Well I say put your money where your mouth is.
I plan to do that. And I apologize if I came off as condescending or arrogant in my previous post. I didn't mean to be; I just was stating a couple facts, and I was only trying to say that the design can be improve. Nothing else.
The tiny little bends may plague you but for me no problem and I still get the distance of a SC.
These bends are tiny. 180 degree is a large bend. I design my guns to avoid this problem because I know how bad it can get. I'll address that below a bit.
So do you care to bet on that and maybe you should ask a plumber about a pressurized system getting a leak. Maybe you could take apart a Supersoaker and realize that most of the guns don't shoot directly off the pressure chamber and have an actual barrel to make the flow more laminar.
Last time I checked, most off-the-shelf water guns perform poorly.
All stock air pressure guns need a bend to work. They just won't work without it. But nowadays, most off the shelf water guns do a pretty good job at laminating the stream. Most CPS guns had straws in the nozzle (which is much better than just a long pipe), and nowadays many off-the-shelf water guns have converging nozzles to speed up the stream right before it exits. And they use ball valves too.
Really you must not understand the simplicity of this gun which is better for me. I dont really care if you think the barrel does nothing or that the bend will ruin the performance. You could not back that up with scientific proof, actual test results and not theoretical math or guesses. I know it works fine you may be able to give other bad info but I know just how many failures and mistakes you have made in the past the fact that some of you act like an authority on the matters is laughable. Seriously dont say anything if you are not sure what your saying is 100% true and verifiable by tests not guesses
Well, I'll go into the math of it.
adronl, I hope you don't think that I'm trying to be mean to you or anything. All I'm doing is explaining why having bends is a bad idea in performance water guns. You've made great guns in the past, and I'm sure you'll make many more. But I'm just explaining the science behind this, and why this is an important thing for water gun performance. I was just pointing out that since you always build performance water guns, you always get the best results. This water gun has a design flaw that makes it perform less than expected. All I was trying to do was point that out.
Moving on: The math I will do is pretty easy, though one equation will look kinda hard, but I avoid the hardness through an assumption.
Let's first list all my assumptions:
- Steady state -- I'm assuming that there is no variation in time.
- Constant velocity profile -- This is to simplify the calculations, mostly to simplify the integral.
- I'm analyzing a section of pipe between the chamber and before the firing valve. This simplifies the diagrams and math. From this assumption, the pressure at the outlet won't be zero since the water hasn't left the gun yet.
Okay, I'll explore too cases for comparison: a straight pipe, and a 180 degree turn. I won't go into the case for a 90 degree bend because the math gets messy (it involves a lot of vectors), but the idea is similar to the 180 degree case.
Better explain my variables:
- Q = Flow rate
- rho = Water Density
- d = Pipe diameter
- V = Velocity
- P = Pressure
- F = Reaction Force from a surface
- KL = Minor Loss Coefficient
Let's start off with conservation of mass:
Okay, so from conservation of mass, the velocity entering my control volume is the same when it leaves. That means no mass will build up inside the control volume.
Let's move onto conservation of momentum:
Okay, this is the trick I did. I assumed the velocity is constant over the area, so it simplified the equation a bit. The integral equation says that changes in the velocity over the control surface mean that there are forces changing them. A change in velocity would result into a force, in other words. Also, the velocities are equal, so the whole left side cancels to zero.
This equation has the reaction force in it. This is a force that the pipe transfers to the water. Why does the pipe transfer it? Because the pipe isn't moving; it's static. So to remain static, it has to transmit a force if the water is pushing against it. It's Newton's Third Law.
No something interesting happens. In the straight pipe case (Case 1), the water isn't pushing against the pipe; it's flowing against it. There is no surface for the water to push against. So the reaction force is zero! The equation simplifies:
This means that there is no pressure drop at all. That's what you expect, ideally, for a straight pipe.
But what about the 180 degree turn? Well, the water pushes against the pipe while it enters, so there is a reaction force. Let's simplify and solve for it.
So a bend means that there is a reaction force from the pipe to the water, so that the pipe doesn't more. That means that there is a pressure drop. A drop in pressure means a drop in performance, just as I said before. Bends hurt performance.
But I'll prove it a second way. I'll move onto conservation of energy using Bernoulli's equation.
Okay, this is much simpler. I assumed that there some pressure drop,
from minor losses, as the water moves forward. These minor losses are considered minor in large-scale problems, but in small-scale they affect the design greatly.
Back to the equation: The equation simplifies to a pressure drop from minor losses. This pressure drop is changed by the
minor loss coefficient. That means that the higher the minor loss coefficient, the more pressure is lost. Simple enough.
But look at that, velocity is in there too. And it's squared. That means when performance increases, more pressure is lost! Obviously, any performance water gun (like yours adronl) needs to have as low a KL as possible.
How do you lower the KL? Simple: Have linear flow. A straight pipe has a KL of 0; a 180 degree bend has a KL of between 0.2 to 1.5 (depending on the situation). And with the velocity tied in, that means that the better the gun performs, the worse it actually will if designed with anything that greatly increases the minor losses -- like bends. Bends hurt performance.
Edit:
I did some quick pressure drop calculations, it seems that, for high flow rates especially, the pressure drop is sizeable. From the picture, I assumed your gun was made off 1" PVC. I then took the worst case scenario (which is how you should design, always) of KL = 1.5, and used the equation from above. Here's the graph:
This is unacceptable. adronl, this is what I was talking about. Small changes in design, like having a bend, greatly change performance. This graphs says that at 500 oz/s (what SuperCannon 2 gets) the pressure drop is 93 psi. That's a lot. Most guns don't even use 93 psi.
Of course, this is the worst case scenario. The best case scenario of KL = 0.2 is a bit better: only a pressure drop of 12 psi. Still, a performance water gun needs all the pressure it can get, so little things like this add up. Removing these little things is an easy way to improve the design and to drastically improve your gun's performance.