Question 4 of 5: An Instantaneous Chloride Pulse, Capillary Rise, and Brooks-Corey Vadose-Zone Moisture

Question 4: An Instantaneous Chloride Pulse, Capillary Rise, and Brooks-Corey Vadose-Zone Moisture (equal value)

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) Darcy velocity $q=0.09\ \text{m/day}$, $n=0.37$, $\alpha_L=26\ \text{m}$, $D^{*}=1.1\times10^{-10}\ \text{m}^2/\text{s}$, instantaneous mass $M=1000\ \text{kg}$ Cl⁻, $t=110\ \text{days}$. (b) Tube radius $r=0.25\ \text{mm}$, interfacial tension $\sigma=72\ \text{dynes/cm}$, completely wetting ($\theta=0$). (c) Depth to water table $z=4\ \text{m}$, Brooks-Corey $\lambda=2.75$, $\Psi_d=0.43\ \text{m}$, $S_{wr}=0.15$, $S_m=1.0$, porosity $n=0.37$, hydraulic head gradient 0.1 (head lower at the water table), $T=10^{\circ}\text{C}$.

Find. (a) Peak chloride concentration and its location after 110 days, and the concentration 50 m from the ditch at the same time. (b) Height of capillary rise. (c) Moisture content and relative humidity at the ground surface.

Approach. (a) Convert the Darcy velocity to a seepage velocity, build the 1-D dispersion coefficient, and evaluate the Gaussian instantaneous-pulse solution at the plume centroid and at $x=50\ \text{m}$. (b) Apply the Young-Laplace/Jurin capillary-rise formula directly. (c) Integrate the given hydraulic-head gradient to get the suction head at the ground surface, convert to saturation via the Brooks-Corey retention curve, and convert suction to relative humidity via the Kelvin equation.

1. Part (a) — seepage velocity and dispersion coefficient. $v=q/n=0.09/0.37=0.2432\ \text{m/day}$. $$D_L=\alpha_Lv+D^{*}=(26)(0.2432)+(1.1\times10^{-10})(86400)=6.324+0.0000095\approx\boxed{6.324\ \text{m}^2/\text{day}}$$ (molecular diffusion is negligible against dispersion here, as usual at typical seepage velocities).
2. Peak concentration and its location. For an instantaneous 1-D pulse the concentration profile is Gaussian, centred on and travelling with the mean seepage velocity: $$x_{peak}=vt=(0.2432)(110)=\boxed{26.8\ \text{m}}\ \text{downgradient of the ditch}.$$ At the plume centroid the exponential term is 1, so $$C_{max}=\frac{M/(nA)}{\sqrt{4\pi D_Lt}}=\frac{1000/(0.37\times1)}{\sqrt{4\pi(6.324)(110)}}=\frac{2703}{93.5}=\boxed{28.9\ \text{kg/m}^3\ \text{(per m}^2\text{ of injection cross-section)}}.$$
3. Concentration 50 m from the ditch. $$C(50,110)=C_{max}\exp\!\left[-\frac{(x-x_{peak})^2}{4D_Lt}\right]=28.9\exp\!\left[-\frac{(50-26.8)^2}{4(6.324)(110)}\right]=28.9\,e^{-0.194}=\boxed{23.8\ \text{kg/m}^3\ \text{(per m}^2\text{)}}.$$ At $x=50\ \text{m}$, still only 23 m past the peak, the plume has spread little relative to its 110-day travel — consistent with the modest dispersion coefficient relative to the mean travel distance.
4. Part (b) — capillary rise. Convert units: $\sigma=72\ \text{dyne/cm}=0.072\ \text{N/m}$. Young-Laplace/Jurin's law with a fully wetting fluid ($\cos\theta=1$): $$h=\frac{2\sigma\cos\theta}{\rho g r}=\frac{2(0.072)}{(998)(9.81)(0.25\times10^{-3})}=\boxed{0.0588\ \text{m}\ (58.8\ \text{mm})}.$$
5. Part (c) — suction head at the ground surface. Taking the water table as datum ($z=0$, total head $H=0$ there, pressure head $=0$), the given uniform head gradient $dH/dz=0.1$ (head increases upward, i.e. water is being drawn/drained upward relative to hydrostatic) gives $H(4)=0.1(4)=0.4\ \text{m}$. Pressure head at the surface is $\psi_p=H-z=0.4-4=-3.6\ \text{m}$, i.e. a matric suction of $$\Psi=\boxed{3.6\ \text{m}}$$ (versus the 4.0 m suction that pure hydrostatic equilibrium, $dH/dz=0$, would give — the applied gradient reduces the suction slightly below the no-flow case).
6. Moisture content via Brooks-Corey. Since $\Psi=3.6\ \text{m}>\Psi_d=0.43\ \text{m}$, the medium is beyond air entry: $$S_e=\left(\frac{\Psi_d}{\Psi}\right)^{\lambda}=\left(\frac{0.43}{3.6}\right)^{2.75}=0.00290,\qquad S=S_{wr}+S_e(S_m-S_{wr})=0.15+0.00290(0.85)=\boxed{0.1525}.$$ Volumetric moisture content: $$\theta_w=Sn=(0.1525)(0.37)=\boxed{0.0564\ (5.64\%)}.$$
7. Relative humidity via the Kelvin equation. The soil-water suction depresses the equilibrium vapour pressure above the pore water: $$\ln(RH)=-\frac{M_w\,g\,\Psi}{R\,T}=-\frac{(0.018015)(9.81)(3.6)}{(8.314)(283.15)}=-2.70\times10^{-4},$$ $$RH=\boxed{99.97\%}.$$ Even several metres of suction barely depresses relative humidity (the Kelvin effect only becomes significant at suctions of hundreds of metres, e.g. near the wilting point), which is why RH is a poor discriminator of moisture state in this range but suction/moisture content are the useful diagnostics.