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CO2 powered water rocketry

by sigma
Tags: powered, rocketry, water
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sigma
#1
Jun17-05, 03:08 PM
P: 22
Heya. Let's say I want to build a water rocket and I want to pressurise it with a 12 or 16 g CO2 canister used for powering airguns. I want to calculate the optimal size of the pressure tank, and I want to predict the thrust of this device at various nozzle diameters.

So here's the principle:
CO2 is sprayed from the canister into the top of a water-filled tank. The water tank has got a hole of some sort in the bottom where water is ejected to produce thrust.

The main problem I assume is to determine the pressure in the tank. The flow (and thus the thrust) is sort of proportional to the pressure.

Any Ideas?

//Cheers
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chroot
#2
Jun17-05, 03:24 PM
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This is actually a very difficult problem to answer from first principles, because the size and shape of the opening at the bottom of the tank has a LOT to do with the rate at which water leaves the tank.

If you know that number, the rest of the design is just kinematics.

Ideally, you want there to be just enough water in the tank. If there's too much, you're pushing all its wasted mass all through the launch. If there's too little, some of your pressurized gas will escape at the end, and you'll be wasting your CO2 resources.

The best thing I could tell you to do is build the nozzle, strap it to a table, and do some experiments.

- Warren
sigma
#3
Jun18-05, 02:38 AM
P: 22
Really? Can't I assume that the thrust from the motor is the pressuere in the tank multiplied by the nozzle area, and then add a correction factor for friction/turbulence loss. Then calculate mass flow from this.

How do i know what pressure I need to design the tank for? How do I know how big the tank should be?

/ J.

enigma
#4
Jun18-05, 12:53 PM
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CO2 powered water rocketry

No, your thrust is (ignoring the pressure correction) mass flow*flow velocity.

To calculate both of those, you need to rely on thermodynamic principles, and water cannot be dealt with as an ideal gas.
sigma
#5
Jun18-05, 03:23 PM
P: 22
Ok. I see. I've done some research and found that I can use this expression:
p + rho * g * h + 1/2 rho * v^2 = constant at any point in the water.
p: pressure
rho: water density
h: water height
v: flow velocity

I understand this is some variant of Bernoullis' law, or a statement of conservation of energy.
This with some geometric conditions and the assumption that water is incompressible would calculate thrust at any pressure.

Now for the CO2:
All I Know is that p*V^k = constant for an ideal gas. k is ratio of specific heats or something that I can probably find in a table, V volume of gas. (Is there a corresponding expression for temperature?)
This can't be true if the amount of CO2 is not constant, remember the CO2 is injected into the tank as the water is expelled.
I could assume that CO2 would vaporize just below critical pressure. But from my experience the CO2 will get really cold and that would lower pressure according to the ideal gas law. Also critical pressure will be lower at lower temperatures. How do I account for this?


/cheers
abercrombiems02
#6
Jun19-05, 12:58 AM
P: 114
For a given pressure in the CO2 tank acting on some area of fluid (Lets call then Ptank and Atank). We have F = PA = mdot*Vexit

mdot = rho*Aexit*Vexit

So

Ptank*Atank = rho*Aexit*Vexit^2

Vexit = sqrt (Ptank/rho*Atank/Aexit)

OK thanks great but we need to know more.....

The thrust is simply the mass flow rate multiplied by the exit velocity
This needs to be equal to the sum of weight and drag forces.
Note that the Ptank is the GAUGE Pressure in the tank with respect to the ambient. rho is the density of water 1000g/L (I''d assume you working in this range of magnitude) If you building a 1000kg water rocket then let me know I want to see it j/k.

So anways the thrust will be

mdot*Vexit = totalmass(t) + Drag(v)

mass is a function of time. Most likely something of the form
M(t) = Mo - mdot*t (this is assuming the flow of water is constant)
For small speeds Drag increases linearly with velocity
It really is a function of the dynamic pressure which is a function of V^2.

Let not worry exactly what the drag is for now, because that requires knowledge of the shape of your rocket, we can just assume its goign to not be big.

Lets let the mass be 10 kg so its wieght is about 100N. Lets let the drag be about 20N.

This means at the heaviest case we need to produce about 120N of thrust

120 = rhoH20*Aexit*Vexit^2. Working in kilograms and meters we have

rho = 1000 kg/m^3

So 0.12 = Aexit*Vexit^2, but Vexit^2 is....

Ptank/rho*Atank/Aexit

So 0.12 = Ptank*Atank/rho. The pressure is a max to begin with and will drop slightly, so all that really matters using this simplified example is area between the CO2 tank and the water.

multplying by density gets us back to

120 = Ptank*Atank, however we know not all of the pressure in the tank will be directly converted into thrust so lets let the force need between these two maybe be 20% higher or about 144N

So now we have 144=Ptank*Atank. Ptank is some number you know and I dont, remember its the gauge pressure between the atmosphere and the tank, not the absolute pressure in the tank. So thats how we then find Atank.

To find the area of the throat we need to ask ourselves how much water do we have and how long do we want to "burn" for. More water is more burn time, however due to friction we cannot make the hole infinitely small and have water jetting out at lightspeed, nor can we have a hole so big that all the water makes a giant water bomb upon launch. Well we know
mdot = rhoH20*Aexit*Vexit

We determined Atank above, so now the exit Velocity is strictly a function of exit area only. This is where the open ended part of the problem comes in. Let's choose a mass of water of 10kg, and lets let the water exit at a desired rate of 1.5 kg/s. This is a burn of about 6 seconds which isnt too bad for a small rocket. So we have

1.5 = rhoH20*Aexit*Vexit

Vexit = sqrt (Ptank/rho*Atank/Aexit)

Ahhhh look now....

1.5 = rhoH20*Aexit*sqrt(Ptank/rhoH20*Atank/Aexit)

everything is a function of

the density of water...known
Ptank....known (use a guage if its not given)
Atank...solved for using F = dP/dt
Aexit...only unknown.

So basically what I did here is that I knew my mass of water to start with and I wanted an appropriate burn time, for this I can determine the mass flow rate by assuming a CONSTANT flow rate (this will never happen, the world isnt that nice) Then I can determine Aexit, knowning all my areas now and the previous junk, this will give me my exit Velocity as well.

So we know everything. Just by looking at this I'd guess an exit area mybe 3-4cm in diameter would be appropriate. Also You will probably have a large Atank. When transitioning to these areas, do it gradually, likea venturi. Just look up what a venturi tube is on google or something if you dont know what that is, kinda like an hourglass. Make inner surfaces as smooth and continuous as possible, this will keep frictional losses at a minimum. Also, flar your nozzle at the end, This slightly reduce the ambient pressure on the boundary of the exit, this effectively increases the gauge pressure in the tank by a small amount. (That means u can make your rocket more slender reducing the drag).

Hope this helps took me abotu an hour or so to write.
GENIERE
#7
Jun19-05, 02:45 AM
Sci Advisor
P: 288
Quote Quote by abercrombiems02
...Hope this helps took me abotu an hour or so to write.
I don't know if it helped the OP but I enjoyed it thoroughly.


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