tilted nozzle 6 reinforcement area
tilted nozzle 6 reinforcement area
(OP)
Hi!
Has anyone experienced problems in verifying reinforcement area requirements in tilted nozzles?
I’ve some troubles with a 45° degrees downward oriented nozzle type 9.
In computing the limits of reinforcement I usually refer to Article D-5 of ASME VII div.2.
So I find a limit in the normal to the vessel axis direction, then accordingly to ASME VIII- div 2 AD-540.2 I consider as reinforcement area the heavy barrel area comprised underneath the normal to the vessel wall limit H, that for a 45° tilted nozzle has the shape of a parallelogram. I believe that compress in calculating this area is taking into account, in favour of safety, a smaller area than the one actually available. In fact, in cross sectional view, Compress computes the area taking the normal to the vessel limit H times the heavy barrel thickness, tn - trn i.e. Area = H * (tn - trn), which is the area of a rectangle smaller than the parallelogram; while I’m used to compute the whole area of the parallelogram i.e. A = (H/sin beta) * (tn - trn), where beta is the inclination angle of the tilted nozzle (in my case 45°).
Can any one explain to me the reason of this choice? Is it a different interpretation of the ASME code?
I remark that this different way of computing the area leads to two different results, what for me is adequate result in a “Not adequate” reinforcement for Compress, with the obvious consequences in the deficiencies report.
Giangiacomo
www.atbrivacalzoni.com
Has anyone experienced problems in verifying reinforcement area requirements in tilted nozzles?
I’ve some troubles with a 45° degrees downward oriented nozzle type 9.
In computing the limits of reinforcement I usually refer to Article D-5 of ASME VII div.2.
So I find a limit in the normal to the vessel axis direction, then accordingly to ASME VIII- div 2 AD-540.2 I consider as reinforcement area the heavy barrel area comprised underneath the normal to the vessel wall limit H, that for a 45° tilted nozzle has the shape of a parallelogram. I believe that compress in calculating this area is taking into account, in favour of safety, a smaller area than the one actually available. In fact, in cross sectional view, Compress computes the area taking the normal to the vessel limit H times the heavy barrel thickness, tn - trn i.e. Area = H * (tn - trn), which is the area of a rectangle smaller than the parallelogram; while I’m used to compute the whole area of the parallelogram i.e. A = (H/sin beta) * (tn - trn), where beta is the inclination angle of the tilted nozzle (in my case 45°).
Can any one explain to me the reason of this choice? Is it a different interpretation of the ASME code?
I remark that this different way of computing the area leads to two different results, what for me is adequate result in a “Not adequate” reinforcement for Compress, with the obvious consequences in the deficiencies report.
Giangiacomo
www.atbrivacalzoni.com





RE: tilted nozzle 6 reinforcement area
http:
Please check it out.
Giangiacomo
www.atbrivacalzoni.com
RE: tilted nozzle 6 reinforcement area
One note: you need to make sure that the limits of reinforcement measured parallel to the vessel wall are not exceeded.
RE: tilted nozzle 6 reinforcement area
I checked the parallel to the vessel wall limits and they're respected.
Actually the reason that is at the base of my interpretation is that the available area comprised inside the limits (parallel and normal to the vessel wall) stated by ASME,can be considered as reinforcement but, now that you say you've never seen such way of calculation, I'll try to ask directly to ASME.
In the meanwhile any reply is welcome.
Giangiacomo
www.atbrivacalzoni.com
RE: tilted nozzle 6 reinforcement area
ASME issues official interpretations. The question is submitted to a committee. Be prepared to wait a long time before you receive an answer.
RE: tilted nozzle 6 reinforcement area
http
and sorry for cross posting...
Giangiacomo
www.atbrivacalzoni.com