# Step-by-Step Solution: Fire Hydrant Discharge Using Bernoulli's Equation ## **Given Data** - **Nozzle diameter (\( D \))**: 2½ inches = 2.5 inches = .2083 ft - **Pressure at nozzle (\( P \))**: 50 psi - **Atmospheric pressure at exit**: gauge (open to atmosphere) - **Assume**: The nozzle is horizontal and the velocity inside the hydrant is negligible compared to the nozzle exit. --- ## **Step 1: Convert Units** - **Pressure:** \( 1 \text{ psi} = 144 \text{ lb/ft}^2 \) \( 50 \text{ psi} = 50 \times 144 = 720 \text{ lb/ft}^2 \) - **Diameter to Area:** \( D = 2.5 \text{ in} = \frac{2.5}{12} = .2083 \text{ ft} \) \( A = \frac{\pi}{4} D^2 = \frac{\pi}{4} (.2083)^2 = .0341 \text{ ft}^2 \) --- ## **Step 2: Bernoulli’s Equation Application** Apply Bernoulli’s Equation between the inside of the hydrant (point 1) and the nozzle exit (point 2): \[ \frac{P_1}{\gamma} + \frac{V_1^2}{2g} = \frac{P_2}{\gamma} + \frac{V_2^2}{2g} \] Where: - \( P_1 = 50 \) psi gauge \( = 720 \text{ lb/ft}^2 \) - \( P_2 \) (atmospheric, gauge) \( = \) - \( V_1 \) (velocity in hydrant) \( \approx \) - \( \gamma \) (unit weight of water) \( = 62.4 \text{ lb/ft}^3 \) - \( g = 32.2 \text{ ft/s}^2 \) \[ \frac{720}{62.4} = \frac{V_2^2}{2 \times 32.2} \] \[ 115.38 = \frac{V_2^2}{64.4} \] \[ V_2^2 = 115.38 \times 64.4 = 7431 \] \[ V_2 = \sqrt{7431} = 86.2 \text{ ft/s} \] --- ## **Step 3: Calculate Discharge \( Q \)** \[ Q = A \times V_2 \] \[ Q = .0341 \text{ ft}^2 \times 86.2 \text{ ft/s} \] \[ Q = 2.94 \text{ ft}^3/\text{s} \] --- ## **Step 4: Final Answer** ### **Discharge from the fire hydrant:** \[ \boxed{2.94\ \text{ft}^3/\text{s}} \] --- ### **Summary Table** | Quantity | Value | |--------------------|-----------------------| | Nozzle Diameter | 2.5 in (.2083 ft) | | Pressure | 50 psi (720 lb/ft²) | | Area (\(A\)) | .0341 ft² | | Exit Velocity (\(V_2\)) | 86.2 ft/s | | Discharge (\(Q\)) | 2.94 ft³/s | --- **Assumptions:** - Neglect velocity inside hydrant. - Neglect elevation change. - Water temperature gives \(\gamma = 62.4 \text{ lb/ft}^3\).