Let's solve the problem step by step. --- ## **Given Data** - **Water temperature:** \(20^\circ C\) - **Hose diameter:** \(2\) in \(= .167\) ft (\(2 \times 1/12\)) - **Height from water surface to point 4:** \(3\) ft up, \(2\) ft over - **Height from water surface to point 2 (bottom):** \(8\) ft down - **Atmospheric pressure (\(P_{atm}\))** (assumed): \(101325\) Pa \(= 101.3\) kPa - **Density of water at 20°C (\(\rho\)):** \(998\) kg/m³ - **Vapor pressure of water at 20°C (\(P_{vap}\))**: \(\approx 2.34\) kPa Let’s use consistent SI units: - **Diameter (D):** \(2\) in = \(.0508\) m, - **Area (A):** \(\pi D^2/4 = \pi (.0508)^2/4 = .002026\) m² - **Height difference (from tank surface to outlet):** \(8 + 3 = 11\) ft = \(3.353\) m --- ## **A) Calculate the Flowrate** Assumptions: - The siphon is running steadily. - Velocity at the tank surface is negligible due to large area. - Neglecting friction and minor losses for simplicity (unless asked). Apply **Bernoulli’s Equation** between free surface (1) and outlet (5): \[ P_1 + \frac{1}{2}\rho v_1^2 + \rho g z_1 = P_5 + \frac{1}{2}\rho v_5^2 + \rho g z_5 \] Where: - At (1): \(P_1 = P_{atm}\), \(v_1 \approx \), \(z_1 = \) (let’s choose tank surface as datum) - At (5): \(P_5 = P_{atm}\), \(v_5 = v\), \(z_5 = -3.353\) m (down from the surface) Plug in: \[ P_{atm} + + = P_{atm} + \frac{1}{2}\rho v^2 - \rho g (3.353) \] \[ = \frac{1}{2}\rho v^2 - \rho g (3.353) \] \[ \frac{1}{2}\rho v^2 = \rho g (3.353) \] \[ v^2 = 2g(3.353) \] \[ v = \sqrt{2 \cdot 9.81 \cdot 3.353} = \sqrt{65.8} = 8.11 \text{ m/s} \] **Flowrate:** \[ Q = v \cdot A = 8.11 \times .002026 = \boxed{.0164 \text{ m}^3/\text{s}} \] or \[ Q = 16.4 \text{ L/s} \] --- ## **B) Gauge Pressure at Points (1)-(4)** Let’s calculate **gauge pressure** (\(P_{gauge} = P_{abs} - P_{atm}\)) at each point. ### **Point 1: Surface of Tank** - At surface, open to atmosphere: \[ P_{gauge,1} = \] ### **Point 2: Bottom of Tank (8 ft below surface)** \[ h = 8 \text{ ft} = 2.438 \text{ m} \] Hydrostatic pressure: \[ P_2 = P_{atm} + \rho g h \] \[ P_{gauge,2} = \rho g h = 998 \times 9.81 \times 2.438 = 23,900 \text{ Pa} = 23.9 \text{ kPa} \] ### **Point 3: Top of Siphon (3 ft above surface)** \[ h = -3 \text{ ft} = -.914 \text{ m} \] We need to consider Bernoulli from (1) to (3): - At (1): \(P_{atm}, v_1 \approx , z_1 = \) - At (3): \(P_3, v_3 = v, z_3 = .914\) m \[ P_{atm} + + = P_3 + \frac{1}{2}\rho v^2 + \rho g (.914) \] \[ P_3 = P_{atm} - \frac{1}{2}\rho v^2 - \rho g (.914) \] \[ P_{gauge,3} = P_3 - P_{atm} = - \frac{1}{2}\rho v^2 - \rho g (.914) \] Plug in values: - \(v = 8.11\) m/s - \(\frac{1}{2}\rho v^2 = .5 \times 998 \times (8.11)^2 = 32,860\) Pa - \(\rho g z = 998 \times 9.81 \times .914 = 8,970\) Pa \[ P_{gauge,3} = -32,860 - 8,970 = \boxed{-41,830 \text{ Pa}} = -41.8 \text{ kPa} \] ### **Point 4: Outlet** - At outlet, open to atmosphere: \[ P_{gauge,4} = \] --- ## **C) Cavitation Check** Cavitation occurs if absolute pressure drops below vapor pressure. \[ P_{abs,3} = P_{atm} + P_{gauge,3} = 101,300 - 41,830 = 59,470 \text{ Pa} = 59.5 \text{ kPa} \] Vapor pressure at 20°C = 2.34 kPa, which is much less than 59.5 kPa. **Conclusion:** **No cavitation will occur** since \(P_{abs,3} > P_{vap}\). --- ## **Final Answers (Summary Table)** | Point | Gauge Pressure (kPa) | |-------|----------------------| | 1 | | | 2 | +23.9 | | 3 | -41.8 | | 4 | | - **Flowrate:** \(\boxed{.0164\ \text{m}^3/\text{s}}\) or \(16.4\) L/s - **Cavitation:** **No cavitation occurs.** --- If you need the same calculations with more detailed steps or with friction losses included, just ask!