23-Chem-A4 Chemical Reactor Engineering · December 2014
Question 2 of 5: Reactor Sizing from Rate-Concentration Data
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams / EGBC — December 2014 — 04-Chem-A4 Chemical Reactor Engineering. Three-hour open-book exam; the designated Fogler textbook (any edition), unit-conversion / mathematical tables and a non-communicating programmable calculator are permitted. Five questions are printed and any four constitute a complete paper (each worth 20 marks); all five are solved below for completeness. No credit is given for re-deriving standard rate expressions, so the batch / PFR / CSTR design equations are quoted and applied. These are Levenspiel-style graphical / analytical reactor problems (enzyme kinetics, rate-data analysis, autocatalytic staging, parallel-reaction selectivity, and CSTR kinetic determination).
Reference texts: O. Levenspiel, Chemical Reaction Engineering (3rd ed., Wiley) — batch/MFR/PFR design equations, the graphical “$1/(-r_A)$ vs $C_A$” area method, autocatalytic optimum staging, product distribution for parallel reactions, and constant-volume CSTR kinetic analysis; H. S. Fogler, Elements of Chemical Reaction Engineering (4th/5th ed., Prentice Hall) — the designated open-book text: stoichiometric tables with the expansion factor $\varepsilon$, gas-phase reactions with change in moles, and reactor sizing; supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.).
Question 2: Reactor Sizing from Rate-Concentration Data (20 marks: a 7, b 6, c 7)
Given. Tabulated rate data (below); the rate rises to a maximum near $C_A=0.4$ then falls, so the design integrals must use the actual curve, not a rate law. Feed rate $F_{A0}=1000$ mol/hr in parts (b) and (c).
$C_A$ (mol/L)
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
1.0
1.3
2.0
$-r_A$ (mol/L·min)
0.1
0.3
0.5
0.6
0.5
0.25
0.1
0.06
0.05
0.045
0.042
Find. (a) batch reaction time; (b) PFR volume for $X=0.80$; (c) MFR volume for $X=0.75$.
Figure 2 — The measured rate rises to a peak at $C_A\approx0.4$ mol/L then falls — an autocatalytic-type curve. Batch time and reactor volumes are areas/heights read from $1/(-r_A)$ across this curve.
Approach. All three follow from the performance equations written as areas or heights on the rate curve: the batch/PFR need $\int dC_A/(-r_A)$ (trapezoidal on the tabulated points), while the single MFR needs only the exit rate.
Batch time (part a). For a constant-volume batch reactor $t=\displaystyle\int_{C_{Af}}^{C_{A0}}\frac{dC_A}{-r_A}$. The trapezoidal area of $1/(-r_A)$ between $C_A=0.3$ and $1.3$ mol/L gives$$\boxed{t=12.7\ \text{min}}.$$Most of the time is spent in the slow high-$C_A$ region ($C_A>0.6$), where $1/(-r_A)$ is large.
PFR volume (part b). $V=F_{A0}\displaystyle\int_0^{X}\frac{dX}{-r_A}=\frac{F_{A0}}{C_{A0}}\int_{C_A}^{C_{A0}}\frac{dC_A}{-r_A}$. With $X=0.80$ the exit is $C_A=C_{A0}(1-X)=1.5(0.2)=0.3$ mol/L, so the integral runs 0.3 → 1.5 mol/L. Using $F_{A0}=1000$ mol/hr and converting the minute-based rate,$$V=\frac{1000}{1.5}\Big(\int_{0.3}^{1.5}\frac{dC_A}{-r_A}\Big)\frac{1\ \text{hr}}{60\ \text{min}}=\boxed{191\ \text{L}}.$$
MFR volume (part c). A mixed-flow reactor reacts entirely at its exit condition, so $V=\dfrac{F_{A0}X}{(-r_A)_{exit}}$. For $X=0.75$ and $C_{A0}=1.2$, the exit is $C_A=1.2(0.25)=0.3$ mol/L where $(-r_A)=0.5$ mol/L·min. Thus$$V=\frac{1000\times0.75}{0.5}\cdot\frac{1}{60}=\boxed{25\ \text{L}}.$$The MFR is far smaller than the PFR because its exit sits near the rate maximum ($C_A=0.3$–0.4), whereas the PFR must also crawl through the slow high-$C_A$ tail.
