allowable stress based on B31.3 2

[lschmid75]

We are working in high pressures in the 11,000 psi range. I want to use 1 1/2" piping. The allowable stress of A312 316 SS is 20K. The allowable stress of A20 pipe (No8020)is 23.3K which is just enough so that when I calculate the wall thickness required, I am Ok with 1 1/2" XXS pipe. My problem is that I cannot get Alloy (carpenter) 20 vaves in such a high pressure range. 316 SS valves are available and they are rated for the pressure I need in Butt weld ends, but if I weld 316 SS valve to Alloy 20 pipe I am afraid that I must use the 20K for the allowable stress rather than the 23.3K in the B31.3 equation. I am afraid that the joint is weaker than the A20 to A20 joints because SS has diluted the metallurgy. Has anyone been through this?

 

[BigInch]

Sorry I have no answer to your question. I'm just curious as to what type of rocket you're building? What is it?

Going the Big Inch!

http://virtualpipeline.spaces.msn.com

 

[Cockroach]

I did a fast computation using your numbers and the tri-axial stress model for Von Mises-Hencky. Carrying a factor of safety of two, I get a wall thickness of 0.239 inches under your said conditions.

In 1 1/2 pipe size, Schedule 160 would be acceptable but Schedule XXS is a tremendously conservative. If you need to drop the factor of safety very slightly, then your application can suit Schedule 80 wall.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[stanier]

Biginch,

It sounds like a coal to oil plant. Such high pressures are used in that process. Gioven the price of oil there is renewed interest.

Geoffrey D Stone FIMechE C.Eng;FIEust CP Eng

http://www.waterhammer.bigblog.com.au

 

[TGS4]

lschmid75,

Per B31.3 article 304.1.2(b), how are you calculating your thickness? Are you including mill under-tolerance? You'll need to talk to your metallurgist/welding engineer for the answer to your dissimilar-weld question. It all depends on what is used as filler.

 

[lschmid75]

cockroach,
I am unsure what material you were using when you said that 1 1/2" sch 160 was sufficient. Was it A20 at 23,300 psi or 316 SS at 20,000 psi allowable stress? The equation I have to sue in this case is the one from B31.3 which is t=PD/2(SE-(P-y))
t= min thickness; P= operating pressure; D=OD; SE= allowable stress from tables (A1); y=factor (~0.4)

TG54,
Are you saying that if I use A 20 weld wire to weld A20 (SE=23.3 KPsi)and 316 SS(SE=20.0 KPsi) that the combination of the 316SS and A 20 base metals melted together may still be able to be viewed as strong as the A20 material only?

 

[TGS4]

lschmid75,
The equation that you are using for calculating the thickness may not be appropriate - please re-read 304.1.2(b), which states:
"For t>=D/6 or for P/SE>0.385, calculation of pressure design thickness for straight pipe requires SPECIAL CONSIDERATION of factors such as theory of failure, effects of fatigue, and thermal stress" (emphasis added).

The equation that you are using is only applicable for t<D/6.

Also, you NEED to include mill under-tolerance and corrosion allowance (plus any other allowances such as erosion allowance, etc).

W.r.t. the dilution zone strength, this is a question that only your welding engineer (and/or metallurgist) can answer. If you don't have one on staff/contract, I would highly recommend that you hire one (at least on contract).

 

[Cockroach]

ISchmid75, ASTM A20.

I fully agree with TGS4. My point is that I think you have some mathematical modelling issues. You may be better off computing the wall stresses due to internal pressure considerations, then applying a factor of safety in order to get a proper material strength.

The Von Mises-Hencky equation was used in my computation, but iteritively in order to obtain the outer to inner diameter ratio. Since you gave me the size of pipe, the ID became a natural consequence of the computation. This is how I arrived at the pipe schedule.

Most mathematical models are constructed based on specific assumptions that are meant to limit uncertainty or difficulties in the computations. The end user must be careful and understand these uncertainties, they may be a physical reality in the case under consideration. I believe TGS4 has successfully point out such a limitation with the equation you are using.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada