Calculating Vacuum Pressure within a Cylinder??? 1

[Superslinky]

Hi all,

lets say I have a tube of material (any) and I cap both ends. Now I drill a hole in one side (end or whatever) and draw a vacuum. How do I determine the strength of the material being used on either end and the "tube".

For the sake of the question let say the cylinder was 6061 alum and measured 20" dia. by 25" tall with a 1/4" wall thickness and each end was capped with 1" thick lexan.

I would be drawing a vacuum to around a negative 10 psi.

Any help would be appriciated. Maybe even a reference website where I could read about it would be helpful as well.

Thank you all...

 

[desertfox]

hi Superslinky

The cylinder a thin walled vessel so the hoop stress in the cylinder can be found by:-

[σ]= P*D/(2*t)

for your case if the vessel internal pressure was 10psi
then:-
P= 10-14.7 psi = 4.7psi

[σ]= 4.7 * 20/(2*0.25) = 188 psi

because the end of the vessels are capped there will be a longitudinal stress this is typically half the hoop stress.

This of course is not the full story because the end caps need stressing and I am not familiar with the Lexan material, the stressing in the end caps will depend on the shape whether flat or spherical etc.
You would be wise to look at ASME 8 or PD5500 which are both pressure vessel codes and should cover vessels under vacuum.


desertfox

 

[Superslinky]

I'm having a hard time grasping the issue with P=10-14.7 psi = 4.7 psi, if my application is a vacuum then does that change? For example, I will be applying around 12psi internally or I should say negatively to the outside atmosphere to create a "suction".

Also if I follow your calculation (assuming it's still simply 12psi wether + or -) then my calculation would be:

?= 12 * 20/(2*0.25) = 480 psi

So all in all if I'm using an aluminum tube it's seeing a hoop stress of 480psi so the material I choose should be greater than that?

 

[desertfox]

Hi Superslinky

Maybe I misunderstood your post Atmospheric pressure is 14.7 psi and I thought you meant, that you would have 10psi Absolute inside the vessel, so the difference between inside and outside the vessel would be 4.7 psi.
Yes to the last part of your last post once you have the stress you need to apply a safety factor and compare it to the allowable stress of the material.

desertfox

 

[desertfox]

Further to my last post, if your applying a vacuum ie zero pressure on the inside and then apply 12psi gauge to the outside then the pressure difference is 12+14.7 = 26.7psi.
We need to be clear on what pressure is applied where and whether your talking gauge pressure or absolute.

desertfox

 

[Cockroach]

There are no "negative" pressures, a vacum means some number less than atmospheric. Remember there is no zero atmosphere much less a negative. Even in the vastness of space there is something like three atoms per cubic litre, so by Brownian Motion, a low positive pressure.

DesertFox uses the thin wall pressure vessel equation which does not account for end cap reaction creating longitudinal stresses through the vessel. Clearly we expect a very low error as a result, since pressures are low and OD large. I prefer the Von Mises-Hencky Equation for the tri-axial state of principle stresses in order to get around the wall thickness less than OD/10 assumption.

Given your dimensions and 10 psi vacumn or 4.7 psig I would get a stress of 321.6 psi.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[Compositepro]

"Negative 10 psi" is a a "gauge" pressure and is a commonly used term in engineering and industry. In this case it is also the only source of stress on the cylinder and Lexan plate. The cylinder is not in outer-space. It is in air and a vacuum is being drawn inside, so it in under compression not tension. There is no point in converting to absolute pressures in this case.

As for which equations to use, I can't say off the top of my head, but these are common calculations that I would expect to find with Google or in a textbook. The cylinder would fail by buckling, not material failure. The dimensions given by the OP seem to be about what I would expect to see in a vacuum chamber of that size.

 

[desertfox]

If the -10psi is a gauge pressure then is that pressure applied inside or outside the shell, if its internal then the 14.7-10psi in my first post would be correct.
The cylinder will not fail by buckling because its length to diameter ratio is to short.

desertfox

 

[jte]

Well, as Compositepro noted, "negative 10 psi" is pretty clear to me as a pressure vessel guy. We can dance around the wording but to the the ordinary guy that means that there is a uniform pressure of ten pounds for every square inch of external surface, applied in an inward direction. Some might call it 10 psig external, some might call it 4.7 psia internal. But those arguing for psia will usually be process engineering types, not mechanical design types.

Now, there are a few failure modes to consider. Hoop stress due to internal or external pressure has been discussed. Not the likely failure mode.

The end caps - and remember that flat plates are not good for pressure - will fail in some combination of membrane and bending stress. This will be a tensile failure.

The failure mode of the cylinder will most likely be compressive - buckling. Not likely Eulerian column buckling since, as desertfox noted, the length to diameter ratio is not prone to column buckling. It will wrinkle. Just like an aluminum can.

But the fun part will be the design of the shell (cylinder) to head (lexan plate) connection. These connections will also serve as the stiffening rings for the shell - and should be sized accordingly. See ASME Section VIII Div. 1 UG-28 and UG-29.

jt

 

[desertfox]

Hi jte

Well I can see the -Ve being a compressive pressure applied externally, however that doesn't mean that the inside of the pressure vessel is less then atmospheric pressure or indeed a vacuum, it just means that the vessel is subject to an external pressure.

desertfox

 

[jte]

I would be drawing a vacuum to around a negative 10 psi.

desertfox I fail to see your point. Yes, I could in theory be talking about SCUBA tanks which are subject to internal pressure due to the compressed air and external pressure due to the water. But I am somehow under the impression that the OP is not describing a submersible device.

In the world of mechanical / structural design of pressure vessels, gage pressures rule. I would severely beat a young engineer who thought that it would be a good idea to specify an internal pressure of say 100 psia along with an external pressure of 14.7 psia and hope that the fabricator along with the rest of the people seeing his spec understand that he really wants a vessel with a design pressure of 85 psig.

Superslinky You started this thread... Care to elaborate?

jt

 

[JStephen]

The pressure vessel codes have formulas for the design of cylinders subject to external pressure and for flat plates/heads, and could be used if available.

Roark's formulas for Stress and Strain also includes critical buckling pressure for a cylinder under external pressure and includes bending of flat plates that could be applied to the item in question.

 

[desertfox]

jte my point was that what you described was an external pressure acting on the outer surface for which as I stated earlier does not require the internal diameter of the shell to be under vacuum to quote:-

""negative 10 psi" is pretty clear to me as a pressure vessel guy. We can dance around the wording but to the the ordinary guy that means that there is a uniform pressure of ten pounds for every square inch of external surface, applied in an inward direction."


My understanding was that and still is that the -10psi was the internal pressure and that to obtain the resultant pressure I merely subtracted the -10psi from atmospheric pressure and therefore as in my first post I ended up with 4.7psi acting over every square inch of external surface applied in an inward direction.

In Superslinky's second post the 10psi is now 12psi and he suggests that he doesn't see why I have subtracted 14.7psi which is atmospheric pressure, I went onto explain how I came by the 4.7psi and stated that I might have misunderstood his post.

desertfox

 

[prex]

Great discussion about vacuum...
To come back to reality:
-I don't think that a 1" lexan flat plate 20" OD can withstand 10 psi
-I don't think that a lexan plate can be used as a stiffening ring for an aluminum shell, due to the difference in elastic moduli
-then the aluminum shell must be regarded as having infinite length, and I'm pretty sure that a 1/4" thick shell 20" OD cannot withstand 10 psi of external pressure.

prex

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

: Magnetic brakes and launchers for fun rides

http://www.levitans.com

: Air bearing pads

 

[desertfox]

Hi

I think this link might clear this discusion:-

http://books.google.co.uk/books?id=...resenting pressures below atmospheric&f=false

I think, also where I subtracted the two pressures I should have added them together because 14.7 - (-10)= 24.7psi

desertfox

 

[Compositepro]

Dessertfox, the pressure gauge reads the difference between the outside atmosphere and the inside of the chamber. That is 10 psi. That is all that is of significance.

 

[jte]

Well put, Compositepro. Put another way, the gage does the math for you. The structure doesn't care whether it has 5 psiA inside and 15 psiA outside or 15 psiA inside and is submerged in about 11 feet of water. In a lake, ocean, or other open body of water. (I said "about". Not trying to start a contest about the density of water.) Either way, the pressure GAGE will read about the same. And the structure will react about the same.

Oh, and links to subscription sites are not terribly useful to those of us without subscriptions...

prex- Good points. My thoughts in my first post were that the Lexan would somehow be held in place with some sort of rings. As a vessel guy, I'd try to design these rings to do double duty as stiffening rings and thus limit the effective length of the cylinder to the span between the rings. Thus, the reference to UG-29.

jt

 

[desertfox]

Compositpro
Your right my mistake.


You don't need a subscription for the last link I posted, as certainly don't subscribe.

desertfox

 

[hydtools]

Here is visual aid that may help.

http://www.engineeringtoolbox.com/pressure-d_587.html

The gauge reading is referenced to atmospheric pressure; therefore -10 is 10 psi below atmospheric.

Ted

 

[chicopee]

Try this formula which I had tucked away fo9r it seems years:
To calculate pipe thickness under external pressure (=ambient pressure- internal pressure)

? = P x [ R^2 - (R-t)^2 ] / [ R^2 - (R-t)^2 ] + ??TE
? = Stress (psi)
P = external pressure (psi)
R = external radius (in)
t = thickness (in)
? = coefficient of thermal expansion (in/°F)
?T = change in temp (°F)
E = modulus of elasticity (psi

N.B. Valid only from 0-100 °C