CivilSolve

Geotechnical

Darcy Seepage Velocity & Discharge

Darcy's law through soil: discharge (Darcy) velocity v = k·i from hydraulic conductivity and gradient (or head loss over flow length), volumetric flow Q = v·A, and true seepage velocity v_s = v/n through the pores. Use for: 'flow through a soil sample/aquifer', seepage quantity, contaminant travel (seepage) velocity, permeameter problems. Not for: flow nets with multiple equipotential drops, well drawdown.

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Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

SI permeameter: k=0.005 cm/s, Δh=2 m over L=10 m, A=25 m², n=0.35

Given

k
0.005 cm/s
head loss
2 m
flow length
10 m
area
25 m^2
porosity
0.35

Assumptions

  • Darcy's law is valid: laminar flow through a saturated soil (fine sands and finer; Reynolds number below about one).
  • Seepage velocity divides by the (effective) porosity as given; if total rather than effective porosity was supplied, v_s is a lower-bound estimate.

Solution steps

  1. Hydraulic gradient

    The gradient is the head lost per unit length of flow path — the driving force of seepage.

    i = Δh / L

    i = 2 m / 10 m = 0.2

    = 0.2

  2. Discharge (Darcy) velocity

    v = k·i is a superficial velocity: the flow rate spread over the TOTAL cross-section, soil grains included. It is the v used for computing Q.

    v = k·i

    v = 5.00e-5 m/s × 0.2 = 1.00e-5 m/s

    = 1.00e-5 m/s

  3. Seepage discharge

    The flow quantity uses the DISCHARGE velocity over the total area (never the seepage velocity — the pore area is smaller by exactly the same factor).

    Q = v·A = k·i·A

    Q = 1.00e-5 m/s × 25 m^2 = 0.00025 m^3/s

    = 0.00025 m^3/s

  4. Seepage (pore) velocity — the classic trap

    Water only flows through the voids, whose area is n·A, so actual particles move FASTER than the Darcy velocity: v_s = v/n. Use v_s for travel time and contaminant transport, v for discharge.

    v_s = v / n

    v_s = 1.00e-5 m/s / 0.35 = 2.86e-5 m/s

    = 2.86e-5 m/s

Results

Hydraulic gradient i

0.2

Discharge (Darcy) velocity v = k·i

1.00e-5m/s

Seepage discharge Q = k·i·A

0.00025m^3/s

Seepage (pore) velocity v_s = v/n

2.86e-5m/s

Where this answer was checked
source
Darcy's law (NCEES FE Reference Handbook): v = k·i, Q = k·i·A, vs = v/n — i from Δh/L, then v, Q and vs
verified by
hand-recomputed
derivation
k = 0.005 cm/s = 5e-5 m/s. i = 2/10 = 0.2. v = 5e-5·0.2 = 1e-5 m/s. Q = 1e-5·25 = 2.5e-4 m³/s. vs = 1e-5/0.35 = 2.8571e-5 m/s.

Example 2

US aquifer: k=10 ft/day, i=0.05, A=200 ft², n=0.30

Given

k
10 ft/day
gradient
0.05
area
200 ft^2
porosity
0.3

Assumptions

  • Darcy's law is valid: laminar flow through a saturated soil (fine sands and finer; Reynolds number below about one).
  • Seepage velocity divides by the (effective) porosity as given; if total rather than effective porosity was supplied, v_s is a lower-bound estimate.

Solution steps

  1. Discharge (Darcy) velocity

    v = k·i is a superficial velocity: the flow rate spread over the TOTAL cross-section, soil grains included. It is the v used for computing Q.

    v = k·i

    v = 10 ft/day × 0.05 = 0.5 ft/day

    = 0.5 ft/day

  2. Seepage discharge

    The flow quantity uses the DISCHARGE velocity over the total area (never the seepage velocity — the pore area is smaller by exactly the same factor).

    Q = v·A = k·i·A

    Q = 0.5 ft/day × 200 ft^2 = 100 ft^3/day

    = 100 ft^3/day

  3. Seepage (pore) velocity — the classic trap

    Water only flows through the voids, whose area is n·A, so actual particles move FASTER than the Darcy velocity: v_s = v/n. Use v_s for travel time and contaminant transport, v for discharge.

    v_s = v / n

    v_s = 0.5 ft/day / 0.3 = 1.67 ft/day

    = 1.67 ft/day

Results

Hydraulic gradient i

0.05

Discharge (Darcy) velocity v = k·i

0.5ft/day

Seepage discharge Q = k·i·A

100ft^3/day

Seepage (pore) velocity v_s = v/n

1.67ft/day

Where this answer was checked
source
Darcy's law (NCEES FE Reference Handbook): v = k·i, Q = k·i·A, vs = v/n — Direct gradient path in US units
verified by
hand-recomputed
derivation
v = 10·0.05 = 0.5 ft/day. Q = 0.5·200 = 100 ft³/day. vs = 0.5/0.30 = 1.6667 ft/day. (SI check: k = 10·0.3048/86400 = 3.5278e-5 m/s; v = 1.7639e-6 m/s = 0.5 ft/day ✓.)

Example 3

SI silty soil, velocities only: k=2e-6 m/s, i=0.8, n=0.40

Given

k
0.000002 m/s
gradient
0.8
porosity
0.4

Assumptions

  • Darcy's law is valid: laminar flow through a saturated soil (fine sands and finer; Reynolds number below about one).
  • Seepage velocity divides by the (effective) porosity as given; if total rather than effective porosity was supplied, v_s is a lower-bound estimate.

Solution steps

  1. Discharge (Darcy) velocity

    v = k·i is a superficial velocity: the flow rate spread over the TOTAL cross-section, soil grains included. It is the v used for computing Q.

    v = k·i

    v = 2.00e-6 m/s × 0.8 = 1.60e-6 m/s

    = 1.60e-6 m/s

  2. Seepage (pore) velocity — the classic trap

    Water only flows through the voids, whose area is n·A, so actual particles move FASTER than the Darcy velocity: v_s = v/n. Use v_s for travel time and contaminant transport, v for discharge.

    v_s = v / n

    v_s = 1.60e-6 m/s / 0.4 = 4.00e-6 m/s

    = 4.00e-6 m/s

Results

Hydraulic gradient i

0.8

Discharge (Darcy) velocity v = k·i

1.60e-6m/s

Seepage (pore) velocity v_s = v/n

4.00e-6m/s

Where this answer was checked
source
Darcy's law (NCEES FE Reference Handbook): v = k·i, vs = v/n — No area given — velocities only
verified by
hand-recomputed
derivation
v = 2e-6·0.8 = 1.6e-6 m/s. vs = 1.6e-6/0.40 = 4e-6 m/s.