Use Darcy's law to find discharge velocity, volumetric flow, and true seepage velocity from hydraulic conductivity, gradient, area, and porosity.
Direct solving is free and needs no account. Have a word problem instead? Submit it as text.
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
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
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
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
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
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
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
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
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
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.