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Darcy's Law Calculator

Groundwater flow through a porous medium: hydraulic gradient, Darcy flux, total flow rate Q = KiA, and the seepage velocity and travel time that actually govern how fast water — and anything dissolved in it — moves.

ft/day
ft
ft
ft²
ft/day
ft³/day
gal/day
ft/day
ft/yr
years

Defaults: clean sand, K = 1×10⁻³ cm/s (2.835 ft/day), 0.5 ft of head loss over 100 ft, 500 ft² of aquifer cross-section, effective porosity 0.30.

$$ Q = K\,i\,A \qquad i = \frac{\Delta h}{L} \qquad q = K\,i \qquad v = \frac{q}{n_e} \qquad t = \frac{L}{v} $$
Q volumetric flow rate · K hydraulic conductivity · i hydraulic gradient · Δh head loss along the flow path · L flow path length · A gross cross-sectional area perpendicular to flow · q Darcy flux (specific discharge) · nₑ effective porosity · v average linear (seepage) velocity · t advective travel time.

Darcy flux is not how fast the water moves

The most common Darcy's law mistake is using the flux q as a velocity. The flux divides flow by the whole cross-section, grains included. Water can only travel through connected pore space, so its average speed is q divided by effective porosity. With nₑ = 0.30 the seepage velocity is 3.3 times the Darcy flux. For contaminant plumes, capture-zone sizing and well-head protection travel times, seepage velocity is the number that matters — and it is still only an average, since dispersion spreads a real plume ahead of it.

Hydraulic conductivity varies over more than ten orders of magnitude between gravel and unfractured clay, which makes K by far the most uncertain input. A factor-of-ten error in K is a factor-of-ten error in Q. Use pumping-test, slug-test or permeameter values where you have them; see the soil permeability reference card for typical ranges by soil type.

Hydraulic conductivity unit conversions

Converting hydraulic conductivity K between common units
1 unit ofcm/sm/dayft/daygpd/ft²
cm/s18642,834.621,205
m/day1.157×10⁻³13.280824.542
ft/day3.528×10⁻⁴0.304817.4805
gpd/ft²4.716×10⁻⁵0.0407460.133681

gpd/ft² is gallons per day through one square foot under unit gradient ("Meinzer unit"), common in US water-well work. Hydraulic conductivity K includes the fluid's density and viscosity; intrinsic permeability (darcys, m²) does not.

Worked examples

Example 1 — Flow and travel time through a sand aquifer

Given: Clean sand, K = 1×10⁻³ cm/s. Water table drops 0.5 ft over 100 ft. Flow cross-section 500 ft². Effective porosity 0.30.
Find: Flow rate and how long water takes to travel the 100 ft.
K = 1×10⁻³ cm/s × 2,834.6 = 2.835 ft/day
i = Δh/L = 0.5/100 = 0.005
q = K·i = 2.835 × 0.005 = 0.01417 ft/day
Q = q·A = 0.01417 × 500 = 7.09 ft³/day = 53.0 gal/day
v = q/nₑ = 0.01417/0.30 = 0.0472 ft/day = 17.3 ft/yr
Q = 53 gal/day · travel time over 100 ft = 100/0.0472 = 2,117 days ≈ 5.8 years

Example 2 — Seepage under a cutoff (SI)

Given: Silty sand, K = 0.5 m/day. 3 m of head across a 25 m seepage path. Section 40 m wide × 2 m thick. nₑ = 0.25.
i = 3/25 = 0.12; A = 40 × 2 = 80 m²
q = 0.5 × 0.12 = 0.060 m/day; Q = 0.060 × 80 = 4.80 m³/day
v = 0.060/0.25 = 0.24 m/day
Q = 4.8 m³/day · water crosses the 25 m path in about 104 days

Where Darcy's law applies — and where it doesn't

Not to be confused with the Darcy-Weisbach equation, which describes friction head loss for flow in pipes.

References: Darcy, H. (1856). Les Fontaines Publiques de la Ville de Dijon. Freeze, R.A., Cherry, J.A. (1979). Groundwater. Prentice-Hall. Fetter, C.W. (2001). Applied Hydrogeology, 4th ed.

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