Question 4 of 5: Fixed-Bed PFR — Catalyst Weight and Pore-Diffusion Effectiveness

Question 4: Fixed-Bed PFR — Catalyst Weight and Pore-Diffusion Effectiveness (25 marks: a 5, b 12, c 8)

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.

Given. Equimolar reaction $A\to B$ so $\delta=0$ and $\varepsilon=0$: the gas volumetric flow is constant and $C_A=C_{A0}(1-X)$.

Find. (a) intrinsic catalyst weight; (b) effectiveness factor at bed inlet and outlet; (c) actual catalyst weight with pore diffusion.

Approach. Integrate the second-order PFR catalyst-weight balance for part (a); build the generalized Thiele modulus from the ring’s $L_c=V_p/S_p$ for part (b); and, recognising strong pore diffusion collapses the apparent order to $3/2$, re-integrate for part (c).

Part (a) — intrinsic catalyst weight

1. Catalyst-weight design equation. $F_{A0}\,dX/dW = -r_A' = kC_{A0}^2(1-X)^2$ with $F_{A0}=C_{A0}v_0 = 12(0.5)=6$ mol/s. Integrating, $W = \dfrac{F_{A0}}{kC_{A0}^2}\dfrac{X}{1-X}$.
2. Evaluate at $X=0.9$. $$W = \frac{6}{(2.5\times10^{-3})(12)^2}\cdot\frac{0.9}{0.1} = 16.67\times 9 = \boxed{150\ \text{kg}}.$$

Part (b) — effectiveness factor at inlet and outlet

1. Characteristic length of the ring. With $R_o=0.01$ m, $R_i=0.005$ m, $L=0.02$ m: pellet volume $V_p=\pi(R_o^2-R_i^2)L = 4.71\times10^{-6}$ m$^3$; external surface (outer + inner walls + two annular faces) $S_p = 2\pi R_o L + 2\pi R_i L + 2\pi(R_o^2-R_i^2) = 2.36\times10^{-3}$ m$^2$. Hence $$L_c = V_p/S_p = 2.0\times10^{-3}\ \text{m} = 2\ \text{mm}.$$
2. Volumetric rate constant. Converting the per-mass constant, $k_v = k\rho_p = (2.5\times10^{-3})(3000) = 7.5$ m$^3$/(mol·s), so the intrinsic rate per unit pellet volume is $k_v C_A^2$.
3. Generalized Thiele modulus (second order). $\phi = L_c\sqrt{\dfrac{n+1}{2}\dfrac{k_v C_A}{D_e}}$ with $n=2$. At the bed inlet $C_A=12$ mol/m$^3$: $\phi_{\text{in}} = 2\times10^{-3}\sqrt{1.5(7.5)(12)/10^{-7}} = 73.5$. At the outlet $C_A=C_{A0}(1-0.9)=1.2$ mol/m$^3$: $\phi_{\text{out}} = 23.2$.
4. Effectiveness factors. For these large moduli the slab result $\eta=\tanh\phi/\phi \approx 1/\phi$ applies: $$\boxed{\eta_{\text{inlet}} \approx 0.014,\qquad \eta_{\text{outlet}} \approx 0.043}.$$ Both are far below unity — the pellet is severely pore-diffusion limited, and the limitation eases toward the outlet only because $C_A$ (and hence $\phi$) has fallen.

Part (c) — catalyst weight with pore diffusion

1. Apparent kinetics under strong diffusion. In the asymptotic regime $\eta=1/\phi \propto C_A^{-1/2}$, so the observed rate is $\eta k C_A^2 = (k/a)\,C_A^{3/2}$ with $a=L_c\sqrt{1.5\,k_v/D_e}=21.2\ (\text{m}^3/\text{mol})^{1/2}$. Pore diffusion turns the true 2nd-order reaction into an apparent $3/2$-order one.
2. Re-integrate the design equation. $W = \dfrac{F_{A0}}{(k/a)C_{A0}^{3/2}}\displaystyle\int_0^{0.9}\dfrac{dX}{(1-X)^{3/2}}$, and $\int_0^{0.9}(1-X)^{-3/2}dX = 2[(0.1)^{-1/2}-1] = 4.32$. Therefore $$W = \frac{6}{(2.5\times10^{-3}/21.2)(12)^{1.5}}\times 4.32 = \boxed{5.3\times10^{3}\ \text{kg}}.$$
3. Interpret. Internal diffusion inflates the required bed roughly $35\times$ (from 150 kg to $\sim$5300 kg). The rings are simply too large; crushing them to smaller $L_c$ (or using a higher-porosity support to raise $D_e$) would recover most of the intrinsic activity.