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23-Chem-A3 Heat and Mass Transfer · May 2016

Question 4 of 7: Effective gas-film thickness; water then ethanol

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

Notes on this paper

Paper format. EGBC 04-CHEM-A3 Mass Transfer Operations, May 2016 — three-hour open-book exam. Seven questions in three parts: Part A (Q1–2, molecular diffusion), Part B (Q3–4, film mass transfer), Part C (Q5–7, staged and equilibrium separations). The candidate answers one of Q1–2, one of Q3–4 and two of Q5–7 (four questions of equal value). All seven are worked in full below.

Reference texts. J.M. Coulson & J.F. Richardson, Chemical Engineering Vol. 1 (Fluid Flow, Heat and Mass Transfer) and Vol. 2 (Particle Technology & Separation Processes) — the source of all seven problems; C.J. Geankoplis, Transport Processes and Separation Process Principles; R.E. Treybal, Mass-Transfer Operations; Bird, Stewart & Lightfoot, Transport Phenomena; Perry’s Chemical Engineers’ Handbook.

Question 4 — Effective gas-film thickness; water then ethanol (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.

Check — source data taken as printed

The printed rate “4.11 × 103” is read as $4.11\times10^{-3}$ kg m⁻² s⁻¹ (a flux of 4110 kg m⁻² s⁻¹ is physically impossible; the minus sign was lost in printing). The printed water vapour pressure, 34 mm Hg, is well below the true saturation value at 349 K (≈ 300 mm Hg); per exam Note 2 the value is used exactly as given, so the “effective” film thickness below is a lumped fitted parameter. Had 304 mm Hg been intended, the film would be ≈ 2.0 mm and the ethanol rate ≈ 0.012 kg m⁻² s⁻¹; parts (b) and (c) use the same film derived in (a), so the method is unchanged.

Given. Evaporation through a stagnant gas film with $p_A(\text{bulk})=0$. Water: rate $=4.11\times10^{-3}\text{ kg m}^{-2}\text{s}^{-1}$, $T=349\text{ K}$, $p^{*}=34\text{ mm Hg}=4.53\text{ kPa}$, $D=2.6\times10^{-5}$. Ethanol: $T=343\text{ K}$, $p^{*}=544\text{ mm Hg}=72.5\text{ kPa}$, $D=1.2\times10^{-5}$. Total $P=101.3\text{ kPa}$.

Find. (a) film thickness $L$; (b) ethanol mass rate at the same $L$; (c) bulk-flow fraction at the ethanol surface.

gas film L (effective)bulk air p_A = 0liquid (p_A = p°)
Stagnant gas film of effective thickness $L$ over the liquid; vapour diffuses through non-diffusing air ($p_A=0$ in the bulk).

Approach. Apply the Stefan (stagnant-air) flux to the water data to back out $L$; reuse that $L$ with ethanol’s properties for its rate; the bulk-flow proportion at the surface is simply the surface mole fraction.

  1. Stefan flux & film thickness (part a). $N_A=\dfrac{D\,P}{RT\,L\,p_{BM}}\,(p_A^{*}-0)$ with $p_{BM}$ the log-mean of $(P-p_A^{*})$ and $P$. For water, $N_A=\dfrac{4.11\times10^{-3}}{0.018}=0.228\text{ mol m}^{-2}\text{s}^{-1}$, $p_{BM}=99.0\text{ kPa}$. Solving, $$L=\frac{D\,P\,p_A^{*}}{RT\,N_A\,p_{BM}}\;\Rightarrow\;\boxed{L\approx0.18\ \text{mm (effective)}.}$$
  2. Ethanol rate at the same film (part b). Now $p_A^{*}=72.5\text{ kPa}$, $p_{BM}=57.6\text{ kPa}$, $D=1.2\times10^{-5}$: $$N_A=\frac{D\,P\,p_A^{*}}{RT\,L\,p_{BM}}=2.95\text{ mol m}^{-2}\text{s}^{-1}\;\Rightarrow\;\dot m=N_A M_{\text{EtOH}}=\boxed{0.14\ \text{kg m}^{-2}\text{s}^{-1}.}$$ Despite ethanol’s lower diffusivity, its far higher vapour pressure makes it evaporate much faster than water.
  3. Bulk-flow proportion at the surface (part c). The total flux splits as $N_A=\underbrace{-C_TD\,dy_A/dz}_{\text{diffusion}}+\underbrace{y_A N_A}_{\text{bulk flow}}$; the bulk-flow fraction is $y_A$, largest at the surface: $$\frac{\text{bulk}}{\text{total}}\Big|_{\text{surface}}=y_{A,1}=\frac{p_A^{*}}{P}=\frac{72.5}{101.3}=\boxed{0.716\;(71.6\%).}$$
QuantityValue
(a) Effective gas-film thickness≈ 0.18 mm
(b) Ethanol evaporation rate0.14 kg m⁻² s⁻¹
(c) Bulk-flow fraction at ethanol surface71.6 %