Humidification of Air Over a Water Container: 23-Chem-A3 Heat and Mass Transfer, May 2013, Q3

Question 3: Humidification of Air Over a Water Container (Part B — 20%)

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. Air at $32\,°\text{C}$, 1.0 atm, 25% RH, $u=0.15$ m/s flows over a water surface $L=1.2$ m long; interface at $20\,°\text{C}$. $\nu=1.51\times10^{-5}\ \text{m}^2/\text{s}$, $D_{AB}=2.77\times10^{-5}\ \text{m}^2/\text{s}$; $p_{\text{sat}}(20)=0.02308$ atm, $p_{\text{sat}}(32)=0.04696$ atm.

Find. (a) the average convective mass-transfer coefficient $k_c$; (b) the water evaporation rate per unit width of the container.

Figure 3 — Laminar external flow over a flat liquid surface. A concentration boundary layer grows along the 1.2 m plate; the average Sherwood number gives k_c, and the surface-to-bulk concentration difference drives the evaporation.

Approach. Check the flow regime with $Re_L$, obtain the average Sherwood number from the laminar flat-plate correlation, convert to $k_c$, then multiply by the interface-to-bulk water-vapour concentration difference and the plate area per unit width.

Reynolds and Schmidt numbers. $$Re_L=\frac{uL}{\nu}=\frac{(0.15)(1.2)}{1.51\times10^{-5}}=1.19\times10^{4}\;(<5\times10^{5}\Rightarrow\text{laminar}),\qquad Sc=\frac{\nu}{D_{AB}}=0.545.$$
Average Sherwood number and coefficient. For a laminar flat plate, $$\overline{Sh}_L=0.664\,Re_L^{1/2}Sc^{1/3}=0.664(109.2)(0.817)=59.2,$$ $$k_c=\frac{\overline{Sh}_L\,D_{AB}}{L}=\frac{(59.2)(2.77\times10^{-5})}{1.2}=\boxed{1.37\times10^{-3}\ \text{m/s}}.$$
Interface and bulk vapour concentrations (ideal gas). At the surface (saturated at 20 °C) and in the bulk (25% RH at 32 °C), $$c_{As}=\frac{p_{\text{sat}}(20)}{RT_s}=\frac{2338.6}{(8.314)(293.15)}=0.960\ \text{mol/m}^3,\quad c_{A\infty}=\frac{0.25\,p_{\text{sat}}(32)}{RT_\infty}=\frac{1189.6}{(8.314)(305.15)}=0.469\ \text{mol/m}^3.$$
Evaporation per unit width. The flux is $N_A=k_c(c_{As}-c_{A\infty})=1.37\times10^{-3}(0.960-0.469)=6.71\times10^{-4}\ \text{mol}/(\text{m}^2\cdot\text{s})$. Over area $L\times1$ m, $$\dot m_A'=N_A L\,M_{\text{H}_2\text{O}}=(6.71\times10^{-4})(1.2)(0.018015)=\boxed{1.45\times10^{-5}\ \text{kg}/(\text{s}\cdot\text{m})}\;(=8.05\times10^{-4}\ \text{mol}/(\text{s}\cdot\text{m})).$$

Quantity	Result
(a) Mass-transfer coefficient, $k_c$	$1.37\times10^{-3}$ m/s
Molar flux, $N_A$	$6.71\times10^{-4}\ \text{mol}/(\text{m}^2\cdot\text{s})$
(b) Water evaporated per unit width	$1.45\times10^{-5}\ \text{kg}/(\text{s}\cdot\text{m})$ ($\approx0.052$ kg/h·m)

Check: surface concentration is evaluated at the interface temperature (20 °C, saturated) and the bulk at 32 °C; using a single film temperature for both shifts the driving force by only a few percent. Laminar flow is confirmed by $Re_L\approx1.2\times10^{4}$.

Concept & Theory

Topic. External forced-convection mass transfer (heat–mass analogy) from a flat liquid surface.

Key relations. $Re_L=uL/\nu$, $Sc=\nu/D_{AB}$, laminar plate $\overline{Sh}_L=0.664Re_L^{1/2}Sc^{1/3}$, $k_c=\overline{Sh}_L D_{AB}/L$; evaporation $N_A=k_c\,\Delta c_A$.

Why this works. The Blasius laminar boundary-layer solution for a flat plate carries over to mass transfer by replacing $Pr$ with $Sc$ and $Nu$ with $Sh$. The average coefficient (factor 0.664, twice the trailing-edge local value) is the correct quantity for the total evaporation over the whole plate.

Common pitfall. Forgetting to convert saturation pressures to molar concentrations at the correct temperatures, or applying a turbulent correlation — here $Re_L$ is well below the $5\times10^{5}$ transition.

Source. Welty et al., Fundamentals of Momentum, Heat and Mass Transfer §30 (convective mass transfer, flat plate); Incropera & DeWitt §7.2 with the heat–mass analogy.

Question 3 of 7: Humidification of Air Over a Water Container

Question 3: Humidification of Air Over a Water Container (Part B — 20%)