| 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 |
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)
After one iteration, flows are adjusted typically by 2–5 L/s until loop head loss sum ≈ zero. Problem 6.1 A town has a daily demand pattern with maximum hourly demand of 400 m³/h and average 200 m³/h. Supply from pumping is constant at 220 m³/h for 20 hours. Determine balancing storage required.
Velocity V = Q/A = 0.25 / (π×0.2²) = 0.25 / 0.12566 = 1.99 m/s
Q_avg = 75,000 × 200 = 15,000,000 L/day = 15,000 m³/day (173.6 L/s)
Maximum surplus = +140 m³ (after low demand) Maximum deficit = –220 m³ (after peak) Balancing storage = max deficit + max surplus = 220 + 140 =
