This problem deals with hydrogenation of acetylene over a supported palladium catalyst in a packed
bed reactor, and we need to start the process by actually evaluating the reaction kinetics,
assuming a power rate law model given here, where we need to find k, alpha and beta, and
to do that, differential reactor data has been collected, where at low conversion, the
amount of A in the feed and B in the feed, or hydrogen in the feed, has been either varied
or kept constant at a fixed temperature, and we've measured the outlet flowrate of ethylene
product.
So the problem mentions that we can just do a quick estimate, and this is obviously made
up data, so it's possible to do a pretty easy quick estimate.
So for example, we see that changing the partial pressure of A doesn't affect the rate of formation
of ethylene at all within the artificial noise here.
So we can quadruple the amount of A and get no real change in the amount of product being
formed, and therefore, alpha, the order with respect to ethylene is zero, because A is
evidently having no effect on the rate of reaction.
Conversely, if we keep the amount of A fixed and vary, these are the three data points
of interest, the amount of hydrogen, we see that doubling the amount of hydrogen doubles
the rate in this case, and then tripling it from the original case also triples the rate.
So therefore we expect that the rate is going to obey first order behavior with respect
to hydrogen, and then we want to do a quick estimate of k, so remember that in the differential-type
reactor that the rate, because of the low conversion, is equal to the outlet flowrate
of the reactant, so we can convert, for example in a material balance, the change in the flowrate
of ethylene with respect to catalyst mass, we can convert that to deltas because the
conversion is so low, and so the partial pressures of each component don't really change with
respect to distance down the reactor.
Therefore, the flowrate of ethylene over the total catalyst mass that it sees is equal
to the reaction rate constant times the partial pressure of hydrogen to the first power, because
the partial pressure of acetylene doesn't matter.
k is what we want to solve for, partial pressure of hydrogen, let's just take this data point
as equal to 4 atmospheres, the outlet flowrate of ethylene for that data point is equal to
0.0121 moles per minute, and the catalyst mass in the reactor is 25 grams.
So we can solve for a value of k of 1.21 times 10 the -4 in units of moles minute gram catalyst
per atmosphere.
Now we're going to use this information in part B in order to size the reactor.
So we have our necessary kinetic expressions, and so we're going to now, we're no longer
operating at differential conversion, so we're actually going to have to integrate the performance
of our packed bed reactor, our tubular reactor.
We'll need to use the fact that we can describe the amount of catalyst in terms of F_A0 times
the integral of dX over the rate of consumption of A, evaluated from a conversion of 0 to
a conversion of X, which is going to be 70%, and we will start to plug in some of our expressions
here.
So this is F_A0, integrating again from 0 to 0.70, dX, and now we have an expression
in terms of the pressure, so we just have k, which we now have a value of, times the
partial pressure of hydrogen.
The remaining trick in solving the problem is just relating that partial pressure of
hydrogen to conversion so that we can actually do this integration.
The amount of hydrogen that we have at any position in the reactor as a function of conversion
is equal to the amount that we started with times 1 minus the conversion.
The amount that we started with we can get from the ideal gas law, because we're dealing
with relatively low pressures here, so the ideal gas law is a pretty good assumption.
So the flow rate of hydrogen is equal to the pressure times the volumetric flow rate over
RT, so what we can actually say is that the partial pressure of hydrogen is equal to the
molar flow rate of hydrogen times RT over the volumetric flow rate down the reactor,
so just an ideal gas law relationship.
So the flow rate of hydrogen in term is equal to the flow rate of hydrogen at the inlet,
we'll call that FH20, times 1 minus the conversion, so here we start to see the conversion coming
in.
This defines how the molar flow rate changes with the length down the reactor, times RT.
Now the volumetric flow rate also changes down the reactor, which is one of the complications
of this problem, and what we can see is that the flow rate as you go to 100% conversion
gets cut, the volumetric flow rate will get cut in half because the total number of moles
is going to decrease going from 2 to 1 as a result of this reaction, since we're feeding
only the pure components at equimolar ratios.
So we'll have a decrease in going from 0 to 100% conversion of the volumetric flow rate,
to half its original value, and that functional dependence is described by the initial volumetric
flow rate times 1 minus 0.5 times x, and this value is negative 0.5 if you're using Fogler,
is equal to what he calls YA0 times delta where YA0 is half, because half the feed is
composed of hydrogen, and delta is -1, because for each mole of hydrogen that reacts, we
lose a mole of total gas.
So this is what we can now substitute in for our partial pressure of hydrogen in terms
of conversion, so I'll just point out that we're using this expression, but we're going
to now start plugging in that pressure of hydrogen, that's going to complicate things
a bit, and when we collect, we can collect the constant terms on the outside of the integral,
so we have an F_A0 divided by a kRT, then we have a volumetric flow rate up here, an
initial flow of hydrogen here, and what I'll point out is that the initial flow rate of
A, which is one of our reactants, hydrogen is the other, those are the same because we're
using an equimolar mixture of reactants.
So those two cancel because F_A0 is equal to F_H20, and we have our integral, which
is from 0 to 0.70, dX, 1 minus 0.5X minus X, and now we can start plugging in for some
of the values that we know here.
I won't go through and do all of the solution, this is a value that can be determined through
the back of a textbook, integrating calculator, the volumetric flow rate is given as being
1000 liters per minute, k we determined earlier in the problem was equal to 1.21 times 10
to the -4, the gas constant that is probably most convenient to use is 0.082 atm-L/mol-K,
and then finally the temperature is 400 Kelvin, and I at least get the value of this quantity
integrated to be equal to 0.952, so this all leads to a catalyst mass of about 240 kilograms.
This problem illustrates how we can start with the kinetic data that we determined from
the first part of the problem, that it's a first order dependence on hydrogen, zero order
dependence on acetylene, and we know the value of the rate constant, and use it solve for
the catalyst mass needed in a packed bed reactor.