Question 8 of 8: m-Xylene Pool Volatilization Time

Question B-3: m-Xylene Pool Volatilization Time (20 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. $V = 5{,}000$ L spilled as a pool of thickness $\delta = 2.5$ mm; density $\rho = 0.867$ g/mL; molar mass of m-xylene ($C_8H_{10}$), $M = 106.16$ g/mol; $T = 25^\circ\text{C} = 298.15$ K; partial (vapour) pressure $p = 0.0109$ atm; air-phase mass-transfer coefficient $k_g = 1500$ cm/h at 3 m/s wind.

Find. The time for the spilled pool to fully volatilize, and the qualitative/quantitative effect of wind speed on that rate.

Approach. Two-film mass-transfer theory: with background air concentration ≈ 0, the volatilization flux is gas-phase controlled and given by $N = k_g p/(RT)$; multiplying by the pool area and total spilled mass gives the time to fully evaporate.

1. Pool area. $A = V/\delta = \dfrac{5.0\ \text{m}^3}{0.0025\ \text{m}} = \boxed{2000}\ \text{m}^2$.
2. Total mass spilled. $m = V\rho = 5{,}000{,}000\ \text{mL} \times 0.867\ \text{g/mL} = \boxed{4335}\ \text{kg}$.
3. Molar volatilization flux. Converting $k_g = 1500$ cm/h $= 15$ m/h, and using $R = 8.206\times10^{-5}\ \text{atm}\cdot\text{m}^3/(\text{mol}\cdot\text{K})$: $N = \dfrac{k_g\,p}{RT} = \dfrac{15 \times 0.0109}{8.206\times10^{-5}\times298.15} = 6.68\ \text{mol/(m}^2\cdot\text{h)}$.
4. Mass flux and total evaporation rate. $n'' = NM = 6.68 \times 106.16 = 709\ \text{g/(m}^2\cdot\text{h)}$; over the whole pool, $\dot m = n''A = 709 \times 2000 = \boxed{1419}\ \text{kg/h}$.
5. Time to fully volatilize. $t = m/\dot m = 4335/1419 = \boxed{3.06}\ \text{h}$ (≈3 h 4 min).

Impact of wind speed. The air-phase mass-transfer coefficient $k_g$ is itself a function of wind speed — correlations such as MacKay & Matsugu's for a flat evaporating pool give roughly $k_g \propto u^{0.78}$, so volatilization is directly wind-driven: faster air movement thins the stagnant boundary layer immediately above the pool surface, steepening the concentration gradient that drives diffusion into the bulk air and raising $k_g$ (and hence the evaporation rate) accordingly. Applying that scaling, doubling the wind speed from 3 to 6 m/s would raise $k_g$ (and the evaporation rate) by a factor of $2^{0.78} \approx 1.72$, cutting the volatilization time to roughly $3.06/1.72 = \boxed{1.78}\ \text{h}$; conversely, calm conditions (low wind) would substantially prolong both the volatilization time and the associated inhalation-hazard/odour footprint around the spill, which is why wind speed and direction are core inputs to any spill-response air-monitoring and evacuation-distance decision.