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CivilGeotechnicaleasyFE ExamCoursework

Seepage Flow Rate by Darcy's Law

Water seeps through a soil sample with hydraulic conductivity k=1.0×10−4 m/sk = 1.0 \times 10^{-4}\ \text{m/s}.

The hydraulic gradient is i=0.50i = 0.50 and the cross-sectional area of flow is A=0.050 m2A = 0.050\ \text{m}^2.

Find the volumetric flow rate.

Given
k0.0001 m/sHydraulic conductivity
i0.5 -Hydraulic gradient
A0.05 m^2Cross-sectional area
Hint 1

Darcy's law relates flow rate to conductivity, gradient, and area.

Hint 2

Q = kiA. All three factors are given directly.

Hint 3

Check your units: the answer should be a volume per time.

Worked solution — try the problem first

Darcy's law:

Q=kiAQ = kiA

Q=(1.0×10−4)(0.50)(0.050)=2.5×10−6 m3/sQ = (1.0\times10^{-4})(0.50)(0.050) = 2.5\times10^{-6}\ \text{m}^3/\text{s}

Q=2.5×10−6 m3/s=2.5 mL/s\boxed{Q = 2.5\times10^{-6}\ \text{m}^3/\text{s} = 2.5\ \text{mL/s}}

Two velocities, often confused. The discharge velocity v=ki=5×10−5v = ki = 5\times10^{-5} m/s is what Darcy's law gives, computed over the gross area. The seepage velocity vs=v/nv_s = v/n is the actual speed of water through the pores and is larger, since only the void fraction nn is open to flow. Use discharge velocity for flow rates and seepage velocity for travel times — for example, in contaminant transport.

Concepts:Darcy's lawPermeabilitySeepage

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