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Prove q = k h Nf/Nd for earth mass

Short answer

Head drop per field Δh = h/Nd, gradient i = h/(Nd a). Flow per channel Δq = k i b = k h b/(Nd a). For square fields a=b, Δq = k h/Nd. Total discharge q = Nf Δq = k h Nf/Nd.

Prove that the discharge through an earth mass is given by q=k×h×(NfNd)q = k \times h \times \left(\frac{N_f}{N_d}\right), where k = coefficient of permeability; h = head; NfN_f = number of flow channels and NdN_d = number of equipotential drops.

Consider a flow net with NfN_f flow channels and NdN_d equipotential drops, and a total head loss hh across the earth mass.

   equipotential 0     1     2   ...  Nd
        |      |      |      |
 ------ +------+------+------+------ flow line
        |  a   |  a   |  a   |   (a = b : square field)
 ------ +------+------+------+------
          b

Derivation

Take one field of length aa (in the direction of flow) and width bb (normal to the flow), per unit thickness perpendicular to the section.

Head drop across each field:

Δh=hNd\Delta h = \frac{h}{N_d}

Hydraulic gradient across the field:

i=Δha=hNd ai = \frac{\Delta h}{a} = \frac{h}{N_d\,a}

Discharge through the channel, by Darcy's law (A=b×1A = b \times 1):

Δq=k i A=k hNd a b\Delta q = k\,i\,A = k\,\frac{h}{N_d\,a}\,b

For square fields, a=ba = b:

Δq=k hNd\Delta q = k\,\frac{h}{N_d}

The same Δq\Delta q flows in each of the NfN_f channels, so the total discharge per unit length is

q=Nf Δq=k h NfNd■q = N_f\,\Delta q = k\,h\,\frac{N_f}{N_d} \qquad \blacksquare

✓ Answer checked by Claude Opus 5.5 (high effort)

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