Okay this problem deals with the sizing of a packed bed reactor and in this reactor we're
dealing with an A plus B goes to C reaction which is the hydrogenation of ethylene to
ethane and this reaction occurs over a solid catalyst in a packed bed tubular type reactor
and we're assuming that it follows elementary kinetics, and you're given information about
the rate constant and the temperature and part a asks you to ignore the pressure drop
and to basically size the reactor to determine the mass of the nickel based catalyst that's
needed in order to get a conversion of 70% under the given conditions. And then in part
b we'll go back and look at what effect pressure drop might have for a particular situation
where we change the flow rate. Okay but to start off with, we're going to have a material
balance and again we've started these in many different forms but we have a packed bed reactor
that we need to write our material balance for and the way that we've commonly written
this is that the mass of catalyst that's needed is equal to the inlet flow rate of component
A in the species for which we're writing this material balance, integrated from the inlet
condition which is no conversion to conversion of x and this is equal to dx over negative
r_A. So we're going to need to solve that material balance in order to solve for the
catalyst mass. Alright but first we're going to need to know what the reaction rate of
A is, the kinetics or if it's elementary and so that means that we'll have a k times a
concentration of A to the first power times the concentration of B to the first power.
Alright and then the only thing that we need to do is that we have our integrating with
respect to x in this expression, reaction rate expression is in terms
of concentration and so we're going to need to express the concentration of A and the
concentration of B in terms of conversion. Okay and there's 1 to 1 stoichiometry here
so the first part of this is simple, alright that the concentration of A, for example,
is equal to the molar flow rate of A at the inlet times 1 minus x for conversion, divided
by the volumetric flow V, but the volumetric flow rate is also a function of the conversion
because the number of moles that's going through the reactor actually decreases with conversion.
Alright so we have 2 moles of reactants on one side and 1 mole of product on the other
and so for every mole of A that reacts, we lose one of the reactants and so we've often
referred to this parameter as delta_A is equal to negative 1. Alright and then if we assume
for example that, let's assume that all of the A that's initially fed is converted.
Alright we initially are feeding equal molar amounts of A and B. Alright if all of the
A is converted and if we started with 2 moles, we'll only have 1 left. Alright and whereas
if half of the A's converted, if we started with 2 moles, we'd have 1.5 for example and
so we can derive an expression for this that is in terms of 1 plus epsilon_A, in Fogler
notation, times X. Our epsilon_A is equal to delta_A times y_A0 which is equal to 0.5
because the feed consists of half A, that's the mole fraction and so therefore epsilon_A
would be equal to negative 1 times 0.5 which is negative 0.5 and we can see that that's
true because if we plug in the value of x which is 1, the volumetric flow rate will
go down to half of its original value as the moles are cut in half above, according to
the ideal gas law. Alright so we go to full conversion and the same you can do other test
cases, for example for half conversion. Alright and so epsilon_A here is equal to negative 0.5 and
we'll just plug that in below and we'll simplify and note that this value is equal to C_A0.
Okay then next, we also need to solve for the concentration of B and we have equal molar
concentrations of A and B. B is consumed at the same rate as A and so C_B is always equal
to C_A. We can everywhere we can replace C_B with C_A whenever it comes up. Okay and so
we can now plug this into our rate expression that this is equal to k times C_A squared
or k times C_A0 squared times 1 minus x, where x is the conversion, divided by 1 minus 0.5
x. So this is our expression for concentration in terms of conversion and we can now use
that, we can now integrate that expression to solve for the catalyst mass. We're going
to need a couple of constants in order to do that, that we don't have yet, one of those
being the the inlet concentration of A is not a given parameter, but we can use
the ideal law again to get that. We have Y_A0 times P divided by RT and Y_A0, as we just
discussed is 0.5, that the inlet mole fraction of A is 1/2 because we have an equal molar feed with
B. The pressure is given as 5 atmospheres, R, we can use the 0.082 value which I should
put units on. The temperature is 400 Kelvin and it's isothermal as given in the problem
statement and therefore the inlet concentration of A is 0.076 moles per liter. Alright now
we can go back to our expression that we need to integrate and start plugging things
in. Catalyst mass is equal to... Alright so we can factor out a few constants so we can
pull out the k over C_A0 squared. Okay and now we just need to look up in an integral
table some of these parameters. I'll just point out, this something you can also just
solve with an integrating calculator and there is an analytical solution. The volumetric
flow rate is 8 liters per second, k is given as being 0.2 liters squared per second kilogram
and the concentration of A we just solve for. Alright that is 0.76 moles per liter. So we
have everything we need to solve for the catalyst mass. If you look in a common integral table,
this quantity ends up being, the constant in front ends up being 526 kilograms of catalyst
and then the... this integrand if you evaluate it on a calculator, as I was lazy and did,
ends up being 1.36. You can also find formulas for this. For example in the back of Fogler
or in a common calculus textbook. In this case the total catalyst mass will end up being
716 kilograms. Okay in part B we're told that the pressure drop down the reactor will be
half an atmosphere, alright we're not going to end up using that data, but another...
the pressure drop that we haven't accounted for is going to tend to increase
the molar flow rate as we go down the reactor and lower the effect of residence time in
the reactor and therefore lead to less conversion. So the pressure drop will have a deleterious
effect and so another engineering company suggests that cutting the entering volumetric
flow rate in half will increase the conversion and lower the pressure drop. Alright so what
you want to do is evaluate this claim. Alright the first issue that we'll consider is the
pressure drop. It's currently half an atmosphere and if we have our packed bed reactor full
of particles and if we lower the flow rate, so we're just decreasing v_0, you could plug
this into the Ergun equation and see what the effect is but we won't actually bother
to do that because you can think about it purely conceptually. If you decrease the volumetric
flow rate then you lower the average velocity of the stream that's going through here. Alright
and therefore you're going to lower the frictional losses as the stream goes around the catalyst
particles. Alright so you're lowering friction because the velocity is going to be lower.
So you're not having to force as much fluid through this catalyst bed and therefore the
delta P, the pressure drop as we go across the catalyst bed, is actually going to go
down. So you'll have less pressure drop as a result of this and because the pressure
drop goes down, that effect should make the conversion go up, but there's another effect
that's likely to be larger and that's that also as we cut the volumetric flow rate the
space time goes up. So again space time is equal to the volume divided by the volumetric
flow rate. We're not changing the volume of the reactor or the amount of catalyst. We're
just decreasing the volumetric flow rate and so the space time goes up and that means that
the reaction is going to have a longer amount of time to occur inside the reactor and that
also is going to make the conversion go up. So both because the delta P is going to go
down and because the effective space time is going to go up, since we decreased the
volumetric flow rate, we're going to have an increase in the conversion. Alright but
there's never a free lunch, there's always a tradeoff for this and the tradeoff is that we will be
producing fewer pounds of C. Alright so we're not feeding A and B as quickly, we're able
to react at higher conversion as a result of that, but the total amount of the product
that we make is going to go down because we're just not flowing as much through our reactor
and we're allowing what's in there to stay in for longer. Alright so we can't produce
C at as fast a rate.