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Assessment of inspection findings

using API 579

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PMTSB 2006

Assessment of inspection findings

Lets try something harder..using API 579

c

s

Given the following;P = 300 psiD = 100 S = 18,000 psiE = 1.0Original thickness = 1.25

Minimum required thickness;

"828.0)3004.0()118000(

503004.0 PSE

PRt

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Assessment of inspection findings

Lets try something harder..using API 579

c

s

Inspection established thefollowing;

Localized thinning area with alongitudinal length (s) 4 & circumferential length (c) 3.7 Lowest reading within area =0.5 Corrosion rate = 0.025 peryear

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Assessment of inspection findings

Lets try something harder..using API 579

c

s

Question : Can the vessel be

safely operated if the nextscheduled inspection is in 5 yearstime?

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Step 1 Find out required radius - R c

Assessment of inspection findings

FCA = Future corrosion allowance= Corrosion rate x time until next shutdown= 0.025 inch/year x 5 years= 0.125 inch

Rc = R + FCA= (100 x ) + 0.125 inch= 50.125 inches

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Step 2 Find out minimum required thickness towithstand load in circumferential direction

Assessment of inspection findings

"844.0)3006.0()118000(

125.503006.0min PSE

PRt cc

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Step 3 Find out minimum required thickness towithstand load in longitudinal direction

Assessment of inspection findings

"416.0

)3004.0()118000(

125.50300

4.0min PSE

PRt c L

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Step 4 Find out t min

Assessment of inspection findings

"844.0andof larger minminminc L t t t

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Step 5 Determine the parameter R t

Assessment of inspection findings

t mm = the minimum thickness measured at the thinned area= 0.5 inch

443.0844.0

125.05.0min

t

FCAt R mmt

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Step 6 Determine the parameter

Assessment of inspection findings

s = defect longitudinal length= 4 inch

559.0844.01004285.1285.1

min Dt

s

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Step 7 Refer to figure 5.6 to determine longitudinalacceptance

Assessment of inspection findings

From this curve, the

minimum value of R t required is about 0.38,which is smaller than thecalculated value of 0.443

PASS

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Step 8 Determine c/D

Assessment of inspection findings

c = defect circumferential length= 3.7 inch

037.0100

7.3 Dc

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Step 9 Refer to figure 5.7 to determinecircumferential acceptance

Assessment of inspection findings

From this curve, theminimum value of R t required is about 0.2,which is smaller than thecalculated value of 0.443

PASS

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Assessment of inspection findings

Conclusion..

c

s

The vessel can be safely operateduntil the next scheduled inspection in5 years time

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Assessment of inspection findings

Lessons learned..

c

s

This is merely a Level 1 FitnessFor Service Assessment thesimplest of all assessments by API

579 Level 2 & Level 3 assessments canget more complicated