I bought a 80 cubic foot 3000 psi aluminum scuba tank and I'm going to use it to fill a 72 cubic inch tank for paintball gun. After filling the smaller tank one time how much pressure should be in each of the tanks? I am using a device that will allow the pressure in each to equalize which will fill the smaller tank.
bonairetrip
Favorite Answer
Starting with the math:
An 80 Cft tank would be 12 x 12 x 80 = 11,520 Cubic Inches. If you multiply that by 3000 psi, you get 34,560,000.
The new combined volume the gas will occupy is 11,592 (11520 plus 72) . So, 34,560,000 divided by 11,592 means that each tank will be about 2981 psi.
That is the theory -- the actual will be somewhat different. Why? As you transfer the gas, the temp in both containers will change and that will change pressures. Second, the hose will contain some gas and, after you equalize pressure in both cylinders, you will clsoe them leaving a pressurized fill hose. You will then ven some of that air.
All of that said:
a) Be certain the 72 inch cylinder is rated to at least 72 psi, that it is within hydro, has been inspected and is in very good condition (i.e., nothing other than purely cosmetic issues). Verify that the fill rate you are considering does not exceed the manufacturer specs.
b) Fill the tank super slowly. I would probably do that by hooking the two to the hose. Open the big tank but keep the small tank closed. Close the big tank. Slowly open the small tank. This will let some of the air in the hose into the small tank. Close the small tank. Repeat as needed.
c) Ideally, keep the small tank submerged in a container of water. This will help keep it cool.
d) Exploding scuba tanks can be lethal -- this is not something to fool around with.
christler
3000 Psi Scuba Tank
Anonymous
The simplest answer is......less. There is tons of math that I can give you and show you how to calculate it but its not worth the time. I have done what you are doing several times and you can fill your paintball tank many times off your scuba tank. Let me make a few suggestions that will give you the most efficient use or your source tank so it will last the longest.
-You want to fill the small tank SLOW.
This is key to both safety and efficiency. As you fill the small tank, it will get hot. This is due to ideal gas law. You must do this slow or the tank will only take about 70% of its max charge. As it cools, the pressure will drop. So you get to fill it again. The main reason for filling slow is safety. A fast fill can shock load the structure of the tank while its also under a large amount of heat stress. I have seen pictures of these tanks when they explode, very messy. A good rule of thumb is a minimum of one minute per 1000 PSI.
-Top off the tank after each game/use. There is far less parasitic heating caused by small fills than from big fills from 0 to the max. Besides, you shouldn't ever go below 700 PSI anyway as almost all paint guns require that to operate (unless modified for low pressure use). You will get far more total shots out of your scuba tank by doing this. Safer too.
-Since you are taking responsibility of filling your tank, you need to know and understand proper inspection and safety of the tank. Not really that complicated. Basically, if it shows ANY signs of wear or damage, do not use it. Never fill over its max rated pressure. Make sure its within its hydro test date. These things are simple and if you have any questions, feel free to ask me.
Ive done this many times, used to own 4 scuba tanks and 2 114ci nitro duck tanks that I filled and maintained myself. Far cheaper than paying to have them filled.
tressa
well, lets start with radius of "r" and see where we get the surface area of a sphere is given as S=4*pi*r^2 the surface of a cylinder is given as S= pi*dia(d)*height(h) the volume of the two sphere ends together is the same as the volume of one sphere which is 4/3*pi*r^3 the volume of the cylinder is pi*r^2*h we know that the total volume has to equal 3000 ft^3 so, 3000=pi*r^2*h+4/3 * pi*r^3 if we solve for h, we will have the only height possible for each chosen radius that will give 3000 ft^3 tank (3000-4/3 *pi*r^3)/pi*r^2=h before we go there we can calculate what size of sphere will have a volume of 3000 ft^3 this will tell us the shortest possible tank (essentially no cylinder). 4/3*pi*r^3=3000 r=8.95 if I did that right so, the biggest possible radius (giving a zero length cylinder between the sphere halves) is just under 9 feet Now if I was doing this for work, I would skip the math, and get out a spreadsheet and just run the surface calculations for the vessels with radii between 0 and 9. We can try to do the math next to make the mathmaticians and teachers happy. with a radius of 1 foot, the height has to be 715 ft to get the 3000 ft^3 volume that gives us a sphere surface of 13 ft^2 and a cylinder surface of 4494 ft^2 if you make a relative cost of surface by multiplying 2*13and adding that to 4494 you get about 4520 so, given shown as a table of form r,h,cost that would be 1,715,4520 continuing through the other radii we get 2,177,2325 3,77,1670 4,41,1427 5,24,1371 6,14,1429 okay, we see that its going up again (and it continues to go up from there as you would expect) remember, the more cylinder the more total surface (the plain sphere is the smallest surface case) but the more cylinder, the LESS expensive sphere surface. Thats why there is minimum cost point. okay, we know the winning answer is a radius between 4 and 6 so we can try some smaller increments on our spread sheet and see where the turning point is about 4.89 so, the low cost vessel has a radius=4.89 diameter=9.78 straight wall height=25.1 now lets see if we can do it the hard way (3000-4/3 *pi*r^3)/pi*r^2=h the relative sphere cost is the sphere surface times 2 plus the cylinder surface, that is 2*4*pi*r^2+pi*2*r*h if you plug in the equation above that is solved for "h" in place of h in this expression then you have the relative cost function all in terms of "r" if you take the derivative with respect to r and set that equal to zero, you can solve for the "r" that will give the minimum cost with the problem stipulation if I have done my arithmetic right it should come out just like the trial and error method above. good luck