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Factor A in Asme section ii , thk for External Pressure 1

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AliGrailly

AliGrailly

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Hi

i could not extract factor A from tabluar values figure G

Material from table HA-1
D/T=69.7
L/D=50
1-A=?

2- if there is not same D/L according to L/D how shall be do , forexample in above there are 50 in L/D 60 and also 80
Ex: D/t=20 , L/D=40 A=0.275-02
D/t=22 , L/D=40 Whats A ?
D/t=25 , L/D=40 Not in Table
 
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AliGrailly, see UG-28 (c)(1), Steps 3 & especially Step 4. You have to interpolate between the tabular values. Multiple interpolations are usually needed. Logarthmic interpolation is generally considered more accurate than linear, but not mandated.

Often more convenient to just pick the values off of Fig. G if a "high degree" of accuracy is not needed.

For your first example, Do/t at L/Do = 50 is about 70, A is about halfway between 0.00015 and 0.0003, roughly 0.00023. A reasonably good clear platic scale is useful here.

Regards,

Mike

The problem with sloppy work is that the supply FAR EXCEEDS the demand
 
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You can get an approximate value for A from Figure G.

If you want a more exact value you will have to interpolate between the ones listed in Table G. Given that Figure G is clearly log-log, you should consider using log-log interpolation for the most accurate value of A.
 
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This is what I get with my spreadsheets.

Dot = 69.7
LDo = 50
A = 0.0002410

I use an excel formula to index the tables I have in excel, then interpolate.

 
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hi

linear 0.23 or 0.24 ? and if log-log interpolate be accurate so what is the final result ? in somewhere i saw cal for 69.7 & 50 >>> 0.25-03
 
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i found following from searching here :

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.

please define whats 1.19733 in above Eq ( is there mistake ??? - should be 1.1874 instead it or this is right in Eq)


D/t = 500

L/D 0.2 0.4
factor A 0.231 ?
factor A 0.645 0.305

 
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I've always kept this example as my starting point. Table G doesn't have nice matching pairs of numbers so a single interpolation won't work.
[ol 1]
[li]Interpolate on the Do/T line above to get the value of A above at your L/Do.[/li]
[li]Interpolate on the Do/T line below to get the value of A below at your L/Do.[/li]
[li]Interpolate between these two A values to get A final at your actual Do/T.[/li]
[/ol]

This is just one way of getting there. DO NOT just blindly follow my sequence. Think about it and convince yourself this is proper.

 
 https://files.engineering.com/getfile.aspx?folder=bcfe6ad0-1f02-4ad0-8ee5-5414e32826a9&file=Log-Log_Interpolation.jpg
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For the OP's first example only a single interpolation is needed since L/Do = 50. This is not generally the case though.

The problem with sloppy work is that the supply FAR EXCEEDS the demand
 
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WITH D/T=69.7 , L/D 50
in tabular values figur G section ii
60 80
0.306 0.172
Factor A General interpolation =0.24101-03
Factor A with exp(use ln in linear interpolation ) = 0.2267-03

Now which ones to be used ?

 
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this one: Factor A General interpolation =0.24101-03
 
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AliGrailly said:
Now which ones to be used ?
Click to expand...

Ref UG-28 (c)(1), Step 4: It's your choice

EDIT: I recommend you perform the calculations using both values and observe if the difference is worth worrying about or not.

The problem with sloppy work is that the supply FAR EXCEEDS the demand
 
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is Factor B in 750 F or 400 C degree
2750psi or 2903 psi ... i am afraid of wrong calculation manual
 
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Again, as long as you are consistent with your method (linear vs logarthmic) it's your choice. You can be more "conservative" or more "accurate". Your choice.

The problem with sloppy work is that the supply FAR EXCEEDS the demand
 
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