# Water guns

1. Apr 25, 2007

### btrettel

[SOLVED] Water guns

Currently on a forum about water guns there is a debate going on about what determines a water gun's performance between pressure and force. On one side are people who say that higher pressure differentials combined with a higher "Cv value" will mean more flow, which translates into greater distances achieved. On the other side are those (mainly myself) who say that Newton's second law is clear and that force determines how much flow (movement) is created.

Now, I will admit that I am just a high school student. I've taken 3 physics classes and two calculus ones and I think I'm pretty good with basic physics. I've built many water guns, including a water gun designed explicitely to have extremely high force applied to the water by having high pressure in addition to a large area to apply the pressure. This water gun currently holds a record for distance shot and I'm working on one to shoot further with more force. Naturally, I'm on the side of force because it's reality as far as I'm seen. And the more I research it the more sure I am, especially after reading about the Navier-Stokes equations

The pressure people's debate essentially states that the equation Q = Cv * sqrt(Pgauge) determines how much flow is created by a pressure. I'm not quite sure that equation is used for that purpose. I've tried to explain to these people that the Cv equation is meant to calculate flow lost through a valve in a simplified manner, but these people don't want to hear it. Can someone explain to me if this equation can be used to calculate flow created from a pressure differential and if it can not be used that way, can someone explain to me why this equation can not be used?

My debate essentially comes down the Newton's second law applied to fluid dynamics. I've read pages that describe the Navier-Stokes equations for fluid flow that state movement (flow) is based upon force. I've brought these up with these people, but they ignore them.

I don't have the time to get into all of the small debates involved, but the larger debate comes down to this. Who's right?

In the next week I intend to build a water gun more powerful than the one I mentioned earlier and I feel this will show that force matters when it comes to water gun performance... but I would like to get the facts straight before then.

2. Apr 25, 2007

### Staff: Mentor

In f=ma, Force determines acceleration if a given mass is constant. But for fluid systems, mass is itself a rate and varies according to pressure and cross sectional area. Using pressure instead of force takes away that complication and allows you to directly find velocity.

This is a common question in an intro fluid dynamics course. Static pressure and Bernoulli's equation tell you how much velocity a SuperSoaker will generate at it's nozzle. For tank and hose sizes much larger than the nozzle size, the nozzle size doesn't matter as the static pressure will remain roughly constant over short timeframes.

Bernoulli's principle for this application says the velocity pressure of the stream is equal to the static pressure of the tank. So....

Static pressure (1)= Velocity Pressure (2)
S.P. (1) = 1/2 rho*$$v^2$$

rho is the density of water.

Last edited: Apr 25, 2007
3. Apr 26, 2007

### btrettel

That sounds like something they've been telling me. But I don't think Bernoulli's equation is what we're looking for here. While it has velocity, it lacks the total flow figure we're looking for. I'm sure velocity and acceleration are proportional, but if they're proportional and velocity is proportional to the width of the pipe, couldn't the flow be the same if the velocity was reduced in a wider diameter pipe? Again, we're looking for flow and not simply velocity. Do I use Bernoulli's equation and multiply the velocity by the area to get a flow rate?

4. Apr 26, 2007

### FredGarvin

What your opponents are saying is correct. The Cv is a measure of the flow capacity of a valve, orifice, etc...The definition of Cv is the flow of water in GPM created by a 1 psig delta p across the valve. It has nothing to do with "flow lost" (I'm not really sure what that means).

What it comes down to, like in all engineering applications, is a trade off. For the distance, you need a higher velocity. For efficiency you want as much flow as possible. Since you have a set maximum pressure differential that your gun can supply, you need to balance between the two desired results. Distance means a smaller orifice to get the velocity up. However, that limits your flow.

5. Apr 26, 2007

### btrettel

Perhaps flow lost was a poor word choice. I know it essentially states the flow limit of the valve. But, using pressure here doesn't match what I've seen in practice. The main limit to flow is the nozzle, and by this argument, all water guns are the same pressure will perform similarly with the same nozzle. This is why I've argued in the force side.

That doesn't match what is seen in practice though, especially considering I got 65 feet of range with one water gun I made at 40 PSI with a 1/2 inch nozzle. Most water guns couldn't shoot 10 feet with a 1/2 inch nozzle at higher pressures. I have come to the conclusion that there is more to water guns than pressure for this reason, so right now I'm wondering exactly what is going on.

The water gun I made before puzzles me given the responses here. Is there anything particularly redeeming about this water gun design? I designed it to create as much force as possible and it so far has outperformed anything else, though, I took many many other factors into consideration as well. Most of you should understand the basic design. Pictures are available here: http://www.sscentral.org/images/supercannon/

4 inch diameter piston
Uses a conical shaped nozzle
73 feet of range @ 100 PSI
65 feet of range @ 40 PSI
77 feet of range @ 80 PSI with 19 parts water, 1 part glycerin
~4 liters per second water output with 1/2 inch nozzle @ 100 PSI
~24 liters per second water output with 1 1/2 inch pipe opening @ 80-100 PSI (for a fraction of a second)

For comparison, no other water gun performs on this level aside from the other 4 inch diameter pressure chamber water gun I made that used a backpack. It shoots about 63 feet at 40 PSI. Any other water guns at 40 PSI wouldn't get over 50 feet.

So basically, I'm not inclined to believe that pressure matters that much given my experience with high performance low pressure water guns. Could someone explain the Navier-Stokes equations to me? I've used the fact that those use force in support of my debate and I'm interested in whether they do or if they do something else. I'm open to what's the truth, but I'd like what made me decide force matters to be explained as well.

Again, thanks for the quick replies.

6. Apr 26, 2007

### Staff: Mentor

A 1/2" nozzle is very large, so at that size you have to consider the size of the tubing and the size of the reservoir and the calculations get a lot more complicated.

Another issue is that for very high pressure and velocity - for a water gun - the integrity of the stream due to air resistance will matter as well, so a thickeer stream will go further than a thinner one even if they start with the same velocity.

Last edited: Apr 26, 2007
7. Apr 26, 2007

### btrettel

Oh, I know the physics of the stream very well. I choose a conical shaped nozzle for several reasons. Larger streams definitely hold together better, but to get a larger stream, you lose velocity. I've plotted range vs. orifice diameter before to approximate an ideal orifice diameter.

What I am saying is that I don't understand the discrepancies between pressure and performance in water guns. For a while I decided force was what made the most sense, but some people recently have pushed me to change that thought. There are many discrepancies between pressure and performance. My homemade water gun example was just one. Another good example of high performance low pressure water guns would be the CPS series: http://www.geocities.com/mikahw/images/1500.jpg

That water gun I just linked to shoots over 40 feet with ease and operates at only 22 PSI. It's system is far less than efficient. Another good example is the SS 300, which according to the patent operates at 30 PSI, yet shoots water around 50 feet. Most water guns, even with higher pressures and more efficient designs, shoot water at around 35 feet.

I don't mind things getting a little over my head, so if you wouldn't mind could you explain the entire thing to me? I've taken two years of calculus, so as long as it doesn't get too advanced I think I will understand the math.