Water Supply Engineering Solved Problems Pdf May 2026

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

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)

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

Make a mass diagram (cumulative supply – cumulative demand):

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

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