# Water Rocket

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 = $\frac{\pi}{2}$PD2
Height, h = ($\frac{Mi}{Mr}$)2($\frac{Pi}{\rho g}$) 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$\frac{dv}{dt}$ = $\alpha$Ve + Fext = $\alpha$Ve + Fg + Fdrag, where $\alpha$ is -($\frac{dM}{dt}$), Ve is the velocity of the water leaving nozzle, and is it safe to suggest that Fd is Stokes' drag at small velocities (Fd $\propto$ -b$\upsilon$)?

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.

haruspex
Homework Helper
Gold Member
2020 Award
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 = $\frac{\pi}{2}$PD2
Height, h = ($\frac{Mi}{Mr}$)2($\frac{Pi}{\rho g}$) 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$\frac{dv}{dt}$ = $\alpha$Ve + Fext = $\alpha$Ve + Fg + Fdrag, where $\alpha$ is -($\frac{dM}{dt}$), Ve is the velocity of the water leaving nozzle, and is it safe to suggest that Fd is Stokes' drag at small velocities (Fd $\propto$ -b$\upsilon$)?

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.
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?

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?

Okay I think I found a better expression for height. It follows,

h = 0.122g(tend)2

What do you think?

haruspex
Homework Helper
Gold Member
2020 Award
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.

K^2
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.

haruspex
Homework Helper
Gold Member
2020 Award
you can get an approximation for flow rate from conservation of momentum.
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?

K^2
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?
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.

haruspex
Homework Helper
Gold Member
2020 Award
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.
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.

mfb
Mentor
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 $p'=p(1-x)^\gamma$ (should be checked).
The average pressure difference is then given by $\overline{p}=\frac{p'+p}{2} - p_{outside}$ and the water velocity is given by $v = \sqrt{2\frac{\overline{p}}{\rho}}$ with the water density rho.
This can be plugged in the ordinary rocket equations to get the maximal velocity: $v_{max}=v \log \frac{m_0}{m_1}$. Note: m0 depends on x.

The height, neglecting air resistance is then given by $h=\frac{v^2}{2g}$.

I tried again last night and since the rocket is in an open system, I got this:

$\frac{d}{dt}$(mbv+$\int$ρ(u + v) dV) = (Pout - Patm)Aout + Fdrag - mtotal g + $\dot{m}$(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, $\dot{m}$ is the rate of change in the mass rocket,

$\dot{m}$ = $\frac{dm_{tot}}{dt}$ = ρ$_{out}$u$_{out}$A$_{out}$

and the external drag force is,

Fdrag= -$\frac{1}{2}$CdρatmAbv$\left|v\right|$.

And the acceleration of the bottle is defined by,

a(t) = $\frac{F_{thrust} + F_{int} + F_{drag}}{m_{total}}$ - 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?

mfb
Mentor
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.

Pardon my naivete, but how would I do that?

mfb
Mentor
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.

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:
haruspex
Homework Helper
Gold Member
2020 Award
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 $p'=p(1-x)^\gamma$ (should be checked).
The average pressure difference is then given by $\overline{p}=\frac{p'+p}{2} - p_{outside}$ and the water velocity is given by $v = \sqrt{2\frac{\overline{p}}{\rho}}$ with the water density rho.
This can be plugged in the ordinary rocket equations to get the maximal velocity: $v_{max}=v \log \frac{m_0}{m_1}$. Note: m0 depends on x.

The height, neglecting air resistance is then given by $h=\frac{v^2}{2g}$.
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.

haruspex
Homework Helper
Gold Member
2020 Award
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?

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.

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.

But putting it together is what I am having trouble with. And the dots, are those supposed to be multiplication signs?

haruspex
Homework Helper
Gold Member
2020 Award
But putting it together is what I am having trouble with. And the dots, are those supposed to be multiplication signs?

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.

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 $\pi$r2

R is $\sqrt{\frac{cV^{-\gamma}+ Pa}{ρ}}$, what is c?

EDIT: Is the expression ARρ actually $\dot{m}$ = $\int$$\int_{A}$ j$_{m}$dA where j$_{m}$ 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?

#### Attachments

• Optimized-IMAG0583.jpg
33.1 KB · Views: 416
Last edited:
haruspex
Homework Helper
Gold Member
2020 Award
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 $\pi$r2

R is $\sqrt{\frac{cV^{-\gamma}+ Pa}{ρ}}$, what is c?

EDIT: Is the expression ARρ actually $\dot{m}$ = $\int$$\int_{A}$ j$_{m}$dA where j$_{m}$ 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?
Whoops - I made a mistake here:
R is $\sqrt{\frac{cV^{-\gamma}+ Pa}{ρ}}$
should be - Pa:
R is $\sqrt{\frac{cV^{-\gamma}- Pa}{ρ}}$

c = P(0).V(0)γ

I didn't understand the question about mass flux and double integral. Yes, ARρ is the mass flux through the jet. What curved surface are you thinking of?

Torricelli's law is not relevant, but the more general form, Bernoulli's principle, is.
In fact, it suggests I have another mistake. Where I have
P - Pa = R2
Bernoulli says I'm missing a factor of 2:
2(P - Pa) = R2
I can't see where I went wrong, but deriving it another way does indeed bring in that factor of 2.
As someone else posted, forget about air resistance. It won't matter at the speeds you'll get.
In the attached photo, I see one error in transcription. You may have misinterpreted this:
Rate of increase of V = A.R
I mean that dV/dt = A.R.
One more error: left out -g in the acceleration.

Please see attached spreadsheet. I had to cut it off after relatively few rows to fit in the forum limit, but you can just copy the last row down to cover a longer time period. All units MKS.

#### Attachments

• waterrocket.xls
32.5 KB · Views: 139
Yeah, interpreted the rate of increase as dV/dt.
Okay, so c is the pressure of air at t = 0 times the volume of remaining air at t = 0 to gamma = 1.4? What exactly is c then?

haruspex
Homework Helper
Gold Member
2020 Award
Yeah, interpreted the rate of increase as dV/dt.
Okay, so c is the pressure of air at t = 0 times the volume of remaining air at t = 0 to gamma = 1.4? What exactly is c then?
It's the quantity that's preserved in adiabatic expansion. AFAIK, it may be called the adiabatic constant for the process. Not aware of any other name or physical interpretation.

Ah okay, okay. Well, thank you so much for everything! You helped me out a lot.

Last edited:
mfb
Mentor
The first steps are a bit rough, and the volume of the air should be larger to keep the high thrust longer.

Here a modified version - I fixed the total volume to 1 liter, but that is arbitrary. On the right side I added the analytic approach - as you can see, the value is too high, but it is not completely off.

#### Attachments

• waterrocket.xls
43.5 KB · Views: 157