# Solution: Forces on Elbow in Pipe Flow Let's break down the given problem and solve it step by step. --- ## **Given Data** - Elbow flow passage volume, \( V_{\text{elbow}} = .37 \) m³ - Flowrate, \( Q = .90 \) m³/s - Inlet pressure, \( P_1 = 200 \) kPa = 200,000 Pa - Outlet pressure, \( P_2 = 130 \) kPa = 130,000 Pa - Elbow weight is negligible - Densities and other properties not specified; assume water, \( \rho = 100 \) kg/m³ - Inlet and outlet diameters: - \( D_1 = 400 \) mm = .4 m - \( D_2 = 200 \) mm = .2 m --- ## **Step 1: Find Velocities at Inlet and Outlet** \[ A_1 = \frac{\pi}{4} D_1^2 = \frac{\pi}{4} (.4)^2 = .126 \text{ m}^2 \] \[ A_2 = \frac{\pi}{4} D_2^2 = \frac{\pi}{4} (.2)^2 = .0314 \text{ m}^2 \] \[ V_1 = \frac{Q}{A_1} = \frac{.90}{.126} = 7.14 \text{ m/s} \] \[ V_2 = \frac{Q}{A_2} = \frac{.90}{.0314} = 28.66 \text{ m/s} \] --- ## **Step 2: Forces on Control Volume** The sum of forces on the control volume in x and y directions (ignoring gravity and elbow weight): \[ \vec{F} = \sum \text{momentum out} - \sum \text{momentum in} + \sum \text{pressure forces} \] ### **a) X-direction (horizontal)** Assume section 1 is horizontal (inlet, left to right), section 2 is vertical (outlet, upwards). - At section 1, flow is in \(+\hat{i}\) direction. - At section 2, flow is in \(+\hat{j}\) direction. #### **Pressure Forces** - At section 1 (left side): \( F_{P1} = P_1 A_1 \) (to the right) - At section 2 (top): \( F_{P2} = P_2 A_2 \) (downwards) #### **Net Force Components** - \( x \)-component: \( F_x \) - \( y \)-component: \( F_y \) #### **Momentum Fluxes** \[ \text{Momentum in at 1 (horizontal)} = \rho Q V_1 \] \[ \text{Momentum out at 2 (vertical)} = \rho Q V_2 \] --- ## **Step 3: Momentum Equation** The fixed control volume momentum equation in x-direction: \[ F_x = \rho Q (V_{2x} - V_{1x}) + (P_1 A_1 - P_2 A_2 \cdot ) \] But since section 2 is vertical, its x component is zero. \[ F_x = -\rho Q V_1 + P_1 A_1 \] ### **Y-direction** \[ F_y = \rho Q (V_{2y} - V_{1y}) + (P_2 A_2 - P_1 A_1 \cdot ) \] Section 1 is horizontal; its y-component is zero. Section 2 is vertical: \[ F_y = \rho Q V_2 - P_2 A_2 \] --- ## **Step 4: Plug in Values** ### **X-Direction** \[ F_x = -\rho Q V_1 + P_1 A_1 \] \[ = -(100)(.90)(7.14) + (200,000)(.126) \] \[ = -6426 + 25,200 \] \[ = 18,774 \text{ N} \quad \rightarrow \] --- ### **Y-Direction** \[ F_y = \rho Q V_2 - P_2 A_2 \] \[ = (100)(.90)(28.66) - (130,000)(.0314) \] \[ = 25,794 - 4,082 \] \[ = 21,712 \text{ N} \quad \uparrow \] --- ## **Step 5: Final Answers in Vector Notation** \[ \boxed{ \vec{F} = (18,774~\hat{i} + 21,712~\hat{j}) ~ \text{N} } \] - **Horizontal Anchoring Force (x-direction):** \( \boxed{18,774~\text{N} \rightarrow} \) - **Vertical Anchoring Force (y-direction):** \( \boxed{21,712~\text{N} \uparrow} \) --- ## **Summary Table** | Direction | Force (N) | Direction | |----------------|-----------|------------| | Horizontal (x) | 18,774 | Right | | Vertical (y) | 21,712 | Up | --- **Assumptions**: - Water is used as the fluid (\(\rho = 100 \text{ kg/m}^3\)). - Flow is steady, incompressible. - Elbow weight is negligible. - No losses or other forces. --- ### **If you need a freebody diagram or want to see the forces broken down further, let me know!**