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safety factor and joint efficiency 2

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Sergio76

Mechanical
Apr 11, 2012
28
I have a non-code vessel, SA-36 material, I am increasing the allowable stress to 23200 psi decreasing the safety factor to 2.5, I'm not performing any RT, so I'm using 70% efficiency, but with the thks I have my MAWP is still below my DP, can I increase my Efficiency to reach my DP keeping my Safety factor over 2.5?
 
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Efficency is based on weld profile/type and the amount of NDE you wish to preform. Minimum thickness required for vacuum design will not be effected by weld efficiency.

I would say if you are already using SA-36, which is for non pressure parts, and reducing the safety factor to 2.5:1; just overlook the MAWP calculation. You are already demonstrating a disregard for ASME rules. Why is this calculation required for non-code?
 
With E of .7, your actual design allowable "S" is 33,143 psi, which gives you a safety factor of 1.75 or 92% of the SMYS.

As per innovation2, since you have fully disregarded ASME VIII design rules, why are you asking the question? Now you really must consider toughness at the design conditions to avoid fracture during service or pressure testing.

Should you plan to pneumatically test, contact your rescue responders to pick up the human remains.

 
I know material and other consideration are overlooking the rules for code vessel; the point of this question is: I have several non-code veseel which i try to validate calculations in order to figure out what is the min SF which could not be less than 2.4 (as min required per ASME div. II), so according with your answers can I use 100% efficiency and just modified the Stress to have a SF=2.5,(in this case 23200 for SA-36 or 28000 for SA-516,70). By the way those vessels are already in service in USA.
So in conclusion if i want 2.5 SF I can use the following stress in ASME formulas:
Smodified=S*E/2.5 with E=0.7 (NO RT)
 
No, you should not and cannot simply modify joint efficiency to make a desired outcome for a non-code vessel. ASME code calculations have conservatisms for material properties and joint design. If you need to increase efficiency, perform volumetric NDT and be done with it. By the way, I presume a non-code vessel means either the Jurisdiction does not regulate the vessel or it operates below 15 psig.
 
once again thks for your answers, i think i have a mistake on the Smod y suggested, I have some confusions...I know per code Sallow=S/SF, now if i have a Spot RT (E=0.85), when I put those values in the thks formula per UG-27 or APPX 1, the results are having a higher SF than 3.5 due to a lack on examination, my point here is how i correlated the SF, the S value and E, supposed I have a code vessel with no RT req'd with min SF=2.5?
Some body could help me!
 
I forgot something metengr said if I want to increase the efficiency, I must perform volumetric NDE, could you help me to figure out what are the calcs for that?
 
I am sorry guys The volumetric NDE are the RT or UT, so disregard my last comment, so at this point I can't not performed any NDE because the vessels are in service already, and they have been working for couple years. I just tried to have some min requirements for future non-code projects, where we can have some liability, and my main corcern is to have a min SF=2.5 for non-code vessels.
 
Were these vessels manufactured to another international standard and if so, to what standard? A number of Jurisdictions permit vessels manufactured to Standards other than ASME.

 
I can't not performed any NDE because the vessels are in service already, and they have been working for couple years.

Yes you can perform NDT on in-service vessels. It is done all of the time and is a matter of proper scheduling.
 
yes, i'm agree to perform any NDE in a service vessel is a matter of scheduling, but again I am thinking in future non-code projects, with min SF=2.5 with no RT, how can i determined if the thks shown by the fabricator have at least this req'd? can i use just E=1 and increase the the Sallow?
 
Sergio76,
Simply calculate the induced stresses based on the pressure, diameter, thickness and joint efficiency of the actual vessels and compare to the minimum specified ultimate tensile strength of the material to determine the actual factor of safety.

 
so you mean solve S from the thickness formula per UG-27 or APPX-1, and divided the S"ult"over this value...it sounds valid..thanks
hey guys any other thoughts
 
Safety factor should be determined from minimum allowable tensiles (SA-36 is 58000psi).

Efficency should be accounted for in the formulas used to calculated stress.

Actual component safety factor can be determined by taking minimum allowable tensiles and deviding by calculated stress.

Try not to cut corners in your design...stick to the rules, that is why they are there.

Honestly you will not save a terrible amount of money on an atomospheric or low pressure tank by using a reduced S.F.
 
The application hasn't been stated, but I note that 23,200 psi is the allowable stress for A36 in API-650.

A couple of thoughts on this. First, I'm not that familiar with ultrasonic testing, but could it be done from the outside while a tank or vessel was in service?

Secondly, the calculation of a particular factor of safety per one code or another also assumes that the vessel was otherwise constructed in accordance with the code. So if you're designing a vessel with 70% joint efficiency due to not having any radiograhy, you're still assuming that the weld procedures and welders were properly qualified and that the welds were full-penetration and that some degree of visual examination was done on them. On the other hand, if you look at a weld after the fact and don't know how it was welded or by whom, you don't know any of those things. You may also find that nozzle reinforcement is what actually limits the pressure rating rather than shell thickness itself. A lot of the non-code vessels that I see also have heads that don't comply with the knuckle radius requirements of ASME, so you don't really have a good basis for design of them.
 
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