## Water Rocket

 Quote by haruspex 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?

Recognitions:
Homework Help
 Quote by Alcubierre 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? Attached Thumbnails

Recognitions:
Homework Help
 Quote by 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 $\pi$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.
Attached Files
 waterrocket.xls (32.5 KB, 13 views)
 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?

Recognitions:
Homework Help
 Quote by 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?
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.

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.
Attached Files
 waterrocket.xls (43.5 KB, 14 views)
 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.

Recognitions:
Homework Help
 Quote by 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.
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.
 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?

Recognitions:
Homework Help
 Quote by 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?
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.
 Mentor 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.

Recognitions:
Homework Help
 Quote by 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.
No, it increases to Vf. Have you found something in the spreadsheet that says otherwise?
 Mentor Oh... sorry, I got confused by the decimal places. You are right.
 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.

Mentor
 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.