Let's solve the problem step by step, based on the provided image and data: ### **Given Data** #### **Soil Properties Table** | Soil Soil Porosity | Hydraulic conductivity (cm/sec) | |------|---------------|----------------------------------| | A | 0.55 | 4 × 10⁻⁴ | | B | 0.65 | 5 × 10⁻⁴ | | C | 0.40 | 7 × 10⁻⁴ | #### **Other Data from Figure** - Diameter of tube = 20 cm ⇒ Radius = 10 cm - Length of each soil section = 300 mm = 30 cm - Total length = 3 × 30 cm = 90 cm - Water levels: - Upstream head \( h_1 = 500 \) mm \( = 50 \) cm - Downstream head \( h_2 = 290 \) mm \( = 29 \) cm --- ## **1. Quantity of Water Flowing Through the Sample per Day** ### **Step 1: Darcy's Law for Series Flow** For layers in series: \[ Q = \frac{K_{eq} \cdot A \cdot \Delta h}{L_{total}} \] Where: - \( K_{eq} \) = equivalent hydraulic conductivity - \( A \) = cross-sectional area - \( \Delta h \) = total head drop = \( h_1 - h_2 \) - \( L_{total} \) = total length But for series, use: \[ \frac{L_{total}}{K_{eq}} = \frac{L_A}{K_A} + \frac{L_B}{K_B} + \frac{L_C}{K_C} \] ### **Step 2: Calculate Area** \[ A = \pi r^2 = \pi (10)^2 = 314.16 \text{ cm}^2 \] ### **Step 3: Layer Lengths and \( K \) Values** - \( L_A = L_B = L_C = 30 \) cm - \( K_A = 4 \times 10^{-4} \) cm/sec - \( K_B = 5 \times 10^{-4} \) cm/sec - \( K_C = 7 \times 10^{-4} \) cm/sec ### **Step 4: Equivalent Hydraulic Conductivity** \[ \frac{L_{total}}{K_{eq}} = \frac{L_A}{K_A} + \frac{L_B}{K_B} + \frac{L_C}{K_C} \] \[ \frac{90}{K_{eq}} = \frac{30}{4 \times 10^{-4}} + \frac{30}{5 \times 10^{-4}} + \frac{30}{7 \times 10^{-4}} \] Calculate each term: - \( \frac{30}{4 \times 10^{-4}} = 75,000 \) - \( \frac{30}{5 \times 10^{-4}} = 60,000 \) - \( \frac{30}{7 \times 10^{-4}} \approx 42,857 \) Sum: \[ \frac{90}{K_{eq}} = 75,000 + 60,000 + 42,857 = 177,857 \] \[ K_{eq} = \frac{90}{177,857} \approx 5.06 \times 10^{-4} \text{ cm/sec} \] ### **Step 5: Calculate \( Q \) (per second)** \[ \Delta h = h_1 - h_2 = 50 - 29 = 21 \text{ cm} \] \[ Q = \frac{K_{eq} \cdot A \cdot \Delta h}{L_{total}} \] \[ Q = \frac{5.06 \times 10^{-4} \times 314.16 \times 21}{90} \] \[ = \frac{5.06 \times 10^{-4} \times 6597.36}{90} \] \[ = \frac{3.34}{90} \] \[ = 0.0371 \text{ cm}^3/\text{sec} \] ### **Step 6: Convert to per day** - 1 day = 86400 sec \[ Q_{day} = 0.0371 \times 86400 \approx 3207 \text{ cm}^3/\text{day} \] --- ## **2. Elevation Head (\( Z \)), Pressure Head (\( u/\gamma_w \)), and Total Head (\( h \)) at Each Layer Interface** - Assume datum at downstream water level (bottom right). - Tube is horizontal, so elevation from datum is the vertical distance above downstream base. **At each interface:** - **Entrance (leftmost):** - Elevation (\( Z \)) = 0 (datum at base of downstream) - Pressure head = water column above = \( 50 \) cm - Total head = \( 50 \) cm - **After Soil A (end of A, start of B):** - Distance from left = 30 cm - Elevation = 0 (since datum is at the same base) - Head loss through A: \( \Delta h_A = Q \frac{L_A}{K_A A} \) - But since A, B, C are in series, head drops linearly. The head drop across each layer: \[ \Delta h = 21 \text{ cm} \] The proportion for each layer: - \( \Delta h_A = \Delta h \cdot \frac{L_A / K_A}{\sum (L_i / K_i)} \) - \( \Delta h_B = \Delta h \cdot \frac{L_B / K_B}{\sum (L_i / K_i)} \) - \( \Delta h_C = \Delta h \cdot \frac{L_C / K_C}{\sum (L_i / K_i)} \) Let’s compute: Total = 177,857 - \( \Delta h_A = 21 \cdot \frac{75,000}{177,857} = 8.85 \) cm - \( \Delta h_B = 21 \cdot \frac{60,000}{177,857} = 7.09 \) cm - \( \Delta h_C = 21 \cdot \frac{42,857}{177,857} = 5.06 \) cm So: - At start (left): **Total head = 50 cm** - After A: \( 50 - 8.85 = 41.15 \) cm - After B: \( 41.15 - 7.09 = 34.06 \) cm - After C: \( 34.06 - 5.06 = 29 \) cm --- ## **3. Discharge Velocity (\( v \)) and Seepage Velocity (\( v_s \))** **Discharge velocity**: \[ v = \frac{Q}{A} \] \[ v = \frac{0.0371}{314.16} = 1.18 \times 10^{-4} \text{ cm/sec} \] **Seepage velocity for each soil**: \[ v_s = \frac{v}{\text{porosity}} \] - Soil A: \( v_{sA} = \frac{1.18 \times 10^{-4}}{0.55} = 2.15 \times 10^{-4} \) cm/sec - Soil B: \( v_{sB} = \frac{1.18 \times 10^{-4}}{0.65} = 1.82 \times 10^{-4} \) cm/sec - Soil C: \( v_{sC} = \frac{1.18 \times 10^{-4}}{0.40} = 2.95 \times 10^{-4} \) cm/sec --- ## **Summary Table** | Location | Total Head (cm) | |------------------|----------------| | Upstream/Entrance| 50 | | After Soil A | 41.15 | | After Soil B | 34.06 | | After Soil C | 29 | | Soil | Porosity | Seepage Velocity (\( v_s \), cm/sec) | |--------|----------|--------------------------------------| | A | 0.55 | \( 2.15 \times 10^{-4} \) | | B | 0.65 | \( 1.82 \times 10^{-4} \) | | C | 0.40 | \( 2.95 \times 10^{-4} \) | --- ## **Final Answers** 1. **Quantity of water flowing per day:** \( \boxed{3207 \ \text{cm}^3/\text{day}} \) 2. **Heads at each interface:** - Upstream: \( 50 \) cm - After A: \( 41.15 \) cm - After B: \( 34.06 \) cm - Downstream: \( 29 \) cm 3. **Discharge velocity:** \( 1.18 \times 10^{-4} \) cm/sec **Seepage velocities:** - Soil A: \( 2.15 \times 10^{-4} \) cm/sec - Soil B: \( 1.82 \times 10^{-4} \) cm/sec - Soil C: \( 2.95 \times 10^{-4} \) cm/sec If you need the calculations in SI units or further explanations, please let me know!