This is an example where we are modeling a
tubular reactor as a laminar flow reactor,
and I am going to use a segregated flow model,
which means we are going to treat the laminar
flow reactor as a bunch of plug flow reactors
in parallel, where the length of the plug
flow reactor is determined by the velocity
profile in laminar flow. So we have a second
order reaction. We are given the rate constant,
isothermal reactor, and we are given feed
concentrations, and feed volumetric flow rates.
The question is, what is the conversion in
this laminar flow reactor for a volume of
1500 liters? So first let's look at what the velocity
profile is that we are talking about, and
how we are going to model this. So this is
just an approximation of the velocity distribution.
The velocity in the center is twice the average
velocity, and of course the velocity goes
to 0 at the walls. So from this velocity distribution
it is possible to derive a residence time
distribution. This is not a straightforward
derivation. It takes a little bit of work.
So what we want to do is use that residence
time distribution, and we are not going to
derive it, so the residence time distribution,
and I will use the symbol p(t), so it's
a function of time. The residence time distribution is 0 for
time less than the space time over 2. So the
space time, the average of this distribution of
velocities is the space time, the volume of
the reactor divided by the volumetric flow
rate, and the center section of the reactor,
of the tube, arise if we do a Paul's tracer experiment
to get to residence time distribution. That center arises first at half of the space time and the residence time
is then equal to space time squared over 2,
time cubed for all taus greater than or equal
to tau over 2. We can plot what this looks
like to get a little better feeling for the
behavior. So here is the RTD plotted as as
function of time. Zero up until tau over 2.
Then this delay is t^-3. So the idea of segregated
flow is that we can calculate the average
concentration of a reactant leaving the
reactor, and I will use this symbol to indicate
average, by integrating over all of these
plug flow reactors in parallel. So integrating
from 0 to infinity, p(t) the residence time distribution determines the different types of plug flow reactors
and residence time we would have, and then
the concentration changes with time, and this
essentially is a batch reactor. So if we use
this model, the concentration we get here is
the concentration at the exit, and we are
using the residence time distribution, and the concentration as a
function of time from a batch reactor. We
can make this a little easier to do the calculation
by just differentiating this equation, and
then we are going to solve this differential
equation, where of course initial conditions
are going to be time equals 0, CA bar equals
0. If we go back to this equation, if we integrate
only to time 0, then the concentration CA
average is 0. Then we need the batch reactor
equation for the concentration, and it is
second order, minus sign, concentration of A to the
second power. We also need initial conditions
for this, and that's that is that CA is equal to 12,
and this is given in our problem statement
of 12 moles per liter. Then of course we have to solve
also in these equations, we have to have p(t), and
this is the format that they would use in
POLYMATH. So if t is greater than tau over
2, then p(t) is tau squared over 2 t cubed.
Otherwise it is 0. So we have 3 equations.
We have the initial conditions. We have to
put in values for k, and for tau. So k is
given, tau is the volume over the volumetric
flow rate, so tau is 3 hours. Then we need
to integrate this initial value problem, 2
differential equations. What we are going
to do is integrate from time equals 0 to some
long time, and turns out that 20 hours is
more then long enough to get sufficiently
accurate answers. We can show this by integrating
to different times. If we use POLYMATH, and
I will show you what the POLYMATH program
looks like, we can get CA leaving the laminar
flow reactor, 1.13 moles per liter. Problem asked for
a conversion. Because it is liquid phase , we
can write conversion in terms of concentrations.
So the feed concentration, exit concentration
over the feed concentration, conversion is
0.906, or significant figures 91% conversion
in this laminar flow reactor. Lets look at
the POLYMATH program. So the POLYMATH program
has differential equation, initial condition,
the batch reactor differential equation, initial
condition, and constants. We integrated to
20 hours. Also did integration to 10 hours,
and within 3 significant figures, the same answer,
and this program calculated the conversion.
So this is a different conversion than we
woud get for plug flow reactor by taking into account
this velocity distribution.
