Pressure differential 3

[DRSEngineer]

I am a fairly new engineer, recently graduated. I have been given a pressure vessel problem and I need some help. We have a dome shaped lens with a small lip which gives us a certain surface area to which we apply the adhesive. Currently the adhesive has been failing at 40,000 ft. I need to find the PSI inside the lens at 40,000 ft above sea level, so I can determine a proper adhesive to use. Can anyone direct me to the proper equations that will help me determine the PSI acting on the domed lens?

 

[btrueblood]

Google "standard atmosphere", or "NACA standard atmosphere" or "International Standard Atmosphere". Should find hits for tables of air pressure, temperature, density vs. altitude for a "standard day". You may also be able to find methods to correct the standard atmosphere to known ground temperature/pressure variations (commonly called "weather").

 

[btrueblood]

From the above search, you should find:

http://www.aviation.ch/tools-atmosphere.asp

whether its results are accurate or not, you should judge for yourself.

 

[DRSEngineer]

Thank you for the quick responses. I think I might have asked the question incorrectly. I determined the outside pressure at 40,000 ft to be ~2.72PSI.What I am needing to calculate is the internal force acting on the lense at that altitude.

 

[BobM3]

If there's no active pressure compensation then you would have to know how well your vessel is sealed and how much it expands with pressure differentials and temperature changes.

 

[rb1957]

DRS, i think you need to know the pressure differential of the plane. most planes maintain a cabin altitude of 8,000 ft, which means that the cabin pressure is equal to 8,000ft (look up the table) and the outside pressure is 2.7psi, so the pressure differential is something like 8-10psi.

 

[Twoballcane]

"What I am needing to calculate is the internal force acting on the lense at that altitude."

drsengineer, are you asking how much force is acting on the lens? If that is the question, then you take the surface area of the lens and multiply it against the pressure differential. This will give you the force acting on the lens.

Tobalcane
"If you avoid failure, you also avoid success."

 

[israelkk]

Twoballcane

The area should be the section area not the surface area. For example if the the lens is half a ball shape with internal radius "R" the area to multiply by the pressure differential is Pi*R^2 and not the surface area of the a half ball.

 

[rb1957]

shouldn't it be both ! the external pressure acts on the external surface, the internal pressure acts on the internal surface ?

 

[Twoballcane]

israelkk,
I don't think so. If you put half of the ball in a tube and then pressurize the tube, the pressure (force) is acting on the surface area not the cross or section area of the ball to push it out.

rb1957, yes you are correct but the pressure differential is the delta force acting on the lens. Think of a sub that dives. On land the pressure inside and out are the same so this is equilibrium, but once it dives, the pressure inside is standard and the pressure outside is rising due to the water pushing. We all know once the pressure outside gets higher it will start to crush the sub. Same thing here, you can calculate the pressure differential, then calculate the surface area of the sub, multiply the two numbers and this is the force acting on the sub. Well, this is how I would approach it.


Tobalcane
"If you avoid failure, you also avoid success."

 

[ivymike]

twoball - so a domed piston would experience more force along the cylinder axis than a flat-top one would, for a given cylinder pressure? not correct. might be worth reviewing basic hydrostatics. try p27 of

http://faculty.washington.edu/tlars...ge/2 Pressure forces on submerged objects.pdf

 

[Twoballcane]

Im not understanding your argument? On pg 30 it is showing the presure along the curved surface and then showing the resultant force due to the pressure. This is not showing pressure at the cross section.

Tobalcane
"If you avoid failure, you also avoid success."

 

[israelkk]

Twoballcane

The pressure is acting on the surface however if you look at the pressure as a vector it can be devided to two vectors one pointing to the axis of the dome and the second is perpendicular to the dome axis. The perpendicular parts cancel each other because the dome is axisymetric around the dome axis.

 

[hydtools]

Here is another illustration of a pressurized vessel.

http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/pressure_vessel.cfm

Ted

 

[Twoballcane]

Well first pressure is a scalar quantity not a vector. The definition of Hydrostatic Pressure is force per surface area. So if you take the surface area of the lens and multiply it by the pressure differential, you will get the force that is exerting against the lens which in turn is pushing it out.

Tobalcane
"If you avoid failure, you also avoid success."

 

[israelkk]

Twoballcane

I used the vector to simplify things. However, I do not know your matmatical background but I will try to explain it with simplified calculus. Assume a small infinitisimal area "dA" on the surface. The pressure on the "dA" area give an infinitisimal force "dF" (which is definitely a vector!! which is in the radial direction from the center point of the axis). This force can be resolved into two infinitisimal forces one is "dFx" force in the dome axis direction and "dFr" (I am using polar coordinate for this force because this is a 3D system) which is a perpendicular force to the dome axis. Now, integrate this force over the surface of the dome and you will find that the total force will be the pressure multiplied by Pi * R^2.

Just a small quizz for you? What will be the total force on a hollow ball (pressure vessel) filled with a compressed gas. Acoording to your logic the force will be the pressure multiplied by the inner surface area of the ball. Therefore, a filled ball shape pressure vessel will never sit still?

 

[zdas04]

Israelkk,
I think that it is safe to assume that most engineers can muddle through high school calculus, or maybe we can get our kids to 'splain it to us.

David

 

[tmacm]

Twoballcane, so you think that if you take a half sphere and place it in a sphere in a cylinder and have atmospheric pressure acting on either side that the half sphere will be pushed out of the cylinder?

 

[eromlignod]

Pressure on a spherical, domed end cap will indeed act outwardly in a normal, radial direction to the inner surface of the dome. But the outward components all have an equal counterforce on the opposite side of the dome (mirror image axially) that cancel them out. These forces only contribute to the wall tension of the dome.

Only the components along the cylindrical axis exert force on the dome in relationship to the tank. You will find that the sum of this distributed force is the pressure multiplied by the cross sectional area of the tank, just like a piston.

Assuming that the sealing lip is holding the vessel together axially, then the distributed force along the lip is this total force divided by the circumference.

As far as the pressure at altitude is concerned, you didn't tell us what the internal pressure is. Assuming that you mean that the interior was at normal barometric pressure when on the ground, then you should probably just design it to withstand an exterior vacuum, so the interior pressure could be considered as 14.7 psi. In fact, I would at least double that for a safety factor.

Don
Kansas City

 

[DRSEngineer]

Thank for everyone's responses.

I am not designing this unit, I am simply trying to figure out what type of expoxy to use on a current design since the original expoxy is failing.
From what I am told, the internal pressure of the lense is ~15PSI and at 40,000 ft I am using external pressure to be ~2.72PSI. I am negating termal expansion/contraction and any airflow, for now.
If I understand correctly, if my dome shaped lense has an internal radius of .821in do I negate the thickness to derive the cross section area? Once I have the cross sectional area then I would multiply it by the pressure differential (15PSI-2.72PSI?) and that will give me the internal force acting on the lense?

From there, would I take the thickness and derive the surface area (which is the area the epoxy is added to) and multiply it by the force acting on the lense to determine the required tensile strength the epoxy will need to have?