Yield from a well using specific capacity A well has specific capacity 0.3183 × 10^-3 per second. For a 4 m diameter well under a depression head (drawdown) of 4 m, estimate the discharge.

Introduction / Context:
Specific capacity is a convenient index in well hydraulics. For some formulations, it is expressed per unit area per unit drawdown, allowing quick discharge estimates when geometry and drawdown are known.

Given Data / Assumptions:

• Specific capacity, Cs = 0.3183 × 10^-3 s^-1.
• Well diameter, d = 4 m → area A = π d^2 / 4.
• Drawdown (depression head), h = 4 m.

Concept / Approach:
Assuming the given Cs represents discharge per unit area per unit drawdown, use Q = Cs * A * h. Compute area and multiply by drawdown to obtain discharge in m^3/s; then convert to L/s by multiplying by 1000.

Step-by-Step Solution:
A = π d^2 / 4 = π * 4^2 / 4 = π * 4 = 12.566 m^2 (approx).Q = Cs * A * h = 0.3183 × 10^-3 * 12.566 * 4.Compute product of area and head: 12.566 * 4 = 50.264.Q ≈ 0.3183 × 10^-3 * 50.264 ≈ 0.015999 m^3/s ≈ 16 L/s.

Verification / Alternative check:
Back-check: 16 L/s corresponds to 0.016 m^3/s. Dividing by A * h (≈ 50.264) gives about 0.318 × 10^-3 s^-1, matching the given specific capacity.

Why Other Options Are Wrong:

• 8–14 L/s underestimate the computed discharge for the stated area and drawdown.

Common Pitfalls:
Misinterpreting the meaning of specific capacity; forgetting to include the well cross-sectional area; unit conversion errors between m^3/s and L/s.

Final Answer:
16 litres/sec