I want to design a pressure vessel whose diameter is2800mm, height 4200mm and pressure inside is2290 psig , 250F Temperature material is SA-516 Gr.70 my calculated thickness in too high around 7in . which is not practically possible . kindly suggest any one.
You'll need to decrease your diameter until it works at a practical thickness, as long as you can increase your length to maintain your volume. You might also make more tanks of a smaller diameter. This pressure will probably get you down around the diameter of a SCUBA tank. (Take a look at the pressures than various SCUBA tank manufacturers claim.SCUBA tanks are 2640 psi, 3000 psi, 3300 psi, 3442 psi, and 3500 psi. The higher pressure tanks are almost impossible to lift) If you really have to build at this diameter, you may be able to put rings around it that are made from I-Beams or channel. It wall look like a submarine that you have turned inside-out. Remember that the ends are going to need a similar treatment, without concentrating these huge stresses at the ends of the cylindrical portion. Best of luck.
According to ASME VIII sect 2 or onother code? Where are the nozzles? Pressure is 158 barg (sorry I am metric orientated) What kind of heads are you using All of above have direct consequences on the material thickness Michael Source
Be very, very careful about mixing your units like that. That's how so many mistakes are made..... Assuming a thin-walled cylindrical pressure vessel, your wall thickness seems about right based on the hoop stress for the dimensions, pressure and materials you've suggested with a factor of safety of about 2. But like Michael suggests there will be other factors.... Simplistically, you could reduce the wall thickness by increasing the strength of the material or by reducing the diameter of the vessel (you'd then need to make it taller to house the same volume, of course). I guess there are other things you could do to allow you to reduce the local thickness (bands, wraps etc). Why is 7 in not practically possible? Andrew Source
I agree with Mike and Andrew. If you have to keep the diameter for whatever reason, reduce the thickness of the material and provide supporting structures connecting both hemispheres. Either longitudinal or lateral. Make sure to calculate surface area with and without the supporting structures(eg: ribs in rectangular blocks), it will give you an understanding of the capacity....good luck! Sabitha Source
I was trying to do a running for your tank and was assuming several variables. The results. Thin Cylindrical Shells Internal pressure Design for Inside Dimensions Stresses in the Circumferential Direction. Refer to paragraph UG-27 of VIII-1 The quantity 0.385*S*E = 0.385* 20000 * 0.85 = 6545 is greater than the Design Pressure of P = 2290 6545 > 2290 Thus: The equation for the internal required thickness in the Circumferential direction due to internal pressure is: t = PRi/(SE - 0.6P) When t < 0.5P or P < 0.385SE The Inside radius in the corroded condition is: Ri = 55.1181 + 0.125 = 55.2431 Thus: t = PRi/(SE - 0.6P) + C t = 2290 * 55.2431 /( 20000 * 0.85 - 0.6* 2290 ) + 0.125 = 8.2209 The required thickness, is t = 8.2209 We have now: 4/16 Now we MUST have at least, t = 8.25 We verify that t < 0.5*R = 0.5 * 55.1181 = 27.55905 thus: 8.2500 <= 27.55905 [Ok] Therefore the equation is applicable Stresses in the Longitudinal Direction The quantity 1.25*S*E = 1.25* 20000 * 0.85 = 21250 is greater than the Design Pressure of P = 2290 21250 > 2290 Thus: The equation for the internal required thickness in the Longitudinal direction due to internal pressure is: t = PRi/(2SE + 0.4P) When t < 0.5P or P < 1.25SE Thus: t = PR/(2SE + 0.4P) + C t = 2290 * 55.2431 /(2* 20000 * 0.85 + 0.4* 2290 ) + 0.125 = 3.7482 The required thickness, is t = 3.7482 We have 12/16 Now we MUST have at least, t = 3.7500 We verify that t < 0.5*Ri = 0.5 * 55.1181 = 27.55905 thus: 3.7500 <= 27.55905 [Ok] Therefore the equation is applicable Thus the required thickness of the vessel is: t = Max( 8.25 , 3.75 ) = 8.2500 Use: 8.2500 Alberto Source
Thanks to all of you tooo much for that support. CODE: ASME VIII Div 2 In my case, ellipsoidal heads shall be used .Yes, sure we can keep the dia or use higher strength material but our client requirement is the same as I mentioned above.But we can use rings around it. Kindly provide me if anyone of you have similar related design /drawings. Kindly also elaborate if I can use the same thickness around 8" then what about the welding details.. Thanks Ahsan
If we use internal attachment clips .. then what will be the effect on the thickness of the vessel? Regards AHSAN
You could also consider a cylindrical vessel, wire wound with external yoke. ABB makes these kinds of vessels.
The inside surface needs to be free of 'edges' or corners. Pressure is really good at using those places to cause problems. Elipsoid, while seemingly is ok in some cases, won't really work out as well. A seemless design with hemisphereical ends would work better. The trick is to make the pressurre equal on all interior surfaces. Failing that, it will fail to meet your critera. You also don't say wether the diameter given is interior or exterior, and the same for the height. I am presuming your client has issues with its overall external size. If you have access to Neutronium, then a inch thick shell might be achieveable, barring that, you might consider a composit route, by using winding the exterior surface with Fiberglass, in all of the appropriate directions, with an high strength epoxy coating on the inside, that has the same expansion and contraction coefficiants as the steel. You might also consider using Metglass windings as well if Fibergalss can't be used. The closer you can get the steel to be glass like on the interior, the less chance the pressure will be able to cause strain on your vessel. Donald Source