# Reaction force for force exiting a tank through a clean hole

1. Apr 22, 2013

### DylanW

1. The problem statement, all variables and given/known data
Determine the reaction force from water flowing through a clean 0.02cm (Diamater assumed) hole in tank depth 3m

2. Relevant equations
P = rho.g.h
A = Pi.R^2
P=F/A

3. The attempt at a solution
P = rho.g.h = 29400 Pa
A = Pi.R^2 = 0.00031415 m^2
P = F/A -> F = PA = 9.236 Newtons

Looks solid to me, but on the internet I found an equation for the reaction force in this case as: F(Reaction) = 10^3 (Rho water) * 2*g*h.A and the 2 is making me unsure. Am I overlooking something or is the formula I found just a dud?

2. Apr 22, 2013

### Staff: Mentor

You calculated the force on a wall, or similar obstacle. If water can flow out, you have to consider its motion. At which velocity does it leave the tank? How is that related to a force (momentum per time)?

3. Apr 22, 2013

### DylanW

Ok so I got it i think. V(jet)=sqrt(2gh) = sqrt(58.8)
Q(mass) = AvP = 0.01^2.Pi.sqrt(58.8)*1000 = 2.409
F = V(jet)Q(mass) = 18.47N which strangely agrees with above formula and is less work than original (wrong) answer

4. Apr 22, 2013

### Staff: Mentor

The agreement is not strange ;), you calculated the same as the formula you found somewhere.

5. Apr 23, 2013

### Staff: Mentor

This analysis seems incorrect to me. The pressure at the outlet of the hole is atmospheric. The fact that you got v = sqrt(2gh) is based on the assumption that the outlet pressure is atmospheric, and equal to the pressure at the top of the tank. So, if the diameter of the jet at the tank outlet is equal to the hole diameter (and remains so after exiting), the reaction force should be zero.

This is a vena contracta problem. Even if the stream exits the tank at the hole diameter, the jet diameter decreases after the exit. Under these circumstances, the pressure at the exit hole is not atmospheric, and the fluid pressure at the exit is exerting a force on the fluid ahead of it to accelerate it up to a higher velocity. Typically, the ratio of the vena contracta area to the hole diameter is about 0.6 to 0.7 (0.64 is often used as a rule of thumb). Because, when a vena contracta is present, the pressure at the exit hole is not atmospheric, there will be a reaction force at the exit.

6. Apr 23, 2013

### Staff: Mentor

Wait, what?

The pressure outside? I agree.
Right. And the pressure inside the tank, at the height of the hole, is larger than the pressure outside.

If water is exiting at one side, momentum conservation forces an acceleration of the tank towards the other side.
If the stream falls down in gravity, and accelerates, before it forms drops due to surface tension. How is that relevant here?
Which effect should maintain a pressure difference (on the outside I think)?

7. Apr 23, 2013

### Staff: Mentor

Inside the tank, the fluid pressure decreases monotonically in the region closely approaching the exit oriface. The streamlines of the fluid flow get closer together, and the fluid velocity increases. This increase in fluid momentum approaching the exit oriface is brought about by a decrease in the fluid pressure. By the time each fluid parcel reaches the exit oriface, it is at full jet velocity, and its pressure has dropped to atmospheric.

To get a better understanding of what happens in the region within the tank near the exit oriface, and in the region of the jet immediately outside the exit oriface, google vena contracta. The vena contracta effect occurs irrespective of whether gravity is present.

8. Apr 24, 2013

### Staff: Mentor

I agree.

I don't see how this changes the resulting velocity (or pressure). Okay, a part of the energy is needed for a radial flow at the hole, so the force could be slightly lower.

9. Apr 24, 2013

### Staff: Mentor

I can see what the issue is. There are two possible interpretations to the question "what is the reaction force."

1. What is the reaction force that the water exerts on the tank?
2. What is the reaction force that the water in the jet exerts on the water directly behind it at the exit cross section of the tank?

mfb interpreted the question as item 1. and I interpreted the question as item 2.

10. Apr 24, 2013

### haruspex

I read it the same way as mfb.
One gripe about the problem statement: in practice, the effective aperture is less than the apparent one. For a simple circular hole the factor is about 2:1. To restore the effective aperture to something approaching the apparent value you need a Borda mouthpiece to shape the flow.