In this screencast we will analyze a process which is a naturally interacting system. An
interacting system in one where in order to get from the input variable to the output variable,
one does not follow a sequential order of intermediate variables. So in order to illustrate
this, we will take a look at a tank, who, because of imperfect mixing, we are going
to approximates the process as a tank split in two parts, each well mixed, and our goal
here is to determine the transfer function that relates how a change in the inlet concentration,
in deviation variables, will affect a change in the outlet concentration of deviation variables.
And to look at our process and to show why it is interacting, whats coming in is a flow
rate f, and a inlet concentration c(in), and in order to go from c(in) to c(out) we need
this intermediate concentration, c(1). However, c(1) is not just a function of c(in), it is
also a function of c(out). Why? Because of the recirculating stream shown here. The recirculating
stream is an equal flow rate, f(r) is going from volume one, the upper part, to volume two,
the lower part, but since this is well mixed the concentration of the outlet coming out
of the lower portion is equivalent to that concentration inside the tank, so therefore
this concentration would be c(out). Now that we see the fact that we have an interacting
system, lets develop the mass balances to describe the process. So here, a total mass
balance on the process will be unnecessary, because what we'll show is that f equals f,
so instead what we are gonna focus on is component balances. Because there is no reaction here,
our balances are going to be pretty straightforward, its going to be accumulation equals in minus
out, and we are going to analyze this for both the upper portion, and the lower portion.
So for the upper portion what is coming in, we have two inlet streams coming in. The first
is whats coming in from the general inlet, f times c(in), and the second is what is coming
from this stream here, which is going to be f(r), multiplied by that concentration already
described as c(out). Exiting we have one exiting stream going to the lower portion, which is
going to be f(r) times c(1). Our accumulation term is going to represent the amount of mass
of this component we have, and that can be represented by multiplying the volume of the
upper portion,V1, by the concentration of the upper portion, c(1). A similar balance
ensues for the lower tank. Here we have one stream coming in, thats the one coming from
the upper portion, f(r)c(1). We have two streams exiting, one which is the one going to the
upper portion, so that will be minus f(r) times c(out), minus the stream exiting the
tank, which will just be f times c(out), and in this case here the amount of material in
the lower portion can be described as its volume, V2, multiplied by its concentration
c(out). So what we now want to do is begin to get this into the Laplace domain, but we
are going to do a couple quick things before that. We are going to do two things with these
balances. We are going to one, factor out the volumes, and two we are going to put into
deviation variables. Here because we are assuming that the flow rates are constant, so the only
things that can vary the concentrations all the terms here are linear, so what this means
is the fact that we can easily replace all of the c values that we have, with c prime,
because there is no difference in the deviation and the non-deviation in this situation if
you were to go through the process and subtract off the steady state equation. So by doing
this, what we are now left with is V1 times dc(1) dt, put the prime in, equals f times
c(in) prime plus f(r) times c(out) prime, minus f(r) times c(1) prime, and again this is for
the upper portion of the tank. And for the lower portion of the tank, it will be something
very similar. From here we'll take the Laplace transform of this. For the upper tank we will
take V1 time, putting a capital C to indicate that we are in the Laplace domain, and then
we will just do something very similar for the capitals for the remaining terms. And
a similar logic ensues for the lower tank. So remember that our goal here is to find
a transfer function that relates c(out) to c(in). We have c(out) as a function of c(1),
and we see from the first balance we see that c(1) is a function of c(in) and c(out), hence
the interacting nature of this process. So what we wanna do here is develop both into
transfer function form, therefore we are going gather the like terms for the term with the s,
so therefore when we rearrange, what we end up with is. And similarly for the lower tank
we end up with. So now we'll put these in transfer function form, and when we do this
we'll have c(1) prime will equal f over v1 s plus f(r) multiplied by c(in) prime, plus
f(r) divided by that same denominator, and this you multiply by c(out) prime. And then
similar here c(out) prime equals f(r) divided by V2s plus f(r) plus f, multiplied by c(1)
prime. Notice that our top equation for the upper portion is an equation explicit for c(1) prime,
and all I need from c(out) prime is that exact term, so therefore we are going to substitute
in. So this is going to be a little messy algebraically, so again in order to get our
final transfer function of c(out) prime over c(in) prime we have to gather the c(out) prime
terms together. So if we factor out the c(out) prime part, and move the c(in) prime over,
we actually have our transfer function. To make this a little cleaner to write right
now I am going to call this term G1 and this term G2. C(out) prime over c(in) prime equals
G2 divided by one minus G1. If you notice, both the G1 and G2 have a least common denominator,
so therefore we can simplify this, and what we'll end up with is f(r)f over V2s plus f(r)
plus f multiplied by V1s plus f(r) minus f(r) squared. Because of the coupledness of the
differential equations slash transfer functions, because of the fact that they are interacting,
it causes out solutions to be a bit more complicated than for non-interacting systems, where the
relationships between the input and output variables are very cleanly sequential.