Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Water Rockets

  1. Jun 9, 2003 #1
    My son & I will soon be building water + air-powered rockets for cub-scouts with soda-bottles and compressed air. You put so much water in a bottle, pressurize it, and release it upwards... Now, if nearly full of water, it would not go far, nor if nearly empty, using only air, but if optimally filled, it will go dozens, maybe hundreds of feet. How to calculate the best fill, or the height reached as a function of percentage of original volume filled with water?? A stumper so far for me, and I love Newtonian physics. Perhaps a good one for the guy who asked what is calculus good for :).
    Assume the bottle will handle a certain pressure (10-15 atm?) and it doesn't have much mass relative to even a small amount of the water it could contain. How do you get started on this? If the energy is stored in the compressed air, what does the water do, anyway? It is something to "push against"? Is the instantaneous thrust equal to the pressure differential times the area of the orifice? If so, again what difference does the water make? Thanks for your thoughts, Eric
     
  2. jcsd
  3. Jun 9, 2003 #2

    enigma

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Water is incompressible, so trying to force more into the bottle will just cause the bottle to break.

    Air is needed to be compressed and store the energy, but it has very little mass.

    From conservation of momentum, you want a lot of mass to get ejected out the bottom. Therefore, the best solution is for the compressed air which is "riding" on top of the water to force the higher mass water out of the nozzle.

    To find the optimal volume, you'd need to use the ideal gas law, a little thermo, etc. It's not an easy problem, by any stretch of the imagination

    Since the air will (optimally) be the last thing out of the bottle, you'd want the air to be something over 1atm of pressure when the last of the water is ejected. From there, you'd need to work back and push as much water in without causing the air pressure to be able to rupture the bottle.
     
    Last edited: Jun 9, 2003
  4. Jun 9, 2003 #3
    It would be practically impossible

    to calculate the best solution because you have to include the hydrodynamics of the water, which won't squirt out as a solid slug but will flow and spread while exiting. Probably best to experiment, maybe starting with half full and see if adding or removing water helps. A dozen or so "test launches" should get you a pretty good value.
     
  5. Jun 9, 2003 #4

    chroot

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Remember Newton's third law of motion: every action has an equal and opposite reaction. A rocket engine uses the third law to achieve propulsion. Effectively, a rocket throws mass out of a rearward nozzle at high velocity. The reaction drives the rocket forwards. (You can also easily think of this as the conservation of momentum.)

    The more mass you throw out the back of your rocket, or the faster you throw it, the more reaction you'll enjoy. So naturally, you want to throw as much mass out as you can, as fast as you can.

    The compressed air provides the energy for the rocket's ascent. Air works well as an energy store in such a rocket because it is highly compressible -- but unfortunately it doesn't have much mass compared to the mass of the rocket, and thus can't impart enough reaction to get it moving well.

    As a result, you combine compressed air with water. The water provides the mass necessary to make the rocket attain good velocities. It is also advantageous that the compressed air always stays above the water, preventing its own premature escape.

    There are two opposing trends in this problem:

    1) The more water you use, the more mass you can throw out the back of the rocket, and thus the faster you can make it move.

    2) The more water you use, the more massive is the rocket, and according to Newton's second law (F=ma), the more slowly it will accelerate.

    Trend 1 provides a lower bound. If the air inside the rocket expands all the way to 1 atm before all the water is pushed out, the excess water is just excess mass. That excess mass has slowed the rocket's acceleration all the way from the launchpad. The conclusion is that you always want to make sure that no matter what mixture you use, all the water in the rocket must be expelled at or before the air chamber inside reaches 1 atm.

    Trend 2 provides an upper bound. You could compress the air so highly so that, even after all the water is expelled, compressed air will continue escaping. The problem is that the more you compress the air, the more water you'll have in your rocket, and thus the more massive will be the rocket on the launchpad.

    The volume problem is easy; it's just Boyle's law: halve the pressure, double the volume. So if you fill your rocket 90% with water, and 10% with 10 atm air, the air will expand to ten times that volume, or 100% of the rocket's volume, as the rocket ascends.

    Let's see if we can do all this with math.

    Let V be the total volume of the rocket. Let Vw be the volume of water in the rocket, and Va be the volume of air. Let Vai and Vwi be the inital volumes of air and water, at the time of launch. Let p be the gauge pressure inside the rocket (i.e. the pressure above atmospheric). Let pi be the pressure at the time of launch. We'll assume static fluids at all times (this is not really true, but at low escape velocities it is inconsequential). Let m be the mass of the rocket (we'll assume that the water is the only important mass -- the mass of the air and rocket body will be considered negligible). Let o be the exhaust hole area at the bottom of the rocket, which is considered a plane hole. Let v be the exit velocity of the water. Let h be the height of the water in the rocket.

    Keep in mind that nearly all of these quantities are time-dependent. The mass, volume of water, pressure, etc. all change with time. I will not explicity write these parameters as p(t) or h(t) or so on for the sake of brevity -- just remember that they all change.

    First, let's derive the exhaust velocity of the water. Simple conservation of energy of a water droplet falling a distance y leads to Torricelli's theorem:

    1/2 mv2 = mgy
    v = [squ](2gy)

    where g is the gravitational acceleration and y is the distance the water droplet fell (from the top of the liquid to the bottom).

    This basic form is only valid, however, when the top of the water column and exit hole are at the same pressure, which is not true in this case. We can simply pretend that the water column is actually higher than it is, and modify Torricelli's theorem to this:

    v = [squ](2gh + 2p/[rho])

    Now if o is the cross-sectional area of the nozzle, the effective area is 0.62 o, assuming again that the nozzle is just a plane hole. (I will not derive this result here.) If the nozzle is strongly tapered, this coefficient gets closer to 0.5. This is really not an extremely important consideration for a crude calculation, so we'll just take the effective area to be half the real area.

    The rate of efflux of water from the rocket is then simply the effective area times the exit velocity, or

    dVw/dt = 0.5 o v

    So, combining all these equations, we have

    dVw/dt = -0.5 o [squ](2gh + 2p/[rho])

    which is the rate of change of volume of water. The minus sign is there to make dVw/dt negative, signifying the volume of water is decreasing.

    When water leaves, air expands:

    dVa/dt = -dVw/dt

    The pressure p(t) at time t is related to the volume of air by Boyle's law.

    p(t) = pi * (Vai/Va)
    p(t) = pi * Vai / (V-Vw)

    The height h(t) at time t is related to the volume of water Vw in the rocket and the cross-sectional area of the rocket, A (assuming, for example, a cylindrical rocket).

    h(t) = Vw / A

    The volume of water, upon which both p(t) and h(t) depend, is just the integral of the rate of change we found above:

    Vw = Vwi + dVw/dt t

    If you combine all these equations, you wind up with a quite ugly differential equation in Vw:

    dVw/dt = -0.5 o [squ](2g (Vw/A) + 2 (pi/[rho] (Vai/(V-Vw)))

    If you would like to solve it, be my guest! I would probably use numerical methods rather than trying to finish this problem analytically.

    Now for the 'rocket science.'

    The thrust (force) generated by a rocket is just the exit velocity times the rate of mass ejection:

    F = -vR

    In this problem, R, the rate of mass ejection, is just the volume rate of change times the density:

    R = dVw/dt * [rho]

    and of course we already know v.

    Therefore, the thrust generated by the rocket is:

    F(t) = -[squ](2gh + 2p/[rho]) * - 0.62 o [squ](2gh + 2p/[rho]) * [rho]

    or simply

    F(t) = 0.5 o [rho] 2gh + 2p/[rho]

    or

    F(t) = o (gh[rho] + p)

    You can see at once how beautiful this expression is -- it includes your two variables, the the height of the water (h) and the pressure (p), and determines the thrust of the rocket. (Of course, you fix the initial conditions for both of these variables, and thrust at time t depends on the pressure and water height at that time) The only other quantities you need to know are the density of water and the cross-sectional area of the nozzle!

    The rest of the exercise is just Newtonian mechanics, and trajectories in a gravitational field.

    The rocket has two forces acting on it at any time -- its thrust pointing up and gravity pointing down -- assuming the rocket stays in a vertical orientation for its entire thrust-producing period. The net force is just

    Fnet = o (gh[rho] + p) - mg

    The acceleration experienced by the rocket is:

    a = Fnet / m

    To calculate the maximum altitude of the rocket's trajectory, we just need to find out when the thrust equals the force due to gravity. At this point, the rocket stops going up and begins coming back down.

    This problem therefore has a 'surface' of solutions, rather than a single solution. You can fix the pressure and find the best fraction of water, or the fraction of water and find the best pressure!

    - Warren
     
  6. Jun 9, 2003 #5

    chroot

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

  7. Jun 10, 2003 #6
    damn, chroot!

    i dont know if that post qualifies you to be a rocket scientist, but it most certainly qualifies you to be a water-rocket scientist!!!
     
  8. Aug 6, 2003 #7
    Agree with On Radioactive Waves statement chroot, but might add that the water will absorb some of the air that is pressured into it, hence cooling the water down to 4 degrees C (max Density) will give a very slight edge on that factor.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Water Rockets
  1. Water Rockets (Replies: 6)

  2. Water Rockets (Replies: 7)

  3. Water Rocket Glider (Replies: 17)

  4. Water Rocket (Replies: 36)

Loading...