Calculating Water Rocket Performance: Thrust, Velocity & More

[Alcubierre]

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} = \alphaV_e + F_ext = \alphaV_e + F_g + F_drag, where \alpha is -(\frac{dM}{dt}), V_e is the velocity of the water leaving nozzle, and is it safe to suggest that F_d is Stokes' drag at small velocities (F_d \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]

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} = \alphaV_e + F_ext = \alphaV_e + F_g + F_drag, where \alpha is -(\frac{dM}{dt}), V_e is the velocity of the water leaving nozzle, and is it safe to suggest that F_d is Stokes' drag at small velocities (F_d \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?

 

[Alcubierre]

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?

 

[Alcubierre]

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

h = 0.122g(t_end)²

What do you think?

 

[haruspex]

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.ρ
V(t) = volume of air remaining
Vf = volume of tank = final volume of air
Expansion of air is (near enough) adiabatic, so P.V^γ = constant c = P(0).V(0)^γ = (R².ρ + Pa).V^γ
where γ is the adiabatic index (1.4 for air).
Rate of increase of V = A.R = A.√((c.V^-γ + 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]

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]

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]

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}.

 

[Alcubierre]

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

\frac{d}{dt}(m_bv+\intρ(u + v) dV) = (P_out - P_atm)A_out + F_drag - m_total g + \dot{m}(u_out + 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, (P_out - P_atm) 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,

F_drag= -\frac{1}{2}C_dρ_atmA_bv\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 F_int 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]

A numerical evaluation of the formulas will give better results, especially when one of my assumptions is not realized in your rocket.
The analytic approach is interesting to evaluate the general performance of the systems.

 

[Alcubierre]

Pardon my naivete, but how would I do that?

 

[mfb]

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.

 

[Alcubierre]

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?

 

[haruspex]

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]

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.

 

[Alcubierre]

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]

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.

 

[Alcubierre]

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 \pir²

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?

 

[haruspex]

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 \pir²

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.ρ
Bernoulli says I'm missing a factor of 2:
2(P - Pa) = R².ρ
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.

 

[Alcubierre]

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]

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.

 

[Alcubierre]

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

 

[mfb]

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.

 

[Alcubierre]

Which one is more correct because the two sets are different in the velocity, time that it takes for the water to fully escape rocket, flow velocity, and the adiabatic constant.

Edit: the change in total volume affected all of the above? And mfb, how did you derive your height equation, I get the same expression but without the square root. And is the area A theoretical because the area of the cross section is pi*r^2 and the radius is 0.025 meters.

 

[haruspex]

Which one is more correct because the two sets are different in the velocity, time that it takes for the water to fully escape rocket, flow velocity, and the adiabatic constant.

mfb, didn't understand what you said about fixing the total volume to one litre. That's what I already had: Vf = 0.001.

You can get a handle on the accuracy of the spreadsheet by making the time steps finer until it doesn't change the result much.
mfb is right that the steps were much too coarse for the early stages. I've change the time step formula in my copy to e.g.:
A5 = A4*1.01 + 0.0001
That way you get a much finer step initially, but the steps broaden as things settle down. Another approach would be to make the step size inversely proportional to the acceleration caclulated in the preceding row.
Determining the accuracy of an analytic solution, given that it necessarily simplifies some equations to be able to solve them, is much harder. OTOH, if it is at all accurate then you could differentiate wrt settings (like initial airspace) to see the effect of changing them.
Of course, you could do the same or better programmatically. Turn the spreadsheet into a C program and wrap it in a loop which hunts for max height.

 

[Alcubierre]

Haruspex, how did you get area A?

I noticed that mfb has different values for V0 (volume of air) and Vf. Vf has another decimal place (8E-4) and it's even smaller than haruspex's (1E-3). Why the discrepancy?

 

[haruspex]

Haruspex, how did you get area A?

I noticed that mfb has different values for V0 (volume of air) and Vf. Vf has another decimal place (8E-4) and it's even smaller than haruspex's (1E-3). Why the discrepancy?

I just plugged in some sensible numbers to see the spreadsheet work. It's up to you to put the right ones in. As I said, I used MKS throughout, so A is in sq metres.
From mfb's earlier comment about fixing the total volume, I think there might be some misunderstanding here. The spreadsheet takes Vf as a fixed total volume with air volume V=V(t) expanding from V(0) towards Vf, and the remaining water at time t is therefore Vf-V(t).
If the initial air pressure is insufficient for the given V(0) then V will never reach Vf.

 

[mfb]

Your initial V_air is V0, and it increases to V0+Vf (assuming the pressure is enough to eject all water). Therefore, the total volume is V0+Vf. I fixed this, which gives the formula V0=Vtotal-Vf with the total volume Vtotal.

I modified the values a bit to increase the height and to increase the similarity with the coke bottle. 0.5kg are still much for the empty rocket, I think that can be reduced a bit - and the total height depends crucially on this parameter.

 

[haruspex]

Your initial V_air is V0, and it increases to V0+Vf (assuming the pressure is enough to eject all water). Therefore, the total volume is V0+Vf. I fixed this, which gives the formula V0=Vtotal-Vf with the total volume Vtotal.

No, it increases to Vf. Have you found something in the spreadsheet that says otherwise?

 

[mfb]

Oh... sorry, I got confused by the decimal places.
You are right.

 

[Alcubierre]

P(t) or P0 is the pressure of the air inside the bottle? If so, that is obtained by the amount of psi used in the experiment, correct, which in my case, will vary from 50 psi to 65 psi (but in pascals)?
And how is the volume of the air, V0 obtained?

I apologize for all the questions, I just want to be completely clear with all the steps used.

 

[mfb]

P(t) or P0 is the pressure of the air inside the bottle? If so, that is obtained by the amount of psi used in the experiment, correct, which in my case, will vary from 50 psi to 65 psi (but in pascals)?

Right

And how is the volume of the air, V0 obtained?

The volume of the bottle minus the volume of the water inside.

 

[Alcubierre]

Okay, so haruspex's spreadsheet is fine except that I need to fix the time and change the variables to the ones I need (mass, etc).

Edit: 2 L is 0.002 cubic meters and the volume of the water is 0.001 cubic meters and that gives me 0.001 cubic meters. When I change V0 to that, all the other variables because the same.

 

[Alcubierre]

Help. Anyone? This is due in a few hours and I don't know how you got the volume of air so low when I get 0.001 cubic meters and that ruins the whole spreadsheet as it must be, apparently, at least 0.0001 cubic meters to take any effect.

 

[haruspex]

Help. Anyone? This is due in a few hours and I don't know how you got the volume of air so low when I get 0.001 cubic meters and that ruins the whole spreadsheet as it must be, apparently, at least 0.0001 cubic meters to take any effect.

This is probably too late. I plugged in Vf = .002, P0 = 4*atmospheric ("=4*I2"), M = 0.5, A = 0.002 (pi*.025^2). I played around with V0 to maximise height. I got nearly 5m at V0 = 0.00113. But go just a fraction over that and you lose big time.
Not sure about P0. If your gauge says 45 psi, is that 3 atmospheres, or 3 atmospheres in addition to the background 1 atmosphere? I'm assuming it's the second.