Let's break down the problem step by step. Given: - **Circular jet** (diameter not given, but radius can be figured from "5 cm" notation for the jet) - **Section (1) pressure:** \( P_1 = 175\,\text{kPa} \) - **Temperature:** \( 20^\circ\text{C} \) (properties of water) - **Neglect losses** - **Jet diameter:** \( d = 5\,\text{cm} = .05\,\text{m} \) - **Section (1) diameter:** \( D = 25\,\text{cm} = .25\,\text{m} \) - **Atmospheric pressure:** \( P_\text{atm} = 101.3\,\text{kPa} \) - **Density of water at 20°C:** \( \rho = 998\,\text{kg/m}^3 \) --- ## a) **Mass flow rate of water (\(\dot{m}\)) in kg/s** ### **Step 1: Find velocity at the jet exit using Bernoulli's Equation** Apply Bernoulli between section (1) and the jet exit (open to atmosphere): \[ P_1 + \frac{1}{2}\rho V_1^2 = P_\text{atm} + \frac{1}{2}\rho V_2^2 \] Assume velocity at (1), \(V_1\), is much smaller than at the nozzle (due to area difference), so \(V_1 \approx \): \[ P_1 = P_\text{atm} + \frac{1}{2}\rho V_2^2 \] \[ P_1 - P_\text{atm} = \frac{1}{2}\rho V_2^2 \] \[ V_2 = \sqrt{\frac{2(P_1 - P_\text{atm})}{\rho}} \] Plug in the numbers: \[ P_1 - P_\text{atm} = 175,000\,\text{Pa} - 101,300\,\text{Pa} = 73,700\,\text{Pa} \] \[ V_2 = \sqrt{\frac{2 \times 73,700}{998}} = \sqrt{\frac{147,400}{998}} \approx \sqrt{147.7} \approx 12.15\,\text{m/s} \] --- ### **Step 2: Calculate mass flow rate** The area of the jet (\(A_2\)): \[ A_2 = \frac{\pi d^2}{4} = \frac{\pi (.05)^2}{4} \approx .00196\,\text{m}^2 \] \[ \dot{m} = \rho A_2 V_2 = 998 \times .00196 \times 12.15 \approx 23.8\,\text{kg/s} \] --- ## b) **Volume flow rate in L/h** \[ Q = A_2 V_2 = .00196 \times 12.15 \approx .0238\,\text{m}^3/\text{s} \] Convert to L/h: \[ .0238\,\text{m}^3/\text{s} \times 100\,\text{L}/\text{m}^3 \times 360\,\text{s}/\text{h} = 85,680\,\text{L}/\text{h} \] --- ## c) **Height of the fluid in the stagnation tube, \(H\)** This is a classic Pitot tube measurement: The height \(H\) is equivalent to the **velocity head**: \[ \Delta P = \rho g H \] \[ H = \frac{P_1 - P_\text{atm}}{\rho g} \] \[ H = \frac{73,700}{998 \times 9.81} \approx \frac{73,700}{9,788} \approx 7.53\,\text{m} \] --- ## **Final Answers** **a) Mass flow rate:** \[ \boxed{23.8\,\text{kg/s}} \] **b) Volume flow rate:** \[ \boxed{85,680\,\text{L/h}} \] **c) Height in stagnation tube, \(H\):** \[ \boxed{7.53\,\text{m}} \] Let me know if you need any clarifications or step breakdowns!