Maximizing a Nozzle's Stream Velocity

[Silence]

The past few days, while I was on vacation, I organized my thoughts from the debate in the Waterzooka LR thread. I wanted to do some testing of water cannons first, but I just had to get this out. My conclusion: for any given nozzle, the only factor that will affect stream velocity, and hence the range, is FLOW.

Q = An * v
v = Q / An

Where Q is the flow, An is the area of the nozzle, and v is the velocity of the stream.

Because the area of the nozzle is constant (we're improving the velocity for a given nozzle), the velocity of the stream is directly proportional to the flow.

Now, there are a couple of ways to increase flow - maximizing the flow coefficient (or efficiency, if you will, of a section of tubing or a system) and increasing the pressure. Increasing the Cv generally involves large tubing. For the most part, we can relate the Cv to the cross-sectional area of tubing.

Cv = Q / sqrt(dP)
Q = Cv * sqrt(dP)
dP = delta P, or change in P = Pinlet - Poutlet
dP = Pinlet - 1 atm = Pgauge
Q = Cv * sqrt(Pgauge)

Where Cv is the flow coefficient, Q is the flow, Pinlet is the inlet pressure, Poutlet is the outlet pressure, and dP is delta pressure (or the gauge pressure, in this case).

The flow coefficient for the entire system, minus drag and all the other cumulative efficiency issues, is the same as the flow coefficient of the least efficient part. We can raise the flow coefficient for nearly all the parts of the system - except for the nozzle. We can't change the nozzle for the purposes of this article.

We can increase the tubing size and make the flow coefficient for the rest of the system tremendous, but the system's flow will still be limited by the smallest flow coefficient - that of the nozzle. Therefore, it is only practical to raise the flow coefficient of each part to be a bit higher than that of the nozzle, just in case.

Places like McMaster-Carr will let you calculate the flow coefficient for nozzles - as per the equations, divide the flow by the square root of the pressure differential. Alternatively, you can just make sure all the tubing has an ID slightly larger than the nozzle aperture.

Achieving sufficient efficiency for an entire system is generally easy enough for smaller or medium sized nozzles - which is why they get decent range. However, most water guns have insufficient flow for larger nozzles - which is why those nozzles have poor range. It's not a lack of pressure (SuperCannon II got mid-70 feet of range at only 100 PSI), but a lack of efficiency, that limits the flow.

SuperCannon II gets excellent range because it's the only gun that has enough efficiency as well as enough pressure to support a large nozzle. Its nozzle is 1/2" wide at the aperture - the rest of the gun comprises 3/2" and 4" tubing.

While there's a limit to the flow coefficient of a system with a certain nozzle, there's no limit to the pressures you can use to achieve high flow.

Which brings us to the eternal pressure vs. force debate...
The Third Force vs. Pressure Debate
I'm fairly confident that pressure is important in determining stream velocity, but that pure force at any place in the system is not important.

I. The formulas above
- None of the above formulas mention force. However, the flow formula does mention pressure.
- Because flow determines the stream velocity and range, pressure is important in determining performance.

II. My Stream Force Tests
Several days ago, I did a few tests on the Orca, with streams of different sizes. A scale showed me that the force used to accelerate a stream is likely proportional to the area of the stream, and that the pressure is constant. Thus:
- The pressure in a water gun, and the given area of whatever nozzle is being used, determines how much force is being used to accelerate the water. You can have a ton of force in a piston, but the amount used on the stream will be proportional to the mass of the water because of the proportional area. Thus, streams of all sizes will have similar acceleration and velocity. (Although that depends on how related flow coefficient and area are.)

III. Constant pressure system (CPS)
- CPS water guns have constant velocity and range.
- As per the name, CPS involves constant pressure. It was called a constant pressure system because the pressure, and not the force, is constant.
- As per science, CPS involves constant pressure. The rubber of a CPS bladder abides by Hooke's Law of elasticity, which is usually only applied to springs.

F = k * x

Where F is the force, k is the spring constant, and x is the deflection distance.

So the rubber applies more force as it is stretched more. Try it with a rubber band - you'll see what I mean.

The area of the rubber also increases as the bladder expands. The following calculations relate the force and area, disregarding lengthwise expansion.

x = dr, or delta r
dA = l * 2 * pi * dr, or delta A
dF = k * dr, or delta F
P = dF / dA = (k * dr) / (l * 2 * pi * dr) = k / (l * 2 * pi) = constant

Where x is the deflection distance, r is the radius, A is the area, l is the length, F is the force (delta Force... ), k is the spring constant, and P is the pressure.

So, disregarding the lengthwise expansion, the pressure is constant because k, l, 2, and pi are all constants. I'm sure a similar thing occurs with the lengthwise expansion.

- Velocity can not be related to force. With CPS water guns, the velocity is constant even though the force from the tubing increases. Note how I always have to refer to force "at the ____," because force is irrelevant to the stream's performance and it isn't even constant throughout a system.
- Velocity can be related to pressure. Both are constants here. And if the rest of the science here holds true, then velocity is related to pressure.

IV. McMaster-Carr's nozzle flow ratings
- All the nozzles in the McMaster-Carr online catalog have rated flows at specific pressures. Thus, flow is related to pressure, according to professionals.
- Because velocity is dependent on flow, and because flow is dependent on pressure, velocity must be dependent on pressure.

V. Future water cannon (WC) tests
Sometime, I'm going to create a PCgH/WC hybrid with a 2" chamber, a 1" barrel, and a 3/8" pump. I'll also build a 3" or 4" WC to compare the hybrid against. Both should have sufficient flow. I'll compare ranges or flow (output, essentially) at equivalent pressures. If the larger cannon has an advantage of less than 10%, it'll probably be because of a better flow coefficient overall due to slower velocities in larger chambers when inside the gun.

New stream theories related to all this - somehow
1. Smaller streams don't have more pressure than larger streams. Pressure throughout a system of incompressible fluid (ie, a liquid) is nearly constant. However, smaller streams will exerts less force when striking a target.

2. All streams will have *similar* velocities if the flow coefficients are suitable. The pressure is constant, so...

v1 / Cv1 = v2 / Cv2 = v3 / Cv3 = etc.

Smaller streams don't have lightning-fast velocities compared to the others - the gun's flow coefficient is good enough, medium streams can also be fast.

3. It's no longer surprising how on many water guns, the small and medium nozzles have similar ranges, but then you have large nozzles with limited range. At some point, you select a nozzle with a higher flow coefficient than that of a part of the actual water gun. After that, increasing nozzle size is going to lose velocity, as the flow coefficient is no longer dependent on the nozzle - it's constant and depends on the gun.

I'm looking for feedback and debates regarding this article...

Hmm, spent hours on the weekend planning the article, took 1-1.5 hours to type up! And I've trimmed it down a lot.

EDIT: Changed CODE tages to QUOTE tags, added more definitions for terms in the second set of formulas

[Ben]

You're making too numerous errors in this to allow it to work.

Yes, flow is what matters. I said it myself before anyone else did. But, flow is created by a force, not a pressure.

First off, dP does not equal gauge pressure. There is a very big difference. dP/dt is the derivative of the pressure and is where dP comes from. You throw around a lot of differentials in here, but you don't know how to use them at all. I'm not even going to comment further on how you don't use the math correctly. If you wanted to calculate the total flow with this equation for all of the pressure dropping to atmospheric pressure, I suppose you could use the pressure itself, but that's not really what's happening here. From what I have heard, this equation is most often used to calculate flow loss over a valve from the pressure loss over the valve. That's what the pressure differential is for... the pressure before the valve and after.

Second off, you know for a fact that you can create high flow with little pressure. I suppose you will ignore all empirical data to the contrary of your theory. Again, I'll repeat this because you are ignoring it: force is the ability to accelerate mass. Therefore, force can be thought of as the ability to create flow, or to accelerate a fluid in a pipe.

Just to show that this equation is not used to calculate how much flow is created from a certain pressure or force, I'll quote Wikipedia:

Originally Posted by Wikipedia

The use of the flow coefficient offers a standard method of comparing valve capacities and sizing valves for specific applications that is widely accepted by industry.

Again, this equation does not show how much flow is created, rather, how much flow is lost.

- None of the above formulas mention force. However, the flow formula does mention pressure.
- Because flow determines the stream velocity and range, pressure is important in determining performance.

Again, you are ignoring points I brought up earlier.

If you wanted to put force into these equations, you could without trouble. All you would have to do is differentiate P = F / A, reorganize the result, and stick it in for dP. Nothing hard. You and joanna seem to think that if the equation doesn't have what you're looking for in it already, it can't be used for what you are looking for. That is severely limited. You yourself plugged in other equations, which I find ironic considering you wrote this after I told you how to plug in force.

The fact that this equation is mainly for lost flow over a valve however doesn't prove that force has nothing to do with creating flow. Again, to reiterate, this equation mainly is used to show lost flow from lost pressure. Lost pressure is easily measurable, while lost flow is not. That is what this equation is used for.

Several days ago, I did a few tests on the Orca, with streams of different sizes. A scale showed me that the force used to accelerate a stream is likely proportional to the area of the stream, and that the pressure is constant. Thus:
- The pressure in a water gun, and the given area of whatever nozzle is being used, determines how much force is being used to accelerate the water. You can have a ton of force in a piston, but the amount used on the stream will be proportional to the mass of the water because of the proportional area. Thus, streams of all sizes will have similar acceleration and velocity. (Although that depends on how related flow coefficient and area are.)

What you were testing for was impact force. It depends mainly on how much water was being thrown (the output) and how fast it was going (the velocity). The nozzle does adjust those values.

However, the nozzle DOES NOT APPLY A FORCE. The nozzle changes the velocity and output. I don't know how many times I have to say what force is. Force is the ability to accelerate a mass. Force is only at parts of the system that are moving things into others. There is a flow, but it is not a push, a pull, or a lift, so it's not force.

- As per science, CPS involves constant pressure. The rubber of a CPS bladder abides by Hooke's Law of elasticity, which is usually only applied to springs.

Very wrong. The way the tubing is inflated is very different than is stretching a rubber bands.

What matters in force is movement. Force is the ability to accelerate mass (I will repeat this endlesslessly). In an inflated CPS chamber, there is only one direction the tubing moves. The tubing collapses in at a constant velocity.

The tubing does not expand simultanously in diameter and length as if the tube was simply enlarging itself. It expands in diameter first and then increases in length. The two situations are very different.

- All the nozzles in the McMaster-Carr online catalog have rated flows at specific pressures. Thus, flow is related to pressure, according to professionals.
- Because velocity is dependent on flow, and because flow is dependent on pressure, velocity must be dependent on pressure.

Those nozzles have flow ratings at GPM and the pressure to go along with it. The pressure is important because it affects the total mass going through the nozzle. It also affects how readily the nozzle/valve will fail. That's something you can't ignore.

What you've just said proves nothing. I find it unusual that you jump to conclusions so easily. McMaster-Carr simply wants people to avoid overstressing their nozzles. And besides, by the equation I posted in the LR thread, flow and force are best buddies.

Sometime, I'm going to create a PCgH/WC hybrid with a 2" chamber, a 1" barrel, and a 3/8" pump. I'll also build a 3" or 4" WC to compare the hybrid against. Both should have sufficient flow. I'll compare ranges or flow (output, essentially) at equivalent pressures. If the larger cannon has an advantage of less than 10%, it'll probably be because of a better flow coefficient overall due to slower velocities in larger chambers when inside the gun.

Good luck with this. I already know the results because I've proved it empirically. I'm quite confident that the 3 inch cannon will shoot at least 5 - 10 feet less than the 4 inch cannon at the same pressure. You can't ignore nearly twice the force. As the only one who has experienced larger diameter water guns, I know what sort of a difference it makes in the force applied.

A normal 3 inch APH should perform similarly to your 3 inch cannon for your information, even factoring in turbulence. Those designs do not even begin to compare to the 4 inch cannon, or even larger ones such as 6 inch.

And flow doesn't matter as much as it will benefit each water gun. What matters mainly is having the potential for sufficient flow. Something with a two inch diameter has way more than enough potential to support a half inch nozzle. I think you're just looking to justify something in advance in case your data doesn't support it.

If you're so confident, definitely make a two inch cannon as well. By what you've been saying, you seem to think that one will perform the best overall. In reality however, it'd be a sorry sight to see.

1. Smaller streams don't have more pressure than larger streams. Pressure throughout a system of incompressible fluid (ie, a liquid) is nearly constant. However, smaller streams will exerts less force when striking a target.

You're ignoring something fundamental here. What matters for cutting (pressure) is higher with a smaller stream due to drop in applied area. So, smaller streams impact with more pressure.

Also, from what I have read the pressure does increase in the nozzle. Saying the pressure is uniform is actually simplifying things greatly. The pressure is relatively uniform and can be treated as uniform in most calculations.

3. It's no longer surprising how on many water guns, the small and medium nozzles have similar ranges, but then you have large nozzles with limited range. At some point, you select a nozzle with a higher flow coefficient than that of a part of the actual water gun. After that, increasing nozzle size is going to lose velocity, as the flow coefficient is no longer dependent on the nozzle - it's constant and depends on the gun.

I don't think anyone before has denied nozzles that are too large can not be supported by water guns that can't create enough output (a synonym for flow). This isn't a new idea really.

I like that you're thinking, but you're making many errors and ignoring what has been learned. I know you probably won't listen to me, so go ahead and build your cannons and test them out. I already know what the results will be.
Just because you seem to like throwing equations around, here's the derivation to flow from force:

F = m * a

m = p * Q
p is density. For water, p is about 1 at most temperatures (let's not get too technical though).

F = p * Q * a
Drop p because it equals one.

Q = A * v

F = A * v * a

a = dv/dt

F = A * v * dv/dt

To use this, you'll have to do at least some nasty differential equation work that I don't care for. But, there's the relationship between applied force and flow. There are absolutely no errors in the math, unlike your application of an equation to calculate flow lost over an orifice. This in fact uses the most basic equations of physics to get the job done.

Interestingly enough, what this does show is that the velocity of the fluid (not the flow rate) is proportional to the pressure. F / A = P, so P = v * dv/dt. Very interesting. I thought they would at least be close since both rely on the area.

[Silence]

Originally Posted by Ben

Yes, flow is what matters. I said it myself before anyone else did. But, flow is created by a force, not a pressure.

Then please tell me why force is mentioned nowhere in the equations. The formulas say the flow is directly proportional to the square of the pressure differential - not anything about force.

Originally Posted by Ben

First off, dP does not equal gauge pressure. There is a very big difference. dP/dt is the derivative of the pressure and is where dP comes from. You throw around a lot of differentials in here, but you don't know how to use them at all. I'm not even going to comment further on how you don't use the math correctly.* If you wanted to calculate the total flow with this equation for all of the pressure dropping to atmospheric pressure, I suppose you could use the pressure itself, but that's not really what's happening here.*

You seem to have misinterpreted the formula. Yes, you'd be right if the delta pressure was the difference in pressure between the beginning of the shot and the end of the shot. But it's not.

I clearly said the delta pressure is the differential between the inlet and the outlet of a valve or a section of tubing. The inlet for a nozzle is the barrel, and its pressure is whatever you've got the gun pressurized to. The outlet is in the atmosphere, and the pressure is 1 atm. Gauge pressure is pressure over 1 atm, or the difference between a pressure and 1 atm. Thus, dP, or the difference between the gun's pressure atmospheric pressure, is the same as the gauge pressure.

Originally Posted by Ben

From what I have heard, this equation is most often used to calculate flow loss over a valve from the pressure loss over the valve. That's what the pressure differential is for... the pressure before the valve and after.

Yes, you could find the flow as a function of pressure and use the flow to model the drop in pressure from a chamber...but that's not what I'm doing. I'm finding the flow at a fixed, maximum value for the pressure. It's really not as complicated as you make it sound.

Originally Posted by Ben

Second off, you know for a fact that you can create high flow with little pressure. I suppose you will ignore all empirical data to the contrary of your theory.*

I stated two ways to increase the flow - increase the Cv and increase the pressure. It is very difficult to increase the Cv past a certain point, as it is limited by the nozzle, which is the control. Pressure, however, is simply another way to increase the flow.

Please tell me what the empirical data that I'm allegedly ignoring is. The only thing I can think of is SuperCannon II, which I brought up anyway to prove my point. SuperCannon II uses 100 PSI, higher than nearly all conventional water guns, and has massive tubing for increased Cv. It is indeed the best example I have of how increased flow - through both Cv and pressure - leads to increased range.

Originally Posted by Wikipedia

The use of the flow coefficient offers a standard method of comparing valve capacities and sizing valves for specific applications that is widely accepted by industry.

Originally Posted by Ben

Again, this equation does not show how much flow is created, rather, how much flow is lost.

Yes, it does. The equation, once again is:

Cv = Q / sqrt(dP)

Where Cv is the flow coefficient, Q is the flow, and dP is delta pressure.

Which can be rewritten to:

Q = Cv * sqrt(dP)

Where Cv is the flow coefficient, Q is the flow, and dP is delta pressure.

Just because the equation is often used to calculate the Cv doesn't mean you can't use a known Cv and the pressure to calculate the force. The equation still holds true.

Originally Posted by Ben

If you wanted to put force into these equations, you could without trouble. All you would have to do is differentiate P = F / A, reorganize the result, and stick it in for dP.

If that were the case, why don't people use force in the equations? Plugging in F/A for P leaves you with the variable for area to deal with as well.

I don't see why you would plug in F/A. Pressure solves the equation perfectly well - just because you could substitute force and area doesn't mean you have to. And in any case, the F/A term would still indicate pressure because of the relationship between force and area.

However, if you do the differentiation I might be able to see where you're going with this.

Originally Posted by Ben

The fact that this equation is mainly for lost flow over a valve however doesn't prove that force has nothing to do with creating flow. Again, to reiterate, this equation mainly is used to show lost flow from lost pressure. Lost pressure is easily measurable, while lost flow is not. That is what this equation is used for.

Once again, to be frank, I couldn't care less how this equation is mainly used. I'm using it in a perfectly correct context. You yourself have said that physics is all about manipulating and combining different formulas.

Originally Posted by Ben

However, the nozzle DOES NOT APPLY A FORCE. The nozzle changes the velocity and output. I don't know how many times I have to say what force is. Force is the ability to accelerate a mass. Force is only at parts of the system that are moving things into others. There is a flow, but it is not a push, a pull, or a lift, so it's not force.

The nozzle changes the mass of water being shot in a certain amount of time because it changes the area of the stream. Thus, the nozzle changes the amount of force used to accelerate the stream - because the force is a second value dependent on the area.

The nozzle does not change the velocity because the amount of force used is proportional to the mass of part of the stream. My tests demonstrated this fact by measuring the force. The force measured is the same as the force used to accelerate the stream, according to the law of conservation of momentum. Momentum is related to the force, as offered by this proof.

p = m * v
F = dp / dt
F = d(m * v) / dt
The mass is constant, therefore
F = m * dv / dt
F = m * a

Where p is momentum, m is mass, v is velocity, F is force, dp is delta momentum, dt is delta time, dv is delta velocity, and a is acceleration.

Force really can't be used to define flow because it depends on the area of the nozzle and stream. Pressure does not depend on the area of the nozzle and stream.

Originally Posted by Ben

Very wrong. The way the tubing is inflated is very different than is stretching a rubber bands.*

Ah, but I did not say the inflation is exactly like stretching a rubber band. All I said is that the relation between force and deflection can be experienced by stretching a rubber band.

The force that CPS rubber applies increases as you stretch it. However, the range is the same.

Originally Posted by Ben

What matters in force is movement. Force is the ability to accelerate mass (I will repeat this endlesslessly). In an inflated CPS chamber, there is only one direction the tubing moves. The tubing collapses in at a [nearly?] constant velocity.

The tubing does not expand simultanously in diameter and length as if the tube was simply enlarging itself. It expands in diameter first and then increases in length. The two situations are very different.

There's still more force applied by the rubber after the tubing has expanded lengthwise. The more you stretch the rubber, the more force it applies. It's the same deal. The area also increases as the rubber expands, both widthwise and then lengthwise. However, the pressure and the range are constant.

Originally Posted by Ben

Those nozzles have flow ratings at GPM and the pressure to go along with it. The pressure is important because it affects the total mass going through the nozzle.

The rest of the information McMaster-Carr presents is constant. It's implied that the flow is based on the pressure. I'll send them an email if you want it see it spelled out.

Originally Posted by Ben

[Pressure] also affects how readily the nozzle/valve will fail. That's something you can't ignore.

What you've just said proves nothing. I find it unusual that you jump to conclusions so easily. McMaster-Carr simply wants people to avoid overstressing their nozzles.

It's true that more pressure can be bad. You don't want to increase it too much. However, the fact is that increasing pressure is going to increase the flow and thus increase the range.

Once again, though, I'll send them an email if necessary. But even if they showed performance at lower pressures for safety reasons, that would still not disprove the pressure and flow correlation.

Originally Posted by Ben

And besides, by the equation I posted in the LR thread, flow and force are best buddies.*

Hmm, just saw your derivation below. I don't think it works though, see my comments below.

Originally Posted by Ben

Good luck with this. I already know the results because I've proved it empirically. I'm quite confident that the 3 inch cannon will shoot at least 5 - 10 feet less than the 4 inch cannon at the same pressure. You can't ignore nearly twice the force. As the only one who has experienced larger diameter water guns, I know what sort of a difference it makes in the force applied.

I'll respect your experience. However, SuperCannon II admittedly has more Cv and more pressure than any other water gun we have access to. The cannon also has not been directly compared with another gun with similar values but smaller force in the chamber, thus, your results are not necessarily conclusive.

Originally Posted by Ben

A normal 3 inch APH should perform similarly to your 3 inch cannon for your information, even factoring in turbulence. Those designs do not even begin to compare to the 4 inch cannon, or even larger ones such as 6 inch.

I would think otherwise. An APH can't generate pressures high enough, an APH has chokepoints in the small tubing diameter and turns, and most APHs use low-quality nozzles with low Cv for their area.

Originally Posted by Ben

And flow doesn't matter as much as it will benefit each water gun. What matters mainly is having the potential for sufficient flow. Something with a two inch diameter has way more than enough potential to support a half inch nozzle.

Hence my explanation of the Cv. 2" tubing has relatively high Cv, but the gun's Cv is still limited by the 1/2" nozzle. Using 3/4" tubing might limit the Cv more. However, the difference between 2" tubing and 4" tubing is probably negligible.

Originally Posted by Ben

I think you're just looking to justify something in advance in case your data doesn't support it.

I'm actually fairly confident the data will support my theory. Actually, we don't have any data so far that points to my theory being incorrect. I've used all the experiments we know of to support the theory of flow, Cv, and pressure being more important than force.

Part of the reason I decided to go ahead and post this is (1) I just wanted to, after spending hours planning and typing it (this response itself has taken over 1.5 hours so far ), and (2) I don't know when I'll be able to build the cannon. I already want to build a CPH, but I frankly don't know when my parents will let me build two other water guns in addition, especially if using parts from McMaster-Carr. It'll have to wait until summer, I'm afraid.

Originally Posted by Ben

You're ignoring something fundamental here. What matters for cutting (pressure) is higher with a smaller stream due to drop in applied area. So, smaller streams impact with more pressure.

That's exactly the opposite of what my tests have shown. If different streams exert force proportional to their area, then they must have equal pressure. (Which makes sense, as the pressure is constant throughout the system, whereas force isn't.) Also, my test in the sand, which was mostly inconclusive, still showed 6" of depth for the large stream's hole and at least a few inches of depth for the small stream. I couldn't see the complete depth of the small stream as the hole caved in, but if it had equal pressure, it'd also hit a similar depth. If it had more depth, as you say, it'll hit greater depth.

The tests appeared to confirm the others' findings. Pressure is constant, force exerted by the streams is not.

On further thought, I'll try the sand test with the medium-sized nozzle(s) and the large one. Maybe the medium-sized holes won't cave in.

Originally Posted by Ben

I don't think anyone before has denied nozzles that are too large can not be supported by water guns that can't create enough output (a synonym for flow). This isn't a new idea really.

I should probably clarify that. Another way to look at it is as follows:

As you increase nozzle size, the range increases because the circumference-to-area ratio, and the effect of drag, decreases, and Cv-to-area increases. It probably has less to do with the drag on the actual stream in the air.

[EDIT=7:43 PM, SSC time]Of course, air resistance does have some role. XN's mod article that Duxburian reposted shows this to some extent. Nozzle lamination is very important as well as size, though.[/EDIT]

However, after some point, the Cv is limited by some other part of the gun. Flow no longer increases as quickly as or more quickly than nozzle area increases. After that point, the velocity decreases instead of being constant or increasing.

The old theory just suggested a bell-curve, with little explanation of why the range increases slowly and then decreases quickly.

Originally Posted by Ben

F = m * a

m = p * Q
p is density. For water, p is about 1 at most temperatures (let's not get too technical though).

F = p * Q * a
Drop p because it equals one.

Q = A * v

F = A * v * a

a = dv/dt

F = A * v * dv/dt

(In case people are confused, note that density's actual designation is the Greek letter ρ, or rho, which looks like p.)

Thanks for showing the derivation, but what strikes me as incorrect is the substitution of mass flow rate for mass. Mass flow rate is new to me, so thanks for the link, but I suppose it's like volume flow rate, but done by mass.

Anyway, I doubt you can substitute units of mass flow rate for units of mass in the force formula. Perhaps the confusion between the symbol for mass flow rate and the "m" for mass led to that? Or is it correct, somehow? At any rate, I wouldn't think you can substitute ρ*Q for mass.

I currently am skeptical of that derivation simply because it appears to confuse mass and mass flow rate due to the symbols. But perhaps I'm wrong, maybe they are interchangeable sometimes...

Also, just wondering, but I don't think you're allowed to "drop ρ because it equals one." Yes, water has a density of 1 g/cm^3, but (1) the units still matter; (2) so far, we haven't used metric units to any extent anyway; (3) even with metric, we could use other units; and (4), water only has a density of 1 g/cm^3 at 4 degrees Celsius (from what I remember, at least. I believe that's the temperature at which is is least dense).

Originally Posted by Ben

To use this, you'll have to do at least some nasty differential equation work that I don't care for. But, there's the relationship between applied force and flow. There are absolutely no errors in the math, unlike your application of an equation to calculate flow lost over an orifice. This in fact uses the most basic equations of physics to get the job done.

See both my response to your critique of my math and my critique of your math above.

I haven't taken physics yet, but it sounds fun. I'm actually learning quite a bit right now.

Originally Posted by Ben

Interestingly enough, what this does show is that the velocity of the fluid (not the flow rate) is proportional to the pressure. F / A = P, so P = v * dv/dt. Very interesting. I thought they would at least be close since both rely on the area.

First of all, I believe the math to be wrong. But even if it wasn't, the modified flow coefficient equation along with the flow equation tells us that only the square of the pressure differential is proportional to the velocity.

All in all, you haven't actually disproved anything I've said. None of the points are valid, as they all incorrectly reinterpret what was said.

*Also, every time you've called me out, you haven't offered an alternative explanation to qualify what you've said. The quotes are littered with asterisks I've added to demonstrate this issue.

[Ben]

SilentGuy, I looked for the equation you are using here in my fluid dynamics book. I didn't find it until page 400 or so. The first chapter on fluid flow never mentioned it once. It did however mention force on the second page of the chapter, defining the flow by the momentum (which is the integral of the force). I can scan in this page when my scanner works.

You might have the same book. I have an old 80s copy of Fox & McDonald's "Introduction to Fluid Dynamics" which my father referred to as the classic text on the subject. It's on page 114 of chapter 4.

I clearly said the delta pressure is the differential between the inlet and the outlet of a valve or a section of tubing. The inlet for a nozzle is the barrel, and its pressure is whatever you've got the gun pressurized to. The outlet is in the atmosphere, and the pressure is 1 atm. Gauge pressure is pressure over 1 atm, or the difference between a pressure and 1 atm. Thus, dP, or the difference between the gun's pressure atmospheric pressure, is the same as the gauge pressure.

I see now that my concerns come more from how well versed I am in calculus than anything else. Delta (the pyramid symbol) P and dP are different. Yes, for the most part, they can be interchanged and often are mainly by those without much experience in calculus. However, dP generally is a much smaller value than delta P would be. Delta P implies simply the change. dP implies a small, in fact very minute change. More often than not numbers are not put in a differential. The differential is the basis of calculus. A formula that has a differential in it can be approximated by treating the differentials as changes, but they can not be thought of as accurate in many cases. Perhaps I should not assume that you realize that using a differential as a change is inaccurate.

My other real problem with this is that it is meant to be used in pipe systems. I do not think it applies outside of a pipe system.

Please tell me what the empirical data that I'm allegedly ignoring is. The only thing I can think of is SuperCannon II, which I brought up anyway to prove my point. SuperCannon II uses 100 PSI, higher than nearly all conventional water guns, and has massive tubing for increased Cv. It is indeed the best example I have of how increased flow - through both Cv and pressure - leads to increased range.

Reread the threads on my most recent homemade water guns. I achieved 65 feet of range at 40 PSI with Supercannon II. Similar range (definitely over 60 feet from the chalk marks, but not measured) was achieved from SuperCAP at 40 PSI as well. SuperCAP had a 3/8 inch nozzle with 1 inch pipe ID compared to Supercannon II's 1 1/2 inch ID and 1/2 inch conical nozzle. I do not think you will stretch the truth enough to say that SuperCAP is as efficient as Supercannon II. What mattered was the force, and extremely high force could not be dampered much by inefficiencies.

Justify this. There is no way you can with your model. 40 PSI is as much pressure as most smaller water guns. And SuperCAP is actually less efficient than any APH style water gun is. By your model, SuperCAP should perform like crap. Reality is very different however.

[quote]The nozzle changes the mass of water being shot in a certain amount of time because it changes the area of the stream. Thus, the nozzle changes the amount of force used to accelerate the stream - because the force is a second value dependent on the area.

The nozzle does not change the velocity because the amount of force used is proportional to the mass of part of the stream. My tests demonstrated this fact by measuring the force. The force measured is the same as the force used to accelerate the stream, according to the law of conservation of momentum. Momentum is related to the force, as offered by this proof.

Please don't tell me things I already know. I appreciate that you are trying to explain things, but I know what momentum is.

Force really can't be used to define flow because it depends on the area of the nozzle and stream. Pressure does not depend on the area of the nozzle and stream.

How is force affected by the area of the nozzle and the stream? Explain.

There's still more force applied by the rubber after the tubing has expanded lengthwise. The more you stretch the rubber, the more force it applies. It's the same deal. The area also increases as the rubber expands, both widthwise and then lengthwise. However, the pressure and the range are constant.

You have to break up the rubber into tiny parts and integrate to get an equation for it's force. We actually derived the equation for the force of a "perfect" spring in class that way.

Only a certain part of the rubber is moving. That applies a limited amount of force. In fact, the tubing we often use has a very high spring constant, yet it only achieves limited range. Why is that? Because only a small part of the tubing is creating a flow at a time.

What matters again is NET FORCE. Not total force. NET FORCE. The fact that certain parts of the tube are not releasing their energy simply shows that not every part is contibuting to the flow. Just because it is stretched does not mean that it is contributing. Saying this system's force increases like a rubber band is a very big error. A rubber band involves the entire system and thus the force increases.

The rest of the information McMaster-Carr presents is constant. It's implied that the flow is based on the pressure. I'll send them an email if you want it see it spelled out.

I don't know which nozzles you are looking at, but the only ones I've seen say things like "gpm @ 100 psi." That doesn't mean force doesn't determine flow. They're giving the total flow there. The GPM is the volume and the PSI is pressure. I don't know what you're talking about, but that sounds like flow to me.

The people behind the scenes there know little about the physics and more about the products also. I have emailed them before.

That's exactly the opposite of what my tests have shown. If different streams exert force proportional to their area, then they must have equal pressure. (Which makes sense, as the pressure is constant throughout the system, whereas force isn't.) Also, my test in the sand, which was mostly inconclusive, still showed 6" of depth for the large stream's hole and at least a few inches of depth for the small stream. I couldn't see the complete depth of the small stream as the hole caved in, but if it had equal pressure, it'd also hit a similar depth. If it had more depth, as you say, it'll hit greater depth.

More total water penetrates deeper. However, smaller streams have more cutting ability. That's how water jet cutters work. They have more pressure. Don't deny reality. A water jet cutter would not work with a larger nozzle.

Anyway, I doubt you can substitute units of mass flow rate for units of mass in the force formula. Perhaps the confusion between the symbol for mass flow rate and the "m" for mass led to that? Or is it correct, somehow? At any rate, I wouldn't think you can substitute ρ*Q for mass.

They call it m with a dot over it for a reason. However, I'm afraid that it probably was not interchangable with mass because it involves time. I am not sure though because there are things such as moment of inertia which can be substituted for mass that are not mass. It definitely is on the right track though.

With that being said, there are errors in the physics, but not in the math.

Also, just wondering, but I don't think you're allowed to "drop ρ because it equals one." Yes, water has a density of 1 g/cm^3, but (1) the units still matter; (2) so far, we haven't used metric units to any extent anyway; (3) even with metric, we could use other units; and (4), water only has a density of 1 g/cm^3 at 4 degrees Celsius (from what I remember, at least. I believe that's the temperature at which is is least dense).

We did stuff like this all the time in physics to keep things simple. The density of water does not change much in our range of temperatures. I mentioned that, but I suppose you like getting technical and complicated.

More to write tomorrow... SilentGuy, get to testing soon. You'll agree with me very quickly. I have absolutely no doubts because I have tested myself.

[Silence]

The only fluid dynamics book I've checked so far is my dad's copy of Physical Fluid Dynamics, by Tritton. It doesn't have much. But I might recall Intro to Fluid Dynamics, I'll check around the house. Thanks.

I probably should have imported the delta symbol from Word or OO. Sorry for the confusion. I see where you're coming from now.

Originally Posted by Ben

My other real problem with this is that it is meant to be used in pipe systems. I do not think it applies outside of a pipe system.

In other words, are you suggesting that the outlet has to be in a closed system, as opposed to the atmosphere? Interesting. I'm thinking it's the drop in pressure that matters more, as the pressure is equalized and the outlet pressure is one atm.

Originally Posted by Ben

Reread the threads on my most recent homemade water guns. I achieved 65 feet of range at 40 PSI with Supercannon II. Similar range (definitely over 60 feet from the chalk marks, but not measured) was achieved from SuperCAP at 40 PSI as well. SuperCAP had a 3/8 inch nozzle with 1 inch pipe ID compared to Supercannon II's 1 1/2 inch ID and 1/2 inch conical nozzle. I do not think you will stretch the truth enough to say that SuperCAP is as efficient as Supercannon II. What mattered was the force, and extremely high force could not be dampered much by inefficiencies.

Justify this. There is no way you can with your model. 40 PSI is as much pressure as most smaller water guns. And SuperCAP is actually less efficient than any APH style water gun is. By your model, SuperCAP should perform like crap. Reality is very different however.

Good, I'd have sworn I'd been hallucinating had you not brought up these figures. The only figure I found was 100 PSI, 1250 pounds of force, and 73 feet of range early in the SuperCannon II thread. These figures are what I remember from the earlier debates.

Could you qualify SuperCAP's poor efficiency? 1" bore ID sounds pretty decent, although you might have to make that 3/4" to factor in the hose barbs (correct me if I'm wrong). The thread also says SuperCAP Mk.2 got a little over 60 feet with an endcap nozzle, perhaps more with a conical nozzle.

I was under the impression that your homemades also got 65 feet, and I'm certain they got 60 feet. That's probably with at least 75 PSI though (tell me if it's more, which it probably is, according to the Boyle's law and pump volume calculations).

So let me try the calculations, assuming the flow was equal for both a regular homemade and SuperCAP. The homemade is 1, SuperCAP is 2.

Q = Cv1 * sqrt(75-15) = Cv1 * sqrt(60) =~ 7.75 * Cv1
Q = Cv2 * sqrt(40-15) = Cv2 * sqrt(25) =~ 5.00 * Cv2
7.75 * Cv1 = 5.00 * Cv2
7.75 / 5.00 = 1.55

In other words, as is obvious, SuperCAP only needed to have 50% more Cv than an APH or CPH. That's mighty impressive. It does tell me, though, that Cv is much larger for a system than the smallest Cv of any part, due to cumulative drag and the like. Both guns have equal nozzles, essentially, but SuperCAP has slightly larger tubing.

In the past, there's been no focus on the square root of the pressure. The square root seems to definitely even things out between extreme pressure differences when comparing two guns.

That's all assuming we're not talking gauge pressure, and it made some assumptions regarding the homemade gun's pressure. Here, I'll use the same stats but I'll assume gauge pressure.

Q = Cv1 * sqrt(75) =~ 8.66 * Cv1
Q = Cv2 * sqrt(40) =~ 6.32 * Cv2
8.66* Cv1 = 6.32 * Cv2
8.66 / 6.32 = 1.37

That's an even smaller Cv difference than what I calculated earlier. But clearly, pressure isn't even as much of a factor as Cv is. I might now scrap my plans for a CPH with a small tubing ID.

The model works, or so it seems. I didn't even expect the square root to be so important.

Originally Posted by Ben

Please don't tell me things I already know. I appreciate that you are trying to explain things, but I know what momentum is.

Sorry. Just trying to make connections, for myself as much as for the next person.

Originally Posted by Ben

How is force affected by the area of the nozzle and the stream? Explain.

Force is pressure times area. Increase the area, and you increase the force. Now I know you'll tell me that the nozzle exerts no force, so I'll word it a different way.

Given an endcap nozzle, force is being exerted over the entire face. However, only the force acting through the aperture is used to accelerate water, translating into work. Increase the nozzle area, and more of the force accelerates more water.

Force in the chamber is not affected by the nozzle area. But picture a similar situation in the reducer. Most of the force is simply being exerted on the wall of the reducer, but no work is being done. Only a small amount of force makes its way into the barrel, and an even smaller amount is used to accelerate the water in the stream.

Originally Posted by Ben

What matters again is NET FORCE. Not total force. NET FORCE. The fact that certain parts of the tube are not releasing their energy simply shows that not every part is contibuting to the flow. Just because it is stretched does not mean that it is contributing. Saying this system's force increases like a rubber band is a very big error. A rubber band involves the entire system and thus the force increases.

Interesting theory. However, this goes back to the force vs. work issue.

Not all the parts of the tube are releasing their energy, but the are still exerting force. Take a rubber band, stretch it, and hold it. It's not releasing energy, but it's still exerting force. Lift a barbell, and hold. Just because it's not releasing energy doesn't mean it's not exerting force.

Originally Posted by Ben

I don't know which nozzles you are looking at, but the only ones I've seen say things like "gpm @ 100 psi." That doesn't mean force doesn't determine flow. They're giving the total flow there. The GPM is the volume and the PSI is pressure. I don't know what you're talking about, but that sounds like flow to me.

That sounds like flow to me. But it also sounds like flow is dependent on pressure. The flow is the dependent variable, the pressure is the independent variable. Given a certain pressure, there is a certain flow. If the site said "gpm @ 20 pounds of force," then I'd agree with you.

Originally Posted by Ben

The people behind the scenes there know little about the physics and more about the products also. I have emailed them before.

Oh well. Thanks for the tip.

Originally Posted by Ben

They call it m with a dot over it for a reason. However, I'm afraid that it probably was not interchangable with mass because it involves time. I am not sure though because there are things such as moment of inertia which can be substituted for mass that are not mass. It definitely is on the right track though.

With that being said, there are errors in the physics, but not in the math.

Definitely the weirdest symbol I've ever seen . None of the symbol menus I check have it.

So...what's the consensus with the derivation? Is it valid or not? I think not, but I can't say for sure. I just can't see how density*flow, or the mass flow, is compatible with mass. Perhaps if you add something to deal with the time, but in its current state, I doubt it. Sounds more like symbol confusion.

Originally Posted by Ben

We did stuff like this all the time in physics to keep things simple. The density of water does not change much in our range of temperatures. I mentioned that, but I suppose you like getting technical and complicated.

I've been found out!

However, here it's less an issue of precision than it is an issue of the actual concept. The technicality was only the fourth thing I pointed out. Earlier, I did indeed replace Poutlet with 1 atm or 14.7 PSI. However, I did not actually remove the term, because it still exists.

I suppose, for simplicity's sake, one may replace all of the constants with a constant k that takes into account all of the units. But the mass flow/mass issue still exists.

Originally Posted by Ben

More to write tomorrow... SilentGuy, get to testing soon. You'll agree with me very quickly. I have absolutely no doubts because I have tested myself.

I'm also anxious to test. I've explained your tests with my model, but there's no way to know for sure.

[DX]

When the dust settles, I'd love to know what I should do for my next water cannon to get more power in a smaller package. These math topics might as well be in a foreign language - tech only makes sense [to me at least] in action.

[Ben]

In other words, are you suggesting that the outlet has to be in a closed system, as opposed to the atmosphere? Interesting. I'm thinking it's the drop in pressure that matters more, as the pressure is equalized and the outlet pressure is one atm.

Yes. It's meant for small changes before and after valves. I suppose it can be used for larger values and outside of a system (if you can call that part of the system), but if it still works theoretically, the value will be inaccurate due to the use of the differential as a change variable.

Could you qualify SuperCAP's poor efficiency? 1" bore ID sounds pretty decent, although you might have to make that 3/4" to factor in the hose barbs (correct me if I'm wrong). The thread also says SuperCAP Mk.2 got a little over 60 feet with an endcap nozzle, perhaps more with a conical nozzle.

From the 4 inch backpack tank the water reduces to 1 1/2 inches, which has to affect efficiency very much. Then there's an immediate 90 degree turn. Then there is the tubing, curving. The tubing connects to the gun. Immediately the diameter reduces to one inch. After the ball valve the nozzle reduces to 1/4 inch or 3/8 inch (not sure on which one, but it's probably 3/8 inch). The nozzle was a flat end cap. Very turbulent and definitely inefficient.

I do not think you can call that anywhere near as efficient as even an APH water gun. You can say "high flow" all you want, but that does not make the difference here. Indeed, I have seen one inch APHs and they do not magically have more power. The stream coming from SuperCAP was noticably turbulent.

I was under the impression that your homemades also got 65 feet, and I'm certain they got 60 feet. That's probably with at least 75 PSI though (tell me if it's more, which it probably is, according to the Boyle's law and pump volume calculations).

My APH got 55 feet (effective range - not to the last drop) as reported in the article. 75 PSI sounds accurate to me, though I once did some math and calculated the pressure to be 80 PSI.

I also have read here and through emails that APH style water guns with 1 inch internals diameters do not perform differently than the 3/4 inch ones. What matters from what I have seen is having the potential for enough flow to have high performance. Once you reach a certain diameter (around 3/4 - 1 inch), there is no benefit from going bigger unless you want your water gun to shoot 80 or more feet or use nozzles larger than the ID. This is why I am designing SuperCAP II (or whatever I will call it) to use 3/4 inch tubing. Another benefit is the fact that such tubing is cheaper, one that can not be ignored.

SilentGuy, I could step the internal diameter down to half an inch and SuperCAP would still outperform any APH at the same pressure. Your model does not adequately explain this. You said before that if the range was within 10% of the LR system, you would consider the performance to be comparable and this the LR system does not work. But, here SuperCAP outperforms other water guns by more than 10% (12 - 18%) in range and you say your model explains it... Keep your qualifications consistent. Adding 10% more range takes much more power. I do not think anyone will disagree with me when I say that much more is needed to get 10% more range in water guns. So, even a 10% increase is very significant.

How does changing the diameter to increase the flow increase performance this much? What if I made a water gun with a two inch chamber, two inch valve, and two inch conical nozzle? Would that perform great because of the high flow? No, it won't because two inches creates low force. Again, as I read in my fluid mechanics book and as I knew, force is the ability to accelerate mass. Accelerating mass in a pipe creates flow.

Force is pressure times area. Increase the area, and you increase the force. Now I know you'll tell me that the nozzle exerts no force, so I'll word it a different way.

Given an endcap nozzle, force is being exerted over the entire face. However, only the force acting through the aperture is used to accelerate water, translating into work. Increase the nozzle area, and more of the force accelerates more water.

Force in the chamber is not affected by the nozzle area. But picture a similar situation in the reducer. Most of the force is simply being exerted on the wall of the reducer, but no work is being done. Only a small amount of force makes its way into the barrel, and an even smaller amount is used to accelerate the water in the stream.

I'm having to read this multiple times because I'm trying to figure what exactly you think is going on.

First off, work is done simply from the fluid moving. It doesn't need to exit the nozzle. It just needs to move.

Second off (and for the 50th time), force is not a property of matter. The fluid has momentum. The fluid DOES NOT CARRY A FORCE. Again, a force is a push or pull. The momentum will push the water into other things, and the impact creates a force. And lastly, aside from the force experienced by the air and the stream when both meet, water going through the nozzle experiences no force.

You are again getting into your thoughts that force is a property of matter and that any increase in diameter changes that. That is wrong. Your impact tests simply showed that larger streams have more mass and therefore more impact. The stream itself does not carry a force. It's impact is a force. The stream itself has momentum. But IT DOES NOT HAVE FORCE. Force is a push, pull, or lift. None of that occurs when the stream goes through the nozzle. There is consequently no increase in force for a larger diameter stream because that has little to do with force aside from the force of the drag.

Not all the parts of the tube are releasing their energy, but the are still exerting force. Take a rubber band, stretch it, and hold it. It's not releasing energy, but it's still exerting force. Lift a barbell, and hold. Just because it's not releasing energy doesn't mean it's not exerting force.

You're not reading me right. By not contributing, I mean cancelling out because another force is equalizing that one.

So...what's the consensus with the derivation? Is it valid or not? I think not, but I can't say for sure. I just can't see how density*flow, or the mass flow, is compatible with mass. Perhaps if you add something to deal with the time, but in its current state, I doubt it. Sounds more like symbol confusion.

From what I'm seeing in my fluid mechanics book, mass is a differential, possibly to avoid this problem. dm represents a small piece of mass. The equation I'm seeing has it's parallels to my simply derivation, but it is different because it is based upon momentum and has some wierd differentials in it.

I'm also anxious to test. I've explained your tests with my model, but there's no way to know for sure.

I definitely do not think anything you've posted explains the huge difference in performance between 4 inch chambers and smaller ones. Why does SuperCAP perform similarly to Supercannon II at 40 PSI? Why can't any other water gun perform on that level at 40 PSI? You can't say "Cv values" and "higher flow" because SuperCAP is bad in Cv value and it's flow isn't much higher than APHs that perform at a much lower level at higher pressure.

Duxburian, SilentGuy's theory has no basis. I am not seeing the equation he posted used in that manner in the book I have. I am completely confident that the equations he's throwing around are meant for other things after looking into the subject more deeply. In the past I had avoided my fluid mechanics/dynamics book because it used partial differential equations and harder differential equations so much, but I find now that I can understand the basics given time. Force is the determining factor in the power of a water gun because the force creates the flow.

SilentGuy, what makes you so sure the equations you posted are what is used here? Again, I looked all over my book and I could only see this equation used in one chapter for a few things. It had a few practice problems and nothing more. And when I looked for the connection between applied force and flow, I found what I was looking for very quickly and early in the book (on the second page about fluids in motion for crying out loud!). I don't think Wikipedia (or whereever you found this equation) knows what it's used for if it said it is used this way.

[Silence]

Originally Posted by Ben

Yes. It's meant for small changes before and after valves. I suppose it can be used for larger values and outside of a system (if you can call that part of the system), but if it still works theoretically, the value will be inaccurate due to the use of the differential as a change variable.

Going back and looking at various definitions, with your calculus interpretation in mind, I can see how you'd interpret "pressure drop." But some sources expound on it, referring to the inlet and outlet pressures. But most sources say "pressure loss across the valve" - ie, the difference between pressure as fluid is forced in from the inlet and equalizes through the outlet.

Google search: flow coefficient

Site 2 on the results page

In a constant, established air flow, the flow coefficient is determined by the volume of air flow through the valve and the inlet and outlet pressures. When holding the inlet and outlet pressures constant, the Cv of a given valve is also constant.

They mention airflow, but nevertheless, it works for water.

A PDF linked from Site 3 on the results page. Has some really cool stuff. And it clearly mentions inlet pressure and outlet pressure.

It also mentions that an undersized valve can lower the outlet pressure. But that's for valves, and not nozzles.

Originally Posted by Pressure drop (delta P) section, middle of the first page

Of course, if the valve is discharging to the atmospher, the pressure drop is equal to the the inlet pressure when dealing with liquids.

Most systems using compressed air (industrial applications usually, and this is meant for industrial systems, I should think) already use gauge pressure, so that statement just verifies what I've already said.

I could go on to other link, but I don't need to. Sites 1 and 4 both say "pressure drop across the valve," and sites 2 and 3 clearly say "inlet pressure and outlet pressure."

Originally Posted by Ben

From the 4 inch backpack tank the water reduces to 1 1/2 inches, which has to affect efficiency very much. Then there's an immediate 90 degree turn. Then there is the tubing, curving. The tubing connects to the gun. Immediately the diameter reduces to one inch. After the ball valve the nozzle reduces to 1/4 inch or 3/8 inch (not sure on which one, but it's probably 3/8 inch). The nozzle was a flat end cap. Very turbulent and definitely inefficient.

I do not think you can call that anywhere near as efficient as even an APH water gun. You can say "high flow" all you want, but that does not make the difference here. Indeed, I have seen one inch APHs and they do not magically have more power. The stream coming from SuperCAP was noticably turbulent.

...

I also have read here and through emails that APH style water guns with 1 inch internals diameters do not perform differently than the 3/4 inch ones.

Turns and tees don't decrease flow too much. It's the size that counts. Note how check valves and many on-off valves have flow coefficients based mainly on their size. Some valves might have more turns than others, but their Cv will be similar to similarly-sized valves that have fewer turns.

I've done 2D CFD work modeling both air and water going through elbows. Yes, pressure increases in the inside of the curve, but the velocity of the fluids was nearly same in the front and in the back of the curve - and since the area was the same, the flow was nearly the same.

That elbow was the first tutorial in the GAMBIT tutorial book I was using, if you ever get a chance to try it out. It's also in the online resources if you get a chance to look at those.

Size matters much more than the number of turns. I'm thinking even a small difference in area leads to a larger difference in flow. Why? Because flow = area * velocity. If you cut the area by a certain amount, the velocity increases by a certain amount. The drag forces increases as the square of the velocity. The velocity even through SuperCAP's long tubing is low.

Thus, it isn't hard to believe that SuperCAP is more efficient. Sure, its tubing is longer, but that's not as important as the wideness. And, once again, I'll point out that SuperCAP only needed 50% more total efficiency than your 55-foot APH. I'm willing to be the nozzles were fairly similar.

A bit of searching revealed that Duxburian's 3rd 21K hit 65 feet of range. That's with poor efficiency but high pressure (probably 80-90 balloons), and of course a great nozzle. There are a lot of factors to keep in mind.

The 1" barrel APH you spoke of could have had any number of problems. A poor pump seal that can't build up enough pressure, just not enough pressure being attained in the first place, a poor nozzle, etc. You got information of that gun through email - and we have no knowledge regarding anything else about it. For all we know, they could have used 3/4" tubing between the PC and the barrel - we just don't know.

The fact is that we have no actual head-to-head performance specs comparing guns' forces and pressures.

Originally Posted by Ben

SilentGuy, I could step the internal diameter down to half an inch and SuperCAP would still outperform any APH at the same pressure. Your model does not adequately explain this. You said before that if the range was within 10% of the LR system, you would consider the performance to be comparable and this the LR system does not work. But, here SuperCAP outperforms other water guns by more than 10% (12 - 18%) in range and you say your model explains it... Keep your qualifications consistent. Adding 10% more range takes much more power. I do not think anyone will disagree with me when I say that much more is needed to get 10% more range in water guns. So, even a 10% increase is very significant.

First, the 10% was for the tests I hope to do. With those tests, the nozzle(s), pressure(s), etc. will all be controlled. Only the force will vary, except perhaps for slight drag force differences - hence the 10% margin of error.

Second, 55/60 is 92%. The APH only got 8% less range than SuperCAP, if the perhaps slightly variable range figures can be compared. Or if you prefer it another way, 60/55, that's 109%. SuperCAP only got 9% more range than the APH.

Originally Posted by Ben

How does changing the diameter to increase the flow increase performance this much? What if I made a water gun with a two inch chamber, two inch valve, and two inch conical nozzle? Would that perform great because of the high flow? No, it won't because two inches creates low force. Again, as I read in my fluid mechanics book and as I knew, force is the ability to accelerate mass. Accelerating mass in a pipe creates flow.

Why didn't you make SuperCannon II's 4" chamber feed into 1/2" Schedule 40 tubing (5/8"), and then feed that into a 1/2" nozzle? The flow coefficient really is quite important. Obviously you can go over a reasonable limit, but most guns apparently don't have enough Cv.

Perhaps I should compare two cases with identical nozzles and identical Cv.

sqrt(40) / V1 = sqrt(100) / V2
6.3 / V1 = 10 / V2
10 / 6.3 = 1.6

Those numbers are acceptable (I have so far not seen a valid critique of my math). The cannon's 100 PSI stream only had 60% more velocity initial than when the cannon was fed 40 PSI. It seems surprising, but it really isn't. The formula explains that with the square root.

As you've said, even a 10% increase is significant. But (73-65)/65 is only an increase of 12%. 12%! I think it's safe to say that a 60% increase, or 1.6 times, the flow would net a 12% increase in range. A reasonable proposition.

However, 100 PSI provides 150% more, or 2.5 times, the force. It's hard to justify that much extra "power" netting only 1.12 times the range.

Originally Posted by Ben

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