of the equation by L

transpose VA IX is anyway 0 and multiply by M inverse. Now when you do this two

operations that is you are subtracting both sides by this L transpose VA right and then you

multiply by M inverse, you will get an equation in the form minus M inverse L transpose

VA it is wrong it is L transpose VA.


Okay this is equal to VX, now we have obtained the expression or equation for VX in

terms of other non-quantities okay. So that what we do is that we substitute the

expression for VX in equation 7.3 right. So that this VX is eliminated and the moment we

eliminate this VX you can write down the equation in the form IA is equal to K times VA

minus LM inverse L times VA. Okay and therefore this equation can be put in the form IA

equal to K minus LM inverse L transpose into VA right that is if I write the equation in

this form, IA equal to K minus LM inverse L transpose this is L transpose into VA.


Here, I and V have the same dimensions and this is a matrix is going to be square matrix

have the same dimensions as that of the K and therefore now I can say that the my model

becomes IA equal to Y bus reduced Y bus reduced into VA. Therefore, this reduced Y bus

this you can say this is Y bus reduced okay and this reduced Y bus can be written in the

form as K minus LM inverse L transpose. Therefore, here the network which initially had

certain number of nodes and those nodes where there was no current injection okay.

These nodes have been eliminated and therefore this is now a reduced order model and

there are no nodes where there is no injection. Actually I will just illustrate this with the

help of a example

Now, let us consider that we have a simple system is the third order system let us say

okay and the Y bus matrix as original system is minus 3.330, 3.330 minus 7.5, 2.5, 3.333

and in, I can put 3 here also 2.5 and minus 10.833. This is let us say is Y bus of a system

which has 3 nodes okay. Now let us assume that the node number 3 is 1where there is no

current injection and I want to eliminate that node, what we do is that we partition this

matrix because there are 3 nodes 1 and 2 and 3, okay therefore we partition it in the

fashion.


So that K becomes now matrix minus 3.3300 minus 7.5 okay. L is a vector in this

particular case 3.33, 2.5, j will be there in all the cases okay and M is minus j times

10.833. Okay now you substitute the value of KL and M in the expression, K minus L M

inverse L transpose.


Now we make the substitution here this is a column column matrix this is simply a scalar

quantity this is a row where when you multiply this 2 it will be a 2 by 2 matrix and then

subtract this form the elements of K matrix and the resultant matrix which comes out to

be j times 2.308, 0.769, 0.769 minus 6.923.

Now I have illustrated through a very a you know small example but this is applicable to

any system which may have ah any number of nodes to be eliminated and this simplifies

the process to a great extent okay. Now our next step will be in solving the multi machine

problem, we have to develop the classical model.


When I say it is a they are using the classical approach we will have to develop the

model. Now in the classical approach for transient stability analysis we make all those

assumptions which are made earlier. These 5 assumptions which we made earlier I will

just reiterate them, first was that mechanical input power remains constant, second was

all asynchronous powers or dumping torques were assumed to be 0, third assumption was

asynchronous generator can be represented by a constant voltage behind direct axis

transient reactance, fourth was that synchronous power can be computed from the steady

state solution of network and the fifth was the phase angle of the voltage behind transient

reactance coincides with the total position with respect to the synchronously rotating

reference plane.

These were the 5 assumptions which we had made, the next assumption which we will

make here is all loads because in any power system we will have loads at various buses

all loads may be considered as shunt impedances to ground with values determined with

values determined by conditions prevailing immediately prior to the transient conditions

that is in the system we have loads connected and we will convert the loads loads by

equivalent shunt impedances okay and while calculating the shunt impedances, we will

consider the condition prevailing prior to the transient.

Okay whatsoever are the condition before the occurrence of transient or steady state

condition, okay using that condition we will find out the equivalent load admittances or

load impedances. Okay this is very important next assumption which is made

Now for analyzing the transient stability or for any a stability problem we always require

the initial operating condition or the steady state operating condition prior to the

occurrence of disturbance.


Now for a multi machine system the steady state pre fault conditions for the for the

system are computed using load flow, there is a mistake by mistake I did not write. The

steady state pre fault conditions for the system are computed using a load flow.

Therefore, a load flow for the system becomes a pre condition for starting the solution of

any stability problem in a multi machine system. Okay now the model for the system.


Okay now the model of the system actually with this all the assumptions which we have

made it looks like this. The transmission network is shown here, to this transmission

network we have generators connected at certain nodes. We can, so that these are the

points at which the generators are connected. There are certain points in the network

where the loads are connected. Okay now these loads are represented by constant

impedances connected to the reference bus.

Now this generators are replaced by ah voltage behind transient reactance direct axis

transient reactance. Now here, here we will not neglect the armature resistance of the

synchronous machine okay. Once we account for the armature resistance then we are

representing the generator by a impedance r1 plus j times xd1 prime r2 plus j times xd2

prime and we find out the voltage behind this impedance, if you neglect this resistance

then it becomes simply the voltage behind direct axis transient reactance but here when

we do the transient stability analysis there is no need to ignore the resistance and we can

account for it.

Once you account for this then these are the voltages behind the transient reactance.

Therefore, now if you see here the network which is enclosed in this rectangle is a

passive network it does not have any source now, okay all the sources have been taken

out that is at this nodes we have connected the voltage sources and all the loads have

been replaced by constant impedances and therefore what we exactly do when solving

this problem is that initially you will perform a load flow analysis okay and once you

perform the load flow analysis you will find out the voltages at all these nodes and all the

nodes where the all the nodes where the loads are connected.

Okay then using using this information the loads will be converted into equivalent

impedances and we will replace the generators by equivalent voltage in series with

impedance. Okay then then this becomes my model which is going to be used for

transient stability analysis.

Now in this case what we have done is that I have shown n generators okay and there

may be ah r nodes where the loads are connected. Okay therefore the system may have

originally n plus r nodes. Now when you perform this load flow analysis you will, you

will find out that the Y bus matrix considering only the line parameters. Okay therefore

now when you perform the transient stability analysis the Y bus matrix will be modified

and it will be obtained considering the machine impedances and this load impedances.

Okay once you obtain this Y matrix accounting for this machine impedances and the load

impedances, the load buses can be eliminated okay and therefore you will be left with

only these n buses where the generators are connected okay all the load buses you

eliminated and the the buses where this generators are connected they are also eliminated

because there is no injection here also but ultimately you will need or you will get a

model which will have the n nodes, where the generators are connected and these are the

constant voltages that is the voltages of constant magnitude their phase angles will very

during the dynamic conditions okay.


Now for this network we can write down the equations in the from I equal to Y bus E,

where I is the current vector are or current injected at all the n buses and Y bus is the

admittance matrix obtained after eliminating all the nodes. Okay therefore using this we

shall be in a position to obtain the equations for for electrical power output from all the

machines because we know actually that where we have to obtain or we have to analyze

the transient stability analysis we need the expressions for electrical power output from

the generators, okay. Therefore using this model which is having only the n nodes where

the generators or internal voltage of the generators are connected and this exercise is a

important exercise for which I have already the explained the methodology.


Now here, we will represent the uh diagonal elements of the Y bus matrix which are

called Yii will be equal to magnitude Yii angle theta ii, which will have real and

imaginary parts Gii plus j times Bii. Similarly, the diagonal elements will be shown as Yij

magnitude Yij angle theta ij, Gij plus j times Bij and these voltages, Ei will be written as Ei

angle delta i.



Now here I am not writing this prime right because once we understand that this is the

voltage behind the direct axis transient reactance and the state of resistance included.

Okay then there because writing every time E prime is little what inconvenient okay

therefore we consider that these are the internal voltages and the magnitudes of these

voltages will remain constant. Now we can find out the electrical power output Phi by

this basic equation that is the real part of Ei Ii star real part of Ei Ii star this is very

important.

Now here you look like this that this is the bus at which the voltage source is connected

say Ei this is i th bus, this Ii is the current which is injected by this voltage source. Okay

and we know actually the the complex power injected is equal to Pi Si equal to Pi plus j

times Qi and this can be obtained by the formula formula always and in this particular

system Ei into Ii conjugate, this star stands for conjugate. Okay and therefore when I take

the real part of this Ei Ii star that becomes the electrical power output from the I th

generator.


Now you we make the we substitute the expression for Ii, the expression for Ii is we have

the vector Ii can be obtained as Yi1 Yi2 Yin multiplied by this vector E1 up to that is you

have this is this is the I th row of the Y bus matrix and when you multiply this with the

voltage vector, bus voltage vector you will get actually this Ii as Yi1 plus Yi2 E2 plus Yin En that is we can write down this Ii as summation of Yij Ej where j varies from 1 to n, this

is the way we will be writing the expression for Ii .

Now when you make this substitution here for the expressions for Ii that is you have to

the conjugate of this current and then perform simplifications you will get the expression

Pei that is electrical power output of I th machine as Ei square Gii plus j equal to 1 to n Ei

Ej cosine of theta ij minus delta i plus delta j and j will not take the value equal to i and

this equation is valid for i varying from 1 to n for all the machines that is you put i equal

to 1, 2, 3, 4. You will find that you will be in a position to get the expression for electrical

power.

This is a very important expression and for n machine system we will get n such

equations. Now you can if you just look this look at this expressions carefully, you will

find actually that this depends upon the magnitudes of all the voltages which are known

to us, they remain constant during the transient stability analysis that is during the

transient period. These are the Gii there is one term which is missing here actually just

please correct it. It is Yij it is there will be Ei Ej, Yij will also be there.

Okay cosine of theta ij minus delta i plus delta j therefore this, Yij is also known to us

only thing which is not known will be the quantity which vary the delta i delta ij

therefore, what happens is that these angles will be computed by using the numerical

technique and at the end of each step of computation these angles will be known.

Therefore, you substitute the value of these angles which are computed when you solve

the swing equations right and you get the expression for Pei at each step. Now, once we

know the expressions for electrical power output of the machines for all the n machines

then we can write down the n swing equations okay.


The swing equations for each machine will be 2 times Hi upon omega s, d omega i by dt

equal to Pmi, that is the mechanical input to i th machine minus the electrical output of the

i th machine again here. Yij term is missing. Okay that is what we have done is that this

expression is same as I have computed this is the expression for electrical power output

from i th machine okay then this is the main equation and we write down this d delta i by

dt equal to omega I, which is equal to the actual speed of the rotor minus synchronous

speed because d delta i by dt is the access speed of the rotor right and this it is access over

synchronous speed. Therefore, I am writing d delta i by dt equal to omega i which is

equal to d theta i by dt minus omega s.

Okay therefore what we see here is that we have written 2 equations, 2 first order

equations for one machine and therefore, if you have n machines in the system then you

will get total 2 n such first order equations. Now we need the information of what is the

mechanical power input to the machines Pmi and we assume that this Pmi is constant over

the transient period that is mechanical power input to all the machines are assumed to be

constant right. Therefore, there is a necessity to find out what will be the mechanical

power input initially okay.

Now how to do it because under steady state conditions the electrical power output is

equal to the mechanical power input and the system is in steady state condition that is the

derivative terms on the right left hand side of the equations are all 0 okay.


Therefore, with this condition we can write down the expression for mechanical power

input as Pmio equal to Ei square Giio, I am putting o here means this is the this is the

element of the Y bus matrix before the occurrence of transient again here you have Yij

okay, that is the complete expression of wherever actually ah angles were delta i, I put

delta io, delta j, delta jo, now theta ijo here you will find actually that when the system is

subjected to disturbance.

Okay therefore during fault conditions the Y bus matrix of the system will be different,

during post fault condition it is again going to be different right and therefore we will

have 3 different ah Y bus matrices one pre fault condition, one during fault condition,

third post fault condition. Now here to compute the mechanical power input initial

mechanical power input Pmio right. We use the conditions which are prevailing the in the

system before the occurrence of disturbance and this is, I am denoting by putting a further

subscript o, okay.


Now the equations which we have written were 2n equations, these 2n equations can be

written in the general form as X dot equal to a function a non-linear function X, Xo, t

where, X is a vector of dimensions 2n into 1X is a vector of dimension 2n into 1 where, n

is the number of machines and these are all first order nonlinear differential equations or

we can say a set of 2n first order couple non-linear differential equations they are all

couple.

Now X transpose is actually the X is the state vector the state variables are omega 1, delta

1, omega 2, delta 2, omega n, delta n right and let me just summarize here, how we obtain

the mathematical model of the system. The first step is therefore a given system you

perform a load flow study and find out the initial steady state operating condition okay,

once you have done this the next step will be to replace all the loads by constant

impedances.

Then next step will be to knowing the information which you have obtained in the steady

state ah condition you replace all the generators by constant voltage behind a impedance r

plus j times Xd prime okay. Once you have done this thing, you can illuminate all those

nodes where they are no current injections or no source is connected okay using the

matrix algebra.

Once you have done this thing you can find out the set of non-linear coupled differential

equations for the machine okay. Once these equations are there our task is to solve these

equations using numerical technique and plot the graph relating delta versus time that is

the, we have to plot swing curves for all the machines and then we examine the swing

curves to see whether the system is stable or not okay this is what is step.


Now here before we do all the transient steady analysis one has to perform some

preliminary computations. The preliminary preliminary computations to be computed

done are, all system data are converted to a common base this is a pre important

requirement that when you assemble the system data you will find actually that data are

given on the basis of there name plate ratings. Suppose you have a generator of 5hundred

MVA, okay then the generator parameters will be given on the basis of name plate rating

of the machine.

Okay but once you are consoling a system you have to convert all the data on a common

base and the common base generally chosen is hundred MVA, okay 100 MVA is

common because it looks to be little convenient for computations okay. Otherwise there

is no such hard and fast rule you choose the common base and when you say the one is

MVA base another is the voltage voltage in a particular circuit you will choose and then

you will compute the per unit values of all the system parameters on common base.

The second step is the loads are converted into equivalent impedances or admittances the

needed for this is obtained from the load flow study, okay this is what I have already told

you. Now how do we convert actually the loads into constant admittances these steps are

also important this we can write down here, that at suppose the load is represented as PL plus j times QL at any bus as load is a PL plus j times QL.


This load can be written as VL into IL star because this is the complex complex load right

and with this can be written as the bus voltage , of the load bus, the voltage of the load

bus into the conjugate of the current drawn by the load IL star. Okay using this expression

we will compute the value of elements of the admittance which is going to replace the

load by a constant admittance. Now the admittance will be written as YL plus j times YL equal to GL plus j times BL, okay.


This can be simply obtained that in this expression in this expression you can substitute

the value of IL, we know that the current drawn by an admittance is equal to admittance

into the voltage and admittance load admittance multiplied by the voltage that will be the

current and admittance is given by this formula, load voltage is VL therefore IL star is

going to be VL star into YL star IL star is VL star into therefore VL into VL star becomes

VL square and IL then YL will become GL minus times j times BL, YL star.

So that we can write down this load as VL square into GL minus j times BL right and

therefore, using this expression equation 7.8 right, we can say that the load admittance is

equal to PL minus VL square minus j times QL by VL square that is you can say that the ah

real part is PL divided by VL square and reactive part is BL divided QL minus VL square

okay and the proper sign is to be consider.

Therefore, this is one important step and this voltage is VL which are substituting in this

expression are the obtained by load flow analysis. Okay another computation which is

required to perform is the voltage behind the transient reactance. Okay now here you can

use this expression Ei prime equal to Vti plus j times Xdi prime into Ii.


Now here I have not considered the armature resistance right but you can modify this by

putting here instead of j times Xdi prime, we can put here Ri plus j times Xdi prime okay

and since the all information is are known, current is known, terminal voltage is known,

you can find out the while finding out this internal voltage we will use this expression,

the current injected by the generator is Pi minus j times Qi divided by Vti star.

This is very standard equation which can be used to compute the value of current and this

current is required in the equation 7.10 to compute the internal voltage. The internal

voltage which you compute, the internal voltage which you compute right will have its

magnitude and phase angle right and this phase angle which you get become the initial

value of deltas, okay and the magnitudes will remain constant during the transient.

Now once you have done these preliminary calculations okay we obtain using this

information, the swing equations. Now you know that we require swing equations under

3 different conditions, one is the pre fault operating condition, second is the during fault

and third is the post fault and therefore when you solve the swing equations suppose I

start solving the equations at time t equal to 0, okay now if the perturbation or

disturbances occur at time t equal to 0 right then immediately from that initial steady state

condition I will be using the swing equations, I will be solving the swing equations which

are applicable during the during the fault on period.

Then once I know that okay after this much time the fault is cleared right it means the

movement I reach that particular point i will change over from the fault on period swing

equations to post fault swing equations and continue to solve it right. Now one should

know actually that suppose the fault occur in the system, how to obtain the ah the swing

equations which will be applicable during fault conditions, pre fault conditions we have

already obtained.

Now the faults can be ah symmetrical faults or unsymmetrical faults. Now in case it is a

symmetrical fault then the problem is bit simple, if it is unsymmetrical then also it is not

difficult because we have already seen how to account for the unsymmetrical fault. Now

if it is suppose I take fault at a particular bus, let us say that fault occurs at a particular

bus then voltage of that bus will collapse it will become 0. Therefore, the occurrence of

fault at the particular bus is accounted by considering or by is by setting that voltage of

that bus to 0.

Okay then once the voltage of that bus becomes 0 we have to modify the Y bus matrix

which will be applicable to this period, this particular fault on condition. This

modification are not difficult one can, once you see that a particular bus is grounded okay

looking into that condition you can modify the Y bus matrix. Now simplest way is that if

you have Y bus matrix for the pre fault condition then if suppose at a particular bus let us

say ah bus number five bus number five if suppose there is a 3-phase short circuit then

the row and column corresponding to the bus number 5 can be deleted from the pre fault

Y bus matrix and the remaining, the remaining Y bus matrix right will become the Y bus

matrix applicable to post fault condition.

In fact actually suppose one particular bus is grounded or since it is sorted it gets get

grounded it they the number of buses which remain now will become less by one right

therefore Y bus matrix also reduce in dimension. Similarly, once you consider the post

fault condition right then, you can obtain the Y bus matrix during the post fault condition

because under post fault condition what happens one line may be out therefore for the

remaining system you have to find out the Y bus matrix.

Now this Y bus matrix will not be assembled fresh one can be modify the existing Y bus

matrix by modifying the elements of Y bus matrix some elements will be affected, for

example actually if suppose there is a Y bus matrix and if the line connecting the node

one and two that is tripped.

Okay then what is that is going to affect the or which elements of the Y bus matrix will

be affected that is Y11 will be affected because that line is out, Y22 will be affected and

Y12 will be affected others will not be affected, right. Therefore once we know that

particular type of line element has been removed right you can easily find out the Y bus

matrix applicable during the post fault period and once we know the Y bus matrix which

are applicable during post fault, during fault and during pre-fault conditions that you

already known because our equations that is the equations for electrical power output are

written using the reduced Y bus matrix therefore this reduction process has to be done for

all the 3 conditions pre-fault fault on condition and post fault condition.


Now I will suggest you to refer to these problems or these two examples which are given

in here, one example is given in the book written by P. M. Anderson and Fouad, power

system control and stability volume one, example 2.6 and 2.7, these two examples this is

example which is very very commonly used for illustrating the transient stability

analysis. It is a 9 bus system, it has total 3 generators, 9 buses okay and in these two

examples all the step by step computations have been carried out to illustrate the steps

involved in solving the transient stability problem.

One more simple example which is given in the ah basic book on power system analysis

that is elements of power system analysis by Stevenson, everybody is aware of this book

Stevenson here he has solved 5 bus system. Okay and in this 5bus system he has

illustrated all the steps which are required to obtain the complete transient stability

solution. Now once the model is obtained and you solve the swing equations, couple a set

of the coupled swing equations, you will get the swing curves.

Okay the swing curves will plot actually as delta one delta two delta three as function of

time. Now to examine whether the system remains stable or not we have to see the

relative variation of the swing curves that is not the absolute values of delta 1, delta 2

delta n that is important what is the relative value that is more important.

Suppose the two machines may accelerate simultaneously okay but the angles may

remain very close to each other or angle difference may remain very close then system is

not losing synchronism therefore, when you examine the stability of the multi machine

system we have to examine the ah the plots of relative ah angles that is we plot actually

delta 12 that is delta 1 minus delta 2.

You plot delta 1 minus delta three suppose there is a three machine system they are only

a 3angles involved and once this curves are plotted in case actually the curves or we can

say delta 12 reaches the maximum value and decreases it shows the stability condition,

in case delta 12 increases continuously it shows unstable condition. Okay with this let me

conclude what we have done today.

We have an we have given or we have studied the basic steps involved in analyzing the

transient stability of a multi machine system. Here we have used a classical model for

analyzing the stability I have mentioned that how do we obtain the required required

models for pre-fault, fault on and post fault operating conditions. Okay and the swing

equations which we obtained for the system are solved by using numerical techniques

and the stability of the system is understood by examining the swing curves which we

plot and particularly the is the swing curves are relating the relative relative angles of the

machines, okay thank you very much.

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