Question 4 of 5: Capillarity – Rise in a Tube, a Non-Wetting Droplet Between Plates, and Total TCE Mass

Question 4: Capillarity – Rise in a Tube, a Non-Wetting Droplet Between Plates, and Total TCE Mass (20 marks: a-4, b-8, c-8)

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 (a). Tube diameter $d=0.1\ \text{mm}$; fluid density $\rho=1400\ \text{kg/m}^3$; capillary rise $h=4\ \text{cm}$; contact angle $\theta=10^\circ$.

Find (a). Air-liquid interfacial tension $\sigma$.

Approach. Rearrange the standard capillary-rise (Jurin's law) equation for $\sigma$.

1. Capillary rise equation, solved for $\sigma$. $$h=\frac{2\sigma\cos\theta}{\rho g r}\ \Rightarrow\ \sigma=\frac{h\rho gr}{2\cos\theta},\qquad r=\frac{d}{2}=5\times10^{-5}\ \text{m}.$$ $$\sigma=\frac{(0.04)(1400)(9.81)(5\times10^{-5})}{2\cos10^\circ}=\boxed{13.9\ \text{mN/m}}\ (=13.9\ \text{dyn/cm}).$$

Given (b). Plate gap $t=10\ \mu\text{m}$; mercury droplet volume $V=0.7\ \text{mL}$, approximately cylindrical; nonwetting (air-Hg interfacial tension $\sigma=480\ \text{dyn/cm}=0.480\ \text{N/m}$).

Find (b). Weight of the upper plate the droplet can support.

Approach. For a droplet sandwiched between parallel plates, the only significant curvature is across the gap (radius $=t/2$); because mercury is non-wetting, the meniscus bulges outward and the internal (Laplace) pressure exceeds atmospheric, pushing the plates apart — that excess pressure, acting over the droplet's footprint, is what supports the upper plate's weight.

1. Capillary (Laplace) pressure across the gap. $$P_c=\frac{2\sigma}{t}=\frac{2(0.480)}{10\times10^{-6}}=\boxed{9.60\times10^{4}\ \text{Pa}}.$$
2. Droplet footprint area. Cylindrical droplet of height $t$: $$A=\frac{V}{t}=\frac{0.7\times10^{-6}\ \text{m}^3}{10\times10^{-6}\ \text{m}}=\boxed{0.0700\ \text{m}^2}.$$
3. Supported weight. $$W=P_c\,A=(9.60\times10^{4})(0.0700)=\boxed{6720\ \text{N}}\ (\approx685\ \text{kgf}).$$

Given (c). TCE vapour (pore-gas) concentration $C_g=100\ \text{mg/L}$; distribution coefficient $K_d=2\ \text{L/kg}$; $n=0.4$; $\rho_b=2\ \text{kg/L}$; $S_w=0.3$, $S_g=0.7$; dimensionless $H=0.42$.

Find (c). Total mass of TCE per m³ of soil (vapour + dissolved + sorbed).

Approach. Back out the pore-water concentration from the given vapour concentration via Henry's law, get the sorbed concentration from $K_d$, then sum all three phases on a per-m³-of-bulk-soil basis using the volumetric water/gas fractions and bulk density.

1. Pore-water and sorbed concentrations. $$C_w=\frac{C_g}{H}=\frac{100}{0.42}=238.1\ \text{mg/L}.$$ $$C_s=K_dC_w=2(238.1)=476.2\ \text{mg/kg}.$$
2. Volumetric water and gas content. $$\theta_w=nS_w=0.4(0.3)=0.12,\qquad \theta_g=nS_g=0.4(0.7)=0.28.$$
3. Mass per m³ of soil, by phase. $$M_{gas}=C_g\theta_g(1000\ \text{L/m}^3)=100(0.28)(1000)=28{,}000\ \text{mg/m}^3.$$ $$M_{water}=C_w\theta_w(1000)=238.1(0.12)(1000)=28{,}571\ \text{mg/m}^3.$$ $$M_{sorbed}=C_s\rho_b(1000\ \text{L/m}^3)=476.2(2)(1000)=952{,}381\ \text{mg/m}^3.$$
4. Total mass per m³ of soil. $$M_{total}=M_{gas}+M_{water}+M_{sorbed}=28{,}000+28{,}571+952{,}381=\boxed{1.009\times10^{6}\ \text{mg/m}^3}\ (\approx1.01\ \text{kg/m}^3).$$ Sorption on the solids accounts for about 94% of the total mass, illustrating why sorbed-phase soil sampling — not soil-gas or groundwater alone — typically dominates a TCE source-zone mass estimate.