How to Calculate Variable Force Acting On Tank Pitting Spots?

None. Pressure's the same. Strength of the thinner sections should be your concern.

Thanks but how? I was assuming if the the surface area varies due to pitting, force and pressure should change accordingly. Would you mind explianing or provide me mathematical relation to justify why pressure is contant? Thanks

Well, maybe I misspoke. Pressure varies with depth, but is constant at any given depth.

So, there may be a SLIGHT difference in pressure in pitted areas due to the elevation change in the pitted area.

Check this web site: absoute pressure

Thanks, it means pressure might be different for different pits at multiple locations but it will remain contant for a single pit irrespective of how severe it is. I was trying to assume as P=F/A and as pitting area is different, so it will definitely impact the pressure.

I'd guess it would depend on the depth of the pitting.

Pressure is one thing, but force is quite different--pressure x area. At any one depth from the liquid surface the pressure will be the same, thus the pressure on a pit at that depth is pretty much a constant (varies slightly with vertical dimension of the pit.) However, the force on a pit depends on the area of the pit.

I don't think you said whether this is a pressurized tank. If so, this pressure might be the biggest thing, and the change in pressure due to depth (and density of the slurry) would be a minor part of the total pressure.

Assume that the internal pressure acts normal ("perpendicular") to the vessel wall, and that the wall thickness has been reduced by the depth of the deepest pits in the area of interest.

if pressure testing is an option, i'ld test at 150% the maximin working pressure. maintain your tank.

If I understand correctly.There could be many reasons.

1) basic-boding of the lining is not of good quality.

2) coating material defective

3) The velocity at which the slurry comes in contact with the lining.

4) The path of slurry flow inside the tank.

5) hardness of rubber

6) selection of type of rubber

7) typical chemical reaction of some ingredient of the slurry with the rubber

8) abrasive particles in slurry.

look for above It may lead you in different direction of thinking

thanks for attention

Assuming that this is not a pressurized vessel (open top exposed to the atmosphere), and is a vertically oriented tank/cylinder:

The pressure acting against (normal to) the tank wall is a function of the fluid column (h) and the unit weight of the fluid, in this case a slurry in suspension. A temperature difference, such as yours, will have very little impact on the overall stresses in the steel (?).

P = φ * h (ie, unit weight of fluid (in pcf) times fluid height (in feet).

P results will be in pounds per sq. ft. (psf) ....convert this to pounds per sq. inch. (psi)

Next, you need to determine the tank's HOOP STRESS within the steel wall....again, making the assumption that this is a "THIN-WALLED" tank. Check to see if inside radius (r) is greater than 5 times the tank wall thickness: (r > 5 * t).

Hoop Stress = (p * r) / t

p denotes the fluid pressure, in psi, acting perpendicular (horizontally) to the tank wall.

r denotes the inside radius of the tank, in inches.

t denotes the tank wall thickness, as measured at the pitted area, in inches.

[make sure you have "like" terms"!]

Compare the calculated stress against the allowable stress pursuant to applicable code(s) based on the Yield Stress of the steel.

If you have a pressurized vessel that is a whole different animal!