Part B: Evaporation of Water into Air Flowing Over a Flat Surface: 23-Chem-A3 Heat and Mass Transfer, December 2017, Q6

Question 6 — Part B: Evaporation of Water into Air Flowing Over a Flat Surface (25 points)

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 2.5 m/s passes over a $L=2$ m water surface. The air holds water vapour at 10 mm Hg with 33.3% relative humidity, so the saturation pressure (and hence the surface, which is saturated) is $p^{*}=10/0.333=30$ mm Hg, corresponding to a surface temperature near 29 °C. The vapour transfers from the saturated surface (30 mm Hg) to the bulk air (10 mm Hg).

Find. The evaporation rate per unit area, $\dot m''$ (kg water per m² per hour).

Figure 6 — Forced-convection evaporation: air sweeps a 2-m water surface, a concentration boundary layer grows along the plate, and water leaves at a rate set by the flat-plate mass-transfer coefficient and the 30 → 10 mm Hg driving force.

Approach. Treat the surface as a flat plate in parallel flow: form $Re_L$ and $Sc$, use the laminar average Sherwood correlation $Sh=0.664\,Re_L^{1/2}Sc^{1/3}$ to get $k_c$, then multiply the molar flux $k_c\,\Delta c$ by the molar mass to get the water loss.

Check

The transport properties as printed are physically impossible: $D=7.0$ cm²/hr and $\nu=1\times10^{-3}$ cm²/hr give a Schmidt number $\sim10^{-4}$, whereas water-vapour–air has $Sc\approx0.6$. Per the exam’s own instruction to proceed with a sound methodology when data are unreliable, standard values at ~29 °C are adopted: $\nu_\text{air}=1.6\times10^{-5}$ m²/s and $D_{\text{H}_2\text{O-air}}=2.6\times10^{-5}$ m²/s ($Sc=0.62$). The method is unaffected; only the numeric answer depends on these substituted values.

Surface state. Relative humidity ties the bulk vapour pressure to saturation: $p^{*}=p_v/\phi = 10/0.333 = 30$ mm Hg, so the saturated surface sits at ≈ 29 °C and the driving pressure difference is $\Delta p = 30-10 = 20$ mm Hg $= 2.67$ kPa.
Flow and property groups. With the adopted air properties, $$Re_L=\frac{uL}{\nu}=\frac{(2.5)(2)}{1.6\times10^{-5}}=3.1\times10^{5}\ (\text{laminar}),\qquad Sc=\frac{\nu}{D_{AB}}=\frac{1.6\times10^{-5}}{2.6\times10^{-5}}=0.62.$$
Average mass-transfer coefficient. The laminar flat-plate analogue of the Blasius heat-transfer result is $Sh=0.664\,Re_L^{1/2}Sc^{1/3}$: $$Sh = 0.664\,(3.1\times10^{5})^{1/2}(0.62)^{1/3}=316,\qquad k_c=\frac{Sh\,D_{AB}}{L}=\frac{(316)(2.6\times10^{-5})}{2}=4.1\times10^{-3}\ \text{m/s}.$$
Molar flux and water loss. The interfacial concentration difference is $\Delta c=\Delta p/(RT)=2670/[(8.314)(302)]=1.06\ \text{mol/m}^3$, so the molar flux is $N_A=k_c\,\Delta c = (4.1\times10^{-3})(1.06)=4.37\times10^{-3}$ mol m$^{-2}$s$^{-1}$. Multiplying by $M=0.01802$ kg/mol and 3600 s/hr, $$\dot m'' = N_A M \times3600 = (4.37\times10^{-3})(0.01802)(3600) = \boxed{0.28\ \text{kg}\,\text{m}^{-2}\text{hr}^{-1}}.$$

Quantity	Result
Surface saturation pressure $p^{*}$	30 mm Hg (≈ 29 °C surface)
$Re_L$ / $Sc$ / $Sh$	$3.1\times10^{5}$ / 0.62 / 316
Mass-transfer coefficient $k_c$	$4.1\times10^{-3}$ m/s
Water evaporation rate	≈ 0.28 kg·m$^{-2}$·hr$^{-1}$

Concept & Theory

Topic. Forced-convection mass transfer (evaporation) from a flat plate, using the heat–mass-transfer analogy for the Sherwood number.

Key relations. Laminar flat plate $Sh=0.664\,Re_L^{1/2}Sc^{1/3}$ (the mass analogue of $Nu=0.664\,Re_L^{1/2}Pr^{1/3}$); $k_c=Sh\,D_{AB}/L$; flux $N_A=k_c(c_s-c_\infty)=k_c\,\Delta p/(RT)$; relative humidity $\phi=p_v/p^{*}$.

Why this works. The concentration boundary layer over the plate is governed by the same convection–diffusion balance as the thermal layer, so the Sherwood correlation is obtained from the Nusselt one by replacing $Pr$ with $Sc$. The water surface is saturated at its own temperature, so its vapour concentration is fixed by $p^{*}$; the humidity of the incoming air sets the far-field value, and their difference drives the flux.

Common pitfall. Forgetting that the surface is saturated (using the bulk 10 mm Hg on both sides gives zero driving force), or checking the transition — here $Re_L<5\times10^{5}$, so the fully laminar coefficient is valid; a longer plate would need the mixed laminar–turbulent form.

Source. Incropera & DeWitt, Fundamentals of Heat and Mass Transfer, Ch. 7 (external flow, the analogy); Treybal, Mass-Transfer Operations, Ch. 3.

Question 6 of 6: Part B: Evaporation of Water into Air Flowing Over a Flat Surface

Question 6 — Part B: Evaporation of Water into Air Flowing Over a Flat Surface (25 points)