Sure! Let's break down the solution step by step for this fluid mechanics problem involving a control volume analysis of a converging elbow. --- ### **Given Data** - **Angle turned by elbow:** \(135^\circ\) - **Inlet diameter, \(D_1\):** \(400\) mm = \(0.4\) m - **Outlet diameter, \(D_2\):** \(200\) mm = \(0.2\) m - **Elbow flow passage volume:** \(V = 0.2\) m³ - **Volume flow rate, \(Q\):** \(0.5\) m³/s - **Inlet pressure, \(P_1\):** \(140\) kPa = \(140,000\) Pa - **Outlet pressure, \(P_2\):** \(110\) kPa = \(110,000\) Pa - **Elbow mass, \(m_{\text{elbow}}\):** \(12\) kg - **Density of water, \(\rho\):** \(1000\) kg/m³ (assumed) - **Gravitational acceleration, \(g\):** \(9.81\) m/s² --- ### **Step 1: Find Velocities at Inlet and Outlet** #### **Area at Inlet and Outlet** \[ A_1 = \frac{\pi}{4} D_1^2 = \frac{\pi}{4} (0.4)^2 = 0.126 \text{ m}^2 \] \[ A_2 = \frac{\pi}{4} D_2^2 = \frac{\pi}{4} (0.2)^2 = 0.0314 \text{ m}^2 \] #### **Velocities** \[ v_1 = \frac{Q}{A_1} = \frac{0.5}{0.126} = 3.97 \text{ m/s} \] \[ v_2 = \frac{Q}{A_2} = \frac{0.5}{0.0314} = 15.9 \text{ m/s} \] --- ### **Step 2: Control Volume Force Balance (Momentum Equation)** The forces required to anchor the elbow are obtained by applying the momentum equation in both \(x\) and \(z\) directions. #### **General Momentum Equation** \[ \sum F = \dot{m} (v_{\text{out}} - v_{\text{in}}) \] where \(\dot{m} = \rho Q\). #### **Component Directions (from geometry):** - **Inlet (1):** Along \(z\)-axis (upward) - **Outlet (2):** At \(135^\circ\) to \(z\)-axis, so: - \(x\)-component: \(v_2 \sin(135^\circ)\) - \(z\)-component: \(v_2 \cos(135^\circ)\) --- ### **Step 3: Calculate Forces** #### **Calculate Mass Flow Rate** \[ \dot{m} = \rho Q = 1000 \times 0.5 = 500~\text{kg/s} \] --- #### **(A) Horizontal Force (\(F_x\))** **Momentum flux in \(x\)-direction:** \[ F_x = \text{Pressure forces} + \text{Momentum fluxes} \] - Pressure at inlet (acts inward, negative \(x\)), outlet (acts outward, positive \(x\)) - Only outlet has \(x\)-component due to angle. \[ F_x = P_1 A_1 \cdot 0 + P_2 A_2 \sin(135^\circ) + \dot{m} (v_2 \sin(135^\circ) - 0) \] \[ \sin(135^\circ) = \sin(45^\circ) = 0.707 \] \[ F_x = (110{,}000)(0.0314)(0.707) + 500 \times (15.9 \times 0.707) \] Calculate: - \(110,000 \times 0.0314 \times 0.707 = 2,448\) N - \(15.9 \times 0.707 = 11.25\) - \(500 \times 11.25 = 5,625\) N \[ F_x = 2,448 + 5,625 = \boxed{8,073~\text{N}} \] --- #### **(B) Vertical Force (\(F_z\))** **Include weight of elbow and water inside:** \[ F_z = P_1 A_1 - P_2 A_2 \cos(135^\circ) + \dot{m} (v_2 \cos(135^\circ) - v_1) + \text{Weight} \] \[ \cos(135^\circ) = -0.707 \] \[ F_z = (140,000 \times 0.126) - (110,000 \times 0.0314 \times -0.707) + 500 \times (15.9 \times -0.707 - 3.97) + (12 + 200) \times 9.81 \] - \(140,000 \times 0.126 = 17,640\) - \(110,000 \times 0.0314 \times -0.707 = -2,448\) (minus sign, so add) - \(15.9 \times -0.707 = -11.25\), so \( -11.25 - 3.97 = -15.22 \) - \(500 \times -15.22 = -7,610\) - Water in elbow = \(0.2 \times 1000 = 200\) kg - Total weight = \(212 \times 9.81 = 2,079\) N \[ F_z = 17,640 + 2,448 - 7,610 + 2,079 = \boxed{14,557~\text{N}} \] --- ## **Summary** ### (a) Horizontal force: \[ \boxed{F_x = 8,073~\text{N}} \] ### (b) Vertical force: \[ \boxed{F_z = 14,557~\text{N}} \] --- **Let me know if you want more explanation on any step!**