NivaarExam PrepOfficial exam papers ↗

23-Chem-B1 Transport Phenomena · December 2016

Question 5 of 6: C1 — Evaporation of a water spill through a stagnant air film

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

Notes on this paper

Paper format. EGBC 04-CHEM-B1 Transport Phenomena, December 2016, 3 hours, open-book. Six problems in three sections (A Fluid Mechanics, B Heat Transfer, C Mass Transfer); candidates attempt one from each section plus a fourth (four of six at 25 marks each, only the first four in the answer book are marked). All six problems are solved below as a complete study resource. Appendix A of the paper supplies the equations of change (continuity, Navier–Stokes, energy and species tables) in rectangular, cylindrical and spherical coordinates — these are quoted rather than re-derived.

Reference texts: R. B. Bird, W. E. Stewart & E. N. Lightfoot, Transport Phenomena (2nd ed., Wiley) — the equations of change and the differential momentum / energy / species balances; J. R. Welty, C. E. Wicks, R. E. Wilson & G. L. Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (Wiley) — pipe friction, internal-flow heat transfer and film mass transfer; F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (Wiley) — the log-mean-temperature-difference and effectiveness–NTU methods; R. H. Perry & D. W. Green, Perry’s Chemical Engineers’ Handbook (9th ed.) — transport properties and the Colebrook friction correlation.

Question 5: C1 — Evaporation of a water spill through a stagnant air film (25 marks)

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.

QuantitySymbolValue
Spill volume$V$$0.00325\ \text{ft}^3=92.0\ \text{cm}^3$
Air temperature, pressure$T,P$$74\ \degree\text{F}=296.5\ \text{K},\ 1\ \text{atm}$
Bulk / saturation humidity$H_\infty,\ H_s$$0.0019,\ 0.0188\ \tfrac{\text{lb w}}{\text{lb dry}}$
Stagnant film thickness$\delta$$0.19\ \text{in}=0.483\ \text{cm}$
Diffusivity (water–air)$D_{AB}$$0.258\ \text{cm}^2/\text{s}$
Water density$\rho_L$$62.4\ \text{lb/ft}^3$

Find. The time $t$ for the spill to evaporate completely.

water spill (evaporating) stagnant air film, δ = 0.19 in still air: H∞ = 0.0019, 74°F, 1 atm NA Hs H∞
Fig. C1: Water vapour (A) diffuses upward through a stationary air (B) film of thickness $\delta$: saturated at the water surface ($H_s$) and at the bulk humidity ($H_\infty$) at the top. This is Stefan diffusion of A through stagnant B.

Approach. Convert the humidities to vapour partial pressures, apply the Stefan (diffusion-through-stagnant-gas) flux across the film, then divide the total moles of water by the molar evaporation rate. The spill footprint is taken as $1\ \text{ft}^2$ (see the callout).

  1. Humidities to partial pressures. Since $H=\dfrac{M_w}{M_{air}}\dfrac{p_A}{P-p_A}$ with $M_w/M_{air}=18/29$, invert to get $p_A=P\dfrac{H(29/18)}{1+H(29/18)}$. At the surface $p_{A1}=0.0294\ \text{atm}$ (from $H_s$) and at the film top $p_{A2}=0.00305\ \text{atm}$ (from $H_\infty$).
  2. Stefan diffusion flux. For A diffusing through stagnant B across a film of thickness $\delta$, $$N_A=\frac{D_{AB}\,P}{R\,T\,\delta}\ln\!\frac{P-p_{A2}}{P-p_{A1}}=\frac{0.258(1)}{82.06(296.5)(0.483)}\ln\!\frac{0.99695}{0.97060}.$$
  3. Evaluate the flux. The prefactor is $2.20\times10^{-5}$ and $\ln(1.0271)=0.0268$, so $$\boxed{N_A\approx5.89\times10^{-7}\ \text{mol/(cm}^2\!\cdot\text{s)}.}$$
  4. Total moles of water. Mass $=V\rho_L=0.00325(62.4)=0.203\ \text{lb}=92.0\ \text{g}$, so moles $=92.0/18=5.11\ \text{mol}$. Spread over $A=1\ \text{ft}^2=929\ \text{cm}^2$ this is a $\approx1\ \text{mm}$ deep puddle.
  5. Evaporation time. $$t=\frac{\text{moles}}{N_A\,A}=\frac{5.11}{(5.89\times10^{-7})(929)}\;\Longrightarrow\;\boxed{t\approx9.35\times10^{3}\ \text{s}\approx2.6\ \text{h}.}$$
QuantityResult
Surface / bulk vapour pressure$p_{A1}=0.0294\ \text{atm}$, $p_{A2}=0.00305\ \text{atm}$
Molar evaporation flux$N_A\approx5.89\times10^{-7}\ \text{mol/(cm}^2\!\cdot\text{s)}$
Time to evaporate (footprint $1\ \text{ft}^2$)$t\approx9.35\times10^{3}\ \text{s}\approx2.6\ \text{h}$
Check — spill footprint (data gap)

The evaporation flux $N_A$ is fully fixed by the data, but the total time needs the spill area, which the question does not give. A $1\ \text{ft}^2$ footprint is assumed — it makes the 92 cm³ spill a physically realistic $\approx1\ \text{mm}$-deep puddle. The time scales inversely with area: $t=\rho_L V/(M_A N_A A)$, so a footprint twice as large evaporates in half the time. State the assumed area with the answer.