Section 4: Soil and Water Conservation Engineering
Soil Mechanics
Subtopic: Permeability and seepage analysis
Question Statement:
A flow net constructed for an earthen dam of length $100\text{ m}$ founded on an impermeable base gives the number of flow channels $N_f = 4$ and the number of equipotential drops $N_d = 16$. The total hydraulic head causing flow is $12\text{ m}$ and the coefficient of permeability of the dam material is $4 \times 10^{-6}\text{ m/s}$. The total seepage rate through the entire length of the dam is ________ $\text{m}^3/\text{day}$ (round off to two decimal places).
Answer Key:
103.68
Step-by-Step Solution & Explanation
1. Seepage discharge per unit width ($q'$):
$q' = k \cdot H \cdot \frac{N_f}{N_d}$
$q' = (4 \times 10^{-6}\text{ m/s}) \times 12\text{ m} \times \left( \frac{4}{16} \right) = 4.8 \times 10^{-5} \times 0.25 = 1.2 \times 10^{-5}\text{ m}^3/\text{s/m}$
2. Total seepage for dam length $L = 100\text{ m}$:
$Q = q' \times L = 1.2 \times 10^{-5} \times 100 = 1.2 \times 10^{-3}\text{ m}^3/\text{s}$
3. Convert to $\text{m}^3/\text{day}$ ($1\text{ day} = 86400\text{ s}$):
$Q_{\text{day}} = 1.2 \times 10^{-3} \times 86400 = 103.68\text{ m}^3/\text{day}$