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Steel is an elastic material, just like rubber. Within the safe working region for a pressure vessel, (the elastic region of the material), the increase in diameter is linearly related to the pressure in the tank. Steel is both a strong and stiff material. Mild steel has a yield point (limit of the elastic region) of about 60000 pounds per square inch. The linear relationship between stress and strain is given by Young's Modulus, which for steel is about 30 Million pounds per square inch. It would be reasonable to design a compressed air tank to have a safety factor of about 6. So for a mild steel tank at maximum working pressure, assume a maximum stress within the steel shell of the tank of 10000 PSI. (Don't confuse this with the air pressure within the tank). (Up to about twice this max stress is plausible -- I am sure there are regulatory constraints as well as standards bodies involved). Dividing the max working stress by Young's Modulus gives us the amount of elongation. 10000/300000 0 = .000333 (inches per inch) For a two foot diameter tank, we get a circumference of 24 inches * PI = 75 inches Applying the strain determined above, the circumference would grow by 0.025 inches, or about 1/32 of an inch. (The diameter grows by only 1/100 of an inch). It would be difficult, though perhaps barely possible to measure this change. Typical hydrostatic testing is done by loading the cylinder to 5/3 of its max working pressure, and measuring the change in volume of the cylinder, not the change in diameter or circumference. The volume change is proportional to the cube of the dimension change, making it more tractable to measure. A one foot diameter tank (from your "small compressor"), should have a thickness of about 0.12 inches to have a max stress in the steel of 10000 PSI with 200 PSI in the tank. So your "10 guage" steel is just about right for this application.
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