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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

The correct answer is
1167 L/day

Calculating Seepage Under a Sheet Pile Wall Using Flow Net

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.

  • Flow Lines: These represent the paths that water particles follow as they move through the soil.
  • Equipotential Lines: These connect points of equal hydraulic head (potential energy). The flow occurs perpendicular to these 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.

Flow Net Parameters and Seepage Formula

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:

  • $k$ is the coefficient of permeability of the soil.
  • $H$ is the total hydraulic head loss across the flow region (the difference in water levels upstream and downstream).
  • $N_f$ is the number of flow paths (or flow channels) in the flow net.
  • $N_d$ is the number of equipotential drops in the flow net.

Applying the Formula to the Sheet Pile Wall Problem

We are given the following parameters for the flow net under the sheet pile wall:

  • Number of flow paths, $N_f = 5$
  • Number of equipotential drops, $N_d = 10$
  • Coefficient of permeability, $k = 6 \times 10^{-3}$ mm/s
  • Total head, $H = 4.5$ m

Unit Conversion

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.

  • $k = 6 \times 10^{-3} \text{ mm/s}$
  • Since 1 m = 1000 mm, we have 1 mm = $10^{-3}$ m.
  • $k = 6 \times 10^{-3} \times 10^{-3} \text{ m/s} = 6 \times 10^{-6} \text{ m/s}$

Calculating the Seepage Rate (Q)

Now, substitute the values into the formula:

$$ Q = k \cdot H \cdot \frac{N_f}{N_d} $$

$$ Q = (6 \times 10^{-6} \text{ m/s}) \cdot (4.5 \text{ m}) \cdot \frac{5}{10} $$

$$ Q = (6 \times 10^{-6}) \cdot (4.5) \cdot (0.5) \text{ m³/s} $$

$$ Q = 13.5 \times 10^{-6} \text{ m³/s} $$

Converting Seepage Rate to Liters per Day (L/day)

The calculated seepage rate is in cubic meters per second (m³/s). We need to convert this to Liters per day (L/day).

  • 1 m³ = 1000 Liters (L)
  • 1 day = 24 hours $\times$ 60 minutes/hour $\times$ 60 seconds/minute = 86400 seconds (s)

So, to convert m³/s to L/day, we multiply by 1000 (to get L/s) and then by 86400 (to get L/day):

$$ Q_{\text{L/day}} = Q_{\text{m³/s}} \times 1000 \text{ L/m³} \times 86400 \text{ s/day} $$

$$ Q_{\text{L/day}} = (13.5 \times 10^{-6} \text{ m³/s}) \times 1000 \times 86400 \text{ L/day} $$

$$ Q_{\text{L/day}} = 13.5 \times 10^{-3} \times 86400 \text{ L/day} $$

$$ Q_{\text{L/day}} = 1166.4 \text{ L/day} $$

Comparing with Options

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.

Revision Table: Flow Net Calculation Summary

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)

Additional Information on Flow Nets and Seepage

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.

  • Orthogonality: Flow lines and equipotential lines intersect at right angles, provided the soil is isotropic. In anisotropic soils, they intersect at angles other than 90 degrees, and the flow net drawing needs adjustment based on transformed sections.
  • Head Loss per Drop: The total head loss $H$ is distributed equally among all equipotential drops. The head loss per drop, $\Delta h$, is given by $\Delta h = H / N_d$.
  • Gradient: The hydraulic gradient ($i$) in any part of the flow net can be approximated by $\Delta h$ divided by the average length of the flow field element in that region.
  • Applications: Flow nets are used to estimate seepage quantity, uplift pressures under structures, exit hydraulic gradients (important for piping analysis), and pore water pressure distribution.
  • Piping: High exit gradients near the downstream toe of structures can lead to soil erosion and instability, a phenomenon known as piping. Flow nets help analyze the risk of piping.

Understanding how to sketch and interpret flow nets is crucial in geotechnical design to ensure the stability and safety of structures interacting with groundwater.

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