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23-Chem-A3 Heat and Mass Transfer · May 2014

Question 1 of 7: Gas-Film Thickness for Ammonia Scrubbing

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

Notes on this paper

National Exams — May 2014 — 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. Property data (diffusivities, vapour pressures, psychrometric enthalpies) are stated in each Given block; the psychrometric appendix supplied with the exam furnishes the saturated-air data used in Q6.

Reference texts: Geankoplis, Transport Processes and Separation Process Principles (4th ed., Prentice Hall) — molecular diffusion, convective mass transfer, absorption and humidification; Treybal, Mass-Transfer Operations (3rd ed., McGraw-Hill) — gas absorption, wetted-wall and cooling-tower design; Welty, Wicks, Wilson & Rorrer, Fundamentals of Momentum, Heat and Mass Transfer — boundary-layer and falling-sphere correlations; supporting property/psychrometric data from Perry's Chemical Engineers' Handbook (9th ed.).

Question 1: Gas-Film Thickness for Ammonia Scrubbing (equal value)

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. Ammonia (A) diffuses through a stagnant air (B) film to the gas–liquid interface; the flux and both boundary partial pressures are known, so the film thickness is the only unknown in the Stefan-diffusion law.

QuantityValue
$T$, $P$295 K, 101.325 kPa
Diffusivity $D_{AB}$0.24 cm²/s $=2.4\times10^{-5}$ m²/s
Bulk-gas NH₃5% ⇒ $p_{A,\text{bulk}}=5.07$ kPa
Interface NH₃$p_{A,i}=66$ kPa (equilibrium with liquid)
Molar flux $N_A$$1.0\times10^{-3}$ kmol/m²·s

Find. The gas-film thickness $z$ (in mm).

bulk gas: 5% NH₃ (p_A = 5.07 kPa)gas–liquid interface: p_A = 66 kPastagnant gas film, thickness zzN_A
Figure 1 — The whole mass-transfer resistance sits in a thin stagnant gas film; ammonia diffuses across it between the bulk-gas and interface partial pressures, with air as the stagnant inert.

Approach. Write the steady Stefan-diffusion flux for A through stagnant B across the film and invert it for the thickness $z$.

  1. Total molar concentration of the gas. $c = P/RT = 101{,}325/(8.314\times295) = 41.3\ \text{mol/m}^3 = 0.0413\ \text{kmol/m}^3.$
  2. Inert (air) partial pressures at the two faces. Air is the stagnant species, so $p_{B,\text{bulk}} = 101.325-5.07 = 96.26$ kPa and $p_{B,i} = 101.325-66 = 35.33$ kPa.
  3. Stefan-diffusion flux across the film. For A through non-diffusing B, $$N_A = \frac{D_{AB}\,c}{z}\,\ln\!\frac{p_{B,i}}{p_{B,\text{bulk}}}.$$ The magnitude of the log term fixes $z$ regardless of the transfer direction.
  4. Solve for the film thickness. $$z = \frac{D_{AB}\,c}{N_A}\,\left|\ln\frac{p_{B,\text{bulk}}}{p_{B,i}}\right| = \frac{(2.4\times10^{-5})(0.0413)}{1.0\times10^{-3}}\,\ln\!\frac{96.26}{35.33} = \boxed{9.9\times10^{-4}\ \text{m} \approx 1.0\ \text{mm}}.$$
QuantityResult
Gas molar concentration $c$0.0413 kmol/m³
Log inert-pressure ratio1.00
Gas-film thickness $z$≈ 0.99 mm (≈ 1.0 mm)
Check
As printed, the interface partial pressure (66 kPa) exceeds the bulk value (5.07 kPa), so the sign of the driving force points from interface to bulk. The film-thickness magnitude is nonetheless pinned by the flux and the (large) inert-pressure ratio; $|z|\approx1$ mm is the physically meaningful result and agrees with the classic textbook value for this problem.
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