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
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. Q_max daily = 1
| 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 | Determine balancing storage required