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Friends we start today, the next topic that
is on solution of swing equation. We have
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derived swing equation or a synchronous machine
we have also derived swing equation for a
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multi-machine system. We have seen that the
swing equation is function of or is a non-linear
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function of the power angles, the there is
no formal solution available or possible because
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the swing equation is a non-linear differential
equation and therefore numerical techniques
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have been developed to solve the swing equation.
The 2 methods which is called method 1 and
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method 2 will discussed today. Before I tell
you about the method 1 and method 2 for solving
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the swing equation, let us consider a simple
case.
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A synchronous machine connected through a
transformer to a double circuit transmission
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line and hence and then connected to an infinite
bus, a large system
on this transmission lines we provide circuit
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breakers at these locations. The purpose of
providing circuit breakers it is location
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is that as and when any fault occurs on the
transmission line by operating the circuit
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breaker this faulting line can be taken out.
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We will consider a disturbance where due to
some unknown reason this line trips, okay.
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Now here before before the occurrence of this
disturbance these 2 lines are in service.
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Now if I draw the power angle characteristic
I will denote this as Pe on this axis, we
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put delta, okay for this case the power angle
characteristic will come out to be a sin curve
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on this path okay.
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Now in case the system is operating under
steady state condition I will represent the
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mechanical input line by a straight line divided
by Pm then this is our operating, now the
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moment a disturbance occurs where out of the
2 lines, 1 lines trips then the post fault
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system will have only 1 line in operation
and the new power angle characteristic will
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be different from the power angle characteristic
before the occurrence of fault or we call
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it pre-fault disturbance or pre-fault power
angle characteristic or we can call it pre-disturbance
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power angle characteristic.
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Now let us represent the power angle characteristic
after the disturbance we will this characteristic
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can be represented as Pe1 equal to Pmax sin
delta and this characteristic may be represented
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as Pe2 equal to, let us say r into P max sin
delta okay. Our initial operating point is
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here, this denoted by this angle delta naught.
Now what we see here is that the moment fault
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has occurred or a disturbance has occurred
the operating point will shift from this position
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a to position b because the moment disturbance
occurs right the angle cannot change instantaneously,
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angle will remains same.
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However, the power angle characteristic becomes
Pe2 equal to r Pmax sin delta r is, r is a
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quantity or a fraction which is less than
1 and it depends upon what where the values
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of the transformer reactance, synchronous
machine, transient reactance and transmission
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line reactance and so on. Now what we see
here is that the moment this disturbance takes
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place there is difference in the mechanical
input and electrical output, electrical output
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determined by this point, this difference
become the acceleration power Pe, we call
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it accelerating power Pa.
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Now this as we will see actually that because
of the accelerating power the rotor will accelerate
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okay and delta will increase, the increase
of delta right as a the function of time right
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or we call it actually the delta versus time
this plot is our swing curve and they are
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interested in plotting the swing curve of
the system okay. Our primary objective is
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to obtain the solution of the swing equation
and that solution is your swing curve. Now
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what I do here is that I plot the accelerating
power Pa as a function of time.
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We start with time t equal to say 0 at time
t equal to 0, the accelerating power is equal
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to ab and it is a positive as time passes
the accelerating power is going to decrease
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okay
Therefore, let us represent the variation
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of accelerating power by this graph I am not
showing the complete one because as far as
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this problem is concerned in this problem
the accelerating power is going to be positive
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all through and the resulting system will
be unstable system.
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Now what we do is that for is solving the
swing equation, we will divide the time into
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small time steps. Let us consider 2 consecutive
time intervals we will represent the nth time
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interval, the let us say this is nth time
interval. Now this nth time interval at the
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beginning of this interval, let us call it
time is t1 at the end of the interval, let
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time is say ah t2. Okay then if this is the
nth interval right then t1 is equal to n minus
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1 delta t and t2 will be equal to n delta
t, right this is the notation which we will
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be adapting. Now in the step by step solution
or we normally call point by point solution
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because these 2 terms are used interchangeably
in the method 1.
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we assume that the accelerating power remains
constant during the time interval and equal
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to equal to its value at the beginning of
the interval that is we calculate the accelerating
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power at the time t1 which is the beginning
of n minus 1th interval and we presume at,
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we that this accelerating power remains constant
during this interval it means the accelerating
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power remains constant, okay and we shall
denote this accelerating power by the symbol
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Pa n minus 1 okay, with this introduction
we can proceed to discuss the method 1 for
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solving the swing equation.
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As I stated in the beginning this, this method
is called point-by-point solution therefore,
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the point-by-point solution of the swing equation
consists of two processes which are carried
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out alternatively which are those two processes
that is what we have to understand.
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The first process is the computation of the
angular positions and angular speeds, I am
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putting the word angular positions and angular
speeds. At the end of each interval from the
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knowledge of from the knowledge of angular
positions and speeds at the beginning of the
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time interval
that is the, that is
and the accelerating power is assumed for
the interval.
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Now this is general statement what will be
the assumed value of the accelerating power
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during that interval right is a subject of
concern and we will see actually that the
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method 1, there is one way of choosing the
accelerating power during that interval method
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2, we have another way of choosing the accelerating
power. However, the accelerating power during
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the time interval remains constant right and
therefore the fist step we understand is that
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we calculate the value of angular positions,
angular speeds at the end of time interval
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with the knowledge of angular positions, angular
speeds at the beginning of the time interval
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okay, this is the first step.
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Now once we have obtained the angular positions
at the end of the time interval then the next
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step we want to second step will be or the
second process we call it the second process
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is the computation of accelerating power of
each machine from the angular positions of
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all the machines of the system because when
we go from one step to the next step right,
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we have to again compute the accelerating
power of each machine, okay and this accelerating
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power can be computed from the knowledge of
network solution.
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In general for a multi-machine system, one
has to solve the network to find out the accelerating
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power of each machine at the beginning of
interval or we can say at the beginning of
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next interval. Now this uh these 2 processes
are carried out alternatively that you can
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understand now that suppose I know the angular
positions, angular speeds at the beginning
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of time interval and I start with the knowledge
of accelerating power at the beginning of
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the time intervals or accelerating power during
that interval, with this knowledge we compute
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the angular positions, angular speeds at the
end of the interval.
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Then using this information we go to second
step where we solve the network and we find
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out what will be the accelerating power to
be used for the next interval right and this
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this these 2 steps are carried out alternatively
that is I discuss in method 1.
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The accelerating power is constant throughout
the time interval delta t and has the value
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computed for the beginning of the interval
that value we have computed at the beginning
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of the interval to presume this to remain
constant during that interval, okay.
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Now here in this diagram, instead of showing
the plot of accelerating power versus time
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what we are showing here is the acceleration
that is d2 delta by dt2 which is equal to
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accelerating power Pa by that is M is constant
you divide this by M. So that instead of plotting
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accelerating power if I plot acceleration
versus the number of time steps. Now instead
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of plotting here ah time in seconds I plot
this as t by delta t. Okay and this plot is
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similar to the plot for accelerating power
except the units will be different because
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the accelerating power has been divided by
M.
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Now if you see here actually that suppose
I compute for time interval n minus 1 then
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this is the assumed value of the accelerating
power that is I compute the accelerating power
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at beginning of the time interval and this
will remain constant. Then when I come here,
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we again compute what is the acceleration
and this acceleration that assumed to remains
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constant here like this okay. Please note
down actually this graph here what we observe
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in this graph. In this method 1, when the
accelerating acceleration is decreasing the
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assumed acceleration is always more than the
actual acceleration and the we will see that
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the error will depend upon what is the time
step which we choose.
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Suppose if I take very small time step then
the assumed value of accelerating power will
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very close to the actual value of acceleration
or accelerating power. We will derive now
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a simple algorithm to implement this method
one. We start with our swing equation. Now
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to illustrate this problem I am considering
a simple system where one machine connected
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to infinite bus and there is only one swing
equation. Now this is our swing equation dt
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delta by dt2 equal to Pa by M, okay.
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Instead of using this coefficient here H by
phi here, we prefer to use this M because
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it is easy to what I told the time then you
integrate this equation assuming that Pa is
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constant accelerating power is constant then
this integration will give you d delta by
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dt equal to omega, this is the definition
of dt delta by d, dt represented as omega
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this is going to be equal to omega O plus
Pa into t by M. Omega O is the initial value
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of angular speed.
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Now here d delta by dt is not the actual speed
of the rotor it is, it is excess speed of
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the rotor over synchronous speed. Initially
when the system is operating under steady
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state condition then what happens Pa will
be 0 and d delta d delta by dt is also 0 because
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there is no acceleration, there is no difference
in the in the rotor actual speed and the synchronous
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speed, the rotor rotates at synchronous speed.
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Now here with finite value of Pa the accelerating
power right d delta by dt is omega and this
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is the excess speed over the synchronous speed,
now if you further integrate this equation
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4.2 with respect to time, we will get delta
equal to delta naught omega O into t Pa into
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t square by 2M. Now you have to very clearly
understand the meaning of all the terms which
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are involved in the these 3 equations 4.1,
4.2 and 4.3. Is there any doubt, this is state
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forward okay. Now what we do is we will use
these equations to write down the speed and
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angular positions at the end of interval in
terms of information at the beginning of the
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interval.
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In the equation 4.2, I will substitute the
value of the speed at the end of the nth interval.
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The speed at the end of nth interval is omega
n, the speed of the rotor at the beginning
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of the nth interval is omega n minus 1, t
delta t is the time, time state and Pa n minus
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1 is the accelerating power at the beginning
of the interval. Okay, therefore from the
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equations which we have obtained from swing
equation by integrating.
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Now what we are trying to do is that we are
implementing or applying this equations to
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a particular time interval. Now if we substitute
the values of angular positions at the beginning
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and at the end of time interval, we get the
equation 4.5, delta n equal to delta n minus
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1,omega n minus 1 into delta t delta t square
by 2M Pa n minus 1 okay
and these equations are valid for any value
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of n, they can start with n equal to 0 that
is time t equal to 0 okay and then we go from
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n equal to 1, 2, 3 like this, that is from
one step to the next step next step to next
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step and so on.
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From the equation 4.5, we can write down the
change in speed during the nth interval as
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omega n minus omega n minus 1 as delta omega
n which is equal to delta t by M Pa n minus
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1.This can be easily understood because this
is the accelerating power accelerating power
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by M is the acceleration and we are assuming
this acceleration to remain constant during
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the time interval therefore deviation in the
speed during the time interval is called delta
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omega n equal to the assumed value of acceleration
multiplied by time period.
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Similarly, we can write down the change in
angular position during the nth time interval
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as omega n minus 1 delta t equal plus delta
t square by 2M Pa n minus 1. Now with the
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information of the quantity at the beginning
of the interval what are the quantity at the
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beginning of the interval omega n minus 1,
delta n minus 1 and Pa n minus 1.
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We can solve this equation 4.6 and 4.7 okay
and obtain this swing curve okay what we have
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to we have to do is that we change, this angular
change during the interval and we can find
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out the actual angular position by adding
this change to the angular position at the
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n minus 1th at the beginning of n minus 1th
interval okay and you can find out the swing
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curve. Now when I said that there are 2 steps
involved, this is the first step, the second
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step will be that you obtain the value of
this angle delta and substitute in the power
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angle characteristic applicable during that
transient period to find out the new value
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of accelerating power that is your second
step.
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Once you obtain a new value of accelerating
power we will substitute here in this equation
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and obtain the new values of angle deviation
and speed deviation. Now do we require the
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information about speed deviation as function
of time for assessing the stability of the
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system, the answer is no. If I know the plot
of delta as function of time then by examining
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the swing curve I can say whether system is
stable or not as we have seen in last time
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and therefore if I am interested only in plotting
the swing curve then what we do is that we
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eliminate this speed term and obtain an expression
which is independent of or which does not
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contain speed term.
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Now this can be easily done easily done, here
equation 4.8 is same as equation 4.5, I will
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just written for the safe of convince. Now
what we do is that we write a similar equation
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for the preceding time interval that is instead
of n we put it n minus 1, n minus instead
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of n minus 1 you put n minus 2.
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So that we can write down the equation of
this form delta n minus 1 equal to delta n
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minus 2 delta t omega n minus 2 delta t square
by 2 M Pa n minus 2. This equation is for
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nth interval this equation is for n minus
1th interval, okay our next step will be that
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you subtract these 2 equations, you will get
delta n minus delta n minus 1 equal to delta
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n minus 1 minus delta n minus 2 plus delta
t into omega n minus 1 minus omega n minus
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2 plus delta t square by 2 M Pa n minus 1
minus Pa n minus 2.
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Now what we do is that we make some substitutions,
we denote the speed change I am sorry, correction
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we denote the angle change or angle deviation
during nth interval as delta delta n. Similarly,
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for n minus 1th interval delta delta n minus
1 and so on now if we make these substitutions
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here in this equation we can write down expression
in the form delta delta n equal to delta delta
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n minus 1.
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Now we had the term delta t into omega n minus
1 minus omega n minus 2, now this term is
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nothing but delta omega n minus 1 this is
the speed deviation in n minus 1th interval
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and
using this equation 4.4, we can write down
that the speed deviation during the n minus
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1th interval can be written as
or nth interval can be written as delta t
by M Pa n minus 1 that is equation 4.4 okay.
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Now you make that substitution here.
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So that now what we find that in this equation
we do not have speed term speed has been eliminated
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okay. This equation these 2 terms can be combined
and you get the resulting expression for delta
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00:34:33,349 --> 00:34:40,349
delta n equal to delta delta n minus 1e plus
delta t square by 2 M Pa n minus 1 plus Pa
202
00:34:45,450 --> 00:34:52,450
n minus 2. Now this is the algorithm for obtaining
the swing curve or a machine connected to
203
00:35:08,260 --> 00:35:10,329
infinite machine.
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00:35:10,329 --> 00:35:17,329
Similar equations can be derived if you have
a multi machine system. A process goes like
205
00:35:27,250 --> 00:35:34,250
this when I want to find out the change in
total angle position during nth interval I
206
00:35:39,099 --> 00:35:44,450
know what was the change in angle rotor angle
position during the n minus 1th interval but
207
00:35:44,450 --> 00:35:51,450
that was that computation has already been
completed, we know this information, we also
208
00:35:52,400 --> 00:35:59,400
know the uh assumed value of accelerating
power at the beginning of n minus 1th interval
209
00:36:03,450 --> 00:36:07,509
and at the beginning of nth interval, this
is the accelerating power at the beginning
210
00:36:07,509 --> 00:36:08,589
of nth interval.
211
00:36:08,589 --> 00:36:14,519
The accelerating power at the beginning of
n minus 1th interval is also known because
212
00:36:14,519 --> 00:36:20,390
we have already done the calculation for that
interval and therefore this algorithm goes
213
00:36:20,390 --> 00:36:27,390
in a iterative manner you know the complete
information on the which is required to compute
214
00:36:28,230 --> 00:36:35,230
the expression on the right hand side of this
equation, you get the value of delta n.
215
00:36:36,509 --> 00:36:43,509
Then the moment you get the value of delta
delta n, you obtain the new value of delta
216
00:36:51,059 --> 00:36:58,059
n as delta n minus 1plus delta right and you
continue to do it because as I told you that
217
00:37:09,710 --> 00:37:16,710
the process has 2 steps, the step one is to
compute the new angular positions and new
218
00:37:17,420 --> 00:37:24,420
angular speeds. However in the method one
if you are interested only in swing equation
219
00:37:27,279 --> 00:37:34,279
then we do not require the information about
the angular speeds right.
220
00:37:34,660 --> 00:37:41,660
Then once we get this new value of angle we
refer back to the power angle characteristic
221
00:37:42,269 --> 00:37:49,269
compute the electrical power output, mechanical
power input is assumed to be constant we compute
222
00:37:51,759 --> 00:37:58,420
what is acceleration power and use that accelerating
power for the next interval okay and therefore
223
00:37:58,420 --> 00:38:01,089
this process is to continue alternatively.
224
00:38:01,089 --> 00:38:08,089
We have discussed actually this technique
now how good this technique is in giving you
225
00:38:16,619 --> 00:38:23,619
the correct solution and as we will see actually
they are numerical techniques are not the
226
00:38:25,650 --> 00:38:32,650
exact solutions they will give you always
approximate solutions because we make some
227
00:38:33,160 --> 00:38:39,950
assumptions. Only this is that you would like
to get the solution which is very close to
228
00:38:39,950 --> 00:38:46,950
the exact solution and to evaluate the
method 1 step by step method 1 there we assume
swing equation of this form.
229
00:39:02,700 --> 00:39:09,700
Let us presume that swing equation is given
by this equation, now this is the swing equation
230
00:39:16,680 --> 00:39:23,680
which we want to solve. Let me know whether
this swing equation is a linear differential
231
00:39:29,069 --> 00:39:36,069
equation or non-linear differential equation.
This is a linear differential equation, I
232
00:39:38,119 --> 00:39:45,119
have chosen the linear differential equation
deliberately to illustrate the we will say
233
00:39:51,589 --> 00:39:58,589
illustrate the effect of, effect of time step
on the accuracy of the solution.
234
00:40:02,369 --> 00:40:09,369
Now so far this swing equation is concern
the initial conditions are given as delta
235
00:40:11,619 --> 00:40:18,619
naught equal to phi by 4 and the initial speed
is 0 d delta by dt initial is 0, with these
236
00:40:19,140 --> 00:40:24,730
initial condition the formal solution of the
swing equation is delta equal to phi by 2
237
00:40:24,730 --> 00:40:31,730
minus phi by 4 cosine square root of 2 f by
H, t. This you can obtain yourself please
238
00:40:35,519 --> 00:40:42,519
do the it is an exercise find out the formal
solution of this linear second order equation.
239
00:40:47,619 --> 00:40:54,619
Now to compare the accuracy of step by step
method one we solve this equation by step
240
00:41:02,690 --> 00:41:08,299
by step method because step by step method
which is applicable are suitable for solving
241
00:41:08,299 --> 00:41:15,299
non-linear equation can also be applied for
solving a linear differential equation, for
242
00:41:15,490 --> 00:41:22,490
solving this H is equal to 2.7 mega joules
per MVA is assumed.
243
00:41:42,130 --> 00:41:49,130
This graph shows ah swing curve obtained using
the formal solution and by using this method
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00:41:51,829 --> 00:41:58,829
1, this firm curve shows this graph 1, a firm
curve it shows the solution obtained using
245
00:42:03,529 --> 00:42:10,529
the formal solution. The graphs which are
shown here in this diagram are obtained for
246
00:42:14,799 --> 00:42:21,799
different values of time step delta T equal
to 0.15 second. Then next graph is for delta
247
00:42:29,220 --> 00:42:36,220
T 0.1 second then the third graph is for 0.05
second and the last one which is obtained
248
00:42:48,259 --> 00:42:55,259
using the method 1 point by point solution
is 0.0167 delta T equal to 0.0167 is one third
249
00:43:03,470 --> 00:43:10,470
of the delta T equal to .05, originally a
delta T equal to 0.05 second is chosen and
250
00:43:14,549 --> 00:43:21,549
in order to understand how much improvement
one gets in the solution delta T is reduce
251
00:43:25,789 --> 00:43:32,789
by a factor of 3 and graph is obtained for
delta T equal to 0.0 second.
252
00:43:33,930 --> 00:43:40,930
This axis shows the time in seconds and y
axis shows delta in electrical degrees, it
253
00:43:45,359 --> 00:43:52,359
can be very clearly seen that as the time
step increases the accuracy of the solution
254
00:43:58,499 --> 00:44:05,499
of the swing equation decreases or deteriorates
and hence one concludes that for getting solution
255
00:44:09,619 --> 00:44:16,619
closer to the accurate solution one has to
use very short time step. However, if we use
256
00:44:20,730 --> 00:44:27,730
a short time step it will require more time
for computation and this is the major drawback
257
00:44:31,190 --> 00:44:34,249
of method 1.
258
00:44:34,249 --> 00:44:41,249
In order to overcome this shortcoming of this
method 1, a method 2 has been developed and
259
00:44:43,849 --> 00:44:50,849
this method two is different from method 1
in terms of the accelerating power which is
260
00:44:57,259 --> 00:45:04,259
assumed to remain constant during the time
step. Now to understand this method 2 let
261
00:45:07,890 --> 00:45:14,890
us look at this graph which shows the acceleration
versus t by delta T, now here on this x axis
262
00:45:19,720 --> 00:45:21,599
we have put t by delta T.
263
00:45:21,599 --> 00:45:28,599
So that it shows the u the time step count,
now this graph shows the variation of acceleration
264
00:45:34,880 --> 00:45:41,880
with t by delta T or with t, now in this method
two instead of assuming the accelerating power
265
00:45:43,460 --> 00:45:50,460
remaining constant at the value equal to the
1 computed at the beginning of the time interval.
266
00:45:56,380 --> 00:46:03,380
We compute the value of the accelerating power
or acceleration at the beginning of the time
267
00:46:03,589 --> 00:46:10,589
interval for example, say time interval starts
at n minus one and ends at n that is the nth
268
00:46:11,490 --> 00:46:18,490
time interval. Now at this time step n minus
1, we compute the acceleration alpha n minus
269
00:46:26,660 --> 00:46:33,660
1 and assume that this acceleration remains
constant from half the preceding interval
270
00:46:38,170 --> 00:46:45,170
to the next half interval that we if we see
in this diagram the accelerating power computed
271
00:46:46,660 --> 00:46:53,660
at the beginning of nth interval remains constant
from n minus 3 by 2 to n minus half it can
272
00:47:01,579 --> 00:47:08,579
be very clearly seen actually that by assuming
this accelerating power in this fashion, the
273
00:47:10,910 --> 00:47:17,910
assumed value of the accelerating power is
equal to the average value over the time step.
274
00:47:25,529 --> 00:47:32,529
Now
we can write down with this assumption the
change in speed during the time step n minus
275
00:47:46,059 --> 00:47:53,059
half this can be written as delta t into alpha
n minus 1 where alpha n minus 1 is the acceleration
276
00:47:56,279 --> 00:48:03,279
computed at at step n minus 1. Now alpha n
minus 1can be replaced by this Pa n minus
277
00:48:08,680 --> 00:48:15,680
1 by M. So that we can say that speed deviation
delta omega n minus 1 by 2 is equal to delta
278
00:48:18,440 --> 00:48:22,999
t Pa n minus 1 by M.
279
00:48:22,999 --> 00:48:29,999
Now we can write down the rotor speed rotor
speed at the end of n minus 1 by 2 interval
280
00:48:37,230 --> 00:48:44,230
as the speed at the beginning of this interval
that is omega n minus 3 by 2 plus delta omega
281
00:48:46,499 --> 00:48:53,499
n minus 1 by 2. This quantity is the change
in speed during the time step delta t.
282
00:49:05,410 --> 00:49:12,410
Further, since the acceleration assumed to
remain constant during this time step speed
283
00:49:20,630 --> 00:49:24,930
is obviously going to vary during this time
interval.
284
00:49:24,930 --> 00:49:31,930
However, the total change in the time and
total change in the speed will be accounted
285
00:49:33,299 --> 00:49:40,299
by considering a step change in the speed
that is the total change in the speed which
286
00:49:42,400 --> 00:49:49,400
is equal to delta omega n minus half which
is equal to delta t into alpha n minus 1,
287
00:49:55,160 --> 00:50:02,160
this changes in the speed is assume to take
place in a step manner at the instant at which
288
00:50:06,630 --> 00:50:13,630
we compute the accelerations of the rotor,
with this assumption with this assumption
289
00:50:16,140 --> 00:50:23,140
the speed remains constant during the nth
interval and it is its value is equal to omega
290
00:50:26,160 --> 00:50:33,160
n minus half, with this speed, with this speed
during this interval the change in rotor angle
291
00:50:40,160 --> 00:50:47,160
is equal to delta t into omega n minus half.
292
00:50:47,670 --> 00:50:54,089
Since omega n ,omega n minus half is constant
during this interval therefore change in angular
293
00:50:54,089 --> 00:51:01,089
position is obtained as the constant speed
a omega n minus half into delta t therefore,
294
00:51:02,680 --> 00:51:09,680
angular position of the rotor at the end of
nth interval is equal to angular position
295
00:51:11,029 --> 00:51:18,029
at the beginning of the nth interval plus
change during the nth interval therefore,
296
00:51:18,359 --> 00:51:25,359
we can write down delta n equal to delta n
minus 1plus delta delta n.
297
00:51:25,499 --> 00:51:32,499
Now when we solve this equations four point
1, 2 to 4.14 we can get the swing curve as
298
00:51:37,869 --> 00:51:44,869
well as we can get the speed of the rotor
as a function of time. However if we want
299
00:51:50,809 --> 00:51:57,809
to obtain swing curve only we may use a formula
for delta delta n from which omega has been
300
00:51:59,239 --> 00:52:06,239
eliminated from the equations 4.12, 4.13 and
4.14.
301
00:52:10,940 --> 00:52:17,819
We get an expression for delta delta n as
delta t omega n minus 3 by 2 plus delta t
302
00:52:17,819 --> 00:52:24,819
square by M Pa n minus 1. By analogy with
equation 4.14, one can write down the change
303
00:52:30,989 --> 00:52:37,989
in angular position during the n minus 1th
interval as delta delta n minus 1 equal delta
304
00:52:41,900 --> 00:52:48,900
t into omega n minus 3 by 2. So, that we can
substitute the value of delta delta n minus
305
00:53:03,190 --> 00:53:10,190
1 that is the change in the rotor angular
position in the equation 4.16 using equation
306
00:53:17,489 --> 00:53:24,489
4.17. We get the expression for delta delta
n equal to delta delta n minus 1 plus delta
307
00:53:26,420 --> 00:53:31,249
t square by M Pa n minus 1.
308
00:53:31,249 --> 00:53:38,249
This is a equation which can be used for computing
the swing curve the left hand side term delta
309
00:53:43,960 --> 00:53:50,960
delta n, shows the change in rotor position
rate rotor angular position during the nth
310
00:53:55,339 --> 00:54:00,789
interval, this is expressed in terms of the
change in angular position during n minus
311
00:54:00,789 --> 00:54:07,789
1th interval plus delta t square by M Pa n
minus 1 and this can be easily programmed
312
00:54:11,950 --> 00:54:18,950
and using this expression the problem which
was solved using method 1 is again solved
313
00:54:21,960 --> 00:54:24,900
here.
314
00:54:24,900 --> 00:54:31,900
Now this graph shows the swing curve obtained
using formal solution of the second order
315
00:54:36,799 --> 00:54:43,799
differential equation and using method 2 considering
the several values of time step. This graph
316
00:54:53,890 --> 00:55:00,890
shows the solution obtained using the this
graph shows the swing curve obtained from
317
00:55:07,140 --> 00:55:14,140
the formal solution, you can say formal solution.
Now I say formal solution means it is the
318
00:55:18,690 --> 00:55:25,690
closed form solution of the second order differential
equation. Now here this swing curve is for
319
00:55:37,739 --> 00:55:44,739
delta t equal to 0.25 second, well this swing
curve is for delta T equal to 0.05 second.
320
00:56:01,299 --> 00:56:08,299
Now if you, if you carry if we carefully examine
this swing curves we can notice that, that
321
00:56:12,339 --> 00:56:19,339
the amplitude of the swing curve or the maximum
deviation of delta obtained using the formal
322
00:56:22,849 --> 00:56:29,849
solution and that obtained using method 2,
have hardly any difference, only difference
323
00:56:30,170 --> 00:56:37,170
which we observe is the time, time period
of the solution obtained within the formal
324
00:56:41,119 --> 00:56:48,119
solution and that obtained using the method
2 have some difference and therefore, I can
325
00:56:48,729 --> 00:56:55,729
conclude here that the method 2 provides the
accurate swing curve and we can use reasonable
326
00:57:01,009 --> 00:57:08,009
value of time step as we see that for time
step equal to delta T equal to 0.05,
327
00:57:08,690 --> 00:57:15,690
the solution obtained by method 2 that is
the step by step method 2 is very close to
328
00:57:15,749 --> 00:57:22,749
the formal solution.
329
00:57:28,440 --> 00:57:35,440
Now we consider how do we account for the
discontinuities in the accelerating power,
330
00:57:39,089 --> 00:57:44,130
discontinuities in the accelerating powers
occurs at the occurrence of disturbance or
331
00:57:44,130 --> 00:57:51,130
at the instance of switching. In case the
discontinuities at the beginning of the time
332
00:57:53,869 --> 00:57:59,170
interval that is the first time interval then
this can be computed delta delta 1 can be
333
00:57:59,170 --> 00:58:06,150
computed by using this expression delta t
square by M, Pa O plus by 2, where Pa O is
334
00:58:06,150 --> 00:58:12,539
the accelerating power just after the disturbance
and by putting 2 we are taking the average
335
00:58:12,539 --> 00:58:17,059
value of the accelerating power half the accelerating
power.
336
00:58:17,059 --> 00:58:24,059
In case the discontinuity occurs at the beginning
of mth interval then this can be the accelerating
337
00:58:26,009 --> 00:58:32,029
power during the m th interval will be computed
by this formula Pa m minus 1 minus plus Pa
338
00:58:32,029 --> 00:58:37,509
m minus 1 plus by 2, this minus indicates
the accelerating power just before the occurrence
339
00:58:37,509 --> 00:58:44,509
of disturbance and Pa m minus 1 plus indicates
the plus sign indicates just after the occurrence
340
00:58:44,779 --> 00:58:45,749
of incidence.
341
00:58:45,749 --> 00:58:52,749
With this, we conclude the ah the computation
of the swing curve using 2 different methods
342
00:58:56,170 --> 00:59:03,170
method one and method 2, method 2 is more
accurate and this can be used. Thank you!