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

Question 1 of 6: Evaporation of a Gasoline Spill Through a Stagnant Air Film

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

Notes on this paper

National Exams — December 2013 — 04-Chem-A3 Mass Transfer Operations. Three-hour, open-book exam; any non-communicating calculator permitted. Format: six questions in two parts — answer two of Q1–Q3 (Part A) and two of Q4–Q6 (Part B); four questions of equal value constitute a complete paper. All six 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 enthalpies used in Q5.

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

Question 1: Evaporation of a Gasoline Spill Through a Stagnant Air Film (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. Gasoline evaporates upward through a stagnant (non-diffusing) air film; air is not absorbed, so this is diffusion of A through stagnant B (Stefan diffusion).

QuantityValue
Spill volume / area50 L over 8.0 m²
Diffusivity $D_{AB}$0.65 m²/h = $1.806\times10^{-4}$ m²/s
Film thickness $z$2.0 m
$T$, $P$298 K, 1.0 atm (101.325 kPa)
Vapour pressure at surface76 mm Hg ($y_{A1}=0.10$); $y_{A2}=0$ at top
$\rho$, $M$ of gasoline720 kg/m³, 200 kg/kmol

Find. The time for all 50 L of gasoline to evaporate.

gasoline pool (8.0 m²)still-air film, thickness = 2.0 mc_A = 0 (p_A = 0) at topp_A = 76 mm Hg at the liquid surfacez
Figure 1 — Gasoline evaporating through a 2.0 m stagnant air film; the vapour mole fraction falls from 0.10 at the surface to 0 at the top of the film.

Approach. Compute the steady Stefan-diffusion molar flux through the film, multiply by the pool area to get the evaporation rate, and divide the moles of spilled gasoline by that rate.

  1. Total molar concentration of the gas. From the ideal-gas law, $c = P/RT = 101{,}325/(8.314\times298) = 40.9\ \text{mol/m}^3.$
  2. Stefan flux through the stagnant film. For A diffusing through non-diffusing B, $$N_A = \frac{D_{AB}\,c}{z}\,\ln\!\frac{1-y_{A2}}{1-y_{A1}} = \frac{(1.806\times10^{-4})(40.9)}{2.0}\,\ln\!\frac{1}{0.90} = \boxed{3.89\times10^{-4}\ \text{mol/m}^2\text{s}}.$$ The logarithmic factor is the “drift” correction for the bulk flow of A through stagnant air.
  3. Evaporation rate over the pool. Multiplying by the area, $\dot n = N_A A = (3.89\times10^{-4})(8.0) = 3.11\times10^{-3}\ \text{mol/s}.$
  4. Moles of gasoline spilled. $n = \dfrac{V\rho}{M} = \dfrac{(0.050\ \text{m}^3)(720\ \text{kg/m}^3)}{0.200\ \text{kg/mol}} = 180\ \text{mol}.$
  5. Time to evaporate. Dividing the inventory by the rate, $$t = \frac{n}{\dot n} = \frac{180}{3.11\times10^{-3}} = 5.78\times10^{4}\ \text{s} = \boxed{16.1\ \text{h}}.$$ The film is treated as quasi-steady because the pool depth (6.25 mm) drains far faster in vapour terms than the 2.0 m film relaxes.
QuantityResult
Stefan molar flux $N_A$$3.89\times10^{-4}$ mol/m²·s
Evaporation rate$3.11\times10^{-3}$ mol/s
Time to evaporate 50 L≈ 5.78×10⁴ s (16.1 h)
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