Engineering Board Exam Reviewer · Fluid Mechanics · medium
Water is siphoned from a tank whose surface is 6 m above the floor. The siphon outlet is 1.5 m above the floor and discharges to the atmosphere. Neglecting losses, what is the outlet velocity?
- 10.85 m/s
- 5.42 m/s
- 12.13 m/s
- 9.40 m/s
Answer
D. 9.40 m/s
Solution
V = sqrt(2gh) with h = 6 - 1.5 = 4.5 m: V = sqrt(2 × 9.81 × 4.5) = 9.40 m/s.
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