2
$\begingroup$

Let'say the steel tank would be submerged deep in the ocean with 136 atm ambient pressure then the other tank with the same material would just be used to contain 136 atm pressure. How would the thickness of the wall differ?

Also, if the tank is designed to be sunk deep in the ocean, would it be cheaper to design a tank that could contain 136 atm pressure then pressurize it at that pressure so that it would have zero forces acting on the tank's walls if the ambient pressure is also 136 atm?

$\endgroup$
4
$\begingroup$

It's the difference between holding up a weight by by hanging it from a cable vs. putting it on top of a column.

A tank that starts out round (ideally, if it starts out spherical) and is pressurized will tend to have its shape held by the pressure of the material inside. Basically, the internal pressure will try to make the tank round, so if that's the starting point, the internal pressure will, to some extent, help support the tank.

Think of a balloon -- the walls essentially don't have any bending strength, yet the balloon can hold pressure.

A tank that's maintaining a vacuum has an unstable shape: it wants to suck down to nothing. A perfectly spherical tank can withstand that, but it's not stable in buckling. Any dent will tend to "want" to get bigger; this tendency will have to be opposed by walls that resist local bending even as they're getting compressed by the outside forces.

And yes, if you were going to sink a tank in the ocean, and if it otherwise made sense to do so, starting with it pressurized before sinking it would be a good idea. Another sensible idea, if it fits what you're doing otherwise, would be to leave it open to the seawater and just let it fill as it sinks, then purge it once it's been installed.

$\endgroup$
2
$\begingroup$

when the tank is pressurized with 185atm it is very simple to calculate the thickness needed to hold the pressure. Say the radius of the spherical tank is R. the max pressure is at a great circle passing through the center and is equal to $ \pi*R^2*186 $.

This pressure has to be supported by the tension stress on the thickness of the peremeter of the tank, t, at that circle.

$t = \frac{\pi R^2*186}{2 \pi R* \text{allowable tension stress of steel}}$

Which is a reasonably small number.

But if the tank will be exposed to 185atm external pressure it is susceptible to catastrophic buckling and explosive crushing. so it has to be designed for moment resistance in case of buckling, which will turn out to be mostly following empirical methods which includes thick wall vessel design and possible use of rib stiffeners or use of thick hardened steel or special materials.

$\endgroup$
  • $\begingroup$ What about for cylindrical vessel with domed endcaps that would contain high pressure? How do you calculate the thickness for the wall and domed endcaps? $\endgroup$ – booger king Oct 9 at 8:46
  • $\begingroup$ @boogerking, the cylindrical tank walls are subject to triaxial stresses, radial compression and circumferential tension and longitudinal tension and analyses of these stresses is not a simple process especially at the interface of the caps and the walls. In Roark's formulas for stress and strain there are formulas for different cylindrical shells and vessels some which are commonly used by the industry are empirical industry adopted and some are analytical deferential equations. $\endgroup$ – kamran Oct 10 at 0:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.