Volume of Oil for Pressurization

69.12m³

68.13m³ (assuming vessel originally filled with air at 1 bar).

29.12m³ more than the oil it contains now, assuming it is full?

In sufficient data.

Pleas do not take the above posts seriously I doubt on what basis the volume has almost become 1.5 times.

Surely the guest has not done a 40*1.2*1.2*1.2 .

You have to calculate the stress based on the wall thickness and based just be bit not bery simple mechanical engineering. and the new volume can be calculated. Of course you are never going to reach the %E stage in case you are in real life and you will be sufficiently below the Yield point, working in the elastic region.

the standard compressiblity of hydraulic fluid for normal conditions is 0.5% per 1000 PSI (14.5 PSI= 1 BAR) for normal conditions this is OK to use. (up to 6000 PSI) at normal temps.

Are you planning on doing destructive testing? Only then would you take the vessel into the plastic range. And even then, it would probably stretch in only two dimensions (circumferentially).

For normal testing you might need the volume of the vessel itself, plus the pump/reservoir/hose, plus say 5-10%.

Good grief--don't nobody know nuthin' no more?

As various posters have said, there's something odd here. If it's anything like an ordinary cylindrical steel vessel, with a design stress ~ 170 MPa, elastic modulus 2*10⁵ MPa, stress in test condition ~ 200 MPa, so strain = percentage elongation = 0.1%. Taking Poisson's ratio into account, change in vessel volume is about 0.2%, or 80 litre. Note that on above assumptions, the actual test pressure doesn't come into it.

There's also the volume needed due to oil compression. Taking Mike L's figure for compressibilty, I make that about 137 liter for 40 m³ to 70 bar. This does depend on the test pressure.

Total about 210 litre.

Cheers.........Codey

You will find that the answer to your question depends on the the shape of the container and the material it is constructed from. The most efficient shape is spherical because the strain is uniformly distributed. However, the material and its thickness are also important. A very high modulus material will exhibit less strain for a given pressure (stress). Thicker material will also result in lower strain for a given pressure (stress). So you need to supply more information before the problem can be solved.