Water Supply Engineering Solved Problems Pdf | Direct

Increases: 10, 13, 16 Increments of increases: 3, 3 Average increment = 3 Average increase = 13 P2030 = 84,000 + (2×13) + [2×3×(2+1)/2] = 84,000 + 26 + 9 = 110,035 2. Problem Set 2: Water Demand & Fire Flow Problem 2.1 A town of 75,000 people has a per capita water supply of 200 L/day. Calculate: (a) Average daily demand (m³/day) (b) Maximum daily demand (assume factor = 1.8) (c) Peak hourly demand (assume factor = 2.7) (d) Fire demand using Kuichling’s formula

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

Average increase per decade = [(55-45)+(68-55)+(84-68)] / 3 = (10+13+16)/3 = 13,000 per decade P2030 = P2010 + 2 × 13,000 = 84,000 + 26,000 = 110,000 Increases: 10, 13, 16 Increments of increases: 3,

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 Problem Set 4: Pump Sizing Problem 4