Fluidized Bed vs Packed Bed — Pore Diffusion and Catalyst Economy: 23-Chem-A4 Chemical Reactor Engineering, December 2017, Q4

Question 4: Fluidized Bed vs Packed Bed — Pore Diffusion and Catalyst Economy (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Check: the printed question does not state the conversion reached in the experimental reactor. It is taken here as equal to the design target, 80% — the natural reading, since the experiment establishes the rate at the design conditions. The conclusion (fluidized bed wins) is robust as long as the 15-mm particles are strongly pore-diffusion limited and the 1-mm particles are not, which the moduli below confirm.

Given. First-order $A\rightarrow R$; experimental MFR at $T = 336\,\text{°C} = 609$ K, $P = 1$ atm, $W = 10$ g, $d_p = 1.2$ mm, $v_0 = 4\ \text{cm}^3/\text{s}$ pure A, experimental $X \approx 0.80$ (see note). Design: 80% conversion; options (i) fluidized bed, $d_p=1$ mm, mixed flow, (ii) packed bed, $d_p=15$ mm, plug flow. $\rho_s = 2000\ \text{kg/m}^3$, $D_e = 1\times10^{-6}\ \text{m}^2/\text{s}$.

Find. The reactor that minimises catalyst mass, and the size of the advantage.

Approach. Extract the intrinsic rate constant from the small-particle experiment (checking it is diffusion-free), then use the Thiele modulus / effectiveness factor to correct each design particle size and size a mixed-flow (fluidized) and a plug-flow (packed) reactor to 80% conversion.

Feed concentration and molar rate. $C_{A0} = \dfrac{P}{RT} = \dfrac{101{,}325}{(8.314)(609)} = 20.0\ \text{mol/m}^3$, and $F_{A0} = C_{A0}v_0 = (20.0)(4\times10^{-6}) = 8.0\times10^{-5}$ mol/s.
Observed rate constant from the MFR experiment. With $X=0.80$, exit $C_A = C_{A0}(1-X) = 4.0$ mol/m$^3$ and $-r'_A = \dfrac{F_{A0}X}{W} = \dfrac{(8.0\times10^{-5})(0.80)}{0.010} = 6.4\times10^{-3}\ \tfrac{\text{mol}}{\text{kg}\cdot\text{s}}.$ Hence $k' = \dfrac{-r'_A}{C_A} = 1.6\times10^{-3}\ \tfrac{\text{m}^3}{\text{kg}\cdot\text{s}}$ and the volumetric constant $k''' = \rho_s k' = 3.2\ \text{s}^{-1}.$
Is the experiment pore-diffusion free? Using characteristic length $L = d_p/6$, the Thiele modulus for the 1.2-mm particles is $M_T = L\sqrt{k'''/D_e} = (2.0\times10^{-4})\sqrt{3.2/10^{-6}} = 0.36 < 0.4.$ Effectiveness $\eta\approx1$, so the measured $k''' = 3.2\ \text{s}^{-1}$ is essentially the intrinsic rate constant.
Effectiveness factors for the two design particles. $$M_T(1\,\text{mm}) = \tfrac{10^{-3}}{6}\sqrt{3.2/10^{-6}} = 0.30 \Rightarrow \eta_{fb}\approx1.0,$$ $$M_T(15\,\text{mm}) = \tfrac{15\times10^{-3}}{6}\sqrt{3.2/10^{-6}} = 4.47 \Rightarrow \eta_{pb}\approx\frac{1}{M_T} = 0.224.$$ The large particles run in the strong pore-diffusion regime, using barely a fifth of their catalyst effectively.
Size the fluidized bed (mixed flow). $\dfrac{W}{F_{A0}} = \dfrac{X}{\eta_{fb}k'C_{A0}(1-X)} = \dfrac{0.80}{(1.0)(1.6\times10^{-3})(20.0)(0.20)} = \boxed{125\ \tfrac{\text{kg}\cdot\text{s}}{\text{mol}}.}$
Size the packed bed (plug flow). $\dfrac{W}{F_{A0}} = \dfrac{-\ln(1-X)}{\eta_{pb}k'C_{A0}} = \dfrac{1.609}{(0.224)(1.6\times10^{-3})(20.0)} = \boxed{225\ \tfrac{\text{kg}\cdot\text{s}}{\text{mol}}.}$
Decide and quantify. The fluidized bed needs $125/225 = 0.56$ of the catalyst — the packed bed requires $\approx1.8\times$ more. Although mixed flow is intrinsically worse than plug flow for a first-order reaction (a $\tfrac{X/(1-X)}{-\ln(1-X)} = 2.5\times$ penalty), the effectiveness advantage of fine particles ($1/0.224 = 4.5\times$) more than compensates.

Figure 4.1 — Catalyst requirement (W/F₀ to 80% conversion). The fine-particle fluidized bed wins despite its mixed-flow penalty because its effectiveness factor (≈1.0) dwarfs that of the 15-mm packed bed (≈0.22).

Decision. Choose the fluidized bed of 1-mm particles; it cuts the catalyst inventory by roughly 45% relative to the packed bed of 1.5-cm particles.

Concept & Theory

Topic. Pore-diffusion resistance in porous catalysts (Thiele modulus, effectiveness factor) and its interaction with reactor flow pattern in reactor selection.

Key relations. Thiele modulus $M_T = L\sqrt{k'''/D_e}$ with $L = d_p/6$; effectiveness $\eta\to1$ for $M_T<0.4$ and $\eta\to1/M_T$ for $M_T>4$; MFR $\tfrac{W}{F_{A0}}=\tfrac{X}{\eta k'C_{A0}(1-X)}$; PFR $\tfrac{W}{F_{A0}}=\tfrac{-\ln(1-X)}{\eta k'C_{A0}}$.

Why this works. Reaction and pore diffusion compete inside each pellet; when diffusion is slow (large particles, $M_T\gg1$) only an outer rind reacts and the observed rate falls as $1/d_p$. Fine fluidized particles remove that resistance, and here that fivefold rate gain outweighs the well-known twofold catalyst penalty of backmixing, so the fluidized bed is the smaller reactor.

Common pitfall. Selecting plug flow (packed bed) on flow-pattern grounds alone and forgetting the effectiveness factor; also mis-defining the characteristic length ($L=d_p/6$ for spheres, not $d_p$ or the radius).

Source. Levenspiel, Chemical Reaction Engineering, 3rd ed., Ch. 18; Fogler, 5th ed., Ch. 15 (diffusion and reaction in porous catalysts).

Question 4 of 5: Fluidized Bed vs Packed Bed — Pore Diffusion and Catalyst Economy

Question 4: Fluidized Bed vs Packed Bed — Pore Diffusion and Catalyst Economy (25 marks)