Question 4 of 5: Instantaneous Surface Reaction on a Cylinder

Question 4: Instantaneous Surface Reaction on a Cylinder (15 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.

Given. Instantaneous surface reaction $\mathrm{A}\rightarrow 2\mathrm{B}$ on a long cylinder of diameter $0.03\ \text{m}$ (radius $R=0.015\ \text{m}$); ambient ~100 mol% A at $T=300\ \text{K}$, $P=1\ \text{atm}$. For part (ii): $D_{AB}=1.6\times10^{-5}\ \text{m}^2/\text{s}$ and a boundary-layer (film) thickness $\delta=1\ \text{cm}=0.01\ \text{m}$.

Find. (i) the radial mole-fraction profile $y_A(r)$; (ii) the rate of formation of B per unit length of cylinder.

Approach. This is a mixed problem: first derive the profile by writing the radial molar flux of A with the stoichiometric bulk-flow coupling ($N_B=-2N_A$) and imposing a constant molar flow through every cylindrical surface; then, because the reaction is instantaneous, set $y_{As}=0$ at the surface and $y_A=1$ at the film edge and evaluate the rate.

Because the reaction is instantaneous, A is consumed the moment it reaches the surface, so the process is diffusion-controlled and the surface mole fraction $y_{As}\rightarrow0$. Two moles of B leave for every mole of A that arrives, so the fluxes do not balance ($N_B=-2N_A$): the diffusion is non-equimolar and the convective (bulk-flow) term must be retained.

1. (i) Flux of A with stoichiometric coupling. The radial molar flux is $N_A=-cD_{AB}\dfrac{dy_A}{dr}+y_A\,(N_A+N_B)$. With $N_B=-2N_A$ the bulk term is $N_A+N_B=-N_A$, so $$N_A\,(1+y_A)=-cD_{AB}\frac{dy_A}{dr}.$$
2. Constant molar flow through each surface. At steady state the molar flow of A crossing any cylindrical surface is constant: $\dot{N}_A=2\pi r L\,N_A=\text{const}$, so $N_A=\dfrac{\dot{N}_A}{2\pi r L}$.
3. Separate the variables. Substituting and grouping, $\dfrac{\dot{N}_A}{2\pi L}\dfrac{dr}{r}=-cD_{AB}\dfrac{dy_A}{1+y_A}$. Integrating from the surface ($r=R$, $y_A=y_{As}$) outward gives $\dfrac{\dot{N}_A}{2\pi L}\ln\dfrac{r}{R}=-cD_{AB}\ln\dfrac{1+y_A}{1+y_{As}}$.
4. Mole-fraction profile. Rearranging into exponential form, $$\frac{1+y_A}{1+y_{As}}=\left(\frac{r}{R}\right)^{-\dot{N}_A/(2\pi L c D_{AB})},$$ the required profile: $y_A$ climbs from $y_{As}$ at the surface to the ambient value at the film edge. Multiplying through by $1+y_{As}$ and subtracting 1 gives the printed form $y_A=(1+y_{As})(r/R)^{-\dot{N}_A/(2\pi LcD_{AB})}-1$. $\boxed{\,\dfrac{1+y_A}{1+y_{As}}=\left(\dfrac{r}{R}\right)^{-\dot{N}_A/(2\pi L c D_{AB})}\,}$
5. (ii) Fix the boundary values. The instantaneous reaction makes the surface a perfect sink, so $y_{As}=0$ at $r=R=0.015\ \text{m}$; the ambient is ~100% A, so $y_A=1$ at the film edge $r=R+\delta=0.025\ \text{m}$.
6. Total molar concentration. From the ideal-gas law, $c=\dfrac{P}{R_gT}=\dfrac{101\,325}{(8.314)(300)}=\boxed{40.6\ \text{mol/m}^3}.$
7. Molar flow of A per unit length. Writing the result per unit length ($W_A=\dot{N}_A/L$) and inserting the boundary values into step 3, $\dfrac{W_A}{2\pi}=-\dfrac{cD_{AB}\ln 2}{\ln[(R+\delta)/R]}$. With $\ln(0.025/0.015)=0.511$, $$W_A=-\frac{2\pi cD_{AB}\ln 2}{\ln[(R+\delta)/R]}=-5.54\times10^{-3}\ \text{mol m}^{-1}\text{s}^{-1},$$ the minus sign indicating that A flows inward (it is consumed).
8. Rate of formation of B. Two moles of B form per mole of A consumed, so per unit length $W_B=2\,|W_A|=\boxed{1.11\times10^{-2}\ \text{mol m}^{-1}\text{s}^{-1}}.$