To show how two first order processes in series can lead to a second order transfer function
we will use the example here, where we have two well insulated tanks in series where the
flow rates entering/exiting the tanks are going to be constants and the same. Here our goal is
to determine the transfer function that relates our output variable, which is the outlet temperature
of the second tank, T3, and by the way this is in deviation variables, or the input to
the first tank, which can vary, T1. So in order to do this analysis we need to conduct
two energy balances. Why? Because from our first balance we'll be able to get T3 as a
function of T2. So tank two will allow us to find T3 as a function of T2. However, because
our goal is to find T3 as a function of T1, the second tank balance is insufficient. Why?
Because T2 is a function of T1, and we can find this relationship by the energy balance
on the first tank. So in this case here we have no chemical reaction happening and the
system is well insulated, so therefore our energy balance for both tanks can be simply
written down as the accumulation of energy is equal to the energy in minus the energy
out. And for tank 1, the accumulation will be in d energy dt, and that will equal the
inlet energy which we can find dividing the enthalpy multiplied by the mass. So we have
a mass flow rate of f coming in multiplied by enthalpy. In this case we see p multiplied
by the temperature T1 minus a reference temperature. We'll subtract that from the out term which
will be the flow rate coming out, f, multiplied by the enthalpy there, which we'll say is
cp times T2 minus Tref. We'll do some analysis here, where we have the energy inside the
tank can be represented as the mass of the fluid which is going to be V1, which is a volume,
multiplied by rho, multiplied by the enthalpy, which will be cp. Here, because we're assuming
the fact that the tanks are well mixed, that means that the temperature of the outlet is
going to be the same as the temperature of the fluid inside the tank, so we can write
this as T2 minus Tref. And what we can also do on the right hand side is the Tref terms
will cancel out, and that leaves us with fcp times T1 minus T2. We can simplify this a
little bit more by factoring out the constants, which in this case will be the volume of the
fluid inside the tank, rho, and the heat capacity, and we can also take advantage of the fact
that since Tref, a reference temperature is a constant, its derivative will be zero. So therefore
when we finish this up our balance on tank 1 will be V1 rho cp dT2 dt equals fcp times
T1 minus T2. And one final simplification we can take advantage of here is the fact
that the heat capacities will cancel. for tank 2 we have an identical energy balance,
with the only exception that the subscripts for the temperatures are different and for the volume are different.
Our next step will be to put this into deviation variables
We have our variables here are T1 prime, T2 prime, and T3 prime.
Fortunately all of our terms are linear, so therefore in order to put these equations
in to deviation variables all we have to do is add primes to these respective variables.
Note there are no primes necessary for f, V, or rho because they're all assumed to be
constant in this question. What we can now do is develop the transfer function for both
tanks. We've shown that on the screen here, but an important point further that we want to
do here is we want to put things into standard form.So when we put things into standard form what that
means is that the T2 prime here and the T3 prime here, the coefficient in front of them
must be one. This can be done by dividing everything by f, and when we do that we end
up with the equations that you can see right here. And then in the next step what we can do is
we can factor out the T2 prime from the tank 1 equation, and the T3 prime from the tank
3 equation. When we complete these we can develop our transfer function, so for tank
1 we'll have that T2 over T1 prime equals one over V1 rho over f s plus one. And similarly
for tank two we have that T3 prime over T2 prime equals one over the V2 rho over f s
plus one. Although we have found these two transfer functions, our overall goal recall
was to find T3 prime over T1 prime, and there are two ways we can do this. One way we could
have done it was substituted in T2 prime from this first equation and substituted it in
to the second equation. Another way we can do this is also recognize the fact that T3
prime over T1 prime is just going to be equivalent to doing T3 prime divided by T2 prime multiplied
by T2 prime over T1 prime, with the T2s cancelling out. Either way will get you to the same place.
So when we do this, we can see the fact that T3 prime over T1 prime equals one divided
by V1 rho over f s plus one multiplied by V2 rho over f times s plus one. And what you'll
notice here is the fact that we don't have to factor this out but we now have an s squared
term in the denominator, which means that we have a second order transfer function.
So this shows the fact that although many chemical processes do not have second order
dynamics naturally, if you put multiple processes together in a control scheme what can happen
is we can develop transfer functions that are second order in spite of the fact that
each of the individual processes is first order.