23-Chem-A4 Chemical Reactor Engineering · May 2017
Question 2 of 5: Parallel First-Order Reactions in Two Mixed-Flow Reactors in Series
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. Chemical Engineering — 23-Chem-A4 Chemical Reactor Engineering, May 2017. Open-book, 3 hours. Five questions; answering any four constitutes a complete paper (each 25 marks). All five are solved below.
Reference texts. O. Levenspiel, Chemical Reaction Engineering, 3rd ed. (Wiley); H. S. Fogler, Elements of Chemical Reaction Engineering, 5th ed. (Pearson); Perry's Chemical Engineers' Handbook, 9th ed.
Question 2: Parallel First-Order Reactions in Two Mixed-Flow Reactors in Series (25 marks)
Given. Constant-density liquid; two parallel first-order reactions sharing reactant A.
Quantity
Value
Feed $C_{A0},\,C_{R0},\,C_{S0}$
$1,\,0,\,0$ mol/L
Space times $\tau_1,\ \tau_2$
2.5 min, 5 min
Reactor-1 exit $C_{A1},C_{R1},C_{S1}$
$0.4,\,0.4,\,0.2$ mol/L
Kinetics
$-r_A=(k_1+k_2)C_A,\ r_R=k_1C_A,\ r_S=k_2C_A$
Find. The exit composition $C_{A2},\,C_{R2},\,C_{S2}$ leaving the second MFR.
Figure 2.1 — Two ideal mixed-flow reactors in series; the first reactor's measured exit sets the two rate constants, which then size the second.
Approach. The first reactor's known exit is one steady-state mixed-flow balance per species, which delivers $k_1$ and $k_2$ directly; those constants then close the three balances on the second reactor.
Extract $k_1+k_2$ from the A-balance on MFR 1. For a first-order disappearance in a mixed-flow reactor, $C_{A1}=\dfrac{C_{A0}}{1+(k_1+k_2)\tau_1}$, so $$k_1+k_2=\frac{1}{\tau_1}\!\left(\frac{C_{A0}}{C_{A1}}-1\right)=\frac{1}{2.5}\!\left(\frac{1}{0.4}-1\right)=\boxed{0.6\ \text{min}^{-1}.}$$
Split the constants using the product balances. R and S are formed (not consumed), so their MFR balances read $C_{R1}=k_1C_{A1}\tau_1$ and $C_{S1}=k_2C_{A1}\tau_1$. Thus $$k_1=\frac{C_{R1}}{C_{A1}\tau_1}=\frac{0.4}{0.4\times2.5}=0.4\ \text{min}^{-1},\quad k_2=\frac{C_{S1}}{C_{A1}\tau_1}=\frac{0.2}{0.4\times2.5}=0.2\ \text{min}^{-1}.$$ These sum to $0.6$ min$^{-1}$, consistent with Step 1, and fix the selectivity $k_1/k_2=2$.
Solve MFR 2 for A. The second reactor's feed is the first reactor's exit: $$C_{A2}=\frac{C_{A1}}{1+(k_1+k_2)\tau_2}=\frac{0.4}{1+0.6\times5}=\frac{0.4}{4}=\boxed{0.1\ \text{mol/L}.}$$
Solve MFR 2 for R and S. Each product balance adds the generation over the second reactor to what entered: $$C_{R2}=C_{R1}+k_1C_{A2}\tau_2=0.4+0.4(0.1)(5)=\boxed{0.6\ \text{mol/L},}$$ $$C_{S2}=C_{S1}+k_2C_{A2}\tau_2=0.2+0.2(0.1)(5)=\boxed{0.3\ \text{mol/L}.}$$
Confirm the overall balance. $C_{A2}+C_{R2}+C_{S2}=0.1+0.6+0.3=1.0=C_{A0}$ — every mole of A is accounted for, and R and S keep the $2:1$ ratio set by $k_1:k_2$.
