Question 4 of 5: Two-Layer Vadose-Zone Saturation Profile via Brooks-Corey and the Kelvin Equation

Question 4: Two-Layer Vadose-Zone Saturation Profile via Brooks-Corey and the Kelvin Equation (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. Water-table datum ($z=0$, $\psi=0$, $H=0$); sand layer $t_{sand}=0.6\ \text{m}$ ($i_{sand}=0.05$, $P_{d,sand}=0.05\ \text{m}$, $S_{wr,sand}=0.10$, $\lambda_{sand}=6.0$) directly above the water table; clay layer $t_{clay}=2.0\ \text{m}$ ($i_{clay}=0.26$, $P_{d,clay}=0.5\ \text{m}$, $S_{wr,clay}=0.25$, $\lambda_{clay}=2.0$) above the sand, up to the ground surface.

Find. (a) Water saturation in each material at the clay-sand interface, and at the ground surface. (b) Given $S_{gs}=0.26$, the depth to the capillary fringe and the new water-table depth. (c) Given 15% relative humidity at the ground surface (in clay, $T=10^{\circ}\text{C}$), the saturation and total hydraulic head there.

Approach. Integrate the piecewise-uniform hydraulic-head gradient upward from the water table through each layer to get total head, subtract elevation to get pressure head (suction), then convert suction to saturation with each material's own Brooks-Corey curve. Part (b) inverts the same chain (saturation → suction → required vadose-zone thickness); part (c) uses the Kelvin equation to convert a stated relative humidity directly to a suction, then proceeds the same way.

1. Part (a) — suction at the clay-sand interface. With $H=0,\ z=0$ at the water table and gradient $dH/dz=i_{sand}=0.05$ constant through the sand: $$H(0.6)=i_{sand}(0.6)=0.030\ \text{m}.$$ Pressure head (and suction) at the interface: $$\psi_{int}=H-z=0.030-0.6=-0.570\ \text{m}\ \Rightarrow\ \boxed{\text{suction}=0.57\ \text{m}}.$$
2. Saturation at the interface, both materials. Since $0.57\ \text{m}>P_{d,sand}=0.05\ \text{m}$, the sand is well past air entry: $$S_e=\left(\frac{P_d}{\psi}\right)^{\lambda}=\left(\frac{0.05}{0.57}\right)^{6.0}\approx0,\qquad S_{sand}=S_{wr}+S_e(1-S_{wr})\approx\boxed{0.100}\ \text{(essentially residual)}.$$ For clay at the SAME suction (continuity of capillary pressure across the interface), using clay's own curve since $0.57\ \text{m}>P_{d,clay}=0.5\ \text{m}$: $$S_e=\left(\frac{0.5}{0.57}\right)^{2.0}=0.7695,\qquad S_{clay,int}=0.25+0.7695(0.75)=\boxed{0.827}.$$ The clay stays much wetter than the sand at the very same suction — exactly the point of a capillary-barrier pairing (developed further in Question 5(a)): coarse sand desaturates almost completely just above its own small air-entry pressure, while fine clay holds onto most of its water at the same suction.
3. Suction and saturation at the ground surface. Continuing the head integration through the clay ($dH/dz=i_{clay}=0.26$, thickness 2.0 m): $$H(2.6)=H(0.6)+i_{clay}(2.0)=0.030+0.26(2.0)=0.550\ \text{m}.$$ $$\psi_{gs}=H-z=0.550-2.6=-2.05\ \text{m}\ \Rightarrow\ \text{suction}=2.05\ \text{m}.$$ $$S_e=\left(\frac{0.5}{2.05}\right)^{2.0}=0.05949,\qquad S_{gs}=0.25+0.05949(0.75)=\boxed{0.295}.$$
4. Part (b) — required suction for $S_{gs}=0.26$. Invert the clay Brooks-Corey curve: $$S_e=\frac{S-S_{wr}}{1-S_{wr}}=\frac{0.26-0.25}{0.75}=0.01333,\qquad \psi_{gs,new}=\frac{P_{d,clay}}{S_e^{1/\lambda}}=\frac{0.5}{(0.01333)^{1/2}}=\boxed{4.33\ \text{m suction}}.$$ This is well beyond the 2.05 m suction reached at the ORIGINAL water-table depth (part (a)), so the water table must now sit deeper — below the whole 2.6 m of already-characterized stratigraphy, into a thicker extent of the same sand unit.
5. New water-table depth. Let $z_1$ be the (now larger) thickness of sand between the new water table and the clay-sand interface. Integrating head through $z_1$ of sand then 2.0 m of clay and solving for the suction at the ground surface, $\text{suction}_{gs}=(1-i_{sand})z_1+(1-i_{clay})(2.0)$, gives $$z_1=\frac{4.33-(1-0.26)(2.0)}{1-0.05}=\frac{4.33-1.48}{0.95}=\boxed{3.00\ \text{m}}.$$ $$D_{new}=z_1+2.0=\boxed{5.00\ \text{m below ground surface}}.$$
6. Depth to the capillary fringe. The capillary fringe is the near-saturated zone immediately above the new water table, bounded by the sand's OWN air-entry suction $P_{d,sand}=0.05\ \text{m}$: $$z_{fringe}=\frac{P_{d,sand}}{1-i_{sand}}=\frac{0.05}{0.95}=0.0526\ \text{m above the new water table},$$ $$\text{depth to fringe}=D_{new}-z_{fringe}=5.00-0.053=\boxed{4.95\ \text{m below ground surface}}.$$ The fringe itself is barely 5 cm thick — coarse sand's tiny air-entry pressure means capillary saturation collapses almost immediately above the free water surface.
7. Part (c) — suction from the Kelvin equation, $RH=15\%$. The Kelvin (psychrometric) relation links matric suction to the equilibrium vapour-phase relative humidity above the pore water: $$\psi=-\frac{RT\ln(RH)}{M_wg}=-\frac{(8.314)(283.15)\ln(0.15)}{(0.018015)(9.81)}=\boxed{2.53\times10^4\ \text{m suction}}.$$ This astronomically large suction is a genuine feature of the Kelvin relation, not an error — vapour-pressure lowering only becomes appreciable at suctions of many thousands of metres (soil this dry is in equilibrium with a nearly-desiccated atmosphere), which is exactly why relative humidity is a poor diagnostic of soil moisture state except in extremely dry conditions.
8. Saturation and hydraulic head at $RH=15\%$. This suction is far beyond clay's air-entry pressure, so the Brooks-Corey curve places the soil essentially at residual saturation: $$S_e=\left(\frac{0.5}{25271}\right)^{2.0}\approx0,\qquad S=\boxed{0.250\ (\approx S_{wr})}.$$ Total hydraulic head at the ground surface (still $z=2.6\ \text{m}$ above the ORIGINAL water-table datum of part (a), since this sub-part explicitly refers back to "the clay in Question 4(a)"): $$H=z+\psi=2.6-25{,}271=\boxed{-25{,}268\ \text{m}}.$$