Question 5 of 6: Gas Absorption in a Laminar Liquid Jet (Higbie)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Reference texts: Incropera, DeWitt, Bergman & Lavine, Fundamentals of Heat and Mass Transfer (Wiley) — conduction resistance networks, critical radius, convection correlations, radiation exchange between surfaces; Coulson & Richardson, Chemical Engineering Vol. 1 (Butterworth-Heinemann) — heat-exchanger sizing and diffusion through stagnant films; Treybal, Mass-Transfer Operations (3rd ed., McGraw-Hill) — differential (Rayleigh) distillation and penetration-theory absorption; Geankoplis, Transport Processes and Separation Process Principles (4th ed., Prentice Hall) — molecular-diffusion flux relations. Property data from Perry's Chemical Engineers' Handbook (9th ed.).
Exam format: six problems worth 25 points each — Part A (Heat Transfer, Q1–Q3) and Part B (Mass Transfer, Q4–Q6). The rubric asks for at least two problems from each part; every problem is solved here for completeness. Open-book, three-hour paper.
Part A — Heat Transfer
Question 5: Gas Absorption in a Laminar Liquid Jet (Higbie) (25 points)
Given. Liquid flow $Q=4$ cm³/s; jet diameter $d=1$ mm, length $L=3$ mm; absorption rate $0.12$ cm³/s (gas at 1 atm, $30$ °C; the printed “mixture cm³” is read as cm³ of the pure gas, and part (b)’s “evaporation” as absorption); solubility $c^{*}=1\times10^{-4}$ mol/cm³ (at 1 atm, pure gas). Penetration theory applies.
Quantity
Value
Liquid flow $Q$
4 cm³/s
Jet diameter / length
1 mm / 3 mm
Absorption rate
0.12 cm³/s (1 atm, 30 °C)
Solubility $c^{*}$
$1\times10^{-4}$ mol/cm³·atm
Find. (a) the gas diffusivity $D$; (b) the effect on absorption rate of shrinking the jet to 0.9 mm.
Figure 5 — Laminar liquid jet issuing from the nozzle into pure gas A. Each fluid element is exposed to the gas for only the transit time $t_e=L/v$; penetration theory gives the closed-form absorption $W_A=4c^{*}\sqrt{DQL}$, in which the jet diameter cancels.
Approach. Use the penetration-theory result for a rod-like jet, $W_A=4c^{*}\sqrt{DQL}$; convert the measured volumetric absorption to a molar rate, then invert for $D$. For (b) inspect the same formula for any diameter dependence.
Molar absorption rate. At $30$ °C, $1$ atm the molar volume is $V_m=RT/P=82.06\times303.15=24{,}882$ cm³/mol, so$$W_A=\frac{0.12}{24{,}882}=4.82\times10^{-6}\ \text{mol/s}.$$For a pure gas the interfacial concentration is $c^{*}=1\times10^{-4}$ mol/cm³.
Penetration-theory jet result. For a rod-like laminar jet the total absorption is$$W_A=4\,c^{*}\sqrt{D\,Q\,L}.$$Note that the surface area ($\propto d$) and the exposure time ($\propto d^{2}/Q$) combine so that the diameter cancels — only $Q$ and $L$ set the contact.
Invert for the diffusivity (part a). Solving,$$D=\frac{1}{Q L}\left(\frac{W_A}{4c^{*}}\right)^{2}=\frac{1}{(4)(0.3)}\left(\frac{4.82\times10^{-6}}{4\times10^{-4}}\right)^{2}=\boxed{1.21\times10^{-4}\ \text{cm}^2/\text{s}}=1.21\times10^{-8}\ \text{m}^2/\text{s}.$$A cross-check through the explicit penetration flux $N_A=2c^{*}\sqrt{D/\pi t_e}$ over the jet surface reproduces the same $0.12$ cm³/s.
Effect of a smaller jet (part b). Since $W_A=4c^{*}\sqrt{DQL}$ contains no diameter, reducing $d$ from 1 mm to 0.9 mm at fixed $Q$ and $L$ leaves the absorption rate unchanged at 0.12 cm³/s. Physically the thinner jet moves faster (shorter exposure, higher instantaneous flux) but presents less surface area, and the two effects cancel exactly.
