Seepage discharge from a flow net (homogeneous earth dam): A homogeneous earth dam is 50 m high with 2 m freeboard (head h = 48 m). A flow net gives number of potential drops Nd = 24 and number of flow channels Nf = 4. If permeability k = 3 × 10^-5 m/s, compute discharge per metre length of dam.

Introduction / Context:Flow nets approximate seepage beneath/through hydraulic structures. For a homogeneous, isotropic medium, discharge per unit thickness is proportional to the product of permeability, head, and the ratio Nf/Nd derived from the net geometry.

Given Data / Assumptions:

• Dam height = 50 m with freeboard 2 m → seepage head h = 48 m.
• Number of flow channels Nf = 4.
• Number of potential drops Nd = 24.
• Permeability k = 3 × 10^-5 m/s (converted from 3 × 10^-3 cm/s).

Concept / Approach:For a unit width (1 m) section, discharge q is given by q = k * h * (Nf / Nd). This stems from integrating Darcy’s law along elementary squares of the flow net.

Step-by-Step Solution:

Verification / Alternative check:Units: k (m/s) × head (m) yields m^2/s; with unit width (1 m) gives m^3/s, dimensionally consistent.

Why Other Options Are Wrong:

• 12 × 10^-5 and 6 × 10^-5 m^3/s understate Nf/Nd or head.
• 24 × 10^-3 m^3/s is two orders too large (uses k in cm/s without conversion or misuses head).
• 9.6 × 10^-5 m^3/s corresponds to Nf/Nd = 0.4 instead of 1/6.

Common Pitfalls:Forgetting to convert cm/s to m/s; using dam height instead of water head above downstream toe; misreading Nf and Nd from the net.

Final Answer:24 × 10^-5 m^3/s