Specifying Eccentric reducers

Posted by: Ajay Kumar

Specifying Eccentric reducers - 02/20/03 09:35 AM

Hi,
How to specify eccentric reducer in CAESARII . It is explained in the application guide but it gives a good picture about concentric one .
In my case i have to specify one eccentric reducer for the vertical run of pipe .
Posted by: Richard Ay

Re: Specifying Eccentric reducers - 02/20/03 09:59 AM

Assuming you are going to use "average properties", here is an example of a concentric reducer model and an eccentric reducer model.

Concentric Model:
D1 = 8.625 t1 = 0.322 L = 16.0
D2 = 6.625 t2 = 0.280

In CAESAR II -> DX = 16, Diam=7.625 Thck=.301

Eccentric Model:
D1 = 8.625 t1 = 0.322 L = 16.0
D2 = 6.625 t2 = 0.280

In CAESAR II -> DX = 16, DY=-1.00
Diam=7.625 Thck=.301
Posted by: Gustavo Cuervo

Re: Specifying Eccentric reducers - 02/24/03 09:17 AM

I want to add that when you specify an eccentric reduction (or concentric too) you have to use SIF’S=2 (the maximum, or use the equation, Applications guide page 6-2) in the node when the reduction starts and in the node when the reduction ends (each end of the reducer element).

Regards

Gustavo Cuervo
Posted by: Samuel

Re: Specifying Eccentric reducers - 03/19/03 01:55 PM

For Eccentric: why not DY = (6.625-8.625)/2 = -1.0? (a half of delta OD rather than a half of Delta thickness)
Posted by: Richard Ay

Re: Specifying Eccentric reducers - 03/19/03 02:11 PM

Oops, you're 100% correct. I'll fix the above post.
Posted by: Andrew Weighell

Re: Specifying Eccentric reducers - 03/19/03 04:01 PM

I've always assumed that the average properties model is complicated for little, if any, improvement of accuracy(?).
A 50% big / 50% small model is probably adequate and within the relatively wide modelling tolerances. More importantly, it has the advantage of correctly reporting the stress at the small end? I came across a calc recently where the the small end didn't show an overstress with notional average properties but was overstressed with correct properties !!!! ?
Posted by: Amit

Re: Specifying Eccentric reducers - 03/26/03 09:49 AM

For either type of Reducer, is it required to put Tapered Transition? I want to understand the significant effect CAESER shows by selecting this option.
Posted by: Edward Klein

Re: Specifying Eccentric reducers - 04/02/03 02:34 PM

Just a reminder, B31.3 does not apply an SIF to reducers the way B31.1 does.

I would also like to get some clarification on the "tapered transition" fitting type. I've seen other people use it in a file to model the two ends of a reducer, but I don't know what the basis for that is and there's no mention that I can find in the manuals as to what it is supposed to mean. I was admittedly disturbed to see it generate an SIF value at all in a B31.3 model, if it is intended to represent a reducer.