Understanding Bernoulli's Principle

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My understanding of Bernoulli's Principle is something like this: Pressure is inversely proportional to velocity. Fluid flowing through smaller cross-sectional area has increase velocity & decrease in pressure.

Also P = F/A.... so would force also decrease for fluid going through small area with increased velocity & decrease pressure.

How does that relate to water coming out of a hose with a nozzle?? By Bernoulli's Principle: speed increases hence pressure decreases. So would the force coming out of the nozzle also decrease due to small area and decreased pressure?? Why is that you feel a greater force from a hose with a nozzle than one with a greater diameter.
 

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  • #2
phinds
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Have you tried a forum search? There have been a lot of discussions here of Bernoulli's Principle
 
  • #3
rcgldr
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There is a net "forwards" force exerted onto the water as the water is accelerated, this in turn coexists with a net "backwards" force on whatever contains the water, in this case the hose.
 
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There is a net "forwards" force exerted onto the water as the water is accelerated, this in turn coexists with a net "backwards" force on whatever contains the water, in this case the hose.
That’s not my understanding. My understanding from a momentum balance on the water is that the water exerts a forward force on the nozzle. Let’s see you momentum balance on the water.
 
  • #5
sophiecentaur
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Why is that you feel a greater force from a hose with a nozzle than one with a greater diameter.
In the case of a simple garden hose, the same volume of water per second is maintained by the pressure of the mains and the basic resistances of the valves in the circuit. It will emerge at a low velocity from a large hole and at a high velocity from a small hole. The same mass of water per second at the higher velocity will have more Momentum so you will feel a greater force on your hand.
This presupposes that the pressure of the water is maintained to keep the flow rate the same.
 
  • #6
rcgldr
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There is a net "forwards" force exerted onto the water as the water is accelerated, this in turn coexists with a net "backwards" force on whatever contains the water, in this case the hose.
That’s not my understanding. My understanding from a momentum balance on the water is that the water exerts a forward force on the nozzle. Let’s see your momentum balance on the water.
Sophiecentaur just answered this. Mass flow is the same everywhere in the system, while velocity and momentum increase. There's an external power source that supplies and maintains the pressure and mass flow inside the hose. Once the water is inside the hose, and ignoring friction, mechanical energy (pressure + kinetic) is conserved, but not momentum. As actual examples, which way does a fire hose or "water wiggle" move if not held?
 
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Sophiecentaur just answered this. Mass flow is the same everywhere in the system, while velocity and momentum increase. There's an external power source that supplies and maintains the pressure inside the hose. Once the water is inside the hose, and ignoring friction, mechanical energy (pressure + kinetic) is conserved, but not momentum. As actual examples, which way does a fire hose or "water wiggle" move if not held?
Of course momentum is conserved. Like I said, let’s see your equations including your momentum balance equation.
 
  • #8
sophiecentaur
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As actual examples, which way does a fire hose or "water wiggle" move if not held?
That's interesting and it seems that the reaction force from the ejected water is exerted on the back of the final bend in the pipe. There is a resultant sideways force on the pipe which causes it to move to the side if it's not controlled, causing a snaking motion. Forces along the rest of the pipe are all balanced by tension , holding the pipe onto the hydrant outlet. This link makes my point. (Aren't firemen just great?) There are a bunch of YouTube movies to find, once you are in there.
See this link for a water jet pack where the force on the rider is from the two downward jets and independent of the route the feed hose takes. (Good fun, eh?)
 
  • #9
sophiecentaur
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Of course momentum is conserved. Like I said, let’s see your equations including your momentum balance equation.
That's ok when you realise that the momentum transferred to the water jet is equal and opposite to the Mass of the Earth times its change in velocity (which is verrrry small).
 
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That's ok when you realise that the momentum transferred to the water jet is equal and opposite to the Mass of the Earth times its change in velocity (which is verrrry small).
I have no idea what you’re talking about. Are you saying that, I, as an engineer with formal training in fluid mechanics, am unable to do a proper macroscopic momentum balance on water flowing through a hose nozzle?
 
  • #11
rcgldr
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That's ok when you realise that the momentum transferred to the water jet is equal and opposite to the Mass of the Earth times its change in velocity (which is verrrry small).
I assume the earth reaction is related to the pump or whatever power source that supplies the pressure and mass flow into the hose? An alternative power source could be a large tank that collects rain water, which would require that the momentum of the system include the rain in addition to the other components of the system.
 
  • #12
rcgldr
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I have no idea what you’re talking about. Are you saying that, I, as an engineer with formal training in fluid mechanics, am unable to do a proper macroscopic momentum balance on water flowing through a hose nozzle?
It's not a closed system unless you include the power source and whatever that power source effects.
 
  • #13
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It's not a closed system unless you include the power source and whatever that power source effects.
I still haven’t seem a single equation out of you. And who said anything about a closed system? I’m not impressed by what you’ve said so far. If you can’t prove what you are saying using equations, i’m not buying it.

On the other hand, here is an analysis of a hose and nozzle that I have presented in Physics Forums just this past December, including the Bernoulli equation and the macroscopic momentum balance on the fluid. It shows that the force exerted by the fluid on the nozzle is in the same direction as the flow (not backwards). https://www.physicsforums.com/threads/force-required-to-hold-a-fire-hose.769442/#post-5906921
 
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  • #14
sophiecentaur
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I have no idea what you’re talking about. Are you saying that, I, as an engineer with formal training in fluid mechanics, am unable to do a proper macroscopic momentum balance on water flowing through a hose nozzle?
I seem to have upset you a bit and I'm sorry. But to know the force on your hand, you need know nothing about the internals of the pipe and nozzle. Newton 3 tells us that the momentum given to the exiting water (mass per second) must be balanced by an equal an opposite momentum on the Earth.
I suspect that you are of the big-endian persuasion that would say an aeroplane stays up because of Bernoilli and not because of the (little-endian) net downward flowing mass of air. There is quite enough room for each theory as long as they are applied appropriately. If the volume flow rates are the same then the 'easy' solution applies.
 
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I seem to have upset you a bit and I'm sorry. But to know the force on your hand, you need know nothing about the internals of the pipe and nozzle. Newton 3 tells us that the momentum given to the exiting water (mass per second) must be balanced by an equal an opposite momentum on the Earth.
I suspect that you are of the big-endian persuasion that would say an aeroplane stays up because of Bernoilli and not because of the (little-endian) net downward flowing mass of air. There is quite enough room for each theory as long as they are applied appropriately. If the volume flow rates are the same then the 'easy' solution applies.
Are you saying you disagree with the routine fluid mechanics control volume macroscopic momentum balance (in conjunction with the Bernoulli equation) I presented in post #13? If so, please point out where you think my error was in determining the force exerted by the flowing fluid on the section of hose between some arbitrary location and the nozzle exit. This same approach is presented in Section 7.2 THE MACROSCOPIC MOMENTUM BALANCE in Transport Phenomena by Bird, Stewart, and Lightfoot (a text that has stood the test of time and abundant usage in colleges and universities for over 70 years).

As far as the force on your hand, there is no need for it (aside from keeping the hose from curving) since the hose itself is under tension. This system is nothing remotely approaching the complexity of an airfoil, and its behavior is not a matter of opinion.
 
  • #16
sophiecentaur
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Are you saying you disagree with the routine fluid mechanics control volume macroscopic momentum balance (in conjunction with the Bernoulli equation) I presented in post #13?
I am not disagreeing with it at all. I am just pointing out that it is not necessary to do any of that if all you want it to calculate the force on a hand, given the volume flow and the speed. I was making the point that the momentum of this water is balanced by the momentum of the rest of the Earth. That was the only issue I was answering, way back. I was trying to avoid using a sledgehammer to crack a simple nut, which is never a bad thing, I asm sure you would agree.
The reaction force on the bend in the hose or, in the case of a straight hose, the first bend inside the hydrant would be what is accelerating the Earth 'backwards'. What else goes on in there is of interest to you and others but it is not basically relevant to the water jet, once it has emerged.
As far as the force on your hand, there is no need for it (aside from keeping the hose from curving) since the hose itself is under tension.
I'd agree with that but the case where a hose emerges from the hydrant and is straight up until the nozzle is only fit for a text book example. The reaction force against the first curve is highly relevant (especially to a fireman).
 
  • #17
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I am not disagreeing with it at all. I am just pointing out that it is not necessary to do any of that if all you want it to calculate the force on a hand, given the volume flow and the speed. I was making the point that the momentum of this water is balanced by the momentum of the rest of the Earth. That was the only issue I was answering, way back. I was trying to avoid using a sledgehammer to crack a simple nut, which is never a bad thing, I asm sure you would agree.
The reaction force on the bend in the hose or, in the case of a straight hose, the first bend inside the hydrant would be what is accelerating the Earth 'backwards'. What else goes on in there is of interest to you and others but it is not basically relevant to the water jet, once it has emerged.

I'd agree with that but the case where a hose emerges from the hydrant and is straight up until the nozzle is only fit for a text book example. The reaction force against the first curve is highly relevant (especially to a fireman).
I agree that for a curved hose, the hose must exert a force in the fluid toward the inside of the curve to change its direction, and the fluid then exerts a force on the hose toward the outside of the curve. So, to prevent the hose from moving in that direction, one would have to exert a force toward the inside of the curve to hold it in place.

Of course, to have a straight hose, it would not have to be point straight up in the air. I'm focusing only on the downstream part of hose that is straight, and where this section of the hose is going to be under tension. The section of hose is under tension because the pressure force exerted by the internal pressure on the converging section of the nozzle exceeds the change in axial momentum (jet) imparted to the fluid (as the analysis referenced in post #13 clearly shows).
 
  • #18
sophiecentaur
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I agree that for a curved hose, the hose must exert a force in the fluid toward the inside of the curve to change its direction, and the fluid then exerts a force on the hose toward the outside of the curve. So, to prevent the hose from moving in that direction, one would have to exert a force toward the inside of the curve to hold it in place.
All hoses plus pipes are curved somewhere. With multiple curves, it's mostly only the last curve that counts. Earlier curves will give a net balance forward and backward. It's the last one that a fireman makes sure to rest on the ground to counter the reaction force.
I'm focusing only on the downstream part of hose that is straight, and where this section of the hose is going to be under tension.
I have no problem with that but you are producing a reason for the pressure at the outlet of the nozzle. If the supply has a constant volume flow then how are the details of the internal geometry relevant to this? They can not affect it.
It's like considering the current through a circuit with a constant current source. Assuming there's no 'limiting', the current through the last resistance the chain is what the current supply wants it to be.
In practice, there is usually an optimum adjustment of the screw-in nozzle to get the maximum force. Screwing the control in tighter will prevent a constant flow and a powerful jet will become a high pressure jet with low volume flow.
 
  • #19
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All hoses plus pipes are curved somewhere. With multiple curves, it's mostly only the last curve that counts. Earlier curves will give a net balance forward and backward. It's the last one that a fireman makes sure to rest on the ground to counter the reaction force.

I have no problem with that but you are producing a reason for the pressure at the outlet of the nozzle. If the supply has a constant volume flow then how are the details of the internal geometry relevant to this? They can not affect it.
It's like considering the current through a circuit with a constant current source. Assuming there's no 'limiting', the current through the last resistance the chain is what the current supply wants it to be.
In practice, there is usually an optimum adjustment of the screw-in nozzle to get the maximum force. Screwing the control in tighter will prevent a constant flow and a powerful jet will become a high pressure jet with low volume flow.
I'm sorry. I don't follow this at all. There is a huge difference between current flow in a circuit and fluid flow through ducts. I stand by what I said.
 
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  • #20
sophiecentaur
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I'm sorry. I don't follow this at all. There is a huge difference between current flow in a circuit and fluid flow through ducts. I guess we'll just have to agree to disagree.
It was a loose analogy - hardly worth arguing over but conservations laws apply. By definition, constant flow rate means constant flow rate and it's set largely by the mains supply and the pipe diameter.
We all realise that this is an approximation and that there is a limit which is where your detailed formulae have to come in.
But whatever the values of all the variables, the reaction force is still equal to the rate of change of momentum of the water hitting your hand. (Assuming the water gives it all up.
 
  • #21
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It was a loose analogy - hardly worth arguing over but conservations laws apply. By definition, constant flow rate means constant flow rate and it's set largely by the mains supply and the pipe diameter.
We all realise that this is an approximation and that there is a limit which is where your detailed formulae have to come in.
But whatever the values of all the variables, the reaction force is still equal to the rate of change of momentum of the water hitting your hand. (Assuming the water gives it all up.
This makes no sense to me. I stand by what I said and double down on it. If there was an error in the analysis I referenced in post #13, I'm still waiting for you to point it out. otherwise, it stands as correct, and no amount of hand-waving will get you past it.

Chet
 
  • #22
sophiecentaur
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no amount of hand-waving will get you past it.
I wouldn't have classed Newton's Third Law as arm waving.
PS I never said you were wrong. You came to that conclusion on your own.
 
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  • #23
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I wouldn't have classed Newton's Third Law as arm waving.
PS I never said you were wrong. You came to that conclusion on your own.
As an expert on Newton's third law, are you saying that I applied this law incorrectly in the analysis I referenced. I said that the force that the hose and nozzle exert on the fluid in the control volume is equal in magnitude and opposite in direction to the force that the fluid exerts on the hose and nozzle. Is that incorrect? And who says I had concluded that I'm wrong. I assert that what I said is totally correct, and that the hose is under tension.
 
  • #24
sophiecentaur
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As an expert on Newton's third law, are you saying that I applied this law incorrectly in the analysis I referenced. I said that the force that the hose and nozzle exert on the fluid in the control volume is equal in magnitude and opposite in direction to the force that the fluid exerts on the hose and nozzle. Is that incorrect? And who says I had concluded that I'm wrong. I assert that what I said is totally correct, and that the hose is under tension.
I think you need to calm down @Chestermiller. At no time was I telling you that you were wrong. You even managed to 'mis-parse' my last post, in which I was saying that you had decided that I had told you you were wrong. I am sorry that you should feel offended. The only way in which we are disagreeing is that you have been implying that you need to whole analysis, even when the stipulation is constant volume flow rate. The basic Newton 3 for the whole system is all that's necessary.
You really shouldn't be so quick to take offence. Absolutely none was intended at any stage.
Yes, the hose is under tension but that is not the point here. If things were exactly the way you have been putting it, a water jet pack would not work - in fact no jet would actually work if you ignore the reaction right at the back of the system. But of course you know that the tension in the hose is not enough to give a net forward pull or we could have a reactionless drive. We have been talking at cross purposes and wasting a lot of time.
 
  • #25
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Here is a solved problem involving flow through a fire hose (that I paid $15 for) from Chegg Textbook Solutions.

Fire Hose Statement.png

Fire Hose.png

In the above solution, see the 4th equation below the text "Calculate the force transmitted by the coupling between the nozzle and hose." Aside from the notation, it is exactly the same macroscopic momentum balance equation that I presented in my analysis in the link referenced in post #13, including the term of the right hand side that has caused some people some consternation:
$$P_1A_1+F=\rho v_2A_2(v_2-v_1)$$
The final answer from Clegg shows tension at the coupling (and thus in the hose), in agreement with my analysis.

In this problem, the Clegg statement specified the upstream pressure (rather than in our problem where the upstream pressure was not specified), so I wanted to see how the results would compare with what I would determine using the Bernoulli equation to determine the upstream pressure. Here is the calculation of the upstream pressure using their jet velocity and their velocity inside the hose:
$$P=\frac{\rho}{2}(v_2^2-v_1^2)=\frac{1000}{2}(32^2-3.56^2)=505679\ Pa=507 kPa$$
This is practically the exact same pressure as specified in the Chegg problem (which gave an upstream pressure of 510 kPa), so the final tension in the hose is what I would calculate. The final tension comes out to be about 2kN.

I hope this internet example (in which Chegg puts their reputation on the line) convinces members of the validity of my analysis, including the rate of change of momentum terms in the momentum balance equation, and the fact that the hose is under tension.

Now, the only questions that need to be dealt with are:
1. Why do fire fighters have to pull on the hose to keep it extended (if it is under tension)
2. What happens if the hose isn't perfectly straight
 

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