Nope, not volume, it’s area (Pi x r squared, 3.14 x 12" x 12") X pressure (100 pounds/square inch) = force on the end cap of the tank. The bigger the tank, the bigger the force, for a given pressure.
Tester
I’ve seen the aftermath of a split rim tire going off. I took my truck to a tire shop to have put new tires on. When I arrived, I saw a HUGE hole in the cinder block wall and an ambulance outside. Luckily the tire guy wasn’t seriously injured.
Yeah they have been using cages for split rims for years. Very nasty. Some people like to spray starter fluid in the tire and light it up to seat the rims. You should have to have a license to work, in a tre shop.
I understand pi. But seems to me the area of ag container is volume, like a five gallon pail. At any rate when I pump the compressor up and hit the blow nozzle, it sure doesn’t seem like 46 thousand pounds.
In 1989 I worked at a tire store that sold commercial truck tires, the manager told customers: “we don’t install tires on split rims, replace your wheels”. All of those old rusty split rims must have been sent to Minnesota.
It was about 1972 that the tire shop I went to bought their first cage. Never survived to 1989 but that’s another story.
The end of the pressure vessel has 12 x 12 x 3.14 (pi r squared) square inches of surface area. Each one of those square inches has 100 pounds per square inch (psi) pushing against it. The larger the area the more force pushing against it.
Same thing works for tires. You don’t think your car is pushing down with 30 pounds on the tire, do you? Each square inch of the tire that touches the ground is pushing with 30 pounds (at 30 psi). so each tire supports 30 pounds times the number of square inches for the contact patch.
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One more try: say you have a piston in a closed cylinder with 30 psi air in it. How hard would you have to push down on that piston to keep it from popping out? If I don’t tell you the diameter (hence the area) of the piston you can’t give me an answer. If it’s a 1.0 square inch piston, then it’s 1 square inch x 30 pounds/square inch = 30 pounds of force. If it’s a 2.0 square inch piston, then it’s 60 pounds of force, right?
Now imagine it’s a 100 square inch piston and 100 psi - you’d have to push with 10,000 pounds of force. That’s the force pushing out the failed end cap on a small (11.3" diameter) air tank with 100 psi air in it.
The formula for the surface area of a sphere is 4 * pi * r^2. r is radius. The end cap seems like it is roughly one half of a complete sphere (tank has a diameter of 2 feet, or a radius of 1 foot or 12 inches) , so wouldn’t the end cap’s surface area be half the surface area of a sphere , or 2 * pi * r^2? There seems to be a “2” missing.
Note that “unit” or “dimensional analysis” confirms the formula is for an area, not a volume.
The total force on the tank is indeed as claimed, 100 pounds per square inch * the number of square inches (with perhaps the formula needing a missing 2); but the force is not all pushing in the same direction. Forces in pressured gasses push at 90 degrees to the surface. But there is a net force tending to push the end cap away from the rest of the tank. Presumably the folks who designed the tank were aware of this , and constructed the tank accordingly. Over time tanks could rust, but you’d think in most case of tank failure where the tank was only pressured to 100 psi or so, it would start as a small leak, a hissing sound, not a major explosive event without any warning.
Home bbq-use propane tanks are also inflated a little over 100 psi. The place where I buy propane first confirms the tank is within a certain age limit. There’s a manufacturing date stamped on the tank. If too old, they’ll refuse to recharge it. So there must be some concern for a dangerous propane tank failure. Hard to compare the hazard of a propane tank to an air tank that fails though, b/c propane is explosive, air isn’t.
Good point, George, except that now we need to get into some additional complexity. Say your hemispherical is welded to a cylindrical tank. The force on the weld is not a function of the end cap shape, just the area enclosed by the end of the tank. So pi r squared is the correct formula.
Think of it this way-the force on the weld is the same if it’s a flat end cap or a domed one.
I’m still stuck on telling the interviewer his formula was bogus. Reminded me, in my youth I interviewed for a bank examiner job. Don’t remember anymore if it was fdic or not. But they had about ten people peppering questions. Then after a while they started in with insults of one kind or another. I figured they were just trying to see if they could get me upset. I would have walked out if they brought my mother into it, but at the end I thought who would want to work with these jerks?