Let's break down the problem step by step: --- ## **Given Data** - **Pressure at the nozzle:** \( P = 80 \) psi - **Water flows without viscous effects** (Assume ideal fluid) - **Gravity:** \( g = 9.81 \) m/s² - **Water density:** \( \rho = 1000 \) kg/m³ (standard for water) - **Atmospheric pressure at exit** - **Need to find:** - **Flowrate** (L/s) - **Maximum height (\( h \))** water can reach --- ## **Step 1: Convert Pressure Units** Convert psi to Pascals (SI units): \[ 1\,\text{psi} = 6894.76\,\text{Pa} \] \[ P = 80\,\text{psi} \times 6894.76\,\text{Pa/psi} = 551,580.8\,\text{Pa} \] --- ## **Step 2: Use Bernoulli’s Equation** Assume the nozzle exit is open to the atmosphere (so pressure at exit = atmospheric). Bernoulli’s equation between gauge and nozzle exit: \[ P_1 + \frac{1}{2}\rho v_1^2 + \rho g z_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g z_2 \] Assume: - Point 1: at gauge (pipe, upstream, large area, velocity \( v_1 \approx 0 \)) - Point 2: at nozzle exit (open to atmosphere, at same height, \( z_1 = z_2 \)) So, \[ P_1 = P_2 + \frac{1}{2}\rho v_2^2 \] Since \( P_2 \) is atmospheric, and \( P_1 \) is the gauge pressure (above atmospheric): \[ P_{\text{gauge}} = \frac{1}{2}\rho v_2^2 \] Solve for \( v_2 \): \[ v_2 = \sqrt{\frac{2 P_{\text{gauge}}}{\rho}} \] \[ v_2 = \sqrt{\frac{2 \times 551,580.8}{1000}} \] \[ v_2 = \sqrt{1103.1616} \] \[ v_2 \approx 33.23\,\text{m/s} \] --- ## **Step 3: Flowrate Calculation** \[ Q = A v_2 \] But the nozzle area (\( A \)) is not given. Typically, you'd need to know the diameter. **If not specified, let's assume a standard small fountain nozzle, say diameter \( d = 1 \) cm = 0.01 m:** \[ A = \frac{\pi d^2}{4} = \frac{\pi (0.01)^2}{4} = 7.854 \times 10^{-5}\,\text{m}^2 \] So, \[ Q = 7.854 \times 10^{-5}\,\text{m}^2 \times 33.23\,\text{m/s} \approx 0.00261\,\text{m}^3/\text{s} \] Convert to L/s (\(1\,\text{m}^3 = 1000\) L): \[ Q \approx 2.61\,\text{L/s} \] --- ## **Step 4: Maximum Height (\( h \))** Use kinematics (projectile motion, vertical): \[ v^2 = 2gh \implies h = \frac{v^2}{2g} \] \[ h = \frac{(33.23)^2}{2 \times 9.81} = \frac{1104.2}{19.62} \approx 56.3\,\text{m} \] --- ## **Final Answers** ### **1. Flowrate** \[ \boxed{2.61~\text{L/s}} \] (for a 1 cm diameter nozzle; use the actual nozzle size if known) ### **2. Maximum Height** \[ \boxed{56.3~\text{m}} \] --- ### **Summary Table** | Quantity | Formula | Value | |------------------|-------------------------------------------------|--------------------| | Exit velocity | \( v = \sqrt{2P/\rho} \) | 33.2 m/s | | Flowrate | \( Q = A v \) | 2.61 L/s (d=1 cm) | | Max height | \( h = v^2/2g \) | 56.3 m | > **Note:** If the nozzle diameter is different, recalculate \( Q \) using \( d \) provided. --- Let me know if you have the nozzle diameter or need more details!