Question 6 of 6: C2 — Falling-level pool evaporation

Question 6: C2 — Falling-level pool evaporation (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. An open pool (constant plan area, vertical walls, total depth $H$) losing water by evaporation into the air above. As the level falls, the length $z$ of the stagnant air column between the water surface and the pool rim grows: at fill fraction $\phi$ the air column is $z=(1-\phi)H$. The breeze holds the conditions above the rim constant.

Find. (a) the modelling assumptions; (b) the average flux $N_A$; (c) $t_1$, $t_2$ and the ratio $t_2/t_1$.

Part (a) — assumptions

To reduce this moving-surface problem to a tractable Stefan-type diffusion calculation, we assume: (i) evaporation is controlled by steady one-dimensional diffusion of water vapour through the stagnant air column of length $z$ above the surface (pseudo-steady state — the level falls slowly compared with the time for the vapour profile to establish); (ii) the constant breeze fixes the vapour concentration at the rim, so the concentration difference driving diffusion, from saturation $C_{As}$ at the surface to the bulk $C_{A\infty}$ at the top, is constant in time; (iii) constant, uniform temperature and pressure (constant $D_{AB}$ and $c$); (iv) constant plan area (vertical walls), so the fill fraction maps linearly to the air-column length $z=(1-\phi)H$; and (v) dilute vapour, so the flux takes the simple Fickian form $N_A\approx D_{AB}(C_{As}-C_{A\infty})/z$.

Part (b) — average flux

1. Pseudo-steady diffusion through the air column. One-dimensional Fickian diffusion across the stagnant column of current length $z$ gives $$N_A=\frac{D_{AB}\,(C_{As}-C_{A\infty})}{z}\equiv\frac{K}{z},\qquad K\equiv D_{AB}(C_{As}-C_{A\infty}),$$ where $K$ is constant by assumption (ii)–(iii). The flux therefore falls as the column lengthens.

Part (c) — time intervals and their ratio

1. Liquid mass balance links flux to a moving level. Each mole leaving the surface lengthens the air column: with liquid molar density $c_L$, $c_L\dfrac{dz}{dt}=N_A=\dfrac{K}{z}.$
2. Separate and integrate — the $z^2$ law. $z\,dz=\dfrac{K}{c_L}\,dt$ integrates over one step from $z_i$ to $z_f$ to $$t=\frac{c_L}{2K}\big(z_f^2-z_i^2\big).$$ Each interval’s duration depends only on the difference of squares of the air-column lengths.
3. Evaluate the two steps. Writing $z=(1-\phi)H$: for $t_1$ (80 %→70 %) $z$ goes $0.2H\!\to\!0.3H$; for $t_2$ (70 %→60 %) $z$ goes $0.3H\!\to\!0.4H$. Thus $$t_1=\frac{c_LH^2}{2K}\big(0.3^2-0.2^2\big)=\frac{c_LH^2}{2K}(0.05),\quad t_2=\frac{c_LH^2}{2K}\big(0.4^2-0.3^2\big)=\frac{c_LH^2}{2K}(0.07).$$
4. Ratio. The prefactors cancel: $$\boxed{\;\frac{t_2}{t_1}=\frac{0.07}{0.05}=1.4\;}$$ The second equal-mass step takes 40 % longer than the first, because the vapour must diffuse across a longer air column.