How to determine Factor A by interpolation for external pressure ASME 1

[Mac111]

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

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

 

[SnTMan]

Mac111

Do/T = 500, L/Do = 0.231:
A = 0.645-(0.231-0.2)/(0.4-0.2)*(0.645-0.302) = 0.592^-3

Do/T = 600, L/Do = 0.231:
A = 0.486-(0.231-0.2)/(0.4-0.2)*(0.486-0.231) = 0.446^-3

Do/T = 540, L/Do = 0.231:
A = 0.592*(540-500)/(600-500)*(0.592-0.446) = 0.534^-3

This is a linear scheme, have seen schemes using logs of the data as well.

Frankly I have often found I can get an acceptable degree of accuracy much faster by reading off the graphs.

Regards,

Mike

 

[prex]

All interpolations should be done by logs, as clearly the graph has a logarithmic behavior in both variables and in the result.
By doing this I obtained A=0.490059E-3, closer to the value in the example.

prex

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[Mac111]

Hi Prex,

Thanks for your reply.

Will you please attach step-by-step calculations, because I could not remember "How to do interpolation with Logarithmic...."?

Thanks

Mac

 

[prex]

Simply follow the scheme by SnTMan above, replacing all terms with the log (natural or base 10, is the same).
At the end take exp() of the result (for natural logs) or 10^result (for base 10).
Example for first row (natural logs):
Do/T = 500, (log=6.2146), L/Do = 0.231 (log=-1.46534):
log(A) = -0.4385-(-1.46534+1.60944)/(-0.91629+1.60944)*(-0.4385+1.19733) = -0.59625
This would give (excluding the factor 10^-3) A=exp(-0.59625)=0.551 , quite different wrt the linear interpolation.

prex

http://www.xcalcs.com

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