Problem dealing with the material and energy balance being conducted on a continuously
stirred tank reactor.
I'll let you read the problem statement yourselves and just launch into it here.
Alright, so here we have, in this problem statement we have equimolar amounts of A and
B, so in a way it doesn't matter which one we track.
For the purposes of this problem statement I'm just going to track component B somewhat
arbitrarily here, and we have equimolar, we're given information on the volumetric flow rate
of the total stream, and most of this stream is actually, as expected, solvent, since the
concentrations of A and B are relatively dilute.
So for example, if water was the solvent, then its concentration would be roughly 55
moles per liter.
The reactor temperature is held constant at 350 K, although as we'll see later in the
problem, the reactor feed temperature is going to be 300 Kelvin.
So part A asks, what reactor volume do we need to achieve an outlet flow rate of C of
4.5 moles per minute?
And the first thing that we want to do in just about any reactor problem is to solve
a material balance, and there are many ways that we could set this up.
I'll express this in terms of conversion, with the volume, which we want to solve for
in this case, is equal to the inlet molar flow rate of B times the fractional conversion
of B, divided by the reaction rate of component B. And we can furthermore express that inlet
flow rate of B in terms of variables that we actually have, which is the inlet volumetric
flow rate of the total stream times the concentration of B times the fractional conversion, and
dividing this by the reaction rate of B, which we're given the reaction rate constant in
terms of A, and in the reaction we consume 2 moles of B for each mole of A that we consume,
so we consume B at twice the rate at which we consume A. So we can write down the rate
of consumption as 2 times k_A times the concentration of A times the concentration of B. Now, the
next step is that we want to try to express these concentrations in terms of conversion,
and so I'll do that directly in this expression, but again we carry along these terms, but
now the concentration of B at the inlet is equal to the concentration of B, and so we
can write this down as a C_B0 and multiply it by a conversion expression, where the amount
of A that's remaining in the reactor, that's equal to 1 minus x over 2, then a C_B0, and
you have to understand that the concentration of B that's left over is a function of conversion,
1 minus x, so that's the amount of B that's left over after a certain fraction x has been
converted.
Alright, the reason we divided by 2 here is, again, for each mole of B that we convert,
we only convert half a mole of A because of that 2 to 1 stoichiometry, so that's how we
know to multiply the conversion by a factor of 1/2.
Alright, so now in this expression, we want to solve for the volume, V, and we know alot
of these numbers, so V_0 is equal to 18 liters per minute, inlet concentration of B is equal
to 0.8 moles per liter, so we know that here and here and here, we are given a value of
k_A, so we know that value, so the only thing we need in order to solve for the volume is
the conversion, and we need to get that from our other piece of information, which is that
the outlet flow rate of C is equal to 4.5 moles per minute.
So if we take that, of course we can express this in terms of conversion, and so again
for each mole of C that we form, we had to react 2 moles of B, and so therefore, this
outlet flow rate has to be equal to 1/2 the amount of B that we reacted, which we can
express as the B that we started with times the fractional conversion times 1/2.
Alright and so we know how much B we started with, so this is equal again to 1/2 times
the inlet volumetric flow rate times the inlet concentration of B times x, and so we again
know this number and know this number and we can solve for a fractional conversion of
0.625.
Alright, now we know everything in our expression for the volume, and we can furthermore solve
for the volume, and when I plug in those numbers, if I don't make a mistake, then I get a volume
of around 160 liters.
Okay, so the next part of the problem, part B, asks for how much heat we're going to have
to add or remove, and we have to determine which we need to do, if we're sending the
feed into the reactor at a temperature of 300 K, but we want to operate the reactor
at 350 K, and here we're given that this reaction has a heat of reaction of -24 kcals per mole,
and I should specify that that's per mole of A reacted, alright and so this is an exothermic
reaction, and so that will tend to release heat, but we're also going to be absorbing
heat through both the solvent and the reactants.
And so the way it will be, the way in which that balance is will determine how much heat
we actually have to add to the reactor as we'll solve for quantitatively in just a minute
here.
So we want to set up an energy balance for this equation, and I've gone ahead and written
this here, and so any heat that we add will need to be offset by heat that is brought
in, or by the sensible heat that's brought in by the reactants that are heated up from
a temperature of 300K to a temperature of 350 K, and the amount of heat that they absorb
is determined by their heat capacity, shown here, times the molar flow rate, so how much
you're actually bringing through the reactor.
The more you bring through, the more heat you'll need to add in order to heat up those
reactants.
Then, furthermore, the heat of reaction can play a role, the fact that this is an exothermic
reaction, and so the heat of reaction times the fractional conversion times the molar
flow rate of B into the reactor will also have to be a term in this equation.
And here we need to divide by 2, because if you remember I modified the problem statement
to say that the heat of reaction is given in units of per mole of A, but we've tracked
our conversion and all our other variables, really, in terms of B, and for each mole of
A that reacts, 2 moles of B react, so that means that for each mole of B that reacts,
we only will liberate 12 kilocalories per mole, so we can just do a unit conversion
of moles of A to moles of B, and you can see that that will give us our 1/2, so our heat
of reaction will effectively be 12 kcals per mole of B, exothermic.
Okay, and so again we can start plugging in some of what we know here, and I should point
out a very key assumption that I am making in this solution, one that you would find
holds, and that's that I'm only taking into account the solvent-specific terms in solving
this problem, and that's because again, this is a very dilute system, alright and so this
solvent makes up most of the fluid, and I'm assuming that you can ignore the components,
the heat capacity of A and B as if we were just including the solvent, and that will
give you a very good answer for the amount of heat that you need to add.
Alright, if that answer wasn't good enough for you, or you were dealing with a system
in which components A and B made up a significant amount of the fluid, then you would need to
include additional terms here.
It would be the flow rates of each of those compounds times the temperature difference
times their heat capacities.
So we just need to sum up other terms that look like this, but in many liquid-phase systems
in fact, this solvent makes up most of the system, and so this is a fairly realistic
problem.
Okay, and so now we actually have a fairly straightforward problem solution.
Again, F_B0 here we can break down, and that's being equal to the total volumetric flow rate
times the inlet concentration of B, and so we know both of these two terms as we described
in the beginning of the problem.
We solved in part A for the conversion, which was 62.5 %. We know that this number is 24.5
kcals per mole, we know our reactor temperature and our inlet temperature, we're given in
the problem statement a flow rate of solvent of 720 moles per minute, and then finally
we're given the heat capacity of the solvent, 40 joules per mole K, and the only thing that
you have to be careful of here is, recall that these are given in joules and this number
is given in kilojoules, and so you'll want to convert your heat capacity, for example,
to 0.04 kilojoules per mole K in order to keep your units consistent.
We can basically plug in the numbers and solve for Q, and if I do that and again, didn't
make any math mistakes, I get a Q of about 1300 kcals per minute that need to be added
to this reaction, and so we know that it needs to be added because this is a plus sign, and
inputs correspond to positive values here, so for example, an exothermic reaction is
a heat input, and that leads to a positive term in this equation, another heat input
is the heat capacity times the temperature of the inlet stream, and so that is a positive
number here, and so we see that that's another heat input that's correlated with a positive
term, whereas we have a negative number for the heat capacity times the flow rate times
the outlet temperature, T, and so that corresponds with the heat output, and it was negative,
and so again, this corresponds to heat that we have to add to the system in order to bring
it up to 350 Kelvin.
So in other words, the exotherm here isn't enough to increase the temperature all the
way to 350 Kelvin and we need to provide a little bit of extra heat.