Let's solve the problem step by step: --- ## **Given Data** - **Diameter of specimen:** \( d = 110 \) mm \( = .110 \) m - **Height of specimen:** \( L = 270 \) mm \( = .270 \) m - **Upper reservoir elevation:** \( h_1 = 201 \) mm - **Lower reservoir elevation:** \( h_2 = 1671 \) mm - **Piezometer readings (200 mm apart):** - \( h_a = 1809 \) mm - \( h_b = 1578 \) mm - **Collected volume:** \( V = 910 \) ml \( = 9.10 \times 10^{-4} \) m³ - **Time:** \( t = 25 \) min \( 15 \) s \( = 1515 \) s --- ## **Step 1: Calculate Cross-sectional Area** \[ A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{.110}{2}\right)^2 = \pi \times (.055)^2 = \pi \times .003025 = .00951 \text{ m}^2 \] --- ## **Step 2: Calculate Discharge Rate (\( Q \))** \[ Q = \frac{V}{t} = \frac{9.10 \times 10^{-4}}{1515} = 6.01 \times 10^{-7} \text{ m}^3/\text{s} \] --- ## **Step 3: Calculate Hydraulic Gradient (\( i \))** Piezometer difference: \( h_a - h_b = 1809 - 1578 = 231 \) mm \( = .231 \) m Distance between piezometers: \( \Delta L = 200 \) mm \( = .200 \) m \[ i = \frac{\Delta h}{\Delta L} = \frac{.231}{.200} = 1.155 \] --- ## **Step 4: Use Darcy's Law to Find Hydraulic Conductivity (\( k \))** Darcy's Law: \[ Q = k \cdot i \cdot A \] So, \[ k = \frac{Q}{i \cdot A} \] Plug in values: \[ k = \frac{6.01 \times 10^{-7}}{1.155 \times .00951} \] \[ k = \frac{6.01 \times 10^{-7}}{.01099} \] \[ k = 5.47 \times 10^{-5} \text{ m/s} \] --- ## **Step 5: Reasonableness Check** - **For fine sand:** Typical values for hydraulic conductivity (\( k \)) range from \( 10^{-5} \) to \( 10^{-4} \) m/s. - Our result, \( 5.47 \times 10^{-5} \) m/s, falls within this range, so it seems reasonable. --- ## **Final Answer** ### **Hydraulic conductivity, \( k \), is:** \[ \boxed{5.5 \times 10^{-5} \ \text{m/s}} \] ### **Reasonableness:** Yes, this value is reasonable for fine sand, as it falls within the expected range for such materials.