Is there an equal and opposite force from gas movement due to pressure gradient force?

[Sensei]

Conservation of momentum is even visible at quantum level. During decay of unstable particle, newly created particles are ejected in opposite directions with velocities which depend on difference between rest-masses of particles. In Cloud Chamber it is visible by naked eye.

[Gerrard]

On 1/22/2020 at 9:45 PM, MigL said:

You have never clarified the original question.
You have allowed us to interpret the OP as we see it, and once we offer a solution to OUR interpretation, you tell us we are wrong ?

How about reposting the original question, as YOU interpret it ( but in a clear manner, not like previously ), and then we can discuss it properly.

The question was pretty clear. Let’s start with the first hypothetical scenario. I have a tote full of water. I use a pump to remove the water from the tote and add it to a lake. The flex hose is connected from a valve on the side of the tote to the pump. The flex hose has a lot of slack. When the pump removes the water, is there a force on the tote in the opposite direction of water flow? Would there be tension on the flex hose? Let’s say the draw side of the pump is 0 psig. 
 

Bullfrog understood the original question and stated that there wouldn’t be force on the tote. Do you agree?

On 1/22/2020 at 11:38 PM, Sensei said:

Conservation of momentum is even visible at quantum level. During decay of unstable particle, newly created particles are ejected in opposite directions with velocities which depend on difference between rest-masses of particles. In Cloud Chamber it is visible by naked eye.

I didn’t say that conservation of momentum doesn’t exist. I stated it is incorrect to apply it to rockets in the manner stated by others.

 

 

Edited January 24, 2020 by Gerrard

[Bufofrog]

17 minutes ago, Gerrard said:

The question was pretty clear. Let’s start with the first hypothetical scenario. I have a tote full of water. I use a pump to remove the water from the tote and add it to a lake. The flex hose is connected from a valve on the side of the tote to the pump. The flex hose has a lot of slack. When the pump removes the water, is there a force on the tote in the opposite direction of water flow? Would there be tension on the flex hose? Let’s say the draw side of the pump is 0 psig. 
 

Bullfrog understood the original question and stated that there wouldn’t be force on the tote. Do you agree?

This Bullfrog fellow may have misunderstood you.  In the scenario above there will be a force on the tote.  The force would be due to the mass exiting the tote based on Newton's second law.

Edited January 24, 2020 by Bufofrog

[MigL]

I' would agree.
( and disagree with Bufofrog )

In this scenario there is no force on the tote, as it is connected by a slack flex-hose, but the pump, which is accelerating the contents of the tote by exerting a force on the water in a specific direction, will feel an equal and opposite force in the other direction.

Assuming the tote is open to atmosphere, only gravity and atm pressure are acting on the contents.
Gravity can be disregarded, while atm pressure acts equally from all sides.

If the pump was integral to the tote ( a solid connection ), the pump-tote system would experience a force in the opposite direction as the flow, but you specified a slack flex-hose.

[Bufofrog]

34 minutes ago, MigL said:

I' would agree.
( and disagree with Bufofrog )

Hmm, I certainly could be wrong on this!  It seems that there is a differential pressure across the opening in the tote that accelerates the water through the nozzle creating a resulting force.  What am I missing here?

[MigL]

The tote, on its own, would feel a trivial force due to redirecting the flow due to gravity.
It is the ( isolated ) pump which is setting up the pumping condition ( suction on inlet/pressure of outlet ).
It does this by exerting a force on the water ( whether by centrifugal or positive displacement means ), imparting an acceleration to it, and a change in momentum.
Conservation of momentum then requires the water to do the same to the pump in the opposite direction.

I am interested in any arguments opposing this view, though; so if you have one, I'd be glad to discuss it.

[Bufofrog]

58 minutes ago, MigL said:

I am interested in any arguments opposing this view, though; so if you have one, I'd be glad to discuss it.

My reasoning is as follows.  The pump pulls a suction on the line that goes to the tote.  Ignoring head loss in the tube and cavitation in the pump, let"s say the suction side of the pump is at 0.7 psi absolute and the tote is at 14.7 psi absolute.  If we further say the tubing has a much larger ID than the opening in the tote then the pressure drop would be 14 psi across the opening.  The water in the tote would accelerate through the opening and due to the conservation of momentum that should result in a force that is opposite to the direction of the flow through the hole.  This seems exactly the same as if the tote was pressurized to 28.7 psig and a hole was put into the tote, which I think we can agree would put a force on the tote opposite to the direction of the water leaving the hole.

What am I missing, here?

[MigL]

My viewpoint is that it is the pump generating the vacuum on the inlet, By accelerating the water.
So it is the pump that will feel the countereffect.

I have previously used a diaphragm pump ( positive displacement ), and if not properly fixed, you can see the pump 'dance' as it is pumping.
You could alternatively, feed the pomp from the top opening of the tote, with a dip-leg.
Wold you then expect the tote to be accelerated down, and get weigh more, as you are pumping water out of it ?

Edited January 24, 2020 by MigL

[Bufofrog]

39 minutes ago, MigL said:

Would you then expect the tote to be accelerated down, and get weigh more, as you are pumping water out of it ?

Yes, obviously as you pump out water the tote weight will decrease, but the instant you turn on the pump the tote weight would increase.  

[J.C.MacSwell]

1 hour ago, Bufofrog said:

Yes, obviously as you pump out water the tote weight will decrease, but the instant you turn on the pump the tote weight would increase.  

No. It would decrease.

There are a number of effects going on.

It would decrease as the pump created the pressure X area reduction on the tote/water system.

The reaction due to the water being accelerated upward would decreased that reduction, so have the opposite effect, but not enough to offset it. (the response of the water would act to decrease the pressure drop over the area/maintain higher pressure in absolute terms even though below atmospheric)

...and of course any removal of water would reduce the weight over time.

 

 

[Gerrard]

7 hours ago, MigL said:

My viewpoint is that it is the pump generating the vacuum on the inlet, By accelerating the water.
So it is the pump that will feel the countereffect.

I have previously used a diaphragm pump ( positive displacement ), and if not properly fixed, you can see the pump 'dance' as it is pumping.
You could alternatively, feed the pomp from the top opening of the tote, with a dip-leg.
Wold you then expect the tote to be accelerated down, and get weigh more, as you are pumping water out of it ?

Wouldn’t the imbalanced for be felt in the tote. The pump causes the fluid in the tote to come out the side. The fluid is pushing on all side of the tote equally except the side with the hole. Thus there should be an imbalanced force?

[MigL]

Minimal, and due solely to head ( height of liquid ) pressure.

I'll put it another way...

Say you are an astronaut floating in space while holding a bucket full of rocks.
You take rocks out of the bucket, and toss them behind you.
Does the bucket experience forward acceleration, out of your hand, or do you, the astronaut ?
( in this case the bucket would be the tote, and the astronaut would be the pump )

[Gerrard]

10 hours ago, MigL said:

Minimal, and due solely to head ( height of liquid ) pressure.

I'll put it another way...

Say you are an astronaut floating in space while holding a bucket full of rocks.
You take rocks out of the bucket, and toss them behind you.
Does the bucket experience forward acceleration, out of your hand, or do you, the astronaut ?
( in this case the bucket would be the tote, and the astronaut would be the pump )

The pump merely provides negative pressure or suction. Why wouldn’t there be an imbalanced force in tote? Like in a balloon, there is negative pressure in the atmosphere compared to the balloon. Let’s say you had a vacuum at the end of the balloon, the balloon would still move forward. The vacuum merely creates additional negative pressure? 
 

In other words, there is pressure pushing equally on all sides of the container except for the side with the hole. How is the scenario different if there was a pump attached? 
 

An astronaut throwing rocks is not really a good example because the pump only creates a pressure difference and pressure gradient force is what actually moves the fluid right?

[MigL]

The relationship is simple...
F=ma

The force is exerted by that which causes the acceleration.
If the water is accelerated by the pump, it is the pump which feels the 'equal and opposite reaction'.

[Gerrard]

8 minutes ago, MigL said:

The relationship is simple...
F=ma

The force is exerted by that which causes the acceleration.
If the water is accelerated by the pump, it is the pump which feels the 'equal and opposite reaction'.

Force is provided by pressure gradient force and is calculated as below: 

the pump merely creates a pressure difference and then pressure gradient force takes over.

pressure differential is a potential energy like gravity. When you drop a ball from a height, gravity provides the force and there is no opposite force on you. The pump is like a person climbing stairs to gain height giving rise to gravitational potential energy. When you drop the the ball, there is no opposite force on the person dropping the ball. the pump creates a pressure differential giving rise to pressure gradient potential energy. 
 

having said that, there may force on the pump due to imbalanced force just like on the tote. 

Perhaps Bufofrog is correct?

[MigL]

You realise that a Jet engine is basically an air pump, don't you ?

A large turbofan engine has a low pressure in front of it and a higher pressure behind the large fan ( from which it gets most of its thrust ).
This air that it accelerates backwards provides an equal reaction foreward on the engine ( which is attached to the plane )

Now I could say, get a pump, and see how it behaves.
But I'm sure you've seen planes flying.

Edited January 25, 2020 by MigL

[J.C.MacSwell]

Wrong thread...

 

 

Edited January 25, 2020 by J.C.MacSwell

[Gerrard]

39 minutes ago, MigL said:

You realise that a Jet engine is basically an air pump, don't you ?

A large turbofan engine has a low pressure in front of it and a higher pressure behind the large fan ( from which it gets most of its thrust ).
This air that it accelerates backwards provides an equal reaction foreward on the engine ( which is attached to the plane )

Now I could say, get a pump, and see how it behaves.
But I'm sure you've seen planes flying.

Yes just like a propeller plane. The explanation was already given on how that works. An explanation on how imbalanced forces are applied to rockets was also explained. 
Are you saying these explanations are wrong?

Are you saying that pressure is not a potential energy?

Are you saying that pressure gradient force is incorrect as well as the equation to calculate it?

 

Edited January 25, 2020 by Gerrard

[MigL]

OK, one last time...

If the pump stops pumping there might be some flow, depending on the kind of pump,  but usually if the pump can create a suction, therw will be very little.
That is the amount of momentum that is imparted to the tote as a result of pressure flow ( in the opposing direction, of course ).

When the pump is pumping, there is a large mass flow of water, at a much greater rate.
That is the amount of momentum that is imparted to the pump, as a result of pump flow.

One is almost negligible compared to the other.

[Gerrard]

4 hours ago, MigL said:

OK, one last time...

If the pump stops pumping there might be some flow, depending on the kind of pump,  but usually if the pump can create a suction, therw will be very little.
That is the amount of momentum that is imparted to the tote as a result of pressure flow ( in the opposing direction, of course ).

When the pump is pumping, there is a large mass flow of water, at a much greater rate.
That is the amount of momentum that is imparted to the pump, as a result of pump flow.

One is almost negligible compared to the other.

“If the pump can create suction, there will be very little” 

how do you suppose a pump works if it doesn’t create a pressure difference to create flow?

[MigL]

If the pump can create a suction, it has very tight tolerances.
IOW it is not a flow fed pump.

So if the pump isn't pumping, there will be very little flow, due to the tight tolerances.

Try to read in context; not isolated groups of words.

[Gerrard]

2 hours ago, MigL said:

If the pump can create a suction, it has very tight tolerances.
IOW it is not a flow fed pump.

So if the pump isn't pumping, there will be very little flow, due to the tight tolerances.

Try to read in context; not isolated groups of words.

When the pump is not pumping, the flow is because of the fluid inertia that was obtained when the pump was pumping. Like a car when you accelerate to 60mph and then you let go of the accelerator, the car will still move 60mph even thought there are no forces being applied assuming no friction 

when the pump is pumping it moves the fluid because of pressure gradient 

Edited January 26, 2020 by Gerrard

[guidoLamoto]

I'm a firm believer in the KISS system:

Newton's Law states that the total  momentum vector of a system remains unchanged in the absence of outside forces.....In the balloon rocket  system, the total F vector is 0 as the balloon sits at rest. Release the potential energy of the pressurized air and the balloon of course shoots off  in one direction with momentum M(1) = +m(1) v (1), while the air expelled shoots off in the opposite direction with momentum vector M(2) = -m(2)v(2) and M(1) + M(2) = 0 The total force vector is still 0 with +F of the balloon balanced by -F of the expelled gas. Total force & momentum of the system are conserved.

Back to  the original question about a pump. Let's change the example slightly to better illustrate the workings of the system: let's visualize a nurse filling a syringe from a bottle of medicine-- you know-- how they hold the bottle up and draw back on the plunger of the syringe....The nurse is supplying a new force to the resting syringe/bottle of liquid system. If she held the bottle with syringe stuck in it in only one hand and pulled, the whole thing would move and no liquid would be drawn into the syringe...Instead, she holds the bottle in one hand and the plunger in the other. As she draws back on the plunger (action) she has to push back on the syringe with the other (equal & opposite reaction). ..The fluid pressure, or momentum of the fluid in the bottle, if you prefer, goes down as the momentum/pressure goes up in the syringe. Total system state still conserved...For a mechanical pump and tote system, the reactionary force acts via the pump's anchoring bolts or maybe just friction on its base.

Edited February 9, 2020 by guidoLamoto

[joigus]

On 1/18/2020 at 3:16 AM, Gerrard said:

Would there be an opposite force on the tote moving away from the pump and perhaps even create tension on the flex hose?

I agree that there's some indefiniteness in your problem. AFAIKS, e.g., what is at the other end of the hose?

1) An infinite reservoir of water (or another fluid)

2) A finite, or relatively small, and closed, container of water (or another fluid)

3) It is a dead-end hose

4) Another pump, sucking water in whatever direction

Some hydrodynamics questions are extremely difficult, because pressure gradients do not distribute homogeneously, neither do they transmit the action as a symmetric reaction at the other end. Part of the pressure would be transmitted to the walls.  The detailed answer depends on how the pressure gradients distribute according to differential equations. The most general one is the Navier-Stokes PDE system, which is notable for its difficulty. For a static regime and an incompressible fluid with no vortices (perfect fluid,) I think that your question could be worked out, even if qualitatively, if you gave some more details.

People have been racking their brains with similar problems for decades. See, e.g., the Feynman sprinkler. One I particularly like (because I was able to qualitatively solve it, I think, although never to the satisfaction of the OP) was this:

Does a fly hovering on a plane really weigh?

But then this particular person was extremely difficult to satisfy.

[sethoflagos]

On 1/25/2020 at 3:41 PM, Gerrard said:

.......
In other words, there is pressure pushing equally on all sides of the container except for the side with the hole. How is the scenario different if there was a pump attached? 
......

This.

Rule 1) There is no such force as suck. All the energy for accelerating water out of the container comes from the container and its contents, so you can forget about everything downstream of the hole - it's irrelevant.

In the vicinity of the hole the water converts some of its (undirected) internal energy into directed momentum heading left let's say. This results in a localised drop in internal pressure on the left hand side of the container while the right hand side of the container continues to see the higher local fluid pressure, and is hence kicked to the right by a nett reaction force equal and opposite to the momentum of the exiting water stream. 

In a small system, it's a small effect. Industrially, say when a large pressure relief valve opens - these reaction forces can be many tonnesf and require serious structural support.