Seepage Velocity from Discharge Velocity

vs=vnv_s = \frac{v}{n}

Worked example: v 0.02 cm/s in n = 0.40 → vs = 0.05 cm/s — press Try an example to run it live, then adjust anything.

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Seepage Velocity from Discharge Velocity explained

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Darcy's velocity v = ki is discharge divided by the gross area, which pretends the water flows through the grains as well as around them. It does not: only the pore space conducts, so the water that actually gets through has to move faster by exactly the reciprocal of the porosity. A discharge velocity of 0.02 cm/s in a soil with n = 0.40 means real particles of water are moving at 0.05 cm/s — two and a half times faster.

The trap is contaminant travel time. Use v instead of vsv_s to work out when a plume reaches the property line and you will be late by a factor of two or three, which is the difference between a monitoring programme and a lawsuit. The second trap is tortuosity: even vsv_s is an average along a straight line, while a water molecule actually threads a winding path perhaps 1.2–1.5 times longer, so true particle speeds are higher again. For a first-pass arrival time vsv_s is the right number; for a defensible one, add a dispersion model.

Seepage Velocity from Discharge Velocity formula

vs=vnv_s = \frac{v}{n}
Where
  • vsv_s= Seepage velocity (cm/s)
  • vv= Discharge velocity (cm/s)
  • nn= Porosity