I Nozzle Reaction Forces: Explaining the Resultant Force

[Chestermiller]

OK I can buy your analysis of the straight situation and I accept that there is no 'backward force' on the nozzle. (I have a problem with the mechanics of how Equations 4 and 5 give you 6 but I guess you have been into that and that it's OK)
The OP is about a fire hose, though and the situation is always different because there is always some curvature just behind the hose; the picture of a fireman holding a hose, is pretty typical. Hoses do snake around and they need a firm hold to prevent it. That means there really is a force, even though it may not be 'backwards'. What could be causing it? Could it be the due to the centripetal force which constrains the water to move in a curve?

Sure. For this to happen, the hose has to exert a higher pressure on the outside of the curve than the inside. This results in an equal and opposite force on the hose.

 

[Michael W]

Chester's equations above look correct, and I think they describe a real force, but they zero out when the area at the beginning of the nozzle is equal to the area at the end.
That violates our intuition that it would still be difficult to hold a hose with very fast moving water even if there was no area reduction.
Is our intuition wrong?
I still think sophiecentaur's instinct to use the rocket equation is correct. If you're concerned that the water is not de-compressing or chemically changing, but maintaining its density unlike rocket fuel, I agree, but I think that a similar but opposite reaction is happening at the water inlets to the pump. If the "jet pack" was fed by a submersible pump hanging from a free-floating barge, of course we would expect the barge to drift opposite the direction of the flow at the pump inlet.

 

[sophiecentaur]

Chester's equations above look correct, and I think they describe a real force, but they zero out when the area at the beginning of the nozzle is equal to the area at the end.
That violates our intuition that it would still be difficult to hold a hose with very fast moving water even if there was no area reduction.
Is our intuition wrong?
I still think sophiecentaur's instinct to use the rocket equation is correct. If you're concerned that the water is not de-compressing or chemically changing, but maintaining its density unlike rocket fuel, I agree, but I think that a similar but opposite reaction is happening at the water inlets to the pump. If the "jet pack" was fed by a submersible pump hanging from a free-floating barge, of course we would expect the barge to drift opposite the direction of the flow at the pump inlet.

Resurrecting a two year old thread - but why not?
Whatever happens to the fluid that's being pumped, the mass flow rate is conserved but the velocity can be altered in a taper (and if the fluid density changes due to Bernoulli). If the pump is in a straight line with the nozzle then there is a force 'somewhere' on the water of mv, where m is the m is flow per second and v is the velocity. The reaction force would be on the blades / pistons or whatever of the pump. I think it's important to identify which of the forces are actually relevant and also which (lateral) forces 'cancel out'. But Newton 3 must always apply or all our lives could change.

 

[Chestermiller]

Chester's equations above look correct, and I think they describe a real force, but they zero out when the area at the beginning of the nozzle is equal to the area at the end.
That violates our intuition that it would still be difficult to hold a hose with very fast moving water even if there was no area reduction.
Is our intuition wrong?
I still think sophiecentaur's instinct to use the rocket equation is correct. If you're concerned that the water is not de-compressing or chemically changing, but maintaining its density unlike rocket fuel, I agree, but I think that a similar but opposite reaction is happening at the water inlets to the pump. If the "jet pack" was fed by a submersible pump hanging from a free-floating barge, of course we would expect the barge to drift opposite the direction of the flow at the pump inlet.

If the area at the beginning and end of the nozzle is the same, then you basically have a straight pipe in static equilibrium, with a fluid flowing at constant velocity through it. So let's consider the forces acting on the pipe. What force is the atmosphere exerting on the very exit end of the pipe (exposed to the air)? What is the nature of the horizontal force that the flowing fluid exerts on the inside surface of the pipe (assuming that the fluid is not totally inviscid)? What direction is this force acting?

 

[gmax137]

I did some googling on fire hose nozzle reactions. Here's a link
https://www.fireengineering.com/articles/print/volume-159/issue-4/features/measuring-and-demonstrating-nozzle-reaction.html

it shows a measurement of reaction force using a fish scale

 

[Chestermiller]

I did some googling on fire hose nozzle reactions. Here's a link
https://www.fireengineering.com/articles/print/volume-159/issue-4/features/measuring-and-demonstrating-nozzle-reaction.html

it shows a measurement of reaction force using a fish scale

View attachment 234858

In this photo, the hose is obviously curved, which is the reason for a constraining force to be needed. If the hose were perfectly straight, no such force would be required.

 

[sophiecentaur]

This discussion is reminding me strongly of the 'how does an aeroplane fly?' question. There are the bigendians who say it's because of Beernoilli and the littleendians who say it's because of Newton's Third Law.
But, whatever calculations that the bigendians introduce, there is no way that a plane flies because of a reaction less force. Similarly, with the fireman's hose, there has to be a reaction force but where it applies and what other forces are involved is a rich source of ideas and disagreements. For a hose with a vertical curve, the lateral forces will balance. For an in-line short hose and pump, there will be a force on the back of the pump. If the hose has an S in it then there will be forces on the 'back' internal face of the hose and possibly some force still on the pump itself. This will constitute a couple which will cause the hose to snake up and down. The centripetal force on the water flowing round the curve can also be relevant. If there is any deviation from straightness of hose, there will be some component of force on it that is partly balancing the Impulse force on the ejected water.

 

[sophiecentaur]

What direction is this force acting?

The net force backwards has to be the same as the force propelling the water. There will be a component of force in the line of the nozzle on every internal part of the water system. Many of these forces within the water path will cancel (as with symmetrical tapers). In particular, as a nozzle will be tapered from the wide hose to the narrow outlet, that resultant force there will be forwards, If you take the simplest form of pump, it could be a simple piston in a very long cylinder. If cylinder and hose are the same diameter and everything is straight in line, up to the nozzle taper, the reaction force on the piston will be the same as the propulsive force on the emerging water plus the forward component of the force on the taper. Other tapers and bends will introduce more forces in the system.

 

[Chestermiller]

The net force backwards has to be the same as the force propelling the water.

Actually, in a straight pipe of constant cross section, the only force interaction between the flowing fluid and the pipe (in the horizontal direction) is the frictional drag at the wall. The wall exerts a viscous shear stress in the negative flow direction on the fluid, and the fluid exerts an equal and opposite viscous shear stress in the positive flow direction on the inner wall of the pipe.

If the fluid were inviscid, there would be no horizontal force interaction between the straight pipe and the fluid at all. The fluid would just be traveling down the pipe at constant velocity, and, since it is not accelerating, no forces would be needed to maintain this. The only forces that would come into play in this kind of inviscid flow situation would be if there were curvature to the pipe. Such forces would be necessary to change the direction of the flowing fluid. We are all familiar with the inertial forces between a flowing fluid and the pipe when 90 degree and 180 degree bends are involved. In such cases, the pipe would have to be properly supported.

 

[sophiecentaur]

Actually, in a straight pipe of constant cross section, the only force interaction between the flowing fluid and the pipe (in the horizontal direction) is the frictional drag at the wall.

I have a feeling that we are talking slightly in quadrature here. Have I missed bit where you consider the reaction force that I am batting on about?
I have no problem with the detail of what you are saying but a 'complete' explanation must include a Newton3 reaction force against 'something', which will usually have contributions at many different points. If the hose is curved then there is some force component from the 'outside face' of the curve.
The friction forces you introduced are internal and will be balanced by other internal forces such as tension in the hose and forces in the pump. (And in bends, as you say).
But the ejected water has gained Momentum and that must have had forces on it during its passage through the system. Forget the water supply pipe and replace it with a large tank of water and you can say the initial velocity is zero. The accelerating forces must be acting, ultimately, either as a force against the Earth or a rocket-like force to accelerate the pump equipment.
The water jet harness, mentioned earlier, is an extreme case where a reaction force on the upper face of the hose loop provides the force to support the driver.

 

[CWatters]

Would be interesting to consider a straight rigid but telescopic pipe with a tapered nozzle...

When you turn the water on does it..

a) extend (due to pressure on the inside of the nozzle) or
b) retract (due to the water accelerating through the nozzle and conservation of momentum)
c) remain at constant length
d) both effects a) and b) exist but one is greater?

 

[jbriggs444]

Would be interesting to consider a straight rigid but telescopic pipe with a tapered nozzle...

When you turn the water on does it..

a) extend (due to pressure on the inside of the nozzle) or
b) retract (due to the water accelerating through the nozzle and conservation of momentum)
c) remain at constant length
d) both effects a) and b) exist but one is greater?

Seems quite obvious. There are only two forces involved in the interaction between nozzle and water: pressure and viscous friction. Both act to push the nozzle in the direction of flow. There are only two forces involved in the interaction between the telescoping pipe walls and the water: pressure and viscous friction. The former has no effect in the direction of the flow. The latter acts to drag the pipe in the direction of the flow.

One end of the pipe+nozzle is anchored to the tap. The other end is free. The net horizontal force on nozzle plus pipe causes them to extend away from the tap. So the nozzle and pipe extend when you turn on the valve.

If we include gravity and assume a horizontal pipe, there is a vertical pressure gradient which acts to compensate for it and keeps the flow horizontal.

Given a freely telescoping pipe and nozzle with negligible mass and unlimited extent, the egress flow rate is also negligible. The pipe extends out almost as fast as water comes in and only a dribble actually comes out of the nozzle. We may have to consider an equilibrium condition involving air resistance on the rapidly extending pipe.

Edit: the usual deployment is in a vertical orientation on a timer so that the pop-up watering nozzles can irrigate one's lawn. (Though the flow diverter at the nozzle makes this less than an ideal example).

 

[russ_watters]

This discussion is reminding me strongly of the 'how does an aeroplane fly?' question. There are the bigendians who say it's because of Beernoilli and the littleendians who say it's because of Newton's Third Law.
But, whatever calculations that the bigendians introduce, there is no way that a plane flies because of a reaction less force. Similarly, with the fireman's hose, there has to be a reaction force but where it applies and what other forces are involved is a rich source of ideas and disagreements.

I kind of agree from a science standpoint, but the main problem with these threads is that many people are not answering the question being asked.

My speculation on why is that when one points out that the firefighter is holding onto the hose, leaning forward and asks why, it triggers an analysis of what the water is doing to the nozzle, and a set of blinders that obscures the original question. Conclusion: the water is pushing the nozzle forward. But the question being asked is not "what is the direction fo the force the water is applying to the nozzle?", the question is "what force is pushing the firefighter backwards?"

The exact numerical answer is dependent on the setup and assumptions (as always), but the answer will be a combination of one or more of the following (in no particular order):

1. The elasticity of the hose.
2. The static pressure of the water in the hose.
3. The momentum of the water entering and exiting the hose.

In addition, it is notable that for some scenarios the firefighter is indeed being pushed forwards, not backwards. That's interesting, but not what is generally being asked about.

Speaking of backwards, I think a lot of the reason for the rhetorical problem in these threads is that they are in fact backwards. Most of what we do here is homework help and for most such problems, the scenario is clearly defined and you calculate what happens. But this thread starts with what happens and asks why. Then immediately, people construct a scenario, calculate a correct answer for their own scenario and never go back to check if that answers the OP's question. Many construct the wrong scenario and as such answer the OP's question wrong.

 

[sophiecentaur]

I kind of agree from a science standpoint, but the main problem with these threads is that many people are not answering the question being asked.

Not for the first time on PF, i think.
I like your post. The original question is hiding a paradox / misconception in people's intuition. The OP gives two apparently opposing ideas when they are actually not mutually exclusive. A narrowing taper will produce a forwards force on the hose but there are other forces at work, which can push / pull firefighters in many directions - including backwards. It's not surprising that there's confusion because we all know about rockets. But a bit of thought will remind us that rocket engines tend to have an widening taper and not a narrowing one. N3 comes along and he has to be right. The perfect storm for a PF thread.

The action-reaction pair you are asking about is (a) the forward pressure force that the fluid exerts on the nozzle body and (b) the backward force exerted by the nozzle body on the fluid.

Not that pair; there are any number of pairs that could be considered and I am considering the overall forces on the whole pump / hose / nozzle kit, due to the expelled water. I realize that half of the OP content asks about the force on the nozzle (the one you correctly refer to and deal with) but also the op asks for an explanation of the forces on the firefighter. I'm talking about the force that must have existed to accelerate the fluid out of the nozzle and the necessary N3 force must act on a number of parts of the restraint of the system. Whenever Momentum is changing, there will be a force and a curve on the hose is one example. The fighter needs to minimise that force or at least make sure that it's in a controllable direction (down towards the ground and not sideways or up).
I cannot see how people still seem to be ignoring the need for the overall N3 constraint. As I said above - this is just like the flying aircraft wing discussions. Both approaches are valid as far as they go.

 

[sophiecentaur]

Though the flow diverter at the nozzle makes this less than an ideal example).

I just spotted this on re-reading. If you mean by "flow diverter", a mechanism for spreading the water around, then a rotary flow diverter (Catherine wheel style) follows exactly my explanation for the fire hose. The reaction against the force on the departing water causes the arm on the end to spin in the opposite direction.

 

[glennr]

An old thread, but this is relevant:

https://link.springer.com/article/10.1007/s10694-014-0430-5

 

[cjl]

Not for the first time on PF, i think.
I like your post. The original question is hiding a paradox / misconception in people's intuition. The OP gives two apparently opposing ideas when they are actually not mutually exclusive. A narrowing taper will produce a forwards force on the hose but there are other forces at work, which can push / pull firefighters in many directions - including backwards. It's not surprising that there's confusion because we all know about rockets. But a bit of thought will remind us that rocket engines tend to have an widening taper and not a narrowing one. N3 comes along and he has to be right. The perfect storm for a PF thread.

The rocket nozzle is expanding because that is the way to achieve the highest exhaust velocity. Supersonic flows behave inversely to subsonic flows when it comes to changes in flow area, so expansion increases velocity and contraction decreases it. This is not necessary to generate thrust though. A rocket with a convergent-only nozzle would still produce a very considerable amount of thrust. Indeed, this is frequently the case on jet engines.

Not that pair; there are any number of pairs that could be considered and I am considering the overall forces on the whole pump / hose / nozzle kit, due to the expelled water. I realize that half of the OP content asks about the force on the nozzle (the one you correctly refer to and deal with) but also the op asks for an explanation of the forces on the firefighter. I'm talking about the force that must have existed to accelerate the fluid out of the nozzle and the necessary N3 force must act on a number of parts of the restraint of the system. Whenever Momentum is changing, there will be a force and a curve on the hose is one example. The fighter needs to minimise that force or at least make sure that it's in a controllable direction (down towards the ground and not sideways or up).
I cannot see how people still seem to be ignoring the need for the overall N3 constraint. As I said above - this is just like the flying aircraft wing discussions. Both approaches are valid as far as they go.

The overall N3 constraint isn't ignored, but it can be somewhat nonintuitive. In the case of a completely straight pipe/hose with a nozzle on the end, the only force acting on the entire pipe/nozzle system will be in the same direction as the flow. The N3 reaction force from the fact that the water is coming out the end will only be felt in the backwards force exerted on the pump itself (which can alternatively be seen as the force caused by the pressure differential between the pump inlet and outlet). For your water tank example, there's a slight force on the tank itself caused by the fact that the opening out of which the water is flowing is an opening, and thus, the water is not exerting hydrostatic pressure on the tank at that point. There's an imbalance because on one side of the water tank, the water is pushing against a region that is not mirrored on the other side due to the opening. As such, the tank will experience a force away from the direction of the outlet.

(I realize this is now the second time this has been resurrected, but it felt like it had never really come to a satisfactory answer, so hopefully this is helpful)

 

[sophiecentaur]

The rocket nozzle is expanding because that is the way to achieve the highest exhaust velocity.

I would say that, because of Bernouli, the pressure after the constriction goes up and the gas speed drops. But that doesn't actually go against what you say because the combustion conditions are optimised.

There's an imbalance because on one side of the water tank, the water is pushing against a region that is not mirrored on the other side due to the opening. As such, the tank will experience a force away from the direction of the outlet.

I agree that looking for N3 involves looking everywhere. People only want to look at the nozzle and that accounts for the wrong thinking that we read about.

 

[cjl]

I would say that, because of Bernouli, the pressure after the constriction goes up and the gas speed drops. But that doesn't actually go against what you say because the combustion conditions are optimised.

This is false. After the constriction, the speed rises dramatically (and the pressure continues to drop). Bernoulli is not applicable. For a fairly common rocket nozzle where the exit pressure is 1/125 of chamber pressure (so, for example, an exit pressure of 10 PSI and a chamber at 1250, which corresponds to an area expansion ratio of just under 10:1), the mach number at the exit will be 3.85, and the exhaust velocity will be 2.12 times higher than the velocity in the throat.

The reason bernoulli (and a lot of common sense about how fluids behave in general) doesn't apply is because of the compressibility of the flow. In compressible flows, a constriction only accelerates a flow if the flow is subsonic. Once you hit a mach number of 1, you actually have to expand the flow to continue accelerating it. The pressure drop as the flow accelerates is so dramatic that the volumetric flow rate is increasing faster than the linear flow rate is (due to the drop in density of the gas as the pressure falls), so a converging-diverging nozzle is actually the best way to achieve maximum exhaust velocity when you have a large pressure gradient across the nozzle (and your working fluid is compressible).

 

[sophiecentaur]

The reason bernoulli (and a lot of common sense about how fluids behave in general) doesn't apply is because of the compressibility of the flow. In compressible flows, a constriction only accelerates a flow if the flow is subsonic.

Brilliant - thanks for that.

 

[Chestermiller]

See this literature article which backs up everything I was saying in this thread: https://link.springer.com/article/10.1007/s10694-018-0719-x Particularly, that there is no reaction force at the nozzle exit and that a straight hose is under tension. Also, it is the curvature of the hose in actual practice that gives rise to what firefighters falsely perceive as a reaction force at the nozzle.

 

[sophiecentaur]

I totally agree. In videos, firefighters keep an S shape in the hose, to have the nozzle at waist height. Friction of the length of hose on the ground is probably what provides support against the reaction force you describe.
When a hose is let go it snakes with the motion dictated by an ‘obvious’ force on the curve that forms some distance behind the nozzle.
You have to be fair to firefighters when they accept the story about the nozzle. Newton’s Laws are not uppermost in their minds when wrestling with a live snake.