# Solution to Curved Pipe & Nozzle Problem Let's solve each subpart step-by-step. --- ## **Given Data** - **Inlet velocity:** \( V_1 \) - **Inlet area:** \( A_1 \) - **Outlet velocity:** \( V_2 \) - **Outlet area:** \( A_2 \) - **Inlet pressure:** \( p_1 \) - **The pipe is bent 180°.** - **Bernoulli’s equation applies.** --- ## **(i) Expression for the Horizontal Force Required to Hold the Pipe in Place** ### **Step 1: Apply the Conservation of Mass (Continuity)** \[ \rho A_1 V_1 = \rho A_2 V_2 \implies V_2 = \frac{A_1}{A_2} V_1 \] --- ### **Step 2: Apply the Linear Momentum Equation** For a control volume around the pipe: \[ \vec{F}_{\text{ext}} + \int p \vec{n}\, dA = \int \rho \vec{V} (\vec{V} \cdot \vec{n})\, dA \] - **Horizontal direction:** Take right as positive. #### **Control Surfaces:** - **Inlet:** Left to right (\(+x\)) - **Outlet:** Right to left (\(-x\)), because of the 180° bend #### **Force Components:** Let \( F_x \) be the required horizontal force (positive to the right). #### **Pressure Forces:** - Inlet: \( p_1 A_1 \) (to the right) - Outlet: Atmospheric pressure, can be set to gauge (open to atmosphere) #### **Momentum Flux:** - Inlet: \( \rho Q V_1 \) (to the right) - Outlet: \( \rho Q V_2 \) (to the left; negative in \( +x \)) #### **Total Horizontal Force:** \[ F_x + p_1 A_1 = \rho Q V_1 - \rho Q V_2 \] where \( Q = V_1 A_1 = V_2 A_2 \). \[ F_x = \rho Q (V_1 - V_2) - p_1 A_1 \] \[ \boxed{ F_x = \rho V_1 A_1 (V_1 - V_2) - p_1 A_1 } \] With \( V_2 = \frac{A_1}{A_2} V_1 \): \[ F_x = \rho V_1 A_1 \left[V_1 - \left(\frac{A_1}{A_2} V_1\right)\right] - p_1 A_1 \] \[ F_x = \rho V_1^2 A_1 \left(1 - \frac{A_1}{A_2}\right) - p_1 A_1 \] --- ## **(ii) Evaluate the Force for Given Values** Given: - \( V_1 = .1\ \text{m/s} \) - \( A_1 = 6\ \text{cm}^2 = 6 \times 10^{-4}\ \text{m}^2 \) - \( A_2 = 2\ \text{cm}^2 = 2 \times 10^{-4}\ \text{m}^2 \) - \( p_1 = \) (gauge, assume open to atm) - \( \rho_{\text{water}} = 100\ \text{kg/m}^3 \) Calculate \( V_2 \): \[ V_2 = \frac{A_1}{A_2} V_1 = \frac{6 \times 10^{-4}}{2 \times 10^{-4}} \times .1 = 3 \times .1 = .3\,\text{m/s} \] Now plug in: \[ F_x = 100 \times .1^2 \times 6 \times 10^{-4} \left[1 - \frac{6 \times 10^{-4}}{2 \times 10^{-4}}\right] \] \[ = 100 \times .01 \times 6 \times 10^{-4} \left[1 - 3\right] \] \[ = 10 \times 6 \times 10^{-4} \times (-2) \] \[ = 6 \times 10^{-3} \times (-2) \] \[ = -12 \times 10^{-3} = -.012\ \text{N} \] **The negative sign means the force is to the left.** \[ \boxed{F_x = -.012\ \text{N}} \] --- ## **(iii) Vertical Force Component** ### **Analysis:** - The fluid enters and exits horizontally. - **There is no vertical momentum change** (assuming pipe is horizontal and neglecting gravity). \[ F_y = \] \[ \boxed{F_y = } \] --- ## **Summary Table** | Part | Expression / Value | |------|---------------------------------------------| | (i) | \( F_x = \rho V_1^2 A_1 \left(1 - \frac{A_1}{A_2}\right) - p_1 A_1 \) | | (ii) | \( F_x = -.012\ \text{N} \) | | (iii)| \( F_y = \) | Let me know if you need any clarifications or derivation details!