Calculation of wall thickness for atmospheric pressure vessels
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Atmospheric pressure vessel; medium: pyridine; medium density: 1.5; diameter: 4000; cylinder height: 7200; material: 304 stainless steel. Why does the wall thickness calculated using the formula come out to just a few tenths of a unit? Could some expert please explain how to calculate the wall thickness?
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For the calculation of pressure vessels at normal pressure, the key lies in details beyond what’s obvious; for a vessel with a diameter of 4 meters, 8 mm is sufficient if it is supported by vertical skirt supports. If it is a lug, the thickness of the cylinder at the lug location must be calculated; if the equipment is horizontal, calculating the saddle is sufficient.
The stiffness requirements also need to be considered, and the design pressure can be increased appropriately
Under normal circumstances, the wall thickness of pressure vessels at atmospheric pressure is determined to be not less than the minimum thickness specified in the standards. For example, in the case of cylinders made of Q235B material, the minimum thickness is 3 mm (excluding the corrosion allowance). If the corrosion allowance is 2 mm, then the minimum thickness becomes 3 + 2 + the negative deviation of the steel plate thickness; for a 6 mm thick plate, this value is 0.6, and for an 8 mm thick plate it is 0.8. Thus, the minimum thickness is 6 mm.