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| | #1 |
| Senior Member Join Date: Feb 2007 Location: Central Vermont
Posts: 132
UserID: 976 | Just a small query: Is pressure and surface area directly proportional for force? EG: You have two chambers, both are half filled with water. 3" diameter chamber at 100 PSI 6" diameter chamber at 50 PSI. Assuming a reduction down to 1" output, will they both have the same level of power? I've been looking around for formulas on this but haven't had much luck, though from just guessing I would say that they would result in an about equal force of water out the nozzle.
__________________ Captain-Canis: Founder of the Maple-Mountain-Marines. Terrifying, but oddly refreshing. -B.D. |
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| | #2 |
| Founder Join Date: Mar 2003 Location: Maryland
Posts: 5,793
UserID: 1 | It's pressure that matters, not force, so the 3 inch chamber will be stronger. It should have approximately 87% more water output from the empirical data I've measured combined with the Bernoulli equation. Also, the answer the first question, Force = Pressure * Area, so pressure and surface area are proportional to force.
__________________ Ben Trettel / email: ben at sscentral dot org Please read this page before emailing me. Last edited by Ben : 03-02-2008 at 01:58 PM. |
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| | #3 |
| Administrator Join Date: Apr 2006 Location: Virginia
Posts: 3,073
UserID: 576 | The pressure determines the range. Thin tubing for the barrel and valve could reduce the flow coefficient, but that's not really an issue in the pressure chamber.
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| | #4 |
| Super Moderator Join Date: Oct 2004 Location: Somewhere
Posts: 241
UserID: 301 | Something just feels amiss here. If one has the following situation: A) 3" diameter chamber at 50 PSI B) 6" diameter chamber at 50 PSI wouldn't B offer better performance/power thanks to good ol' F=P*A? If diameter/volume of the PC were irrelevant to force, that'd mean builders would be better off making smaller PCs to maximize force, but I do not believe that to be the case. A larger PC should be able to yield greater stream force at lower total pressures thanks to increased area. ![]()
__________________ :: Leave NO one dry! :: iSoaker.com :: |
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| | #5 |
| Founder Join Date: Mar 2003 Location: Maryland
Posts: 5,793
UserID: 1 | The increase in PC diameter would not change the performance dramatically. Yes, I used to be be all about force, but it's not what matters after reading into it. Once you have substantial volume, changing the diameter would not affect much. I would suggest reading what was discussed in this thread even though I'm not done with it: http://forums.sscentral.org/t4795/ An article with a complete derivation of everything I did in my modified Bernoulli equation will be available eventually. Drenchenator also was working with some other more advanced equations to approximate flow with turbulence approximations... but I do not know how far he went with that.
__________________ Ben Trettel / email: ben at sscentral dot org Please read this page before emailing me. |
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| | #6 |
| Administrator Join Date: Jun 2004 Location: Maryland
Posts: 677
UserID: 320 | From my understanding, the 6" chamber would only perform better in output because the cross-sectional area is far greater. This could also increase the range because the overall flow rate is increased, though it of course depends on the nozzle. It's not the force: A greater cross-section would aid performance because it has a greater potential for flow. Edit: Well, all the above is good if the pressures are the same. I was doing some rather simple calculations with the Navier-Stokes Equations, but not in the way that would actually model flow in a Super Soaker. I was just inputing in velocity fields with zero divergence (incompressible flow) and solving for the pressure, P(x,y,z,t) in this case. This is really the limit I can do; the only partial differential equations that I can solve are the ones involved in calculating a function from its gradient. You would have to do the exact opposite of what I did--input a pressure equation and solve for the velocity field--to solve for the overall flow; at that point, calculating the flow rate at certain times and points is straightforward with a "simple" flux integral. I can't solve the partial derivatives that way though.
__________________ The Drenchenator, also known as Lt. Col. Drench. Last edited by Drenchenator : 03-02-2008 at 05:32 PM. |
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| | #7 |
| Administrator Join Date: Apr 2006 Location: Virginia
Posts: 3,073
UserID: 576 | Re-read the first post...the 3" chamber is at 100 PSI, not at 50 PSI. Yes, F = P * A, but F doesn't even matter. Flow analysis uses pressure, not an arbitrary force. (In this case, the force is arbitrary since it acts within the chamber, not at the nozzle.)
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| | #8 | |
| Super Moderator Join Date: Oct 2004 Location: Somewhere
Posts: 241
UserID: 301 | Quote:
Sidenote: my head hurts Edit: forget my babble above. Pressure is all (sort of) that matters. Well, it's pressure versus size of PC opening. Only the pressure pointing in the direction of the opening is what will determine the available force/flow. If the exit pipe is larger, you can get more force and flow. If exit pipes are the same, a larger PC will have more volume at a given pressure, but a PC at higher pressure will have more available force through the opening. ![]()
__________________ :: Leave NO one dry! :: iSoaker.com :: Last edited by isoaker_com : 03-03-2008 at 12:28 PM. | |
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| | #9 | ||
| Administrator Join Date: Apr 2006 Location: Virginia
Posts: 3,073
UserID: 576 | Yeah, for the most part, it doesn't matter how large the pressure chambers are because the flow is going to be constricted elsewhere anyway. It might help a bit though. And there's always the capacity issue. Quote:
Quote:
Beyond that, you need to consider how the nozzle and tubing increases friction and drag, as well as how the stream will perform once it's actually in the air.
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