# Problem Overview An adiabatic pump moves water vertically within a building. The goal is to determine the outlet velocity of water at the faucet and the shaft power supplied by the pump. --- ## Given Data | Parameter | Value | Units | | --- | --- | --- | | Mass flow rate, \(\dot{m}\) | .25 | kg/s | | Density of water, \(\rho\) | 100 | kg/m\(^3\) | | Height difference, \(h\) | 10 | m | | Pressure at faucet exit, \(P_{exit}\) | 100 | kPa | | Pressure before pump, \(P_{in}\) | 110 | kPa | | Diameter of faucet, \(D\) | 2 | cm | | Gravitational acceleration, \(g\) | 9.8 | m/s\(^2\) | | Negligible inlet velocity | Yes | - | --- ## Part (a): Find the outlet velocity of water ### Step 1: Convert pressure difference to Pascals \[ \Delta P = P_{in} - P_{exit} = (110 - 100) \text{ kPa} = 10\, \text{kPa} = 10,000\, \text{Pa} \] ### Step 2: Cross-sectional area of the faucet \[ A = \frac{\pi}{4} D^2 = \frac{\pi}{4} (.02)^2 \approx 3.14 \times 10^{-4}\, \text{m}^2 \] ### Step 3: Mass flow rate relation \[ \dot{m} = \rho Q \quad \Rightarrow \quad Q = \frac{\dot{m}}{\rho} = \frac{.25}{100} = 2.5 \times 10^{-4}\, \text{m}^3/\text{s} \] ### Step 4: Outlet velocity \(v_{out}\) Using the volumetric flow rate: \[ v_{out} = \frac{Q}{A} = \frac{2.5 \times 10^{-4}}{3.14 \times 10^{-4}} \approx .796\, \text{m/s} \] **Answer (a):** \[ \boxed{ v_{out} \approx .80\, \text{m/s} } \] --- ## Part (b): Find the shaft power supplied by the pump ### Step 1: Mechanical energy considerations The power supplied by the pump involves elevating the water and overcoming pressure differences: \[ P_{shaft} = \dot{m} \left( \frac{v_{out}^2}{2} + g h + \frac{\Delta P}{\rho} \right) \] ### Step 2: Calculate each term - **Kinetic energy per unit mass:** \[ \frac{v_{out}^2}{2} = \frac{(.80)^2}{2} = .32\, \text{m}^2/\text{s}^2 \] - **Potential energy per unit mass:** \[ g h = 9.8 \times 10 = 98\, \text{m}^2/\text{s}^2 \] - **Pressure energy per unit mass:** \[ \frac{\Delta P}{\rho} = \frac{10,000}{100} = 10\, \text{m}^2/\text{s}^2 \] ### Step 3: Total energy per unit mass \[ E_{total} = .32 + 98 + 10 = 108.32\, \text{m}^2/\text{s}^2 \] ### Step 4: Power calculation \[ P_{shaft} = \dot{m} \times E_{total} = .25 \times 108.32 \approx 27.08\, \text{W} \] --- # **Final Answers** ## (a) Outlet velocity of water \[ \boxed{ v_{out} \approx .80\, \text{m/s} } \] ## (b) Shaft power supplied by the pump \[ \boxed{ P_{shaft} \approx 27\, \text{W} } \]