Alternatively, use the Rippl method: storage = 360 m³ (rounded to 400 m³ design). These solved problems illustrate the core computational skills required for water supply engineering: population forecasting, demand estimation, friction loss calculation, pump selection, network balancing, and storage sizing. Mastery of these methods ensures reliable and cost-effective water distribution system design.
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% water supply engineering solved problems pdf
Using h_f = K × Q^1.852 with K = 10.67×L / (C^1.852×D^4.87) D=0.25m → D^4.87 = 0.25^4.87 = 0.25^4 × 0.25^0.87 = 0.003906 × 0.305 = 0.001191 C^1.852 = 100^1.852 = 5120 (approx) Alternatively, use the Rippl method: storage = 360
Velocity V = Q/A = 0.25 / (π×0.2²) = 0.25 / 0.12566 = 1.99 m/s h_f = 10