Question 5 of 7: Length of an aerated remediation trench (TCE stripping)
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 5: Length of an aerated remediation trench (TCE stripping) (Part C — equal value)
Given. A plug-flow open channel in which sparged air bubbles strip dissolved TCE to the atmosphere. Because the stripped solute is carried away by a large excess of air (and vented), the equilibrium liquid concentration is effectively $C^\ast\approx0$, so the trench acts as a first-order stripper.
Find. The trench length $L$ required for 50 → 0.05 mg/L.
Check — method and assumption
No bubble rise velocity is given, and none is needed: the data supplied (bubble diameter, liquid and air densities, liquid viscosity, diffusivity) are exactly the inputs of the Calderbank & Moo-Young correlation for mass transfer from swarms of gas bubbles, which is driven by buoyancy through a Grashof number. The gas flow rate is not given either, so we assume the continuously sparged, vented air keeps the TCE partial pressure in the bubbles negligible ($C^\ast\approx0$) — stated per the exam’s “state any assumptions” clause. (“Diffusivity of the air in wastewater” is read as the diffusivity of TCE in the wastewater.)
Figure 5. Plug-flow stripping trench: TCE transfers from water into rising air bubbles and vents; interfacial area $a=6\varepsilon_g/d_b$.
Approach. Compute the specific interfacial area from the gas holdup, get the bubble-swarm $k_L$ from the Calderbank–Moo-Young correlation (buoyancy-driven, via $Gr$), confirm liquid-film control, form $k_La$, then integrate the plug-flow mass balance (first-order stripping, $C^\ast=0$) for the length.
Interfacial area. For spherical bubbles, $$a=\frac{6\varepsilon_g}{d_b}=\frac{6(0.02)}{0.01}=12\ \text{m}^2/\text{m}^3.$$
Grashof and Schmidt numbers. $$Gr=\frac{d_b^3\rho_L g(\rho_L-\rho_G)}{\mu_L^2}=\frac{(0.01)^3(998.2)(9.81)(998.2-1.19)}{(9.93\times10^{-4})^2}=9.90\times10^{6},\quad Sc=\frac{\mu_L}{\rho_L D_{AB}}=\frac{9.93\times10^{-4}}{(998.2)(8.9\times10^{-10})}=1118.$$
Liquid-film coefficient (Calderbank & Moo-Young, $d_b\ge2.5$ mm). $$Sh=\frac{k_Ld_b}{D_{AB}}=0.42\,Gr^{1/3}Sc^{1/2}=0.42(214.7)(33.43)=3.02\times10^{3},\qquad k_L=\frac{(3.02\times10^{3})(8.9\times10^{-10})}{0.01}=2.68\times10^{-4}\ \text{m/s}.$$ Hence $k_La=(2.68\times10^{-4})(12)=\boxed{3.22\times10^{-3}\ \text{s}^{-1}}.$ A Higbie penetration estimate with a typical 0.25 m/s rise velocity, $k_L=2\sqrt{D_{AB}u_b/\pi d_b}=1.7\times10^{-4}$ m/s, is the same order, which supports the value.
Liquid-film control. The dimensionless Henry’s constant is $H/RT=9.97/[(0.08206)(293)]=0.415$: TCE is volatile, so the gas-side resistance at the mobile bubble surface is small and the overall coefficient is $K_L\approx k_L$.
Plug-flow length. The steady mass balance $Q\,dC=-k_La(C-C^\ast)A_{cs}\,dz$ with $C^\ast=0$ integrates to $$L=\frac{Q}{k_La\,A_{cs}}\ln\frac{C_{in}}{C_{out}}=\frac{0.1}{(3.22\times10^{-3})(2)}\ln\frac{50}{0.05}=(15.5\ \text{m})(6.91)=\boxed{107\ \text{m}}.$$ Each factor-of-$e$ reduction needs about 15.5 m of trench, and a 1000-fold reduction needs about seven of them — a long but practical channel, consistent with the problem’s note that remediation trenches can run from a holding pond to the discharge point.
