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* There may be some math errors in what follows but I don't have the time right now to get things right, but you will see them. Sorry for the sloppyness* Re: Water Powered Rocket. Date: 1997/04/26
In article <862018574.4174@dejanews.com> dmoe@luminet.net writes:
> In my physics class, we were firing rockets which were made from a two > liter bottle with water in the bottom of the bottle(we could pick the > amount) and then filled with air to 80 psi. Now to the question(s)... A. > What is the ideal amount of water to put in the bottle(to make it go a > long distance)? or B. How would one figure out the ideal amount of > water to put in the bottle? BTW, my rocket went about 169m give or take. > Thanks John Moe
I will try to get you started on question B.
Equations that will be helpful:
Let the rocket be filled with 1 liter of water which is under a pressure of 80 psi and consider what happens in say the first 1/1000th of a second after the rocket is set off. Use .05 Kg for an estimate of the empty mass of the rocket. Use Bernoulli's eq. to determine the initial velocity of the water (v2), the term rho g(y_2 - y_1) can be ignored for now.
Now figure out how much water escapes in 1/1000 seconds, it will simply be the velocity of the water times the area of the bottle opening times 1/1000 seconds. Using an opening area estimate of 3.49E-4 m^2 i get a differential volume of water expelled of,
Now use this information to determine the change in momentum of that small volume of water. It is simply dV times the density of water times the velocity of the water:
Now use F = dp/dt to get the force due to the rocket thrust. dp is given above and this change in momentum occurred in 1/1000 seconds so,
Quite impressive, two such rockets working together would lift the average person off the ground! It would be an interesting experiment to determine what the actual thrust is by say placing weights on the rocket and determining how much weight the rocket could just move off the ground.
Use this force plus the force due gravity (mg = 1.05 Kg * 9.8 m/s^2 = 62.8 N) to determine the change in momentum of the rocket (let V_r stand for the velocity of the rocket),
Now if we use one half of this number as the average velocity we will determine about how far the rocket went in 1/1000 seconds,
Now repeat this process to determine the velocity of the rocket after 2/1000 seconds. The new velocity should be averaged with the previous estimate and this value multiplied by the time interval gives you an estimate for the distance traveled in that time interval. Now continue this process until the velocity changes sign and ask the computer to "spit" out the distance traveled. Repeat this process for different initial loads of "rocket fuel" to come up with an ideal set-up. Keep the following things in mind. After the first 1/1000 of a second the rocket will be a little lighter. How much, well we figured that out above, it is the volume of water expelled in 1/1000 seconds times the density of water and is,
Also the water is now under less pressure, just use PV = constant to determine the new pressure keeping in mind the initial volume of air is 1E-3 m^3 and the change in volume is dV above so the final volume of air (after 1/1000 seconds) is,
With a little thought you should be able to set up a computer program to do the above computations.
What we have left out:
Aerodynamic forces which will be important near the end of the flight. The force will be proportional to the area of the rocket and to the velocity of the rocket squared. The proportionality constant can be determined by estimating the terminal velocity of the rocket. At the terminal velocity the drag force and the gravitational force will be equal. Say for example the rocket is estimated to have a terminal velocity of 20 mph or 29 ft/s or 8.8 m/s. Then we know,
Using an estimate of .05 Kg for the mass of the empty rocket the constant is then about,
So the drag force on the rocket is approximately,
This force must be added to the other two forces acting on the rocket to determine its change in momentum and therefore its change in velocity.
For better results use a smaller time interval.
All this won't be easy to do but may be fun. If there are any glaring mistakes above i hope they are pointed out to you and myself. Sorry if the above is not too coherent but if you are clever you should get the general idea.
Water rocket launcher parts list and offer. Date: 1997/05/19
With the following parts i have constructed a water rocket launcher. The rocket bodies are any of a type of disposable plastic soda container with common size bottle neck of .840 inches inside diameter. The "rocket fuel" is water and a bicycle pump "supplies the energy". Only common tools are need in construction.
Tools Needed:
If you would like a photocopy of detailed plans for the above send me your address and a reason why i should send to you the plans at no cost.
Re: Water Powered Rocket Date: 1997/04/26
In article <5js7tc$1cmq@r02n01.cac.psu.edu> ale2@psu.edu (ale2) writes:
> All this won't be easy to do but may be fun. If there are any glaring > mistakes above i hope they are pointed out to you and myself.
Well i found one mistake i made in listing the equations i thought would be useful: > > Equations that will be helpful: > > F = ma > F = dp/dt > p_1 - p_2 = rho(v2^2 - v1^2)/2 + rho g(y_2 - y_1) A form of > Bernoulli's eq. > PV = nRT the ideal gas law ^^^^^^^^
What is needed for this problem is the pressure-volume relationship for a gas which does work on expanding and for which no heat flows in or out of the gas, it is,
PV^gamma = constant
For air gamma is about 1.40.
Later i write > > Also the water is now under less pressure, just use PV = constant to ^^^^^^^^^^^^^ this should not be used but the above PV^gamma = constant
> determine the new pressure keeping in mind the initial volume of air is > 1E-3 m^3 and the change in volume is dV above so the final volume of > air (after 1/1000 seconds) is, > > 1E-3 m^3 + 1.16E-5 m^3 = 1.01E-3 m^3 so the new pressure is > > P2 = P1V1/V2 = [80 psi * 1E-3 m^3]/1.01E-3 m^3 = 79.2 psi ^^^^ wrong
So P2 = P1[V1/V2]^gamma = 78.9 psi
Re: Water Powered Rocket. Date: 1997/05/06
In article <19970506203501.QAA27529@ladder01.news.aol.com> dgoncz@aol.com (DGoncz) writes:
> OOPS! > > PV^gamma # 1.4 > > gamma = about 1.4 > > PV^gamma = nRT > > Sorry.
I still think you have it wrong %^)
For reversible expansion of a gas the work done by the gas translates into the temperature of the gas dropping. PV = nRT always works for an ideal gas but in the water rocket we don't know the final temperature of the gas. In this case we plug and chug with the following:
PV^gamma = a constant
But i could be wrong %^(
Link to thread of next article. When viruses reenter the Earth's atmosphere, do they get hot? Date: 1997/05/05
Say i am on a spacewalk and just as i sneeze my helmet flies off and my germs enter outer-space and i die. As the germs are in low earth orbit they will eventually reenter the earth's atmosphere. Can someone give me an order of magnitude estimate for how hot such germs (viruses and bacteria) will get on their trip back to the surface of the earth? Exclude any heating due to the sun and consider only the heating due to the frictional heating of entering the atmosphere.
Re: When viruses reenter the Earth's atmosphere, do they get hot? Date: 1997/05/06
In article <01bc59b7$c02b9460$1b47dbcf@bfielder.quadrant.net> "Bruce C. Fielder" <bfielder@quadrant.net> writes: > > Been watching "Invader", have we?
No, just wondered how many plants, animals, and other life forms get onto the surface of Mars when the earth is struck by the biggest of meteorites. Seems like at the very least soil bacteria must make it to the surface of Mars in small clumps of dirt. That is why I'm kind of curious how big something can be and not get too hot when it reenters the atmosphere. For example if we put a rat into orbit and let it reenter the earth's atmosphere will the bacteria in the gut of the rat make it back to the surface of the earth alive? I know, Mars does not have much of an atmosphere, but it is once thought to have had one. > > Thinking on my keyboard, I suspect that a first approximation may be > achieved by a ratio of molecular mass. A virus, with a "molecular" mass of > 1,000,000 or so, could be expected to heat up to a ratio of 1,000,000/28 > (mass of N2) x (whatever the upper atmosphere temperature is, not that it > makes a difference with this ratio). As a second order estimate, take into > account radiation cooling, which would take place between collisions with > air molecules. The initial speed of the virus would obviously make a huge > difference. Even at comet speeds, I would suspect (minus _any_ math) that > a virus would quickly burn up. If we assume that they have been 'laid' so > that the earth will sweep through them, then any answer you want is > possible. Can anyone add some math to this? > > ale2 <ale2@psu.edu> wrote in article <5kkok1$1g8o@r02n01.cac.psu.edu>... > > Say i am on a spacewalk and just as i sneeze my helmet flies off and my > > germs enter outer-space and i die. As the germs are in low earth orbit > > they will eventually reenter the earth's atmosphere. Can someone give > > me an order of magnitude estimate for how hot such germs (viruses and > > bacteria) will get on their trip back to the surface of the earth? > > Exclude any heating due to the sun and consider only the heating due to > > the frictional heating of entering the atmosphere.
Re: How the Hell is this possible? Date: 1997/05/16
In article <jjl-ya02408000R1505971028580001@news.knoware.nl> jjl@knoware.nl (J. J. Lodder) writes:
> With the sun higher than 42^o no rainbow is possible.
If you are atop a mountain you can look down below the "horizon" so all bets are off.
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