Question

For the flow setup shown, hydraulic conductivities are $k_1=10$ mm/s (Soil 1) and $k_2=1$ mm/s (Soil 2). Unit weight of water $=10$ kN/m$^3$. Ignore velocity head. At steady state, what is the total head (in m, rounded to two decimals) at the junction of the two samples? \includegraphics[width=0.5\linewidth]{image111.png}

Hint:

For steady 1D flow through layered soils in series, use resistance $L/k$ to partition the total head loss:
\(\Delta h_i = \dfrac{(L_i/k_i)}{\sum (L/k)} \times \text{(total head drop)}\).
Include pressure head from applied pressure at boundaries when computing total head.

Answer 1

Step 1: Boundary total heads.
Top reservoir: applied pressure $=10$ kPa $\Rightarrow$ pressure head $=10/10=1$ m; elevation head $=4$ m
$\Rightarrow$ total head at top $H_{\text{top}}=4+1=5$ m.
Bottom reservoir: open to atmosphere at elevation $0$ m $\Rightarrow H_{\text{bot}}=0$ m.

Step 2: Use Darcy's law in series layers.
Head loss partitions in proportion to hydraulic resistance $L/k$.
Soil 1: $L_1=1$ m, $k_1=10$ mm/s $=0.01$ m/s $\Rightarrow R_1=L_1/k_1=1/0.01=100$.
Soil 2: $L_2=1$ m, $k_2=1$ mm/s $=0.001$ m/s $\Rightarrow R_2=L_2/k_2=1/0.001=1000$.
Total resistance $R=R_1+R_2=1100$. Total head drop $=H_{\text{top}}-H_{\text{bot}}=5$ m.

Step 3: Head at the junction (between Soil 1 and Soil 2).
Head drop across Soil 1
$\displaystyle \Delta h_1=\frac{R_1}{R}\times 5=\frac{100}{1100}\times 5=0.4545\ \text{m}$.
Hence head at junction
$H_{\text{junction}}=H_{\text{top}}-\Delta h_1=5-0.4545=4.5455\ \text{m}\approx \boxed{4.55\ \text{m}}$.