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Fitness for ServiceIN THE CASE OF A NEW VESSEL WE

CAN DESIGN IT TO A CODE

IN-SERVICE VESSEL THAT ISCORRODED WHAT DO WE DO

THIS IS A PUBLIC SAFETY ISSUE

Corrosion Corrosion

Corrosion Pitting

Examples of In-Service Defects

Images from www.materialsengineer.com

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STATED OBJECTIVES

Quantitative engineering evaluations which areperformed to demonstrate the structural integrity ofan in-service component containing a flaw.

Brittle Fracture Section 3

General Metal Loss Section 4

Local Metal Loss Section 5

Pitting Corrosion Section 6

Blisters Section 7

Weld Misalignment and Shell Distortions Section 8

Crack-like Flaws Section 9

Creep Damage Section 10

Fire Damage Section 11

API RP 579 Flaw Coverage

Section 4, 5 & 6 arecurrently available inCodeCalc (and PVElite)

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General FFS Procedures

Identify flaw type and damage mechanism

Check applicability and limitations of the specific flawtype procedure.

Review data requirement and gather the data.

Apply the assessment techniques and compare theresults to the acceptance criteria.

Estimate remaining life for the inspection interval.

Apply remediation as appropriate.

Apply in-service monitoring as appropriate.

Document the results

Level of Assessment Level 1 Simplified methods using charts and

conservative assumptions.

Level 2 More detail evaluations with less-conservative

results.

Level 3 Uses more sophisticated methods such as

FEA.

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API 579 Example 4.11

Pressure Vessel Information:

Design Conditions = 300 psig @ 350 OF

Inside (new) Diameter = 48 inches D

Nominal Thickness = 0.75 inches

Uniform Metal Loss = 0 inches (no general corrosion)

Future Corrosion Allowance = 0.10 inches FCA

Material = SA 516 Grade 70

Joint Efficiency = 0.85

PERFORM THE INITIAL CALCULATIONS

CALCULATE THE MINIMUM REQUIRED THICKNESS:

Using the ASME VIII, Division 1 Code

In the Circumferential (hoop) Direction

tCmin =P x ( R + FCA )

SE 0.60 x P

=300 x ( 24 + 0.10 )

17500 x 0.85 0.60 x 300= 0.492 inches

In the Longitudinal ( axial ) Direction

tLmin =

P x ( R + FCA )

2SE + 0.40 x P

=300 x ( 24 + 0.10 )

2 x 17500 x 0.85 + 0.40 x 300= 0.242 inches

tmin = Max ( tCmin, tLmin) = 0.492 inches

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WE NOW KNOW THE REQUIRED MINIMUM THICKNESS

This thickness which is tmin = 0.492 inches

We must now consider the actual current condition of the vessel wall toassess its future use

During a routine inspection, an area of corrosion is found

1 we determine the area of this corrosion

2 we measure the various thicknesses in this area

3 let us now see how we do this

WHAT DO WE DO NOW ?

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C1 C2 C3 C4 C5 C6 C7 C8

M1

M2

M3

M4

M5

M6

M7

SURVEY METHOD

1 Find the corrosion area Identify it

2 Overlay a Grid with nodes at the intersections 1.5 inches pitch

4 Measure thicknesses at every NODE tabulate the results3 Label the grid Circumferential and Longitudinal Gridlines

NOW TABULATE THE VALUES

0.48

0.75

0.55

0.36

0.48

049

0.75

0.360.75 0.48 0.47 0.55 0.48 0.49 0.75

2 Find the CTP for the Circumference Thicknesses

1 Set out the table of node thicknesses

3 Find the CTP for the Longitudinal Thicknesses

5 Determine minimum thickness tmm = 0.36 inches

4 Complete the CTP values

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0.48

0.75

0.55

0.36

0.48

049

0.75

0.360.75 0.48 0.47 0.55 0.48 0.49 0.75

WE NOW HAVE THE COMPLETED CTP TABLE

Minimum measured thickness tmm = 0.36

CONTINUE WITH THE CALCULATION

Find the Remaining Thickness Ratio (Rt):

Rt =tmm - FCA

tmin=

0.36 - 0.10

0.492= 0.528

Allowable Remaining Strength Factor (RSFa):

Section 2.4.2.2(d) states: Recommended value forRSFa is 0.90 which is conservative..

RSFa = 0.90

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Remaining Thickness Ratio Rt = 0.528

Allowable Remaining Strength Factor RSFa = 0.90

Calculate the Q factor (or read from Table 4.4):

Q = 1.123 [ ( )2 - 1 ]0.51 - Rt

1 - Rt / RSFa

= 1.123 [ ( )2 - 1 ]0.51 - 0.528

1 - 0.528 / 0.90

= 0.619


Remaining Thickness Ratio R = 0.528


Factor Q = 0.619

Calculate Length for thickness averaging (L):

L = Q D x tmin

= 0.619 48 x 0.492

= 3.0 inches

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Allowable Remaining Strength Factor RSFa = 0,90Factor Q = 0.619

Length for thickness averaging L = 3.0 inches

tmin = 0.492

s = flaw dimension

Longitudinal Critical Thickness Profile CTP

FCA

Flaw dimension s = 8.71 inches

L

L is considered for the 3 thinnest points (see later)




Factor Q = 0.619


Flaw dimension s = 8.71 inches

We need not consider the Circumferential CTP because

tLmin < tmm - FCA

0.242 < 0.36 - 0.10 ( = 0.26)We are now only concerned with the profile within theflaw distance L taking the three thinnest adjacent points

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1,5 1,5 1,5 1,5

L = 3.0

Longitudinal Critical Thickness Profile (CTP) within length L

0.470.55

0.36

0.48 0.49tmm

Divide into equal slices using the Grid PitchDimension the L length and pitches


1,5 1,5 1,5 1,5

L = 3.0

Longitudinal Critical Thickness Profile (CTP) within length L

0.470.55

0.36

0.48 0.49tmm

Calculate the Average Thickness (tam) over length L

This is done by taking the area and dividing by L:

tam =(0.55 + 0.36 + 0.36 + 0.48) x 1.5

2 x L ( = 3.0)= 0.438 inches

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Per paragraph 4.4.2.1.f.1:

tam - FCA = 0.438 - 0.10 = 0.338 inchestCmin = 0.492 inches

tam - FCA > tCmin is FALSE

Per paragraph 4.4.2.1.f.2:

tmm - FCA = 0.36 - 0.10 = 0.26 inches

Max(0.5tmin, 0.10) = Max(0.246, 0.10) = 0.246 inches

tmm - FCA > Max(0.246, 0.10) is TRUE

BECAUSE ONE CONDITION IS FALSE, LEVEL 1ASSESSMENT CRITERIA ARE NOT SATISFIED !

BECAUSE ONE CONDITION IS FALSE, LEVEL 1ASSESSMENT CRITERIA ARE NOT SATISFIED !


We can derate the MAWP based on Level 1 assessment

Using the formula in ASME for pressure:

Set t = tam - FCA in the equation = 0.338 inches

MAWP =SEt

Rc + 0.60.t

= 17500 x 0.85 x 0.33824.10 + 0.60 x 0.338

= 207 psi

Therefore MAWP is derated from 300 psi down to 207 psi

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This is based upon the LEVEL 1 assessment !

WE CAN PERFORM A LEVEL 2 ASSESSMENT TOPERFORM A MORE DETAILED ANALYSIS

Per paragraph 4.4.3.2.1.a.1:

tam - FCA = 0.438 - 0.10 = 0.338 inches

RSFa x tCmin = 0.90 x 0.492 = 0.443 inches

tam - FCA > RSFa x tCmin is FALSE

Per paragraph 4.4.3.2.1.b:

tamm - FCA = 0.36 - 0.10 = 0.260 inches

Max(0.5.tmin, 0,1) = Max(0.246, 0.1) = 0.246 inches

tamm - FCA > Max(0.5.tmin,0.1) is TRUE


Because LEVEL 2 assessment criteria are not met, wemust re-evaluate the MAWP according to LEVEL 2requirements.

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We can derate the MAWP based on Level 2 assessment

Using the formula in ASME for pressure:

Set tc = (tam - FCA) / RSFa in the equation

MAWP =SEtc

Rc + 0.60.tc

=17500 x 0.85 x 0.376

24.10 + 0.60 x 0.376

= 230 psi



= (0.438 - 0.10) / 0.90 = 0.376 inches

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API579 Ray Ms