Question

A homogenous anisotropic earthen dam of height 52 m with a free board of 2 m is constructed on an impermeable foundation. The horizontal and vertical hydraulic conductivities of soil are \(K_h = 4.5 \times 10^{-8} \, m/s\) and \(K_v = 2.0 \times 10^{-8} \, m/s\). There are 6 flow channels and 25 equipotential drops in a square flownet drawn in the transformed dam section. If the downstream dam side is dry, the quantity of seepage per unit length through the dam in m\(^3\)/day/m is \underline{\hspace{3cm}}.

Hint:

For anisotropic dams, use equivalent permeability \(K = \sqrt{K_h K_v}\). Seepage = \(K H (N_f/N_d)\).

Answer 1

Step 1: Effective permeability. For anisotropic soil: \[ K = \sqrt{K_h \cdot K_v} = \sqrt{4.5 \times 10^{-8} \times 2.0 \times 10^{-8}} = 3.0 \times 10^{-8} \, m/s \]

Step 2: Head causing seepage. Head = height of water above base = \(52 - 2 = 50 \, m\).

Step 3: Discharge per unit length (Darcy's law with flownet). \[ q = K H \frac{N_f}{N_d} = 3.0 \times 10^{-8} \times 50 \times \frac{6}{25} \] \[ q = 3.6 \times 10^{-8} \, m^3/s/m \]

Step 4: Convert to per day. \[ q_{day} = 3.6 \times 10^{-8} \times 86400 = 0.097 \, m^3/day/m \] \[ \boxed{0.097 \, m^3/day/m} \]