---
title: "Prove q = k h Nf/Nd for earth mass"
subject: "Soil Mechanics"
chapter: 7
key: "semester_4/soil_mechanics:flow-net-discharge-formula-proof"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/flow-net-discharge-formula-proof
site: BCT Engineering
---

# Prove q = k h Nf/Nd for earth mass

Soil Mechanics · Chapter 7: Seepage through Soils

- Asked in: 2079 Asoj, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/flow-net-discharge-formula-proof
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-7

## Question

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

## 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.

## Answer

Consider a flow net with $N_f$ flow channels and $N_d$ equipotential drops, and a total head loss $h$ across the earth mass.

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

#### Derivation

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

Head drop across each field:

$$
\Delta h = \frac{h}{N_d}
$$

Hydraulic gradient across the field:

$$
i = \frac{\Delta h}{a} = \frac{h}{N_d\,a}
$$

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

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

For square fields, $a = b$:

$$
\Delta q = k\,\frac{h}{N_d}
$$

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

$$
q = N_f\,\Delta q = k\,h\,\frac{N_f}{N_d} \qquad \blacksquare
$$

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With total head loss h and N_d equipotential drops, what is the head loss across one field?
   A. h x N_d
   B. h / N_d
   C. h / N_f
   D. h N_f / N_d
   Answer: B. The total head h is lost equally in N_d drops, so each drop is h/N_d. (Key point: Head drop per field)

2. For a field of length a (along flow), what is the hydraulic gradient?
   A. h / b
   B. h / (N_f a)
   C. h / (N_d a)
   D. h N_d / a
   Answer: C. i = (h/N_d)/a = h/(N_d a). (Key point: Gradient across a field)

3. For a square field (a = b), what is the discharge through one flow channel?
   A. k h N_d
   B. k h / N_d
   C. k h / N_f
   D. k h N_f
   Answer: B. Delta q = k (h/(N_d a)) b, and a = b cancels to k h / N_d. (Key point: Square field simplification)

4. How is the total discharge obtained from a single channel?
   A. Divide by N_f
   B. Multiply the channel flow by N_d
   C. Multiply the channel flow by N_f, since each channel carries the same flow
   D. Add the head drops
   Answer: C. All channels carry equal flow, so q = N_f x Delta q = k h N_f / N_d. (Key point: Total discharge)

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