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
- A1367 L/day
- B1223 L/day
- C1167 L/dayCorrect
- D1023 L/day
Solution
Concept
Seepage per unit length of the wall
Formula
q = k H Nf / Nd
Step-by-step solution
1. k = 6×10⁻³ mm/s = 6×10⁻⁶ m/s. 2. q = 6×10⁻⁶ × 4.5 × 5/10 = 1.35×10⁻⁵ m³/s per metre. 3. Per day: 1.35×10⁻⁵ × 86 400 = 1.166 m³ = 1166 L/day ≈ 1167 L/day per metre length.
Explanation
The seepage per unit length of the wall is the permeability times the head times the ratio of flow channels to potential drops.
Why the other options are wrong
The other values do not match 1.35×10⁻⁵ m³/s.
Shortcut
q = kH(Nf/Nd); convert mm/s → m/s.
Common mistake
Forgetting the unit conversion.
Correct answer: C) 1167 L/day