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Discussion in 'The main mechanical design forum' started by ahsan, Aug 8, 2011.

1. ahsanNew Member

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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.

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3. Tom SherlockNew Member

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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.

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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

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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

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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

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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

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8. ahsanNew Member

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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

9. ahsanNew Member

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If we use internal attachment clips .. then what will be the effect on the thickness of the vessel?

Regards

AHSAN

10. bcvanbelleNew Member

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You could also consider a cylindrical vessel, wire wound with external yoke. ABB makes these kinds of vessels.

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