Maximum surplus = +140 m³ (after low demand) Maximum deficit = –220 m³ (after peak) Balancing storage = max deficit + max surplus = 220 + 140 =
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 water supply engineering solved problems pdf
Q_max daily = 1.8 × 15,000 = 27,000 m³/day (312.5 L/s) Maximum surplus = +140 m³ (after low demand)
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) 000 = 27
Average increase per decade = [(55-45)+(68-55)+(84-68)] / 3 = (10+13+16)/3 = 13,000 per decade P2030 = P2010 + 2 × 13,000 = 84,000 + 26,000 = 110,000
