Water gun power

[DX]

Hehe, the relativity of water guns strikes yet again. *cue Imperial March*

[Drenchenator]

I think what this whole discussion has shown is that the qualities of a water gun can't be reduced to one single number, no matter how much one might want to. Different people want different things. Some favour range, others output, others will want a balance of both, still others care about shot time, or want a weapon that's small and light, and so on.

But there should be an overall measure of how powerful something is. Power is a measure of work per time, so the the higher the number the more work it can do in less time. The overall "goodness" of a gun might be impossible to measure, but we should be able to find out how much work a gun can do in a certain amount of time.

[SSCBen]

If you want to get into literal power from a physics standpoint, you could figure out how much work the system is doing as a function of time and take the derivative. That shouldn't be that hard with basic thermodynamics. You didn't take that class though Drenchenator so for the sake of enlightenment I'd suggest taking a look at that Wikipedia page.

I might figure out some actual power numbers to compare them against what's been posted.

[Drenchenator]

I actually based the equation I came up with from that one in a similar method to what Ben said. Since taking the derivative is basically a funky form of division (think about it), I divided by time on both sides to get power from work and flow rate from volume. Not to hard to understand. I also used the pressure-volume-work equation to get the energy put into an air pressure gun in this article. There's no time component, so it's pretty much impossible to take the derivative to get the power. If you knew how the pressure dropped with respect to time, it could be done though.

[Silence]

If you knew how the pressure dropped with respect to time, it could be done though.

Well, that's dependent on the type of pressure chamber, but it could be done. Tie flow to the change in volume and use Boyle's Law to get to the change in pressure. Unfortunately, things get weird with precharging and "constant" pressure, plus the volume of the chamber makes a difference, too.

[SSCBen]

It'd really only be reasonable to figure out literal power in theoretical approximations of systems, and the most reasonable system would be "constant " pressure. You'd still have to do 3D integrals but if the pressure is constant with respect to time the math would be much simpler. It'd probably be interested to do for some example CAP systems to see what actual power numbers will come out.

Needless to say figuring out the pressure with respect to time would nearly always involved inaccurate approximations, and the limits of the integral would change with time as well so it gets rather messy and inaccurate.

If anyone doesn't understand that, that's okay, take Calculus 3 and you'll follow. A year ago I wouldn't understand what I said here.

I guess this is why simpler measures of water gun power are necessary.

[Hunt_and_Annoy]

Athlete peak power ~ 1000 Watts. If you could harness your full power for a water gun (by connecting a pump to a bike or rowing machine I guess), think what you might be able to get.

It should be noted that peak power is different from sustained power. Even Lance Armstrong couldn't sustain more than 500 watts for very long, and the peak power of a "very fit" regular human is quite a bit lower than that of superathletes.

[cantab]

True. Peak power for a normal person is likely to be lower, and sustained power lower still. Even so, mine are about 400 and 200 Watts respectively (measured on a concept 2 rowing machine). And I'm hardly the most powerful person around.

[Hunt_and_Annoy]

True. But carrying around any device which would allow one to use that power would be very difficult anyway, so moot point.

[cantab]

Haha. And now you have set the new challenge - build a portable, lightweight, device that allows a person to generate a lot of power to power a water pump.

Of course you'd likely end up having to stay put while using it - but you'd probably be sucking water from a lake or river anyway.

You may as well just carry an engine mind - but where's the fun in that?

In any case, since you're by the water, there's a much simpler way of soaking someone, and more effective than any other - push them in!

[Hunt_and_Annoy]

Stationary water guns? God, soon the helicopters outfitted with WBL's that I dreamed up years ago will be reality...

BTW, you had 140 posts and your userid is 1400, just thought it was kind cool.

[cantab]

Helicopters with WBL's? Pah! Just get one of those firefighting planes to airstrike the opposition!

[Twisted]

Hmmm, sounds like a good addition to the car project

[SSCBen]

I just had a "doh" moment about actual water gun power. Finding the average power (yes, true power) of a water is pretty easy if we assume constant pressure or linearly dropping pressure. All you have to do is find the total amount of work necessary to charge the water gun (W = the integral of P*dV), realize that's the amount of energy stored, and then divide by the shot time to get average power. It's that simple.

For some reason before I assumed finding the total work would be much more difficult than it actually is. I don't know what I was thinking but it's probably safe to assume I wasn't thinking, especially considering I've done the math before.

I'd love to make a derivation of a formula for air pressure and constant pressure but I need to get back to studying for a thermodynamics midterm.

Edit: For the constant pressure case power is simple.

power = (pressure*volume)/(shot time)

With SI units you have (Pa)*(m^3)/(s) which is the same as Watts.

With truly constant pressure the average power is the same as the instantaneous power.

Edit #2: Some approximate average power figures for the largest nozzles of various CPS water guns are below:

Supercannon II - ~7800 Watts (I need to find the exact numbers and do a non-constant pressure calculation)
CPS 2000 - 247.5 Watts
CPS 2500 - 103.4 Watts
Monster XL - 101.0 Watts (assuming pressure of 23 psi)
Arctic Blast - 90.6 Watts (assuming pressure of 23 psi)
CPS 3000 - 67.6 Watts
CPS 1500 - 47.7 Watts
CPS 1000 - 32.9 Watts
SC 600 - 8.0 Watts

Some data is assumed, some came from memory, some from iSoaker.com, and some came from a website that had pictures of Super Soaker boxes that listed pressures.

Edit #3: I just realized that this method is the same as what Drenchenator used before, just taken from a different angle and more general. Drenchenator's numbers are eerily similar. Remember that flow rate = volume/(shot time) so what he had is the same in the constant flow case.

So... for CPS... P = p*Q

Edit #4: I took a few minutes to derive the average power of a more general system. If we assume the pressure drops linearly with volume (as it does in most cases) then we can simply use the % pressure drop figures I had described before in the LPD thread.

P = p0*V*(1-0.5*b)/d = p0*Q*(1-0.5*b)

where
P is power
p0 is the initial pressure
V is the total volume of water being shot
b is the fraction that describes pressure drop (i.e. 25% drop is 0.25)
d is the shot duration
Q is average flow

As you can see this is a simple variation of the formula I derived earlier. If you make b=0 so the pressure is truly constant the equation is the same.

Edit #4: I think it should go without saying that total energy stored = p0*V*(1-0.5*b).

[cantab]

I have yet another measure of power. Well, I thought I did, but actually it reduces to isoaker's!

Basically, my original method used the ACTUAL velocity of the stream.

If, however, I instead use the THEORETICAL MINIMUM velocity, vmin, according to simple projectile motion, then I get what I shall call 'Effective Power, Pe'

For the 45 degree angle, range

R = vmin^2/g

Then power

P = 0.5 p Q R g

Where p is the density of water, Q is the output, R is the range, and g is the acceleration due to gravity.

************************************

So now we have three dimensionally correct measures:

Ben's new method, based on the release of the stored energy (as pressure). I'll call this internal power, Pi

My original method, based on the kinetic energy of the stream exiting the nozzle. I'll call this stream power, Ps.

My new method, based on projectile motion. I'll call this effective power, Pe.

Pi > Ps > Pe
This is ALWAYS the case or energy is added from nowhere and you have a perpetual energy machine.

From these, we can calculate three measures of efficiency:

Internal efficiency, Ei = Ps/Pi
This measures energy losses within the gun

Stream efficiency, Es = Pe/Ps
This measures energy losses in the air, eg from atmospheric drag.

Overall efficiency, Eo = Pe/Pi
This measures all losses.