Water Supply Engineering Solved Problems Pdf < 2K · 480p >

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)

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

Static head = 95 – 50 = 45 m Velocity V = Q/A = 0.05 / (π×0.1²) = 0.05 / 0.0314 = 1.59 m/s Friction loss h_f = f × (L/D) × (V²/2g) = 0.02 × (1200/0.2) × (1.59²/19.62) = 0.02 × 6000 × (2.528/19.62) = 0.02 × 6000 × 0.1288 = 15.46 m Total head H = 45 + 15.46 = 60.46 m Using h_f = K × Q^1

Growth rates: r1 = (55-45)/45 = 0.2222 (22.22%) r2 = (68-55)/55 = 0.2364 r3 = (84-68)/68 = 0.2353 Average r = 0.2313 (23.13%) P2030 = 84,000 × (1+0.2313)² = 84,000 × 1.515 = 127,260 Pipe diameter = 200 mm, length = 1200 m, f = 0

Velocity V = Q/A = 0.25 / (π×0.2²) = 0.25 / 0.12566 = 1.99 m/s

Q_max daily = 1.8 × 15,000 = 27,000 m³/day (312.5 L/s)

| Time | Demand (m³/h) | Cumulative Demand | Supply (cumulative) | Difference (Supply-Demand) | |------|---------------|-------------------|---------------------|----------------------------| | 0-1 | 200 | 200 | 220 | +20 | | 1-2 | 200 | 400 | 440 | +40 | | … (peak hour 7-8) | 400 | … | … | -180 at hour 8 |