Let's break down the solution step by step: --- ## **Step 1: List Given Data** - Inlet diameter, \( D_1 = 10.3 \) in - Outlet diameter, \( D_2 = 5.43 \) in - Inlet pressure, \( P_1 = 38.7 \) psi (absolute) - Outlet velocity, \( V_2 = 5.7 \) ft/s - Atmospheric pressure at outlet, \( p_a = 15 \) psi (absolute) We'll need to convert all units to SI later, but let's keep working in imperial units for now. --- ## **Step 2: Find Areas** \[ A_1 = \frac{\pi}{4} D_1^2, \quad A_2 = \frac{\pi}{4} D_2^2 \] Convert inches to feet: - \( D_1 = 10.3 \) in \( = \frac{10.3}{12} = 0.858 \) ft - \( D_2 = 5.43 \) in \( = \frac{5.43}{12} = 0.4525 \) ft Calculate areas: - \( A_1 = \frac{\pi}{4} (0.858)^2 = 0.578 \) ft² - \( A_2 = \frac{\pi}{4} (0.4525)^2 = 0.161 \) ft² --- ## **Step 3: Find \( V_1 \) Using Continuity** \[ Q = A_1 V_1 = A_2 V_2 \implies V_1 = \frac{A_2}{A_1} V_2 \] \[ V_1 = \frac{0.161}{0.578} \times 5.7 = 1.59 \text{ ft/s} \] --- ## **Step 4: Find Mass Flow Rate** For water, \( \rho = 62.4 \) lbm/ft³. \[ \dot{m} = \rho Q = \rho A_2 V_2 = 62.4 \times 0.161 \times 5.7 = 57.2 \text{ lbm/s} \] --- ## **Step 5: Apply Momentum Equation (Control Volume Analysis)** The sum of forces in the x-direction: \[ \sum F_x = \dot{m} (V_{2x} - V_{1x}) \] Forces include: - Force by the flange (\( F \)) **to the left** (what we solve for) - Pressure force at inlet: \( P_1 A_1 \) **to the right** - Pressure force at exit: \( -p_a A_2 \) **to the left** (since atmosphere pushes back) \[ F + p_a A_2 - P_1 A_1 = \dot{m} (V_2 - V_1) \] \[ F = \dot{m}(V_2 - V_1) + P_1 A_1 - p_a A_2 \] --- ## **Step 6: Plug in the Numbers** Convert pressures to lb/ft²: - \( 1 \) psi \( = 144 \) lb/ft² - \( P_1 = 38.7 \) psi \( = 5570 \) lb/ft² - \( p_a = 15 \) psi \( = 2160 \) lb/ft² Now calculate each term: **Momentum term:** \[ \dot{m}(V_2 - V_1) = 57.2 \times (5.7 - 1.59) = 57.2 \times 4.11 = 235.3 \text{ lbm} \cdot \text{ft/s}^2 \] But to get lbf, divide by \( g_c = 32.174 \) lbm·ft/lbf·s²: \[ \frac{235.3}{32.174} = 7.32 \text{ lbf} \] **Pressure terms:** \[ P_1 A_1 = 5570 \times 0.578 = 3221 \text{ lbf} \] \[ p_a A_2 = 2160 \times 0.161 = 348 \text{ lbf} \] --- ## **Step 7: Sum Up** \[ F = 7.32 + 3221 - 348 = 2880 \text{ lbf} \] --- ## **Step 8: Convert to Newtons** \( 1 \) lbf = \( 4.448 \) N \[ F = 2880 \times 4.448 = 12,819 \text{ N} \] --- ## **Final Answer** \[ \boxed{ F = 12,\!800 \ \text{N} \ (\text{rounded to 3 significant digits}) } \] --- ### **Summary Table** | Quantity | Value | Units | |--------------------|--------------|-------------| | \( V_1 \) | 1.59 | ft/s | | \( \dot{m} \) | 57.2 | lbm/s | | \( P_1 A_1 \) | 3221 | lbf | | \( p_a A_2 \) | 348 | lbf | | \( F \) | 2880 | lbf | | \( F \) | 12,800 | N | Let me know if you have questions about any specific step!