Here’s a summary and breakdown of the solution steps for the given pipe flow problem: --- ### **Given Data** - Pressure difference (\( \Delta P \)) = 20 psi - Roughness (\( k_s \)) = 0.0005 ft - Loss coefficient at contraction (\( k \)) = 0.05 - Temperature (\( T \)) = 70°F - Pipe details: - 6 in diameter, 1000 ft - 10 in diameter, 850 ft - 8 in diameter, 1300 ft --- ## **Step 1: Continuity Equation** **Conservation of mass flow:** \[ Q = A_1 V_1 = A_2 V_2 = A_3 V_3 \] - \( \frac{\pi}{4} \cdot 6^2 \cdot V_1 = \frac{\pi}{4} \cdot 8^2 \cdot V_2 = \frac{\pi}{4} \cdot 10^2 \cdot V_3 \) - Simplifying: - \( 36 V_1 = 64 V_2 = 100 V_3 \) - \( V_1 = \frac{16}{9} V_2 \) - \( V_3 = \frac{16}{25} V_2 \) --- ## **Step 2: Friction Factor Calculation** **Roughness and turbulent flow:** - Assume turbulent flow. - Use Nikuradse equation: \[ \frac{1}{\sqrt{f}} = 2 \log_{10} \left( \frac{d}{2 k_s} \right) + 1.74 \] - For 6 in (\( f_1 \)): \( f_1 = 0.02 \) - For 8 in (\( f_2 \)): \( f_2 = 0.018 \) - For 10 in (\( f_3 \)): \( f_3 = 0.017 \) --- ## **Step 3: Energy Equation Application** **Apply energy equation between points 1 and 2:** \[ \frac{P_1}{\rho g} + \frac{V_1^2}{2g} + Z_1 = \frac{P_2}{\rho g} + \frac{V_2^2}{2g} + Z_2 + h_f \] - \( h_f = h_{f1} + h_{f2} + h_{f3} \) - Calculate each head loss: - \( h_{f1} = \frac{f_1 L_1 V_1^2}{2gd_1} \) - \( h_{f2} = \frac{f_2 L_2 V_2^2}{2gd_2} \) - \( h_{f3} = \frac{f_3 L_3 V_3^2}{2gd_3} \) - Include minor loss (sudden contraction): \( k \frac{(V_1 - V_3)^2}{2g} \) - Substitute velocities in terms of \( V_2 \). **Combine all terms and solve for \( V_2 \):** - Substitute known values: - Pressure difference in ft: \( \frac{P_1 - P_2}{\rho g} = \frac{20 \times 144}{62.3} = 46.25 \) ft - Equation: \[ \frac{P_1 - P_2}{\rho g} = 25.052 \frac{V_2^2}{g} + 0.647 \frac{V_2^2}{g} + 0.05 \frac{V_2^2}{g} \] \[ 46.25 = 25.749 \frac{V_2^2}{g} \] \[ V_2 = 7.772 \text{ ft/s} \] --- ## **Step 4: Flow Rate Calculation** \[ Q = A_2 V_2 = \frac{\pi}{4} \left(\frac{8}{12}\right)^2 \cdot 7.772 = 2.713 \text{ ft}^3/\text{s} \] --- ## **Step 5: Reynolds Number Check** \[ Re = \frac{V d}{\nu} = \frac{7.772 \times \frac{8}{12}}{0.8428 \times 10^{-5}} = 6.148 \times 10^5 \] - Confirms turbulent flow (>4000). --- ## **Final Answer** ### **The flow rate in the system is:** \[ \boxed{Q = 2.713 \ \text{ft}^3/\text{s}} \] --- **All calculations and steps are shown clearly in the provided solution image. If you need a detailed explanation of any step, let me know!**