Ref : Non-mandatory Appendix L-3.1 (Page No. 549 ASME Section VIII Div. 1 - 2005 addenda July 1, 2005)

L/DO = 0.231
DO/T = 540

Now as per the TABLE G. of Section II Part D the values are as under.
DO/T L/DO A
500 0.16 0.830 - 03
0.2 0.645
0.4 0.305

600 0.12 0.868 - 03
0.2 0.486
0.4 0.231
I tried to interpolate the values for Factor A for L/DO = 0.230 (instead of exact 0.231).
They are 0.475 - 03 for DO/T = 500
(0.645 - 0.305)/2 = 0.17 then I deducted 0.17 from 0.645 (for L/Do = 0.2) and I got it 0.475 - 03

And similarly I got 0.3585 - 03 for DO/T = 600

So Now, if we interpolate these 2 values to get the value for Do/T = 540 it comes out to be 0.4285 - 03 (i.e. 0.0004285) , whereas in the example given in this appendix L-3.1 that value given is A = 0.0005.

As such I found it difficult to determine the “correct” value of A from the chart (so minutely), I opted to use Tables and get the value by interpolation.
Can anyone please explain why I got different answer than that given in Appendix L-3.1 OR is there any "FUNDAMENTAL" error in my interpolation steps?

After that, one more question. Is this the "ONLY" way to get/determine the value of Factor A (from Chart and from TABLES by interpolation) OR any other source?

Thanks

Mac

Hi Mac,

You are somewhere wrong.
My calculated value of A with Do/t= 540 and L/Do= 0.231 is exactly the same as given in the Code (example L-3.1).
I used Figure G in Subpart 3 of Section II Part D, see page 748 and 749 in 2010 Edition of II-A.
Alternatively, you may use tabulated values in Table G (subpart 3 of Section II-D) in lieu of using Figure G, as you have already done.
Check your calculations again carefully, do not forget to use exponents as given in the Table G.

If you still need clarifications, please let me know.

Good luck.

_________________
Thanks & Regards,
Nasir
Welding Engineer
DESCON Engineering Limited.
Lahore, Pakistan.
Mobile # 0092-336-4145402

Hi Nasir,

Thanks for your prompt reply.

First of all, let me inform you what I did.

used ASME Section VIII Div. 1 (2005 addenda July 1, 2005)..... I am not sure whether any of the values between 2005 Code and 2010 Code have been changed or not.

First interpolated the value for Factor A for the L/DO = 0.230 (from the value at L/Do = 0.2 (which is 0.645 - 03) at DO/T = 500

Then I interpolated the value for Factor A for the L/DO = 0.230 (from the value at L/DO = 0.2 (which is 0.305 - 03) at DO/T = 600

And that value (i.e. For DO/T = 500 and L/DO = 0.230 Factor A = 0.475 - 03 (Please check whether this calculaton is correct or NOT, Thanks)

And that value (i.e. For DO/T = 600 and L/DO = 0.230 Factor A = 0.3585 - 03 (Please chech this too...)

Then I interpolated these 2 values to get the value for DO/T = 540 and I got Factor A = 0.4285 - 03 (i.e. 0.0004285 instead of 0.0005).

OR will you please attach a word file showing these ( 1+ 1 + 1) interpolations, ten I can compare wit what I did and understand where I made mistake?

Thanks & Regards..

Mac

Hi Mac,

I did try many times to attach file on onetb but never succeeded, please send me your email address at my personnel address Javednasirjilani@yahoo.com,

_________________
Thanks & Regards,
Nasir
Welding Engineer
DESCON Engineering Limited.
Lahore, Pakistan.
Mobile # 0092-336-4145402

Nasir,

I have sent a request to your email id.

Please try to attach the file you mentioned in your last reply (How to determine the value of Factor A through interpolation".

Thanks

Mac

THIS MSG IS FOR GONZAGA

Hi Gonzaga,

Wll you please help me to show my error/mistake in performimg interpolation to determine Factor A for external pressure by using the tables?

Thanks & Regards...

Mac

Hi,
My calculations also gives me different values
First, at DO/T = 500
Interpolated value for Factor A for the L/DO = 0.231 (at L/Do = 0.2, A = 0.645 - 03 and at L/Do = 0.4, A = 0.305 - 03)
My calculated answer is 0.5923 - 3.
Then, at DO/T = 600
Interpolated value for Factor A for the L/DO = 0.231 (at L/Do = 0.2, A = 0.486 - 03 and at L/Do = 0.4, A = 0.231 - 03)
My calculated answer is 0.446475 - 3.

Then I interpolated these 2 values to get the value for DO/T = 540
(at DO/T = 500, A = 0.5923 - 3 and at DO/T = 600, A = 0.446475 - 3)

and I got Factor A = 0.53397 - 03 (i.e. 0.00053397 = approx. 0.00053 instead of 0.0005).

Thanks & Regards..
Shrinath Desai

Dear all,

value of factor A for L/Do of 0.231 at Do/t = 500 is 0.0005923

value of factor A for L/Do of 0.231 at Do/t = 600 is 0.00044647

value of factor A for Do/t = 540 is 0.00053396

Regards
ASD23789

So, the question still remains.
What should we refer to when determine the value of factor A? The chart or the tables (with interpolation)?
OR is there any alternate methods?

Dear Mr.Desai,

I think we should adhere strictly to table only and do the linear interpolation method.
Regards
ASD23789