Let's solve the problem step by step using the **energy equation for incompressible flow** (modified Bernoulli Equation for machines and head loss): --- ## **Given Data** - ρ (oil) = 795 kg/m³ - \(\dot{m}\) (mass flow rate) = 14.5 kg/s - \(d_1\) = 76.2 mm = .0762 m - \(d_2\) = 152.4 mm = .1524 m - \(p_1\) = 68.9 kPa = 68,900 Pa - \(p_2\) = 34.5 kPa = 34,500 Pa - Head loss, \(h_L = 1.2 \dfrac{V_1^2}{2g}\) - Acceleration due to gravity, \(g = 9.81~\text{m/s}^2\) --- # **Step 1: Find Velocities \(V_1\) and \(V_2\)** ### **Area Calculations** \[ A_1 = \frac{\pi}{4} d_1^2 = \frac{\pi}{4} (.0762)^2 = .00456~\text{m}^2 \] \[ A_2 = \frac{\pi}{4} d_2^2 = \frac{\pi}{4} (.1524)^2 = .01825~\text{m}^2 \] ### **Continuity Equation:** \[ \dot{m} = \rho A_1 V_1 = \rho A_2 V_2 \] \[ V_1 = \frac{\dot{m}}{\rho A_1} = \frac{14.5}{795 \times .00456} = 4.08~\text{m/s} \] \[ V_2 = \frac{\dot{m}}{\rho A_2} = \frac{14.5}{795 \times .01825} = 1.00~\text{m/s} \] --- # **Step 2: Apply the Energy Equation** The general energy equation for a system with a pump/turbine and head loss, ignoring elevation and internal energy changes: \[ \left( \frac{p_1}{\rho g} + \frac{V_1^2}{2g} \right) + h_{\text{pump/turbine}} = \left( \frac{p_2}{\rho g} + \frac{V_2^2}{2g} \right) + h_L \] --- ## **a) Find the head loss if for this setup it equals \(1.2 \dfrac{V_1^2}{2g}\)** Given directly: \[ h_L = 1.2 \frac{V_1^2}{2g} \] \[ = 1.2 \times \frac{(4.08)^2}{2 \times 9.81} = 1.2 \times \frac{16.6464}{19.62} = 1.2 \times .849 = 1.019~\text{m} \] **Final Answer (a):** \[ \boxed{h_L = 1.02~\text{m}} \] --- ## **b) What is the outlet velocity?** Already calculated above: \[ \boxed{V_2 = 1.00~\text{m/s}} \] --- ## **c) Is the machine a pump or a turbine?** Let's solve for \(h_{\text{pump/turbine}}\): \[ h_{\text{pump/turbine}} = \left( \frac{p_2}{\rho g} + \frac{V_2^2}{2g} \right) + h_L - \left( \frac{p_1}{\rho g} + \frac{V_1^2}{2g} \right) \] Plug in the numbers: ### **Pressure Terms:** \[ \frac{p_1}{\rho g} = \frac{68,900}{795 \times 9.81} = 8.85~\text{m} \] \[ \frac{p_2}{\rho g} = \frac{34,500}{795 \times 9.81} = 4.43~\text{m} \] ### **Velocity Head:** \[ \frac{V_1^2}{2g} = \frac{(4.08)^2}{2 \times 9.81} = .849~\text{m} \] \[ \frac{V_2^2}{2g} = \frac{(1.00)^2}{2 \times 9.81} = .051~\text{m} \] ### **Head Loss:** \[ h_L = 1.02~\text{m} \] ### **Substitute all values:** \[ h_{\text{pump/turbine}} = \left(4.43 + .051\right) + 1.02 - \left(8.85 + .849\right) \] \[ = (4.481) + 1.02 - (9.699) \] \[ = 5.501 - 9.699 = -4.20~\text{m} \] ### **Interpretation:** - If positive (\(h_{\text{pump/turbine}} > \)): **PUMP** (adds energy) - If negative (\(h_{\text{pump/turbine}} < \)): **TURBINE** (removes energy) Here, it's negative: \[ \boxed{\text{The machine is a turbine.}} \] --- ## **Summary Table** | Part | Answer | |------|------------------------| | (a) | \(h_L = 1.02~\text{m}\) | | (b) | \(V_2 = 1.00~\text{m/s}\) | | (c) | The machine is a **turbine** | --- If you need the full energy equation derivation or more explanation, let me know!