Water Supply Engineering Solved Problems Pdf [extra Quality] -

h_f = 10.67 × L × Q^1.852 / (C^1.852 × D^4.87) D = 0.4 m, Q = 0.25 m³/s h_f = 10.67 × 800 × (0.25^1.852) / (120^1.852 × 0.4^4.87) 0.25^1.852 = 0.065, 120^1.852 = 7061, 0.4^4.87 = 0.4^4 × 0.4^0.87 = 0.0256 × 0.459 = 0.01175 h_f = (10.67×800×0.065) / (7061×0.01175) = 555 / 83.0 = 6.69 m 4. Problem Set 4: Pump Sizing Problem 4.1 A pump delivers water from a lower reservoir (EL 50.0 m) to an elevated tank (EL 95.0 m). Discharge = 50 L/s. Pipe diameter = 200 mm, length = 1200 m, f = 0.02. Calculate: (a) Total dynamic head (b) Hydraulic power required (c) Brake horsepower if pump efficiency = 75%

Q_max daily = 1.8 × 15,000 = 27,000 m³/day (312.5 L/s) water supply engineering solved problems pdf

Reynolds Re = V×D/ν = 1.99×0.4 / 1e-6 = 796,000 (turbulent) Relative roughness = ε/D = 0.045/400 = 0.0001125 From Moody chart: f ≈ 0.014 Head loss h_f = f × (L/D) × (V²/(2g)) = 0.014 × (800/0.4) × (1.99²/(2×9.81)) = 0.014 × 2000 × (3.96/19.62) = 0.014 × 2000 × 0.202 = 5.66 m h_f = 10

Q_avg = 75,000 × 200 = 15,000,000 L/day = 15,000 m³/day (173.6 L/s) Pipe diameter = 200 mm, length = 1200 m, f = 0

Maximum surplus = +140 m³ (after low demand) Maximum deficit = –220 m³ (after peak) Balancing storage = max deficit + max surplus = 220 + 140 =