23-Chem-A4 Chemical Reactor Engineering · December 2017
Question 3 of 5: Packed-Bed Rate Data — Rate, Plug-Flow and Recycle Catalyst Loads
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, December 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, 1999); H. S. Fogler, Elements of Chemical Reaction Engineering, 5th ed. (Pearson, 2016); Perry's Chemical Engineers' Handbook, 9th ed.
Question 3: Packed-Bed Rate Data — Rate, Plug-Flow and Recycle Catalyst Loads (25 marks: a 15, b 5, c 5)
Given. Packed-bed data for $A\rightarrow R$ at $F_{A0} = 10$ kmol/hr:
$W$ (kg)
1
2
3
4
5
6
7
$X_A$
0.12
0.20
0.27
0.33
0.37
0.41
0.44
Find. (a) $-r'_A$ at $X_A = 0.40$; (b) catalyst for $X_A=0.40$ at $F_{A0}=400$ kmol/hr in a plug-flow packed bed; (c) catalyst for the same duty with a very large recycle.
Approach. For a packed bed the differential design equation is $F_{A0}\,dX_A = (-r'_A)\,dW$, so the rate at any conversion is $F_{A0}$ times the local slope of the $X_A$–$W$ curve; plug-flow scale-up preserves $W/F_{A0}$ at fixed $X$, while an infinite recycle turns the bed into a mixed-flow reactor.
Figure 3.1 — Measured X₃₊ vs catalyst weight. The dashed tangent at X = 0.40 (W = 5.75 kg) has slope dX/dW ≈ 0.04 kg⁻¹, which sets the local reaction rate.
Differential rate from the slope (part a). The packed-bed (plug-flow) design equation $\dfrac{dX_A}{d(W/F_{A0})} = -r'_A$ rearranges to $-r'_A = F_{A0}\dfrac{dX_A}{dW}.$ From the data, $X_A = 0.40$ lies between $W=5$ ($0.37$) and $W=6$ ($0.41$), so $W\big|_{X=0.40} = 5.75$ kg and the local slope is $\dfrac{dX_A}{dW}\approx\dfrac{0.41-0.37}{6-5} = 0.04\ \text{kg}^{-1}.$
Evaluate the rate. $$-r'_A\big|_{X=0.40} = (10)(0.04) = \boxed{0.40\ \tfrac{\text{kmol}}{\text{kg cat}\cdot\text{hr}}.}$$
Plug-flow scale-up (part b). For plug flow the sizing depends only on the ratio $W/F_{A0}$ at the target conversion, which is fixed by the kinetics. From the small unit, $W\big|_{X=0.40} = 5.75$ kg at $F_{A0}=10$, so $W/F_{A0} = 0.575\ \tfrac{\text{kg}\cdot\text{hr}}{\text{kmol}}$. At $F_{A0}=400$: $$W_{PFR} = 0.575\times400 = \boxed{230\ \text{kg}.}$$
Very large recycle = mixed flow (part c). An infinite recycle ratio makes the packed bed behave as a mixed-flow (backmix) reactor operating at the exit conversion, for which $\dfrac{W}{F_{A0}} = \dfrac{X_A}{-r'_A}$ evaluated at $X_A=0.40$. Thus $$W_{MFR} = F_{A0}\frac{X_A}{-r'_A} = 400\times\frac{0.40}{0.40} = \boxed{400\ \text{kg}.}$$
Compare. The recycle (mixed-flow) reactor needs $400/230 \approx 1.7\times$ more catalyst than the plug-flow bed — the expected penalty of backmixing for a positive-order reaction, which forces the whole bed to react at the low exit rate.
