Ethanol Evaporation from a Falling Film into a Cross-flow of Air: 23-Chem-A3 Heat and Mass Transfer, December 2016, Q3

Question 3: Ethanol Evaporation from a Falling Film into a Cross-flow of Air (Part B — 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. Plate 2 m (flow width) × 4 m (down-slope), evaporating area $A=8$ m². Air at $u=3$ m/s, 303 K, 1 atm sweeps along the 2 m dimension. $D=1.32\times10^{-5}$ m²/s, $\nu=1.533\times10^{-5}$ m²/s, ethanol vapour pressure $p_s=6.45\times10^{-2}$ atm at the 289 K surface.

Find. the liquid ethanol feed rate that is exactly consumed by evaporation over the plate (so the film just dries out at the bottom).

Figure 3 — Ethanol film runs down the 4 m slope while 3 m/s air crosses the 2 m width; the developing concentration boundary layer sets an average flat-plate mass-transfer coefficient.

Check (data typo): the printed kinematic viscosity $1.533\times10^{-6}$ m²/s gives a non-physical Schmidt number $Sc=\nu/D=0.12$; the physically consistent value for the ethanol–air gas film is $\nu=1.533\times10^{-5}$ m²/s, giving $Sc=1.16$, which is used here.

Approach. Treat the plate as a flat plate in parallel flow: form $Re_L$ and $Sc$, get the average Sherwood number, hence the mass-transfer coefficient, then multiply the surface concentration driving force by the area.

Reynolds and Schmidt numbers. With flow length $L=2$ m, $Re_L=\dfrac{uL}{\nu}=\dfrac{3(2)}{1.533\times10^{-5}}=3.91\times10^{5}$ (<$5\times10^5$, laminar), and $Sc=\dfrac{\nu}{D}=1.16$.
Average Sherwood number (laminar flat plate). $$\overline{Sh}=0.664\,Re_L^{1/2}Sc^{1/3}=0.664(625.6)(1.050)=\boxed{437}.$$
Mass-transfer coefficient. $k_c=\dfrac{\overline{Sh}\,D}{L}=\dfrac{437(1.32\times10^{-5})}{2}=2.88\times10^{-3}$ m/s.
Surface concentration driving force. The film temperature is $\tfrac12(289+303)=296$ K; the ethanol-free air gives $c_{A\infty}=0$, so $$c_{As}=\frac{p_s}{RT_f}=\frac{6.45\times10^{-2}(101325)}{8.314(296)}=2.66\ \text{mol/m}^3.$$
Flux and required feed. $N_A=k_c\,c_{As}=7.65\times10^{-3}$ mol/m²s. Over $A=8$ m² the ethanol removed — which must equal the liquid supplied — is $$\dot m=N_A A M=7.65\times10^{-3}(8)(0.04607)=\boxed{2.82\times10^{-3}\ \text{kg/s}=10.2\ \text{kg/h}}.$$

Quantity	Result
$Re_L$ / $Sc$ / $\overline{Sh}$	$3.91\times10^{5}$ / 1.16 / 437
$k_c$	$2.88\times10^{-3}$ m/s
Ethanol feed rate	$2.82\times10^{-3}$ kg/s = 10.2 kg/h

Concept & Theory

Topic. Convective mass transfer from a flat plate in parallel laminar flow (heat–mass analogy).

Key relations. $\overline{Sh}=0.664\,Re_L^{1/2}Sc^{1/3}$ (analogue of $\overline{Nu}=0.664\,Re^{1/2}Pr^{1/3}$); $N_A=k_c(c_{As}-c_{A\infty})$.

Why this works. If exactly the amount that can evaporate is fed, the film thins to zero at the bottom edge; the plate-average coefficient (from the average Sherwood number) times the full area gives that maximum evaporation rate. Ethanol is dilute ($y_s\approx0.065$), so the low-flux coefficient applies without a bulk-flow correction.

Common pitfall. Using the 4 m length (down-slope) as the flow dimension — the air travels across the 2 m width — or evaluating $c_{As}$ at the bulk 303 K rather than the film temperature.

Source. Geankoplis §7.3 (mass transfer to a flat plate); Treybal §3.1 (mass-transfer coefficients).

Question 3 of 7: Ethanol Evaporation from a Falling Film into a Cross-flow of Air

Question 3: Ethanol Evaporation from a Falling Film into a Cross-flow of Air (Part B — equal value)