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

Question 2 of 7: SO₂ Diffusion Through a Tapering Conduit

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 2: SO₂ Diffusion Through a Tapering Conduit (Part A — 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. SO₂ (A) diffuses through stagnant O₂ (B) along a duct of length $L=2.0$ m at $P=10\ \text{bar}=10^{6}$ Pa, $T=598$ K, $D_{AB}=7.61\times10^{-6}\ \text{m}^2/\text{s}$. One side stays $0.300$ m; the other tapers $0.400\to0.200$ m, so $S(x)=0.3\,(0.4-0.1x)\ \text{m}^2$. Partial pressures: $p_{A1}=0.22$ bar (large end, $x=0$), $p_{A2}=0.055$ bar (small end, $x=2$ m).

Find. The local molar flux of SO₂ at the midpoint, $N_A(x=1\ \text{m})$, in $\text{mol}/(\text{m}^2\cdot\text{s})$.

400 mm200 mmSO₂ flux N_A →L = 2.0 mp_A = 0.22 barp_A = 0.055 barWidth constant at 300 mm; height tapers 400 → 200 mmP = 10 bar, T = 598 K, D_AB = 7.61×10⁻⁶ m²/s
Figure 2 — The duct area changes with position, so the flux NA varies while the molar flow WA = NAS(x) is constant. Integrate over the variable area, then evaluate the flux at the midpoint.

Approach. Conservation makes the molar flow $W_A=N_A\,S(x)$ constant even though the flux is not; separate variables and integrate the stagnant-B flux relation over the varying area, solve for $W_A$, then divide by the midpoint area.

  1. Write the constant-flow diffusion relation. For A through stagnant B with variable area, $$W_A\,\frac{dx}{S(x)}=\frac{D_{AB}P}{RT}\,\frac{-dp_A}{P-p_A}\quad\Rightarrow\quad W_A\!\int_0^{L}\!\frac{dx}{S(x)}=\frac{D_{AB}P}{RT}\,\ln\!\frac{P-p_{A2}}{P-p_{A1}}.$$
  2. Evaluate the area integral. With $S(x)=0.3(0.4-0.1x)$, $$\int_0^{2}\frac{dx}{0.3(0.4-0.1x)}=\frac{1}{0.3}\Big[\tfrac{-1}{0.1}\ln(0.4-0.1x)\Big]_0^{2}=\frac{-10}{0.3}\ln\frac{0.2}{0.4}=23.10\ \text{m}^{-1}.$$
  3. Evaluate the pressure and diffusion group. $p_{B2}=P-p_{A2}=9.945\times10^{5}$ Pa, $p_{B1}=P-p_{A1}=9.780\times10^{5}$ Pa, and $$\frac{D_{AB}P}{RT}=\frac{(7.61\times10^{-6})(10^{6})}{(8.314)(598)}=1.531\times10^{-3}\ \text{mol}/(\text{m}\cdot\text{s}).$$ So $\displaystyle W_A=\frac{(1.531\times10^{-3})\ln(9.945/9.780)}{23.10}=\boxed{1.11\times10^{-6}\ \text{mol/s}}.$
  4. Flux at the midpoint. At $x=1$ m, $S=0.3(0.4-0.1)=0.09\ \text{m}^2$, hence $$N_A=\frac{W_A}{S}=\frac{1.11\times10^{-6}}{0.09}=\boxed{1.23\times10^{-5}\ \text{mol}/(\text{m}^2\cdot\text{s})}.$$
QuantityResult
Molar flow of SO₂, $W_A$ (constant)$1.11\times10^{-6}$ mol/s
Midpoint area, $S(1\,\text{m})$$0.090\ \text{m}^2$
Midpoint flux, $N_A$$1.23\times10^{-5}\ \text{mol}/(\text{m}^2\cdot\text{s})$

Check: the printed dimensions "300 mm by 400 mm to 300 mm by 200 mm" are read as one constant 300 mm side with the other tapering 400→200 mm (midpoint area $0.09\ \text{m}^2$). If instead both sides tapered, the midpoint area and hence the flux would differ.