Water Supply Engineering Solved Problems Pdf -

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

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

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. water supply engineering solved problems pdf

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): Reynolds Re = V×D/ν = 1

Maximum surplus = +140 m³ (after low demand) Maximum deficit = –220 m³ (after peak) Balancing storage = max deficit + max surplus = 220 + 140 =

Q_peak hourly = 2.7 × 15,000 = 40,500 m³/day (468.75 L/s) Increases: 10, 13, 16 Increments of increases: 3,

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