Question 6 of 6: Optional Questions

Question 6: Optional Questions (5 marks each; 3 of 5 required by the source, all 5 answered below)

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.

(6.1) Darcy's Law

Given/Find. State Darcy's Law and describe every component, illustrated with a diagram.

$$q=kiA=k\frac{\Delta h}{L}A,\qquad v=ki,\qquad v_s=\frac{v}{n}$$

Where $q$ is the volumetric flow rate; $k$ is the hydraulic conductivity of the soil (a material property, units of velocity); $i=\Delta h/L$ is the hydraulic gradient, the total head lost per unit length of flow path; $A$ is the gross cross-sectional area of soil (voids and solids together) perpendicular to flow; $v=q/A$ is the discharge (superficial/Darcy) velocity, a fictitious average velocity over the full area $A$; and $v_s=v/n$ is the true, higher seepage velocity through the void space alone ($n$ = porosity), since flow can only occur through the pores. Darcy's Law is empirical and valid for laminar (low Reynolds number) flow, which covers essentially all groundwater seepage.

(6.2) Effective stress between two grains

Given/Find. Derive the effective stress equation from a conceptual grain-contact model, with a diagram.

Consider a horizontal plane of gross area $A$ passing through the soil along the tortuous, wavy path that follows the actual grain-to-grain contacts. A total vertical force $P=\sigma A$ acts across this plane. That force is carried by two mechanisms in parallel: the inter-granular contact forces $P'$ (summed over the very small total contact area $a_c=\sum a_i$), and the pore water pressure $u$ acting on the remaining area $(A-a_c)$:

$$\sigma A = P' + u(A-a_c)$$

Defining the effective stress as the contact force per unit gross area, $\sigma'=P'/A$, and noting that for real soils $a_c/A$ is only a few percent (so $u(A-a_c)/A\approx u$):

$$\sigma A \approx \sigma' A + uA \quad\Rightarrow\quad \boxed{\sigma'=\sigma-u}$$

(6.3) Capillary rise and water retention curves

Given/Find. Describe capillary rise in a tube and relate it to the soil-water characteristic (retention) curve for unsaturated soils, with a diagram.

Surface tension $T_s$ acting around the tube's wetted perimeter $\pi d$, at contact angle $\alpha$ to the tube wall, supports the weight of the raised water column of height $h_c$: equating the vertical component of the surface-tension force to the column weight, $T_s\cos\alpha(\pi d)=\gamma_w\left(\frac{\pi d^2}{4}\right)h_c$, gives

$$h_c=\frac{4T_s\cos\alpha}{\gamma_w d}$$

Rise is greater in a narrower tube. In an unsaturated soil, the interconnected pores behave like a bundle of capillary tubes of varying diameter, so the matric suction $\psi\approx\gamma_w h_c$ needed to hold water in a given pore is inversely related to that pore's size. The soil-water characteristic curve (suction vs. degree of saturation) is the macroscopic expression of this: a soil dominated by small pores (clay) desaturates only at high suction and holds significant water at high suction, giving a flat, high-suction curve; a soil dominated by large pores (sand) desaturates abruptly at low suction (a low, sharp air-entry value) then drains rapidly, giving a steep curve — the same capillary-tube physics, integrated over the soil's pore-size distribution.

(6.4) Sand cone field compaction test

Given. Volume of hole excavated $V=1165\text{ cm}^3$; soil mass wet $=2600$ g; soil mass dry $=1645$ g.

Find. (a) field dry density (b) field water content.

1. (a) Field dry density. $\rho_d=\dfrac{M_{dry}}{V}=\dfrac{1645\text{ g}}{1165\text{ cm}^3}=1.412\text{ g/cm}^3=1412\text{ kg/m}^3$, i.e. $\gamma_d=\rho_d g=1412(9.81)/1000=\boxed{13.85\text{ kN/m}^3}$.
2. (b) Field water content. $w=\dfrac{M_{wet}-M_{dry}}{M_{dry}}=\dfrac{2600-1645}{1645}=\boxed{58.1\%}$.

(6.5) Groundwater table and total head profile

Given/Find. Define the groundwater table; plot elevation head, pressure head, and total head with depth for a 5 m sand layer with the water table 1.5 m below the surface.

The groundwater table is the level in the ground at which pore water pressure equals atmospheric pressure ($u=0$); it is the free surface of the saturated zone, found in practice as the level to which water rises in an open standpipe/observation well. Above it lies the (typically unsaturated) vadose zone; below it the soil is saturated and $u$ increases hydrostatically with depth (under no-flow conditions).

Taking the datum at the base of the layer: at the water table (depth 1.5 m), elevation head $=3.5$ m, pressure head $=0$, total head $=3.5$ m. Above the water table (assumed dry, $u=0$), pressure head stays $0$ and total head simply tracks elevation head, rising to $5.0$ m at the ground surface. Below the water table, pore pressure grows hydrostatically with depth exactly as fast as elevation head falls, so their sum — total head — stays constant at 3.5 m all the way to the base: e.g. at the base (depth 5 m), elevation head $=0$, pressure head $=\gamma_w(3.5)/\gamma_w=3.5$ m, total head $=0+3.5=3.5$ m, matching the water table. This constancy is exactly the no-flow (hydrostatic equilibrium) condition; any deviation from a horizontal total-head line would indicate seepage.