A tank filled with water to a depth of h has a small round hole at the bottom of area A. The density of water is p, and acceleration of gravity is g. A small plate is placed over the hole. The force required to keep back the water is:
F = Apgh
If the water is allowed to flow out of the hole, then its velocity, per Bernoulli, is √(2gh). Then the mass flow is Ap√(2gh), and the thrust force is
F' = 2Apgh
which is twice the force required to keep the plate in place. Experiment easily shows this to be wrong. Where's the mistake?
2007-05-11 14:46:45 · 3 answers · asked by Scythian1950 7 in Science & Mathematics ➔ Physics
theanswerman, F' = Ap(2gh) + pghA = 3Apgh? Looks worse than ever. The term Ap*Sqrt(2gh) doesn't have the dimensions of force.
2007-05-11 15:34:14 · update #1
JJ, the force required to "plug the hole" is based on water pressure at depth h, times hole area A. The thrust force of of the emerging stream of water (unplugged hole) is figured in the same way rocket scientists do it.
2007-05-12 12:53:55 · update #2
Alexander, your answer came closer to the real answer than anyone else. Feynman did a short one on this subject in his Lectures on Physics. What actually happens is that fluid streaming out of the orifice of area A actually shrinks to a bore of cross section (1/2)A, so that both F = F'.
2007-05-14 13:00:21 · update #3
Actually, experiment will show that thrust is indeed
F' = 2Aρgh.
The extra pressure comes from the area around the
orifice, where water flows with non-zero speed, and
therefore has pressure lower than ρgh.
Net decrease of the weight of the tank
is due to the lid removed plus pressure decrease:
Thrust = ΔWeight
Aρgh + integral of ΔP dS =
Aρgh + integral of ρv²(r)/2 2πr dr
2007-05-14 05:52:18 · answer #1 · answered by Alexander 6 · 2⤊ 0⤋
Revised answer
For water rockets...
F’ = m dot * V , where m dot is mass flow
= pA * V^2
V = â(2P/p) as per Bernoulli where P is internal over-pressure and p, water density
Thus...
F’ = 2PA
Fallacy lies in that over-pressure must exist for such to apply, a tank filled with water isn’t effectively a rocket.
2007-05-12 03:47:30 · answer #2 · answered by JJ 2 · 1⤊ 0⤋
the calculated force is correct as is the mass flow...the error is in the thrust force
thrust force (F') = mass flow (m) * velocity (V) + pressure differential (Pe - Po)* Area (A)
therefore
F' = Ap * sqrt(2gh) + pghA
2007-05-11 15:24:33 · answer #3 · answered by theanswerman 3 · 0⤊ 0⤋