Darcy's Law & Groundwater Flow

Pick a material and a head difference to see how much water crosses it, how fast the water is really moving, and where the law stops holding.

Worked example — Darcy's Law & Groundwater Flow

Hydraulic gradient	0.0010
Darcy flux	1.000 × 10⁻⁸ m/s
Flow rate	1.000 × 10⁻⁸ m³/s
Pore water velocity	2.857 × 10⁻⁸ m/s
Water outruns the flux by	2.86×
Travel time along the path	1,109 years
Pore Reynolds number	1.4949 × 10⁻⁶
Steepest gradient that keeps Re ≤ 1	668.9374

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Darcy's Law & Groundwater Flow

Darcy's law says the flow through a porous medium is proportional to the head lost across it and inversely proportional to the distance it is lost over: Q = K A Δh/L. The constant K is the hydraulic conductivity, and it belongs to the medium and the fluid jointly rather than to the rock. Strip the fluid back out with k = Kμ/(ρg) and what is left is the intrinsic permeability, a property of the pore geometry alone. That is why the same sandstone is quoted in m/s by a hydrogeologist and in darcys by a petroleum engineer: they are measuring the same pores with different fluids in them.

Assumes: Saturated, steady, laminar flow through a medium that is uniform at the scale of the flow path. The six conductivities here are representative round numbers rather than measurements: each of these materials spans at least two orders of magnitude in the published compilations, because sand names a grain size and not a rock. The porosities are total porosities, which makes the pore velocity the slowest one the data allows — transport follows the connected fraction, and that is smaller.

324 days through gravel, 634 million years through granite 🖖

Press Sand aquifer: one metre of head across a kilometre of fine sand, and the travel-time row reads 1,109 years. Now click along the material buttons and leave everything else alone. Gravel does the same kilometre in 324.1 days. Silt takes 1.426 × 10⁵ years, clay 1.584 × 10⁷, and unfractured granite 6.338 × 10⁸.

One column is doing all of that. Hydraulic conductivity runs from 10⁻² m/s in the gravel to 10⁻¹² m/s in the granite, ten powers of ten, and the flux, the velocity, the travel time and the Reynolds number all follow from it. Henry Darcy was not hunting for a law when he found the relation. He was building the municipal water supply of Dijon and needed to know how fast his sand filters would run, and it went into an appendix of the 1856 report on the city's fountains.

That span is why a landfill engineer and a well driller use the same equation and never need to talk. One of them is paying to get K as low as compaction can take it. The other is paying to find the largest K he can drill into.

The water is 2,060 years ahead of the flux 🖖

Press Sand aquifer again. The travel time reads 1,109 years. The Darcy flux row above it reads 1.000 × 10⁻⁸ m/s, and a kilometre divided by that is 3,169 years. Both are correct arithmetic and only one of them is a travel time.

The flux is the flow divided by the whole cross-section, grains included, so it is a flow per unit of ground rather than the speed of anything: most of that cross-section is quartz. Push the same flow through the 35% that is pore and the water moves 1/n times faster, which is what the Water outruns the flux by row prints. It says 2.86 here, and 2.86 is exactly the ratio between the two travel times.

Where the factor is biggest is not where you would guess. Gravel has the lowest porosity of the five sediments at 0.28, so its water runs 3.57 times the flux; clay, holding the most water at 0.50, runs only twice. The extreme is the granite, where 2% porosity sends what little water there is fifty times faster than the flux implies.

The travel time here is the late estimate, deliberately. Transport follows the connected pore space rather than the total, and in clay a large share of the water is held against mineral surfaces and never moves. A non-sorbing tracer like chloride arrives sooner than this page says. Something that sticks to the clay can take decades longer, and neither correction is computed here.

A gravel drain on a one-in-twenty fall is already past the law 🖖

Press Gravel drain: one metre of head over twenty metres of clean gravel, which is an unremarkable land drain. The Reynolds row reads 4.9830 against a ceiling of 1, and the right-hand panel shades everything past the gradient where that ceiling falls.

Darcy's law is linear, and the linearity is an assumption with a price of admission. Substituting q = Ki into Re = ρqd/μ leaves Re(i) = ρKdi/μ, which is linear in i, so the gradient at which any threshold is reached can be written down rather than searched for: i_max = Re_maxμ/(ρKd). For this gravel the row says 0.0100, a fall of one metre in a hundred. Run the same metre of head over 100 m and Re is 0.9966, just inside. Over 20 m it is 4.9830. Over 1 m, 99.6605.

Gravel is the only one of the six materials whose ceiling sits at a gradient anyone would build. Coarse sand reaches Re = 1 at a gradient of 2.0068, which you meet in a laboratory column rather than in an aquifer; fine sand at 668.9374; clay at 5.0170 × 10⁸, a figure whose only honest reading is that clay will never leave Darcy's law.

Past the ceiling the flow keeps climbing with the gradient and stops climbing in proportion to it, because inertia in the pores starts taking a share of the pressure that viscous drag had to itself. So the flow rate this page prints for the gravel drain is too large. The Forchheimer equation adds a quadratic term to Darcy's linear one, and that is what a drainage engineer reaches for instead.

Problem solved in full

What gets through a landfill liner 7 steps
A landfill has a 600 mm compacted clay liner: hydraulic conductivity 10⁻⁹ m/s, porosity 0.50. The collection system holds 300 mm of leachate on top of it and the base of the liner drains freely. Find how long water takes to cross the clay, then find how much leaves a hectare of it over a thirty-year life.

Open this problem in the tool
1. The gradient is where this problem is won or lost. Head is elevation plus pressure, so measure both from the base of the liner. The top of the leachate sits 0.6 m of clay plus 0.3 m of liquid above that datum, which is 0.9 m of head, and the base drains freely at atmospheric pressure, which is nothing. So 0.9 m of head is lost across a path 0.6 m long. Write 0.3 divided by 0.6 instead and you have counted the liquid and thrown away the weight of the water standing in the clay itself, which is two thirds of what drives the flow.
  
  $i = \frac{Δ h}{L} = \frac{0 . 3 + 0 . 6}{0 . 6} = \frac{0 . 9 m}{0 . 6 m} = 1.5$
2. Darcy's law in its flux form: multiply the conductivity by the gradient. The Darcy flux row reads 1.500 × 10⁻⁹ m/s, a number nobody can feel, so give it a year. Multiplied by 3.15576 × 10⁷ seconds it becomes 0.047336 m of water a year through every square metre, which is 47.34 litres per square metre per year. A regulator writes the specification as a conductivity; this is what the specification buys.
  
  $q = K i = 1 0^{- 9} \cdot 1.5 = 1.500 \times 1 0^{- 9} m s^{- 1} ⟹ q \cdot 3.15576 \times 1 0^{7} = 0.047336 m y r^{- 1} = 47.34 L m^{- 2} y r^{- 1}$
3. That flux is not a speed. It divides the flow by the whole square metre, solid included, and half of a clay this porous is solid. The water is pushed through the pore half and moves at twice the flux: Pore water velocity reads 3.000 × 10⁻⁹ m/s and the row beneath it reads 2.00×. Clay gives the smallest such factor of any material on this page, because it has the most pore space to spread the same flow through.
  
  $v = \frac{q}{n} = \frac{1 . 500 \times 1 0 ^{- 9}}{0 . 50} = 3.000 \times 1 0^{- 9} m s^{- 1}, \frac{v}{q} = \frac{1}{n} = 2.00$
4. Now the crossing. Six hundred millimetres at 3 × 10⁻⁹ m per second is 2 × 10⁸ seconds, and Travel time along the path turns that into 6.34 years. Use the flux in place of the pore velocity and you get 12.68 years, which is the same mistake as before and is worth exactly double the answer.
  
  $t = \frac{L}{v} = \frac{0 . 6}{3 . 000 \times 1 0 ^{- 9}} = 2.000 \times 1 0^{8} s = 6.34 yr$
5. Ask whether the law was entitled to answer, because it will answer either way. The pore Reynolds number is built on the Darcy flux and a representative grain diameter, 2 × 10⁻⁶ m for clay, and it comes to 2.9898 × 10⁻⁹ against a ceiling of 1. The gradient that would reach that ceiling is 5.0170 × 10⁸: five hundred million metres of head per metre of liner. Nothing you can do to a clay on this page takes it out of Darcy's law.
  
  $Re = \frac{ρ q d}{μ} = \frac{998 . 2 \cdot 1 . 500 \times 1 0 ^{- 9} \cdot 2 \times 1 0 ^{- 6}}{1 . 0016 \times 1 0 ^{- 3}} = 2.9898 \times 1 0^{- 9} ≪ 1$
6. Swap the fill and change nothing else. The same 600 mm under the same 0.9 m of head, filled with each material the page offers, gives the column below. Silt is one step coarser than clay and still nothing anyone would call permeable, and it is crossed in three weeks. The conductivity ratio between the two is exactly a hundred; the travel-time ratio is 111, the extra tenth coming from clay holding more water per cubic metre to push through. Gravel manages eleven seconds, and does it at Re = 149.49, where the figure means nothing.
  
  $fill gravel coarse sand fine sand silt clay granite q (m s^{- 1}) 1.50 \times 1 0^{- 2} 7.50 \times 1 0^{- 4} 1.50 \times 1 0^{- 5} 1.50 \times 1 0^{- 7} 1.50 \times 1 0^{- 9} 1.50 \times 1 0^{- 12} t 11.2 s 4.3 min 3.9 h 20.8 d 6.34 yr 254 yr Re 149.49 0.747 2.24 \times 1 0^{- 3} 2.99 \times 1 0^{- 6} 2.99 \times 1 0^{- 9} 1.49 \times 1 0^{- 12}$
7. The page has now told you everything it can, and a designer still has no number, because flux is per square metre and a landfill is not one square metre. A hectare of that liner is 10⁴ m², so it passes 1.5 × 10⁻⁵ m³ every second, 473.4 m³ a year, and 14,200 m³ over a thirty-year operating life. That is close to six Olympic pools of leachate, arriving through a liner that is performing exactly to specification.
  
  $V = q A t = 1.500 \times 1 0^{- 9} \cdot 1 0^{4} \cdot 30 \cdot 3.15576 \times 1 0^{7} = 1.42 \times 1 0^{4} m^{3}$
Answer

Water crosses the 600 mm liner in 6.34 years, and a hectare of it passes 14,200 m³ over thirty years. Neither figure is a failure. The liner is meeting its specification to the digit, and 47 litres per square metre per year is what that specification means once somebody converts it into a unit a person can picture.

What the arithmetic will not survive is the head. Put 3 m of leachate on the liner instead of 0.3 m, which is what a clogged collection layer does, and the gradient goes from 1.5 to 6. Flux is linear in gradient, so every figure above multiplies by four: 6.34 years becomes 1.58, and the thirty-year volume becomes 56,800 m³. The 300 mm limit in the regulation is not really a statement about the liner at all; once the clay is in the ground, the head is the only term left in the equation that anyone can still change.

$Δ h : 0.9 \to 3.6 m ⟹ i : 1.5 \to 6, t = \frac{6 . 34}{4} = 1.58 yr, V = 4 V_{0} = 5.68 \times 1 0^{4} m^{3}$

References (3)

The experiment itself, in an appendix to a report about a city's water supply: H. Darcy, Les Fontaines Publiques de la Ville de Dijon. Victor Dalmont, Paris, 1856.
Where the conductivity and porosity ranges for real materials are compiled: R. A. Freeze and J. A. Cherry, Groundwater. Prentice-Hall, Englewood Cliffs, New Jersey, 1979.
The Reynolds-number convention used here, and where the linear law stops: J. Bear, Dynamics of Fluids in Porous Media. American Elsevier, New York, 1972. Reprinted by Dover, 1988.

By the 42Math team · Reviewed against published sources. Corrections welcome.