Firstly.. let me introduce my self
my name is indra, i'm new comer in this forum also in "Pressure vessel world", but i have engineering background (mechanical engineering). My company got tender of Pressure Vessel fabrication, the client given us some mechanical data but excluded shell & head thickness, so i must calculated it my self pursuant to dennis moss & bednar's books. but some people said that the thickness is too thin/small. now i want to share this problem to all of you, n hoping the accurate answers..
this is the data:
A. Operating condition
1. Fluid : hot water
2. Pressure : 300 kPa.G
3. Temp : 60 deg.C
4. Liquid density : 1000 kg/m3
5. Total volume : 1.4 m3

B. Design Condition
1. I/D : 1000 mm
2. TL-TL length : 1800 mm
3. Design pressure & temp. : 500 kPaG & 80 deg.C
4. Radiographing : no
5. Head type : 2:1 ellips

C. Material
1. Shell : SS400 internal (removable) : SS400
2. Head : SS 400 internal (non removable) : SS 400
3. Support, nossle flange, bolt.nut : SS400
4. Gasket : V#7035 or eq

the result of my calculations:
Shell thickness = 0.1352 in by using eq, t = PD / (2SE-0.2P)
Head thickness = 0.0677 in by using eq, t = PR / (2SE-0.6P)
How do you think? please give me flow of calculation..
thanks before

Best Regards,

Indra K
PT Arkon Prima Indonesia
Global Steel Fabricator

Hello Indra,
because of the formulas you have used I moved your topic into the Div. 1 forum, probably it´s the better place.

Regarding your questions, please be aware that SS400 is not a Code material, so you can´t use it for a stamped vessel. Regarding the units in your post, please stay with one system. Regarding the fluid you should look into Code paragraph U-1(f), maybe this will solve the above mentioned problems.

With the information on NDE, SA-516 Gr. 60 and 500 kPa the results would look like this:
SHELL:
t = P*R / (S*E - 0,60*P) + Corrosion = 500*500 / (118.000*0,70 - 0,60*500) + 0 = 3,04 mm
HEAD:
t = P*D / (2*S*E - 0,2*P) + Corrosion = 500*1.000 / (2*118.000*0,85 - 0,2*500) + 0 = 2,49 mm

Now you have to add thickness for corrosion, reinforcement of openings, external loads, ...

Michael

Hello Michael

I'm interested how you arrive at 118Mpa design stress for SA-516 Gr. 60 ?
Please explain the basis for my understanding.
Thanks

Michael, I figure it out myself: 118 approx = (210 x .85 )/1.5 where 210 is yield stress , 0.85 is weld factor for no RT.

mgeorgiadis

Dear friend ( I could not know your name)

Please note that when we use code we shall use allowable stresses from code. Mr Michael has used 118 MPa which is obviously from Table:1 A of sec II part D (metric). Ofcourse the basis of Table:1 A is provided in mandatory appendix:1

With regards

SRINIVASAN

_________________
R.SRINIVASAN
Sr. VICE PRESIDENT
Head, Design and Engineering
PED,ISGEC,
YAMUNANAGAR
HARYANA -135001
INDIA

Shrinivasan, thank you for showing me that also!