Flow Net Discharge

q = k H \frac{N _{f}}{N _{d}} (per unit width)

Drawn flow nets: Nf flow channels, Nd equipotential drops. Head loss per drop is H/Nd. For anisotropic soil transform coordinates. Exit gradient iE = (H/Nd)/ℓ_last; piping if iE → icr ≈ (Gs−1)/(1+e).

$k$ (m/s): Permeability
$H$ (m): Total head loss
$N_{f}, N_{d}$: Numbers of flow channels and drops

Step-by-step solved example

k=1.0×10⁻⁵ m/s, H=4.0 m, Nf=3, Nd=6. q per metre width?

q = k H (Nf/Nd) per unit width from the flow net.

1. Ratio
Nf/Nd=0.50.
2. q
q=1e-5×4×0.50=2.0×10⁻⁵ m³/s per m.

Answer: 2.0×10⁻⁵ m³/s per m

10 practice questions

0/10 correct

1.q per unit width equals

2.If you draw more squares, Nf/Nd

3.An equipotential drop of H/Nd is

4.Piping risk is when exit gradient approaches

5.A cutoff wall that lengthens the flow path

6.k of clay vs sand is

7.Anisotropic kx ≠ kz: you transform

8.Confined aquifer Dupuit q is related but flow nets are for

9.If H doubles, q

10.Nf and Nd are