Water Supply | Engineering Solved Problems Pdf

| 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 |

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) water supply engineering solved problems pdf

Make a mass diagram (cumulative supply – cumulative demand): | Time | Demand (m³/h) | Cumulative Demand

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