Good day people of math.stackexchange.com

UPDATE: Version 2 can be found here: https://physics.stackexchange.com/questions/275284/modelling-a-water-bottle-rocket-version-2-long-post-warning.

This is a pet project that I plan to use to convince my Prof that I would rather try something similar to this than to do the prescribed project.

Edit : There are clearly enough views, but as of yet, not 1 reply. I believe the question is stated to late. So here it is : Am I mathematically correct? Should the $\frac{d(v^2t)}{dt}$ be left as is, then simplified like I did in my post, or should I differentiate it, then solve further? The answer is not very different, but it will surely affect the outcome.


The modelling of a Water Rocket's flight profile for a set of predefined variables.

Assumptions to start off with

  • No drag force on the rocket
  • No shear stresses inside the tank due to the flowing water
  • No buoyancy issues
  • No phase changes in either fluid
  • Temperature of everything is at $25^{\circ}C$
  • The nozzle is just a hole in the bottom of the rocket. ( depicted as a cone for clarity )
  • I am using SI-units
  • 3 decimals are acceptable for this problem

Basically, I should only use the conservation of mass and momentum on this. The reason for this is that we are only working through it now. We will start working with Navier-Stokes equations in about 5 weeks time. I have had this subject before, but didn't make it the first time. I plan to incorporate Navier-Stokes for my semester project in this subject. The subject is Transport Theory 1, a third year's Chemical Engineering subject.

Concerns/Things to take into account :

As the rocket starts to lift off, ( assuming that with the specs I have, it can ), water is ejected through a nozzle at the bottom of the rocket. This causes a pressure decrease of the air in the tank and an increase in volume of the air. This in turn causes the force applied by the air to lessen over time and as a result, the acceleration of the rocket decreases.

I have design the rocket's size to ensure that once all the water is ejected, the air inside the tank will have a pressure of 1 atmosphere to ensure that no air escapes and that a back-draft is not created. Bare in mind that this model will only be valid as long as there is water in the tank. The rest of the flight profile is quite easy.

So basically we have :

$ P_{air}:V_{air} : M_{water} : \rho{_{air}} : acceleration(a) : velocity(v) : displacement(x) $ are time dependant

$ M_{air} : Temperature(T) : Atmospheric P(P_{atm}) : \rho{_{water}} $ and rocket dimensions are constant.

Equations available to me :

Text book used : Fundamentals of Momentum, Heat and Mass Transfer by W,W,W,R

Conservation of Mass

$$ ∫∫_{c.s} {\rho}(\mathbf{v \bullet n})dA + \frac{\partial}{{\partial}t}∫∫∫_{c.v}{\rho}dV = 0 $$

Where the first term is the mass change in and out of the system, and the second term is the accumulation of mass. C.S and C.V are the control surface and volume respectively.

Conservation of Momentum :

$$ ∫∫_{c.s} \mathbf{v}{\rho}(\mathbf{v \bullet n})dA + \frac{\partial}{{\partial}t}∫∫∫_{c.v}\mathbf{v}{\rho}dV = \sum F $$

This first term is the net rate of momentum efflux from the control volume, the second term is the rate of accumulation of momentum within the control volume and the right hand side is the sum of forces acting on control volume.



$V_1$ indicates the volume of pressurised air and $V_2$ the volume of water. $M_1 : n_1 $ indicates the mass and mole amount respectively of air and $M_2$ the mass of water in the tank. $M_T$ is the total mass of the rocket. $d_1/A_1 $and $d_2/A_2$ indicate the diameter/Area of the rocket and the nozzle respectively.

Further assumptions:

$V_3 = V_2+V_1$

$h_3 = 5d_1$

$d_1 = 10d_2$

subscript (i,f) are used to define initial and final values whereas no subscript denotes a variable's value at a certain point in time. e.g $P_1 + P_{1,f} - P_{1,i} $

Let's begin, shall we?

$h_3 = 3$

$d_1 = 0.6 : A_1 = 0.283 $

$d_2 = 0.06 : A_1 = 0.003 : $

Calculating the mole amount of air need to fill the empty tank.

$$ V_3 = A_1h_3 = 0.848 $$

$$ n_1 = \frac{P_{atm}V_3}{RT} = 34.655 $$

The Volume that the air occupies with 1.5MPa $( P_{1,i} )$ of pressure :

$$V_1 = \frac{nRT}{P_1} = 0.057 $$

The Volume that the water occupies :

$$V_2 = V_3 - V_1 = 0.791 $$

The mass of all things :

$M_1 = 1 $

$M_2 = 791 $

$M_3 = 108 : mass of the rocket's shell ( chosen to make a round number for total mass )

$M_T = M_1 + M_2 + M_R = 791+1+108 = 900$ kg ;

Finding air pressure as a function of water mass

Because the water is ejected, it creates a pressure drop. It seems logical to formulate $P_1$ like this, for now :

$$ P_1 = \frac{nRT}{V_1} : V_1 = A_1h_1 $$ $$ P_1 = \frac{nRT}{h_1A_1} $$

but recall that $h_1 = h_3 - h_2$ and that $V_2 = A_1h_2$ also, $V_2 = M_2/{\rho}_2$


$$ h_2 = \frac{M_2}{A_1{\rho}} = 0.0035M_2 $$

$$ h_1 = h_3 - h_2 = 3 - 0.0035M_2$$

Plugging this into the pressure equation :

$$ P_1 = \frac{nRT}{A_1(3 - 0.0035M_2)} $$

The Mass Balance

$$ ∫∫_{c.s} {\rho}(\mathbf{v \bullet n})dA + \frac{\partial}{{\partial}t}∫∫∫_{c.v}{\rho}dV = 0 $$

**The first term can be written as ** $$ ∫∫_{c.s} {\rho}(\mathbf{v \bullet n})dA = ∫∫_{A_2} {\rho}vdA = {\rho_{2}}vA_2 $$

Since there is only an efflux of mass here, there is only one term on the right hand side. The sign is positive because the direction of the efflux and the velocity is the same, downward, cancelling each other out.

**The second term **

$$\frac{\partial}{{\partial}t}∫∫∫_{c.v}{\rho}dV = \frac{d}{dt}∫^{M_2}_{M_{2,i}}dM =\frac{d}{dt}(M_2-M_{2,i}) $$

Recall that : $M_2$ is the mass of water at any given time, and $M_{2,i}$ is the mass of water at $ t = 0 $, which is known. Since the mass is only a function of time, the partial derivative can be written as a total derivative because all variables in $M_2$ are only dependent on time.

**Plugging these two terms back into the original equation and simplifying a bit **

$$ {d}(M_2-M_{2,i}) = -{\rho_{2}}vA_2dt $$

Integrating on both sides

$$ ∫{d}(M_2-M_{2,i}) = -∫^{t}_{0}{\rho_{2}}vA_2dt $$

Evaluating the left side as an indefinite integral and solving for $M_2$ produces

$$ M_2 = M_{2,i} - {\rho}vA_2t $$

The Momentum Balance

$$ ∫∫_{c.s} \mathbf{v}{\rho}(\mathbf{v \bullet n})dA + \frac{\partial}{{\partial}t}∫∫∫_{c.v}\mathbf{v}{\rho}dV = \sum F $$

**The first term **

$$ ∫∫_{c.s} \mathbf{v}{\rho}(\mathbf{v \bullet n})dA = -v{\rho}vA_2 $$

The same reasoning behind the first term of the mass balance applies here, but the $-v$ is because $v$ is in a downward direction.

**The second term **

$$ \frac{\partial}{{\partial}t}∫∫∫_{c.v}\mathbf{v}{\rho}dV = \frac{d}{dt}(M_2-M_{2,i}) $$

The same calculation happened here as in the mass balance. Substituting $M_2$ results in :

$$ \frac{d}{dt}(M_2-M_{2,i}) = -{\rho}A_2\frac{d(v^2t)}{dt} $$

**The right hand side of the Momentum Balance **

$$ \sum F = P_{atm}A_2 - P_1A_1 - M_Tg $$

Where the first term is the force exerted by the atmosphere on the nozzle opening in an upward direction. The second term is the force exerted by the compressed air onto the water over $A_1$ in a downward direction. The third term is the gravitational force in a downward direction on the total mass, $M_T$, of the system. Recalling that $M_T$ = M_R + M_1 + $M_2 = 109 + M_2$.

Using the above and substituting $M_T$ and $M_2$ :

$$ \sum F = P_{atm}A_2 - P_1A_1 - (109 + M_{2,i} - {\rho}vA_2t)g $$

**Now, combining the three main terms to produce the completed Momentum Balance **

$$ -{\rho}v^2A_2 -{\rho}A_2\frac{d(v^2t)}{dt} = P_{atm}A_2 - P_1A_1 - (109 + M_{2,i} - {\rho}vA_2t)g $$

Simplifying the Momentum Balance

$$ -{\rho}A_2\frac{d(v^2t)}{dt} = (109 + M_{2,i} - {\rho}vA_2)g + P_1A_1 -{\rho}v^2A_2t - P_{atm}A_2 $$

This gives

$$ \frac{d(v^2t)}{dt} = \frac{(109 + M_{2,i} - {\rho}vA_2)g + P_1A_1 -{\rho}v^2A_2t - P_{atm}A_2}{-{\rho}A_2} $$

Using the chain rule on the left hand side :

$$ \frac{d(v^2t)}{dt} = 2vt\frac{dv}{dt} $$

Substituting this and simplifying :

$$ \frac{dv}{dt} = \frac{(109 + M_{2,i} - {\rho}vA_2t)g + P_1A_1 -{\rho}v^2A_2 - P_{atm}A_2}{-2{\rho}A_2vt} $$

**Plugging in values for all the constants and simplifying each term **

$$ \frac{dv}{dt} = \frac{(8829 - 29.43vt) + (0.283P_1) - (3v^2) - (303.9)}{-6vt} $$

On to the variable $P_1$

Recall that

$$ P_1 = \frac{nRT}{A_1(3 - 0.0035M_2)} $$


$$ M_2 = M_{2,i} - {\rho}vA_2t = 791 -3vt $$

These two formulae combine to form :

$$ P_1 = \frac{303400}{3 - 0.0035(791 -3vt)} = \frac{303400}{0.232 + 0.011vt} $$

In the above equation, $ nR$ , $mol.\frac{J}{molK}$ was replaced with $ M_1R$ , $ kg.\frac{J}{kgK}$

Plugging $P_1$ into the Momentum Balance and simplifying a bit

$$ \frac{dv}{dt} = \frac{3v^2 + 29.43vt -8525.1 }{6vt} - \frac{0.283P_1}{6vt} $$

Substituting $P_1$

$$ \frac{0.283P_1}{-6vt} = \frac{85862.2}{0.232 + 0.011vt}\frac{1}{6vt} $$

Simplifying again

$$ \frac{85862.2}{0.232 + 0.011vt}\frac{1}{6vt} = \frac{85862.2}{1.392vt + 0.066v^2t^2} $$

The Final, Simplified Momentum Balance

$$ \frac{dv}{dt} = \frac{3v^2 + 29.43vt -8525.1 }{6vt} - \frac{85862.2}{1.392vt + 0.066v^2t^2} $$

The Final, Non-Simplified Momentum Balance

$$ \frac{dv}{dt} = \frac{{\rho}A_2v^2 + (M_R + M_1 + M_{2,i} - {\rho}A_2v)g - P_2A_2 }{2{\rho}A_2vt} - \frac{{\rho}A_2nRT}{{\rho}A_2h_3-M_{2,i}+{\rho}A_2vt}\frac{1}{2{\rho}A_2vt} $$

Questions and Concerns

I hope that if you read this horribly long peace of work, that you found my work and explanations adequate enough.

The main Reason I posted this work here

Is my work correct? I believe that this equation is not explicitly solvable for $v$, unless someone has a way? I plan to use MatLab Simulink to graph the function $v$ and $x$, does it seem possible? It does to me, anyway.

I Thank You Sincerely for Your Time in This Matter!

  • 5
    $\begingroup$ Might this be more appropriately placed on the physics site? $\endgroup$ – Maroon Feb 20 '14 at 23:58
  • $\begingroup$ I believe the physics involved here are a bit irrelevant. I require verification of the mathematics done in it. Am I allowed to post it to the physics site as well in addition to this one? $\endgroup$ – 22134484 Feb 21 '14 at 0:00
  • 1
    $\begingroup$ No, you're not. You should pick one site, the one where your question is on topic, and post it there. Seeing as you've already cross-posted this to Physics, you should delete one or the other. $\endgroup$ – David Z Feb 21 '14 at 4:52
  • 1
    $\begingroup$ Thank you David Z for the reply. I have deleted the other post. $\endgroup$ – 22134484 Feb 21 '14 at 7:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.