Question 5 of 5: Vadose-Zone Moisture Profiles, Brooks-Corey Parameter Fitting, and Wetting-Front Advance

Question 5: Vadose-Zone Moisture Profiles, Brooks-Corey Parameter Fitting, and Wetting-Front Advance (20 marks: a-6, b-7, c-7)

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). $k=10^{-12}\ \text{m}^2$, $n=0.4$; Brooks-Corey $\lambda=3.0$, $p_d=40$ cm, $S_{wr}=0.1$, $S_{max}=1.0$; water table 2 m below the surface, zero vertical flux.

Find (a). Moisture content at the ground surface; then, for $RH=50\%$ at the surface (still zero flux), the depth to the water table.

Approach. Under static (zero-flux) equilibrium the matric suction head equals the height above the water table; converting that suction to saturation uses the Brooks-Corey retention curve. For the RH case, the Kelvin equation converts relative humidity directly to an equivalent suction head, which under the same static assumption equals the depth to the water table.

1. Suction head at the surface (2 m water table). Under zero flux, $\psi(z)=$ height above the water table $=200$ cm at the surface.
2. Effective and actual saturation (Brooks-Corey). $$S_e=\left(\frac{p_d}{\psi}\right)^{\lambda}=\left(\frac{40}{200}\right)^{3.0}=0.00800.$$ $$S_w=S_{wr}+S_e(S_{max}-S_{wr})=0.1+0.008(0.9)=0.1072.$$
3. Volumetric moisture content. $$\theta=nS_w=0.4(0.1072)=\boxed{0.0429}\ (4.29\%\ \text{by volume}).$$
4. Kelvin-equation suction at $RH=50\%$. $$\psi_{RH}=-\frac{RT}{M_wg}\ln(RH)=-\frac{(8.314)(293.15)}{(0.018)(9.81)}\ln(0.5)=\boxed{9560\ \text{m}}.$$

Given (b). $(S_w,\psi)$ pairs from TDR + tensiometer: surface ($z=0$) $S_w=0.10$, $\psi=2.0$ m; $z=1$ m, $S_w=0.50$, $\psi=1.0$ m. $S_{wr}=0.05$, $S_{max}=1.0$, $n=0.4$, $K_{sat}=10^{-3}\ \text{cm/s}$.

Find (b). Brooks-Corey $p_d$ and $\lambda$; depth to the water table (no flow).

Approach. Two independent $(S_e,\psi)$ points on the Brooks-Corey curve give two equations in the two unknowns $p_d,\lambda$; solve simultaneously by taking logs. The same zero-flux, suction-equals-height-above-water-table relation used in part (a) gives the water-table depth, checked for consistency at both measurement points.

1. Effective saturations. $$S_{e,1}=\frac{0.10-0.05}{1-0.05}=0.0526\ (\text{surface}),\qquad S_{e,2}=\frac{0.50-0.05}{1-0.05}=0.4737\ (1\ \text{m depth}).$$
2. Solve for $\lambda$ (log-linear Brooks-Corey form, $\ln S_e=\lambda\ln p_d-\lambda\ln\psi$). $$\lambda=\frac{\ln S_{e,1}-\ln S_{e,2}}{\ln\psi_2-\ln\psi_1}=\frac{\ln(0.0526)-\ln(0.4737)}{\ln(100)-\ln(200)}=\boxed{3.17}.$$ (heads in cm: $\psi_1=200$, $\psi_2=100$.)
3. Solve for $p_d$. $$\ln p_d=\frac{\ln S_{e,1}}{\lambda}+\ln\psi_1=\frac{-2.944}{3.17}+5.298=4.369\ \Rightarrow\ p_d=\boxed{79.0\ \text{cm}}.$$ (Confirmed identically from the $z=1$ m point.)
4. Depth to the water table. Under zero flux, $\psi(z)=D_{wt}-z$, so $D_{wt}=z+\psi(z)$. Surface: $D_{wt}=0+2.0=2.0$ m. At $z=1$ m: $D_{wt}=1+1.0=2.0$ m — both points agree exactly. $$D_{wt}=\boxed{2.0\ \text{m}}.$$

Given (c). Soil of part (b) ($p_d=79.0$ cm, $\lambda=3.17$, $n=0.4$, $K_{sat}=10^{-3}\ \text{cm/s}$), initially uniform $S_{w,i}=0.05$ to 10 m depth; infiltrating pulse $S_{w,wet}=0.9$; target wetting-front depth $L=30$ cm.

Find (c). Time for the wetting front to reach 30 cm depth.

Approach. A sharp, near-saturated wetting front advancing into much drier soil is the classic Green-Ampt scenario; use the Brooks-Corey air-entry pressure $p_d$ from part (b) as the effective wetting-front suction and solve the (here non-iterative, since the target depth is specified directly) Green-Ampt time-infiltration relation.

1. Moisture deficit across the front. $$\Delta\theta=n(S_{w,wet}-S_{w,i})=0.4(0.9-0.05)=\boxed{0.340}.$$
2. Cumulative infiltration at $L=30$ cm. $$F=L\,\Delta\theta=30(0.340)=10.2\ \text{cm}.$$
3. Green-Ampt time-infiltration relation. $$t=\frac{F-\psi_f\Delta\theta\,\ln\!\left(1+\dfrac{F}{\psi_f\Delta\theta}\right)}{K_{sat}},\qquad \psi_f\approx p_d=79.0\ \text{cm}.$$ $$\psi_f\Delta\theta=79.0(0.340)=26.9\ \text{cm};\quad \ln\!\left(1+\frac{10.2}{26.9}\right)=0.322.$$ $$t=\frac{10.2-26.9(0.322)}{10^{-3}}=\boxed{1554\ \text{s}}\ (\approx25.9\ \text{min}).$$