Maximum surplus = +140 m³ (after low demand) Maximum deficit = –220 m³ (after peak) Balancing storage = max deficit + max surplus = 220 + 140 =
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 water supply engineering solved problems pdf
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% Maximum surplus = +140 m³ (after low demand)
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. Pipe diameter = 200 mm, length = 1200 m, f = 0
Q_max daily = 1.8 × 15,000 = 27,000 m³/day (312.5 L/s)
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)
Kuichling’s formula: Fire flow (L/min) = 3182 × √P (P in thousands) P = 75 Fire flow = 3182 × √75 = 3182 × 8.66 = 27,556 L/min Convert to m³/day = 27,556 × 1.44 = 39,680 m³/day (459 L/s) 3. Problem Set 3: Pipe Flow – Darcy-Weisbach & Hazen-Williams Problem 3.1 A 400 mm diameter steel pipe (ε = 0.045 mm) carries water at 20°C (ν = 1×10⁻⁶ m²/s) over a length of 800 m. Flow rate = 0.25 m³/s. Calculate head loss using: (a) Darcy-Weisbach equation (b) Hazen-Williams (C=120)