# Which explodes first? Vacuumed tank or pressurized tank?

We have been arguing with a friend and couldn't agree. Here is the question:

We have two systems. In both of the systems, we have an identical sphere tank. In the first system, we are vacuuming the tank and in the other one, we are pumping air. Let's say in both cases tank explodes due to pressure difference.

If we started the process at the same time, in other words, pressure difference changes equally in time, which tank explodes first? or do they explode at the same time?

edit: I guess it is called implosion. So, basically would implosion of a body and explosion of a body due to pressure happens differently? How?

• Normally, the tank walls should be able to take more tension than compression. My guess is that the vacuum tank fails first. – David White Sep 19 at 22:50

Adding on to what others have said, this is very non-straightforward.

As others mention, the failure criteria for a vacuum vessel and a pressurized vessel is quite different. There is one very large factor that no one seems to have mentioned yet.

When the spherical vessel is under a vacuum, it develops compression, and if it's a ductile material, it is susceptible to buckling. This becomes more an issue as the vessel walls become thinner relative to the vessel radius.

This is a completely different failure mode than tensile failure, and due to the nature of buckling, it can happen at a relatively lower compressive stress than an equal strength material would fail at tensile stress.

As others mention, with a vacuum you can only reach some maximum pressure difference, then your vacuum is perfect and you will have the maximum pressure difference. So in the case of waiting, the vacuum vessel may reach a maximum before failure, while the pressure vessel could keep increasing in pressure over time and would eventually fail.

If you were to lay out these ideal conditions:

• Both pressure differentials can increase indefinitely over time, one internally pressurized and the other externally pressurized;
• The material of each vessel is equally strong in tension and compression

I would expect the externally pressurized vessel to fail first more than the internally pressurized, because it has another failure mode that can occur below the typical failure stress of the material (depending on the thickness of the vessel walls).

Basically, tensile stress in the vessel provides negative feedback loop against failure, but compressive stress creates an instability where a small deformation can cause a positive feedback loop which leads to failure.

Also, the vacuum vessel might just crumple up depending on what it's made of, while the pressure vessel would be more like an actual explosion, basically regardless of what it's made of.

There's no physics principle that gives a single answer here.

First of all, a vacuum can never achieve more than a 1atm pressure difference across a vessel (assuming we're doing this here in a normal workshop). So if you have a vessel that can withstand 5atm without any yield, then it will never have a problem with a vacuum, but would fail if pressurized far above that point.

Assuming you had a vessel that cannot withstand a vacuum, then "which happens first" is hard to tell. The speed that you can fill the container depends on the pressure that you supply. Again, evacuating a container has maximum speeds because you can't apply arbitrarily large pressure differences across the valve.

Finally, the vessel shape and material will make a difference. The particular pressure that will cause failure depends on the vessel's shape and the material properties, especially the compressive and tensile strength. Different materials could fail at different pressure levels. There's no simple answer that would hold universally.

• I think the maximum pressure difference is the biggest argument. One could design a material suitable for either case. But for a vacuum, you only need to hold back 1 atm. Whereas under pressure, you need to hold an undetermined pressure difference. It's easier to design for a known value than a potentially unknown one. – tpg2114 Sep 19 at 23:59
• That's a very fair point. Thanks a lot for the explanation. – Ekrem Sep 20 at 18:08

Though failure with the sphere evacuated is highly unlikely given a maximum pressure difference of one atmosphere, if there were no limit to the external pressure, then It would probably depend on the material.

With positive pressure inside a spherical tank, the walls are subjected to tensile stress. If negative, compressive stress. Failure may depend on whether the material is stronger in compression or tension.

Steel, for example, I understand is equally strong in compression and tension. So for steel, they may both fail at the same pressure difference. Ductile materials, such as aluminum, experience large strains under tension. An aluminum tank may fail first with air pumped in. And then again, the rate of pressure increase may be a factor.

Hope this helps.

Expanding tank vs contracting tank is the comparison. Since the collapsing tank may not tear, and therefore technically not be destroyed in that it could be reinflated and still be usable to some extent (search for images of 'tank collapse under vacuum' on web), a different way to look at the problem is useful. Two questions can be posed:

"Which tank would be first to be damage beyond normal use (as a tank)?"

"Which tank would be first to be damaged beyond any use whatsoever (as a tank)?"

The expanding tank is stable during expansion in that the pressure force acts on all internal points uniformly. It will fail fairly close to the failure strength of its weakest part (weld, joint, etc). Once it fails however, it fails beyond use since the material will tear.

The contracting tank is relatively unstable in that any non-uniformities in the strengths of its various parts will cause one part to collapse faster than another. This will result in crumpling/buckle lines, which will damage the structure by introducing stress concentrations. These buckle lines will become a weak point in the tank, whether it is subsequently caused to contract or expand. It should still be usable as weaker tank however after the first collapse, since it will generally not be torn (no hole), so it will be possible to re-inflate it.

Therefore, the contracting tank is first to be damaged beyond normal use, and the expanding tank is first to be damaged beyond any use whatsoever.

As explained in other answers, a tank can fail under internal or external pressure if the material strength is exceeded. The critical pressure differential of a thin spherical shell for this failure mode can be calculated using the following formula (to derive it, one can consider equilibrium of a half of the shell): $$2\pi R h\sigma=\pi R^2 \Delta p,$$ where $$R$$ is the radius of the shell, $$h$$ is the thickness of the shell, $$\Delta p$$ is the magnitude of the pressure differential, $$\sigma$$ is the maximum stress the material can withstand (it can be different for external pressure (compression) and internal pressure (tension)).

It was also mentioned in other answers that a shell can also fail under external pressure due to buckling. The critical pressure differential of a thin spherical shell for buckling can be calculated using the following formula (Timoshenko, S. P., Gere, J. M., Theory of Elastic Stability, McGraw Hill, New York, 1961, p. 512): $$\Delta p=\frac{2 E h^2}{\sqrt{3 (1-\mu^2)}R^2},$$ where $$E$$ is the modulus of elasticity of the material, $$\mu$$ is its Poisson's ratio (typically, $$\mu$$ is about 0.3).

Thus, for thin shells, buckling can be the more dangerous mode of failure. In general, the shell can withstand higher pressure differential under internal or external pressure depending on the shell material properties and shell dimensions.