GATE 2023 CE (CE2) – Question 60

Geotechnical Engineering · Permeability and seepage: flow nets, uplift pressure, piping, capillarity · 2 marks · Numerical answer

For the flow setup shown in the figure (not to scale), the hydraulic conductivities of the two soil samples, Soil 1 and Soil 2, are 10 mm/s and 1 mm/s, respectively. Assume the unit weight of water as 10 kN/m$^3$ and ignore the velocity head. At steady state, what is the total head (in m, rounded off to two decimal places) at any point located at the junction of the two samples?

upper reservoir surface z = 4 m has applied pressure 10 kPa; lower reservoir surface z = 0 m is open. Equal-area Soil 1 extends from z = 3 to 2 m and Soil 2 from z = 2 to 1 m.

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Correct answer: 4.53 to 4.57

Explanation

**Step 1: total heads at the two ends.** Ignore velocity head, so the total head is pressure head plus elevation head ($\gamma_w=10$ kN/m³).
- Top: $h_{top}=z+\dfrac{p}{\gamma_w}=4+\dfrac{10}{10}=5.0$ m.
- Bottom (open): $h_{bot}=0+0=0$ m.

Total head loss across the two samples $=5.0$ m.

**Step 2: head loss in each sample.** The samples are in series with the same discharge per unit area ($v=ki=k\,\Delta h/L$ must be the same in both) and equal lengths $L$ (1 m each), so
$$k_1\frac{\Delta h_1}{L}=k_2\frac{\Delta h_2}{L}\;\Rightarrow\;\frac{\Delta h_1}{\Delta h_2}=\frac{k_2}{k_1}=\frac{1}{10}.$$

**Step 3: split the 5 m.**
$$\Delta h_1=\frac{1}{11}\times5=0.4545\text{ m},\qquad\Delta h_2=\frac{10}{11}\times5=4.5455\text{ m}.$$

**Step 4: head at the junction** (after Soil 1):
$$h_J=5-0.4545=\mathbf{4.55\ m}.$$