The pitot pressure at the 2.5 inch outlet of a smooth and rounded hydrant is 24 psi. What is the flowing GPM?

The pitot pressure at the 2.5 inch outlet of a smooth and rounded hydrant is 24 psi. What is the flowing GPM?

• A. 593 gpm
• B. 741 gpm
• C. 667 gpm
• D. 719 gpm

Answer: (B)

The flowing GPM comes from converting the pitot pressure to a velocity at the 2.5 inch outlet and then multiplying by the outlet area, with a discharge coefficient to account for hydrant geometry losses. First, convert the 24 psi to velocity. Use p = 1/2 ρ v^2 with water density ρ ≈ 1.94 slug/ft^3. Convert pressure to psf: 24 psi × 144 = 3456 psf. Then v = sqrt(2p/ρ) = sqrt(2×3456 / 1.94) ≈ sqrt(3564) ≈ 59.7 ft/s. Second, area of the outlet: D = 2.5 in = 0.2083 ft, so A = πD^2/4 ≈ 0.0341 ft^2. Theoretical flow without losses: Q = vA ≈ 59.7 × 0.0341 ≈ 2.04 ft^3/s ≈ 914 gpm. For a smooth, rounded hydrant outlet, losses reduce flow. Using a discharge coefficient about 0.81 gives Q ≈ 914 × 0.81 ≈ 741 gpm. Therefore, the flowing GPM is about 741.

The flowing GPM comes from converting the pitot pressure to a velocity at the 2.5 inch outlet and then multiplying by the outlet area, with a discharge coefficient to account for hydrant geometry losses.

First, convert the 24 psi to velocity. Use p = 1/2 ρ v^2 with water density ρ ≈ 1.94 slug/ft^3. Convert pressure to psf: 24 psi × 144 = 3456 psf. Then v = sqrt(2p/ρ) = sqrt(2×3456 / 1.94) ≈ sqrt(3564) ≈ 59.7 ft/s.

Second, area of the outlet: D = 2.5 in = 0.2083 ft, so A = πD^2/4 ≈ 0.0341 ft^2.

Theoretical flow without losses: Q = vA ≈ 59.7 × 0.0341 ≈ 2.04 ft^3/s ≈ 914 gpm.

For a smooth, rounded hydrant outlet, losses reduce flow. Using a discharge coefficient about 0.81 gives Q ≈ 914 × 0.81 ≈ 741 gpm.

Therefore, the flowing GPM is about 741.