MARTHA: In this video, we'll consider
the Laplace Transform Method for solving
linear differential equations.
And we'll use the blending process as our example.
In the blending process we have two inlet streams
coming together in a well-mixed tank.
Those two streams are blended
and then we have a single stream that is leaving the tank.
The Laplace Transform Method applies to linear
differential equations, so we will use the linear
approximate model for the mass fraction in the tank.
So the first thing that we're gonna do here
is to take the Laplace Transform
of each term here, the Laplace Transform
of the time derivative is S capital X prime of S
minus any initial deviation in mass fraction
from the steady state.
And here we're going to just assume zero deviation.
And then that is equal to minus four times 10
to the minus four capital X prime of S
plus 0.075
times 10 to the minus four
W one prime of S
minus 0.025 times
10 to the minus four W two prime of S.
And now that we have taken the Laplace Transform
of each of our terms on the left and right side
of the differential equation,
we can combine terms for our known
to algebraically solve for the mass fraction in the tank.
So now we have a general expression where we can solve
for X prime of S as a function of our two inlet streams.
Now we can consider a specific step input
to stream one
and we can flat that.
Where we have a deviation in W1
that begins suddenly at time zero
and has an amplitude of 0.1.
So we're gonna jump up by 0.1
from our previous steady state value
when time equals zero.
So W one prime of S is 0.1,
the amplitude of the step over S.
One over S is the Laplace Transform of a step function.
And then we'll assume no deviation in W2,
so that will just be zero.
So now we can go ahead and plug that in to our
S prime of S expression.
And we'll divide through by S plus
four times 10 to the minus four.
We'll have an enumerator 0.075
times 10 to the minus four and then we'll multiply that
by W1 prime of S which is 0.1 over S.
And we will also divide through by four times 10
to the minus four, we're gonna divide through both sides.
And the reason to do that is when we look up
transfer functions in the Laplace Transform table,
we often have them in a particular form.
So the denominator here will often be in a form
Tau S plus one,
not S plus some constant B.
So now we can rewrite this expression.
So this form, Tau S plus one
where Tau is 2500 seconds
is going to be a more convenient form for us to use.
Now that we have solved for X prime of S,
we can perform an inverse Laplace Transform
to solve for little X prime of T,
the deviation in the mass fraction in the tank.
We can pull out this constant term
and then look up the remaining term
in the Laplace Transform table to be one
minus E for the minus T divided by 2500.
And now maybe you can see why we refer to this
2500 as the time constant of the process.
It's the typical decay time in that exponential term
in fact, if we evaluate X prime at the time constant
2500, we see that it is equal to
1.875
times 10 to the minus three which is the long term value
at infinite T times one minus E to the minus one.
And that quantity is about 0.63.
So we can see that we've reached 63%
of that long term behavior in 2500 seconds.
And we can also sketch the time behavior of the system,
and in fact we can do that in terms of
our original variable X, which is X bar
plus X prime of T, such that at times zero,
we start at our original value, 0.025,
the steady state value.
And then over time, which is in units of seconds here
we approach this long term value which is 1.875
times 10 to the minus three higher.
So that would be 0.0268.
So we would reach about 63% of that value
after 2500 seconds and we would continue
to exponentially approach that long term value
as time goes to infinity.
As we consider the solution X of T,
it is important to consider that we were solving
for X prime of T using a linearized version
of the original non-linear dynamics.
This may or may not be a good approximation
to the original dynamics.
It really depends on how far the system deviates
from its steady state value.
Now here we had a step of .1, deviation of .1
going from one to 1.1 in our inlet flow rate W1.
That's a 10% change and then here we saw
similarly a small but non-negligible change in X.