Let's solve the problem step by step: --- # **Given Data** - **Pressure at gauge (P₁):** 80 psi - **Height to nozzle (z₂):** 0.9 m above gauge - **Diameter of nozzle (d₂):** 0.06 m - **Diameter of pipe (d₁):** 0.1 m - **Neglect viscous effects (ideal flow).** - **Atmospheric pressure at nozzle exit and plume.** - **Gravity (g):** 9.81 m/s² - **Density of water (ρ):** 1000 kg/m³ --- ## **Step 1: Convert Pressure to SI Units** \[ 1~\text{psi} = 6894.76~\text{Pa} \] \[ P_1 = 80~\text{psi} \times 6894.76~\frac{\text{Pa}}{\text{psi}} = 551,580.8~\text{Pa} \] --- ## **Step 2: Apply Bernoulli’s Equation** Between the gauge (1) and the nozzle exit (2): \[ \frac{P_1}{\rho g} + \frac{V_1^2}{2g} + z_1 = \frac{P_2}{\rho g} + \frac{V_2^2}{2g} + z_2 \] At nozzle exit (2): - \( P_2 = \) atmospheric, so \( P_2 = 0 \) gauge - Let \( z_1 = 0 \), \( z_2 = 0.9~m \) \[ \frac{P_1}{\rho g} + \frac{V_1^2}{2g} = \frac{V_2^2}{2g} + 0.9 \] --- ## **Step 3: Relate Velocities with Continuity** \[ Q = V_1 A_1 = V_2 A_2 \] \[ A_1 = \frac{\pi}{4} d_1^2 = \frac{\pi}{4}(0.1)^2 = 0.00785~m^2 \] \[ A_2 = \frac{\pi}{4} d_2^2 = \frac{\pi}{4}(0.06)^2 = 0.002827~m^2 \] \[ V_1 = \frac{A_2}{A_1} V_2 = \frac{0.002827}{0.00785} V_2 = 0.36 V_2 \] --- ## **Step 4: Substitute Values in Bernoulli’s Equation** \[ \frac{P_1}{\rho g} + \frac{(0.36 V_2)^2}{2g} = \frac{V_2^2}{2g} + 0.9 \] \[ \frac{551580.8}{1000 \times 9.81} + \frac{0.1296 V_2^2}{2 \times 9.81} = \frac{V_2^2}{2 \times 9.81} + 0.9 \] \[ 56.25 + \frac{0.1296 V_2^2}{19.62} = \frac{V_2^2}{19.62} + 0.9 \] \[ 56.25 + 0.00661 V_2^2 = 0.05097 V_2^2 + 0.9 \] \[ 56.25 - 0.9 = 0.05097 V_2^2 - 0.00661 V_2^2 \] \[ 55.35 = 0.04436 V_2^2 \] \[ V_2^2 = \frac{55.35}{0.04436} = 1248.3 \] \[ V_2 = \sqrt{1248.3} \approx 35.35~\text{m/s} \] --- ## **Step 5: Find Flow Rate** \[ Q = V_2 \cdot A_2 = 35.35~\text{m/s} \times 0.002827~\text{m}^2 = 0.0999~\text{m}^3/\text{s} \] \[ Q = 99.9~\text{L/s} \] --- ## **Step 6: Find Height \(h\) the Jet Reaches** At point (3), the water leaves the nozzle and rises up. Use energy conservation: \[ \frac{V_2^2}{2g} = h \] \[ h = \frac{(35.35)^2}{2 \times 9.81} = \frac{1248.3}{19.62} \approx 63.7~\text{m} \] --- # **Final Answers** **Flow rate, \(Q\):** \[ \boxed{99.9~\text{L/s}} \] **Maximum height, \(h\):** \[ \boxed{63.7~\text{m}} \] --- Let me know if you want a detailed explanation of any step!