Let's break down the question step by step: --- ## **Question Recap** - **Water flow rate**: \( Q = .130 \, \text{kg/s} \) *(mass flow rate)* - **Inlet velocity**: \( v_1 = 1.5 \, \text{m/s} \) - **Outlet velocity**: \( v_2 = 15 \, \text{m/s} \) - **Gauge pressure at outlet**: \( P_2 = 15 \, \text{psi} = 103.4 \, \text{kPa} \) *(1 psi = 6.895 kPa)* - **Correction factor**: \( \beta = .94 \) - **Density of water**: \( \rho = 100 \, \text{kg/m}^3 \) - **Neglect elevation and weight of hose** Let's answer each part step by step. --- ## **(a) Calculate the Inlet Area** We know: \[ Q = \dot{m} = \rho \cdot A_1 \cdot v_1 \] \[ A_1 = \frac{\dot{m}}{\rho v_1} \] Where: - \( \dot{m} = .130\, \text{kg/s} \) - \( \rho = 100\, \text{kg/m}^3 \) - \( v_1 = 1.5\, \text{m/s} \) Plug in the values: \[ A_1 = \frac{.130}{100 \times 1.5} = \frac{.130}{150} = 8.67 \times 10^{-5} \, \text{m}^2 \] --- ## **(b) Calculate the Gauge Pressure at the Hose Inlet** Apply the **steady-flow energy (Bernoulli) equation** between inlet (1) and outlet (2): \[ \frac{P_1}{\rho} + \frac{v_1^2}{2} = \frac{P_2}{\rho} + \frac{v_2^2}{2} \] (Since elevation head and velocity correction factors are negligible/absent.) Solve for \( P_1 \): \[ P_1 = P_2 + \frac{\rho}{2}(v_2^2 - v_1^2) \] Given: - \( P_2 = 103,425 \, \text{Pa} \) - \( v_2 = 15 \, \text{m/s} \) - \( v_1 = 1.5 \, \text{m/s} \) - \( \rho = 100 \, \text{kg/m}^3 \) \[ P_1 = 103,425 + \frac{100}{2} (15^2 - 1.5^2) = 103,425 + 500 (225 - 2.25) = 103,425 + 500 \times 222.75 = 103,425 + 111,375 = 214,800 \, \text{Pa} \] \[ \boxed{P_1 = 214.8 \, \text{kPa}} \] --- ## **(c) Calculate the Anchoring Force (Magnitude and Direction)** Use **momentum equation** for a control volume: \[ \sum F_x = \dot{m} (v_{2x} - v_{1x}) \] Where: - \(\sum F_x\) = net force acting on the control volume (hose) - \(v_{2x}\) and \(v_{1x}\) = velocities at outlet and inlet Forces acting (in x-direction): 1. Pressure at inlet (\(P_1 A_1\), acts right) 2. Pressure at outlet (\(P_2 A_2\), acts left) 3. Anchoring force (\(F\), acts left) So, \[ F + P_2 A_2 - P_1 A_1 = \dot{m}(v_2 - v_1) \] \[ F = \dot{m}(v_2 - v_1) + P_1 A_1 - P_2 A_2 \] Calculate \(A_2\): \[ Q = \dot{m}/\rho = A_2 v_2 \implies A_2 = \frac{\dot{m}}{\rho v_2} = \frac{.130}{100 \times 15} = \frac{.130}{15,000} = 8.67 \times 10^{-6} \, \text{m}^2 \] Now, plug in: - \(\dot{m} = .130\, \text{kg/s}\) - \(v_2 - v_1 = 15 - 1.5 = 13.5\, \text{m/s}\) - \(P_1 = 214,800\, \text{Pa}\) - \(A_1 = 8.67 \times 10^{-5}\, \text{m}^2\) - \(P_2 = 103,425\, \text{Pa}\) - \(A_2 = 8.67 \times 10^{-6}\, \text{m}^2\) \[ F = (.130 \times 13.5) + (214,800 \times 8.67 \times 10^{-5}) - (103,425 \times 8.67 \times 10^{-6}) \] \[ F = 1.755 + 18.62 - .897 \] \[ F = 1.755 + 18.62 - .897 = 19.478\, \text{N} \] **Direction:** The force acts **to the left** (to hold the hose in place). --- ## **Summary of Answers** **(a) Inlet area:** \[ \boxed{A_1 = 8.67 \times 10^{-5} \, \text{m}^2} \] **(b) Gauge pressure at inlet:** \[ \boxed{P_1 = 214.8 \, \text{kPa}} \] **(c) Anchoring force (magnitude and direction):** \[ \boxed{19.5 \, \text{N}, \text{ to the left}} \] --- If you need *any* step explained further, let me know!