# Calculating max Vacuum for 55 gallon drum

• kevin99
In summary, the vacuum pump needed to pull the steel drum down to 1/3 bar would need to be able to generate a compression of around 3300 psi. It is uncertain if a plain carbon steel drum would be able to withstand this pressure, as it would buckle under the vacuum.
kevin99
I'm trying to make a small Vacuum chamber for some hobby experiments to simulate high altitudes..

I really don't know that much about Vacuum physics so was hoping someone could comment.

My idea is to get a vacuum pump and pump out a 55 gallon steel or plastic drum to 1/3 bar.

My calculation is that the hoop stress, o=P*t/d

If we have 1 bar on the outside and 1/3 bar on the inside, then we have P=2/3bar=67e3 Pa.

The drums are 22.5 inches (.572 m) in diameter and 33.5 inches (.850 m) high.

So if we assume the steel drum is A36 with yield stress of 250MPa then,

250e6=67e3*t/.572 -> t=2.13e3 m

so min thickness is 2.13e3 m?

I'm not sure how thick these steel drums are but after googling, http://www.skolnik.com/blog/steel-drums-thickness-can-preclude-re-use/

says drums could be .6mm (24 gauge).

So that means crushing pressure=2.4e11 Pa = 240 000 bar?

Something seems to be wrong with my calculations.

Can someone comment if my sealed 55gallon steel drum would hold up to being pumped down to 1/3 bar?

Also, I don't know much about vacuum pumps, can I find an appropriate pump cheap on ebay that I can run 24/7 and maintain a vacuum of 1/3 bar?

Thanks in advance for all the help.

First, let's examine the forces in the steel when the inside is overpressured by 10 psi (I will use English units here, because engineers think in psi (pounds per square inch). With a 22.5 inch diameter drum, and steel 0.032 inch thick, the tensile stress in the steel wall is 10 psi x 22/.064 = 3440 psi or 229 bar, a very reasonable tensile stress for steel. The yield strength for mild steel is about 50,000 psi.

However, for a tank holding vacuum, I found the following paragraph in an engineering design book:
"For a plain carbon steel tank to be able to withstand one atmosphere of external pressure (or full internal vacuum), with a safety factor of 4, it has to have walls about 1 percent as thick as the diameter of the tank. Even without no safety factor, the walls must be almost that thick, because the rigidity of the walls increases with the cube of the thickness---a wall half as thick is only 1/8 as rigid, and will collapse under about half an atmosphere (around 8 PSI) of vacuum."
So roughly, for a 22 inch diameter tank, the walls should be about 0.22 or 1/4 inch thick.

Hmm seems like my idea of using a thin walled steel oil drum wouldn't work. A pitty because I buy these things for $25. A quick look on ebay shows some big pressure vessels large enough for someone to crawl into for around$1000.

What kind of vacuum pump would I need to pull it down to 1/3 atmosphere? Most lab pumps I see can get very low Torr but 1/3 bar seems much easier to achieve.

The problem with a thin walled vessel for a vacuum is that it may buckle and that will lead to collapse. This is an instability, rather than a case of overstress.

55-gallon drums come in different gauges. The ones meeting UN standards are usually 18 gauge. The following site gives specs for one type:
http://www.ornl.gov/sci/tpm/pdf/CSCompDrum/110-5885pscreen.pdf
They are tested to 250kPa (36 lbs/sq in). Note: the rolling rings help to prevent buckling.
As for pumps the following site shows some pumps. They may be of some help.

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At one atmosphere (14.7 psi external) the compression in a longitudinal seam is only 3300 psi for 18 gauge (0.050") steel. It certainly shouldn't fail due to this stress.

The axial stress on the rolling rings is another matter, and far higher.

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The rolling rings will actually experience bending stresses as well as axial and circumferential stresses. This is because the tendency to elongate or compress the barrel creates bending in the wall at the rings.

Dr.D said:
The rolling rings will actually experience bending stresses as well as axial and circumferential stresses. This is because the tendency to elongate or compress the barrel creates bending in the wall at the rings.

What's the bending stress?

As the barrel tends to collapse under the vacuum, the rolling rings will experience bending which will induce a bending stress in the material of the wall of the ring. It is an Mc/I type stress, although the calculation will be more complicated than that.

How much stress?

Well, you know, that is a rather complicated shell bending problem that I just don't happen to have on the tip of my tongue. Sorry to disappoint you on that.

I rather doubt that it has ever been solved in general because it will depend on the exact shape of the rolling ring. If you really want to know, I would suggest an FEA model.

Video on YouTube shows one sample surviving -10 in/Hg and collapsing at -12 in/Hg.

If not instructive, it's fun to watch.

It would be extremely interesting to know how reproducible that experimental result is. Bipolar, thanks for posting that link!

## What is the formula for calculating max vacuum for a 55 gallon drum?

The formula for calculating max vacuum for a 55 gallon drum is:

Max Vacuum = (Volume of Drum x Atmospheric Pressure) / (Surface Area of Drum x Height of Drum)

## What is the typical atmospheric pressure used for this calculation?

The typical atmospheric pressure used for this calculation is 14.7 psi (pounds per square inch).

## How do I determine the volume of a 55 gallon drum?

To determine the volume of a 55 gallon drum, use the formula:

Volume = (π x Diameter² x Height) / 4

For a 55 gallon drum, the diameter is typically 22.5 inches and the height is 33.5 inches, so the volume would be approximately 12,523.8 cubic inches.

## What is the surface area of a 55 gallon drum?

The surface area of a 55 gallon drum can be calculated using the formula:

Surface Area = (π x Diameter x Height) + (π x Diameter²) / 2

For a 55 gallon drum with a diameter of 22.5 inches and a height of 33.5 inches, the surface area would be approximately 2,151.5 square inches.

## How do I convert the max vacuum in psi to other units of pressure?

To convert the max vacuum in psi to other units of pressure, use the following conversions:

• 1 psi = 6,894.76 Pascals
• 1 psi = 68.948 millibars
• 1 psi = 51.715 millimeters of mercury

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