2 replies to this topic

[aqassam]

I have another documentation question. Per sub-note ( b ) on Table 9 in API 2000 (6th edition), the F factors are calculated based on the following equation in API 521:

 

F = (InsCond * (Tmax - Tfluid))/ (66570 * InsThk)

 

One discrepancy is that API 521's F factor equation uses 66570 W/m2 instead of 66200 W/m2 which is mentioned in subnote b. 

 

Another issue is that I cannot match the calculated F factor values using either constant. For example (for the first row)

 

F = (22.7 W/m²K * 887.9 K)/ (66570 * 0.025 m) = 12.11

 

How, then, was the value 0.3 in the table calculated? 

 

I thought maybe 887.9 K was not the temperature differential as stated in note ( b ), but the relieving temperature:

 

F = ( 22.7 W/m²K * 16.1) / (66570 * 0.025 m) = .022  

Close but no cigar. 

 

Neither of the methodologies work if I use 66200 W/m²K, either. 

 

Can someone help me validate the calculated F factors?

Edited by aqassam, 25 February 2014 - 03:11 PM.

[christopher.choa]

Hi aqassam,

 

Unit discrepancy.

 

The formula given in API 520 has the parameter insulation conductivity (W/m-K). The insulation conductance in API 2000 Table 9 already incorporates the insulation thickness in it.

 

 

F = 22.7*887.9/(66200) = 0.3

 

Hope this helps!

[ankur2061]

aqassam,

 

The number 22.7 is insulation conductance value in W/m2-K , whereas equation 13 uses insulation thermal conductivity in W/m-K as an input for the equation which as per the explanatory notes provided at the bottom of Table 6 uses a standard thermal conductivity of 0.58 W/m-K for insulation

 

Correcting for inputs the actual calculation as per eqn 13 becomes:

 

F = 0.58*887.9 / (66570*0.025) = 0.3

 

Similarly if you use an insulation thickness of 0.05 m in eqn 13 with all other terms unchanged, the F value calculated becomes 0.15, which corresponds to the insulation conductance value of 11.36 given in table 6 of API STD 521.

 

Regards,

Ankur.