In a flow net for a sheet pile wall, the number of flow paths is 5 and the number of equipotential drops is 10. If the coefficient of permeability is 6×10-3 mm/s and the head is 4.5 m, the seepage under the wall will be nearly
A flow net is a graphical representation used in geotechnical engineering and hydrogeology to visualize the flow of groundwater through a porous medium. It consists of two sets of orthogonal curves: flow lines and equipotential lines.
The space between two adjacent flow lines is called a flow channel, and the drop in hydraulic head between two adjacent equipotential lines is called an equipotential drop.
The rate of seepage ($Q$) through a soil mass can be determined from a flow net using the following formula, derived from Darcy's Law:
$$ Q = k \cdot H \cdot \frac{N_f}{N_d} $$
Where:
We are given the following parameters for the flow net under the sheet pile wall:
Before calculating the seepage rate, we need to ensure all units are consistent. The permeability is given in mm/s, the head in meters, and we need the final answer in Liters per day (L/day). Let's convert the permeability to meters per second (m/s) first, and then convert the final seepage rate to L/day.
Now, substitute the values into the formula:
$$ Q = k \cdot H \cdot \frac{N_f}{N_d} $$
$$ Q = (6 \times 10^{-6} \text{ m/s}) \cdot (4.5 \text{ m}) \cdot \frac{5}{10} $$
$$ Q = (6 \times 10^{-6}) \cdot (4.5) \cdot (0.5) \text{ m³/s} $$
$$ Q = 13.5 \times 10^{-6} \text{ m³/s} $$
The calculated seepage rate is in cubic meters per second (m³/s). We need to convert this to Liters per day (L/day).
So, to convert m³/s to L/day, we multiply by 1000 (to get L/s) and then by 86400 (to get L/day):
$$ Q_{\text{L/day}} = Q_{\text{m³/s}} \times 1000 \text{ L/m³} \times 86400 \text{ s/day} $$
$$ Q_{\text{L/day}} = (13.5 \times 10^{-6} \text{ m³/s}) \times 1000 \times 86400 \text{ L/day} $$
$$ Q_{\text{L/day}} = 13.5 \times 10^{-3} \times 86400 \text{ L/day} $$
$$ Q_{\text{L/day}} = 1166.4 \text{ L/day} $$
The calculated seepage rate is approximately 1166.4 L/day. Let's compare this value with the given options:
| Option | Seepage Rate (L/day) |
|---|---|
| 1 | 1367 |
| 2 | 1223 |
| 3 | 1167 |
| 4 | 1023 |
The calculated value of 1166.4 L/day is nearest to 1167 L/day.
| Parameter | Symbol | Given Value | Units |
|---|---|---|---|
| Coefficient of Permeability | $k$ | $6 \times 10^{-3}$ | mm/s ($6 \times 10^{-6}$ m/s) |
| Total Head | $H$ | 4.5 | m |
| Number of Flow Paths | $N_f$ | 5 | - |
| Number of Equipotential Drops | $N_d$ | 10 | - |
| Formula for Seepage Rate | $Q$ | $k \cdot H \cdot (N_f/N_d)$ | - |
| Calculated Seepage Rate | $Q$ | 1166.4 | L/day (Approx 1167) |
Flow nets are valuable tools for analyzing groundwater flow and seepage quantities, especially in complex boundary conditions like those around dams, sheet pile walls, or excavations.
Understanding how to sketch and interpret flow nets is crucial in geotechnical design to ensure the stability and safety of structures interacting with groundwater.
When once a pocket of smoke, containing air pollutants, is released into the atmosphere from a source like an automobile or a factory chimney, it gets dispersed into the atmosphere into various directions depending upon the
1. prevailing winds
2. temperature
3. pressure conditions
Select the correct answer.
During the compaction test, the weight of compacted soil specimen along with mould is 38.2 N. The volume and weight of mould are 0.95×10-3 m³ and 20.5 N respectively and the water content is 12%. The dry unit weight of the compacted specimen will be nearly