Hydraulic Gradient

i = \frac{\Delta h}{L}

Worked example: 1.2 m of head over 8 m → i = 0.15 — press Try an example to run it live, then adjust anything.

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Hydraulic Gradient explained

Water in soil moves because total head — elevation plus pressure head — is higher at one end of the path than the other, and the gradient is simply that loss spread over the distance travelled. A 1.2 m head difference across an 8 m long path gives i = 0.15, meaning fifteen centimetres of head are consumed by friction for every metre of travel. Both quantities are lengths, so i has no units at all and is the same number in metres or feet.

The trap is the flow path length, not the straight-line distance. Water going under a sheet pile wall travels down one side, around the toe and back up, and using the horizontal wall spacing instead of the true path length will overstate the gradient badly. The second trap is direction: upward seepage subtracts from effective stress, and when the upward gradient reaches the critical value $i_c = \gamma'/\gamma_w \approx 1.0$ , the effective stress hits zero and the sand boils. That is quicksand, and it is why cofferdam design checks exit gradient long before it checks anything else.

Hydraulic Gradient formula

i = \frac{\Delta h}{L}

Where

$i$ = Hydraulic gradient (m/m)
$\Delta h$ = Head loss (m)
$L$ = Flow path length (m)

Missing one of these? Work it out first, then come back

Hydraulic gradient — Darcy's Law for Groundwater Flow, Hazen–Williams Velocity
Head loss — Darcy–Weisbach Head Loss, Hazen–Williams Head Loss
Flow path length — Kirpich Time of Concentration, Beer–Lambert Law (A = εbc)

Part of this formula set

Darcy's law and seepage

4 formulasGroundwater through soil: hydraulic gradient, Darcy discharge, the faster real velocity in the pores, and layered permeability.