Regarding the calculation of wall thickness
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Parameter conditions: operating pressure is at atmospheric pressure, operating temperature is 60°, dimensions are Φ2600×4000, material is 304, and the concentration of ammonia solution used is 24%. Using SW06-2011, I calculated the wall thickness of the cylinder to be 3, and the wall thickness of the head to be 4. I would appreciate it if experts could help me check whether there are any mistakes. The parameters I entered are: design pressure: 0.11 Mpa, design temperature: 60°, test pressure: 0.14 Mpa, corrosion margin: 0, welding joint factor: 0.9, standard elliptical head.
Using SW6-2011 for calculating pressure vessels at atmospheric pressure is not very suitable; there are specialized software programs for such calculations.
This post was last edited by zhangjuhua on 2015-7-26 18:53. Corrosion allowance: 0, weld joint coefficient: 0.9? 0.9 doesn’t seem to be very common for OD2400; with such a thin wall thickness, stiffness is likely to play a dominant role. However, I’ve seen those double-layer hollow panels that are bulgy and quite thin, yet still possess good stiffness
For stainless steel, the corrosion allowance should be 0, right? Should the welding joint coefficient be chosen for a single-sided weld without a gasket in a butt joint? Should it be 0.7 or something else? I’m not quite sure about this one
What software can be used to calculate atmospheric pressure vessels
The weld joint factor is determined by taking into account both the proportion of non-destructive testing and the welding method
Even if you use SW6-2011 for the calculations, I gave it a try as well – a minimum thickness of 5 is required for the end cap to meet the standards; a thickness of 4 is not sufficient and results in failure.
The strength check for SW6 shows that it meets the strength requirements, but I also think that the three sections of DN2400 likely won’t have sufficient stiffness.
The strength is taken into account; stiffness also needs to be considered. How is stiffness calculated?
I want to ask that too; I haven’t done the calculations, but ever since, for DN2000 and above, we’ve stopped using 6 plates of thick thickness. Regardless of the strength values obtained from the calculations, we simply use 8 plates of thick thickness. A standard vertical container.
This post was last edited by sikahuang on 2015-7-27 at 13:25. Generally, the wall thickness required for a cylinder to meet stiffness requirements is: (2Di/1000)+C. Therefore, a cylinder with a wall thickness of 3 mm does not meet the stiffness requirements. The calculated wall thickness of the cylinder must be at least 5.2 mm in order to meet the stiffness requirements. The nominal thickness of the cylinder is: 5.2 + 0.3 + 0 = 5.5 mm; the nominal thickness is rounded to 6 mm