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Darcy’s Law (Seepage)

q = k i A, i = \frac{Δ h}{L}

Discharge q through a saturated soil is hydraulic conductivity k times hydraulic gradient i times area A normal to flow. Laminar flow (Re small) is assumed. Seepage (discharge) velocity vs = q/A = ki; true average velocity through voids is ki/n. k of sands is orders of magnitude larger than clays.

q

(m³/s)

Seepage discharge

k

(m/s)

Hydraulic conductivity

i

Hydraulic gradient Δh/L

A

(m²)

Area normal to flow

Step-by-step solved example

k = 2.0×10⁻⁵ m/s, head loss 0.80 m over 20 m, A = 5.0 m². Find q and seepage velocity.

q = k i A. i = Δh/L along the flow path.

1. Gradient
i = 0.80/20 = 0.040.
$i = 0.040$
2. Discharge
q = kiA = (2.0×10⁻⁵)(0.040)(5.0) = 4.0×10⁻⁶ m³/s. vs = ki = 8.0×10⁻⁷ m/s.
$q = 4.0 \times 1 0^{- 6} m^{3} / s$

Answer: q = 4.0×10⁻⁶ m³/s, vs = 8.0×10⁻⁷ m/s

10 practice questions

0/10 correct

1.If k = 1×10⁻⁴ m/s, i = 0.02, A = 3 m², q =

2.Hydraulic gradient i is

3.True average velocity through voids is

4.k of intact clay vs clean sand is typically

5.Darcy’s law assumes flow is

6.Flow nets: q per unit width = k H Nf/Nd. If H = 4 m, Nf = 3, Nd = 6, k = 10⁻⁵ m/s, q' =

7.Critical hydraulic gradient i_c for heave is about

8.Constant-head permeability tests are suited to

9.Seepage force on a soil element acts

10.If L doubles and Δh is unchanged, q