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23-Chem-A3 Heat and Mass Transfer · May 2013

Question 4 of 7: Oxygenation of a Deep Mountain Lake

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — May 2013 — 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 constitute a complete paper (Parts A/B carry 20% each, Part C 30% each). All seven are solved below for completeness. Property data are stated in each Given block. Two chart appendices (SI humidity–temperature charts) accompany Q6.

Reference texts: Geankoplis, Transport Processes and Separation Process Principles (4th ed., Prentice Hall) — molecular diffusion, mass-transfer coefficients, absorption, humidification; Welty, Wicks, Wilson & Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (5th/6th ed., Wiley) — transient diffusion, boundary-layer mass transfer; Treybal, Mass-Transfer Operations (3rd ed., McGraw-Hill) — packed-tower and cooling-tower design; supporting data from Perry's Chemical Engineers' Handbook (9th ed.) and Incropera & DeWitt, Fundamentals of Heat and Mass Transfer.

Question 4: Oxygenation of a Deep Mountain Lake (Part B — 20%)

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. A very deep, quiescent lake at uniform $5\,°\text{C}$, initially $C_{A0}=3.0\times10^{-5}\ \text{kmol/m}^3$ dissolved O₂. At $t=0$ the ice clears and the surface reaches air-equilibrium. Atmospheric $P=0.769$ atm, air is 21% O₂; Henry constant $k_H=2.91\times10^{4}\ \text{atm}/(\text{kmol O}_2/\text{kmol soln})$; diffusivity of O₂ in water $D_{AB}(5\,°\text{C})=1.58\times10^{-9}\ \text{m}^2/\text{s}$ (given). Depth of interest $x=0.06$ m.

Find. The dissolved-O₂ concentration at $x=0.06$ m after (a) 1 day, (b) 3 days, (c) 30 days.

air–water interface (C_As = 3.08×10⁻⁴ kmol/m³)atmosphere 0.769 atm, 21% O₂O₂ diffuses downwardx = 0.06 mdeep lake, uniform 5°C; initial C_A0 = 3.0×10⁻⁵ kmol/m³
Figure 4 — With the surface held at its air-equilibrium value and the lake effectively infinitely deep, this is transient one-dimensional diffusion into a semi-infinite medium; the penetration front advances as √(Dt).

Approach. Fix the surface concentration from Henry's law, model the lake as a semi-infinite medium with a step change in surface concentration, and apply the complementary-error-function solution at each time.

  1. Surface (interfacial) concentration from Henry's law. Air O₂ partial pressure $p_{A}=0.21(0.769)=0.1615$ atm, so the surface mole fraction is $x_A=p_A/k_H=0.1615/2.91\times10^{4}=5.55\times10^{-6}$. With $c_{\text{tot}}=\rho/M=1000/18.02=55.5\ \text{kmol/m}^3$, $$C_{As}=x_A\,c_{\text{tot}}=\boxed{3.08\times10^{-4}\ \text{kmol/m}^3}.$$
  2. Semi-infinite transient solution. For a step surface concentration, $$\frac{C_A-C_{A0}}{C_{As}-C_{A0}}=\operatorname{erfc}\!\left(\frac{x}{2\sqrt{D_{AB}t}}\right),\qquad C_{As}-C_{A0}=2.78\times10^{-4}\ \text{kmol/m}^3.$$ Use the printed $D_{AB}=1.58\times10^{-9}\ \text{m}^2/\text{s}$ for O₂ in water at $5\,°\text{C}$.
  3. Evaluate at each time. With $\eta=x/(2\sqrt{D_{AB}t})$: $$\begin{aligned} \text{(a) 1 day: }&\eta=2.57,\ \operatorname{erfc}=2.8\times10^{-4}\ \Rightarrow\ C_A=\boxed{3.01\times10^{-5}\ \text{kmol/m}^3}\ (\text{essentially unchanged}),\\ \text{(b) 3 days: }&\eta=1.48,\ \operatorname{erfc}=0.0360\ \Rightarrow\ C_A=\boxed{4.00\times10^{-5}\ \text{kmol/m}^3},\\ \text{(c) 30 days: }&\eta=0.469,\ \operatorname{erfc}=0.507\ \Rightarrow\ C_A=\boxed{1.71\times10^{-4}\ \text{kmol/m}^3}. \end{aligned}$$
  4. Interpretation. After one day the 6 cm depth is untouched; only after weeks does re-oxygenation reach it — consistent with the slow spring recovery described in the problem.
Time$\eta=x/2\sqrt{Dt}$$C_A$ at 0.06 m (kmol/m³)
(a) 1 day2.57$3.01\times10^{-5}$
(b) 3 days1.48$4.00\times10^{-5}$
(c) 30 days0.469$1.71\times10^{-4}$

Measured against the possible rise to the surface value $C_{As}=3.08\times10^{-4}$ kmol/m³, the 6 cm depth has gained only about 4% of that rise after three days and about half after thirty days. The diffusion length $\sqrt{D_{AB}t}$ is roughly 1.2 cm after one day and 6.4 cm after thirty days, which is why the one-day value is indistinguishable from the initial concentration.