Let's solve the problem step by step: --- ## **Step 1: List the Given Data** - **Elevation difference:** \( \Delta z = 31.00\, \mathrm{m} \) - **Flow rate:** \( Q = .004\, \mathrm{m}^3/\mathrm{s} \) - **Density:** \( \rho = 998\, \mathrm{kg}/\mathrm{m}^3 \) - **Dynamic viscosity:** \( \mu = .001\, \mathrm{kg}/(\mathrm{m}\cdot\mathrm{s}) \) - **Pipe length:** \( L = 125\, \mathrm{m} \) - **Pipe diameter:** \( D = .05\, \mathrm{m} \) - **Pipe roughness (cast iron):** \( \epsilon = .26\, \mathrm{mm} = .00026\, \mathrm{m} \) - **Minor loss coefficients:** - Entrance: \( K_{\text{ent}} = .5 \) - Globe valve: \( K_{\text{valve}} = 6.9 \) - **Take \( \pi = \frac{22}{7} \)** - **Atmospheric discharge** --- ## **Step 2: Find the Velocity in the Pipe** \[ Q = A \cdot v \implies v = \frac{Q}{A} \] \[ A = \frac{\pi D^2}{4} = \frac{22}{7} \times \frac{(.05)^2}{4} = .001963\, \mathrm{m}^2 \] \[ v = \frac{.004}{.001963} \approx 2.039\, \mathrm{m/s} \] --- ## **Step 3: Calculate Reynolds Number** \[ \text{Re} = \frac{\rho v D}{\mu} = \frac{998 \times 2.039 \times .05}{.001} = 101,870 \] This is turbulent flow. --- ## **Step 4: Find Relative Roughness** \[ \frac{\epsilon}{D} = \frac{.00026}{.05} = .0052 \] --- ## **Step 5: Estimate Friction Factor (\( f \))** Use the **Swamee-Jain equation** for turbulent flow: \[ f = .25 \left[ \log_{10} \left( \frac{\epsilon}{3.7D} + \frac{5.74}{\text{Re}^{.9}} \right) \right]^{-2} \] Calculate: - \( \frac{\epsilon}{3.7D} = \frac{.00026}{3.7 \times .05} = .001405 \) - \( \frac{5.74}{\text{Re}^{.9}} = \frac{5.74}{(101870)^{.9}} \approx .000104 \) Sum: \( .001405 + .000104 = .001509 \) \[ f = .25 \left[ \log_{10} (.001509) \right]^{-2} \] \[ \log_{10}(.001509) \approx -2.821 \] \[ f = .25 \left[ -2.821 \right]^{-2} = .25 \times (1/7.96) \approx .0314 \] --- ## **Step 6: Head Losses** ### **a) Entrance Loss** \[ h_{\text{ent}} = K_{\text{ent}} \frac{v^2}{2g} = .5 \frac{(2.039)^2}{2 \times 9.81} \approx .106\, \mathrm{m} \] ### **b) Globe Valve Loss** \[ h_{\text{valve}} = K_{\text{valve}} \frac{v^2}{2g} = 6.9 \frac{(2.039)^2}{2 \times 9.81} \approx 1.46\, \mathrm{m} \] ### **c) Pipe Friction Loss** \[ h_{\text{f}} = f \frac{L}{D} \frac{v^2}{2g} \] \[ \frac{L}{D} = \frac{125}{.05} = 250 \] \[ h_{\text{f}} = .0314 \times 250 \times \frac{(2.039)^2}{2 \times 9.81} \] \[ = 78.5 \times \frac{4.16}{19.62} = 78.5 \times .212 = 16.64\, \mathrm{m} \] --- ## **Step 7: Total Head Loss** \[ h_{\text{loss,tot}} = h_{\text{ent}} + h_{\text{f}} + h_{\text{valve}} = .106 + 16.64 + 1.46 = 18.206\, \mathrm{m} \] --- ## **Step 8: Apply the Energy Equation (Bernoulli’s with Turbine and Losses)** From the surface (1) to the outlet (2): \[ z_1 + \frac{P_1}{\gamma} = z_2 + \frac{P_2}{\gamma} + h_{\text{loss}} + \frac{W_{\text{turbine}}}{\gamma} \] At both points, velocity is negligible except in the pipe, and both are open to atmosphere, so pressures cancel out. \( z_1 - z_2 = \Delta z \). \[ \Delta z = h_{\text{loss}} + \frac{W_{\text{turbine}}}{\gamma} \] \[ W_{\text{turbine}} = \gamma (\Delta z - h_{\text{loss}}) \] Where \( \gamma = \rho g \). --- ## **Step 9: Calculate the Power Extracted** \[ P_{\text{turbine}} = \dot{m}g(\Delta z - h_{\text{loss}}) \] Where \( \dot{m} = \rho Q \): \[ P_{\text{turbine}} = \rho Q g (\Delta z - h_{\text{loss}}) \] \[ = 998 \times .004 \times 9.81 \times (31 - 18.206) \] \[ = 998 \times .004 \times 9.81 \times 12.794 \] \[ = 3.992 \times 9.81 \times 12.794 \] \[ = 39.15 \times 12.794 \] \[ \approx 501\, \text{W} \] --- ## **Final Answer** \[ \boxed{501\, \text{W}} \] --- **The power extracted by the turbine is approximately \( \mathbf{501\, W} \).**