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Water Rocket

  1. May 26, 2012 #1
    I am building a water rocket for my physics class and I want to make my calculations as precise as possible. I want to calculate throttle, velocity, acceleration, momentum, and max height.

    With that said, what should I take into account? Gravity, wind velocity, shape of cone of the rocket, density, anything else? And would temperature be a factor?

    The rocket will be made out of a coke bottle (2 L) and it will start with about 300-800 mL of water (I'm not so sure on the amount) so I will have to take into account the change of momentum over time, and we will use 45-60 psi.

    This is what I have so far:

    For thrust, T = [itex]\frac{\pi}{2}[/itex]PD2
    Height, h = ([itex]\frac{Mi}{Mr}[/itex])2([itex]\frac{Pi}{\rho g}[/itex]) where Mi is the mass of the water only, Mr is the mass of the rocket when empty, Pi is the initial gauge pressure inside the rocket, and ρ is the air density.

    External forces, M[itex]\frac{dv}{dt}[/itex] = [itex]\alpha[/itex]Ve + Fext = [itex]\alpha[/itex]Ve + Fg + Fdrag, where [itex]\alpha[/itex] is -([itex]\frac{dM}{dt}[/itex]), Ve is the velocity of the water leaving nozzle, and is it safe to suggest that Fd is Stokes' drag at small velocities (Fd [itex]\propto[/itex] -b[itex]\upsilon[/itex])?

    And I don't know how to account the shape of wings and all that. Any suggestion?
    And what else should I add?

    In essence, I want to be able to calculate a model that would fly the highest and fastest.
     
  2. jcsd
  3. May 27, 2012 #2

    haruspex

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    How do you get your expression for h? I don't see air volume in there. That is surely a crucial quantity (given the pressure). And I don't see why the air density matters.
    The thrust will decline as the air expands; the mass to be accelerated also declines. Have you taken those into account?
     
  4. May 27, 2012 #3
    What would be a better expression for h?
    For the thrust that is declined as air expands and mass that declines, how should I take those into account?
     
  5. May 27, 2012 #4
    Okay I think I found a better expression for height. It follows,

    h = 0.122g(tend)2

    What do you think?
     
  6. May 28, 2012 #5

    haruspex

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    R(t) = velocity of water at jet
    A.R = volumetric rate
    ρ = density of water
    A.R.ρ = mass rate
    A.R2.ρ = thrust
    P(t) = pressure of air
    Pa = atmospheric pressure
    A.(P - Pa) = thrust (also)
    so P - Pa = R2
    V(t) = volume of air remaining
    Vf = volume of tank = final volume of air
    Expansion of air is (near enough) adiabatic, so P.Vγ = constant c = P(0).V(0)γ = (R2.ρ + Pa).Vγ
    where γ is the adiabatic index (1.4 for air).
    Rate of increase of V = A.R = A.√((c.V + Pa)/ρ)
    I see no prospect of integrating that in closed form, so we're looking at numerical methods.
    Mass of rocket = M
    Mass of remaining water = (Vf-V).ρ
    Accn = thrust / total mass
    Numerically we can integrate that twice to arrive at the height when either V = Vf or P = Pa, whichever comes first.
    Might be worth considering a drag term:
    Accn = thrust / total mass - drag
    e.g. drag = k.(velocity of rocket)2
    The hard part would be determining k for the rocket.
     
  7. May 28, 2012 #6

    K^2

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    Drag can be ignored. You'll pick up greater errors from other factors. You can also ignore viscosity of water, so you can get an approximation for flow rate from conservation of momentum.
     
  8. May 28, 2012 #7

    haruspex

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    How does that work? Surely you need to calculate flow rate from the pressure differential and then use conservation of momentum to get the acceleration?
     
  9. May 28, 2012 #8

    K^2

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    If you ignore viscosity, you already know the thrust of the rocket. It's PA, where P is pressure and A is area of the nozzle. So all you need to do is find uniform flow velocity v, such that rate of impulse transfer is equal to force.

    Erm. Let me put this into formuale.

    F = dp/dt = PA.

    dp/dt = v dm/dt, if you eject water at some instantaneous velocity v.

    dm/dt = ρ dV/dt = ρ A dx/dt = ρAv

    I hope this last bit is clear. Water has to flow through nozzle, so Av gives you volume flow rate, and with density, ρ in place, you have mass flow rate.

    Putting it all together.

    PA = ρAv²

    v²=P/ρ

    For total momentum the rocket receives you want to integrate -v(P)dm from m-initial to m-final. Or, realizing that P(V) is a known function and dm=-ρdV, you can integrate v(P(V))ρdV for V from V initial to V final. I'm not sure if this integrates well analytically, but it's trivial to get this computation done numerically.

    Edit: Of course, this doesn't give you altitude directly. The integral for that is going to be quite a bit more complex. Still, it does help with water/gas ratio optimization.
     
  10. May 28, 2012 #9

    haruspex

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    I think you'll find all that is already in my previous post, except that I went into details of how the pressure, air volume and rocket mass will change over time. I don't think that can be avoided.
     
  11. May 28, 2012 #10

    mfb

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    With some approximations, it is possible to evaluate this:

    - assume that all the water gets ejected, neglect viscosity and other stuff
    - determine the initial volume and pressure
    - calculate the final air pressure (adiabatic process)
    - take the mean value, subtract the air pressure, assume that all water gets ejected with a velocity corresponding to this pressure (this avoids the integral - WolframAlpha cannot solve it, even for gamma=1).
    - Assume that most of the water is ejected before the rocket reaches a significant height. This might be wrong, but it allows to avoid the time-dependence of the acceleration.

    With the initial pressure p and a fraction x of water in the bottle, the final internal pressure p' is given by [itex]p'=p(1-x)^\gamma[/itex] (should be checked).
    The average pressure difference is then given by [itex]\overline{p}=\frac{p'+p}{2} - p_{outside}[/itex] and the water velocity is given by [itex] v = \sqrt{2\frac{\overline{p}}{\rho}}[/itex] with the water density rho.
    This can be plugged in the ordinary rocket equations to get the maximal velocity: [itex]v_{max}=v \log \frac{m_0}{m_1}[/itex]. Note: m0 depends on x.

    The height, neglecting air resistance is then given by [itex]h=\frac{v^2}{2g}[/itex].
     
  12. May 28, 2012 #11
    I tried again last night and since the rocket is in an open system, I got this:

    [itex]\frac{d}{dt}[/itex](mbv+[itex]\int[/itex]ρ(u + v) dV) = (Pout - Patm)Aout + Fdrag - mtotal g + [itex]\dot{m}[/itex](uout + v).

    The first part is the change in momentum equals external forces plus momentum flow through outlet, and where (u+v) is the velocity of the fluid relative to the ground, (Pout - Patm) is the pressure differential between exiting fluid and the atmosphere, [itex]\dot{m}[/itex] is the rate of change in the mass rocket,

    [itex]\dot{m}[/itex] = [itex]\frac{dm_{tot}}{dt}[/itex] = ρ[itex]_{out}[/itex]u[itex]_{out}[/itex]A[itex]_{out}[/itex]

    and the external drag force is,

    Fdrag= -[itex]\frac{1}{2}[/itex]CdρatmAbv[itex]\left|v\right|[/itex].

    And the acceleration of the bottle is defined by,

    a(t) = [itex]\frac{F_{thrust} + F_{int} + F_{drag}}{m_{total}}[/itex] - g

    where Fint is the reaction force due to internal acceleration of mass in the rocket. So in essence, I could integrate a(t) to get v(t) or should I just use Tsiolkovsky rocket equation as suggested by mfb?
     
  13. May 28, 2012 #12

    mfb

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    A numerical evaluation of the formulas will give better results, especially when one of my assumptions is not realised in your rocket.
    The analytic approach is interesting to evaluate the general performance of the systems.
     
  14. May 28, 2012 #13
    Pardon my naivete, but how would I do that?
     
  15. May 28, 2012 #14

    mfb

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    Do what? For a numerical approach, see haruspex's post. For each point in time, keep track of the amount of water in the rocket, its velocity and its height. The other variables can be calculated based on that.
    For an analytic approach, put my equations together.
     
  16. May 28, 2012 #15
    Edit: Nevermind.

    I'm having difficulties putting haruspex's expressions together. Could you provide a hint to lead me in the right direction, perhaps an example, even?
     
    Last edited: May 28, 2012
  17. May 28, 2012 #16

    haruspex

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    Yes, I looked at the total energy approach but decided there was no way to get a sufficiently accurate idea of the average height to which the water was carried. The standard rocket equations aren't quite right here because the rate of expulsion of water declines, whereas rocket fuel mass goes down linearly. But arguably they provide a lower bound.
    Also, I wasn't completely sure of the formula for the total energy available. I think it's P(0)V(0) - Pf.Vf. That's certainly the reduction in energy in the air in the tank, but I worry that this might not be taking into account work that's 'wasted' overcoming external air pressure. OTOH, it doesn't look right to subtract another Pa.Vf.
    Btw, gamma=1 can't work. It would mean there's no available energy.
     
  18. May 28, 2012 #17

    haruspex

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    Try it in a spreadsheet first. Have a column for time, going up in some small interval, and columns for each of the other variables with formulae reflecting the step-to-step relationships. You can play around with the time step size to see how small you have to make it before it stops having a major effect on the answer.
     
  19. May 28, 2012 #18
    But putting it together is what I am having trouble with. And the dots, are those supposed to be multiplication signs?
     
  20. May 28, 2012 #19

    haruspex

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    Yes. It's a reasonably standard algebraic notation. If you can have a go at turning them into expressions in a form you understand (spreadsheet formulae even) I'm happy to check them.
     
  21. May 28, 2012 #20
    I'm sorry for extending this thread for so long, but I want to understand this.

    Okay, well, I never really used spreadsheet for this so I am stuck.
    I made the A column from time 0 to 10 in .2 intervals.
    And now what?

    Attached are the formulae and variables I got from you.
    To clear up,

    A is the cross-sectional area [itex]\pi[/itex]r2

    R is [itex]\sqrt{\frac{cV^{-\gamma}+ Pa}{ρ}}[/itex], what is c?

    EDIT: Is the expression ARρ actually [itex]\dot{m}[/itex] = [itex]\int[/itex][itex]\int_{A}[/itex] j[itex]_{m}[/itex]dA where j[itex]_{m}[/itex] is the mass flux and it's a double integral since the surface is curved and not planar?

    EDIT 2: Is Torricelli's law useful in this case?
     

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    Last edited: May 28, 2012
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