Velocity V = Q/A = 0.25 / (π×0.2²) = 0.25 / 0.12566 = 1.99 m/s
Maximum surplus = +140 m³ (after low demand) Maximum deficit = –220 m³ (after peak) Balancing storage = max deficit + max surplus = 220 + 140 = water supply engineering solved problems pdf
Q_avg = 75,000 × 200 = 15,000,000 L/day = 15,000 m³/day (173.6 L/s) Velocity V = Q/A = 0
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 000 × 200 = 15
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