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Power System Dynamics
Prof. M. L. Kothari
Department of Electrical Engineering
Indian Institute of Technology, Delhi
Lecture - 19
Small Signal Stability of a Single Machine Infinite Bus System (Contd….)
Today, we continue our study on small signal stability of a single machine infinite bus system.
(Refer Slide Time: 01:17)
Now today we will include the effect of synchronous machine field circuit dynamics, in the small
perturbation model. Earlier the the transfer function model which we developed earlier in this
transfer function model. We assumed the constant field flux model, okay or the voltage behind
direct axis transient reactance were assumed to be constant that is the classical model now we go
one step further and include the dynamics of the synchronous synchronous machine field
circuits.
Okay now when we developed the synchronous machine model, okay at that time I emphasis
actually that if we neglect the amortisseurs then the synchronous generator can be represented by
a first order model that is the synchronous machine equations there was only one differential
equation. The differential equation was corresponding to the field flux linkages and the equation
when it is written written considering time in seconds is put as P psi fd equal to omega naught
into efd minus Rfd ifd.
(Refer Slide Time: 01:36)
Now this term omega naught comes because we are now considering the time in seconds rather
than in per unit. Okay when we developed this equation in per unit it was simply d by dt of psi fd
equal to efd minus Rfd ifd. Okay now here we will do once that transformation because this this
equation is written considering this efd and ifd in reciprocal per unit system. Okay we replace this
efd by this term using this equation Efd equal to Lad by Rfd efd that is where we derived the
synchronous machine model.
(Refer Slide Time: 03:47)
We introduced this term efd and basically this introduction was to to link the non reciprocal per
unit system to reciprocal per unit system and here when I replace this efd by the field flux, field
voltage that is Efd is the field voltage or output of the exciter right. The equation becomes p psi fd
equal to omega naught Rfd divided by Ladu Efd minus omega naught Rfd ifd right. Therefore, this is
the additional equation that has to be considered while accounting for the field flux dynamics or
we can say the field circuit dynamics.
(Refer Slide Time: 04:57)
The complete model considering the rotor dynamics and the field flux dynamics can be written
in this form that is the these two first two equations represent the dynamics of the synchronous
machine rotor, okay these two equations are well known to us. Now to this these two equations
when I add this third equation that is d by dt of psi fd equal to omega naught Rfd Ladu into Efd
minus omega naught Rfd ifd. Okay now these three equations equations right describe the the
complete behavior of the synchronous machine under dynamic conditions including the field flux
dynamics while while writing these equations we have made one assumption and that
assumption is that the the amortisseurs are ignored , another assumption which we have made is
that the transformer the transformer voltage terms are also neglected and the speed deviation that
is the omega r term which appears in the stator equations, stator voltage equations that omega r is
also assumed to be equal to 1 okay.
Now when we look at these equations, we will clearly see that the state variables are delta omega
r delta and psi fd that is these are the 3state variables. Okay but when you see on the right hand
side of these equations we have a term T that is the air gap torque right. Now this air gap torque
is neither a state variable nor it is a input to the system similarly, this ifd that is the field current is
not a state variable or nor it is a input to the system because when we write the equation this Tm
is the input to the system, okay Efd is the input to the system that is the mechanical torque applied
to the system and the field voltage applied.
Therefore these are the two input quantities and these are three state variables while this Te and
ifd these are not the state variables as well as they are not the input quantities and therefore there
is a necessity to to express these two quantities Te and ifd in terms of state variables. Okay the
moment we we succeed in expressing the air gap torque and the field current in terms of state
variables we can write down the complete model of the system because these equations are
concerned in these equations right. We have these quantities which are to be expressed in terms
of state variables.
(Refer Slide Time: 08:24)
Now to express this quantities in terms of state variables we we make use of the the synchronous
machine equations and we will also we make use of the network equations because here we are
considering a machine connected to infinite bus right and therefore the transmission line right,
we represent the network that we will be in a position to write down the network equation. We
will be in position to write down the synchronous machine equations and since this these two
components are connected right we will be in a position to correlate them. Okay that is we will
find actually the network constant equation okay.
Now these are the three basic equations which represent the d axis field flux linkage q axis field
flux linkage and the field winding flux linkage psi fd, okay psi fd is a state variable while this psi
d and psi q are not the state variable but we can see here that the psi d can be expressed in terms
of these inductances id id that is the direct axis component of the synchronous machine stator
current and ifd. Now these equations are known to you further actually this Lffd this is the this is
the self inductance of field winding this will be expressed as Lad plus Lfd, Lad is the mutual
inductance and in the reciprocal per unit system right this mutual inductances are same for direct
axis stator winding, the field winding. Okay these are the same we have made equations same
then we will make use of this equation that is the air gap torque is equal to psi d iq minus psi q id.
Okay therefore with this we will be in a position to simplify the model, now these steps are
clearly given in the text book which is prescribed for your course okay.
(Refer Slide Time: 10:41)
To start with what we did write that this psi d this the direct axis flux linkage can be expressed as
minus Ll id plus Lads into minus id plus ifd that is if we make use of this equation right then what
we do here is that this Lad Lad these two terms are combined and further further, we will make
use of this the saturated value of the mutual inductance instead of using the unsaturated value,
we will be using the saturated value and the saturated value and unsaturated value are related by
the saturation coefficient okay and therefore the equations now which I am writing here for psi d
and psi q this Lad is replaced by Lads that is we are using the saturated value. Okay therefore this
this term we can see this term that is Lads into minus id plus ifd this term represent the flux linkage
the inductance into current and this we will denote as psi ad psi ad.
Similarly for psi q, we will represent this as minus Ll into iq plus psi aq. Now here we are using
these terms which are normally called the intermediate variables which will ultimately be
eliminated. Okay similarly this equation that is the field flux linkage equation psi fd right. Now
for for this Lffd you substitute Lad plus Lfd. Okay and then these Lad terms will be combined the
terms which have the coefficient Lad.
(Refer Slide Time: 12:58)
You combine and we can write the equation for field flux linkage in this form psi ad plus Lfd ifd
right. Now when we write this field flux linkage in terms of psi ad which is which is also
appearing in our other two equation that is psi d equal to minus Ll id plus psi ad. Okay therefore
the first step here is that now I express this ifd that is the field current in terms of psi fd which is
my state variable psi ad and Lfd Lfd is a parameter of the machine is a machine parameter, okay
synchronous machine parameter it is a known quantity.
(Refer Slide Time: 14:15)
Therefore, this is this is one step where we are now expressing the field current in terms of the
one of the state variables and psi ad and Lf that is this term is additional term which again we
have to eliminate. Now these steps you can perform yourself that is the psi ad was defined as
minus Lads id plus Lads ifd that is this term when we put in the expanded form right it was minus
Lads into id plus Lads into ifd that is it was written as Lads into minus id plus ifd. Okay this can be
simplified and now expressed in terms of id that is the direct axis current id is with this derivation
one can perform and obtain the expression for psi ad in terms of direct axis current id, direct axis
component of the synchronous machine current id and psi fd therefore, if you see this equation
that this psi ad has been expressed in terms of the terminal current a component of the terminal
current and the psi fd which is the state variable okay.
(Refer Slide Time: 15:38)
Similarly, we can express psi aq in terms of iq that is what we are doing here is that these two
terms which are the intermediate variables psi ad and psi aq have now been expressed in terms of
the state variable and the the direct axis and quadrature axis component of stator current okay.
Now these two steps are these steps are very important then we will be further making use of the
stator circuit equations now because the stator circuit is connected to the infinite bus through a
transmission line that is it is a machine infinite bus system right and we have the standard
equation that is the ed equal to minus psi q into Ra id, eq equal to psi d minus Ra iq these are the
stator circuit equations these are the algebraic equation, a stator circuit equations. Okay where
this psi d psi d and psi q can be written in terms of currents that is iq and there is a little mistake
here id okay.
(Refer Slide Time: 16:07)
Now in these equations we see here very interesting thing that psi q and and psi d are expressed
in terms of the our intermediate variables psi ad and psi aq and iq and id. Okay now what we do is
we write down the network equations, okay and then the once we write down the network
equation we will be in a position to write we will be a position to eliminate this ed and eq.
(Refer Slide Time: 17:47)
Okay now to do this thing you, we write down the equation for ed and eq which where earlier
written in this form that is this psi q and psi d terms will be replaced by these terms. Okay so that
ed and eq are written in terms of psi ad and psi and aq. Okay therefore the direct axis and
quadrature axis component of the stator voltage is now written in terms of the currents and these
intermediate variables. Okay Ra Ll these are the circuit or the synchronous machine constants.
Now let us see what are the what are the network constant equations. Okay now to understand
this network constant equations what we have to do is we we make use of our reference axis
right because in a machine infinite bus system right. The infinite bus voltage can be represented
as a reference and the rotor position can be described in terms of position of the q axis right.
(Refer Slide Time: 19:33)
The position of the rotor with respect to the infinite bus voltage phasor can be expressed in terms
of q axis that is here if I represent this as my infinite bus voltage okay and this is the terminal
voltage of the machine synchronous machine right, if this is my q axis this is ed that is the d axis
that is this is your q axis. Okay this all known to you okay and this is your d axis okay. Then the
definition of this power angle delta is the angle by which q axis leads EB the delta is positive
when q axis leads the infinite bus voltage by an angle delta while the internal angle is the angle
by which by which q axis leads terminal voltage the delta i is the internal angle and delta is the
power angle, total power angle okay.
Now here you can write down the components of infinite bus voltage that is if you resolve this
infinite bus voltage EB into 2 components one along q axis another along d axis then the
component along q axis is EBq which is going to be EB cos delta, component along d axis will be
EBd equal to EB sin delta okay and when you express this terminal voltage and infinite bus
voltage as phasor we can write down terminal voltage as ed plus j times eq and infinite bus
voltage as EBd plus j times EBq, okay this is very straight forward nothing to be explain.
(Refer Slide Time: 21:12)
(Refer Slide Time: 21:52)
Now let us write down the ah network constant equation this your synchronous machine,
terminal voltage is Et phaser, the line impedance is RE plus j times XE and here is our infinite bus
the phase the voltage here is EB phasor right and It the current here is written and the phaser
formula right. Therefore our network constant equation is is Et is equal to EB plus RE plus j times
XE into It this is the network constant equation a very simple equation which relates the terminal
voltage of the synchronous generator to the infinite bus voltage in terms of the line impedence
and current okay. Now what we will do here is that we will replace now Et by ed plus j times Eq
EB will be replaced by EBd plus j times Ebq, It will be replaced by Id plus j times Iq and then we
will, we will equate real part with the real part and imaginary part with the imaginary part to get
two two algebraic equations, okay in terms of the real variables.
(Refer Slide Time: 23:53)
(Refer Slide Time: 24:18)
Now this exercise has been done here. This is our network constant equation okay, I am just
introducing here you substitute in this equation ah replace this Et It and EB by the components
and you perform simple algebraic simplifications and we will get these two equations that is ed as
ed equal to EBd plus id into RE minus Iq into XE eq equal to EBq plus id into XE plus Iq into RE
okay this is very there is nothing very difficult, okay anybody can do it.
(Refer Slide Time: 25:07)
Now once we have done this, we can we can now express this current Id and Iq, Id and Iq in terms
of our state variables that is you make use of see this this is one these are the two equations
where ed and eq are expressed in terms of these our intermediate variables id and iq. Okay now
this ed and eq will be replaced by ed and eq will be replaced by this ed and eq right you replace
these by the expressions given in the right hand side of these equations right.
Therefore, once you do this thing you will get equation right where you have only id and iq and
they will be written in terms of these terms these are intermediate variables psi aq psi ad and other
machine parameters and ultimately when you do all the simplifications algebraic simplifications,
you can write down the expressions for id, id in terms of the state variable this is our state
variable this is also a state variable and other synchronous machine constants that is all the all the
previous steps which we have made are basically to come to a stage where we can replace we
can express the obtain the expression for the id and iq in terms of these state variables okay.
Similarly, the expression for iq is given like this which you can derive yourself you have to sit
and derive these equations. The definitions of the terms which appear in those equations are also
expressed here. Okay and further when you write these terms in a per unit system, per unit
system the line inductance or synchronous machine inductances are equal to the synchronous
machine reactances and line reactance. Okay therefore we do not make any difference between
the inductance and reactance because in per unit system our omega o is 1 okay.
(Refer Slide Time: 27:09)
(Refer Slide Time: 27:29)
Once we have obtained the expression for id and iq in terms of the state variables okay the model
is complete it is a it becomes out to be non-linear model if you look it very carefully carefully
then you will find actually that the psi ad and psi aq terms are expressed in terms of these
currents. Okay and this currents are now expressed in terms of state variables we have also seen
actually that the the air gap torque is expressed in terms of in terms of the flux linkages and
currents right therefore ultimately what happens is that we are in a position to express the the
torque term which comes actually in the equation for the synchronous machine that is the swing
equation in terms of the state variables, torque is expressed in terms of the state variables right.
Now these are the non-linear equations we are not done any linearization. Now this we can
obtain now the linear equations considering small perturbations the first step which is considered
for linearization is to obtain expression for incremental changes in id and iq to obtain the
expression for incremental changes in id and iq this id which is function of these state variables.
(Refer Slide Time: 30:14)
Okay you can obtain the expression for incremental changes in terms of the state variables that is
delta id can be written as m one times delta delta plus m 2 times delta psi fd, these m1 and m2 right
these are the quantities which depend upon the synchronous machine parameters, the line
parameters, the initial operating angle of the system that is delta o that is operating condition of
system that is that is this these two equations which are the incremental equations for currents
direct axis current and quadrature axis current are now written in terms of the state variables and
these coefficients are function of system operating condition and system parameters which
includes the synchronous machine parameters and the line parameters, okay.
Now here actually I have derived the expressions for m1, n1, m2, n2 and it can be clearly seen that
these constants are function of function of the machine parameters and initial operating condition
that is delta is delta naught here sin delta o where you find cos delta o term all these line
reactances the infinite bus voltage okay.
(Refer Slide Time: 31:11)
(Refer Slide Time: 31:46)
Now to complete our ah hum model we also obtain now the incremental changes in these
intermediate state variable that is delta psi ad and delta psi aq in terms of our state variables for
example delta psi ad can be written in terms of the incremental changes in the state variables that
is delta psi fd and delta delta delta psi aq can be written in terms of again in terms of these state
variables and the coefficients of these terms are again function of the machine parameters,
system parameters and operating condition.
(Refer Slide Time: 32:57)
We had written the expression for ifd, if you just see this equation you see the equation here this
equation ifd was express as psi fd minus psi ad Lfd. Okay now since psi ad is expressed in terms of
the state variables I can now express incremental change in the field current in terms of the state
variables.
(Refer Slide Time: 33:16)
The expression comes out to be as delta ifd is equal to this coefficient into delta psi fd and this
term into delta delta that is the incremental change in field current is expressed in terms of the
state variables incremental changes in the intermediate variables that is psi ad and psi aq are
expressed in terms of the these variables because psi ad and psi aq have to be ultimately
eliminated. Now we come to our main equation that is the expression for for air gap torque.
(Refer Slide Time: 34:02)
(Refer Slide Time: 34:38)
We have to write down the incremental change in air gap torque that is delta Te, delta Te as we
have seen earlier delta Te is written as I am sorry not do this later it is confusing first we will
write down del Te not delta Te, Te is written as psi d iq minus psi q id. Okay now this psi d right if
you consider the leakage inductance then to obtain the expression for air gap torque this psi d has
to be replaced by psi ads iq minus psi aqs id we will be considering the saturated value of these flux
linkages right and difference between this and this term is the leakage flux which is not
contributing towards the air gap torque. Okay now we want to linearize this equation to linearize
this equation the simplest way will be you write down this Te as Teo plus delta Te express this
term as psi adso plus delta psi ads iq will be expressed as iqo plus delta iq minus this is also perturb
psi aqso plus delta psi aq right into id is expressed as ido id naught plus delta id.
Now under steady state condition Teo is equal to psi adso iqo minus psi aqo so into ido okay and we
will neglect the the terms terms which are the product of two incremental terms like when I
multiply this you know expressions or simplify I will come across the term incremental change
in the psi ed psi ad multiplied with incremental change in current this two incremental changes
when they are multiplied will give you very small quantity which can neglected. Now when you
do this exercise the ultimately you can write down delta Te equal to psi ado delta iq plus psi iqo
delta psi ad minus psi aqo delta id minus ido delta psi aq, here actually the we are considering the
saturated values s is not written just for the simplification okay.
Now we have the expressions for delta id and delta iq, we have also obtained the expression for
incremental changes in the air gap flux linkages psi ad and psi aq right. Therefore, when you
when you substitute these incremental quantities that is delta iq delta psi ad delta id delta psi aq
right and these are the initial values. Okay we will we can write down the expression for
incremental change in air gap torque as K1 time delta delta plus K2 times delta psi fd that is the
the all the derivations which we have done right from beginning right were basically aimed to
obtain an expression for change in air gap torque in terms of state variables and we get the
expression for incremental change in air gap torque this delta Te right equal to K1 time delta delta
plus K2 time delta psi fd.
(Refer Slide Time: 40:02)
Okay this is very a important step that what we find here is that here that the the delta Te that is
change in air gap torque depends upon the change in rotor angle that is the power angle and
change in field flux linkages that is delta psi fd okay. The the model which I had derived earlier
where we assume assumed the field flux linkage is constant right.
(Refer Slide Time: 40:48)
(Refer Slide Time: 42:00)
The the incremental change in torque that is air gap torque was simply equal to Ks into delta
delta while now here the incremental change depends upon the change in flux linkages in the
field winding, field circuit and delta delta. Okay therefore now instead of representing the
coefficient of delta delta as Ks, I am representing this as a K1. Okay this is now going to be the
difference in what we are writing. The expressions for these constants can now be written in the
desired form which can be computed from the knowledge of initial operating condition
synchronous machine parameters and line parameters.
Okay these expressions have been derived because they are only the algebraic simplifications
which you have to do. Okay just, now let us look again at our mathematical model. This was the
starting point for our further derivations. Okay now this equation right has been linearized in the
previous case we have linearized it where mechanic change in mechanical torque we have put as
a delta Tm change in electrical order is delta Te and this is our damping torque term right and
therefore this equation which is going to be a non-linear differential equation the moment you
linearize this this becomes a linear differential equation right.
Now we have now to linearize this equation that is the field circuit equation, we have not done
the linearization of the equation this can be done very easily because here we will consider the
change in field voltage as delta Efd. Okay and change in field current as delta ifd, okay but this
delta ifd see this is not a state variable is replaced by the incremental changes in terms of the
other state variables.
(Refer Slide Time: 44:39)
Let us now examine the complete model of the system, these 3equations represents the complete
model of the system, first two equations represents the the dynamics of the synchronous machine
rotor or rotor dynamics. The third equation represents the the dynamics of the synchronous
machine field winding. This these two equations were linearized earlier and we obtain a simple
stats pay simple model in the transfer function form. This third equation is now linialize
linearized and when we linearize this equation we will have the incremental change in field flux
linkage, incremental change in field winding voltage and the incremental change in field current.
We have earlier obtained the expression for delta ifd in terms of state variables.
(Refer Slide Time: 45:19)
(Refer Slide Time: 45:49)
Now when we when we substitute the expression for delta ifd in this equation the the linearized
equation can be now added to the first two state space equations and the linear equations appear
in the form delta psi fd dot equal to a32 delta omega r a3 plus a33 delta I am sorry this has
correction here. The delta psi fd dot appears as a32 delta delta plus a33 delta psi fd plus b32 delta
Efd okay. This can be written in this form here as delta psi fd dot equal to a32 delta delta plus a33
delta psi fd plus b32 delta Efd. Now we we take the Laplace transform of this equation and obtain
the transfer function model to represent this equation in the transfer function form. Once we take
the Laplace transform the equation can be written in the in terms of the Laplace transform
variables that is delta psi fd is equal to K3 divided by 1 plus pT3 into delta Efd minus K4 delta
delta. Now p is equal to s, okay now we will incorporate the transfer function relating, we will
incorporate the the transfer function relating the change in field winding flux linkage is to change
in rotor angle and field winding voltage.
(Refer Slide Time: 46:38)
Earlier, we have developed the transfer function model relating the variables variable
considering the first two equations in the of the system. Now to this model we will add the
equation relating the field winding variables, now while deriving this model we can do small
simplification that is this portion of the model that is this portion of the model can be simplified
by using the simple block diagram simplification and this portion can be represented by an
equivalent model of the form 1 upon 2 times Hs plus Kd where output of this will be delta omega
r and input to this will be delta Tm minus delta Te.
With this simplification, with this simplification incorporated the complete transfer function
model incorporating the field circuit dynamics is shown here this portion of the model is the one
which was developed earlier. Now since we have to incorporate the affect of field circuit
dynamics since, delta Te is equal to K1 delta delta plus K2 delta psi fd right therefore at this
summation point at this summation point, we to obtain the expression for delta Te we obtain a
quantity equal to K1 into delta delta and to this we add a quantity corresponding to K2 into delta
delta delta psi fd.
(Refer Slide Time: 47:28)
(Refer Slide Time: 48:08)
Now this sum is realized by using the summation block okay therefore in this transfer function
delta Te is obtained by summing the 2 terms K1 delta delta and K2 delta psi fd. Okay then we
have the transfer function to relate delta psi fd to delta efd and delta delta. Now this is shown in
this portion of the block diagram where input to this transfer function K3 over 1 plus s times T3 is
delta Efd minus K4 delta delta right.
Now this k4 and K4, K3 and T3 are again constants and they depend upon the system operating
condition and parameters. Now therefore this is the model which accounts for the the
synchronous machine field winding dynamics. Now we will try to understand the affect of affect
of the including the synchronous machine field dynamics on damping and synchronizing torque
coefficients. To to examine this aspect let us write down, let us write down the delta Te that is the
change in change in air gap torque produced due to the change in field flux linkages only right.
(Refer Slide Time: 51:00)
Therefore delta Te due to the change field flux linkages can be related to the change in delta delta
by this expression where delta Te divided by delta delta due to delta psi fd equal to minus K3, K2
K3, K4 divided by 1 plus s times T3, okay. Now here we can examine the affect of frequency of
oscillation of the rotor, in case the in case the frequency oscillation is 0 right then we can
substitute here in this equation or in this expression s equal to j omega equal to 0. Then delta Te
can be written as minus K2, K 3, K4 delta delta that is the change in air gap torque due to change
in delta psi fd is expressed in the form given by this expression okay that is delta Te is equal to
minus K2, K3, K4 delta delta.
Further, we know that the change in change in the air gap torque due to change in delta delta
which operates directly is written as delta Te equal to K1 delta delta right. Therefore therefore we
can say here the the delta Te is written as K1 minus K2, K3, K4 into delta delta. When we consider
the affect of change in field flux linkages also now for any practical system the constant K2, K3
and K4 are positive and we can see here that this term represents the net synchronizing torque
coefficient right and when the field flux linkages were assumed to be constant the synchronizing
torque coefficient was equal to K1 right but the movement we account for the the the change in
field flux linkages due to change in rotor angle delta right. The net synchronizing torque
coefficient decreases it becomes K1 minus K2, K3, K4 therefore for steady state stability limit the
condition is K1 equal to K2, K3, K4 okay.
(Refer Slide Time: 52:16)
Now let me conclude my presentation today here that we have developed the small pertition
small perturbation model of the machine infinemet infinite bus system including the synchronous
machine field winding dynamics. This model has four constants K1, K2, K3, K4 and we have also
seen that when we account for the the synchronous machine field dynamics then the
synchronizing torque coefficient under steady state operating condition decreases that is we can
say that the demagnetizing affect has a has an adverse affect on the synchronizing torque
coefficient. Thank you!