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Friends, today we shall continue to study
about the modelling of synchronous machine.
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We have developed in the previous lectures,
the basic model for the synchronous machine
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and we found actually that if we use the model
in the original form then the coefficient
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that is the inductances were all function
of rotor angular position. In order to simplify
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the model the dqo transformation was introduced
and with this dqo transformation we found
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actually that the, the in the mathematical
model synchronous machine the all inductances
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become constant.
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Now we will further discuss now the stator
voltage equation in dqo components. We will
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develop the equations for electrical power
and torque. We shall discuss the dqO transformation
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further and its physical interpretation. Any
synchronous machine model, we always prefer
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to use per unit system of representation and
therefore, we will address how do we convert
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the synchronous machine quantities into per
unit quantities then I will also discuss an
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alternative transformation which has all through
been discussed in the literature on synchronous
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machine modelling. Let us again look at the
stator voltage equations in terms of, in terms
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of the dqO components.
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The stator voltage equations were written
in the form of ed equal to p psi d minus psi
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q p theta minus Ra id eq equal to pq p psi
q plus psi d p theta minus Ra iq and eo equal
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to p psi naught minus Ra i naught. Now here,
if you look at this first equation we can
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see here actually the means identify these
terms, the first term p psi d is the rate
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of change of flux linkages. Okay therefore
similarly, in the second equation p psi q
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and in the third equation we have the term
p psi naught these are, these terms are known
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as the the transformer voltages because they
are taking place due to the rate of change
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flux linking the circuit. Then we have next
term as psi q p theta p theta is d theta by
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dt that is the angular speed of the rotor
that is omega R, we call it now here this
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term. Similarly, this term psi d p theta these
two terms are denoted as speed voltages.
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We will further see that in in this equation
similarly, in this equation the speed voltage
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terms dominate over the transformer voltage
terms and in many simplifications we may neglect
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the transformer voltage terms as compared
to the speed voltage terms the moment you
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neglect this then this equation these equations
that is the stator voltage equations will
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become algebraic equations because if you
just look here in this 3 equation then we
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have this as a derivative term and the moment
you can neglect this derivative term right
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the remaining expression that is ed equal
to minus psi q p theta minus Ra id Ra id this
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becomes a algebraic equation right.
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Now we will develop the expression for electrical
power and torque in terms of dqO components.
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The electrical power output we denote it by
the symbol Pt here, as can be obtained as
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product, some of the products of instantaneous
voltage and instantaneous currents that is
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for phase a it is ea into ia, phase b eb into
ib, phase e ec into ic right. These are the
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instantaneous powers you add these 3instantaneous
powers you get the total instantaneous power.
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We all know actually that in case the system
is having the stator currents and stator voltage
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very sinusoidally right, then the some of
the instantaneous powers come out to be constant.
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Okay this is denoted by the symbol Pt, now
what we do is that we apply dqO transformation.
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We can apply the dqo transformation on this
voltage as well as the current currents and
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when you apply this dqo transformation then
the power output Pt can be written as 3 by
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2 times ed id plus eq iq plus 2 times eo io
that is here the terminal power is expressed
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in terms of dq components of voltage and currents
okay.
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Now we will see here actually in this that
instead of having the term e naught, i naught
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we have a term 2 times e naught, i naught
when we talk about another type of transformation
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that is called alternative transformation.
We will find actually that the the Pt will
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be expressed simply as ed into id plus eq
into iq plus eo i naught it is something like
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this the that becomes a power invariant transformation.
Here this transformation is not power invariant
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because you can see there these are 3 by 2
terms then this term is having the coefficient
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2 anyway this is resulting because of the
type of transformation.
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However when we convert these equations into
per unit quantities, the per unit quantities
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will be so chosen, so that this becomes a
simple expression and this term 3 by 2 and
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3 can be eliminated we will find this thing
okay. Now if the system is balanced we do
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not have zero sequence quantities a balance
system this e naught, i naught will be 0 and
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therefore we can write down the total power
Pt as 3 by 2 times ed id plus eq iq okay.
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Now to give further insight into the power
expression what we do is that we substitute
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the expression for ed and eq, ed and eq in
terms of the flux, flux linkages okay that
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is I substitute ed equal to p psi d minus
psi q p theta minus Ra id that is in this
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expression for this power wherever ed and
eq are present what we do is that we substitute
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the expression for ed okay.
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Now if we make this substitution and simplify
then we can write down the output power Pt
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as 3 by 2 term remains as it is id p psi d
plus iq p psi q plus 2 times io p psi naught
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that is in this expression we have this derivative
terms. Okay the second term is psi d iq minus
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psi q id into omega r that the omega r is
your d theta by dt that is the angular speed
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of the rotor minus id square plus iq square
plus 2 times io square into Ra.
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Now this 3 terms which we have in the expression
for Pt can be identified as the first term
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can be identified as rate of change of armature
magnetic energy, rate of change armature magnetic
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energy, second term is identified as power
transferred across the air gap and third term
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is armature resistance losses that is we have
this 3 terms which can be identified as
the rate of change of armature magnetic energy
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that is in these in the armature there is
some magnetic energy stored because of flux
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linkages okay and this magnetic energy which
is stored right is varying at a certain rate
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right, that magnetic energy is that are increases
or decreases it dependents upon whether p
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psi d p psi q and p psi o are positive or
negative right and from that point of view
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this is considered as the rate of change of
armature magnetic energy and therefore when
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we are looking for stability analysis right.
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The the power which is transferred across
the air gap is of concern that is in our swing
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equation which we had written right the electrical
power pe which we write that is the power
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transferred across the air gap because that
is responsible for producing the torque and
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therefore now we write down Te the electrical
torque as three by 2 psi d iq minus psi q
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id omega r divided by omega mechanical now
this torque is written in Newton meters okay
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therefore we had this expression 3 by 2 psi
d iq minus psi q id omega r, we are dividing
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by this omega mechanical.
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We know that this this term was the electromagnetic
power that is the power transferred across
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the air gap. You divide this power by speed
that will give you the torque okay and therefore,
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the expression for torque electrical torque
is given by this expression now we know the
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relationship that the omega r, the rotor speed
in electrical radiance per second divided
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by the mechanical speed in radiance per second
this ratio is equal to Pf by 2, where Pf is
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the number of filed faults.
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Okay therefore, we can write down the expression
for Te the electrical torque as 3 by 2 psi
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d iq minus psi q id Pf by 2 this is the most
important expression which we have to keep
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in our mind. Now here the next step which
we will be studying will be the how do we
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interpret the dqO quantities that this transformation
what is the physical interpretation associated
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with this transformation.
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Now we all know that ah the stator of the
synchronous generator when it carries balance
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three phase currents it produces resultant
mmf wave, this mmf wave is sinusoidally distributed
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in the space that is the stator of a 3 phase
synchronous generator produces sinusoidally
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distributed mmf wave in the air gap okay.
Now any quantity which is sinusoidally distributed
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can be resolved into 2 sign terms.
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Okay now here what is done is that this resultant
mmf which are sinusoidally distributed is
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resolved into 2 quantities one along the d
axis another along the q axis okay and therefore
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what is done is that the armature of the synchronous
generator is replaced by 2 fictitious windings,
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one is along the d axis of the generator synchronous
generator another is across the q axis of
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the synchronous generator and when these windings
are made to carry the current id and iq right
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and they rotate at the same speed as the rotor
it means it means the these two fictitious
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windings are stationary with respect to the
rotor and therefore the mmf produced by this
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fictitious windings act on constant permeance
path and therefore the inductances which we
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come across are constant.
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We will see that under steady state conditions,
the currents which are flowing in these two
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fictitious winding id that is direct axis
fictitious winding and quadrature axis fictitious
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winding these two current that id and iq will
come out to be constant in magnitude they
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will be just DC currents, okay this we will
just now establish.
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To establish this thing we will start with
that let us assume that the armature of the
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synchronous generator is carrying balance
steady 3 phase currents that is system is
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operating under steady state condition, okay
no dynamics is involved let us assume that
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ia is equal to im sin omega s t plus phi this
phi is some initial phase angle of the current
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ia this is arbitrary it can be 0 also ib is
equal to im sin omega s t plus phi minus 2
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phi by 3 and ic equal to im sin omega st plus
phi plus 2 phi by 3 that is these 3 currents
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ia, ib and ic form balanced balanced set of
currents okay.
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Now what we do is we apply dqo transformation
and when you apply the dqo transformation
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we will be in a position to write down id,
iq and this io will be 0 because we are considering
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the balanced steady state conditions that
is in balance system the zero sequence quantities
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are absent. Now here I will just add one thing
that when you talk about the zero sequence
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currents you always talk in terms of zero
sequence current to be the phasor right here
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the zero sequence term we are putting is as
it is the instantaneous component, instantaneous
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that is we are writing zero sequence as ia
plus ib plus ic by 3 is the instantaneous
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quantity but this is applicable here.
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Now after applying this transformation this
exercise has to be done that is in between
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you have lot of trigonometric simplification
to be done after you do this trigonometric
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simplification the id the current in the du
axis, d axis fictitious winding comes out
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to be equal to im sin omega st plus phi minus
theta iq equal to im cos omega st plus phi
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minus theta and i naught is equal to 0. Okay
now here what is this theta, theta is the
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{ angular position of the rotor and angular
position at any time is equal to omega r into
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t, omega r is the rotor speed under under
steady state operating condition the value
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of omega r comes out to be same as the synchronous
speed.
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Therefore, now we substitute here omega rt
equal to omega s into t right the moment you
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make this substitution here in this expression
that is theta is to be put as omega s into
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t this term vanishes and now we will have
the id equal to im sin phi similarly, iq equal
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to im cos phi right and therefore you can
easily see here that for a given stator currents
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im is constant it is a peak or it is the amplitude
of the phase current as if id can be written
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as im sin phi iq equal to im cos therefore
what we see here that is id and iq these are
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constant they are not varying as a function
of time therefore they become DC current right.
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However, under dynamic conditions when the
system is dynamic condition rotor speed is
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not exactly equal to the synchronous speed
right in that case the id and iq will vary
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but the variation will be a low frequency
variation, it is not going to be the 50 hertz
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are 60 hertz variation.
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Okay now some of the advantages which we get
by this transformation can be summarized although
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we have seen these advantages but I will just
summarize the advantages of dqO transformation.
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The first advantage is that the dynamic performance
equations have constant inductances we have
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seen we have seen the mathematical model which
we have written in terms of dqo transformations
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okay and in this dqo transformation by after
applying this dqo transformation, the equations
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have constant inductance this is the one of
the biggest merit of applying the dqO transformation,
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second is that under balance conditions zero
sequence quantities disappear that I am not
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present the id and iq under steady state conditions
are constant in magnitude.
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However, when we perform stability studies
then stability studies involves slow variations
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of id and iq the frequencies will be below
2 to 3 hertz that is the frequency of oscillation
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of the rotor which will come across will be
of the order of, order of point 5 to say 3
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hertz or we can say that the frequency of
oscillation of these currents will be less
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than 2 to 3 hertz is very low frequency variation
this is most important point we have to keep
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in mind and another very interesting advantage
which comes is that that d and q axis quantities
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particularly which will inductance ld, mutual
inductance lq these can be measured by performing
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experiments or by making some measurements
on the terminals of the machine.
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Since all of you must have done experiment
in your laboratory to measure the direct and
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quadrature axis reactance of a synchronous
machine. One important test which you perform
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in on is the slip test right. Now we come
to another important aspect so for what we
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have done is we have developed the expressions
for the stator voltage equations, rotor voltage
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equations. We have also developed expressions
for the stator and rotor flux linkage equations
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we have also developed the expressions for
electrical torque and given some meaning to
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the transformation, physical meaning to represent
transformation.
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Now in any power system studies per unit system
of calculation is the most convenient one
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and merits of per unit calculations are well
known to all of us and therefore now we will
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study how do we transform, transform our system
equations using per unit quantities. Now whenever
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you transform the actual quantities into per
unit quantities we have to choose base quantities.
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For example, in any circuit we may choose
MVA base and the voltage base then we can
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find out the base quantities in different
parts of the circuit MVA base remains same
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while the KV base depends upon the circuit
voltage that is the transformers which are
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involved.
Now here, we have two parts, one is the stator
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another is the rotor. Okay therefore first
we will define how we obtain obtained the
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per unit representation of stator quantities.
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Now to start with here we will define the
base quantities, now in any system we can
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define some base quantities and other quantities
can be derived from those chosen base quantities
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like we start with stator quantity we will
use this subscript s to denote that these
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are the stator quantities or the base quantities
for the stator. We will define esbase as peak
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value of the rated line to neutral voltage
a slight variation is here that is instead
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of defining the RMS value as the base value
here we are defining the peak value okay.
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Similarly, the isbase that is the current
in the stator circuit or the base current
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in the stator circuit is equal to peak value
of the rated current in amperes. The additional
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quantity which we have to refer here is the
base frequency. Generally, we do not require
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actually referring the base frequency when
we talk about the power system network calculations
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but here this is also an important quantity,
therefore these 3 terms are primarily decided
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therefore decision comes like this that rated
frequency becomes my base frequency.
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Now with this 3 quantities defined other quantities
are defined in terms of the basic base quantities
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like say omega base becomes 2 phi f base okay,
omega m base that is the the mechanical speed
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right. The base value of the mechanical speed
is written as the base value omega base into
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2 by pf further the impedance base Zsbase
is equal to es base upon is base but we can
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define. The base value of the inductance in
terms of stator base quantities that is Lsbase
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00:28:59,200 --> 00:29:06,200
is equal to Zsbase upon omega base. We can
also define the flux linkage or base value
199
00:29:11,100 --> 00:29:18,100
of the flux linkage that is psi s base equal
to Lsbase into isbase. We know that flux linkage
200
00:29:19,049 --> 00:29:26,049
can be written as L into i Lsbase can be replaced
by this quantity the Lsbase into is base can
201
00:29:30,240 --> 00:29:36,970
be replaced as esbase upon omega s base that
is what we do here is this Lsbase can be written
202
00:29:36,970 --> 00:29:43,970
as Zsbase upon omega base okay and then Zsbase
into is base can be written as esbase. So
203
00:29:48,080 --> 00:29:53,460
that the flux linkage the base value of the
flux linkage can be written as esbase upon
204
00:29:53,460 --> 00:29:57,659
omega s Weber turns.
205
00:29:57,659 --> 00:30:04,659
Now let us see actually that how we define
the three phase VAbase volt ampere base because
206
00:30:08,820 --> 00:30:15,820
as I told you that your power system computations.
We start with the 3 phase MVAbase, since we
207
00:30:22,580 --> 00:30:26,840
started with the voltage and current bases.
Okay therefore we write down the three phase
208
00:30:26,840 --> 00:30:33,840
VAbase is defined as three times Ermsbase
into Irmsbase that is the phase voltage its
209
00:30:35,549 --> 00:30:42,549
rms value multiplied by stator current and
its rms value this can be written as now 3
210
00:30:44,429 --> 00:30:49,779
rms value is equal to base quantity by root
2 because base quantity is chosen as the peak
211
00:30:49,779 --> 00:30:50,889
quantity here.
212
00:30:50,889 --> 00:30:57,889
Similarly, isbase by root 2. So that what
we see here is very very important relationship
213
00:30:58,220 --> 00:31:05,220
that is 3 phases VAbase comes out to be equal
to 3 by 2 times the voltage base into current
214
00:31:07,549 --> 00:31:14,549
base. Okay this relationship is very important
because we always make use of the three phases
215
00:31:19,629 --> 00:31:24,759
VAbase for our computations right but the
relationship between the quantities which
216
00:31:24,759 --> 00:31:31,759
you have chosen as the base quantities and
the 3 phase VA comes out to be like this.
217
00:31:31,830 --> 00:31:38,830
Then the torque base the torque base can be
always written as three phase VAbase divided
218
00:31:42,289 --> 00:31:49,289
by omega m base. Now when you talk about the
base quantities or base power we always say
219
00:31:53,289 --> 00:32:00,289
base power is equal base MVA if you recollect
in your power system calculations or computations
220
00:32:01,539 --> 00:32:08,539
then whatsoever base MVA is there that is
equal to base power in megawatts and therefore
221
00:32:08,590 --> 00:32:14,690
power divided by the base speed will give
you the torque base therefore here the torque
222
00:32:14,690 --> 00:32:19,919
base is equal to 3 phase VA base that is omega
m okay.
223
00:32:19,919 --> 00:32:26,919
Now when you make the substitutions here in
this expression for 3 phase VAbase equal to
224
00:32:29,600 --> 00:32:36,600
3 by 2 esbase, isbase and this esbase and
isbase, these quantities can be further expressed
225
00:32:36,820 --> 00:32:43,820
in terms of flux linkage base. Therefore,
torque is now expressed as 3 by 2 Pf by 2
226
00:32:45,039 --> 00:32:50,620
psi s base into isbase because we have seen
earlier actually the torque is written in
227
00:32:50,620 --> 00:32:57,620
terms of flux linkages and currents right
therefore we prefer to specify the base or
228
00:32:57,909 --> 00:33:04,909
torque base in terms of flux linkages and
current base quantities. Okay this is in Newton
229
00:33:10,799 --> 00:33:12,330
meters.
230
00:33:12,330 --> 00:33:19,330
Now what till now what we have done is that
started initially the definition of base quantities
231
00:33:22,809 --> 00:33:29,809
in the stator we have chosen the stator voltage
and stator current and the system frequency
232
00:33:30,470 --> 00:33:37,470
as the base quantities okay and then we have
derived that what will be the base quantities
233
00:33:38,340 --> 00:33:44,679
for other variables, other variables like
say impedance, base impedance, base torque
234
00:33:44,679 --> 00:33:51,679
base flux linkages, base flux base in base
inductance and so on.
235
00:33:54,250 --> 00:34:01,250
Now we start with transforming our stator
voltage equations into per unit terms or per
236
00:34:08,570 --> 00:34:15,570
in terms of per unit quantities is it not
okay now again let us start or stator voltage
237
00:34:15,960 --> 00:34:22,960
equation is ed equal p psi d psi q omega r
minus ia, id into Ra okay this is the equation
238
00:34:27,720 --> 00:34:33,530
which we have derived in terms of dqO components
we want to transform this equation in terms
239
00:34:33,530 --> 00:34:40,530
of per unit quantities how do we do it you
divide this whole expression by voltage base
240
00:34:43,910 --> 00:34:49,350
because these all these terms are the voltage
terms therefore what you divide you divide
241
00:34:49,350 --> 00:34:53,440
this by esbase.
242
00:34:53,440 --> 00:35:00,000
Now when you divide this by esbase we make
use of this relationship that esbase is equal
243
00:35:00,000 --> 00:35:07,000
to is base Zsbase it is also equal to omega
s base into psi s base this relationships
244
00:35:07,270 --> 00:35:14,270
we have already derived that is this 3 the
voltage can be written as the product of impedance
245
00:35:14,390 --> 00:35:21,390
and current can also be written as product
of omega and flux linkages okay. Therefore,
246
00:35:21,980 --> 00:35:27,880
what we are doing here is that ed is divided
by esbase p remains as it is we are not touching
247
00:35:27,880 --> 00:35:34,430
p right now psi d since it is a flux term
what we do is that we divide this flux term
248
00:35:34,430 --> 00:35:41,430
instead of es base i put omega s base, phi
s base here because esbase is same as this
249
00:35:45,240 --> 00:35:46,840
quantum.
250
00:35:46,840 --> 00:35:53,840
Similarly, here we put omega s base, omega
s base okay and in the third term where I
251
00:35:57,340 --> 00:36:03,620
have Ra into id I put here not in terms of
the flux linkages and speed by put errors
252
00:36:03,620 --> 00:36:10,620
Zsbase into isbase because this the denominator
terms which I have put here are same as esbase
253
00:36:12,280 --> 00:36:18,000
this can be done okay. Now by doing this what
happens is that this ed becomes now per unit
254
00:36:18,000 --> 00:36:25,000
value. Okay that is the direct axis voltage
ed can now it becomes a per unit quantity.
255
00:36:29,430 --> 00:36:36,430
Similarly, the flux linkage becomes per unit,
the resistance becomes per unit resistance,
256
00:36:37,040 --> 00:36:43,180
now the current becomes per unit current,
okay now to distinguish the per unit quantities
257
00:36:43,180 --> 00:36:50,180
from the quantities a real quantities in amperes
or ohms or watts right what we do is that
258
00:36:50,980 --> 00:36:57,980
we will put a put a super script bar, okay
a super bar on the top that is I will denote
259
00:37:02,390 --> 00:37:09,390
this as ed bar equal to 1 upon omega base
p psi d bar minus psi q bar omega r bar minus
260
00:37:14,240 --> 00:37:21,240
Ra bar id bar as we will see subsequently
the moment we develop a mathematical model
261
00:37:23,780 --> 00:37:29,060
of the system in terms of per unit quantities
right.
262
00:37:29,060 --> 00:37:36,060
We will drop this super bar, drop it because
every time writing this bar is not very convenient
263
00:37:37,200 --> 00:37:41,940
but when you are trying to derive it first
we will maintain this bar and then at the
264
00:37:41,940 --> 00:37:48,940
end we will drop it for further further use
of this model. Now here in this equation p
265
00:37:56,060 --> 00:38:03,060
is d by dt and time is in seconds, okay therefore
we can also define define the base value of
266
00:38:05,520 --> 00:38:12,480
time and the definition of base value of time
is the time in seconds required for the rotor
267
00:38:12,480 --> 00:38:19,480
to rotor to rotate by 1 radian, 1electrical
radian per second electrical radian turn per
268
00:38:22,980 --> 00:38:28,900
second electrical radian that when rotor makes
one electrical radian and the time which is
269
00:38:28,900 --> 00:38:35,900
required to make 1 radian rotation that is
called tbase that is 1 radial divided by omega
270
00:38:36,690 --> 00:38:43,690
base that is when rotor rotates through a
angular displacement of one radian the speed
271
00:38:44,940 --> 00:38:51,940
is omega base base speed then the that particular
time is called tbase base time okay.
272
00:38:52,870 --> 00:38:59,870
Therefore, this base time can be written as
1 upon 2 phi fbase and if I now use this base
273
00:39:00,160 --> 00:39:07,160
time then this term can be replaced by p bar,
psi d bar minus psi q bar, omega r bar minus
274
00:39:11,140 --> 00:39:18,140
Ra bar that is this expression is now in terms
of per unit quantities that is p bar is also
275
00:39:23,730 --> 00:39:30,730
expressed in per unit and all other terms
are in per unit.
276
00:39:31,010 --> 00:39:38,010
Similarly, you can write down for the q axis
voltage and zero sequence voltage. We have
277
00:39:44,590 --> 00:39:51,590
just now mentioned that p bar was defined
as d by dt bar where dt bar is a per unit
278
00:39:53,820 --> 00:40:00,180
current which comes out to be p bar comes
out to be one upon omega base into p okay
279
00:40:00,180 --> 00:40:07,180
and we have seen here actually that one upon
omega base into p is the it comes out to be
280
00:40:08,770 --> 00:40:15,770
equal to p bar that is why we have come to
now this model, this model where ed bar is
281
00:40:23,700 --> 00:40:30,490
equal to p bar, psi d bar minus psi q bar,
omega Ra bar, id bar and similarly for eq
282
00:40:30,490 --> 00:40:37,490
and eo that is we have developed the stator
circuit equations in per unit terms okay.
283
00:40:42,660 --> 00:40:49,660
Now the next step will be to develop the rotor
circuit equations also in per unit quantities.
284
00:40:56,110 --> 00:41:03,110
Now here to start with what I will do is we
will assume assume the rotor circuit voltage,
285
00:41:13,290 --> 00:41:20,290
rotor circuit currents as the base quantities,
I am not I am not addressing at present how
286
00:41:22,790 --> 00:41:28,830
do we choose that we will see later on but
assuming that the for the rotor circuit, the
287
00:41:28,830 --> 00:41:34,780
field circuit because in the rotor we have
3 circuits one is the field circuit another
288
00:41:34,780 --> 00:41:41,630
is the direct axis amortisseur circuit, third
is the quadrature axis amortisseur circuit
289
00:41:41,630 --> 00:41:48,630
and therefore for the field circuit a base
voltage will be efd base which will be written
290
00:41:50,520 --> 00:41:57,520
as omega base into psi fd base that is psi
fd is the flux linkage in the field circuit
291
00:41:58,990 --> 00:42:01,490
a base quantity in the field circuit.
292
00:42:01,490 --> 00:42:08,330
Similarly, we will have Zfdbase into ifdbase
okay that is as we have done in the case of
293
00:42:08,330 --> 00:42:15,330
stator similarly we specify the quantities
in the rotor also. Okay but how do we choose
294
00:42:17,570 --> 00:42:24,570
this base quantities in the rotor that we
will establish in few minutes. Therefore,
295
00:42:24,940 --> 00:42:31,530
now following the same approach as we have
done for the stator we can write down efd
296
00:42:31,530 --> 00:42:38,530
bar equal to p bar psi fd bar, Rfd bar, ifd
bar that is in this equation where actually
297
00:42:40,290 --> 00:42:47,290
we have field circuit applied voltage is efd
okay the current flowing is ifd flux linkage
298
00:42:48,080 --> 00:42:54,540
is psi fd therefore we can establish the equation
in the per unit quantity they are very simple
299
00:42:54,540 --> 00:43:01,230
as state forward the approach is exactly same
you divide by efd base the actual quantities
300
00:43:01,230 --> 00:43:08,230
and then wherever you have this Rfd, ifd divided
Zsbase Zfdbase ifdbase.
301
00:43:08,830 --> 00:43:15,830
Okay and you will get this equation other
two equations also in the same form because
302
00:43:16,600 --> 00:43:20,440
here we do not have term because they are
all close circuits amortisseur closed circuit
303
00:43:20,440 --> 00:43:27,440
therefore 0 equal to okay.
304
00:43:29,790 --> 00:43:36,790
Now we come to very ah crucial thing actually
that is how do we express the stator flux
305
00:43:41,020 --> 00:43:48,020
linkages in per unit. Okay because so far
what we have done is the stator voltage circuit
306
00:43:48,780 --> 00:43:53,340
equation, stator circuit voltage equations,
rotor circuit voltage equations. We have expressed
307
00:43:53,340 --> 00:44:00,340
in per unit terms therefore now our next step
is to express the stator flux linkage equations
308
00:44:03,250 --> 00:44:05,140
in per unit.
309
00:44:05,140 --> 00:44:12,140
Now to understand this let me ah rewrite here
I have written again psi d is equal to minus
310
00:44:17,530 --> 00:44:24,530
Ld id plus La fd ifd Lakd, ikd this equation
was derived this is the direct axis flux linkages
311
00:44:30,250 --> 00:44:36,580
psi d equal to minus Ld id plus Lafd what
is Lf this is the mutual inductance between
312
00:44:36,580 --> 00:44:43,580
the d axis and field winding. This is actually
amortisseur and d axis winding that is why
313
00:44:49,000 --> 00:44:49,980
Lakd.
314
00:44:49,980 --> 00:44:55,830
Okay now what you do is that you divide by
this whole equation by base, when you divide
315
00:44:55,830 --> 00:45:02,830
this by base psi d will be divided by psi
s base, psi s base is equal to Lsbase into
316
00:45:04,330 --> 00:45:11,330
isbase. Okay therefore you can you can the
divide the whole quantities by this base quantity
317
00:45:11,370 --> 00:45:18,370
right. Therefore you will have Ld divided
by id divided by Ld into id divided by Lsbase
318
00:45:19,610 --> 00:45:26,200
is base then Lafd into ifd divided by Lsbase
into isbase.
319
00:45:26,200 --> 00:45:33,200
Now here the problem comes that the per unit
value, per unit value of the field current
320
00:45:39,770 --> 00:45:46,770
is expressed in terms of field current base
not in terms of the stator current base okay
321
00:45:47,520 --> 00:45:54,000
therefore what you do is that you multiply
this term is okay you multiply this term by
322
00:45:54,000 --> 00:46:01,000
ifdbase into ifdbase. Okay then in this term
what we can do here is that this ifd divided
323
00:46:04,380 --> 00:46:11,380
by ifdbase can be written as per unit current
in the field circuit then remaining quantities
324
00:46:12,120 --> 00:46:19,120
Lafd divided by Lsbase into ifdbase divided
by Lsbase this quantity will be denoted by
325
00:46:24,820 --> 00:46:31,820
a term Lafd bar that is you can see very interesting
thing that we have the now this 3 ratios.
326
00:46:35,490 --> 00:46:42,490
Okay we can identity this ratio ifd upon ifdbase
as ifd bar right then we have the term Lafd
327
00:46:45,090 --> 00:46:50,490
by Lsbase then ifd base upon isbase okay.
328
00:46:50,490 --> 00:46:54,310
Now this these are when you divide these two
base currents it comes to be a multiplying
329
00:46:54,310 --> 00:47:01,090
factor only and therefore what we do is that
this Lafd will be denoted by a symbol Lafd
330
00:47:01,090 --> 00:47:08,090
bar to ifd bar. Similarly, you will have Lakd
bar ikd bar okay is it understood. Now once
331
00:47:12,470 --> 00:47:19,470
you understand this thing actually the I have
done it for one direct axis plus linkage equation.
332
00:47:20,170 --> 00:47:27,170
Similarly. We have to do it for other flux
linkage equations that is by doing this exercise
333
00:47:27,700 --> 00:47:33,180
we have written the expression for stator
winding flux linkage equations that is the
334
00:47:33,180 --> 00:47:38,030
mutual flux linkage this psi d is the mutual
flux linkage psi d bar equal to minus Ld bar
335
00:47:38,030 --> 00:47:45,030
id bar plus Lafd bar, ifd bar plus Lakd bar
ikd bar.
336
00:47:47,420 --> 00:47:54,420
Now these terms are defined as Lafd bar is
defined Lafd bar is defined as
now we have just now seen that Lafd bar is
337
00:48:13,940 --> 00:48:20,940
defined as Lafd upon Lsbase into ifd base
upon is base right. Similarly, Lakd bar Lakq
338
00:48:24,190 --> 00:48:31,190
bar right these terms were defined in terms
of the mutual inductances, base quantities,
339
00:48:35,600 --> 00:48:42,460
the stator and base quantities in the field
circuit. This this is very important actually
340
00:48:42,460 --> 00:48:48,670
speaking here here actually it is not state
forward where we have a base actual quantity
341
00:48:48,670 --> 00:48:55,670
divided by base gives you a per unit quantity
here that is let me just summarize here that
342
00:49:06,120 --> 00:49:13,120
the direct axis flux linkages are written
in this form where these quantities are to
343
00:49:14,020 --> 00:49:16,040
be defined okay.
344
00:49:16,040 --> 00:49:22,850
Similarly, psi q is written in this form and
psi naught is written in this form, okay they
345
00:49:22,850 --> 00:49:29,850
are the three flux linkage equations in the
stator circuit okay and where these terms
346
00:49:31,780 --> 00:49:36,170
are defined as Lafd bar by this expression
Lakd by this expression Lakq by this expression
347
00:49:36,170 --> 00:49:43,170
do you understood. Now after we have expressed
express the stator flux linkages psi d, psi
348
00:49:55,510 --> 00:50:02,510
q and psi naught in terms of or or converted
the stator flux linkages equations into per
349
00:50:05,410 --> 00:50:12,410
unit quantities, next step is to convert the
rotor flux linkage equations.
350
00:50:13,810 --> 00:50:20,810
Now when you convert the rotor flux linkage
equations, we will come across the psi bar
351
00:50:27,180 --> 00:50:34,180
fd that is actually this is the flux linkage
of the field circuit you can you again start
352
00:50:34,700 --> 00:50:41,700
with the original equations, original equations
and then divide by the base quantities right
353
00:50:43,920 --> 00:50:50,920
and then then these terms may have to be again
defined. These terms will not be required
354
00:50:59,120 --> 00:51:06,120
to be defined that is Lffd will be simply
equal to Lffd is the self-inductance of the
355
00:51:10,550 --> 00:51:17,550
field winding right therefore, you divide
this by the Lfdbase right where this is the
356
00:51:18,460 --> 00:51:24,000
rotor quantity divided by the rotor base quantity
it becomes a per unit quantity therefore there
357
00:51:24,000 --> 00:51:24,970
is no problem here.
358
00:51:24,970 --> 00:51:28,980
Similarly, for the current also there will
be no problem but when we talk about this
359
00:51:28,980 --> 00:51:35,980
ikd the La, Lfkd divided by ikd. Okay you
will find this similar problem will come right
360
00:51:38,460 --> 00:51:43,690
and therefore these quantities have to be
expressed in terms of for example ikd is to
361
00:51:43,690 --> 00:51:50,690
be expressed in terms of ikdbase. Similarly,
ikq has to be expressed in terms of its own
362
00:51:55,790 --> 00:51:56,540
base right.
363
00:51:56,540 --> 00:52:03,540
Therefore, when you do this exercise you will
find that we can express the rotor circuit
364
00:52:04,310 --> 00:52:11,310
to flux linkages in terms of per unit quantities
where these terms Lfda bar, Lfkd bar are to
365
00:52:17,400 --> 00:52:23,850
be defined as all of the all this 5 terms
I mentioned this 5 terms will appear in those
366
00:52:23,850 --> 00:52:30,850
three equations are to be defined in this
base.
367
00:52:31,930 --> 00:52:38,930
Therefore, now what we see here is that that
we have established, establish the rotor circuit
368
00:52:43,810 --> 00:52:49,760
equations that is rotor circuit voltage equations,
rotor circuit flux linkage equations, okay
369
00:52:49,760 --> 00:52:56,760
in per unit terms while the basic problem
here remains is here remains is that how do
370
00:53:00,530 --> 00:53:07,530
we choose the rotor circuit base quantities
because synchronous machine is having stator
371
00:53:09,150 --> 00:53:14,230
and rotor parts, what is the relationship
of base quantities in the rotor circuit with
372
00:53:14,230 --> 00:53:21,230
the stator circuit base quantities right and
for that we have to establish some basic rules
373
00:53:31,310 --> 00:53:38,310
and I will just state what rules are to be
established is that when we choose the base
374
00:53:40,260 --> 00:53:46,040
quantities for the rotor and we want our our
intention is our our intention or the objective
375
00:53:46,040 --> 00:53:51,280
is we want to simplify the equations make
this equations as simple as possible.
376
00:53:51,280 --> 00:53:58,280
Therefore we have 2 major assumptions or requirements
but in the per unit mutual inductances between
377
00:54:03,800 --> 00:54:10,290
different winding are to reciprocal. This
is one requirement that we would like to choose
378
00:54:10,290 --> 00:54:16,980
the per unit quantities in such a fashion,
so that the mutual inductances become reciprocal
379
00:54:16,980 --> 00:54:23,980
right that is mutual inductance between the
d axis field winding d axis winding that is
380
00:54:25,420 --> 00:54:32,420
end the the field winding there should be
equal therefore if I write down that Lafd
381
00:54:32,780 --> 00:54:37,340
should be equal to Lfda.
382
00:54:37,340 --> 00:54:42,330
Okay another thing which would like to do
is that all per unit mutual inductances between
383
00:54:42,330 --> 00:54:49,330
stator and rotor circuit in each axis are
to be equal that that is another way of simplifying
384
00:54:51,060 --> 00:54:58,060
that we would like to make this inductances
mutual inductances on each axis equal right
385
00:54:58,340 --> 00:55:05,340
with this I conclude my presentation today
and let me summarize what we have done today.
386
00:55:06,660 --> 00:55:13,660
We have discussed the stator circuit, voltage
circuit equations in terms of dqO components.
387
00:55:18,420 --> 00:55:24,770
We have also transformed the stator and rotor
circuit voltage and flux linkage equations
388
00:55:24,770 --> 00:55:31,770
in per unit quantities. Thank you!