Water Supply Engineering Solved Problems Pdf May 2026

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) water supply engineering solved problems pdf

Maximum surplus = +140 m³ (after low demand) Maximum deficit = –220 m³ (after peak) Balancing storage = max deficit + max surplus = 220 + 140 = Q_max daily = 1

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) Calculate head loss using: (a) Darcy-Weisbach equation (b)