Question 6 of 7: Gas-film coefficient on a spinning catalyst paddle
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2015 — 04-Chem-A3 (Mass Transfer Operations). Three-hour, open-book exam; any non-communicating calculator permitted. Format: seven questions in three parts — answer one of Q1–Q2 (Part A), one of Q3–Q4 (Part B) and two of Q5–Q7 (Part C); four questions of equal value constitute a complete paper. All seven are solved below for completeness. Every property datum (diffusivities, Henry’s constants, solubilities, packing constants) is supplied in the question or its data table, so each answer is self-contained.
Reference texts: Geankoplis, Transport Processes and Separation Process Principles (4th ed., Prentice Hall) — Knudsen/molecular diffusion, convective mass-transfer coefficients, gas absorption in packed towers; Welty, Wicks, Wilson & Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (6th ed., Wiley) — boundary-layer analogies and dimensional analysis; Bird, Stewart & Lightfoot, Transport Phenomena (2nd ed., Wiley) — Chapman–Enskog kinetic theory; Treybal, Mass-Transfer Operations (3rd ed., McGraw-Hill) — Sherwood–Holloway packed-tower correlation; supporting data from Perry’s Chemical Engineers’ Handbook (9th ed.).
Question 6: Gas-film coefficient on a spinning catalyst paddle (Part C — equal value)
Given. Each pellet is a sphere in cross-flow; the local gas velocity varies with radial position because $u=\omega r$, with $r$ from 1 cm (inner paddle edge) to 6 cm (outer edge, 1 cm + 5 cm width). The gas-film coefficient is averaged over the paddle width. Diffusivities are computed from Chapman–Enskog kinetic theory using the exam’s Lennard–Jones table.
Quantity
Value
Temperature / pressure
600 K / 1 atm
Pellet diameter $d_p$
0.5 cm
Rotation rate $\omega$
4 rad/s
Radial span (inner–outer)
1 cm → 6 cm
L–J: CO ($\varepsilon/k$=110 K, $\sigma$=3.590 Å)
O₂ (113 K, 3.433 Å), CO₂ (190 K, 3.996 Å)
Find. The width-averaged gas-film $k_c$ for (a) CO and (b) CO₂.
Check — assumption
The carrier-gas (O₂) viscosity and density at 600 K are not tabulated in the paper. We take $\mu_{O_2}\approx\mu_{O_2}(300\,\text{K})(600/300)^{0.7}$ (kinetic-theory temperature scaling) and $\rho_{O_2}=PM/RT$ at 600 K, 1 atm, giving $\nu\approx0.51$ cm²/s. These standard estimates are stated per the exam’s assumptions clause and used identically for both species.
Figure 6. Pellets in the paddle experience a radially varying velocity $u=\omega r$; the gas-film coefficient is averaged across the 1–6 cm span.
Approach. Compute $D_{AB}$ (CO–O₂ and CO₂–O₂) by Chapman–Enskog, form the local sphere $Sh=2+0.6Re^{1/2}Sc^{1/3}$ with $Re=d_p\omega r/\nu$, and average $k_c$ over the paddle width $r=1\to6$ cm.
Chapman–Enskog diffusivity. $$D_{AB}=\frac{0.0018583\sqrt{T^3(1/M_A+1/M_B)}}{P\,\sigma_{AB}^2\,\Omega_D},$$ with $\sigma_{AB}=(\sigma_A+\sigma_B)/2$ and $\varepsilon_{AB}/k=\sqrt{(\varepsilon_A/k)(\varepsilon_B/k)}$. For CO–O₂: $\sigma_{AB}=3.51$ Å, $\varepsilon_{AB}/k=111.5$ K, $kT/\varepsilon=5.38\Rightarrow\Omega_D=0.8422+0.38(0.8124-0.8422)=0.831$ (interpolating the exam table between 5.0 and 6), giving $D_{CO\text{-}O_2}=\boxed{0.690\ \text{cm}^2/\text{s}}$. For CO₂–O₂: $\sigma_{AB}=3.71$ Å, $\varepsilon_{AB}/k=146.5$ K, $kT/\varepsilon=4.09\Rightarrow\Omega_D=0.879$, $D_{CO_2\text{-}O_2}=0.523$ cm²/s.
Carrier-gas kinematic viscosity. Using the assumed O₂ properties, $\mu_{O_2}\approx3.31\times10^{-4}$ g/cm·s and $\rho_{O_2}=PM/RT=6.50\times10^{-4}$ g/cm³, so $\nu=\mu/\rho\approx0.51$ cm²/s at 600 K; hence $Sc_{CO}=\nu/D_{CO}=0.74$ and $Sc_{CO_2}=0.97$.
Local and mean sphere coefficient. Local $Sh(r)=2+0.6\,Re(r)^{1/2}Sc^{1/3}$ with $Re(r)=d_p\omega r/\nu$, and $k_c(r)=Sh(r)D_{AB}/d_p$. Averaging over the width, $$\bar k_c=\frac{1}{r_2-r_1}\int_{r_1}^{r_2}\frac{D_{AB}}{d_p}\Big[2+0.6\Big(\tfrac{d_p\omega r}{\nu}\Big)^{1/2}Sc^{1/3}\Big]dr.$$
(a) CO, (b) CO₂. Only $r^{1/2}$ varies, and its width-average is $\frac{2}{3}\frac{r_2^{3/2}-r_1^{3/2}}{r_2-r_1}=1.826$ cm$^{1/2}$, so the mean $Re^{1/2}=\sqrt{d_p\omega/\nu}\,(1.826)=3.62$ and the mean $Sh$ is 3.96 (CO) and 4.15 (CO₂). This gives $$\bar k_{c,CO}=\boxed{5.47\ \text{cm/s}},\qquad \bar k_{c,CO_2}=\boxed{4.34\ \text{cm/s}}.$$ CO transfers faster: it is the lighter, smaller molecule, so its higher diffusivity raises $k_c$ directly and, through its lower $Sc$, also slightly thins the concentration boundary layer.
