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Water Rocket |
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| May28-12, 05:08 PM | #18 |
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Water Rocket |
| May28-12, 06:08 PM | #19 |
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| May28-12, 08:15 PM | #20 |
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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? |
| May29-12, 02:16 AM | #21 |
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R is [itex]\sqrt{\frac{cV^{-\gamma}+ Pa}{ρ}}[/itex] should be - Pa: R is [itex]\sqrt{\frac{cV^{-\gamma}- Pa}{ρ}}[/itex] 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. |
| May29-12, 03:02 AM | #22 |
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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? |
| May29-12, 03:10 AM | #23 |
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| May29-12, 03:17 AM | #24 |
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Ah okay, okay. Well, thank you so much for everything! You helped me out a lot.
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| May29-12, 10:49 AM | #25 |
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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. |
| May29-12, 12:33 PM | #26 |
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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. |
| May29-12, 04:36 PM | #27 |
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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. |
| May29-12, 05:42 PM | #28 |
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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? |
| May29-12, 07:06 PM | #29 |
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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. |
| May30-12, 08:47 AM | #30 |
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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. |
| May30-12, 04:31 PM | #31 |
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| May30-12, 05:49 PM | #32 |
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Oh... sorry, I got confused by the decimal places.
You are right. |
| May30-12, 05:59 PM | #33 |
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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. |
| May30-12, 06:06 PM | #34 |
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