Question 5 of 6: C1 — Gas-phase mass-transfer coefficient in a wetted-wall column
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. EGBC 16-CHEM-B1 Transport Phenomena, May 2018, 3 hours, open-book (one textbook permitted, no loose notes). Six problems in three sections (A Fluid Mechanics, B Heat Transfer, C Mass Transfer), each worth 25 marks; candidates attempt one problem from each section plus a fourth from any section, and only the first four in the answer book are marked. All six problems are solved below as a complete study resource. Appendix A of the paper supplies the equations of change (continuity, Navier–Stokes, energy and species-continuity tables) in rectangular, cylindrical and spherical coordinates; these are quoted rather than re-derived.
Reference texts: R. B. Bird, W. E. Stewart & E. N. Lightfoot, Transport Phenomena (2nd ed., Wiley) — the equations of change and the differential momentum / energy / species balances; J. R. Welty, C. E. Wicks, R. E. Wilson & G. L. Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (Wiley) — pipe friction, tank draining, the internal-flow and wetted-wall correlations; F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (Wiley) — the Sieder–Tate laminar-tube correlation and cylindrical conduction; C. J. Geankoplis, Transport Processes and Separation Process Principles (Prentice Hall) — the Gilliland–Sherwood wetted-wall correlation and diffusion with heterogeneous reaction; R. H. Perry & D. W. Green, Perry’s Chemical Engineers’ Handbook — transport properties.
Question 5: C1 — Gas-phase mass-transfer coefficient in a wetted-wall column (25 marks: 22 + 3)
Find. (a) the gas-phase mass-transfer coefficient ($k_c$ and $k_G$); (b) the flux of $CO_2$ at this point, as a molar flux $N_A$ and as a mass flux $n_A=M_AN_A$.
Fig. C1: In the wetted-wall column, air rises through the core while water films the wall. The liquid is supersaturated in $CO_2$ relative to the gas ($p_A^{*}=8.2$ atm vs. $0.1$ atm), so $CO_2$ desorbs (strips) from the film into the air — a gas-phase transfer set by the turbulent core.
Approach. Compute the gas density from the ideal-gas law and the operating diffusivity (which falls as $1/P$), form the Reynolds and Schmidt numbers, apply the Gilliland–Sherwood wetted-wall correlation for the Sherwood number, and convert to $k_c$ and $k_G$. Because the interface gas is far from dilute here, the coefficient for $CO_2$ transferring through stagnant air carries the $P/p_{B,lm}$ factor of the Gilliland–Sherwood correlation; multiply by the interfacial-to-bulk partial-pressure driving force for the flux.
Gas density and operating diffusivity. $$\rho=\frac{PM}{RT}=\frac{(10\cdot101325)(0.02897)}{(8.314)(300.15)}=11.76\ \text{kg/m}^3.$$ Diffusivity scales inversely with pressure: $D_{AB}=D_{AB,1}/P=1.36\times10^{-5}/10=1.36\times10^{-6}\ \text{m}^2/\text{s}$.
Reynolds number of the gas core. $$Re=\frac{\rho v d}{\mu}=\frac{(11.76)(0.91)(0.127)}{1.85\times10^{-5}}=7.35\times10^{4}.$$
Schmidt number. Since $\rho D_{AB}$ is pressure-independent, $$Sc=\frac{\mu}{\rho D_{AB}}=\frac{1.85\times10^{-5}}{(11.76)(1.36\times10^{-6})}=1.16.$$
Gilliland–Sherwood correlation. For gas-phase transfer in a wetted-wall column, $$Sh=\frac{k_c d}{D_{AB}}=0.023\,Re^{0.83}\,Sc^{0.44}=0.023\,(7.35\times10^4)^{0.83}(1.16)^{0.44}\ \Longrightarrow\ \boxed{Sh\approx268.}$$
Dilute-basis coefficient. $$k_c^{\circ}=\frac{Sh\,D_{AB}}{d}=\frac{268\,(1.36\times10^{-6})}{0.127}=2.87\times10^{-3}\ \text{m/s},\qquad k_G^{\circ}=\frac{k_c^{\circ}}{RT}=\frac{2.87\times10^{-3}}{(8.206\times10^{-5})(300.15)}=0.117\ \frac{\text{mol}}{\text{m}^2\text{s}\cdot\text{atm}}.$$ This is the coefficient for a dilute (or equimolar) gas film.
Interface condition and the stagnant-air correction. The liquid at $0.5\ \text{mol\%}$ is in equilibrium with an interfacial partial pressure $p_{A,i}=Hx_A=16.4(0.5)=8.2\ \text{atm}$ (gas film taken as controlling), far above the bulk-gas value $p_{A,b}=y_AP=0.01(10)=0.1\ \text{atm}$ — hence $CO_2$ strips from liquid to gas. Air is essentially insoluble, so $CO_2$ transfers through stagnant air, and at $y_{A,i}=0.82$ the film is far from dilute. The air partial pressures are $p_{B,i}=10-8.2=1.8$ atm and $p_{B,b}=10-0.1=9.9$ atm, so $$p_{B,lm}=\frac{9.9-1.8}{\ln(9.9/1.8)}=\frac{8.1}{1.705}=4.75\ \text{atm},\qquad \frac{P}{p_{B,lm}}=2.10.$$ In the Gilliland–Sherwood form $Sh=\dfrac{k_c\,d}{D_{AB}}\dfrac{p_{B,lm}}{P}=0.023Re^{0.83}Sc^{0.44}$, the gas-phase coefficient at this point is therefore $$\boxed{k_c=k_c^{\circ}\frac{P}{p_{B,lm}}=6.04\times10^{-3}\ \text{m/s},\qquad k_G=\frac{k_c}{RT}=0.245\ \frac{\text{mol}}{\text{m}^2\text{s}\cdot\text{atm}}\ (2.42\times10^{-6}\ \text{mol/m}^2\text{s}\cdot\text{Pa}).}$$
(b) Mass flux. With the gas film controlling, $$N_A=k_G\,(p_{A,i}-p_{A,b})=0.245\,(8.2-0.1)\ \Longrightarrow\ \boxed{N_A\approx1.99\ \frac{\text{mol}}{\text{m}^2\text{s}}.}$$ The same value follows from the exact stagnant-film integral $N_A=c\,k_c^{\circ}\ln\dfrac{1-y_{A,b}}{1-y_{A,i}}=(406)(2.87\times10^{-3})\ln\dfrac{0.99}{0.18}=1.99$, with $c=P/RT=406\ \text{mol/m}^3$. As a mass flux ($M_{CO_2}=44.01$ g/mol), $$\boxed{n_A=M_AN_A=0.04401(1.99)\approx0.0875\ \text{kg/m}^2\text{s}.}$$ Ignoring the bulk-flow correction (dilute film, $k_G^{\circ}$) would give only $0.94\ \text{mol/m}^2\text{s}$ — an underestimate by the factor $P/p_{B,lm}=2.1$.
