Sure! Let's break down the problem step by step: --- ### **Step 1: Understand the Problem** We have a pipe elbow with water flowing through it. We are to find the force (**magnitude and direction**) needed to **anchor** the elbow so it doesn't move due to the water flow. We are given: - **Inlet (1):** - Pressure, \( p_1 = 104\, \text{kPa} \) (gage) - Velocity, \( V_1 = 2\, \text{m/s} \) - Diameter, \( D_1 = 300\, \text{mm} = 0.3\, \text{m} \) - **Outlet (2):** - Diameter, \( D_2 = 160\, \text{mm} = 0.16\, \text{m} \) - Pressure is atmospheric (since it's open to atmosphere) - **Atmospheric Pressure:** \( p_{\text{atm}} = 96.6\, \text{kPa} \) --- ### **Step 2: Find velocities and areas** #### **Areas:** \[ A_1 = \frac{\pi}{4} D_1^2 = \frac{\pi}{4} (0.3)^2 = 0.0707\, \text{m}^2 \] \[ A_2 = \frac{\pi}{4} D_2^2 = \frac{\pi}{4} (0.16)^2 = 0.0201\, \text{m}^2 \] #### **Flow Rate (Q):** \[ Q = A_1 V_1 = 0.0707 \times 2 = 0.1414\, \text{m}^3/\text{s} \] #### **Velocity at outlet:** \[ Q = A_2 V_2 \implies V_2 = \frac{Q}{A_2} = \frac{0.1414}{0.0201} = 7.04\, \text{m/s} \] --- ### **Step 3: Control Volume and Forces** We'll apply the **momentum equation** in the x-direction for the control volume. #### **Forces acting:** - Pressure at inlet (\(P_1\)) - Pressure at outlet (atmospheric, cancels with outside atmospheric pressure) - Reaction (anchoring) force \(F_x\) - Momentum change #### **Momentum equation (x-direction):** \[ \sum F_x = \dot{m} (V_{2x} - V_{1x}) \] #### **Mass Flow Rate:** \[ \dot{m} = \rho Q \] Assume water: \(\rho = 1000\, \text{kg/m}^3\) --- #### **Inlet (1):** - \(V_{1x} = V_1\) (to the right, positive x direction) #### **Outlet (2):** - \(V_{2x} = V_2\) (to the left, negative x direction) --- #### **Pressure Force at Inlet:** \[ F_{p1} = (p_1 + p_{\text{atm}}) \cdot A_1 \] But \(p_1\) is **gauge pressure**, so **absolute pressure** at inlet: \[ p_{1,\text{abs}} = p_1 + p_{\text{atm}} = 104 + 96.6 = 200.6\, \text{kPa} \] So, \[ F_{p1} = 200600 \cdot 0.0707 = 14188\, \text{N} \] --- #### **No pressure force at the outlet** (since it's open to atmosphere, cancels with outside). --- ### **Step 4: Plug into Momentum Equation** \[ F_x = F_{p1} - \dot{m}(V_{2x} - V_{1x}) \] But since \(V_{2x}\) is to the left (negative x), and \(V_{1x}\) is to the right (positive x): \[ F_x = F_{p1} - \dot{m} [(-V_2) - (V_1)] \] \[ F_x = F_{p1} + \dot{m} (V_2 + V_1) \] Plug in values: - \(F_{p1} = 14188\, \text{N}\) - \(\dot{m} = 1000 \times 0.1414 = 141.4\, \text{kg/s}\) - \(V_2 = 7.04\, \text{m/s}\) - \(V_1 = 2.0\, \text{m/s}\) \[ F_x = 14188 + 141.4 \times (7.04 + 2.0) \] \[ F_x = 14188 + 141.4 \times 9.04 \] \[ F_x = 14188 + 1278.3 = 15466.3\, \text{N} \] --- ### **Step 5: Direction** Since the result is **positive**, the force is to the **right** (positive x-direction). --- ## **Final Answers:** ### **(a) \( K_x = \boxed{15466}\, \text{N} \)** ### **(b) The force is to the right (positive x-direction).** --- Let me know if you want to see the detailed steps for the y-direction or any other part!