Introduction to Chemical Engineering Processes/Reactions with Recycle

Introduction to Reactions with Recycle

Reactions with recycle are very useful for a number of reasons, most notably because they can be used to improve the selectivity of multiple reactions, push a reaction beyond its equilibrium conversion, or speed up a catalytic reaction by removing products. A recycle loop coupled with a reactor will generally contain a separation process in which unused reactants are (partially) separated from products. These reactants are then fed back into the reactor along with the fresh feed.

Example Reactor with Recycle

Example:

Consider a system designed for the hydrogenation of ethylene into ethane:

• $2H_{2}+C_{2}H_{2}\rightarrow C_{2}H_{6}$
• $(2A+B\rightarrow C)$

The reaction takes too long to go to completion (and releases too much heat) so the designers decided to implement a recycle system in which, after only part of the reaction had finished, the mixture was sent into a membrane separator. There, most of the ethylene was separated out, with little hydrogen or ethylene contamination. After this separation, the cleaned stream entered a splitter, where some of the remaining mixture was returned to the reactor and the remainder discarded.

The system specifications for this process were as follows:

• Feed: 584 kg/h ethylene, 200 kg/h hydrogen gas
• Outlet stream from reactor contains 15% hydrogen by mass
• Mass flows from membrane separator: 100 kg/h, 5% Hydrogen and 93% ethane
• Splitter: 30% reject and 70% reflux

What was the extent of reaction for this system? What would the extent of reaction be if there was no separation/recycle process after (assume that the mass percent of hydrogen leaving the reactor is the same)? What limits how effective this process can be?

Solution:

Let's first draw our flowchart as usual:

DOF Analysis

• On reactor: 6 unknowns $({\dot {m}}_{5},x_{A5},x_{B5},{\dot {m}}_{3},x_{B3},X)$ - 3 equations = 3 DOF
• On separator: 5 unknowns $({\dot {m}}_{3},x_{B3},{\dot {m}}_{5},x_{A5},x_{B5})$ - 3 equations = 2 DOF
• On splitter: 3 unknowns - 0 equations (we used all of them in labeling the chart) -> 3 DOF
• Duplicate variables: 8 (${\dot {m}}_{5},x_{A5},x_{B5}$ twice each and ${\dot {m}}_{3},x_{B3}$ once)
• Total DOF = 8 - 8 = 0 DOF

Plan and Solution

Generally, though not always, it is easiest to deal with the reactor itself last because it usually has the most unknowns. Lets begin by looking at the overall system because we can often get some valuable information from that.

Overall System DOF(overall system) = 4 unknowns (${\dot {m}}_{5},x_{A5},X,x_{B5}$) - 3 equations = 1 DOF.

Let's go ahead and solve for m5 though because that'll be useful later.

$784=100+0.3({\dot {m}}_{5})$

${\dot {m}}_{5}=2280{\mbox{ kg/h}}$

We can't do anything else with the overall system without knowing the conversion so lets look elsewhere.

DOF(separator) = 4 unknowns (${\dot {m}}_{3},x_{B3},x_{A5},x_{B5}$) - 3 equations = 1 DOF. Let's solve for those variables we can though.

We can solve for m3 because from the overall material balance on the separator:

• ${\dot {m}}_{3}={\dot {m}}_{4}+{\dot {m}}_{5}$
• ${\dot {m}}_{3}=100+2280$

${\dot {m}}_{3}=2380{\mbox{ kg/h}}$

Then we can do a mass balance on A to solve for xA5:

• ${\dot {m}}_{3}x_{A3}={\dot {m}}_{4}x_{A4}+{\dot {m}}_{5}x_{A5}$
• $2380(0.15)=100(0.05)+2380(x_{A5})$

$x_{A5}=.1544$

Since we don't know $x_{B5}$ or $x_{B3}$, we cannot use the mass balance on B or C for the separator, so lets move on. Let's now turn to the reactor:

Reactor Analysis

DOF: 3 unknowns remaining ($x_{B3},x_{B5},andX$) - 2 equations (because the overall balance is already solved!) = 1 DOF. Therefore we still cannot solve the reactor completely. However, we can solve for the conversion and generation terms given what we know at this point. Lets start by writing a mole balance on A in the reactor.

${\dot {n}}_{A1}+{\dot {n}}_{A,recycle}-X*a={\dot {n}}_{A3}$

To find the three nA terms we need to convert from mass to moles (since A is hydrogen, H2, the molecular weight is ${\frac {1{\mbox{ mol}}}{0.002016{\mbox{ g}}}}$):

• ${\dot {n}}_{A1}=200{\frac {kg}{h}}*{\frac {1{\mbox{ mol}}}{0.002016{\mbox{ kg}}}}=99206{\frac {\mbox{mol A}}{h}}$
• ${\dot {n}}_{A,recycle}=0.7*{\frac {m_{5}*x_{A5}}{MW_{A}}}={\frac {0.7(2280)(0.1544)}{0.002016}}=122000{\frac {\mbox{mol A}}{h}}$

Thus the total amount of A entering the reactor is:

• ${\dot {n}}_{A,in}=99206+122000=221428{\frac {\mbox{mol A}}{h}}$

The amount exiting is:

• ${\dot {n}}_{A,out}={\frac {{\dot {m}}_{3}*x_{A3}}{MM_{A}}}={\frac {2380*0.15}{0.002016}}=177083{\frac {\mbox{mol A}}{h}}$

Therefore we have the following from the mole balance:

• $221428-2X=177083$

$X=22173{\frac {\mbox{ moles}}{h}}$

Now that we have this we can calculate the mass of B and C generated:

• $m_{B,gen}=-Xb*MW_{B}=22173{\frac {\mbox{ mol B}}{h}}*0.026{\frac {kg}{\mbox{ mol B}}}=-576.5{\frac {\mbox{ kg B}}{h}}$
• $m_{C,gen}=+Xc*MW_{C}=22173{\frac {\mbox{ mol C}}{h}}*0.030{\frac {kg}{\mbox{ mol C}}}=+665.2{\frac {\mbox{ kg C}}{h}}$

At this point you may want to calculate the amount of B and C leaving the reactor with the mass balances on B and C:

• $584+0.7*x_{B5}*2280-576.5=x_{B3}*2380$
• (1) $0.7*(1-0.1544-x_{B5})*2280+665.2=(1-0.15-x_{B3})*2380$

However, these equations are exactly the same! Therefore, we have proven our assertion that there is still 1 DOF in the reactor. So we need to look elsewhere for something to calculate xB5. That place is the separator balance on B:

• ${\dot {m}}_{3}*x_{B3}={\dot {m}}_{4}*x_{B4}+{\dot {m}}_{5}*x_{B5}$
• (2) $2380x_{B3}=0.02(100)+2280x_{B5}$

Solving these two equations (1) and (2) yields the final two variables in the system:

$x_{B3}=0.00856,x_{B5}=0.008058$

Note that this means the predominant species in stream 5 is also C ($x_{C5}=0.838$). However, the separator/recycle setup does make a big difference, as we'll see next.

Comparison to the situation without the separator/recycle system

Now that we know how much ethane we can obtain from the reactor after separating, let's compare to what would happen without any of the recycle systems in place. With the same data as in the first part of this problem, the new flowchart looks like this:

There are three unknowns (${\dot {m}}_{3},x_{B3},X$) and three independent material balances, so the problem can be solved. Starting with an overall mass balance because total mass is conserved:

• ${\dot {m}}_{1}+{\dot {m}}_{2}={\dot {m}}_{3}$
• ${\dot {m}}_{3}=789{\frac {kg}{h}}$

We can carry out the same sort of analysis on the reactor as we did in the previous section to find the conversion and mass percent of C in the exit stream, which is left as an exercise to the reader. The result is that:

• $X=20250{\mbox{ moles}},x_{C3}=0.77$

Compare this to the two exit streams in the recycle setup. Both of the streams were richer in C than 77%, even the reject stream. This occurred because the unreacted A and B was allowed to re-enter the reactor and form more C, and the separator was able to separate almost all the C that formed from the unreacted A and B.