Pressure vessels
What is the formula for the stresses in a spherical pressure vessel? Specifically, one that has a higher inside pressure than an outside pressure. I'd expect the material to be in tension, but I don't know how to calculate the magnitude.

Quote:
[tex] \sigma=\frac{pd}{4t} [/tex] ([tex]p[/tex] pressure, [tex]d[/tex] inner diameter, [tex]t[/tex] wall thickness) for both circumferential & hoop components. 
The formula provided by Perennial is the one most commonly provided by textbooks. But if you're designing a "pressure vessel" (which has a legal definition in most countries) then you need to use the formulas provided by that country's pressure vessel code. In addition, pressure vessels are inspected, tested and "stamped" or authorized for use by a governing body. In the US, that governing body is the "National Board" and the requirements are provided by ASME Boiler and Pressure Vessel codes (ASME BPV).
Per ASME BPV, the equation to be used for spherical pressure vessels is: t = PR / ( 2SE  0.2P) where t = shell thickness P = pressure R = inside radius S = maximum allowable stress E = joint efficiency (for shells fabricated from more than 1 section) 
I'm actually interested in a bizarre physics point, so the theoretical answer is fine.
I wanted to see if the intergal of the trace of the classical stressenergy tensor (T_11+T_22+T_33)*dV (dV = volume element) was zero for the entire pressure vessel (interior + exterior). It looks like it is. For the interior of the pressure vessel T_11 = T_22 = T_33 = P so we have volume*3P = 4/3 * Pi * (d/2)^3 * 3*P For the shell T_11 = T_22 = P*d/4*t so we have volume*2*(P*d/4*t) = 4*Pi*(d/2)^2*t*2*(P*d/4*t) both of which simplify to Pi*d^3*P/2, and since one is pressure and the other is tension, the sum is zero. 
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