Maximizing a Nozzle's Stream Velocity

[Ben]

Change the "carrying force" to "carrying momentum" if you want to be picky. Jo's point, however, is clear.

And joanna's point was what? That a fluid does not carry a force? That was my point too you know.

You're dabbling in the waters of the pressure theory.

The volume of water in a length of tubing is dependent on the area. The force necessary to accelerate that volume is also dependent on the area. Force / area = pressure.

The whole point is that pressure tells you how much force there is for a certain amount of water. For a given nozzle area (or stream volume), pressure tells you how much force is being used to accelerate the stream.

I've no doubt it's force that accelerates the water. All my point is is that pressure tells you how much force is accelerating a volume of water. (And in this example, volume * ~constant density = mass)

This goes back to your faulty thought that the force is reduced somehow by any change in diameter. That is incorrect and I don't feel I should tell you it any longer.

And, on a wider diameter, yes, the water definitely is accelerated less. However, that does not mean the flow is not the same or at least comparable. The velocity should be slower anyway to get the same flow from a larger diameter. So, pressure is proportional to the acceleration. An applied force evens out for the most part for this reason, however, as I speculated before, I do not think the relationship between volume and mass is completely even or I believe there is something else we are missing.

I don't get what you're saying in the last paragraph either. If force accelerates the water, then how does pressure tell you how much force is accelerating a volume of water? As far as I'm concerned what you first said proves the force theory because the entire volume of water is moving. That's flow. We're accelerating the entire system, not a small part of it. Then again, you may not understand the concept of an integral.

You seem to be reverting to the arguments you made in the LR thread about how the force is reduced at the nozzle... not true... and you know it.

You imply that positive pressure only exists when you first move the piston. However, you have to keep applying force to keep on shooting. Even though the pump's velocity is constant, the cylinder of the stream is accelerating. That requires constant force, which involves constant pressure.

Whatever you meant, as long as you apply force, you create no pressure. Tell me how you can have instantaneous pressure with constant force.

Perhaps I should have explained differently. What I meant was that any pressure created was not powering the system and that it was nearly instantaneously equalized. I do not think you will disagree with that statement. No pressure differential between the outside and inside.

Gee, that's a convincing argument...

I'll once again point out Larami's name for CPS: Constant pressure system. Say what you will about propaganda, but you imply that because it's Larami's patent, it's authoritative.

If I ever bother to epoxy a pressure gauge to my CPS 4100, I'll tell you whether or not that pressure is constant. The stats you had regarding a handful of water guns implied there was a constant pressure, although I might be wrong.

Propaganda? Who said that? I said marketing. And I never said they use the correct physics in their patents. I simply said they are more likely to use the correct physics due to their (supposed) qualifications. Look at the later Buzz Bee Toys ones if you want some better physics ones.

The pressure should be fairly constant. Then again, I only have tested pressure over the entire fill with air pressure. Water may work differently.

Pressure differentials create net force. Pressure alone creates total force. Equalized pressure in a chamber creates zero net force, because there's equal total force but in opposite vectors.

There's a gradient between the internals and the atmosphere, but no gradient inside the chamber.

I'm not talking about inside of the chamber. I was talking about the difference between inside and outside. And this is another case of where I think you are arguing for the sake of argument, stating nothing worthwhile.

You're beating a dead force. How can you say that the following quote is NOT saying low pressure is bad?

Beating a dead force?

There's a qualifier. According to you, low pressure is universally bad. I know that more pressure is better, but their qualifier indicates there's more to the picture than pressure alone.

This is very different than my repeating that force is not a property of matter because you repeatedly make mistakes assuming that force is a property of matter. I know more pressure is benefitial. But, there is a qualifier, and that definitely indicates that there is more to the picture.

And if number of mentions were important to the validity of any theory, yours would lose huge where it matters. 9 pages out of over 680. No pages in the PDF book I quoted at the end. In fact, that entire PDF was based mostly on force.

Nearing the end of my last post, which you hadn't gotten to before writing this part (unless you read the posts first), I mentioned how Mooney uses both force and the area of the nozzle. Your model completely ignores whatever area the force is applied across. Mooney used both, so he was essentially talking about pressure.

Look, he's adding up forces to get the net force. One force is the force of the pressure differential between the inside and the outside. Turns out the outside has higher pressure, so he has to calculate the force pushing upward. Not hard. I don't know how many times I have to say it.

That's true, but keep in mind half of what I post is response to your claims of poor sources and documentation.

Even if I were to win a debate over who has the best sources and examples, I'd be concerned over the actual science instead.

I understand you've been challenged a lot by people who claim to be physics professors, engineers, and all that. I don't think you've cared about qualifications though. I don't need to bring up examples of this - you know it better than anyone else.

If you were concerned about the actual physics, you wouldn't argue what you are arguing. I have seen no actual use of the equation you say controls flow in the manner you use it in aside from people on the internet who don't know how to use it. I try to find highly authoritative sources so you don't just shrug them off. And I have found great sources that state exactly what I have been saying. You can not find a single .edu or .gov website that support what you say.

You were responding to my argument that used the patent's information about a pump-powered water gun.

I can see that you were confused about what I was referring to. I was referring to all of the forum links you posted.

We can also assume that, if he fired the LR five times, that he only posted the maximum range of 57 feet. I highly doubt 57 was an average - he'd have said so. And who would not have posted the maximum?

A true statistical study would average several shots. The maximum is not an accurate view of how the water gun performs on average. Take any basic statistics class. And 50 - 55 was a estimation of the range. It is not the range of several values. That is why I averaged it.

If you want to bias your results, I certainly could do the same. But, I don't have to because my physics are correct. I have seen it in real life, I have seen it on paper. I have absolutely no doubts because I've done it all before.

Just wondering, how did you find out the pressures for these guns? I think I recall hearing about something Larami said once. The funny thing is the large difference between the CPS 2000's and the CPS 2500's pressures.

The SS 300 was an approximation based upon where I thought it should be. The CPS 2000 was what I was told once. The CPS 2500 is the same pressure as the CPS 1000, which was somewhat of an assumption, but I think the pressure is at most 25 PSI and I wouldn't consider something off by 2 PSI to be wrong.

But, back on topic, I think the assumption that the SS 300's overall flow coefficient is so poor is the problem. Just because it's narrow in one place doesn't mean it has such a terrible flow coefficient.

Sure. But you should realize that what you just said is in opposition to much of what you said before. You claim that Supercannon II performs so well because it has very high potential for flow. You claim SuperCAP performs so well because it has a very high potential for flow, despite the fact that it only shoots 2 feet less than Supercannon II and has about half the flow.

Overall, though, there are just too many variables and we know too little about stock soakers for me to answer your question. A controlled test is the only solution.

This is what you say when you know the data doesn't support what you say.

Then, in a purely education manner, I ask you why and how you would use flow coefficient in an equation.

The Cv value (the coefficient of velocity) represents how much flow velocity is lost when a flow passes through a valve. The velocity lost is proportional to the square root of the pressure. Multiplying the two gives the lost flow. That is how you use it.

The situations are very similar. In both, you have a constant flow coefficient, and you find flow for a certain pressure. Simple as that.

Just because the source of the flow is not a pressure chamber in the PDF doesn't mean a pressure chamber can't be the source.

I never said that the pressure chamber can't be the source. That's putting words in my mouth that I never said. I said that the equation is not used to calculate flow created by a pressure.

How does it drop all the factors? What factors does it drop? "All of the factors" is far too vague.

Saying I can't rearrange this equation is like saying I can't rearrange the force equation to find the value of 1 G.

I don't know all of the factors that are removed, but I do know that some of them are cross sectional area, all of the stuff about turbulence and vorticity, among other things. I don't know exactly what it drops because I am not an expect in the equation, but I do know that as a coefficient, it does drop something at the very least.

And no, saying that you can't rearrange this is completely different than reorganizing the force equation. This one is a simplified equation meant for one purpose. It's like using an equation I derived for the moment of inertia for a certain shape for a completely different shape (if you understand moment of inertia).

You say that the piston applies a massive amount of force. However, that force is a vector value. If there is a turn in the tubing, then the tubing walls must exert a force in a different vector in order for the final vector to be aligned inside the new tube.

I know force isn't a property of matter - thanks. But you are right in that moving water can push into things and exert force. In my diagram, most of the force exerted by the piston onto the water in turn gets exerted on the walls.

That's why my stream tests showed very little of the water's momentum turning into force applied to the scale. Very little of the chamber's force is going to be used to accelerate the stream.

Pressure is not a vector value, but force is. The force does not, in general, concentrate into the thinner tube - apply the conservation of linear momentum.

The amount of force used to accelerate the stream is dependent on the area of the stream. Which ultimately leads you to realize that it's the pressure that counts...the force applied to a certain amount of water.

What applies a force are the molecules that don't turn fast enough. Certainly, some do bump into the wall and create a force, but it's not as much as you make it out to be. There would be a lot of lost flow if it decreased it as much as you make it out to be.

What do you mean, "concentrate into the thinner tube." Again, the only thing I can think of is that you assume that the water is carrying a force. That is of course wrong. There is nothing in the water but flow and pressure.

And your tests don't show anything. There's no way that a water stream would push anything near how much force is applied simply due to water gun physics. That's like me saying when I throw a baseball, it'll hit as hard as I threw it. There are a lot of things that affect how much force is imparted. Much of that has to do with the fact that we're shooting a stream. The stream breaks up on impact and not all of it is hitting a scale at once. It mainly comes down to the weight of the water. Other things, such as drag, also affect imparted force. Those tests do not show that "force is reduced at the nozzle" as you like to say.

The real reason the situation works is that they're only looking for when there's no pressure difference. When there's no pressure difference, there's no net force, and vice versa. If you were measure applied force or pressure or something, then you'd really need the area in order to convert from one to another.

All you can really do is limit, or find the maximum for a system. Flow is constant throughout a system - you can't say there's X amount of flow here but X-Y amount of flow here because flow is lost. What you'd actually have is X-Y amount of flow at all locations in the system.

Nowhere in the page does it say that they're looking for when there is no pressure difference. They're finding the net force applied to the water to see how it flows. I don't what you are arguing here...

As I've already said, tube diameter at one place does not indicate total Cv for the system. For example, you used large tube sizes for a lot of the guns, but most of them have nozzles between 1/8" and 3/8".

Once again, there are too many controls to provide any real correlation. You've also rejected the pressure*bladder area method of determining the (varying) force in CPS systems, so we'd have to rule that out. At any rate, it would only be fair to plot a force graph.

I don't have time to plot either at the moment.

SuperCAP and SuperCannon II get similar ranges because of the range-velocity function and because of the point of diminishing returns when it comes down to flow coefficient. Speaking of which, how would the force model justify the similar ranges? The water guns have equal forces on the piston/water surface.

Again, you doubt the accuracy of my data when it doesn't support your theory. I think you know as well as I do that the main factor in flow coefficients is cross sectional area.

And the force model works perfectly with Supercannon II and SuperCAP. It appears that again you again are not paying attention to what I say or are forgetful. Both initially have similar flow, but SuperCAP has pipe turns increasing turbulence, slightly smaller diameter tubing reducing flow (but there still is well more than enough potential for flow), and a much worse nozzle (ejecting a turbulent stream). Consequently, it shoots slightly less far at the same pressure. And I have explained this to you before.

Pressure is still generated, even though nothing is compressed. The spring applies force over area. The pressure differential determines the flow through the nozzle.

And yes, regarding what you and DX were discussing right above this post (haven't gotten down to those replies yet), there would be worse performance if the same spring were pushing a larger piston. It's like your paragraph that says 500 pounds of force against a piston in a 4" tube is worse than 500 pounds against a piston in a 2" tube - because there's less pressure.

Okay, you again are not understanding what I am saying. I definitely know that pressure is created. But you seem to think the pressure, not the spring that is pushing, is pushing the water out. That's wrong.

And I didn't say that the 2 inch piston would perform better than the 4 inch one because of pressure. I did once say that higher pressures would be an indicator that one water gun will perform better at the same force. What I mainly was saying was that more water to move per distance is harder to move. 4 inch chambers have more volume per distance.

The pressure gauge will not register any gauge pressure if you close off the first valve. The closed off section would be an independent system - with no source of pressure. It's the air in the PC that stores the pressure.

However, if we had a system so well-built and precise that the liquid's compression could be maintained, then yes, there would still be pressure. There would be flow, but the pressure disappears nearly instantaneously, so there wouldn't be flow for long. Such a scenario is impractical with our materials. Maybe with a massive steel system at incredible pressures.

It will have pressure. I've done stuff like that before with Nerf guns. You yourself say that water carries a pressure. Isolating it does not remove that pressure. Try this out for yourself. I have no doubts about what you'll see.

Where did I say slightly? I punched it into googlebar (third-party, not from Google) and pressed highlight, but nothing came up in the section of my reply that you quoted.

The performance boost would be massive. However, if I indeed did say "slightly," I'd probably have said a system with a small tube and a single piston would probably perform slightly better than this. (Provided you can reach similar pressures in the smaller tube). One less piston means less friction.

You never said slightly. I wasn't quoting. I was just saying that by your argument, a water gun with notably higher pressure will peform more than slightly better than one with the same force but lower pressure.

Tell me how it proves the force theory. You're saying that the same force can achieve different performances.

To accelerate a certain volume of water, you need a certain amount of force. The volume depends on the area. So you need a certain amount of force for a certain amount of area. That's known as pressure.

They're achieve similar performances. They'll be within a certain range, as always.

And your logic in the second paragraph is faulty because pressure doesn't move anything. Pressure does not accelerate anything. Pressure can make things expand, but it doesn't in the case of water at our pressures.

The force is also making the movement in a water gun, yes. But pressure is more important because it essentially tells you how much force you are applying per unit mass of water. In this scenario, you could be applying force to a small clay brick or you could be applying the same amount of force to an enormous concrete pillar.

Now you're changing your argument again back to the "force is reduced because it is per unit area" thing you had going before.

And your example is one using huge differences. What about two pipes that have the same volume, and same pressure? One is one inch in diameter and one is four inches. Which will perform better?

The problem is I still don't see the connection between flow and force. That would be like me saying "F = Δp/Δt, so pressure is related to flow." Maybe you could use the equation in the derivation, but it doesn't prove any connection.

The equation in the book is fine. I just can't see how it could be applied to connect force and flow. If you do it, something could very well turn up that makes such a connection impossible.

Don't see the connection between flow and force? Do you pay attention to anything? Flow is movement. Force accelerates objects, or changes movement. Force can create movement. And have you read the extremely blunt PDF link I gave you? It explicitely states that force creates a flow.

Yes, force does generate motion as it is the rate of change of momentum. But it does not tell you how much mass there is (which you get from the area, volume, and density), and thus it does not tell you the resulting velocity.

They mention some interesting things, like a "blob of water" .

I've heard of the Navier-Stokes equations, but they're way above my head. I'm positive we won't even cover those in AP Physics next year, although I might do some advanced Physics class at UVA for AP credit in my senior year. I dunno.

Do you even know physics? Force, momentum, velocity, acceleration, and movement are all directly related. A force can and does tell you how much something else will be accelerated from Newton's second law, which can calculate the velocity it moves at and from there the flow. Navier-Stokes is from what I know the central equations of fluid dynamics and has everything you need to calculate the velocity of flow.

That and if more mass is involved, the system simply will accelerate slower, following Newton's second law. With more mass involves comes wider chambers, where a lower velocity is needed to achieve the same flow. So, technically the flows should be comparable. I do think there is some small difference over a large range of diameters however.

Though, I suppose you'll say anything to deny that force matters when creating flow.

I'm getting a little confused reading some of your new responses. You took your time posting and I think you were seriously considering much of what I said. Which equation is it? Navier-Stokes or the Cv one? One uses force, while the other uses pressure. You have not said that Navier-Stokes is wrong. But, you argue that in Navier-Stokes it actually is going to use pressure, essentially. However, that is wrong. Why would they use force if they were just going to calculate the pressure created by it?

Navier-Stokes uses partial differential equations and apparently is some of the hardest physics around. I doubt you will use it at any time during high school, even if going to college at the time. You'll need to take Calc I, II, and a differential equations class. Sometimes Calc III is required for differential equations, so you might as well take that as well. That'd be a pretty packed junior and senior year.

Of course, I'd rather not go into what a multivariate derivative is or any other details. Feel free to go more in depth with the flow and force though, although I probably won't understand it.

I won't go over your head. And if you quoted that because somehow it gave creedence to your theory, realize that just because something is proportional to another does not mean that that one thing controls everything about the other.

SilentGuy, I don't know why you keep arguing that pressure determines how a water gun performs. I know for a fact that it does not. I continually pull up highly authoritative sources that say that force creates flow, but you continually deny it matters or put some spin on it that "the force is reduced per unit area, so pressure matters." No, it isn't.

It's perfectly fine to be wrong SilentGuy. Believe me, you're just wasting your time arguing with me over this. The more I research it, the more sure I am of the force model. I can keep bringing up points in contention with what you post and you can keep ignoring them or putting some spin on them about "how the force is reduced" as you continually have. At this point, I believe nothing is being accomplished by my arguing with you and I will not respond to whatever you bring up. It seems that you are more concerned with justifying something than with changing your view.

Go and build your devices to test, and soon. Any day sooner you build would be a day sooner that you realize that smaller diameter water guns at the same pressure simply don't work as well. It's force that matters because of that. I have seen it in real life and I know it is true more than ever after doing some research to find sources to post for you.

As for your edits...

Technically, there actually is flow. The problem is as you say - the water does not expand much (because it did not compress much in the first place). Thus, the volumetric flow (volume/time) yields very little volume pushed out because the the time is so small. The compression of water isn't even noticeable here, so any tests we do wouldn't work.

Using air just allows for a way to preserve the pressure for a little while, and using low pressures at that. Same with a spring. A piston-powered gun still has pressure when used, but it cannot be stored with out crude systems.

There is flow. But, by your argument and your Cv equation, the flow should be THE SAME as if it were not isolated. Which is it?

Take a look my quote of one of Wikipedia's passages. Apparently the N-S equations are used to correlate pressure and flow.

Again, just because some things are proportional does not mean that that is the only variable that controls it. That is a huge error. And is that the best quote you can find? How is that more important than a blunt statement that force pushes water?

Duxburian, force is what gets the water moving, but it does not account for the area of the stream (and thus the volume and thus the mass) that is being accelerated. When you increase the area, there's more mass to be accelerated, so there's less acceleration. Pressure is constant no matter how much area there is - and my tests have shown that the force used to accelerate a stream depends on the area. In other words, the pressure is what counts because then it doesn't matter what how much water you're moving. Plain logic.

Newton's second law DOES account for the mass SilentGuy. Stop ignoring what it actually says. There's less acceleration, but for wider diameters it does not need as much velocity to achieve the same flow. It's just like the Q = V * A equations you posted before. You can differentiate it to find the relationship between dQ and acceleration (small a). There's less acceleration, but the flow should still be comparable.

And your tests are inconclusive as I stated several times.

[joannaardway]

I don't know how you will respond to this, and it's possible it's already been answered, but I experimented last night just to prove the matter to myself.

If I take Vertigo (fairly unique in it's way, but certainly not inefficent like the 3xA), and 3/4 fill the chambers with water, then bring the pressure up to 100 psi with a bike pump, fix the firing angle in a workmate I get 52 feet of range to the last puddle.

I then did the same thing with 1/2 full chambers, and used the same 100 psi pressure.
Now - The force is the same, as is the pressure, yet less water is being accelerated.
If force is truely responsible, then the acceleration is larger, and thus the range should be a little higher. Moving across to a bit of dry ground and trying again, I get the same 52 foot measurement.

I repeated again with 1/4 full chambers, but again, 52 feet.

Unless I'm totally misunderstanding, if force were responsible, should the range not change in the different circumstances?

[Ben]

From what I have experienced and read, I do not think that it's total volume of the system that matters, rather, it is more the cross sectional area. So I don't think the range should change. By that argument, air pressure water guns would be more close to constant flow because of the reduction in volume. But it doesn't work that way.

[Ben]

I'm not referencing just any old textbook here. I'm working from universally accepted flow equations.

As these equations are dependant on pressure, and are defined by water flow (not gas flow), I can only believe that pressure is the cause of water gun performance.

Using only known equations, Pressure can be translated to flow, flow can give stream velocity, and from that range can be found.

Find me highly authoritative examples of these equations being used to create flow from a pressure. You can't. The Cv stuff is not used for that. It's used solely to calculate flow lost and it doesn't work backwards. The pressure drop is proportional to the flow drop.

And Navier-Stokes is dependent on force. Not pressure. SilentGuy went off about how the force can be converted into a pressure, but then why doesn't the equation take pressure to begin with? Navier-Stokes is THE ESSENTIAL FLUID FLOW EQUATION. And you're saying the Cv ones are more important? I see no sense in that.



Pressurize the water gun with valve 1 open and valve 2 closed. Close valve 1. Fire the water gun with valve 2. Are you also saying that this situation will work the same as if valve 1 was not closed because it has a pressure difference? Water's density does not change much, so there won't even be flow due to expansion. With no applied force, there is no flow. This water gun will not work anywhere near where it would had valve 1 open.

SilentGuy tried to post a response to this, but his explanations made no sense. I would like to hear more about this situation from you two because I feel it is one that completely breaks the pressure theory. SilentGuy in fact changed his earlier arguments. Pressure is carried by a fluid as you both have stated. Once pressurized, if it is separated, it still is pressurized unless it is given the opportunity to equalize. A separation and sealing off does not equalize. And his other justification about how the pressure would drop too quickly is simply made up. Let's say we seal off a large area. Would that perform the same then? The answer is no because the applied force is creating the flow. With the force being isolated, no flow occurs.

Edit: I've taken a look at many of the equations and every single one I see is based upon dV, not the actual volume. dV can be thought of as a small cross section. No problem there.

Also, I will be building a 2 inch water cannon the same size as Supercannon II in the next few weeks and have side-by-side testing with videos. I will show that while the guns have the same pressure, same nozzle, same valve, and same everything else, the larger diameter increases the force applied and also increases the power. I have absolutely no doubts about this and I want to do it before SilentGuy does because I feel he might skew his results from his past interpretations of posted statistics. These videos will show undeniably that pressure does not determine a water gun's power. I know you pressure diehards want to believe pressure is the main determining factor, but I know for a fact it is not.

[Silence]

Sorry for not replying - I lost the one I had been working on through Monday, and I haven't rewritten it for a number of reasons - time constraints and laziness chief among them. I'll respond on Sunday or Monday if you don't mind - I've got a few band performances this weekend, including the Dogwood Parade (huge in Charlottesville) and the spring concert. My neighbors also got confirmed recently, and they're holding a party...

Yes, I did change my explanation for that problem. I'll reiterate what I said last.

Technically, there is an equal amount of flow as there would be had valve #1 been open. Problem is, water barely compresses, so the excess flows out in a fraction of a second. The flow stops nearly immediately, so the total amount of water displaced is unnoticeable.

Practically, it isn't feasible to obtain or measure flow in such a scenario - not with our crude PVC. The water's compression is minimal, and the systems are to imprecise to get any flow at all.

I glanced at a few of the Navier-Stokes equations, and none of the symbols stood for force. I can't recall if pressure was there, but I'll look again for my next response.

Saying Cv doesn't work backwards isn't qualified at all. The equations are used properly and it solves for something. You can use the force equation to find force or you can use it to find acceleration. Why can force work backwards, but not this?

I'll wait to see the results from your test. I don't appreciate your comments about bias though - I've explained why I've used the numbers I've used. Although it's true that I won't get around to building anything until the summer at least.

Reply to this or not, but I probably won't be able to do any reply longer than 15 minutes until my full response on Sunday or whenever.

[Ben]

Yes, I did change my explanation for that problem. I'll reiterate what I said last.

Technically, there is an equal amount of flow as there would be had valve #1 been open. Problem is, water barely compresses, so the excess flows out in a fraction of a second. The flow stops nearly immediately, so the total amount of water displaced is unnoticeable.

Practically, it isn't feasible to obtain or measure flow in such a scenario - not with our crude PVC. The water's compression is minimal, and the systems are to imprecise to get any flow at all.

This is not an adequate explanation as I said earlier. According to you two, a fluid under pressure always moves toward where there is less pressure. That is, I'm assuming, how your theory operates. The problem is that is not what is happening.

Your explanation makes no sense in compliance with your theories. In absense of the air, what are you lacking? The applied force that creates the pressure. Of course, when the other area is separate, it still is pressurized. Are you verifing the force model by stating little would occur? Of course you are. The applied force is what actually pushes the water out the nozzle.

By what you're saying after I brought this up, it would be as powerful, but for a fraction of a second. But what if the area was very large to see the full extent of the reduction? Would you say that wouldn't work? And be careful about what you say because if you say that it would be as powerful for a longer peroid than the smaller chamber, you're ignoring reality, and if you say it would be as powerful as the smaller chamber, you're simply verifying what I have been stating all along.

And why does the fact that the water is separated from it's pressure source matter in your theory? There is a pressure differential, which is all that is needed for your theory. Why won't it create significant flow? Because it lacks the applied force? That's the right answer, but you won't say it.

I glanced at a few of the Navier-Stokes equations, and none of the symbols stood for force. I can't recall if pressure was there, but I'll look again for my next response.

Again, this goes back to your flawed belief that if one equation does not have the variable you are looking for, it can not be used for that purpose. That is completely incorrect. From what I have seen and read, most of the Navier-Stokes equations are based upon momentum, which is directly related to the force. The PDF I linked to states there is a relationship between applied force and flow in an introduction to to Navier-Stokes. Want more proof of the relationship? Here's another great link: http://www.allstar.fiu.edu/aero/Flow2.htm

The other governing equation is the Momentum Equation, or Navier-Stokes Equation, and may be thought of as a "momentum balance." As will be seen later, the Navier-Stokes equations are the fluid dynamic equivalent of Newton’s second law, force equals mass times acceleration.

Again, this verifies exactly what I have been saying all along. And the best part is that this link is 1) on a .edu website and 2) partially funded by NASA. Want to argue with a university and NASA? I don't think so.

Another great thing about this link is that the Cv equation is nowhere to be found on a page completely about fluid flow. Are you saying that the Cv equation is right and Navier-Stokes is wrong? Or are you saying that both are right and Navier-Stokes actually uses pressure? Surely, these ones are more complicated, but once someone derives everything to follow a water gun-like situation, Navier-Stokes would be the way to go. And as I've said many times, the Cv equation is a simplification of what is actually going on and does not cover all of the variables.

Saying Cv doesn't work backwards isn't qualified at all. The equations are used properly and it solves for something. You can use the force equation to find force or you can use it to find acceleration. Why can force work backwards, but not this?

I've already explained how it does not work backwards. First, the equation is essentially used to calculate flow lost, not flow created. Second, pressure is proportional to the flow, so pressure drop is proportional to the flow loss. The flow loss restricts total flow. That has been stated many times by more people than myself. Remember the word "proportional" is used. That does not mean that pressure is the only factor. For that reason alone the equation can not be used backwards.

And you know for a fact that the Cv equation doesn't actually work the way you are using it. By your math, regardless of the PC diameter, if I had a wide diameter pipe going out to the nozzle, the water gun would work many times better than if there was a pipe that was smaller. You used this to explain why Supercannon II and SuperCAP perform well. That thought is missing what is actually occuring.

The Cv value that reduces flow the most is the nozzle. The pipe diameter and such really don't matter too much given that they are significant because the nozzle is doing most of the flow loss. Again, the word is flow loss, not "force loss" which has other errors I've already gone into great detail with.

I've already stated that I'm willing to put a long half inch pipe on Supercannon II to show exactly how much that reduces performance. You seem to think that would reduce performance very significantly, but it ignores the fact that the nozzle actually is one half inch in diameter anyway. Not much performance loss...

For that reason, let's say that the nozzle area is proportional to the Cv value. I had done this previously, but after you analyzed my data in more detail you didn't want to believe them. But, Cv values are largely determined by the diameter as this page shows (it is the first listed thing). By this logic, water guns at the same pressure will work similarly on the same diameter nozzles. That again is not true. Justify this if you can.

I'll wait to see the results from your test. I don't appreciate your comments about bias though - I've explained why I've used the numbers I've used. Although it's true that I won't get around to building anything until the summer at least.

Regardless of how you justify why you used those numbers, the numbers you used are not what was reported. Waterzooka said 50 - 55 feet. An estimation. Not a measured range. He didn't say "I shot the gun and measured the shots' distance at anywhere from 50 to 55 feet." Note how "square" the numbers are, all being divisible by five. It was an estimation, not a range of values. If you believe he was saying a range of values, I would highly suggest you ask him. This all makes me worried that you will skew the results or perhaps offer excuses as to why the performance was different or how the test was not representative (or something else I'm not creative enough to think of).

What I don't appreciate though is your ignoring many points I bring up. I take serious effort to find highly authoritative sources that say exactly what I do. You shrug off much of what I say and that is very discouraging. Before I had said that I am not willing to argue if you are not willing to compromise because nothing will be accomplished. I am only trying to tell you what I have seen to be the truth. And the more I research it, the more I am sure that force is the determining factor in water gun performance. This argument has only made me more sure of what I have been saying all along.

[joannaardway]

Originally Posted by Ben

Pressurize the water gun with valve 1 open and valve 2 closed. Close valve 1. Fire the water gun with valve 2. Are you also saying that this situation will work the same as if valve 1 was not closed because it has a pressure difference? Water's density does not change much, so there won't even be flow due to expansion. With no applied force, there is no flow. This water gun will not work anywhere near where it would had valve 1 open.
SilentGuy tried to post a response to this, but his explanations made no sense. I would like to hear more about this situation from you two because I feel it is one that completely breaks the pressure theory. SilentGuy in fact changed his earlier arguments. Pressure is carried by a fluid as you both have stated. Once pressurized, if it is separated, it still is pressurized unless it is given the opportunity to equalize. A separation and sealing off does not equalize. And his other justification about how the pressure would drop too quickly is simply made up. Let's say we seal off a large area. Would that perform the same then? The answer is no because the applied force is creating the flow. With the force being isolated, no flow occurs.

You know why this is. Water cannot expand (more than a few fractions of a fraction of a tiny bit of a percent of a sliver of a bit so small you couldn't see it anyway), so any pressure is immedately lost after valve 2 is opened.
Even if you had a 1000 gallon tank and tried this, the expansion available would be so incredibly small that there would be no visible result.

The fluid does move to the lower pressure, but it equalises under no result. But unless you're measuring it with a laser measurer accurate to the micron, you will see no result.

As water can not be compressed, it cannot flow from one location to the other without something to fill the space it has left. (Take a straw, fill it with water, and put your finger over the top. The water cannot flow out, as there is nothing to fill the space it would have to leave. Nature abhors a vacuum).
This thing that fills the space is either your LRT or air. As there is not anything to expand in your example, indeed there is no flow.

Ok, if you took the system into a vacuum, it could then work, as at exceptionally low pressures (or high temperatures for that matter), water can become vapour, which can then fill the space left.

In the end, this doesn't constitute a legitimate argument.

While we're here, in your mind, if instead of instead of loading your saoker with water you used a gas (and appropriate seperation), what would define flow here? Force again?
I think not - and I know all fluids follow the same flow laws.

Originally Posted by Ben

Also, I will be building a 2 inch water cannon the same size as Supercannon II in the next few weeks and have side-by-side testing with videos. I will show that while the guns have the same pressure, same nozzle, same valve, and same everything else, the larger diameter increases the force applied and also increases the power. I have absolutely no doubts about this and I want to do it before SilentGuy does because I feel he might skew his results from his past interpretations of posted statistics. These videos will show undeniably that pressure does not determine a water gun's power. I know you pressure diehards want to believe pressure is the main determining factor, but I know for a fact it is not.

I do not doubt that larger diameter cannons fire further. However, I also do not doubt that the cause for this is something other than force.

Please refer back to my 1/4, 1/2 and 3/4 full chamber experiments. You didn't really successfully counter that argument.

You said you didn't think it would happen. I want to know exactly WHY you don't think it will happen, and the raw physics to explain the situation.

You said (as way of part of your reply) we don't see water guns closer to constant flow. I thought about this:

Say you have a 1 litre chamber, the mass/pressure (For this case, read pressure as force as chamber diameter is constant) relates as such:

Fill/mass Pressure (rounded a bit)

0% 0 psig
50% 15 psig
66% 30 psig
75% 45 psig
80% 60 psig
83% 75 psig
85% 90 psig

As you can see, this is not a linear relationship anyway. The pressure (and thus the force) falls way faster than the fill and mass. So you'd still see a drop off. The question is, is water gun drop-off dampened by the loss of mass? It would seem not.

However, if you can prove that drop-off damping occurs in some form (and the curve you get fits with equations I can agree with), then the only (current) explanation would have to be force, and I will happily accept that.

However, unfortunately for you, explaining only why drop-off damping doesn't work won't be enough to persuade me.

So, in the end, I either want the physics of why it doesn't happen, or proof that it does.

[isoaker_com]

Reading some of the discussion going suggests lots of partial understandings but no complete picture.

Now, I can't offer a complete picture, but there are some things that do need clarifying.

If there is pressure, there is force. Pressure is force exerted over an area. If you compress water to 40PSI by compressing it alone (water can be compressed; just not very much), then offer an opening by switching a valve, it will flow out the opening initially just as fast as if it were being pressurized by compressed air. However, the duration of the flow would be remarkably short since it does not take a huge change in volume for the stored pressure in water to drop back to 0 (zero).

Decent references:
http://www.school-for-champions.com/...e/pressure.htm

http://www.physlink.com/Education/AskExperts/ae15.cfm

[Ben]

In the end, this doesn't constitute a legitimate argument.

While we're here, in your mind, if instead of instead of loading your saoker with water you used a gas (and appropriate seperation), what would define flow here? Force again?
I think not - and I know all fluids follow the same flow laws.

I think you're playing down this argument because you don't have an explanation for it in your model.

I know that the water is being pushed out. The pressure has little to do with how it is being pushed out. A push is a force. The water is under pressure and for that reason there is a differential. Pressure differentials don't necessarily mean flow as you two make it seem to be. Again, to reiterate for everyone, THERE IS A PRESSURE DIFFERENTIAL. Why then is this situation different? The entire space is full of fluid. You yourself tried to explain to me that pressure doesn't necessarily mean volume change. If your model uses pressure as it does, the volume change is irrelevant.

After all, all fluids follow the same flow laws. Why is water different? You said before that it won't expand, but by what you two have been saying I am under the impression that it will move simply because it is under pressure. By what you're saying, any fluid under pressure will automatically flow given the opportunity to. This raises the question: what is really making this water flow? The answer is the push (a force) from the air. Remove the push and there is no flow.

And there are situations where there is no pressure differential, but there still is flow. Take the piston water gun with no nozzle that I posted previously. Explain to me how that works under your theory. I hope that you won't say that the pressure is creating the flow. The entire situation comes down to the piston pushing the water out--as does most water gun situations.

As for your second question, I would use a fluid flow equation to find out how quickly the gas would leave. The gas isn't really pushing against anything aside from more gas (or air if it is ejecting into the air). For that reason, there is no applied force and it is not like a water gun. However, had this been something like a projectile weapon, I would use a force equation to calculate how the projectile will move.

Also, I rechecked every link SilentGuy posted from the Physics Forums. Previously I was pressed for time and didn't look very closely. Incredibly, I found that NOT A SINGLE ONE used the Cv equation to calculate flow created. In one thread, someone asked why the square root was used. In another thread, someone was calculating the Cv value from flow. Heck, in one thread, someone said "I will be the first to admit that I never heard of Cv until I got to my first job!" I challenge both of you to find any examples of anyone using the Cv equation to calculate flow created as the water gun scenario does. I know for a fact that that equation is not used for that purpose, and even going through every link anyone posted, I can not find anyone ever using this equation to calculate flow created!

You two are telling me that the Cv equation is used to calculate flow created. Again, I have not seen a single place aside from here use those equations. I don't see why Navier-Stokes is wrong given that it is the essential fluid flow equation. I have posted many highly authoritative websites that explicitely state the (1) force creates flow and (2) the Navier-Stokes equations are the ones used in these situations. Are these websites wrong? Unless NASA is wrong, I don't think so.

So, in the end, I either want the physics of why it doesn't happen, or proof that it does.

As I said previously, the equations I have seen use dV as opposed to the entire volume being pushed (in combination with the density to get the total mass). The two are very different, but are related. dV can be thought of as a small cross section. I will say that I believe more total volume is harder to accelerate, but I think the effect is minimal. There's no "proof" that can be made about this aside from what I have said. The equations I have seen use dV, not the total volume.

Take a look at the book scan I posted earlier (it uses dV with a line through the V to differential it from velocity). Take a look at Wikipedia's derivation of Navier-Stokes, which uses partial differential equations and a different symbol for volume (the ohm symbol). Wikipedia calls that the "bounding surface." Take a look at this link I linked to before as well: http://www.physicsforums.com/archive.../t-148977.html

Notice that Q_Goest (someone who posted in the threads SilentGuy posted) said "If the mass of the piston/liquid system (M2) is small compared to M1, we can neglect M2."

Q_Goest essentially is saying that given that the force (measured in the same units as mass) is much larger than the total mass of the water, the total mass of the water is negligible. For most water guns the force applied is much larger, especially for more powerful ones.

I'm sure I could find more examples if you need them. When it comes down to it, the total mass of the water is for the most part negligible.

[isoaker_com]

I know that the water is being pushed out. The pressure has little to do with how it is being pushed out. A push is a force. The water is under pressure and for that reason there is a differential. Pressure differentials don't necessarily mean flow as you two make it seem to be. Again, to reiterate for everyone, THERE IS A PRESSURE DIFFERENTIAL. Why then is this situation different? The entire space is full of fluid. You yourself tried to explain to me that pressure doesn't necessarily mean volume change. If your model uses pressure as it does, the volume change is irrelevant.

After all, all fluids follow the same flow laws. Why is water different? You said before that it won't expand, but by what you two have been saying I am under the impression that it will move simply because it is under pressure. By what you're saying, any fluid under pressure will automatically flow given the opportunity to. This raises the question: what is really making this water flow? The answer is the push (a force) from the air. Remove the push and there is no flow.

And there are situations where there is no pressure differential, but there still is flow. Take the piston water gun with no nozzle that I posted previously. Explain to me how that works under your theory. I hope that you won't say that the pressure is creating the flow. The entire situation comes down to the piston pushing the water out--as does most water gun situations.

Pressure is related to force. If there is pressure, there is force acting over an area. You are sort of correct in stating that without a pressure differential, there would be no flow. If one has water compressed alone to 40PSI and then a valve is opened to a chamber at the same pressure (leaving gravity and surface tension out of the picture for now), there would be no flow. If you push water with a piston, you are applying force, thus creating pressure, thus leading to a flow as long as there is space for the liquid to move to. As soon as you apply any force onto an object (solid, liquid, or gas), you are applying/creating pressure. Not my theory; this is physics theory.

Pressure, alone, does not mean volume change, but if something is pressurized, it suggests it is compressed (i.e. a smaller volume) than when it would be in a lower pressure situation.

http://en.wikipedia.org/wiki/Pressure

Just to complicate things, you can generate flow in a sealed pressure chamber containing air and water on the Earth simply by tilting the container and allowing gravity to act on the liquid. The force applied is gravity and the pressure that is resulting in flow is the force of gravity over the area occupied by the liquid. It is the pressure in the liquid pushing in all directions that causes water to flow towards the lower end of the PC; the force of gravity exerts strictly downwards, not perpendicular, yet water flows along the surface to the lowest point (i.e. moving in a direction perpendicular to the direction of the force applied onto it). Part of the movement is definitely in the same direction as the force applied (i.e. downwards), but part is sideways to the force. This shows how water converts force into pressure and the pressure differential allows water to flow to the lowest point it has access to, even if all contained within a single, sealed, pressure chamber.



Edit:

Cute reference for pressure vs flow rate:
http://mste.uiuc.edu/murphy/WaterTower/default.html

Another page:
http://www.keidel.com/resource/water/pressure.htm

[Ben]

Pressure is related to force. If there is pressure, there is force acting over an area. You are sort of correct in stating that without a pressure differential, there would be no flow. If one has water compressed alone to 40PSI and then a valve is opened to a chamber at the same pressure (leaving gravity and surface tension out of the picture for now), there would be no flow. If you push water with a piston, you are applying force, thus creating pressure, thus leading to a flow as long as there is space for the liquid to move to. As soon as you apply any force onto an object (solid, liquid, or gas), you are applying/creating pressure. Not my theory; this is physics theory.

Pressure, alone, does not mean volume change, but if something is pressurized, it suggests it is compressed (i.e. a smaller volume) than when it would be in a lower pressure situation.

Okay, I can see now that you have read few of my posts. I find that some people want to explain things that I know to me again annoying... to explain what I do know again...

I never said once that pressure differentials can not create flow or that pressure water not related to flow. Please read any post I made to see that. What I say is that the pressure differential is not alone what determines how much flow there is. Very different. Definitely, a pressure differential can create a force, which will push water out. That is what occurs in air pressure water guns.

However, flow can exist without a pressure differential as in the piston water gun with no nozzle. The pressure differential is not what is creating the flow (i.e. movement), rather, it is the force applied, which is the ability to accelerate mass. A fluid of course is a mass. The pressure differential is not necessary to create flow. The error in these interpretations is the thought that the pressure creates the flow. Newton's second law concerns movement and it only uses force. Basic physics, in fact, the most basic.

This entire debate can be broken down into basic physics. Flow is movement. Acceleration is a change in velocity, that is, movement. Force is the ability to accelerate a mass. Therefore, force determines flow, as stated many times in the links I have given. I don't know why people don't want to believe this.

And for the 10,000th time, I never said that pressure means volume change. Somethings are incompressible, that is, they do not change volume. Generally, gasses change in density and liquids do not. Water carries a pressure (i.e. a force per area), but it does not change significantly in density. I must have said this two or three times already in this thread. I wish people would read my posts.

Just to complicate things, you can generate flow in a sealed pressure chamber containing air and water on the Earth simply by tilting the container and allowing gravity to act on the liquid. The force applied is gravity and the pressure that is resulting in flow is the force of gravity over the area occupied by the liquid. It is the pressure in the liquid pushing in all directions that causes water to flow towards the lower end of the PC; the force of gravity exerts strictly downwards, not perpendicular, yet water flows along the surface to the lowest point (i.e. moving in a direction perpendicular to the direction of the force applied onto it). Part of the movement is definitely in the same direction as the force applied (i.e. downwards), but part is sideways to the force. This shows how water converts force into pressure and the pressure differential allows water to flow to the lowest point it has access to, even if all contained within a single, sealed, pressure chamber.

Again, it's not the pressure created that creates flow. The weight of the water is what is creating the flow. Certainly, the flow is proportional to the pressure, but it is not the sole determining factor. Old mining techniques used giant weights simply pushing down on the water to create flow. The water does not have to follow the direction the force is applied in as well. The interactions of each individual molecule makes fluid mechanics and dynamics differen than other physics. The wall of the pipe can apply a force in another direction, accelerating the water in the opposite direction.

The thought that it is the water that will equalize is also erroneous. Certainly, the water will equalize it's pressure. But as I noted before, a pressure differential along in some situations CAN NOT CREATE FLOW. Why is the situation with pressurized water exempt from the physics people have been arguing? There still is a differential, yet there will not be much flow. It is the air equalizing, applying a force, creating movement, which is flow, that powers water guns.

I again, am explaining things that I have explained several times.

You know, I'm looking at Wikipedia's page for pressure, and the word flow is only mentioned once. This indicates to me that pressure alone does not determine flow. However, the page on the Navier-Stokes equations mentions flow 42 times and force 11 times. I think there's something going on there.

[isoaker_com]

Force is intimately related to pressure. Force moves mass, but to say that there is no pressure occuring is untrue. I agree that force applied to a liquid results in flow, but it also results in applied pressure at the same time. As soon as a force is applied on any object, one is now talking about an application of pressure. Pressure is defined as force per unit area. A piston pushing on water is applying a force over the area of the piston head, thus is it applying a pressure.

Force and pressure should not be discussed as unrelated things as they are extremely related when it comes to physical objects. That is where my gripe about this discussion stems from. If you apply a force, you are applying a pressure. Pressure does not need to be stated in many equations if force is already present since the two are related; having both would double count (sort of) the energy being applied.

As for volume change, I agree that you have been stating that pressure does not mean volume change. What I am stating is that if a gas, liquid, or solid is pressurized, it occupies a smaller volume than it would were it not under pressure. The application of pressure, alone, does not lead to compression; it can also lead to motion. Whether something is compressed or moved by an applied pressure depends on whether it is fixed in location or whether it is free to move.

For the pressure and flow business, take a tall plastic bottle and poke a few equal sized holes along the side, one near the bottom, one half way up, and one near the top. Cover the holes with tape,fill the bottle with water, open the top cap, then take off all of the tape covering the holes. Which hole has more water flow coming out of it?

The external forces acting on the water are gravity and atmospheric pressure. However, within the column of water, there is hydrostatic pressure which depends on the height of the column of water. Water at the base comes out more forcefully (more flow) since water at the bottom of the bottle experiences more hydrostatic pressure than water near the top of the bottle that only has a little bit of water above it. Increased pressure leads to increased force. Increased force increases pressure on something. Talking as if you can have force on an object without applying pressure is misleading.

And please don't use word counts as a justification; otherwise, I might as well type force and pressure and flow a hundred times in my next post.

B)

Edit:
Force = mass x acceleration

mass = volume x density

Therefore,
Force = volume x density x acceleration

volume = area x height

Therefore
Force = area x height x density x acceleration

Force / area = height x density x acceleration

Pressure = Force / area

Therefore,

Pressure = height x density x acceleration

(this explains atmospheric pressure as well as hydrostatic pressure and can thus be converted back to Force as necessary)