The force of water

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1. Dec 23, 2015

nibbel11

i want to calculate the force of a water jet
i know the velocity,13,6m/s
mass or volume/s=52,6kg or dm3/s
the pipe has an area of 2,3dm
and the gained area by flying out the pipe is 3,4dm

i want to know the force it creates by hitting for example a way, how do i calculate that

2. Dec 23, 2015

Simon Bridge

When the water hits something, it changes momentum - the force is the rate that the momentum changes.

3. Dec 23, 2015

nibbel11

how do i calculate that then, and isnt their just a way to use newtons second law of motion eventhough the mass is a little inconsistend

4. Dec 24, 2015

Simon Bridge

If you have mass m per second hitting the wall at speed v, and it is completely stopped by the wall (usually not but say), then the change in momentum per second is mv ... if the water rebounds off the wall to go at the exact same speed but opposite direction, then it is 2mv because the velocity change is 2v ... got it?

There are more complicated things you have to do to take into account if the water drops some height as well (it's going faster) and and some energy is always lost in splashes and hitting the wall not quite head-on and so on. It's usually easier just to measure it - save this sort of calculation for the first approximation.

5. Dec 26, 2015

theodoros.mihos

Bernoulli equation say that the total pressure (total energy density) on the flow direction must be constant.

6. Dec 26, 2015

sophiecentaur

Yes. You don't need to involve all the data you've been given - just the speed and the mass per second - to get a reasonable estimate. As Simon says, the final state of the water can account for a 2:1 difference in your result.

7. Dec 26, 2015

nibbel11

it is as easy as 52,6×13,6=715,36N?

8. Dec 26, 2015

theodoros.mihos

Why problem give you diameters of pipes?

9. Dec 26, 2015

nibbel11

i dont understand
i gave the area of the pipe because i didnt know what i needed to do so i gave averything i know

10. Dec 26, 2015

theodoros.mihos

Areas need because flux find no resistance on the open hole. You need the area difference for correct calculation.

11. Dec 26, 2015

Simon Bridge

Bernoulli equation is not helpful for this problem.

12. Dec 26, 2015

just dani ok

yes

13. Dec 26, 2015

theodoros.mihos

This is the force than the water flow act to a wall.

14. Dec 26, 2015

Staff: Mentor

Yes, it is. Force is stagnation (velocity) pressure times area. Both that and Newton's 2nd law work for this problem (similar to how you can use either to describe the lift of a wing).

15. Dec 26, 2015

sophiecentaur

Newton's Second Law is there all the time. Whatever you do with channeling the water in certain ways, can you change more momentum than is available?
How would necessarily working out the details of the flow of water over the surface of the wall be better (or easier?) than using momentum considerations?
You are clearly one of the big endians who think that the measurements of pressure around the plane in a wind tunnel explain why a plane stays in the air and don't believe that a mass of air must be pushed downwards to maintain the lift force when it's actually flying. Myself, I am a little endian for whom momentum is a much easier thing to deal with.

16. Dec 26, 2015

theodoros.mihos

Bernoulli equation has terms of pressure and say that the total pressure on flow direction must be the same on a flow line.
This pressure is energy density and have a direction, that mean momentum. 2nd Newton's law is here.
We cannot calculate the total force ignoring the difference of fluid kinetic energy before and after the nozzle.

17. Dec 26, 2015

sophiecentaur

I understand that the details of flow can help to improve the estimate for a case between the water exiting through the wall at almost 0 velocity and being channeled back the way it came (twice that momentum change). But it's still the momentum change that counts - surely?

18. Dec 26, 2015

theodoros.mihos

Surely. Bernoulli's equation is a deep result of 2nd Newton's Law. We can use it if we understand what really say.
In this case, the force must depends by the area ratio while fluid flow out by the same $dV/dt$.
Through mine eyes Bernoulli's way is much more easier. I calculate the force by the pressure, not the opposite.

19. Dec 26, 2015

sophiecentaur

But how can you calculate that unless the whole setup is very tightly defined? After all, the detailed design of a boat, car or plane can make a huge difference to the fluid forces on it. The same must apply to the force on the wall.

20. Dec 26, 2015

Staff: Mentor

Sure.
No.
I don't understand what you are asking: you have the water velocity and you have the pipe diameter. That's all you need. Perhaps you are thinking something else is needed, that isn't. Please specify what else you think you need.
Of course: I had that lab in my Wind Tunnel class. It was fun.
Since I explicitly said "you can use either", I can't imagine what could possibly make you have that misunderstanding.
Ok, fine. As I said, in this situation you can use either. All I was doing was correcting the erroneous statement that you can't use Bernoulli's principle.
The setup definition is what it is -- whatever limitations it has will affect any method you use to calculate the force.

You may recall that I'm an HVAC engineer: the reason I prefer Bernoulli's principle here his that I use it every day in situations similar to this one (but for air) and have equations with rolled-up constants at my fingertips for describing it.

Last edited: Dec 27, 2015
21. Dec 27, 2015

sophiecentaur

I find this statement worrying because the details of the wall must count; absorption or reflection make a 2:1 difference. I never did fluid dynamics (except the obvious stuff) so I have never tried to solve a problem like this but you have to involve how the water interacts with the wall or you can get only one answer. (No?)
I recall the statement was that it was "not helpful" - which is not the same thing. I agree with that statement because, as I have pointed out, it seems that you would need more information to use the fluid dynamics.
Can you explain how you can get away with just "Pressure times area"?

22. Dec 27, 2015

nibbel11

thank ya all

23. Dec 27, 2015

sophiecentaur

If I were hitting the wall with little balls, fired from a gun, I would need to know whether I would be expecting elastic or inelastic collisions (or some coeficcient of restitution). Personally, I have no idea about how that relates to a jet of water against an unspecified wall. I also would have needed to find out how to calculate the pressure from the original data. A guy with knowledge about fluid dynamics would find that all to be a piece of cake. I think that explains the difference of approach.
@nibbel11: Sorry if our squabbling has interfered with your enjoyment and edification.

24. Dec 28, 2015

Staff: Mentor

I thought you were still arguing about whether you could use Bernoulli's principle here at all? Yes, those details matter, regardless of which method you use.

In particular, if you have a look at the equations for the two methods you see they are almost identical (ρ*A*V2), except that for Bernoulli's equation having the factor of 1/2 in front of it. That's because of different assumptions about the motion (or lack thereof) of the apparatus. That and the bounciness of the water (water usually doesn't bounce) and acceleration of the source and target can all be taken into account if one wants to.
I'm not sure what you are asking. The OP asked for force. Force is pressure times area.

The way the problem is given to us, it probably makes more sense to use momentum, but in trying to use Bernoulli's, I found a problem: the mass flow, velocity and area should all be proportional to each other, but they aren't. So there's an error in the information we were given.

25. Dec 28, 2015

sophiecentaur

Cheers. I now know a bit more than when I started. Good value.