Calculating the wall thickness of a vacuum chamber

In summary: Yes, I think it would be a good demo for that thread. In summary, the minimum chamber wall thickness to prevent implosion in a vacuum chamber can be calculated using ASME Section VIII. It is necessary to use special materials and expensive pumps in order to achieve a ten-millionth of atmospheric pressure. There is no difference between operating at 1 Pa or 10-9 Pa.
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David761
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TL;DR Summary
I am designing a cylindrical vacuum chamber and need to calculate the minimum wall thickness to prevent implosion.

I have an inner radius of 0.96 m, an internal pressure of 10^-9 Pa and the external pressure is atmospheric (101,325 Pa). The material has a yield stress of 355 MPa.

Is there any simple equation that can be applied here to calculate the minimum chamber wall thickness? Thanks!
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  • #2
David761 said:
Summary:: I am designing a cylindrical vacuum chamber and need to calculate the minimum wall thickness to prevent implosion.

I have an inner radius of 0.96 m, an internal pressure of 10^-9 Pa and the external pressure is atmospheric (101,325 Pa). The material has a yield stress of 355 MPa.

Is there any simple equation that can be applied here to calculate the minimum chamber wall thickness? Thanks!

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Welcome to the PF, David. :smile:

What is your background in Mechanical Engineering? Can you say more about this project? Do you have any help on the project, and more importantly are you working with an experienced Mentor who can ensure that you do this safely?

You mention the inner diameter of around 1 meter, but do not say how long it is. How are you planning on sealing the ends of the cylinder? Where are you going to put the hatch (with the exploding bolts)? :wink:

What vacuum pumps are you using to achieve 10^-9 Pa, and what is your anticipated pump-down time?

https://en.wikipedia.org/wiki/Ultra-high_vacuum

https://vacaero.com/wp-content/uploads/2012/02/figure-2_uhv_lg-1.gif

1586104516603.png
 
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  • #3
David761 said:
an internal pressure of 10^-9 Pa
That's deep into the ultrahigh vacuum. You'll need special low outgassing materials, proper welds, expensive pumps, baking of the chamber, copper gaskets, and on and on. This will take a long time to get right and will be expensive.
 
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  • #4
David761 said:
Is there any simple equation that can be applied here to calculate the minimum chamber wall thickness?
No. If you use search term vacuum vessel design calculations, the very first hit starts of with "External pressure (vacuum) calculations start off more complex than internal pressure calculations, ".

That site also mentions that "The calculations are found in ASME VIII-1 UG-28". The last time I read through ASME Section VIII for pressure vessels, I believed the friend who told me that, in order to properly understand and use it, a person needs to work with Section VIII for a full year under the guidance of somebody who knows their way around that document.

That's in addition to Posts #2 and #3. If this is a real project, get a copy of ASME Section VIII, read it enough to understand that you are over your head, then find somebody that knows how to design vacuum vessels. If this is a homework problem, study your textbook to understand what assumptions and simplifications they are making, then post them along with a request to move this to the homework forum.
 
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I think you can all relax a little. There is something about a hard vacuum that terrifies otherwise sane people. It is the high energy physics that goes on in a vacuum that can be dangerous.
https://en.wikipedia.org/wiki/ASME_..._-_Rules_for_Construction_of_Pressure_Vessels
That applies to vessels “having an internal or external pressure which exceeds 15 psi (100 kPa)”.

It should exempt vacuum vessels in the atmosphere, from requiring certification.

The stored energy in a vacuum chamber is negative. It can implode, away from you. Take care not to pinch your fingers.
 
  • #6
The danger is essentially the same at 1 Pa and 10-9 Pa, that's not the point. It's significant for every internal pressure much lower than atmospheric, especially with a vacuum that large. An implosion can still come with smaller components shooting away at high speed.
 
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  • #7
The walls don't need to be very thick to prevent an implosion. The reason why UHV system has thick walls has to do with the need to be able to attach windows, flanges etc AND the fact that even solid metal will be somewhat porous to gases such as hydrogen.

Mfb is of course correct that there is no difference between 1 Pa and a nPa when it comes to the risk of implosion. Hence, any container that is designed to be operated below 1 Bar should be strong enough.

If you want to design something that can reach 1e-11 mBar the risk of implosion is probably the least of your concern.
 
  • #8
The problem with thin walls is buckling. The cylinder is only strong while it remains cylindrical. For that reason hoops are needed to maintain the surface curvature if the wall is too thin. Any dent or elliptical deformation will cause an unbalance of forces that can initiate a catastrophic collapse.

The American Physical Society gives the following surprising demonstration of what can happen with thin walled vessels.
https://www.aps.org/programs/outreach/guide/demos/cancrush.cfm
 
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  • #9
Baluncore said:
The American Physical Society gives the following surprising demonstration of what can happen with thin walled vessels.
https://www.aps.org/programs/outreach/guide/demos/cancrush.cfm
Cool demo. Maybe should be posted in the "Physics Labs at Home" thread, since it's pretty easy for students to do while studying at home now. :smile:
 
  • #10
Is the hydotest applied for vessel working with vacuum -.9bar only or not
 
  • #11
Oroana said:
Is the hydotest applied for vessel working with vacuum -.9bar only or not
What is "hydotest" ?
Hydro test = hydrostatic pressure test ?
 

1. How do you calculate the wall thickness of a vacuum chamber?

The wall thickness of a vacuum chamber can be calculated using the formula: t = (P x R) / (2 x S), where t is the wall thickness, P is the pressure inside the chamber, R is the radius of the chamber, and S is the maximum allowable stress of the material used to construct the chamber.

2. What is the maximum allowable stress for the material used to construct a vacuum chamber?

The maximum allowable stress for the material used to construct a vacuum chamber depends on the type of material and its properties. Generally, the maximum allowable stress is determined by the yield strength of the material, which is the maximum stress that the material can withstand before permanent deformation occurs.

3. What factors should be considered when calculating the wall thickness of a vacuum chamber?

When calculating the wall thickness of a vacuum chamber, factors such as the pressure inside the chamber, the material properties, the desired safety factor, and the shape and size of the chamber should be considered. These factors will affect the final thickness of the chamber walls.

4. How does the pressure inside the vacuum chamber affect the wall thickness?

The pressure inside the vacuum chamber has a direct impact on the required wall thickness. As the pressure increases, the walls of the chamber must be thicker to withstand the force exerted by the vacuum. Therefore, the higher the pressure, the thicker the walls of the chamber need to be.

5. Are there any safety considerations when calculating the wall thickness of a vacuum chamber?

Yes, safety should always be a top priority when designing a vacuum chamber. It is important to consider the maximum allowable stress of the material and to choose a safety factor that will ensure the chamber can withstand the expected pressure without failing. It is also important to regularly inspect and maintain the chamber to ensure its structural integrity.

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