File Preview Awater main with diameter 8 in and length of 880 ft is laid 40 ft uphill to residential partments. When the discharge is 750 gpm as required for fire protection, the minimum pressure at the end fire hydrant must be 20 psi. Consider the hydrant to be an open gate valve. Calculate the pressure at the inlet of the main required to satisfy these conditions. = = 0.281 C D 2.6350.54 Q gal Q = Flow min/ = 750 gpm C = Pipe Characteristic Coefficient = 140 (Copper or PFTE Plastic) D = Pipe Diameter (in) = 8 in SHydraulic Gradient ( head loss = = length of pipe Rearrange to Solve for head loss hL = 0.002083 L 100 1.85 Q1.85 C D4.8655 L = Pipe Length (ft) Solve 1.85 hL = 0.002083 (880 ft) '100' 7501.85 140/ 84.8655 hL = 8.2568 ft Value Will Differ If a Different C value was chosen Calculate for Pressure Difference P1 V² P2 V2 Z1 + + = Z2 + 62.4 64.4 + +hL 62.4 64.4 Q1 Q2 Where V1 = and V2 = but Q1 = Q2 and A₁ = A₂ A1 v² A2 V₂ So V₁ = V2, so the and 64.4 64.4 cancel out. Additionally, Z1 = 0 due to its height being the reference datum. To keep consistent units, P2 = (20 psi) (144 psf) = 2880 psf Solving for the Pressure Difference Yields P1 = Z₂+ 62.4 P2 62.4 P1 ++h₁ → = 40 ft + 62.4 P₁ = 62.4 1 psi 144 psf 94.41 P₁(5891.224 psf) 2880 psf 62.4 ++8.2568 ft = 40.9 psi

File Preview Awater main with diameter 8 in and length of 880 ft is laid 40 ft uphill to residential partments. When the discharge is 750 gpm as required for fire protection, the minimum pressure at the end fire hydrant must be 20 psi. Consider the hydrant to be an open gate valve. Calculate the pressure at the inlet of the main required to satisfy these conditions. = = 0.281 C D 2.6350.54 Q gal Q = Flow min/ = 750 gpm C = Pipe Characteristic Coefficient = 140 (Copper or PFTE Plastic) D = Pipe Diameter (in) = 8 in SHydraulic Gradient ( head loss = = length of pipe Rearrange to Solve for head loss hL = 0.002083 L 100 1.85 Q1.85 C D4.8655 L = Pipe Length (ft) Solve 1.85 hL = 0.002083 (880 ft) '100' 7501.85 140/ 84.8655 hL = 8.2568 ft Value Will Differ If a Different C value was chosen Calculate for Pressure Difference P1 V² P2 V2 Z1 + + = Z2 + 62.4 64.4 + +hL 62.4 64.4 Q1 Q2 Where V1 = and V2 = but Q1 = Q2 and A₁ = A₂ A1 v² A2 V₂ So V₁ = V2, so the and 64.4 64.4 cancel out. Additionally, Z1 = 0 due to its height being the reference datum. To keep consistent units, P2 = (20 psi) (144 psf) = 2880 psf Solving for the Pressure Difference Yields P1 = Z₂+ 62.4 P2 62.4 P1 ++h₁ → = 40 ft + 62.4 P₁ = 62.4 1 psi 144 psf 94.41 P₁(5891.224 psf) 2880 psf 62.4 ++8.2568 ft = 40.9 psi