How to determine Factor A by interpolation for external pressure ASME
How to determine Factor A by interpolation for external pressure ASME
(OP)
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
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





RE: How to determine Factor A by interpolation for external pressure ASME
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
RE: How to determine Factor A by interpolation for external pressure ASME
By doing this I obtained A=0.490059E-3, closer to the value in the example.
prex
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RE: How to determine Factor A by interpolation for external pressure ASME
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
RE: How to determine Factor A by interpolation for external pressure ASME
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 : Online engineering calculations
http://www.megamag.it : Magnetic brakes and launchers for fun rides
http://www.levitans.com : Air bearing pads