Let's solve the problem **step by step** using **Casagrande's method** for seepage through an earth dam with a horizontal impermeable base. --- ## **Step 1: Gather Given Data** - **Height of water on upstream side, \( h_1 \) = 25 m** - **Height of water on downstream side, \( h_2 \) = m** (assumed, since not shown otherwise) - **Top width of dam = 5 m** - **Upstream slope = 1V:2H** - **Downstream slope = 1V:2H** - **Horizontal length of base = 30 m** - **Coefficient of permeability, \( k = 3 \times 10^{-4} \text{ m/min} \)** - **Length of flow (width perpendicular to section) = 1 m** (Assume per meter length) --- ## **Step 2: Calculate Length of Flow Path (\( L \))** Using **Casagrande’s method** for a homogeneous earth dam with an impermeable base, the length of the seepage path (from upstream toe to downstream toe) is: \[ L = \text{horizontal distance from upstream to downstream} = 30\,\text{m} \] --- ## **Step 3: Find the Number of Equipotential Drops (\( N_d \)) and Flow Channels (\( N_f \))** For a simple case, if not otherwise given, we use the length of flow and the head drop directly: - **Total Head Loss, \( h = h_1 - h_2 = 25\,\text{m} - \,\text{m} = 25\,\text{m} \)** - **Length of Seepage Path, \( L = 30\,\text{m} \)** --- ## **Step 4: Casagrande's Formula for Seepage** \[ q = k \cdot h \cdot \frac{N_f}{N_d} \] But for a simple homogeneous dam with an impermeable base, the rate of seepage per meter width (for a rough estimate): \[ q = k \cdot \frac{h_1 \cdot L_{e}}{L} \] Where \( L_{e} \) is the length of the phreatic line (approximate as dam base for a first estimate), and \( L \) is the length of the seepage path. But often, for simple cases: \[ q = k \cdot \frac{h_1}{L} \] *Since the water percolates through the entire dam base*. --- ## **Step 5: Plug in the Values** \[ q = k \cdot \frac{h_1}{L} \] \[ q = (3 \times 10^{-4}\ \text{m/min}) \cdot \frac{25\ \text{m}}{30\ \text{m}} \] --- ## **Step 6: Calculate** \[ \frac{25}{30} = .833 \] \[ q = 3 \times 10^{-4} \times .833 \] \[ q = 2.5 \times 10^{-4}\ \text{m}^3/\text{min}/\text{m} \] --- ## **Step 7: Final Answer** \[ \boxed{ q = 2.5 \times 10^{-4} \ \text{m}^3/\text{min}/\text{m} } \] This is the **rate of seepage through the dam per meter width** using Casagrande’s method (simplified). --- ### **Summary Table** | Parameter | Value | |-------------------|------------------------| | \( h_1 \) | 25 m | | \( L \) | 30 m | | \( k \) | \( 3 \times 10^{-4} \) m/min | | **q** | \( 2.5 \times 10^{-4} \) m³/min/m | --- If you need the detailed flow net construction or a more precise estimate (including phreatic line curvature), please specify. For many introductory problems, this direct approach is acceptable.