Is there any way to calculate the force required to remove the lid of a chamber under vacuum?

Question

Where I work we have vacuum chambers which are about 20’ in diameter and around 2,000 cubic feet. When under vacuum, they’re under 1 militorr, which is almost perfect vacuum. The lids of the chamber simply rest on top. Someone at my work and myself were curious if someone accidentally tried to lift the lid under vacuum, if the crane would actually lift it (15 ton capacity). I don’t know if there’s a mathematical formula to calculate this?

Morningfox · Answer

Yes, there is a formula.You need the area of the lid, then multiply that by 14 pounds per square inch. So if the area of the lid is 2143 square inches, it would take 15 tons to lift it. For a round lid with a diameter of 52.2 inches, the area would be 2143 square inches.

You *could* use the more precise 14.696 psi, but it really depends on your altitude and local air pressure.

roger · Answer

you  would  need  about  332.   tons to  lift it.
10*10*pi * 144 *14.7  pounds  =665012  lb
divide by  2000  to get   332  tons

the   formula  is the area  of the  lid in  square inches  times  14.7   to get  pounds.

Captain Matticus, LandPiratesInc · Answer

Multiply the difference in pressure values by the area of the lid.  Add in the value of the weight of the the lid and you've got the minimum force needed to break that seal.  You have 14.7 lbs per square inch

A = pi * (10 * 12)^2 = pi * 14400

14.7 * 14400 * pi =>
665012.33291188743271777235137261 lbs

So, can your 15-ton crane pull that?  No.  But it's still not a good idea to try.  If you pushed the crane to capacity and you suddenly lost that vacuum, the lid would accelerate upward in an uncontrolled manner.

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Is there any way to calculate the force required to remove the lid of a chamber under vacuum?

3 Answers