Fitness-For-Service Example Problem Manual

Transcription

1 Fitness-For-Servie Example Problem Manual API 579-/ASME FFS- 009 AUGUST, 009

2 Fitness-For-Servie Example Problem Manual API 579-/ASME FFS- 009 AUGUST, 009

3 SPECIAL NOTES This doument addresses problems of a general nature. With respet to partiular irumstanes, loal, state, and federal laws and regulations should be reviewed. Nothing ontained in this doument is to be onstrued as granting any right, by impliation or otherwise, for the manufature, sale, or use of any method, apparatus, or produt overed by letters patent. Neither should anything ontained in this doument be onstrued as insuring anyone against liability for infringement of letters patent. Neither API nor ASME nor any employees, subontrators, onsultants, ommittees, or other assignees of API or ASME make any warranty or representation, either express or implied, with respet to the auray, ompleteness, or usefulness of the information ontained herein, or assume any liability or responsibility for any use, or the results of suh use, of any information or proess dislosed in this doument. Neither API nor ASME nor any employees, subontrators, onsultants, or other assignees of API or ASME represent that use of this doument would not infringe upon privately owned rights. This doument may be used by anyone desiring to do so. Every effort has been made to assure the auray and reliability of the data ontained herein; however, API and ASME make no representation, warranty, or guarantee in onnetion with this doument and hereby expressly dislaim any liability or responsibility for loss or damage resulting from its use or for the violation of any requirements of authorities having jurisdition with whih this doument may onflit. This doument is published to failitate the broad availability of proven, sound engineering and operating praties. This doument is not intended to obviate the need for applying sound engineering judgment regarding when and where this doument should be utilized. The formulation and publiation of this doument is not intended in any way to inhibit anyone from using any other praties. Classified areas may vary depending on the loation, onditions, equipment, and substanes involved in any given situation. Users of this Standard should onsult with the appropriate authorities having jurisdition. Work sites and equipment operations may differ. Users are solely responsible for assessing their speifi equipment and premises in determining the appropriateness of applying the Instrutions. At all times users should employ sound business, sientifi, engineering, and judgment safety when using this Standard. Users of this Standard should not rely exlusively on the information ontained in this doument. Sound business, sientifi, engineering, and safety judgment should be used in employing the information ontained herein. API and ASME are not undertaking to meet the duties of employers, manufaturers, or suppliers to warn and properly train and equip their employees, and others exposed, onerning health and safety risks and preautions, nor undertaking their obligations to omply with authorities having jurisdition. Information onerning safety and health risks and proper preautions with respet to partiular materials and onditions should be obtained from the employer, the manufaturer or supplier of that material, or the material safety data sheet. All rights reserved. No part of this work may be reprodued, stored in a retrieval system, or transmitted by any means, eletroni, mehanial, photoopying, reording, or otherwise, without prior written permission from the publisher. Contat the Publisher, API Publishing Servies, 0 L Street, N.W., Washington, D.C Copyright 009 by the Amerian Petroleum Institute and The Amerian Soiety of Mehanial Engineers ii

4 API 579-/ASME FFS- 009 Fitness-For-Servie Example Problem Manual FOREWORD The publiation of the standard API 579-/ASME FFS- Fitness-For-Servie, in July 007 provides a ompendium of onsensus methods for reliable assessment of the strutural integrity of industrial equipment ontaining identified flaws or damage. API 579-/ASME FFS- was written to be used in onjuntion with industry s existing odes for pressure vessels, piping and aboveground storage tanks (e.g. API 50, API 570, API 653, and NB-3). The standardized Fitness-For-Servie assessment proedures presented in API 579-/ASME FFS- provide tehnially sound onsensus approahes that ensure the safety of plant personnel and the publi while aging equipment ontinues to operate, and an be used to optimize maintenane and operation praties, maintain availability and enhane the longterm eonomi performane of plant equipment. This publiation is provided to illustrate the alulations used in the assessment proedures in API 579- /ASME FFS- published in July, 007. This publiation is written as a standard. Its words shall and must indiate expliit requirements that are essential for an assessment proedure to be orret. The word should indiates reommendations that are good pratie but not essential. The word may indiates reommendations that are optional. The API/ASME Joint Fitness-For-Servie Committee intends to ontinuously improve this publiation as hanges are made to API 579-/ASME FFS-. All users are enouraged to inform the ommittee if they disover areas in whih these proedures should be orreted, revised or expanded. Suggestions should be submitted to the Seretary, API/ASME Fitness-For-Servie Joint Committee, The Amerian Soiety of Mehanial Engineers, Three Park Avenue, New York, NY 006, or SeretaryFFS@asme.org. Items approved as errata to this edition are published on the ASME Web site under Committee Pages at Under Committee Pages, expand Board on Pressure Tehnology Codes & Standards and selet ASME/API Joint Committee on Fitness-For-Servie. The errata are posted under Publiation Information. This publiation is under the jurisdition of the ASME Board on Pressure Tehnology Codes and Standards and the API Committee on Refinery Equipment and is the diret responsibility of the API/ASME Fitness-For-Servie Joint Committee. The Amerian National Standards Institute approved API 579-/ASME FFS- 009 Fitness-For-Servie Example Problem Manual on August, 009. Although every effort has been made to assure the auray and reliability of the information that is presented in this standard, API and ASME make no representation, warranty, or guarantee in onnetion with this publiation and expressly dislaim any liability or responsibility for loss or damage resulting from its use or for the violation of any regulation with whih this publiation may onflit. iii

5 TABLE OF CONTENTS Speial Notes... ii Foreword... iii Part Introdution. Introdution Sope Organization and Use Referenes... - Part - Fitness-For-Servie Engineering Assessment Proedure. General Example Problem Solutions Tables and Figures... - Part 3 - Assessment Of Existing Equipment For Brittle Frature 3. Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Part 4 - Assessment Of General Metal Loss 4. Example Problem Example Problem Example Problem Example Problem Part 5 Assessment Of Loal Metal Loss 5. Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Part 6 - Assessment Of Pitting Corrosion 6. Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem iv

6 Part 7 - Assessment Of Hydrogen Blisters And Hydrogen Damage Assoiated With HIC And SOHIC 7. Example Problem Example Problem Example Problem Part 8 - Assessment Of Weld Misalignment And Shell Distortions 8. Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Part 9 - Assessment Of Crak-Like Flaws 9. Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Part 0 - Assessment Of Components Operating In The Creep Range 0. Example Problem Example Problem Example Problem Example Problem Part - Assessment Of Fire Damage. Example Problem Example Problem Example Problem Part - Assessment Of Dents, Gouges, And Dent-Gouge Combinations. Example Problem Example Problem Example Problem Example Problem Example Problem Part 3 - Assessment Of Laminations 3. Example Problem Example Problem v

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8 PART INTRODUCTION PART CONTENTS. Introdution Sope Organization and Use Referenes Introdution Fitness-For-Servie (FFS) assessments in API 579-/ASME FFS- Fitness-For-Servie are engineering evaluations that are performed to demonstrate the strutural integrity of an in-servie omponent that may ontain a flaw or damage or that may be operating under speifi onditions that ould produe a failure. API 579-/ASME FFS- provides guidane for onduting FFS assessments using methodologies speifially prepared for pressurized equipment. The guidelines provided in this standard may be used to make runrepair-replae deisions to help determine if pressurized equipment ontaining flaws that have been identified by inspetion an ontinue to operate safely for some period of time. These FFS assessments of API 579- /ASME FFS- are urrently reognized and referened by the API Codes and Standards (50, 570, & 653), and by NB-3 as suitable means for evaluating the strutural integrity of pressure vessels, piping systems and storage tanks where inspetion has revealed degradation and flaws in the equipment or where operating onditions suggest that a risk of failure may be present.. Sope Example problems illustrating the use and alulations required for Fitness-For-Servie Assessments desribed in API 579-/ASME FFS- are provided in this doument. Example problems are provided for all alulation proedures in both SI and US Customary units..3 Organization and Use An introdution to the example problems in this doument is desribed in Part of this Standard. The remaining Parts of this doument ontain the example problems. The Parts in this doument oinide with the Parts in API 579-/ASME FFS-. For example, example problems illustrating alulations for loal thin areas are provided in Part 5 of this doument. This oinides with the assessment proedures for loal thin areas ontained in Part 5 of API 579-/ASME FFS-..4 Referenes API 579-/ASME FFS- Fitness For Servie. -

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10 PART FITNESS-FOR-SERVICE ENGINEERING ASSESSMENT PROCEDURE PART CONTENTS. General Example Problem Solutions Tables and Figures General The Fitness-For-Servie assessment proedures in API 579-/ASME FFS- are organized by flaw type or damage mehanism. A list of flaw types and damage mehanisms and the orresponding Part that provides the FFS assessment methodology is shown in API 579-/ASME FFS-, Table.. In some ases it is required to use the assessment proedures from multiple Parts based on the damage mehanism being evaluated.. Example Problem Solutions.. Overview Example problems are provided for eah Part and for eah assessment level, see API 579-/ASME FFS-, Part. In addition, example problems have also been provided to illustrate the interation among Parts as required by the assessment proedures in API 579-/ASME FFS-. A summary of the example problems is ontained in Tables E- - E.... Calulation Preision The alulation preision used in the example problems is intended for demonstration proposes only; an intended preision is not implied. In general, the alulation preision should be equivalent to that obtained by omputer implementation, rounding of alulations should only be done on the final results. -

11 .3 Tables and Figures Table E- - Part 3 Examples on Assessment for Brittle Frature Example Assessment Level Units Type of Equipment Geometry Type or Desription of Analysis US Pressure Vessel --- MAT alulation with PWHT US Pressure Vessel --- MAT alulation without PWHT 3 US Pressure Vessel --- MAT alulation without PWHT 4 US Pressure Vessel --- MAT alulation with PWHT 5 US Pressure Vessel --- MAT redution vs P/Prating (Pressure Temperature Rating Basis) 6 SI Pressure Vessel Cylinder MAT redution vs S*E*/SE (Stress Basis) 7 and US Pressure Vessel Sphere MAT redution vs S*E*/SE (Stress Basis) 8 US Pressure Vessel Sphere MAT redution vs S*E*/SE (Stress Basis) 9 US Pressure Vessel Sphere MAT redution vs operating pressure / hydrotest pressure 0 3 US Demethanizer tower --- Assessment based on frature mehanis priniples of Part 9 Table E- - Part 4 Examples on Assessment of General Metal Loss Example Assessment Level Units Type of Equipment Geometry Loation of Metal Loss Loading(s) Average Thikness based on and SI Heat exhanger Cylinder Away from msd Internal pressure Full vauum Point thikness reading and US Pressure Vessel Cylinder Away from msd Internal pressure Critial thikness profiles 3 and SI Pressure Vessel Elliptial head Away from msd Internal pressure Critial thikness profiles 4 US Pressure Vessel Nozzle At msd Internal pressure Given in the data -

12 Table E-3 - Part 5 Examples on Assessment of Loal Metal Loss Example Assessment Level Units Type of Equipment Geometry Loation of Metal Loss Loading(s) Type of Metal Loss US Pressure Vessel Cylinder Away from msd Internal pressure LTA and US Pressure Vessel Cylinder Away from msd Internal pressure Grooves 3 US Pressure Vessel Cylinder Away from msd Internal pressure Supplemental loads LTA 4 US Pressure Vessel Cylinder Away from msd Internal pressure LTA 5 SI Pressure Vessel Cylinder Away from msd Internal pressure LTA 6 US Pressure Vessel Nozzle At msd Internal pressure Uniform LTA 7 US Storage Tank Cylinder Away from msd Fill Height LTA 8 US Piping Elbow Away from msd Internal pressure Uniform LTA 9 US Pressure Vessel Cylinder Away from msd Vauum LTA Table E-4 - Part 6 Examples on Assessment of Pitting Corrosion Example Assessment Level Units Type of Equipment Geometry Loading(s) Type of Pitting Comment US Pressure Vessel Cylinder Internal pressure Widespread pitting --- SI Piping Cylinder Internal pressure Widespread pitting US Horizontal Pressure Vessel Cylinder Internal pressure Supplemental loads Widely sattered pitting US Pressure Vessel Cylinder Internal pressure Loalized pitting LTA per Part 5 Level 5 US Pressure Vessel Cylinder Internal pressure Pitting in LTA LTA per Part 5 Level 6 US Pressure Vessel Cylinder Internal pressure Widespread pitting Inside and outside

13 Table E-5 - Part 7 Examples on Assessment of Blisters and HIC and SOHIC Damage HIC Damages Example HIC Area Level Loation in Thikness Comment Servie Condition and Surfae breaking Level per Part 5 Level a Surfae breaking b Sub surfae Combined 3 Surfae breaking --- Equipment will remain in hydrogen harging servie 4 and Sub surfae Level per Part 5 Level 3 Sub surfae --- Equipment will not remain in and Sub surfae Level per Part 5 Level hydrogen harging servie Blisters Example Blister Level Bulge Diretion Craking at Periphery Crown Craking or Venting Comment A and Outside No Crak Level per Part 5 Level B Outside No Vent --- C Inside No Vent --- D and Inside No Crak Level per Part 5 Level E Inside No Vent --- F Inside No No --- G and Outside Yes (Inward) Crak Level per Part 5 Level H and Outside No Vent Level per Part 5 Level Note: Common harateristis: - Type of Equipment: Pressure Vessel - Geometry: Cylinder - Units: US - Loading: Internal pressure -4

14 Table E-6 - Part 8 Examples on Assessment of Weld Misalignment and Shell Distortions Example Assessment Level Units Type of Equipment Geometry Loading(s) Type of Damage Comment and US Piping Cylinder Internal pressure Weld misalignment Peaking US Piping Cylinder Flutuating internal pressure Weld misalignment Peaking Fatigue assessment by: - elasti stress analysis and equivalent stress - elasti stress analysis and strutural stress 3 and US Pressure Vessel Cylinder Internal pressure Out-of-roundness Assessment based on Dmax-Dmin 4 US Pressure Vessel Cylinder Internal pressure Weld misalignment Center line offset and peaking 5 and US Pressure Vessel Cylinder Internal pressure Out-of-roundness Assessment based on radius expressed as a Fourier series 6 and 3 US Pressure Vessel Shell - Heads - Stiffening rings Internal pressure Vauum onditions General shell distortion Level 3 based on Finite Element Analysis: - limit load analysis (elasti perfetly plasti material behavior) - hek of loal strain (elasti-plasti with strain hardening material behavior) - elasti bukling analysis (hek of stability of deformed shell) - hek of fatigue requirements -5

15 Table E-7 - Part 9 Examples on Assessment of Crak-Like Flaws Example Assessment Level Units Type of Equipment Geometry Loading(s) Type of Crak Comment US Pressure Vessel Cylinder Internal pressure - Longitudinal - Semi-elliptial Shallow rak in parallel to weld seam SI Pressure Vessel Sphere Internal pressure - Cirumferential - Semi-elliptial Deep rak perpendiular to weld seam 3 and US Pressure Vessel Cylinder Internal pressure - Semi-elliptial - Oriented at 30 from prinipal diretion Flaw length to be used in assessment 4 and US Pressure Vessel Cylinder Internal pressure - Semi-elliptial - Oriented along bevel angle Flaw depth to be used in assessment 5 and US Pressure Vessel Cylinder Internal pressure - Longitudinal - Semi-elliptial - Residual stresses due to welding based on surfae distribution - Uniform distribution along thikness 6 and SI Piping Cylinder Internal pressure Global bending moment - Cirumferential - 360degree rak - Residual stresses due to welding based on through-thikness distribution - Fourth order polynomial along thikness 7 SI Piping Cylinder Internal pressure Global bending moment - Cirumferential - Semi-elliptial - Residual stresses idential to those of example Coeffiients of polynomial alulated by weight funtion method 8 3 US Pressure Vessel Cylinder Internal pressure - Longitudinal - Semi-elliptial - Residual stresses idential to those of example Subritial fatigue rak growth - Remaining life assessment 9 3 US Failure Assessment Diagram based on atual material properties 0 3 SI Pressure Vessel Nozzle Internal pressure Quarter-elliptial Assessment based on elasti-plasti Finite Element Analysis -6

16 Table E-8 - Part 0 Examples on Assessment of Components Operating in the Creep Range Example Assessment Level Units Type of Equipment Geometry Loading(s) Comment US Pressure Vessel Cylinder Elliptial head Internal pressure - Temperature exursion in the reep range - Chek that damage is below the aeptable one US Heater Tubes Internal pressure - Heater operating in the reep range - Exursion at higher temperature than design one - Calulation of overall damage in the omplete expeted life 3 US Heater Tubes Internal pressure - Same as example 0. with the addition of - Calulation of remaining life using Larson Miller parameters 4 3 US Pressure Vessel Cylinder Internal pressure - Vessel operating in the reep range - Longitudinal semi-elliptial surfae rak - Creep rak growth - Calulation of remaining life using MPC Omega projet data Table E-9 - Part Examples on Assessment of Fire Damage Example Assessment Level Units Type of Equipment Geometry Loading(s) Comment US HEZ from observation after fire Horizontal Internal pressure HEZ from observation after fire US Pressure Cylinder Vessel Supplemental loads Allowable stress from hardness results HEZ from observation after fire 3 Allowable stress from hardness results --- Internal pressure 3 US (+SI) Depropanizer tower - Stress analysis for shell distortion - Testing and metallographi evaluation of material samples -7

17 Table E-0 - Part Examples on Assessment of Dents, Gouges and Dent-Gouge Combinations Example Assessment Level Units Type of Equipment Geometry Type of Damage Loading(s) Comment SI Piping Cylinder Dent Internal pressure --- SI Piping Cylinder Dent Flutuating internal pressure Fatigue analysis 3 SI Piping Cylinder Gouge Internal pressure SI Piping Cylinder Dent-Gouge Internal pressure SI Piping Cylinder Dent-Gouge Internal pressure --- Table E- - Part 3 Examples on Assessment of Laminations Example Assessment Level Units Type of Equipment Geometry Loading(s) Comment US Pressure Vessel Cylinder Internal pressure - laminations away from msd US Pressure Vessel Cylinder Internal pressure - Not in hydrogen harging servie -8

18 PART 3 ASSESSMENT OF EXISTING EQUIPMENT FOR BRITTLE FRACTURE EXAMPLE PROBLEMS 3. Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem A pressure vessel, in thik, fabriated from SA-85 Grade C in austi servie was originally subjet to PWHT at the time of onstrution. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division. Determine the Level MAT for the shell setion. Based on Curve A in Figure 3.4, a MAT of 69 F was established for the vessel shell setion without any allowane for PWHT. The material is a P Group steel; therefore, applying the allowane for PWHT redues the MAT by 30 F and establishes a new MAT of 39 F. 3. Example Problem The ylindrial shell of a horizontal vessel 0.5 in thik is fabriated from SA-53 Grade B seamless pipe. There is no toughness data on the material. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division. Determine the Level MAT. Sine all pipe, fittings, forgings, and tubing not listed for Curves C and D are inluded in the Curve B material group, this urve of Figure 3.4 may be used. In this ase, the MAT for the ylindrial shell is found to be -7 F. 3.3 Example Problem 3 A horizontal drum.5 in thik is fabriated from SA-56 Grade 70 steel that was supplied in the normalized ondition. There is no toughness data on the steel. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division. Determine the Level MAT for the shell setion. Sine SA-56 Grade 70 is manufatured to a fine grain pratie and was supplied in this ase in the normalized ondition, Curve D of Figure 3.4 may be used. In this ase, the MAT for the shell setion is found to be -4 F. 3-

19 3.4 Example Problem 4 A stripper olumn was onstruted following the rules of the ASME B&PV Code, Setion VIII, Division. This vessel has the following material properties and dimensions. Vessel Data Material = SA 56 Grade 65 Year 968 Design Conditions = 50 psi@300 F Allowable Stress = 6,50 psi Inside Diameter = 90 in Operating Pressure = 40 psi Wall Thikness =.00 in Critial Exposure Temperature = 0 F The vessel was PWHT Impat test data is not available. Perform a Level Assessment for the shell setion per paragraph Sine SA-56 Grade 65 used in the onstrution of the stripper is in the non normalized ondition, Curve B of Figure 3.4 may be used. In this ase, the MAT for the shell setion is found to be 3 F. The vessel was PWHT and an ASME P Group material was used. Therefore, the MAT determined before an be redued further using Equation 3.. The redued MAT of this setion is equal to F, whih is lower than the CET 0 F. The Level Assessment Criteria are Satisfied for the shell setion. 3-

20 3.5 Example Problem 5 A reator vessel fabriated from SA-04 Grade B 993 (C-½ Mo) has the following material properties and dimensions. The reator was onstruted to the ASME B&PV Code, Setion VIII, Division. Develop a table of MAT for the shell setion as a funtion of pressure based on paragraph and the allowanes given in Figure 3.7 and Table 3.4. Vessel Data Material = SA 04 Grade B Year 993 Design Conditions = 390 psi@300 F Allowable Stress = 7,500 psi Inside Diameter = 34 in Operating Pressure = 40 psi Wall Thikness =.7 in Startup Pressure = 57 psi Weld Joint Effiieny =.0 Corrosion Allowane = /6 in MAT at Design Pressure = 08 F see Curve A of Figure 3.4 Impat test data is not available. Using this relationship, a table of MAT an be established for the shell setion as a funtion of pressure based on paragraph and the allowanes given in Figure 3.7 and Table 3.4. P ( psi ) rating Table E3.5- P P T ( F ) MAT ( F) R The operating pressures and orresponding values of the shell setion MAT in this table must be ompared to the atual vessel operating onditions to onfirm that the metal temperature ( CET ) annot be below the MAT at the orresponding operating pressure. 3-3

21 3.6 Example Problem 6 A CO storage tank with a 03.0 millimeters ID shell setion with a nominal thikness of 7.5 millimeters, was onstruted in 98 aording to the ASME Code Setion VIII, Division. The material of onstrution was SA-6, whih is a arbon steel. It was designed for a non orrosion servie (orrosion allowane equals zero), with a joint effiieny 00% (full X-ray inspetion), and without post-weld heat treatment. This storage vessel has the following harateristis. Tank Data Material = SA 6 Year 98 Design Conditions =.3744 C Allowable Stress = 39.6 MPa Inside Diameter = 03.0 mm Operating Pressure =.3744 C Wall Thikness = 7.5 mm Weld Joint Effiieny =.0 Corrosion Allowane = None MAT at Design Pressure = - C see Curve B of Figure 3.4M Impat test data is not available. Develop a table of MAT for the shell setion as a funtion of pressure based on paragraph and the allowanes given in Figure 3.7M and Table 3.4. Calulate the membrane stress for a ylindrial pressure vessel as a funtion of pressure (see Annex A): D 03.0 R = + FCA + LOSS = = 06 mm t = t FCA LOSS = = 7.5 mm R 06 S * E* = P. E* P.. =. P t = Using this relationship, a table of MAT an be established as a funtion of pressure based on paragraph and the allowanes given in Figure 3.7 and Table

22 P ( MPa ) S* E* ( MPa ) Table E3.6- R ts * * SE = TR ( C) MAT ( C) SE The operating pressures and orresponding values of the MAT in this table must be ompared to the atual vessel operating onditions to onfirm that the metal temperature ( CET ) annot be below the MAT at the orresponding operating pressure. 3-5

23 3.7 Example Problem 7 A spherial platformer reator was onstruted in 958 aording to the ASME Code, Setion VIII, Division. The material of onstrution is C-½Mo, speifiation SA-04 Grade A. The vessel has the following information available: Vessel Data Material = SA 04 Grade A Year 958 Design Conditions = 650 F Allowable Stress = 6,50 psi Inside Diameter = 44 in Operating Pressure = 390 psig Nominal Thikness =.6875 in Atual Wall Thikness =.765 in Weld Joint Effiieny = 0.95 Corrosion Allowane = in Impat test data is not available. The vessel was PWHT Critial Exposure Temperature = 60 F Perform a Level Assessment for the shell setion per paragraph SA-04 Grade A is one of the low alloy steel plates not listed in Curves B, C, and D. Therefore Curve A of, Figure 3.4 shall be used to determine the MAT. In this ase, the MAT found is equal to 93 F. The reator was PWHT ; however, an ASME P3 Group material was used. Therefore, the MAT determined before annot be redued further using Equation 3.7. The MAT is equal to 93 F, whih is higher than the CET of 60 F. The Level Assessment Criteria are Not Satisfied. Perform a Level Assessment per paragraph and develop a table of MAT as a funtion of pressure based on the allowanes given in Figure 3.7 and Table 3.4. Calulate the membrane stress for a spherial pressure vessel as a funtion of pressure (see Annex A): D 44 R = + FCA + LOSS = = in t = t FCA LOSS = =.560 in P R P = + = + = P S*E* 0. E* t.560 Using this relationship, a table of MAT an be established as a funtion of pressure based on paragraph , the proedure in Table 3.4 and the allowanes given by the appropriate urve in Figure

24 P ( psi ) S*E* ( psi ) Table E3.7- R ts * * SE = TR ( F) MAT (F) SE 650 4, , , , , , , , , The operating pressures and orresponding values of the MAT in this table must be ompared to the atual vessel operating onditions to onfirm that the metal temperature CET annot be below the MAT at the orresponding operating pressure. In this partiular ase the reator is operating at 390 psig, and the CET is equal to 60 F. Aording to this table at 390 psig the redued MAT is equal to 49 F, whih is lower than the CET. Therefore, The Level Assessment Criteria are Satisfied for the operating onditions. 3-7

25 3.8 Example Problem 8 A sphere fabriated from SA-44 Grade G has the following material properties and dimensions. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division. Develop a table of MAT for the shell setion as a funtion of pressure based on paragraph and the allowanes given in Figure 3.7 and Table 3.4. Vessel Data Material = SA 44 Grade G Year 005 Design Conditions = 75.0 F Allowable Stress =,400 psi Inside Diameter = in Wall Thikness =.6 in Weld Joint Effiieny =.0 Corrosion Allowane = in MAT at Design Pressure = 80 F see Curve A of Figure 3.4 Impat test data is not available. Calulate the membrane stress for a spherial pressure vessel as a funtion of pressure (see Annex A): D R = + FCA + LOSS = = in t = t FCA LOSS = =.975 in P R P S * E* =. E*.. =. P t + 0 =

26 Using this relationship, a table of MAT an be established as a funtion of pressure based on paragraph , the proedure in Table 3.4 and the allowanes given by the appropriate urve in Figure 3.7. P ( psi ) S*E* ( psi ) Table E3.8- R ts * * SE = TR ( F) MAT (F) SE 74.86, , , , , , , , , The operating pressures and orresponding values of the MAT in this table must be ompared to the atual sphere operating onditions to onfirm that the metal temperature CET annot be below the MAT at the orresponding operating pressure. 3-9

27 3.9 Example Problem 9 A spherial pressure vessel has the following properties and has experiened the following hydrotest onditions. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division. Using paragraph and Figure 3.8, prepare a table showing the relationship between operating pressure and MAT for the shell setion. Vessel Data Hydrotest pressure = 300 psig or50% of design pressure Design pressure = 00 psig Metal temperature during hydrotest = 50 F The maximum measured metal temperature during hydrotest was 50 F. To be onservative, 0 F is added to this and the analysis is based on a hydrotest metal temperature of 60 F. Table E3.9- Operating Pressure (psig) Operating Pressure Hydrotest Pressure Temperature Redution ( F) MAT ( F) The operating pressures and orresponding values of the MAT in this table must be ompared to the atual sphere operating onditions to onfirm that the metal temperature CET annot be below the MAT at the orresponding operating pressure. 3-0

28 3.0 Example Problem 0 A demethanizer tower in the old end of a ethylene plant typially operates older in the top portion of the tower and warmer at the bottom of the tower. The bottom of the tower is kept warm with a side stream irulated through a reboiler. The top portion of the tower is onstruted from a 3½% Ni steel whih has been impat tested for toughness at -0 C. The lower portion of the tower is onstruted from a fully killed, fine grained and normalized arbon steel whih is impat tested for toughness at -46 C. A potential for brittle frature exists if the reboiler does not operate beause old liquid will flow down the tower into the arbon steel setion resulting in operating temperatures signifiantly lower than -46 C. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division. Perform a brittle frature assessment of ethylene plant demethanizer tower onsidering all aspets of operation. The upset ondition of the reboiler not operating properly should be inluded in the assessment. A brittle frature assessment onsistent with paragraph (Level 3 assessment) an be performed on the demethanizer tower. The approah is illustrated with referene to the demethanizer tower as illustrated in Figure E3.0-. The assessment to be utilized is based on the frature mehanis priniples presented in Part 9. In the assessment, the limiting flaw size in the tower will be established, and a sensitivity study will be performed to determine how the limiting flaw size hanges as the temperature in the tower drops during an exursion. Based on the results of the assessment, a graph of limiting flaw size versus temperature will be onstruted. This graph is referred to as a Frature Tolerane Signature (FTS). The FTS provides an indiation of the safety margin in terms of limiting flaw size. In addition, the FTS an be used to selet a lower thermal exursion limit by establishing a flaw size that an be deteted with suffiient onfidene using an available NDE tehnique. The FTS an then be used to develop a modified MAT diagram, onto whih the exursion limits an be superimposed. An assumption in the assessment is that the tower has been orretly fabriated to ode standards at the time of onstrution. It is also a required that the vessel material speifiations and inspetion history are known and doumented. These are essential to enable reasonable assumptions to be made about the material toughness properties, stress levels, and likelihood of fabriation or servie indued flaws. 3-

29 Material: Carbon Steel Material: 3-/ % Ni API 579-/ASME FFS- 009 Fitness-For-Servie Example Problem Manual Temperature ( o C) Feed Tray 6 Feed Normal Operation Potential Exursion Tray 33 Potential Violation Feed 3 Tray 3 9 mm Tray 4 Original MAT ( based solely on Impat Tests) Tray 300 mm Position Detail A Typial Design Detail B Temperature Profile Along The Length Of The Tower Material: SA-56 Grade 70 (KCS) Minimum Yield Stength at operating onditions 6 MPa Pressure: 3.7 MPa-g Toughness: -46 o C PWHT: Yes Weld Joint Effiieny:.0 Figure E3.0- Shemati Of Demethanizer 3-

30 Assessment Approah The frature analysis part of the assessment is based on the methodology presented in Part 9. In order to perform this analysis a flaw size must be assumed, and the applied stress and material toughness must be known. The frature assessment is limited to the lower arbon steel setion of the tower sine this is the only setion to experiene an MAT violation (see Figure E3.0-). Assumed Flaw Size A onservative yet representative hypothetial surfae breaking elliptial rak with an aspet ratio of 6: (:a) is assumed to be loated on the inside surfae of the vessel. The rak is also assumed to be parallel to a longitudinal weld seam. Other representative flaws elsewhere in the vessel ould also be onsidered. However, as will be seen latter, the relative nature of the results as expressed by the FTS are not signifiantly affeted by suh variations, though the minimum exursion temperature will be. Applied Stress In order to utilize the assessment proedures of Part 9, the applied stress at the loation of the flaw must be omputed and ategorized. Based on the operation sequene of the tower, four load soures are used to desribe the applied stress; the hoop stress from internal pressure, the residual stress in welds, loal stress effets from nozzles and attahments, and thermal transient stresses during the upset. In addition, onsideration should be given to oasional loads suh as wind or earthquake loads. These loads are ignored in this example. Hoop Stress From Internal Pressure The pressure stress is alulated using the ode design equations. This stress is ategorized as a primary membrane stress (see Annexes A and B). Residual Stress In Welds The residual stress an be estimated based on whether post weld heat treatment (PWHT) has been performed (see Annex E). Beause the tower was subjet to PWHT at the time of onstrution, the residual stress is taken as 0% of the weld metal room temperature yield strength plus 69 MPa. This stress is lassified as a seondary membrane stress. Loal Stress Effets From Nozzles And Attahments In this sreening study, a detailed analysis of the loal stresses at the nozzles and attahments were not performed. To aount for a level of stress onentration at these loations a stress onentration fator is used. In this example a stress onentration fator.3 will be applied to all primary membrane and bending stresses. Transient Thermal Stresses These stresses may be evaluated by using losed form solutions or a finite element analysis. In this example, a temperature exursion model onsisting of a "old front" of liquid is assumed to move down the tower. The liquid temperature in the old front is defined by the proess upset ondition. The vessel wall is subsequently ooled from its pre-exursion steady-state temperature to the old liquid temperature. Convetive heat transfer from the old fluid to the vessel shell is assumed to be instantaneous, and heat loss to the atmosphere is negleted. The stress versus time history at a point on the vessel wall omputed using a finite element analysis is shown in Figure E

31 Figure E3.0- Transient Thermal Stress Computed From A Finite Element Stress Analysis The results from the finite element analysis onfirm that the magnitude of the maximum transient stress an be readily evaluated from the following equation: where, Eα ΔT σ = exp β β β = hl k ( ν ) with, E = Modulus of Elastiity, MPa, h = Convetion Coeffiient, W/m - o C, k = Thermal Condutivity of the shell material, W/m- o C, L = Shell Wall Thikness, m. Δ T = Temperature differene; the differene between the steady state wall temperature before the exursion and the temperature of the fluid ausing the exursion, o C, α = Thermal expansion oeffiient, / o C, ν = Poisson s ratio σ = Thermal stress, MPa. 3-4

32 Based on the results of the finite element analysis, the maximum stress is a through thikness bending stress with tension on the inside surfae. The resultant transient stress is onsidered to be a primary stress and for further onservatism in this example, it is ategorized into equal membrane and bending omponents. In this example, a thermal stress of 0 MPa is omputed based on a liquid temperature of -7 C and a shell temperature of -35 C. A summary of the applied stresses is shown in Table E3.0-. Table E3.0- Summary Of Applied Stresses Magnitude And Classifiation Of Applied Stresses Soure Of Stress Magnitude Of Stress Classifiation Of Stress Hoop Stress From Internal Pressure 53 MPa P = 53 MPa m Residual Stress In Welds 67 MPa Q = 67 m MPa Loal Stress Effets From Nozzles And Attahments A stress onentration fator of.3 is used in the analysis. A stress onentration fator of.3 is used in the analysis. Transient Thermal Stresses 0 MPa 0 MPa Pm = = 0 MPa 0 MPa Pb = = 0 MPa Applied Stress Results For Use In Frature Assessment Stress Category Final Stress Result Primary Membrane Stress P m = ( 53 MPa + 0MPa). 3 = MPa Primary Bending Stress P b = ( 0 MPa). 3 = 3MPa Seondary Membrane Stress Q m = 67MPa 3-5

33 Material Frature Toughness Atual frature toughness data is not normally available for proess equipment; therefore, it is neessary to adopt a lower bound approah to desribe the variation of toughness with temperature. The most widely used lower bound is the K IR urve from Figure F.3 in Annex F. This urve is shown in Figure E To use this urve it is neessary to estimate a referene temperature to position the temperature axis on an absolute sale. The referene temperature is typially taken as the Nil Dutility Temperature (NDT). In this example, the temperature at whih a 40 Joules Charpy V-Noth energy is obtained from a longitudinal speimen is seleted as the NDT. It should be noted that Annex F reommends the less onservative value of 0 J. The use of this value would shift the FTS urve shown in Figure E3.0-4 upward. When an impat temperature orresponding to 40 J is not available, atual values are extrapolated to give an effetive 40 J test temperature using the relationship:.5 J/ C. For this assessment the lowest average Charpy value was used for determining the NDT as opposed to the lowest minimum. The use of atual values is illustrated in Figure E Shabbiis (WCAP - 63) Ripling and Crosley HSST, 5th Annaula Information Meeting, 97, Paper No. 9 Unpublished Data MRL Arrest Data 97 HSST Info MIG Temperature Differene ( o C) Notes:. Atual Charpy V-Noth data: 33/3 Joules at -46 o C. Equivalent temperature at 40 Joules from: -46 o C + (40 J 33 J)/.5 J/ o C = -4 o C; therefore, NDT in this figure, indexes to -4 o C. Figure E3.0-3 Toughness Evaluation Using The K IR Curve 3-6

34 Material Properties Atual material properties obtained from equipment reords should be used for yield strength and Charpy impat energy. Other properties an be determined using Annex F. A orretion an be adopted to inrease the value of yield strength at low temperature. While this was used in the example its effet is primarily a higher plasti ollapse limit, whih is not a typial limiting fator for low temperature brittle frature. Frature Tolerane Signature (FTS) The applied stress, material properties, and frature toughness parameter defined above are used to reate a plot of limiting flaw size versus temperature as illustrated in Figure E The ritial flaw depth is in the through thikness diretion and is expressed as a perentage of the wall thikness with a 6: aspet ratio maintained. The absolute fator of safety in the ritial flaw size is undetermined, but is a funtion of the assumptions made with respet to lower bound toughness, stress, stress multiplier, and the NDT indexing temperature For A Design Pressure of 37. Bar-g 60 A Crak Depth = 6% of the wall thikness B 0 0 E D C Temperature ( o C) Figure E3.0-4 Frature Tolerane Signature 3-7

35 The influene of the transient operation on the limiting flaw size is shown in Figure E Line segment A-B represents steady operation and defines the limiting flaw for gradual ool down to -36 C where the limiting flaw is 5% of the wall thikness. The exposure to old liquid at -7 C, begins at B and results in an almost instantaneous drop in limiting flaw size to % of the wall thikness at C. This ours as a result of the applied thermal stress. The initial effet of the thermal transient dereases as the shell ools, whih results in a derease of the temperature differene between the shell and the old liquid. During this period the material toughness is redued, but the thermal stress is also redued, with the net result that the limiting flaw size is redued to 7% of the wall thikness at Point D. At this point the metal temperature reahes equilibrium with the old liquid, and from point D to E a return to steady state ool-down ontinues. The limiting flaw size is % of the wall thikness at Point E where the minimum temperature reahed. The shape of the FTS urve in Figure E3.0-4 follows that of the K IR urve, and is modified only by the transient thermal effet. More or less onservative assumptions on stress and flaw size will lower or raise the urve vertially, respetively. Assuming a lower NDT will move the urve horizontally to the left. For example, using the less onservation K IC urve in plae of the K IR urve in evaluating the toughness would shift the urve in Figure E3.0-4 upward resulting in a higher permitted rak depth. For this reason the urve provides useful insight into brittle frature resistane during an exursion. The flatness of the urve between points C and E makes limiting temperature preditions highly sensitive to the minimum flaw size. This in turn is greatly influened by type and extent of inspetion and fators suh as probability of detetion (POD) of flaws. While work still needs to be done to larify POD issues, appliation of detailed NDE to a vessel should enable a minimum flaw size to be assumed with suffiient onfidene to enable the FTS to be used to speify a minimum exursion temperature. Based on the POD urve shown in Figure E3.0-5, a flaw depth of 4.5 mm should be detetable using a magneti partile examination tehnique (MT) with a onfidene level greater than 90%. For the 6: aspet ratio assumed in developing the FTS, this equates to a rak of length 7 mm. 0.8 UT - Nordtest POD - Probability Of Detetion UT0 Inspetion Method UT - Nordtest AE + UT MT Flaw Depth, mm Figure E3.0-5 Comparison Of Inspetion Methods - Probability Of Detetion Curves 3-8

36 Summary Of Results The evaluation of a potential thermal exursion for the demethanizer tower illustrated in Figure E3.0- is summarized in Figure E The stresses and other fators assumed in onduting the evaluation are shown in Table E3.0-. An important aspet of the required data is a realisti estimate of the ritial exposure temperature (CET). This is the atual metal temperature, or more likely the metal temperature predited by proess simulation programs during an exursion. The exursion temperature in the example illustrates that an MAT violation will not our in the 3.5% Ni setion above tray 33. Hene the evaluation need only onsider the lower arbon steel setion. The exursion temperature plotted in Figure E3.0-6 defines two ases to be onsidered. Case The lowest temperature in the arbon steel setion is at tray 3 with a pre-exursion temperature of -35 C and an exursion delta of -37 C to -7 C. Case The largest delta of -49 C ours from a steady state temperature of - C at tray 4 to give an exursion temperature of -6 C. Figure E3.0-6 Demethanizer MAT Versus Loation 3-9

37 To illustrate the influene of inspetion on the results, it is assumed that the tower has been 00% visually inspeted internally. In addition, it is assumed that all internal weld seams are inspeted by wet fluoresent magneti partile methods, and angle probe ultrasoni, from the bimetalli weld to a irumferential weld between trays 4 and 5. It is further assumed that any flaw indiations would be removed by light grinding. As part of suh an assessment it would also be reasonable to ondut a hydrostati test at 50% of design pressure. These assumptions allow the arbon steel setion to be evaluated by two approahes: The visually inspeted region an be assessed using basi MAT priniples in aordane with the "ode ompliant approah", or The MT/UT inspeted region an be assessed using the more sophistiated FTS approah. The MAT approah for two onstant flaw sizes is shown in Figure E One is % of the wall thikness, and was seleted to pass through original design onditions. For larity, the effet of the transient stress is ignored in Figure E The % urve illustrates that the exursion temperature at tray 4 of -6 C is within the aeptable MAT zone and, provided that additional transient stresses an be aommodated within the exursion margin, the MAT an be set at -66 C based on operating rather than design pressure. This hek is made by evaluating the ritial flaw size during the exursion, using an FTS for tray 4, and ensuring it is always above %. The hek is made using tray 4 temperature and exursion onditions, with operating pressure applied rather than design. The hek onfirms that in this ase -66 C is an aeptable exursion limit below tray 4. Figure E3.0-7 Pressure Temperature Relationship for Constant Defet Size - Killed Carbon Steel Setion 3-0

38 The seond feature apparent from the % urve is that a violation still exists at tray 3. Tray 3 is however, loated in the setion of the tower that was subjet to MT/UT inspetion. Thus it an be assessed on the basis of a smaller flaw size. The 6% of the wall thikness urve in Figure E3.0-7 represents this riterion as proposed earlier. It is lear that the -7 C exursion is aommodated, even at design pressure. The FTS urve in Figure E3.0-4, indiates that a 4.5 mm limiting flaw is ritial below -80 C when analyzed at full design pressure. In pratie the ontingeny is unlikely to violate design onditions, hene there is an inherent onservatism over the more realisti operating ase. An FTS for the operating ase results in - C as the limiting temperature. To be of value to operating personnel, and to ompare it with the exursion temperature, it is useful to express the result in the form of an exursion limit for the tower, as shown in Figure E This allows a diret omparison of normal operation, exursion temperature, MAT and exursion limits. The distintion between the MAT and the exursion limits is to differentiate between the "ode ompliant" and non ode ompliant aspets of the assessment. The purpose of the analysis is to establish reasonable exursion limits and to quantify the risk assoiated with exursions below the MAT. It is not meant to enourage normal operation at temperatures lower than the MAT. Reommendations and Conlusions For this partiular type of Level 3 assessment only, the equipment to be evaluated should satisfy the following riteria: Meets the design and fabriation requirements of a reognized ode of onstrution, Demonstrates, by measured values, minimum toughness of weld, HAZ and plate materials, and An appropriate NDE tehnique is used to prelude the existene of flaws with suffiient onfidene based on a risk assessment. When a Level 3 assessment is made, its aeptability should be subjeted to suitable riteria suh as the following: ) Where no additional detailed inspetion for a surfae breaking flaw is performed by an appropriate NDE tehnique, the exursion limits should be no lower than the MAT as developed by using the assessment proedures in this part. ) Where MT examination or equivalent is arried out around nozzles and attahments, the MAT may be based on a ¼-t or 6.4 mm deep flaw, whihever is the smaller, with a 6: aspet ratio. 3) Where an appropriate NDE tehnique is used to prelude the existene of flaws with suffiient onfidene, the exursion limit an be based on a Frature Tolerane Signature FTS approah. 4) The assessment is only valid if the servie onditions in the vessel are essentially unhanged or less severe than those experiened in the past. 5) Poor operation in terms of ontrol tehniques leading to frequent yling or proess upsets should be disouraged by limiting the number of exursions allowed during the life of the vessel. 6) Hydrostati testing at a temperature where the material toughness is above the lower shelf is reommended. This is an example of a Level 3 Assessment. It is not intended to be a "prototype" for all Level 3 assessments, sine there are many different approahes whih an be used suessfully at this level. 3-

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40 PART 4 ASSESSMENT OF GENERAL METAL LOSS EXAMPLE PROBLEMS 4. Example Problem Example Problem Example Problem Example Problem Example Problem Internal orrosion on the ylindrial shell of a heat exhanger has been found during an inspetion. Details regarding the heat exhanger and inspetion data are given below. The heat exhanger was onstruted to the ASME B&PV Code, Setion VIII, Division, Edition 989. Determine if the heat exhanger is suitable for ontinued operation. Vessel Data Material = SA 56 Grade 60 Year 989 Design Conditions = 3.85 C and full C Inside Diameter = 484 mm Nominal Thikness = 6 mm Future Corrosion Allowane = mm Weld Joint Effiieny =.0 Tubesheet to tubesheet distane = 54 mm 4-

41 Inspetion Data Based on a visual inspetion, the orrosion loss is haraterized as general, and point thikness readings will be used in the assessment (see paragraph and ). Point thikness readings were taken in aordane with paragraph Loation Table E4.- Thikness Reading tmm,

42 Perform a Level assessment for internal pressure per paragraph a) STEP Use the point thikness readings shown above; and determine the minimum measured thikness, t mm, the average measured thikness, t am, and the Coeffiient of Variation, COV. A template for omputing the COV is provided in Table 4.3 and is used in Table E4.-. Table E4.- Loation Thikness Reading trd, i, i = ton trd, i tam ( trd, i tam) t N N = t =.0667 S = ( t t ) =.9333 am rd, i N i = 0.5 S COV = t = N am b) STEP The COV equals 8.0%, whih is less than 0%; therefore, the average thikness to be used in the alulation is the average thikness of the thikness distribution, or tam =.0667 mm LOSS = t t = = mm nom am ) STEP 3 Calulate the minimum required thikness (see Annex A). t min t L min ( SE P) ( SE P) i= PR 3.85( ) = = = mm (.0) 0.6(3.85) PR 3.85( ) = = = 4.9 mm (96.96)(.0) + 0.4(3.85) L tmin = max tmin, t min = max[0.670, 4.9] = mm d) STEP 4 Determine if the omponent is aeptable for ontinued operation. Perform a Level assessment using Table 4.4. ( t FCA= mm) ( t = mm) False am min Alternatively, the maximum allowable working pressure MAWP based on the average thikness ( t am ompared to the design pressure with the design pressure as the riterion. rd, i am ) an be 4-3

43 t = t FCA= mm am ( 96.96)( )( ) ( )( ) SEt MAWP = = = 3.83 MPa R+ FCA+ LOSS+ 0.6 t ( ) MPa 3.85 MPa False The Level assessment riteria are not satisfied. Perform a Level assessment for internal pressure using Table 4.4. min ( )( ) ( t FCA = mm) ( RSF t = = 9.58 mm) True am a Alternatively, the maximum allowable working pressure ( MAWP ) based on the average thikness ( t am be ompared to the design pressure with the design pressure as the riterion. ( tam FCA) t = = =.85 mm RSF 0.9 a ( 96.96)( )(.85) ( )( ) SEt MAWP = = = 4.5 MPa R+ FCA+ LOSS t ( ) MPa 3.85 MPa True Chek the minimum measured thikness riterion. ( tmm FCA= 8 mm) max[0.5tmin = 5.065, tlim ] tlim = max[0.tnom = ( 0.)( 6) = 3.00,.500] = 3.00 mm 8 mm (max[5.065,3.00] = mm) True The minimum measured thikness riterion is satisfied. The Level assessment riteria for internal pressure are satisfied. ) an 4-4

44 Perform Level assessment for full vauum ondition. For this example, the unsupported length of the vessel is given as mm. The thikness used for the alulation is mm omputed in STEP 4, above. The alulations below follow the steps shown in Annex A.4.4. E S y y = 7(0) 3 = 57 MPa R = R+ t = = 58 mm o D = R = 56 mm M o x o L 54 = = = Rt o ( 58)( ) Do 56 = = t Do h =. x 3 < x < he MPa C M for M F.058 ( )( ) = = ( )( )( )( ) 3.6ChEt y = = = MPa D 56 o 0.4 F he Fhe Fi = 0.7Sy for 0.55 < =.0533 <.439 S y Sy = ( 0.7)( 57) =.05 MPa 57 F i FS = for 0.55S y< Fi < S y S y.05 = = Fi.05 Fha = = = MPa FS.8774 t Pa = Fha = ( )( ) =.33 MPa Do 56.33MPa > 0.0 MPa 0.4 t The assessment riterion for full vauum ondition is satisfied. 4-5

45 4. Example Problem Internal orrosion at a longitudinal weld seam in a pressure vessel has been found during an inspetion. Details regarding the pressure vessel and inspetion data are given below. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division, Edition 998 with the 999 Addenda. Evaluate if the vessel shell is fit-for-servie. Vessel Data Material = SA 56Grade70 Year 999 Design Conditions = 300 F Inside Diameter = 48 in Nominal Thikness = 0.75 in Uniform metal loss = 0.0 in Future Corrosion Allowane = 0. in Weld Joint Effiieny = 0.85 Inspetion Data The grid used for the inspetion and the thikness readings are shown below. The grid spaing set by the Inspetor in the irumferential and longitudinal diretions is.5 in based on the orrosion profile. Vessel Shell M M M3 M4 M5 M6 M7 C C C3 C4 C5 C6 C7 C8 Inspetion Grid Weld Seam Figure E4.- Inspetion Grid 4-6

46 Table E4.- Inspetion Data (in) Longitudinal Cirumferential Inspetion Planes Inspetion Cirumferential Planes C C C3 C4 C5 C6 C7 C8 CTP M M M M M M M Longitudinal CTP Perform a Level assessment for internal pressure per paragraph 4.4. a) STEP Calulate the minimum required thikness. t C min t L min PR 300( ) = = = in (0.850) 0.600(300) ( SE P) PR 300( ) = = = 0. in (0000)(0.850) (300) ( SE P) [ ] L tmin = max tmin, t min = max 0.430, 0. = in b) STEP Thikness profiles are provided, the data for thikness readings is in the above table. tmm = in ) STEP 3 Determine wall thikness to be used in the assessment. t = t FCA= = in rd d) STEP 4 Compute the remaining thikness ratio, R t R t = = e) STEP 5 Compute the length for thikness averaging from Table 4.5 with t 0.4 Q = 0.46 is read from the table or by the equation: 0.40 Q =.3 = D= 48 + ( LOSS+ FCA) = 48 + ( ) = 48.0 in 0.5 ( )( ) L= Q Dt = =.564 in R = and RSF = 0.9, f) STEP 6 Establish the Critial Thikness Profiles (CTP s) from the thikness profile data (see paragraph ). Determine the average measured thikness t based on the longitudinal CTP and the average measured thikness s am t am based on the irumferential CTP. a 4-7

47 Longitudinal CTP Figure E4.- Longitudinal Critial Thikness Profile.8 t = ( ) = 0.5 in t = ( ) = in.500 The area method is used to determine the average thikness A = (.8) = in A = (.8) = 0.58 in A + A =.093 in t s am Cirumferential CTP.093 = = 0.46 in in 0.55 in 0.48 in 0.49 in t 3 t in.5 in.5 in.5 in.5 in L=.564 in t t 3 4 Figure E4.-3 Cirumferential Critial Thikness Profile.8 = ( ) = 0.5 in = ( ) = in

48 A3 = (.8) = in A4 = (.8) = 0.58 in A + A =.093 in t 3 4 am.093 = = 0.46 in.564 g) STEP 7 Determine the aeptability for ontinued operation using Level riteria in Table 4.4. The averaged measured thikness aeptane is used in this example. Use averaged measured thikness. s tam FCA= = 0.36 in tmin = in fromstep s tam FCA> tmin False tam FCA= = 0.36 in L tmin = 0. in fromstep t FCA> t True am L min The Level assessment riteria are not satisfied due to the average measured thikness in the longitudinal CTP. Chek the minimum thikness riteria in Table 4.4 ( )( ) tlim = max , 0.0 = 0.50 in t FCA= = 0.60 in mm ( )( ) max , 0.50 = 0.5 in tmm FCA max[0.5 tmin, tlim ] 0.60 in > 0.5 in True The minimum thikness riteria are satisfied. The Level assessment riteria are not satisfied. Perform a Level Assessment using Table 4.4. min ( )( ) s ( t FCA = 0.36 in) ( RSF t = = in) False am The Level Assessment riteria are not satisfied. a 4-9

49 4.3 Example Problem 3 A loalized region of internal orrosion on a : elliptial head has been found during an inspetion. The orroded region is within the spherial portion of the elliptial head within 0.8D entered on the head enterline. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division, Edition 989. Determine if the vessel head is suitable for ontinued operation. Vessel Data Material = SA 56Grade70 Year 989 Design Conditions =.850 C Head Inside Diameter = 03 mm Nominal Thikness = 9 mm Uniform Metal Loss = 0 mm Future Corrosion Allowane = 3 mm Weld Joint Effiieny =.0 Seamless Head Inspetion Data The grid used for the inspetion and the thikness readings are shown below. The grid spaing is 00 mm. Table E4.3- Inspetion Data (mm) Meridional Cirumferential Inspetion Planes Inspetion Cirumferential Planes C C C3 C4 C5 C6 C7 C8 CTP M M M M M M M M Meridional CTP Perform a Level assessment per paragraph 4.4. a) STEP Calulate the minimum required thikness using an equivalent radius based on the parameter K for the spherial portion of an elliptial head and the spherial shell design equation, (see Annex A). R ell 3 = ell ell ell min = ( ) ( ) 3 ( ) ( ) ( ) (.850)( 038)( ) 4.09 ( )( ) ( ) K R R R K t = = PDK = = = SE 0.P b) STEP Thikness profiles are provided, the data for thikness readings is in the above table. Determine the minimum measured thikness, t mm. mm tmm = 4 mm 4-0

50 ) STEP 3 Determine the wall thikness to be used in the assessment using Equation. 4. or Equation. 4.3 t = tnom LOSS FCA t = = 6.0 mm d) STEP 4 Compute the remaining thikness ratio, R t R t tmm FCA = = = t 6.0 e) STEP 5 Determine the length for thikness averaging, L. R = and RSF = 0.9, and Q.0, or by equation: From Table 4.5 with t Q = = R = K D= = mm ( )( ) ( ) D= R = = mm a 0.5 ( ) L= Q Dt = = 35.95mm f) STEP 6 Thikness profiles were taken; therefore, determine the longitudinal and irumferential CTP The thikness readings for the ritial inspetion planes are shown in the above table. Meridional CTP Sine in this example the meridional CTP is idential to irumferential CTP, only the assessment of irumferential CTP is performed below. The assessment results of irumferential CTP an be applied for merindional CTP. 4-

51 Cirumferential CTP Cirumferential Distane (mm) Table E4.3- Determine Cirumferential CTP Thikness Reading (mm) Thikness FCA (mm) Figure E4.3- Critial Thikness Profile The average thikness an be determined using the area method. 3 4 ( ) A = (7.974) = 3.3 mm 00 A = ( 3 )( 00 ) + 3(7.974) = 03.9 mm A = A = = mm ( )( ) A5 = ()( 00 ) + (7.974) = mm 4( 7.974) A6 = (7.974) = 6.46 mm 00 A = A + A + A + A + A + A TOT = = s ATOT tam = tam = = = 5.93 mm L mm 4-

52 g) STEP 7 Determine the aeptability for ontinued operation using Level riteria in Table 4.4. t FCA= =.93 mm am.93 mm ( t = mm) False The Level assessment riteria are not satisfied. Perform a Level assessment using Table 4.4. min min ( ) ( t FCA =.93 mm) ( RSF t = =.68 mm) False am Chek the minimum measured thikness riterion. a ( [ min ] ) t FCA= 4 3 = mm max 0.5 t,3 mm = mm True mm The minimum measured thikness riterion is satisfied The Level assessment riteria are not satisfied. 4-3

53 4.4 Example Problem 4 A region of internal orrosion on a inh Class 300 long weld nek nozzle has been found during the inspetion of a pressure vessel. The orroded region inludes the nozzle bore and a portion of the vessel ylindrial shell (see inspetion data). The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division, Edition 999. Determine if the vessel nozzle is suitable for ontinued operation. Vessel Data Shell Material = SA 56Grade70 Year 999 Design Conditions = F Shell Inside Diameter = 60 in Shell Thikness = 0.60 in Shell Weld Joint Effiieny =.0 Shell FCA = 0.5 in Nozzle Inside Diameter =.0 in Nozzle Thikness =.375 in Nozzle Material = SA 05 Year 999 Nozzle Weld Joint Effiieny =.0 Nozzle FCA = 0.5 in Reinforing Pad Material = SA 56Grade70 Year

54 Inspetion Data A sketh of the nozzle and metal loss are shown below. C L.375" Reinforement Zone 0.375" L no Metal Loss Reinforing Pad 8" OD x 0.50" Thik 0.60" L v Figure E4.4- Nozzle Metal Loss From the inspetion data: The average shell thikness in the nozzle reinforement zone is 0.50 in. The average nozzle thikness in the nozzle reinforement zone is 0.90 in. The orrosion is uniform for eah inspetion plane. The thikness for the shell and nozzle to be used in the assessment were determined by averaging thiknesses within the nozzle reinforement zone (see paragraph and Figure 4.9). Perform a Level assessment beause the orrosion is at a major strutural disontinuity From the inspetion data: t t nozzle am shell am = 0.90 in = 0.50 in Required thikness of the shell: t r PR (85)( ) = = = 0.8 in SE 0.6 P (0000)(.0) (0.6)(85) 4-5

55 Required thikness of the nozzle: t rn PR (85)( ) = = = in SE 0.6 P (0000)() (0.6)(85) Chek the nozzle reinforement (see Annex A): Required Area: d r =.0 + ( ) = 3. in F = f = B= 0.0 ( )( )( ) A= = in Available area: fr = E =.0 fr3 = fr 4 = s = = 0.5 in n = = 0.60 in wn = in wp = in Dp = 8 in te = 0.50 in h = 0.0 A = 0.0, w = 0.0 and A = h 43 { d( E ( t s) Ftr) B} { ( t+ tn s n)( E ( t s) Ftr) B} { 3.(( ) (0.8)) 0} =.39 in { } { 5( tn n trn) fr( t s) } { ( tn n trn)(.5( tn n) + te) fr} { 5( )()( ) } =.338 in { ( )(.5( ) + 0.5)} = in A = max A = max =.39 in (( )(( ) (0.8)) 0 = 0.60 in A = min A = min =.338in ( )() ( )() A = w f = = 0.4 in 4 n r A = w f = = 0.4 in 4 p r4 5 p n n e r4 ( )( ) A = [ D d ( t )] t f = [8 3. ( )] 0.5 =.65 in 4-6

56 Reinforement hek: A + A + A + A + A = = in ( ) in A = in True Analysis Results: The area reinforement alulation per the original onstrution ode is satisfied using the average thiknesses for the shell and nozzle in the nozzle reinforement zone. The Level assessment riterion is satisfied. 4-7

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58 PART 5 ASSESSMENT OF LOCAL THIN AREAS EXAMPLE PROBLEMS 5. Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem A region of loalized orrosion has been found on the inside surfae of a pressure vessel during a sheduled turnaround. The vessel and inspetion data are provided below. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division, 989 Edition. Determine if the vessel is aeptable for ontinued operation using a Level Assessment. Vessel Data Material = SA 56Grade70 Year 989 Design Conditions = F Inside Diameter = 96 in Fabriated Thikness =.5 in Uniform Metal Loss (Internal) = 0.0in FCA = 0.5in Longitudinal Weld Joint Effiieny =.0 Cirumferential Weld Joint Effiieny =.0 Supplemental Loads = 0 negligible 5-

59 Inspetion Data The thikness data and the grid used for the inspetion are shown below. The distane from the region of loal metal loss to the nearest strutural disontinuity is 60 in. Another region of loal metal loss with a smaller amount of metal loss is loated 6 in from the region shown below. Pressure Vessel Shell M5 C C C3 C4 C5 C6 C7 C8 C9 Inspetion Grid M4 M3 M M Figure E5.- Weld Seam Table E5.- Inspetion Data (in) Longitudinal Inspetion Cirumferential Inspetion Planes Cirumferential Planes C C C3 C4 C5 C6 C7 C8 C9 CTP M M M M M Longitudinal CTP Notes:. Spaing of thikness readings in longitudinal diretion is ½ in.. Spaing of thikness readings in irumferential diretion is.0 in. 3. The loalized orrosion is loated away from all weld seams. 5-

60 Perform a Level Assessment per paragraph a) STEP Determine the CTP (Critial Thikness Profiles) (see paragraph ) the thikness readings for the ritial inspetion planes are indiated in Figure E5.- and Table E5.- above. b) STEP Determine the wall thikness to be used in the assessment using equation (5.3). tnom =.5 in LOSS = 0. in FCA = 0.5 in trd = tnom LOSS =.5 0. =.5 in t = t LOSS FCA= =.05 in nom ) STEP 3 Determine the minimum measured thikness, t mm, and the dimension, s, for the longitudinal CTP. The LTA being evaluated satisfies the spaing riteria in Part 4, paragraph f.3; therefore, the dimensions of the LTA do not need to be adjusted (see Figure E5.-). Figure E5.- From inspetion data table, the minimum measured thikness is 5-3

61 tmm = 0.47 in From longitudinal CTP, the longitudinal extent of the metal loss is the length between the two end points where the metal loss profile rosses t =.5 in. Linear interpolation is used to determine the length. s = 8 0.5= 4 in rd d) STEP 4 Determine the remaining thikness ratio and the longitudinal flaw length parameter. λ using equations (5.5) and (5.6). tmm FCA Rt = = = t.05 D= 96 + ( LOSS + FCA) = 96 + ( ) = in ( ) ( ).85s λ = = = Dt e) STEP 5 Chek the limiting flaw size riteria using equations (5.7), (5.8), and (5.9). ( Rt = ) 0.0 ( mm = = ) 0.0 ( ) ( ) True t FCA in in True ( ) L = 60 in.8 Dt = = in True msd f) STEP 6 Chek the riterion for a groove-like flaw. This step is not appliable beause the region of loalized metal loss is ategorized as an LTA. g) STEP 7 Determine the MAWP for the omponent using equations (A.0), (A.6), and (A.). Note that E = 0. sine the LTA is remote from weld seams (see paragraph A..5.b) of Annex A. D R= = = 48.5 in C SEt MAWP = = R+ + = psi L MAWP = = R t t = psi ( 7500)(.0)(.05 ) t ( 48.5) 0.6(.05) SE ( t tsl) ( )( )( )( ) 0.4( sl) ( 48.5) 0.4( ) [ ] psi MAWP = min , = h) STEP 8 Evaluate the longitudinal extent of the flaw. From Figure 5.6 with and equation (5.): M t =.0595 λ = R t = , the longitudinal extent of the flaw is aeptable. Using Table 5. R t RSF = = = a = 0.9 ( Rt ) ( ) M t.0595 ( RSF ) 5-4

62 The longitudinal extent of the flaw is aeptable. i) STEP 9 Evaluate irumferential extent of the flaw. ) STEP 9. From the irumferential CTP, determine λc using equation (5.) = 4 = 4.0 in.85( 4.0) ( 96.45)(.05).85 λc = = = 0.57 Dt ) STEP 9. Chek the following onditions (equations (5.3) to (5.7)). ( λ = ) D = = t = = =.0 ( RSF ) ( EL ) ( E ) C 3) STEP 9.3 Calulate tensile strength fator using equation (5.8), E 4 3E C L 4 3 TSF = + = + =.06 RSF E L From Figure 5.8 with Table 5.4, R t _min = 0. λ C = 0.57 R t = ( Rt = ) > ( Rt_min = 0.) The irumferential extent of the flaw is aeptable. The Level Assessment Criteria are satisfied. ( MAWP = psi) > ( PDesign = 300 psi) The equipment is aeptable for ontinued operation. True True True True True, the irumferential extent of the flaw is aeptable. From 5-5

63 5. Example Problem A pressure vessel shell has two groove-like flaws with the following dimensions. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division, 989 Edition. Determine if the vessel is aeptable for ontinued operation. Vessel Data Material = SA 56Grade70 Year 989 Design Conditions = F Inside Diameter = 90 in Measured Uniform Thikness =.5 in Uniform Metal Loss = 0.0 in FCA = 0.5 in Longitudinal Weld Joint Effiieny =.0 Cirumferential Weld Joint Effiieny =.0 Supplemental Loads = 0 negligible Inspetion Data Groove & Orientation = longitudinal Groove & Width =.5 in Groove Depth = 0.45 in Groove Depth = 0.65 in Groove & Length = 8.0 in Groove Radius = 0.60 in Groove Radius = 0.0 in The groove-like flaws are loated 0 in apart from eah other. Eah of the groove-like flaws is loated a minimum distane of 36 in away from the nearest strutural disontinuity or weld. Based on proess onditions and a visual examination, it was determined that both of the grooves were aused by fluid erosion; therefore, both of the groove-like flaws are haraterized as a groove per paragraph 5...b.). Perform a Level Assessment per paragraph Groove a) STEP Determine the Critial Thikness Profiles(s) (see paragraph ). b) STEP Determine the wall thikness to be used in the assessment using equation (5.4). trd = tnom LOSS = =.5 in t = t FCA= =.0 in rd ) STEP 3 Determine the minimum measured thikness, t mm, and the dimension, s, for the longitudinal CTP. The groove-like flaw being evaluated satisfies the spaing riteria in Part 4, paragraph f.3; therefore, the dimensions of the groove-like flaw do not need to be adjusted (see Figure E5.-). 5-6

64 Figure E5.- t mm = = in s= g = 8 in β = 0.0 l o Figure E

65 d) STEP 4 Determine the remaining thikness ratio and the longitudinal flaw length parameter. λ using equations (5.5) and (5.6). tmm FCA Rt = = = 0.55 t.0 D= 90 + FCA= = 90.5 in ( ) ( ).85s λ = = =.08 Dt e) STEP 5 Chek the limiting flaw size riteria for a Level Assessment using equations (5.7), (5.8), and (5.9). ( Rt = 0.55) 0.0 ( mm = = ) ( ) ( ) True t FCA in 0.0 in True ( ) L = 36 in.8 Dt = = 7. in True msd f) STEP 6 Chek the riterion for a groove-like flaw using equation (5.0). ( ) ( ) ( )( ) g r = 0.6 in Rt t = 0.55 = 0.45 in True The groove satisfies the equation. Proeed to STEP 7. g) STEP 7 Determine the MAWP for the omponent using equations (A.0), (A.6), and (A.). Note that E = 0. sine the LTA is remote from weld seams (see paragraph A..5.b of Annex A). D 90.5 R= = = 45.5 in C SEt MAWP = = = R+ + MAWP L ( 7500)(.0)(.0 ) t ( 45.5) 0.6(.0) SE ( t tsl) ( )( )( )( ) 0.4( sl) ( 45.5) 0.4(.0 0.0) [ ] psi = = = psi R t t MAWP = min 38.78, = h) STEP 8 Evaluate the longitudinal extent of the flaw. From Figure 5.6 with λ =.08 R t = 0.55 psi, the longitudinal extent of the flaw is unaeptable at the urrent MAWP determined in STEP 7. Using Table 5. and equations (5.) and (.) to determine the redued maximum allowable working pressure MAWP : r 5-8

66 M t API 579-/ASME FFS- 009 Fitness-For-Servie Example Problem Manual =.87 R 0.55 t RSF = = = < a = 0.9 ( Rt ) ( 0.55) M t.87 RSF MAWPr = MAWP = ( 38.78) = psi RSF 0.9 a ( MAWPr = psi) > ( PDesign = 300 psi) The longitudinal extent of the flaw is aeptable. i) STEP 9 Evaluate irumferential extent of the flaw. ( RSF ) ) STEP 9. From the irumferential CTP, determine λc using equation (5.). = g =.5 in w.85(.5) ( 90.5)(.0) λ = = 0.09 C ) STEP 9. Chek the following onditions (equations (5.3) to (5.7)). ( λ = ) D 90.5 = = t 0.7 = ( RSF ) ( EL ) ( E ) 0.7 = =.0 C 3) STEP 9.3 Calulate tensile strength fator using equation (5.8), True True True True True From Figure 5.7 with E 4 3E C L 4 3 TSF = + = + =.53 RSF E L R t _min = 0. λ C = 0.09 R t = 0.55 ( Rt = 0.55) > ( Rt_min = 0.), the irumferential extent of the flaw is aeptable. From Table 5.4, The irumferential extent of the flaw is aeptable. The Level Assessment Criteria are Satisfied with a design pressure). MAWP r of psi (greater than 300 psi 5-9

67 Perform a Level Assessment per paragraph Groove a) STEP Determine the Critial Thikness Profiles(s) (see paragraph ). b) STEP Determine the wall thikness to be used in the assessment using equation (5.4). t = t LOSS = =.5 in rd nom t = t FCA= =.0 in rd ) STEP 3 Determine the minimum measured thikness, t mm, and the dimension, s, for the longitudinal CTP. The groove-like flaw being evaluated satisfies the spaing riteria in Part 4, paragraph f.3; therefore, the dimensions of the groove-like flaw do not need to be adjusted (see Figure E5.-). t mm = = in s= g = 8 in β = 0.0 l o Figure E5.-3 d) STEP 4 Determine the remaining thikness ratio and the longitudinal flaw length parameter. λ using equations (5.5) and (5.6). tmm FCA Rt = = = 0.35 t.0 D= 90 + FCA= = 90.5 in ( ) ( ).85s λ = = =.08 Dt e) STEP 5 Chek the limiting flaw size riteria for a Level Assessment using equations (5.7), (5.8), and (5.9). ( Rt = 0.35) 0.0 ( mm = = ) ( ) ( ) True t FCA in. 0.0 in True ( ) L = 36 in.8 Dt = = 7. in True msd f) STEP 6 Chek the riterion for a groove-like flaw. ( ) ( ) ( )( ) g r = 0. in < Rt t = 0.35 = 0.65 in False The groove is not aeptable per Level. Groove is not aeptable per Part 5 Level Assessment Criteria. 5-0

68 Perform a Level Assessment per paragraph Groove The Level sreening riteria for groove-like flaws are the same as the riteria in Level proedure; therefore, this groove does not satisfy the Level Assessment riteria. Groove is not aeptable per Part 5 Level Assessment Criteria. The vessel is unaeptable for ontinued operation. Alternatively, Groove an be evaluated as a rak-like flaw using the proedures in Part 9. 5-

69 5.3 Example Problem 3 Inspetion of a proess vessel indiates a region of loal orrosion on the inside surfae in the lower shell setion. In addition to internal pressure, the vessel is also subjeted to axial fores and bending moments. The vessel data is shown below. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division, 989 Edition. Evaluate the region of loalized metal loss for pressure plus supplemental loads and determine aeptability for ontinued operation without repairs. Vessel Data Material = SA 56Grade70 Year989 Design Conditions = 0 F Nominal Thikness = 0.50 in Inside Diameter = 4 in Uniform Metal Loss = 0.0 in FCA = 0.06 in Longitudinal Weld Joint Effiieny =.0 Cirumferential Weld Joint Effiieny =.0 Weight Case Loads (see Figure 5. for definition of applied loads) Applied Axial Fore = lbs 6 M x Applied Bending Moment =.79(0) in lb M y Applied Bending Moment = 0.0 in lb Applied Shear Fore = lbs Applied Torsional Moment = 5.63(0) in lb Thermal Case Loads (see Figure 5. for definition of applied loads) Applied Axial Fore = lbs 6 M x Applied Bending Moment = 3.8(0) in lb M y Applied Bending Moment = 0.0 in lb Applied Shear Fore = lbs 5.59(0) in lb Applied Torsional Moment = Note: The weight ase and thermal ase loads are typially obtained from a stress analysis. The applied fores and moments were omputed at the loation of maximum metal loss. 5-

70 Inspetion Data The thikness data and the grid used for the inspetion are shown below. This is the only region of loalized metal loss found on the vessel during the inspetion. The distane from the region of loal metal loss to the nearest strutural disontinuity is 8 in. Figure E5.3- Table E5.3- Inspetion Data (in) Longitudinal Cirumferential Inspetion Planes Inspetion Cirumferential Planes C C C3 C4 C5 C6 C7 CTP M M M M M M M Longitudinal CTP Notes:. Spaing of thikness readings in longitudinal diretion is.0 in.. Spaing of thikness readings in irumferential diretion is 3.0 in. 5-3

71 Perform a Level Assessment per paragraph beause of the presene of external loads. a) STEP Determine the Critial Thikness Profiles (see paragraph ) (same as STEP for the Level Assessment) - the thikness readings for the ritial inspetion planes are indiated in Figure E5.3- and Table E5.3- above. b) STEP Determine the wall thikness to be used in the assessment using equation (5.3). t = t LOSS = = 0.5 in rd nom t = t LOSS FCA= = 0.44 in nom ) STEP 3 Determine the minimum measured thikness, t mm, and the dimension, s, for the longitudinal CTP. There is only one LTA in the vessel; therefore, the spaing riteria in Part 4, paragraph f.3 do not need to be heked. tmm = 0.6 in s= 6 = 6 in Figure E5.3- Longitudinal CTP d) STEP 4 Determine the remaining thikness ratio and the longitudinal flaw length parameter. λ using equations (5.5) and (5.6). tmm FCA Rt = = = t 0.44 D= 4 + FCA= = 4. in ( ) ( ).85s λ = = =.79 Dt e) STEP 5 Chek the limiting flaw size riteria for a Level Assessment using equations (5.7), (5.8), and (5.9). ( Rt = ) 0.0 ( mm = = 0.0 ) 0.0 ( ) ( ) True t FCA in in True ( ) L = 8 in.8 Dt = = in True msd f) STEP 6 Chek the riterion for a groove-like flaw. This step is not appliable beause the region of loalized metal loss is ategorized as an LTA. 5-4

72 g) STEP 7 Determine the MAWP for the omponent using equations (A.0), (A.6), and (A.). R t sl m D D FCA + tnom = = = =.8 in ( )( ) ( ) 6.79( 0) ( )( ) ( ) F M 500 = + = + = 0.07 in SEπR SEπR π π m m D 4. R= = =.06 in C SEt MAWP = = = R+ + MAWP L ( 7500)(.0)( 0.44 ) psi 0.6t (.06) 0.6( 0.44) SE ( t tsl) ( )( )( )( ) 0.4( sl) (.06) 0.4( ) [ ] psi = = = psi R t t MAWP = min , = h) STEP 8 Determine the Remaining Strength Fator for the longitudinal CTP. The remaining strength fator is based on the Level Assessment proedure. This will provide onservative estimates of the RSF. In general, the RSF should be omputed using the Level assessment proedure. Using Table 5. and equation (5.): M t =.5 R t RSF = = = < a = 0.9 ( Rt ) ( ) M t.5 i) STEP 9 Evaluate the longitudinal extent of the flaw. Sine RSF < RSF, the redued MAWP an be alulated as a RSF MAWPr = MAWP = ( ) = psi RSF 0.9 a ( MAWPr = 84. psi ) > ( PDesign = 0 psi) ( RSF ) The longitudinal extent of the flaw is aeptable. Therefore, a remaining strength fator based on Level Assessment is not neessary. j) STEP 0 Evaluate the irumferential extent of the flaw Beause of the presene of external loads, the extent of the flaw in the irumferential diretion must be evaluated using the proedure in paragraph ) STEP 0. Determine the Critial Thikness Profile (CTP) in the irumferential diretion. This is done in STEP. ) STEP 0. For the irumferential inspetion plane being evaluated, approximate the irumferential extent of metal loss on the plane under evaluation as a retangular shape (Figure 5.). Calulate D f using equation (5.3) and θ using equation (5.5). 5-5

73 ( nom ) ( )( ) ( ) ( ) D = D+ t FCA = = 43 in o D = D t FCA = = 4.6 in f o mm = 6 3= 8.0 in 8.0 θ = = = 0.45 radians D 4.6 f 3) STEP 0.3 Determine the remaining strength fator, RSF, the redued maximum allowable working pressure, and supplemental loads on the irumferential plane. RSF = MAWP = 84 psi F = lbs r Weight Case Supplemental Loads ( ) Mx =.79 0 in lbs M y = 0.0 in lbs V = lbs ( ) M =.63 0 in lbs T ( ) 6 5 Thermal Case Supplemental Loads F = lbs Mx = in lbs M y = 0.0 in lbs V = lbs 6 ( ) M =.59 0 in lbs T 5 4) STEP 0.4 Compute the omponents of the resultant longitudinal bending moment (i.e., exluding torsion) in the plane of the defet relative to the region of metal loss. In this ase, the moments stated in the problem were aligned with the flaw. In general, the moments will not be aligned with the flaw, and the moments results obtained from a stress analysis will need to be resolved to the axis of the flaw as shown in Figure 5.. Weight Case ( ) M =.79 0 in lbs x M = 0.0 in lbs y Thermal Case ( ) M = in lbs x M = 0.0 in lbs y 6 6 5) STEP 0.5 Compute the irumferential stress resulting from pressure for both weight and weight plus thermal load ases at points A and B in the ross setion (Figure 5.) using equation (5.6). ( ) MAWP r D 4. σ m= = = psi RSF Do D

74 6) STEP 0.6 Compute setion properties (use equations in Table 5.3) and the longitudinal membrane stress and shear stress for the weight and weight plus thermal load ases at points A and B in the ross setion. i) STEP 0.6. The irumferential plane of the metal loss an be approximated by a retangular area. Compute setion properties of a ylinder without an LTA. π π Aa = D = ( 4.) = in 4 4 π π Am = ( Do D ) = ( 43.0) ( 4.) = in 4 4 π π I X = ( Do D ) = ( 43.0) ( 4.) = in I = I = in Y X ii) STEP 0.6. Compute setion properties for ylinder with LTA on inside surfae. θ 0.45 Af = ( Df D ) = ( 4.6) ( 4.) = in 4 4 A = A + A = = in w a f 3 3 [ θ ]( Df D ) [ ] ( ) 3 ( ) ( ) sin y = = sin 0.45 =.69 in A A x A = 0.0 in m f Do 43.0 ya = y + =.69+ = 3.9 in Do 43.0 xb = sin[ θ ] = sin[ 0.45] = in Do 43.0 yb = y + os[ θ ] =.69+ os[ 0.45] =.83 in 3 3 sin[ θ ]( D ) [ ] ( 4.6 ) 3 ( 4. ) 3 f D b= = sin 0.45 = in A + A D f R = a 4. 6 = =.3 in ( Df D) f d = = = 0.4 in A tf ( o + Df ) 8.0( ) D = = = 9.6 in

75 I LX y with, LX [ θ ] Rsin d = + 3θ R d R (.3) sin[ 0.45] 0.4 = in 3( 0.45) =.3 = = 3 R d 3 ( ) ( ) [ θ ] 3 3d d d sin + sin 3 θ + [ θ] os[ θ] + R R 4R θ d sin [ θ ] d d + 3R θ ( d R) R 6R 3 3( 0.4) ( 0.4) ( 0.4) + 3 (.3 ) (.3) 4(.3) ( 0.45 ). + sin[ 0.45] os[ 0.45] sin [ 0.45] ( 0.45) ( 0.4) sin [ 0.45] 0.4 ( 0.4) ( ) ( 0.45) 6(.3).3 =.35 in X 4 X m LX f LX ( ) I = I + A y I A y + y ( )( ) ( )( ) 4 = = in I LY 3 3 3d d d = R d + sin os 3 R R 4R = in 4 ( θ [ θ] [ θ] ) 3 3( 0.4) ( 0.4) ( 0.4) + 3 (.3 ) (.3) 4(.3) ( 0.45) sin[ 0.45] os[ 0.45] 3 = (.3) ( 0.4) I = I I = = 3.7 in Y Y LY 4 5-8

76 iii) with, ( D+ D ) ( D+ D ) 0.5π o o At = 8 0.5π = = 3.38 in 8 ( + ) [ + ] STEP Compute the longitudinal membrane stress and shear stress for the weight and weight plus thermal load ases at points A and B in the ross setion using equations (5.7) to (5.3). For the Weight Case, points A and B (.85)( 8.0) ( 4.)( 0.44).85 λc = = = Dt M M C t C s ( λc) ( λc) ( λ ) ( λ ) ( ) ( ) ( ) ( ) = = =.657 C 4 4 d 0.4 C Mt t = = =.4573 d 0.4 t 0.44 C 5-9

77 σ σ A lm B lm A F w T ( ) C MAWP + + r M A A A A s m f m f = E y x A A [ F y + T ( y + b)( MAWPr) A + M w x] + M y I I X Y ( 84.09) = ( 0) 6 ( 0 ) + ( 0) 3.7 = psi ( 500)(.69) + ( )( 84.09)( ) A F w T ( ) C MAWP + + r M A A A A s m f m f = E y x B B [ F y + T ( y + b)( MAWPr) A + M w x] + M y I I X Y ( 84.09) ( 500)(.69) + ( )( 84.09)( ) = ( 0) ( ) + ( 0) in = psi MT V τ = + A A ( At + Atf )( tmm FCA) 5.63 ( 0) ( + )( ) ( ) = = psi m f 5-0

78 σ σ For the Weight plus Thermal Case, points A and B A lm B lm Aw FT ( r ) C MAWP + + A s m Af Am A M f = E ya xa Fy T + ( y+ b)( MAWPr) Aw + Mx + M y I I X Y ( ) ( 84.09) ( )(.69) = + ( )( 84.09) ( ) 6 6 ( ( ) ( ) ) ( 0) + ( 0) 3.7 = psi Aw FT ( r ) C MAWP + + A s m Af Am A M f = E yb xb Fy T + ( y+ b)( MAWPr) Aw + Mx + M y I I X Y ( ) ( 84.09) ( )(.69) = + ( )( 84.09)( ) (.79 ( 0) 3.8 ( 0) + + ) ( 8.866) + ( 0) 3.7 = psi MT V τ = + A A ( At + Atf )( tmm FCA) ( 0) +.59 ( 0) ( + )( ) ( ) = = psi m f 5-

79 7) STEP 0.7 Compute the equivalent membrane stress for both the weight and weight plus thermal load ases at points A and B in the ross setion using equations (5.33) and (5.34). Weight Case ( ) ( )( ) ( ) A A A e = m m lm + lm + 3 σ σ σ σ σ τ ( ) ( )( ) ( ) 3( ) = psi = + + ( ) ( )( ) ( ) B B B e = m m lm + lm + 3 σ σ σ σ σ τ ( ) ( )( 734.0) ( 734.0) 3( ) = psi = + + Weight plus Thermal Case ( ) ( )( ) ( ) A A A e = m m lm + lm + 3 σ σ σ σ σ τ ( ) ( )( ) ( ) 3( ) = + + = psi ( ) ( )( ) ( ) B B B e = m m lm + lm + 3 σ σ σ σ σ τ ( ) ( )( ) ( ) 3( ) = + + = psi 8) STEP 0.8 Evaluate the results using equation (5.35). Weight Case A B { max σe, σe = max [ , ] = psi} S a 7500 H f = (.0) = psi RSFa 0.9 Weight plus Thermal Case A B { max σe, σe = max [ 67.89, ] = 67.8 psi} S a 7500 H f = ( 3.0) = psi RSFa 0.9 The irumferential extent of the flaw is aeptable. Therefore, the equipment is aeptable for ontinued operation without repair 5-

80 5.4 Example Problem 4 Inspetion of a ylindrial pressure vessel indiates a region of loalized orrosion. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division, 99 Edition. Perform a Level Assessment to evaluate the aeptability for ontinued operation. Vessel Data Material = SA 56Grade70 Year 99 Design Conditions = F Inside Diameter = 60 in Wall Thikness =.0 in Uniform Metal Loss = 0.0 in FCA = 0.0 in Longitudinal Weld Joint Effiieny =.0 Cirumferential Weld Joint Effiieny =.0 Supplemental Loads = 0.0 negligible Inspetion Data The ritial thikness profile for the longitudinal plane is shown in the following table. The ritial thikness profile for the irumferential plane an be approximated as a retangular area of metal loss with a length of 0 in. This is the only region of loalized metal loss found on the vessel during the inspetion. The region of metal loss is loated 7 in away from the nearest strutural disontinuity. Inspetion Loation Table E5.4- Longitudinal Loation (in) Measured Thikness (in) Perform a Level Assessment per paragraph a) STEP Determine the Critial Thikness Profiles (see paragraph ) (see Table E5.4-). b) STEP Determine the wall thikness to be used in the assessment using equation (5.3). trd = tnom LOSS = =.0 in t = t LOSS FCA= =.0 in nom 5-3

81 ) STEP 3 Determine the minimum measured thikness, t mm, and the dimension, s, for the longitudinal CTP. There is only one LTA in the vessel; therefore, the spaing riteria in Part 4, paragraph f.3 do not need to be heked. t mm = 0.30 in s = 0.0 in based on t =.0 in rd d) STEP 4 Determine the remaining thikness ratio, R t, and the shell parameter, λ using equations (5.5) and (5.6). tmm FCA Rt = = = 0.3 t.0 D= 60 + FCA= = 60 in ( ) ( ).85s.85 0 λ = = = Dt 60.0 e) STEP 5 Chek the limiting flaw size riteria for a Level Assessment using equations (5.7), (5.8), and (5.9). ( Rt = 0.3) 0.0 ( mm = = 0.3 ) 0.0 ( ) ( ) True t FCA in in True ( ) L = 7 in.8 Dt = = in True msd f) STEP 6 Chek the riteria for a groove-like flaw. This step is not appliable beause the region of loalized metal loss is ategorized as an LTA. g) STEP 7 Determine the MAWP for the omponent using equations (A.0), (A.6), and (A.). D 60 R= = = 30 in C SEt MAWP = = R+ + = psi L MAWP = = R t t = 8.4 psi ( 7500)(.0)(.0 ) t ( 30) 0.6(.0) SE ( t tsl) ( )( )( )( ) 0.4( sl) ( 30) 0.4(.0 0.0) [ ] psi MAWP = min , 8.4 = h) STEP 8 Determine the Remaining Strength Fator for the longitudinal CTP ) STEP 8. Rank the thikness readings in asending order based on metal loss based on the CTP data, inspetion loation 6 would be the starting point for the assessment. 5-4

82 ) STEP 8. Set the initial evaluation starting point as the loation of maximum metal loss, this is the loation in the thikness profile where t mm is reorded inspetion loation 6 has the minimum thikness equals to 0.30 in. Subsequent starting points should be in aordane with the ranking in STEP 8. 3) STEP 8.3 At the urrent evaluation starting point, subdivide the thikness profile into a series of subsetions the thikness profile will be subdivided into 0 setions eah inhes in length. 4) STEP 8.4 For eah subsetion, ompute the Remaining Strength Fator using Equation (5.9) and the data tabulated in Table E5.4-. Table E5.4- Data For Starting Point At Loation 6 Of The Longitudinal CTP Subsetion I s s i () s e i () s i (3) λ i (4) A i (5) A o i (6) M t i (7) RSF i (8) Notes:. Starting loation of metal loss region under onsideration.. Ending loation of metal loss region under onsideration. 3. Length of metal loss for the region under onsideration. 4. Shell parameter evaluated using Equation (5.6) integration with s s i 5. Area of metal loss evaluated using a numerial proedure. 6. Original metal area evaluated using Equation (5.0). 7. Folias fator evaluated using Table 5. with λ = λ. 8. Remaining strength fator; evaluated using Equations (5.9). =. Figure E5.4- Thikness Profile 5-5

83 5) STEP 8.5 Determine the minimum value of the Remaining Strength Fators, RSF i, found in STEP 8.4 for all subsetions. The minimum value of the Remaining Strength Fator for the urrent evaluation is found to be at subsetion 5 when point 6 is used as the subdivision starting point. RSF min = ) STEP 8.6 Repeat STEPs 8.3 through 8.5 of this alulation for the next evaluation point that orresponds to the next thikness reading loation in the ranked thikness profile list; this step is not shown here. 7) STEP 8.7 After the alulation has been ompleted for all thikness reading loations (or evaluation points), determine the minimum value of the Remaining Strength Fator from all the alulated RSFs. It is found that the minimum RSF is assoiated with subsetion 5 using point 6 as the subdivision starting point with a value of: RSF = i) STEP 9 Evaluate the longitudinal extent of the flaw and use equation (.) for alulating MAWP r. Sine ( ) ( 0.9) RSF = < RSF a =, the redued MAWP an be alulated as RSF MAWPr = MAWP = ( ) = psi RSF 0.9 a ( MAWP = psi) < ( PDesign = 570 psi) Therefore the longitudinal extent of the flaw is unaeptable for the stated design onditions and a de-rate to psi is required if no repair is done. j) STEP 0 Evaluate irumferential extent of the flaw. In this example, the Level Assessment method is used beause supplemental loads are negligible. ) STEP 0. From the irumferential CTP, determine λc using equation (5.). = 0.0 in based on trd =.0 in λc = = = Dt ( ) ( 60)(.0) ) STEP 0. Chek the following onditions (equations (5.3) to (5.7)). ( λ = ) D 60 = = 60 0 t = ( RSF ) ( EL ) ( E ) 0.7 = =.0 C 3) STEP 0.3 Calulate tensile strength fator using equation (5.8), True True True True True TSF E 4 3E C L 4 3 = + = + =.3057 RSF E L

84 From Figure 5.8 with Table 5.4, R t _min = 0.4 λ C = R t = 0.3, the irumferential extent of the flaw is unaeptable. From Or from Table 5.4, alulate R t _min for TSF =. and TSF =.4, then find t _min TSF =.3057 through interpolation. R R C C C C C = C t_ min TSF= λ λ λ λ λ R for = ( ) = C C C C C = C ( ) ( ) ( ) ( ) t_ min TSF= λ λ λ λ λ R t _ min TSF =.304 = ( ) ( ) ( ) ( ) ( ) = = (.3057.) = ( Rt = 0.3) < ( Rt_min = 0.498) The irumferential extent of the flaw is unaeptable. Therefore, the Level Assessment Criteria are not satisfied. The equipment is unaeptable for ontinued operation under the design onditions, but may be operated at the redued MAWP of 486 psi per this assessment. Note that the irumferential extent an be re-assessed using the Level proedure. The results will be improved beause a ertain level of supplemental loads is inluded in the Level riteria whih makes Level proedure more onservative. 5-7

85 5.5 Example Problem 5 A region of loal metal loss has been found on the inside surfae of a ylindrial pressure vessel during an inspetion. The vessel and inspetion data are shown below. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division, 989 Edition. Determine if the vessel is aeptable for ontinued operation. Vessel Data Material = SA 56Grade60 Year 989 Design Conditions =.068 C Inside Diameter = 438 mm Fabriated Thikness = 3 mm Uniform Metal Loss =.5 mm Future Corrosion Allowane = 3. mm Longitudinal Weld Joint Effiieny =.0 Cirumferential Weld Joint Effiieny =.0 Supplemental Loads = 0.0 negligible Inspetion Data Based on the inspetion data, the ritial thikness profile in the longitudinal diretion has a length s= 9 mm and has a uniform measured thikness of 6 mm. The ritial thikness profile in the irumferential diretion has a length = 50 mm with the same uniform thikness. The region of loal metal loss is loated 50 mm away from the nearest strutural disontinuity. This is the only region of loal metal loss found in the vessel during the inspetion. Perform a Level Assessment per paragraph a) STEP Determine the CTP (Critial Thikness Profiles) (See Inspetion Data above). b) STEP Determine the wall thikness to be used in the assessment using equation (5.3). tnom = 3 mm LOSS =.5 mm FCA = 3. mm trd = tnom LOSS = 3.5 = 9.5 mm t = t LOSS FCA= = 6.3 mm nom ) STEP 3 Determine the minimum measured thikness, t mm, and the dimension, s, for the longitudinal CTP. There is only one LTA in the vessel; therefore, the spaing riteria in Part 4, paragraph f.3 do not need to be heked. tmm = 6 mm s= 9 mm 5-8

86 d) STEP 4 Determine the remaining thikness ratio and the longitudinal flaw length parameter, λ using equations (5.5) and (5.6). tmm FCA 6 3. Rt = = = t 6.3 D = ( LOSS + FCA) = ( ) = mm ( ) ( ).85s.85 9 λ = = = Dt e) STEP 5 Chek the limiting flaw size riteria for a Level Assessment using equations (5.7), (5.8), and (5.9). ( Rt = ) 0.0 ( mm = 6 3. =.8 ).5 ( ) ( ) True t FCA mm mm True ( ) L = 50 mm.8 Dt = = mm True msd f) STEP 6 Chek the riteria for a groove-like flaw. This step is not appliable beause the region of loalized metal loss is ategorized as an LTA. g) STEP 7 Determine the MAWP for the omponent (see A.3.4) using equations (A.0), (A.6), and (A.). D R= = = 4.7 mm C SEt MAWP = = R+ + = MPa L MAWP = = R t t = MPa ( 03.4)(.0)( 6.3 ) t ( 4.7) 0.6( 6.3) SE ( t tsl) ( )( )( )( ) 0.4( sl) ( 4.7) 0.4( ) [ ] MPa MAWP = min.97, =.97 h) STEP 8 Evaluate the longitudinal extent of the flaw. From Figure 5.6 with and equation (5.): Sine M t =.88 λ = R t = , the longitudinal extent of the flaw is aeptable. Using Table 5. R t RSF = = = < a = 0.9 ( Rt ) ( ) M t.88 a ( RSF ) RSF < RSF, the redued MAWP an be alulated using equation (.) RSF MAWPr = MAWP = (.97) =.0878 MPa RSF 0.9 a ( MAWPr =.0878 MPa) > ( PDesign =.068 MPa) 5-9

87 The longitudinal extent of the flaw is aeptable. i) STEP 9 Evaluate irumferential extent of the flaw. ) STEP 9. From the irumferential CTP, determine λc using equation (5.). = 50 mm based on t = 9.5 mm.85( 50) ( 449.4)( 6.3).85 λc = = =.657 Dt rd ) STEP 9. Chek the following onditions (equations (5.3) to (5.7)). ( λ = ) D = = t = ( RSF ) ( EL ) ( E ) 0.7 = =.0 C 3) STEP 9.3 Calulate tensile strength fator using equation (5.8), True True True True True E 4 3E C L 4 3 TSF = + = + =.669 RSF E L From Figure 5.8 with Table 5.4, R t _min = 0. λ C =.657 R t = ( Rt = ) > ( Rt_min = 0.) The irumferential extent of the flaw is aeptable. The Level Assessment Criteria are satisfied. The equipment is aeptable for ontinued operation., the irumferential extent of the flaw is aeptable. From 5-30

88 5.6 Example Problem 6 A region of orrosion in a NPS-4, Shedule 40 nozzle has been found during the inspetion of a pressure vessel. The orroded region is loated in the nozzle (see inspetion data). The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division, 988 Edition and is not in the fatigue servie. Determine if the vessel is aeptable for ontinued operation. Vessel Data Shell Material = SA 56Grade70 Year 988 Design Conditions = 85 Shell Inside Diameter = 60 in Shell Thikness = 0.60 in Shell Weld Joint Effiieny =.0 Shell Uniform Metal Loss = 0.0 in Shell FCA = 0.5 in Nozzle Outside Diameter = 4.0 in Nozzle Thikness =.5 in Nozzle Nek Length = 5.0 in Nozzle Material = SA 06GradeC Year 988 Nozzle Weld Joint Effiieny =.0 Nozzle Uniform Metal Loss = 0.0 in Nozzle FCA = 0.5 in Nozzle loads = negligible F 5-3

89 Inspetion Data The region of loalized metal loss is shown in the following figure. The opening is loated 45 in from the nearest major strutural disontinuity whih is a NPS- nozzle. Figure E5.6- From the inspetion data: The average thikness in the nozzle reinforement zone is in The orrosion is uniform for all inspetion planes. Perform a Level Assessment per paragraph beause the orrosion is at a nozzle. The assessment proedure in Part 4, paragraph is used. From the inspetion data: t t shell am nozzle am = 0.60 in = in Required thikness of the shell: t r ( 85)( ) ( )( ) ( ) PR ( s + LOSSs + FCAs) = = = in SE 0.6P

90 Determine the orroded shell and nozzle mean diameters: Do = 60 + t = = 6. in D = 60 + ( LOSSs + FCAs) = 60 + ( ) = 60.5 in Do + D Dm = = = in do = 4 in nozzle d = do tam + FCAn = 4 ( 0.875) + ( 0.5) =.5 in do + d dm = = = 3.5 in Perform the assessment using the limit analysis method (see paragraph A.3..b) of Annex A) (equations (A.0) to (A.6)) 5-33

91 Chek the limitations: i. Nozzle material = SA06, Gr. C is a arbon steel Temperature = 650 F < 800 F Limit intable A. True YS 38 ii. Shell = = < 0.8 UTS 70 YS 40 Nozzle = = < 0.8 UTS 70 True iii. Nozzleis NPS 4< NPS 4 True d 3.5 m D m iv. = = , then = = D t 0.6 m True v. The opening is not subjet to yli loading True vi. The opening is in a ylindrial vessel = LOSS + FCA = = 0.5 in s s s ( L = 45 in) >.8 D ( t ).8 6 msd ( = 0.75 m s ( ) = in) True True d + d vii. ( Spaing between two openings = 45 in) > 3 = 3 = 40.5 in True viii. The opening is irular in ross setion withits axis normal tothe surfae of the ylindrial vessel True ix. No signifiant nozzleloads True nozzle ( = ) > am ( = = std ) x. t in 0.875t in nozzle ( 0.5 = =.705 m am ) through an axial length of d t in True nozzle ( ) ( ) nozzle xi. L = 5 in > 0.5 d t = =.705 in t = t =0.875 in m am n am 5-34

92 = LOSS + FCA = = 0.5 in n n n nozzle t = = 0.75 in am n shell t = = in t am nozzle am shell tam s s n 0.75 = =.5789 A= 54 & B = dm Dm λ = = =.467 D t m s 3/ / nozzle d m tam n +.5 shell + λ Dm tam s / 3/ nozzle d m tam n + shell Dm tam s 3/ 3.5 / + (.5789) +.5(.467) = =.775 / 3.5 3/ + (.5789) shell tam s =.95 = 4.37 tr True nozzle nozzle tam n tam n d m A 8 55 shell + B shell + λ + tam s tam s D m d m 08λ λ+ 5 D m (.5789) + 8(.5789) + 38 (.467) = = (.467) (.467) t r [ λ] = (.467 shell ) tam = s Analysis Results: The area reinforement alulation using the limit analysis approah is aeptable based on the orroded dimension of the nozzle onfiguration and the stated design onditions. The Level Assessment riteria are satisfied. The vessel is aeptable for ontinued operation. True 5-35

93 5.7 Example Problem 7 A region of orrosion in an atmospheri storage tank has been found during the inspetion. The tank was onstruted to API 650. Determine if the tank is aeptable for ontinued operation. Tank Data Material = ASTM A85 Grade C Design Temperature = Ambient Design Liquid Height = 40 ft Diameter = 80 ft Shell Height = 40 ft Speifi Gravity =.0 Nominal Thikness = 0.58 in Uniform Metal Loss = 0. in FCA = 0.05 in Weld Joint Effiieny =.0 Inspetion Data The grid and data used for the inspetion are shown below. The region of orrosion is loated 57 inhes from the nearest major strutural disontinuity. Table E5.7- Inspetion Data (in) Cirumferential Inspetion Planes Meridional Inspetion Planes (ft -in) M M M3 M4 M5 M6 M7 M8 M9 M0 M M Meridional CTP C C C C C C C C C C Cirumferential CTP Notes:. Spaing of thikness readings in meridional or longitudinal diretion is 3.0 in. Spaing of thikness readings in irumferential diretion is 6.0 in 5-36

94 Perform a Level Assessment per paragraph a) STEP Determine the CTP (Critial Thikness Profiles) (see Table E5.7-) b) STEP Determine the wall thikness to be used in the assessment using equation (5.3). tnom = 0.58 in LOSS = 0. in FCA = 0.05 in trd = tnom LOSS = = 0.47 in t = t LOSS FCA= = 0.4 in nom ) STEP 3 Determine the minimum measured thikness, t mm, and the dimension, s, for the longitudinal CTP. There is only one LTA in the tank; therefore, the flaw-to-flaw spaing riteria do not need to be heked. tmm = 0.3 in s= 9 3= 7 in d) STEP 4 Determine the remaining thikness ratio and the longitudinal flaw length parameter. λ using equations (5.5) and (5.6). tmm FCA Rt = = = t 0.4 D= ( LOSS + FCA) = ( ) = in.85s.85( 7) λ = = =.776 Dt ( ) e) STEP 5 Chek the limiting flaw size riteria using equations (5.7), (5.8), and (5.9). ( Rt = 0.486) 0.0 ( mm = = 0.8 ) 0.0 ( ) ( ) True t FCA in in True ( ) L = 57 in.8 Dt = = in True msd f) STEP 6 Chek the riteria for a groove-like flaw. This step is not appliable beause the region of loalized metal loss is ategorized as an LTA. g) STEP 7 Determine the MFH for the omponent (see A.6.3) using equation (A.3). ( 0.4)( 3595) ( )( ) t S MFH = + = + = ft.6gd

95 h) STEP 8 Evaluate the longitudinal extent of the flaw. From Figure 5.6 with and equation (5.): λ =.776 R t = M =.4937 based ontable5.equation t, the longitudinal extent of the flaw is aeptable. Using Table 5. R t RSF = = = < a = 0.9 ( Rt ) ( 0.486) M t.4937 RSF MFH r = MFH = ( ) = ft RSF 0.9 a ( MFH r = ft) < ( MFHDesign = 40 ft ) ( RSF ) Therefore the longitudinal extent of the flaw is unaeptable for the stated design onditions. i) STEP 9 Evaluate irumferential extent of the flaw. ) STEP 9. From the irumferential CTP, determine λc using equation (5.) = 6 = 66 in.85( 66) ( 960.3)( 0.4).85 λc = = = 4.9 Dt ) STEP 9. Chek the following onditions (equations (5.3) to (5.7)). ( λ = ) D = = t = = =.0 ( RSF ) ( EL ) ( E ) C The irumferential extent of the flaw is unaeptable. The Level Assessment riteria are not satisfied. The tank is unaeptable for ontinued operation. True True False True True 5-38

96 5.8 Example Problem 8 A region of internal orrosion and/or erosion has been found on the extrados of a seamless long radius piping elbow (90º bend) during an inspetion. A piping stress analysis has been performed on this system and the results indiate that the fores and moments from the weight and thermal load ases whih at on the elbow are negligible. The piping system was onstruted to ASME B Edition. Determine if the pipe bend is aeptable for ontinued operation. Piping Data Material = ASTM A34 GradeWPB Year 980 Design Conditions = F Pipe Diameter = NPS Wall Thikness = Shedule 40 Uniform Metal Loss = 0.0 in FCA = 0.05 in Inspetion Data Thikness readings have been taken based on an inspetion grid on the extrados of the elbow. The spaing to the nearest strutural disontinuity is 3 in. The thikness readings indiate that the LTA is loated in the middle one-third setion of the elbow. The ritial thikness profiles in the longitudinal and irumferential diretions are 6.5 in and 3.0 in in length, respetively. Thikness readings indiate that the metal loss an be assumed to be uniform with the following minimum thikness reading. tmm = 0.8 in Perform a Level Assessment per paragraph Note that a Level Assessment may be performed for piping bends subjet to pressure loading only. In this example, it has been stated that the results of a piping stress analysis indiated that the fores and moments on the pipe bend are negligible. a) STEP Determine the CTP (Critial Thikness Profiles) (see Inspetion Data above) the problem states that the metal loss is uniform with tmm = 0.8 in b) STEP Determine the wall thikness to be used in the assessment using equation (5.3). Do =.75 in tnom = in Outside diameter Shedule 40 LOSS = 0.0 in FCA = 0.05 in D= Do tnom + ( LOSS+ FCA) =.75 ( 0.406) + ( ) =.038 in trd = tnom LOSS = = in t = t LOSS FCA= = in nom ) STEP 3 Determine the minimum measured thikness, t mm, and the dimension, s, for the longitudinal CTP. There is only one LTA in the elbow; therefore, the flaw-to-flaw spaing riteria do not need to be heked. 5-39

97 tmm = 0.8 in s = 6.5 in d) STEP 4 Determine the remaining thikness ratio and the longitudinal flaw length parameter. λ using equations (5.5) and (5.6). tmm FCA Rt = = = t ( ) ( ).85s λ = = = Dt e) STEP 5 Chek the limiting flaw size riteria using equations (5.7), (5.8), and (5.9). ( Rt = 0.365) 0.0 ( mm = = 0.3 ) 0.0 ( ) ( ) True t FCA in in True ( ) L = 3 in.8 Dt = = in True msd f) STEP 6 Chek the riteria for a groove-like flaw. This step is not appliable beause the region of loalized metal loss is ategorized as an LTA. g) STEP 7 Determine the MAWP for the omponent equation (A.30). Sine this is a long radius elbow, the bend radius is.5 times of the pipe diameter. Calulate the Lorenz fator using Eqn. (A.305) for extrados. Rb = 8 in Do + D Rm = = = Rb Rm L 6.97 f = = = Rb Rm 6.97 SE ( 6500)(.0) t ( L ) C f MAWP = = = psi D Y t o B3 C MAWP = MAWP = psi ( )( ) h) STEP 8 Evaluate the longitudinal extent of the flaw. From Figure 5.6 with 5. and equation (5.): λ = R t = 0.365, the longitudinal extent of the flaw is aeptable. Using Table 5-40

98 M t API 579-/ASME FFS- 009 Fitness-For-Servie Example Problem Manual =.67 R t RSF = = = 0.48 < a = 0.9 ( Rt ) ( 0.365) M t.67 RSF 0.48 MAWPr = MAWP = ( ) = psi RSF 0.9 a ( MAWPr = psi) < ( PDesign = 600 psi) ( RSF ) Therefore the longitudinal extent of the flaw is unaeptable for the stated design onditions. i) STEP 9 Evaluate irumferential extent of the flaw. ) STEP 9. From the irumferential CTP, determine λc using equation (5.). = 3.0 in.85( 3) (.038)( 0.356).85 λc = = =.86 Dt ) STEP 9. Chek the following onditions (equations (5.3) to (5.7)). ( λ = ).86 9 D.038 = = t = ( RSF ) ( EL = ) ( E = ) C The irumferential extent of the flaw is unaeptable. The Level Assessment riteria are not satisfied. The pipe bend is unaeptable for ontinued operation. True True False True True 5-4

99 5.9 Example Problem 9 A region of loalized orrosion has been found in a pressure vessel in vauum servie during a sheduled turnaround. The orrosion is loated on the inside surfae of the vessel and between stiffening rings that are 80 ft apart. The vessel and inspetion data are provided below. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division, 989 Edition. Determine if the vessel is aeptable for ontinued operation using a Level Assessment. Vessel Data Material = SA 56Grade70 Year 989 Design Conditions = F ( External Pr essure) Outside Diameter = 00.0 in Fabriated Thikness =.0 in Uniform Metal Loss (Internal) = 0.0 in FCA = 0. in Weld Joint Effiieny =.0 Supplemental Loads = 0.0 negligible Out-of-roundness = 0.0 negligible Inspetion Data Pressure Vessel Shell M5 C C C3 C4 C5 C6 C7 C8 C9 Inspetion Grid M4 M3 M M Figure E5.9- Inspetion Grid Weld Seam 5-4

100 The thikness data and the grid used for the inspetion are shown below. Longitudinal Inspetion Table E5.9- Inspetion Data (in) Cirumferential Inspetion Planes Cirumferential Planes C C C3 C4 C5 C6 C7 C8 C9 CTP M M M M M Longitudinal CTP Notes:. Spaing of thikness readings in longitudinal diretion is 0 in.. Spaing of thikness readings in irumferential diretion is 3.0 in. 3. These readings represent the minimum thikness reading within eah 3 in X 0 in grid after sanning the entire grid area. The distane from the edge of the metal loss to the nearest stiffening ring in the longitudinal diretion is 30 in on one side and 580 in on another side. Figure E5.9- Thikness Profile Perform a Level Assessment per paragraph a) STEP Determine the CTP (Critial Thikness Profiles) (see paragraph ) the thikness readings for the ritial inspetion planes are indiated in Table E5.9- and Figure E5.9-. b) STEP Subdivide the CTP in the longitudinal diretion using a series of ylindrial shells that approximate the atual metal loss (see Figure E5.9-). Determine the thikness and length of eah of these ylindrial shells and designate them t i and L i. The metal loss an be subdivided into 9 regions in the longitudinal diretion based on the table below. 5-43

101 Table E5.9- Subdivision, i t i (in) L i (in) ) STEP 3 Use the method in Annex A, Paragraph A.4 to alulate the allowable external pressure, for eah subdivision (see Table E5.9-3). tnom =.0 in LOSS = 0.0 in FCA = 0. in Do = 00 in Do 00 Ro = = = 50 in L = 960 in T ( ) 6 E = F y S = 8.48 ksi@ 650 F y e P i, Chek the appliability of the method (see paragraph A.4.) ( = = 0 0. = 0.9 ) ( 3/6 = ) t t LOSS FCA in in in True D t o nom 00 = =. 000 True 0.9 Temperature = 650 F 700 F for arbon steel withuts 60 ksi True SA56, Gr.70is arbon steel True e = 0( Out of roundness) True Supplemental load is negligible and does not need to be onsidered. e Detailed alulation of P 4 for Subdivision 4 is given below. 5-44

102 Calulate the predited elasti bukling stress, t4 = 0.7 in t,4 = t4 FCA= = 0.6 in Do = 00 in L 960 Mx = = = 75.7 in Rt o, D o 00 = = t, D o Sine 3 ( M x = 75.7) < = t,4 C.7 = F h he.058.m x. 75 = = ( ) ( ) F he (equations (A.76) (A.8)) ( ) ( ) 3 6.6CEt h y,4.6 ( ) ( ) = = = D 00 o ( ) 3 psi Calulate the predited inelasti bukling stress, F S F i he y ( ) 4 ( ) = = < ( ) = psi F i (equations (A.8) (A.84)). Calulate the in-servie margin, FS, (equations (A.63) (A.65)). ( i = ( ) ) y = ( ) 3 4 F.85 0 psi (0.55S psi) SF = Calulate the allowable external pressure, ( ) 3 F.85 0 i Fha = = = psi FS e t,4 0.6 P = F (59.590) 7. 4 ha = = psi Do 00 e P, (equations (A.85) and (A.86))

103 Calulated e Pi for all subdivisions are given in Table E5.9-3 below. Table E5.9-3 Subdivision, i e P i (psi) d) STEP 4 Determine the allowable external pressure using equation (5.). MAWP r = 9 i= LT L P i e i 960 = = psi e) STEP 5 Compare MAWPr to design pressure ( ) MAWP = psi > ( P = 4.7 psi) r design The Level Assessment Criteria are satisfied. The equipment is aeptable for ontinued operation. 5-46

104 PART 6 ASSESSMENT OF PITTING CORROSION EXAMPLE PROBLEMS 6. Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Widespread pitting on the ID surfae has been disovered on the ylindrial setion of a pressure vessel during an inspetion. The vessel and inspetion data are shown below. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division, 985. Determine if the vessel is aeptable for ontinued operation at the urrent MAWP and temperature. Perform a Level assessment. Consider the pitting damage to be arrested. Vessel Data Material = SA 56Grade70Year985 Design Conditions = 300 psi@50 F Inside Diameter = 60 in Wall Thikness = 0.75 in Uniform Metal Loss = 0.05 in Future Corrosion Allowane = 0.07 in Allowable Stress = 7500 psi Weld Joint Effiieny = 0.85 There are no supplemental loads on the setion. 6-

105 Perform a Level Assessment per paragraph 6.4. Figure E6.- Example Problem E6. Pitting Damage a) STEP - Determine the following parameters: D, DO, FCAand either trd or tnom and LOSS D = 60 in FCA = 0.07 in tnom = 0.75 in LOSS = 0.05 in b) STEP - Determine the wall thikness to be used in the assessment using Equation (6.) or Equation (6.), as appliable. t = t LOSS FCA= = 0.63 in nom ) STEP 3 - Loate the area on the omponent that has the highest density of pitting damage based on the number of pits. Obtain photographs (inlude referene sale), or rubbings of this area to reord the amount of surfae damage. See Figure E6.-. d) STEP 4 - Determine the maximum pit depth, w max, in the region of pitting damage being evaluated. The maximum depth of pitting has been determined as w max = 0.3 in 6-

106 e) STEP 5 - Determine the ratio of the remaining wall thikness to the future wall thikness in the pitted region using Equation 6.3. In Equation (6.3), t rd an be replaed by t nom LOSS. If R wt < 0. the Level assessment riteria are not met. R wt t + FCA wmax = = = t 0.63 Is R 0.? wt Yes f) STEP 6 - Determine the MAWP for the omponent (see Annex A, paragraph A.) using the thikness from STEP 60 Ri = = 30 in R = R + LOSS + FCA= = 30. in i ( 7500)( 0.85)( 0.63) ( )( ) SEt a MAWP = = = 307 psi R t g) STEP 7 - Compare the surfae damage from the photographs or rubbings to the standard pit harts shown in Figures 6.3 through 6.0. Selet a pit hart that has a measure of surfae damage that approximates the atual damage on the omponent. If the pitting damage is more extensive than that shown in Figure 6.0, then ompute the RSF using Equation 6.4 and proeed to STEP 9. Based on the piture, the losest Level pitting hart is Figure E

107 Note: The sale of this figure is 50 mm by 50 mm (6 in by 6 in) R wt (See Equation 6.3) Cylinder Level RSF Sphere Figure E6.- Pitting Chart for Grade Pitting (API 579 Figure 6.4) 6-4

108 h) STEP 8 - Determine the RSF from the table shown at the bottom of the pit hart that was hosen in STEP 7 using the value of wt values of R. wt R alulated in STEP 5. Interpolation of the RSF is aeptable for intermediate Calulations show interpolation in Figure 6.4. Given R wt = 0.635, from Figure 6.4 From Figure 6.4, when R = 0.8 RSF = 0.97 and when R = 0.6 RSF = 0.95 wt thus the differene in RSF = = 0.0 and the differene in R = = 0. Solving for the RSF RSF = ( 0.0) = wt RSF RSF wt i) STEP 9 - Sine the a, then the pitting damage is aeptable for operation at the MAWP determined in STEP 6. To illustrate the Part alulation, determine MAWP for the ase of RSF < RSF. Using the equations in Part, paragraph.4... The MAWP from STEP 6 shall be used in this alulation. MAWPr = 307 psi The Design Pressure is 300 psi, and the MAWP = 307 psi ; therefore, the vessel passes the Level assessment and is aeptable for the design pressure. r r a 6-5

109 6. Example Problem Widespread pitting on the outside surfae has been disovered on the ylindrial straight setion of a piping omponent during an external inspetion. The piping and inspetion data are shown below. The pipe was onstruted to the ASME B3.3 ode 99. Determine if the pipe is aeptable for ontinued operation at the urrent MAWP and temperature. Consider the pitting damage to be arrested. Pipe Data Material = SA 06 Grade B Year 99 Design Conditions = C Outside Diameter = 68.3mm Wall Thikness = 0.97 mm Uniform Metal Loss = 0.0 mm Future Corrosion Allowane = 0.76 mm Allowable Stress = MPa Weld Joint Effiieny = Maximum pitting depth = 5.6 mm There are no supplemental loads on the setion. 6-6

110 Perform a Level Assessment per paragraph 6.4. Figure E6.- Example Problem E6. Pitting Damage a) STEP - Determine the following parameters: DO, FCAand either trd or tnomand LOSS DO = 68.3mm FCA = 0.76mm LOSS = 0.0mm trd = tnom LOSS t = = 0.97mm rd b) STEP - Determine the wall thikness to be used in the assessment using Equation 6. or Equation 6. as appliable. t = trd FCA t = = 0.mm ) STEP 3 - Loate the area on the omponent that has the highest density of pitting damage based on the number of pits. Obtain photographs (inlude referene sale), or rubbings of this area to reord the amount of surfae damage. See Figure E

111 d) STEP 4 - Determine the maximum pit depth, w max w max = 5.6 mm, in the region of pitting damage e) STEP 5 - Determine the ratio of the remaining wall thikness to the future wall thikness in the pitted region using Equation 6.3. In Equation (6.3), t rd an be replaed by t nom LOSS. If R wt < 0. the Level assessment riteria are not met. R wt t + FCA wmax = = = t 0. Is R 0.? wt Yes f) STEP 6 - Determine the MAWP for the omponent (see Annex A, paragraph A.5) using the thikness from STEP. o B3 ( )( ) D = D t = = mm o nom D R = + LOSS + FCA= = mm irumferential MAWP C longitudinal MAWP L ( MA) ( ) SEt a ( tsl MA) 4Y ( t t MA) B3 ( )( )( ) SEt a = = = 7.583bar D Y t MA 68.3 (0.4)(0. 0) 4 = D o ( )( )( ) = = bar (0.4)(0. 0 0) sl C L MAWP = min( MAWP, MAWP ) = min(7.583, ) = bar g) STEP 7 - Compare the surfae damage from the photographs or rubbings to the standard pit harts shown in Figures 6.3 through 6.0. Selet a pit hart that has a measure of surfae damage that approximates the atual damage on the omponent. If the pitting damage is more extensive than that shown in Figure 6.0, then ompute the RSF using Equation 6.4 and proeed to STEP 9. Based on the piture, the losest Level pitting hart is Figure E

112 Note: The sale of this figure is 50 mm by 50 mm (6 in by 6 in) R wt (See Equation 6.3) Cylinder Level RSF Sphere Figure E6.- Pitting Chart for Grade 4 Pitting (API 579 Figure 6.6) 6-9

113 h) STEP 8 - Determine the RSF from the table shown at the bottom of the pit hart that was hosen in STEP 7 using the value of wt values of R wt = 0.56 R alulated in STEP 5. Interpolation of the RSF is aeptable for intermediate Calulations show interpolation in Figure 6.6. Given When Rwt = 0.6 RSF = 0.9 and when Rwt = 0.4 RSF = 0.85 thus the differene in RSF = = 0.05 when the differene in R = = 0. Solve for RSF RSF = ( 0.05) = wt i) STEP 9 - Sine RSF < RSFa, alulate the MAWP r as appliable using the equations in Part, paragraph.4... Aeptability for ontinued servie is determined from RSF MAWPr = MAWP = = 7.bar RSF 0.9 a MAWP. Sine the Design Pressure = 7.4 bar and the MAWP = 7. bar, the pipe fails the Level assessment. r r 6-0

114 6.3 Example Problem 3 Widely sattered pitting has been disovered on the bottom ylindrial setion of a pressure vessel midway between two saddle loations during an internal inspetion. The vessel and inspetion data are shown below. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division, 980. Determine if the vessel is aeptable for ontinued operation at the urrent MAWP and temperature. Vessel Data Material = SA 56Grade70 Year 980 Design Conditions = 500 psi@450 F Inside Diameter = 60 in Wall Thikness =.5 in Uniform Metal Loss = 0.03 in Future Corrosion Allowane = 0.05 in Allowable Stress = 7500 psi Weld Joint Effiieny = 0.85 Saddle Reation Fore = 34690lbf Mid Span Bending Moment = 3600 in lbf Tangent-to-Tangent Length = 30 ft Depth of Head = 5 in Distane from Support to Tangent = 4 ft The region of pitting extends through a girth weld. A Level assessment is required sine the equipment has supplemental loads. 6-

115 Pit Couple, k, k Table E6.3- Inspetion Data P in θ Degrees d k,, in w,, in d,, in w,, in ik ik jk jk Perform a Level Assessment per paragraph Determine aeptability for the Cirumferential Stress Diretion per a) STEP Determine the following parameters. tnom =.5 in D = 60 in Qs = lbf LOSS = 0.03 in Sa = 7500 psi L = 30 ft FCA = 0.05 in E = 0.85 H = 5 in P = 500 psi RSF = 0.9 A= 4 ft a ( ) EL = 0.85 M =.3 0 in lbf α = 0deg trd = tnom LOSS = =.095 in DO = D+ tnom =60 + ( )(.5) = 6.5 in D R = + LOSS + FCA= = in 6 6-

116 b) STEP Determine the wall thikness to be used in the assessment using Equation (6.) or Equation (6.), as appliable. t = t FCA= =.045 in rd ) STEP 3 Determine the pit-ouple sample for the assessment (see ), and the following parameters for eah pit-ouple, kd, ik,, djk,, Pk, wik,, andw jk,. In addition, determine the orientation of the pit-ouple measured from the diretion of the σ stress omponent, θ k (see Figure 6.) For the first pit-ouple (Other alulations will be summarized in Table E6.3-). θ = 0 d = 0.5 in w = 0.5 in P = 3.5 in d = 0.6 in w = 0.4 in d) STEP 4 Determine the depth of eah pit below t in all pit-ouples, w ik, and w j, k (see Figure 6..b) and ompute the average pit depth,, onsidering all readings. In Equation (6.5), the subsript k w avg represents a alulation for pit-ouple k. w w + w avg = = = 0.45 in e) STEP 5 Calulate the omponents of the membrane stress field, σ and σ (see Figure 6.). Membrane stress equations for shell omponents are inluded in Annex A. R m D0 6.5 R = = = in For the loation of the defets given in the example whih is at the enter setion of the lower shell entered between the two saddle supports. Using Annex A, determine the membrane stress values onsidering supplemental loads. 6-3

117 Using Annex A.7.3 Horizontal Vessels Subjet to Weight Loads t t sl sl ( )( Rm H ) ( 3) QL + s L A = ( 4) SE ( ) 4H a L π R m L + 3L ( ) ( ) ( ) ( 5) + ( 3)( 34690)( 30) ( 30) 4 = ( 4) = in ( 7500)( 0.85)( π )( ) ( 4)( 5 ) 30 + ( 3)( 30) P R σ = E t σ = = 785. psi P R σ = 0.4 EL t tsl σ = 0.4 = psi ( 0.85) f) STEP 6 Determine the MAWP for the omponent (see Annex A, paragraph A.) using the thikness from STEP. MAWP MAWP ( 7500)( 0.85)(.045) 506. psi ( )( ) ( 7500) 0.85( ) ( ) SEt = = = R + 0.6t C a SE( t t ) = = = 680.3psi R 0.4( t t ) L a sl sl The MAWP is the lowest of the longitudinal and irumferential MAWPs [ ] MAWP = min 506., = 506 psi 6-4

118 g) STEP 7 For pit-ouple k, alulate the Remaining Strength Fator: Single Layer Analysis This analysis an be used when the pitting ours on one side of the omponent (see Figure 6.). For pit-ouple d d + d P d in avg, avg = = = 0.55 μavg = = = P 3.5 σ 785 σ 860 ρ = = = psi ρ = = = psi μavg μavg 4 ( os [ θ] sin [ θ] )( ρ) 3sin [ θ ]( ρ )( ρ ) + ψ = + 4 ( sin [( 0) ] + sin [( )( 0) ])( ρ ) 4 ( os [( 0) ] + sin [( )( 0) ])( ) 3sin [( )( 0) ]( )( ) ψ = 4.73(0) 8 psi + = 4 ( sin [( 0) ] + sin [( )( 0) ])( ) Φ = μavg max ρ, ρ, ρ ρ Φ = max , , = 785. psi E = min Φ, w RSF = E E ( avg ) avg avg Ψ t = min, = RSF = ( 0.846) = ( 0 ).045 avg 8 h) STEP 8 Repeat STEP 7 for all pit-ouples, n, reorded at the time of the inspetion. Determine the average value of the Remaining Strength Fators, RSF, determined in STEP 7 and designate this value as RSF pit for the region of pitting. The alulation results for all pit-ouples are shown in Table E6.3-. k 6-5

119 Table E6.3- Pit-Couple Results Pit Couple, k w avg, k d avg, k μ avg, k ρ,k ρ,k Ψ k E Φ k avg, k RSF k E E E E E E E E E E E E E E+04.9E E E E E E E E E E E E E E E E E E E+04.57E E E E E E E E E E E E E E E E E E E E E E E E+04.57E+04.95E E E E E E E+04.73E E E E E E E E+04.57E E E RSF pit 9 = RSFk = k = i) STEP 9 Evaluate results based on the type of pitting damage: Widespread Pitting For widespread pitting that ours over a signifiant region of the omponent, if, then the pitting damage is aeptable for operation at the MAWP determined in RSF pit RSF STEP 6. If pit a a RSF < RSF, then the region of pitting damage is aeptable for operation at MAWP r, MAWP r is omputed using the equations in Part, paragraph.4... The MAWP from where STEP 6 shall be used in this alulation. Sine RSFpit RSFa <, determine the redued MAWP for the average RSF RSF MAWPr = MAWP = 506. = psi RSF 0.9 See alulations following STEP 0. a 6-6

120 j) STEP 0 Chek the reommended limitations on the individual pit dimensions: () Pit Diameter If the following equation is not satisfied for an individual pit, then the pit should be evaluated as a loal thin area using the assessment methods of Part 5. The size of the loal thin area is the pit diameter and the remaining thikness ratio is defined below. This hek is required for larger pits to ensure that a loal ligament failure at the base of the pit does not our. In this example, the hek is performed at the pit-ouple with the maximum average diameter. d Q Dt The value of Q in Equation (6.8) shall be determined using Part 4, Table 4.4 and is a funtion of the remaining thikness ratio, R t, for eah pit as given by either of the following equations where, w i, k is the depth of the pit under evaluation. R t + FCA w i, k t = t () Pit Depth The following limit on the remaining thikness ratio is reommended to prevent a loal failure haraterized by pinhole type leakage. The riterion is expressed in terms of the remaining thikness ratio as follows: R 0.0 t Calulations For the first pit t + FCA wi, k R t Rt = Q (.3) = t R t RSF a R t = = Q = (.3) = ( ) ( )( ) Q Dt = = in Is diameter less than allowable? D Q Dt k Yes All the pit-ouple alulations are presented in Table E

121 Table E6.3-3 Limitations on Individual Pit Sizes Pit Couple, k R t, Q,k Q,k Dt Single Pit Diameter Ok? R t, Q,k Q D t Single Pit Diameter Ok? Is R 0. t Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Determine aeptability for the LONGITUDINAL Stress Diretion per paragraph a) STEP Determine the following parameters: D, DO, FCA, either t rd or nom D = 60 in DO = 60 + ( )(.5) = 6.5 in FCA = 0.05 in LOSS = 0.03 in trd =.095 in R = = in t and LOSS. b) STEP Determine the wall thikness to be used in the assessment using Equation (6.) or Equation (6.), as appliable. t =.045 in 6-8

122 ) STEP 3 Determine the remaining strength fator, RSF, the allowable remaining strength fator, RSF a, the permissible maximum allowable working pressure, MAWP r, and supplemental loads on the irumferential plane. The remaining strength fator, allowable remaining strength fator, and the permissible maximum allowable working pressure for the region with pitting damage an be established using the proedures in paragraph The supplemental loads are determined in aordane with paragraphs a and b. RSF pit = RSF = 0.9 MAWP = psi t = in Weight Case r sl a ( ) 6 Q = 34690lbf M =.3 0 in lbf s Thermal Case There are no thermal loads d) STEP 4 Compute the equivalent thikness of the ylinder with pitting damage RSFpit B = min,.0 RSFa B = min,.0 = teq = Bt t = = in eq ( )( ) e) STEP 5 For the supplemental loads determined in STEP 3, ompute the omponents of the resultant bending moment and torsion. This should be done for the weight and the weight plus thermal load ases. There is no thermal load ase. For the weight ase a Zik analysis was performed to determine the reation load and maximum bending load at the midspan. These values are: Weight Case ( ) 6 Q = 34690lbf M =.3 0 in lbf s f) STEP 6 Compute the maximum irumferential stress. σ σ m m MAWP r R = RSFpit osα t eq = psi + = ( 0.856) os[ 0]

123 g) STEP 7 Compute the maximum setion longitudinal membrane stress and the shear stress for both the weight and the weight plus thermal load ases. All redible load ombinations should be onsidered in the alulation. The setion properties required for the alulations are provided in Table 6.. π 4 4 Df = Do ( ) teq Ix = ( Do Df ) 64 π Df = 6.5 ( )( 0.959) = in Ix = ( ) = in 64 π π Am = ( Do Df ) At = ( Do + Df ) 4 6 π π Am = ( ( 6.5) ( 60.33) ) = in At = ( ) = in 4 6 D0 6.5 π π a = = = 3.50 in Aa = ( Df ) = ( 60.33) = in 4 4 Shear Stress There is no torsion loading and the shear load at the midspan is zero M = 0 in lbf V = 0 lbf T MT V τ = + At A t eq m 0 0 τ = + = ( )( )( ) Longitudinal Membrane Stress F is the applied setion axial fore for the weight or weight plus thermal load ase, as appliable. F = 0 lbf Tensile σ σ lmt lmt A a F Ma = MAWPr Eos[ α ] + + Am Am I x ( 0) ( 3.3) = ( 464.3) + + = psi ( 0.85) os[ 0] Compressive σ σ lm lm A a F Ma = MAWP r Eos[ α ] + Am Am I x ( 0) ( 3.3) = ( 464.3) + = psi ( 0.85) os[ 0]

124 h) STEP 8 Compute the equivalent membrane stress for the weight and the weight plus thermal load ases Weight Case - Tensile ( 3 ) σ = σ σ σ + σ + τ et m m lmt lmt et (( ) ( )( ) ( ) ( )( ) ) 0.5 σ = = psi Weight Case - Compressive ( 3 ) σ = σ σ σ + σ + τ e m m lm lm e (( ) ( )( ) ( ) ( )( ) ) 0.5 σ = = psi Thermal Case There are no thermal loads i) STEP 9 Evaluate the results as follows: The following relationship should be satisfied for either a tensile and ompressive longitudinal stress for both the weight and the weight plus thermal load ases: S a σ e H f RSF a H =.0 for the weight ase f 7500 σ e [ σet σe ] [ ] max, max , = psi S a H f RSF a = psi 0.9 The maximum of the tensile or ompressive equivalent stress must be less than or equal to the S a allowable stress H f RSF a S a Aeptable if max [ σet, σe ] H f RSF a 7500 Is max [ , ].0 " Yes" 0.9 If the maximum longitudinal stress omputed in STEP 7 is ompressive, then this stress should be less than or equal to the allowable ompressive stress omputed using the methodology in Annex A, paragraph A.4.4 or the allowable tensile stress, whihever is smaller. When using this methodology to establish an allowable ompressive stress, an average thikness representative of the region of pitting damage in the ompressive stress zone should be used in the alulations. 6-

125 The maximum longitudinal stress in STEP 7 is NOT ompressive. j) STEP 0 If the equivalent stress riterion of STEP 9 is not satisfied, the MAWP and/or supplemental loads determined in STEP 3 should be redued, and the evaluation outlined in STEPs through 9 should be repeated. Alternatively, a Level 3 Assessment an be performed. SUMMARY MAWPr = 464 psi The longitudinal stress is aeptable. The equipment fails the level assessment at500 psig,but it is fit for servie at a redued MAWP of 464 psig. 6-

126 6.4 Example Problem 4 Loalized pitting has been disovered on the ylindrial shell setion of a pressure vessel during a orrosion under insulation external inspetion. There is no internal orrosion on this vessel. The vessel and inspetion data are shown below. The vessel was onstruted to the ASME B&PV Code, Setion VIII Division, 986. Determine if the vessel is aeptable for ontinued operation at the urrent MAWP and temperature. There are no supplemental loads. Perform a Level Assessment. Vessel Data Material = SA 56 Grade70 Year 986 Design Conditions = 30 psi@450 F Inside Diameter = 84 in Wall Thikness =.0 in Uniform Metal Loss = 0.0 in Future Corrosion Allowane = in Allowable Stress = 7500 psi Weld Joint Effiieny = 0.85 Distane to Nearest Disontinuity = 37 in The region of pitting extends through a girth weld and is 5 in longitudinal by 5 in irumferential. Table E6.4- Inspetion Data Pit Couple, k P, in θ,deg dik,, in wik,, in djk,, in wjk,, in k k

127 a) STEP - Determine the following parameters: D, DO, FCA, LOSS, RSF a, and either rd LOSS D= 84 in RSFa = 0.9 FCA = in LOSS = 0.0 in tnom = in trd = tnom LOSS = 0.0 =.0 in D = D+ t = 84 + = 86 in O ( ) ( )( ) nom t or t nom and b) STEP - Determine the wall thikness to be used in the assessment using Equation (6.) or Equation (6.), as appliable. t = trd FCA t = = in ) STEP 3 - Determine the pit-ouple sample for the assessment (see ), and the following parameters for eah pit-ouple. In addition, determine the orientation of the pit-ouple measured from the diretion of the σ stress omponent, θ k (see Figure 6.) For the first pit-ouple. θ = 0 d = 0.7 in w = 0.7 in P = 3.5 in d = 0.6 in w = 0.5 in d) STEP 4 - Determine the depth of eah pit in all pit-ouples, w i, kand w j, k (See Figure 6.b) and ompute the average pit depth, w avg, onsidering all readings. In the following equations the subsript represents a alulation for pit-ouple. The remaining alulations are performed in an embedded matrix w + w wavg = = = in e) STEP 5 - Calulate the omponents of the membrane stress field σ and σ (see Figure 6.). Membrane stress equations for shell omponents are inluded in Annex A. D R = = 4 in External metal loss only There are no supplemental loads 6-4

128 t sl API 579-/ASME FFS- 009 Fitness-For-Servie Example Problem Manual = 0 in P R σ = E t 30 4 σ = = psi P R σ = 0.4 EL t tsl 30 4 σ = 0.4 = psi ( )( 0.85) f) STEP 6 - Determine the MAWP for the omponent using the thikness from STEP. See Annex A paragraph A.. MAWP MAWP MAWP SEt = R + 0.6t C a C ( 7500)( 0.85)( ) 4 + ( 0.6)( ) = = psi SE( t t ) = R 0.4( t t ) L a L sl sl ( )( 7500)( 0.85)( ) 4 ( 0.4)( ) L MAWP = = psi The MAWP is the lowest of the longitudinal and irumferential MAWPs [ ] MAWP = min , = g) STEP 7 - For pit-ouple, alulate the Remaining Strength Fator Show the individual alulations for the first pit-ouple. Remainder of the pit-ouples are shown in Table E6.4-. Single Layer Analysis - This analysis an be used when the pitting ours on one side of the omponent (See Figure 6.). d d + d P d = = = μ = = = avg avg 0.65 in avg P 3.5 μavg μavg psi σ σ ρ = = = psi ρ = = = psi

129 4 ( os [ α] sin [ α] )( ρ) 3sin [ α]( ρ )( ρ ) + ψ = + 4 ( sin [ α] + sin ( α) )( ρ) 4 ( os ( 0) sin ( 0) )( ) 3sin ( 0) ( )( ) + ψ = + = (0) 8 psi 4 ( sin ( ( 0) ) + sin ( ( )( 0) ) )( ) Φ = μavg max ρ, ρ, ρ ρ Φ = 0.843max , , = psi Φ w avg Eavg = min, RSF = ( Eavg) Ψ t Eavg = min, = RSF = ( 0.843) = ( 0) h) STEP 8 - Repeat STEP 7 for all pit-ouples, n, reorded at the time of the inspetion. Results are shown in Table E

130 Table E6.4- Pit-Couple Calulations w avg, k d avg, k avg, k Pit Couple, k μ ρ k ρ k Ψ k E Φ k avg, k RSF k E E E E E E E E E E E E E E+04.35E E E+04.05E E E E E E E E+04.54E E E E E E E E E E E E E E E E E+04.40E E E E E E E E+04.0E E E E E E E E E E RSF pit 5 = RSFk = k = i) STEP 9 - Evaluate results based on the type of pitting damage (see Figure 6.). Loalized Pitting The pitting damage is loalized, then the damaged area is evaluated as an equivalent region of loalized metal loss ( LTA, see Part 5 and Figure 5.3). The meridional and irumferential dimensions of the equivalent LTA should be based on the physial bounds of the observed pitting. The equivalent thikness, t eq, for the LTA an be established using the following equation. To omplete the analysis, the LTA is then evaluated using the Level or Level assessment proedures in Part 5 with tmm = teq, where t eq is given by Equation (6.6). eq ( ) ( )( ) teq = RSFpit t t = = in 6-7

131 Determine if the vessel is aeptable for the urrent MAWP using a Part 5 Level paragraph 5.4. Assessment. a) STEP Determine the CTP (Critial Thikness Profiles) the thikness and the size of the loal thin area is given as: t = in s= 5 in = 5 in eq b) STEP Determine the wall thikness to be used in the assessment. This is the same as Level STEP t = in ) STEP 3 Determine the minimum measured thikness in the LTA and the dimension, s, (see paragraph b) for the CTP. tmm = teq = in s= 5 in d) STEP 4 Determine the remaining thikness ratio using Equation (5.5) and the longitudinal flaw length parameter using Equation (5.6). Note in this ase t eq is based on t and already inludes the FCA. tmm FCA= teq tmm FCA Rt = = = t s λ = = = 3.60 Dt (.85)( 5) ( 84)( ) e) STEP 5 Chek the limiting flaw size riteria; if the following requirements are satisfied, proeed to STEP 6; otherwise, the flaw is not aeptable per the Level Assessment proedure. ( Rt = ) 0. ( mm = ) 0. ( ) ( ) ( )( ) True t FCA in in True ( ) L = 3.5 in = in True msd f) STEP 6 The region of metal loss is ategorized as an LTA, so proeed to STEP 7 g) STEP 7 Determine the MAWP for the omponent (see Annex A, paragraph A.) using the thikness from STEP. The MAWP alulation has been performed in STEP 6 of the pitting evaluation. MAWP = 37 psi h) STEP 8 Enter Figure 5.6 for a ylindrial shell or Figure 5.7 for a spherial shell with the alulated values of λ and R. t 6-8

132 Figure E6.4- Level Sreening Curve (API 579 Figure 5.6) The RSF an be determined by Equation 5. with R t = and λ = 3.60 M t λ λ λ λ λ ( 0 ) λ = = ( 0 ) λ ( 0 ) λ ( 0 ) λ ( 0 ) λ Rt RSF = M t ( R ) t RSF = = ( )

133 i) STEP 9 The omponent is a ylindrial shell so that the irumferential extent of the flaw must be evaluated using the following proedure. ) STEP 9. Determine the irumferential flaw length parameter λ = (.85) (.85)( 5) ( 84)( ) λ = =.70 Dt ) STEP 9. If all of the following onditions are satisfied, proeed to STEP 9.3; otherwise, the flaw is not aeptable per the Level Assessment proedure. ( λ = ).70 9 D = t 0.7 = = = 0.85 ( RSF ) ( EL ) ( E ) True True True True True 3) STEP 9.3 Determine the tensile stress fator using Equation (5.8). E 4 3 E L TSF = RSF + EL ( 3)( 0.85) TSF = + =.775 ( )( 0.865) ) STEP 9.4 Determine the sreening urve in Figure 5.8 based on TSF. Enter Figure 5.8 with the alulated values of λ and R. If the point defined by the intersetion of these values is on or above t the sreening urve, then the irumferential extent of the flaw is aeptable per Level TSF =.775 R = λ =.70 t 6-30

134 Level Sreening Curve for the Maximum Allowable Cirumferential Extent of Loal Metal Loss in a Cylinder. 0.8 Rt Lamda TSF=0.7 TSF=0.75 TSF=0.8 TSF=0.9 TSF=.0 TSF=. TSF=.4 TSF=.8 TSF=.3 USER INPUT Figure E6.4- Level Sreening Curve Cirumferential Extent The point is between the interpolated TSF = 0.75 / TSF = 0.8 sreening urve, so that the irumferential extent is aeptable. SUMMARY The equipment fails the longitudinal extent of the Part 5 Level riteria. The rerated MAWP is RSF = RSF MAWPr = MAWP = = psi RSF 0.9 Return to and omplete the assessment. a 6-3

135 j) STEP 0 - Chek the reommended limitations on the individual pit diameters ) For the first pit t + FCA w Rt = = = t R t Q == (.3) = (.3) =.46 R t RSF a 0.9 Q Dt =.46 (84)(0.9375) =.9759in ( ) Is diameter less than allowable? D Q Dt 0.5 =.9759 Yes ) Pit Depth The following limit on the remaining thikness ratio is reommended to prevent a loal failure haraterized by pinhole type leakage. The riterion is expressed in terms of the remaining thikness ratio as follows: R 0.0 t The alulations are summarized in Table E6.4-3 for all pit-ouples. 6-3

136 Pit Couple, k ti, R Q,k Q Dt Table E6.4-3 Pit-Couple Calulations Single Pit da Ok? R t, j Q,k Q Dt Single Pit da Ok? R Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes t The equipment fails the Level assessment; the re-rated pressure is 33 psig 6-33

137 6.5 Example Problem 5 Pitting in a loal thin area has been disovered on the ylindrial setion on a horizontal pressure vessel during an external inspetion. The loal thin area is inhes longitudinal by 8 inhes irumferential and is entered on the bottom of the vessel with a minimum thikness of 0.39 inhes. The region is loated midway between the two saddles. The vessel is insulated and filled with an oil produt with speifi gravity of 0.9. The vessel and inspetion data are shown below. The pitting depths are measured from the undamaged surfae ( 0.50 inhes) t =. Figure E6.5- shows a sketh of the pitting / LTA damage. The vessel was rd onstruted to the ASME B&PV Setion VIII Division ode 999. Determine if the vessel is aeptable for ontinued operation at the urrent MAWP and temperature. Sine there are supplemental loads a Level Assessment is required. There is no internal orrosion. Vessel Data Material = SA 56 Grade 70 Year 999 Design Conditions = 5 psi@450 F Inside Diameter = 0 in Wall Thikness = 0.5 in Uniform Metal Loss = 0.0 in Future Corrosion Allowane (on OD) = 0. in Allowable Stress = 0000 psi Weld Joint Effiieny = Tangent to Tangent Distane = 70 in Total Weight = lbf Saddle Reation Fore = lbf Length from the tangent line of the = 8 in horizontal vessel to the enterline of a saddle support Height of the Horizontal Vessel Head = 30 in A Zik analysis has determined the supplemental loads ating at the pitting / LTA. 6-34

138 The inspetion data is shown in Table E6.5-. Table E6.5- Inspetion Data Pit Couple, k P, in θ,deg dik,, in wik,, in djk,, in wjk,, in k k Figure E6.5-: Sketh of LTA and Pitting (longitudinal diretion) 6-35

139 a) STEP - Determine the following parameters: D, DO, FCA, LOSS, RSF a, and either rd LOSS D= 0 in t nom = 0.5 in FCA = 0. in LOSS = 0.0 in RSF a = 0.9 ( )( ) D = D+ t = = in O nom t = t LOSS = = 0.5 in rd nom Additional Required Variables S = 0000 psi E = E = a L P= 5 psi α = 0deg L= 70 in s = in = 8 in A= 8 in H = 30 in Q = lb t mm R L msd = 0.39 in minimum thikness in the LTA D = = 60 in 70 = = 35 in s t or t nom and b) STEP - Determine the RSF for the loal thin area per Part 5 Level paragraph 5.4. Assessment. a) PART 5 STEP Determine the CTP (Critial Thikness Profiles) the thikness and the size of the loal thin area is given as: s = in = 8 in b) PART 5 STEP Determine the wall thikness to be used in the assessment. t = t FCA= = 0.4 in rd ) PART 5 STEP 3 Determine the minimum measured thikness in the LTA, and the dimension, s, (see paragraph b) for the CTP. t = 0.39 in s= in mm d) PART 5 STEP 4 Determine the remaining thikness ratio using Equation (5.5) and the longitudinal flaw length parameter using Equation (5.6) tmm FCA Rt = = = 0.75 t 0.4 (.85) s (.85)( ) λ = = =.57 Dt ( )( ) 6-36

140 e) PART 5 STEP 5 Chek the limiting flaw size riteria; if the following requirements are satisfied, proeed to PART 5 STEP 6; otherwise, the flaw is not aeptable per the Level Assessment proedure. ( Rt = 0.750) 0. ( mm = ) ( ) ( ) ( )( ) True t FCA in in True ( ) L = 35 in =.4708 in True msd f) PART 5 STEP 6 The region of metal loss is ategorized as an LTA, so proeed to PART 5 STEP 7 g) PART 5 STEP 7 - Determine the MAWP for the omponent using the thikness from PART 5 STEP. See Annex A paragraph A.5 D Do 0 + LOSS FCA Rm = = = 60.in ( Rm H ) + QL L 4A tsl = 4SEπ R 4H m L + 3L ((60.) (30) ) 8000(70) + (70) ) 4(5) = = 0.00 in 4(0000)() π (60.) 4(30) (70) C SEt ( 0000)( )( 0.4 a ) MAWP = = = 3.80 psi R + 0.6t MAWP L ( ) ( ) ( )( ) ( )( )( )( ) ( )( ) SE a L t tsl = = R 0.4 t t sl = psi The MAWP is the lowest of the longitudinal and irumferential MAWPs [ ] MAWP = min 3.80, = 3.80 psi h) PART 5 STEP 8 Enter Figure 5.6 for a ylindrial shell or Figure 5.7 for a spherial shell with the alulated values of λ and R. t This is not required sine the RSF for the LTA is only needed for the ombined assessment. The RSF an be determined by Equation 5.. Using R t = 0.75 and λ =.57, determine the RSF for the LTA 6-37

141 Determine M t using Table λ λ λ λ M t = λ ( 0) λ.846( 0) λ + = ( 0) λ.563( 0) λ.4656( 0) λ + Rt RSFlta = = = ( R ) 0.7 t ( 50) M.745 t Determine the equivalent thikness for the pitting assessment t = t i RSF = 0.5(0.865) = 0.433in eq rd lta i) PART 5 STEP 9 The omponent is a ylindrial shell so that the irumferential extent of the flaw must be evaluated using the following proedure. NOTE: This needs to be assessed with the ombined RSF due to pitting and LTA and will be performed after the RSF for pitting is determined. ) STEP 3 - Assess the Pitting Damage using the equivalent thikness from PART 5 STEP 8 as t. a) PART 6 STEP - Determine the following parameters: D, DO, FCA, LOSS, RSF a, and either rd t nom and LOSS. These values have been alulated in STEP t or b) PART 6 STEP - Determine the wall thikness to be used in the assessment using Equation (6.) or Equation (6.) as appliable. Use the equivalent thikness alulated in PART 5 STEP 8. Do not adjust pit depth to the depth below the equivalent thikness, pit depth is measured from the orroded surfae. t = t = 0.433in eq ) PART 6 STEP 3 - Determine the pit-ouple sample for the assessment (see ), and the following parameters for eah pit-ouple. In addition, determine the orientation of the pit-ouple measured from the diretion of the σ stress omponent, θ k (see Figure 6.) For the first pit-ouple. θ 0 = 0 d = 0.3 in w = 0. in P = 3.5 in d = 0.4 in w = 0.8 in d) PART 6 STEP 4 - Determine the depth of eah pit in all pit-ouples, w ik and and ompute the average pit depth, w avg w jk (See Figure 6.b), onsidering all readings. In the following equations the subsript represents a alulation for pit-ouple. The remaining alulations are shown in Table E6.5-. w + w wavg = = = 0.95 in 6-38

142 e) PART 6 STEP 5 - Calulate the omponents of the membrane stress field σ and σ (see Figure 6.). Membrane stress equations for shell omponents are inluded in Annex A. P R 5 60 σ = σ = = psi E t P R 5 60 σ = 0.4 σ = p EL t t = sl ( )( ) si f) PART 6 STEP 6 - Determine the MAWP for the omponent using the thikness from PART 5 STEP. See Annex A paragraph A.5. This is not required sine the RSF for the LTA is only needed for the ombined assessment. g) PART 6 STEP 7 - For pit-ouple, alulate the Remaining Strength Fator Show the individual alulations for the first pit-ouple. Single Layer Analysis - This analysis an be used when the pitting ours on one side of the omponent (See Figure 6.). d d + d P d μ avg = = = 0.35 in = = = 0.9 avg avg P 3.5 σ σ ρ = = = psi ρ = = = psi μ μ avg avg 4 ( os [ α] sin [ α] )( ρ ) 3sin [ α]( ρ )( ρ ) + ψ = + 4 ( sin [ α] + sin [( α) ])( ρ ) 4 ( os [( 0) ] sin [( )( 0) ])( ) 3sin [( )( 0) ]( )( ) ψ = + 4 ( sin [( 0) ] + sin [( )( 0) ])( ) Φ = 0.9 max,, [ ρ ρ ρ ρ ] [ ] + = (0) psi 8 Φ = 0.9 max , , = psi Φ w E = RSF = E ψ t E ( avg ) avg min, avg = min, = RSF = ( 0.900) = (0) avg

143 h) PART 6 STEP 8 - Repeat PART 6 STEP 7 for all pit-ouples, n, reorded at the time of the inspetion. Results are in Table E6.5-. Pit Couple, k w avg, k d avg, k Table E6.5- Pit-Couple Calulations μ avg, k ρ k ρ k ψ k E Φ k avg, k RSF k E E E E E E E E E E E E E E E Determine the average RSF for all the pit-ouples speified RSF pit 5 = RSFk = k = i) PART 6 STEP 9 - Evaluate results based on the type of pitting damage (see Figure 6. ): In this ase pitting is onfined within a region of loalized metal loss. () Pitting Confined Within A Region Of Loalized Metal Loss If the pitting damage is onfined within a region of loalized metal loss (see Figure 6.4), then the results an be evaluated using the methodology in subparagraph ) (below) () Region Of Loal Metal Loss Loated In An Area Of Widespread Pitting If a region of loal metal loss ( LTA ) is loated in an area of widespread pitting, then a ombined Remaining Strength Fator an be determined using the following equation. RSF = RSF RSF omb pit lta Combined Analysis RSF RSF RSF pit lta = = = RSF RSF = = 0.74 omb pit lta ( )( ) 6-40

144 The redued MAWP of the damaged omponent is RSF omb MAWPr = min MAWP, MAWP RSFa 0.74 = min 3.80,3.80 = psi 0.9 k) PART 6 STEP 0 - Chek the reommended limitations on the individual pit dimensions For the first pit of the first pit-ouple ) Pit Diameter t + FCA w Rt = = = t R t Q = (.3) = (.3) =.59 R t RSF a 0.9 Q Dt = (.3) ( 0)( 0.433) = 8.89 in Is diameter less than allowable? D Q Dt Yes ) Pit Depth The following limit on the remaining thikness ratio is reommended to prevent a loal failure haraterized by pinhole type leakage. The riterion is expressed in terms of the remaining thikness ratio as follows: R 0.0 t Repeat for all pit-ouples. The alulations are summarized in Table E

145 Pit Couple, k ti, R Q,k Q Dt Table E6.5-3 Pit-Couple Sizing Calulations Single Pit diameter Ok? R t, j Q,k Q Dt Single Pit diameter Ok? R Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes d) STEP 4 - Return to the PART 5 analysis to hek the irumferential extent of the flaw j) PART 5 STEP 9 The omponent is a ylindrial shell so that the irumferential extent of the flaw must be evaluated using the following proedure. ) STEP 9. Determine the irumferential flaw length parameter (.85)( 8) ( 0)( 0.433).85 λ = = = 3.5 Dt ) STEP 9. If all of the following onditions are satisfied, proeed to STEP 9.3; otherwise, the flaw is not aeptable per the Level Assessment proedure. ( λ = ) D = t True 0.7 = True 0.7 = True 0.7 = True ( RSFomb ) ( EL ) ( E ) True Use the ombined RSF for the irumferential extent t 6-4

146 3) STEP 9.3 Determine the tensile stress fator using Equation (5.8). 4 3E L TSF = + RSFomb EL 4 ( 3)( ) TSF = + =.384 ( )( 0.74) 4) STEP 9.4 Determine the sreening urve in Figure 5.8 based ontsf. Enter Figure 5.8 with the alulated values of λ and R (See Figure E6.5-). If the point defined by the intersetion of t these values is on or above the sreening urve, then the irumferential extent of the flaw is aeptable per Level. TSF =.384 R = λ = 3.5 t Level Sreening Curve for the Maximum Allowable Cirumferential Extent of Loal Metal Loss in a Cylinder. 0.8 Rt Lamda TSF=0.7 TSF=0.75 TSF=0.8 TSF=0.9 TSF=.0 TSF=. TSF=.4 TSF=.8 TSF=.3 USER INPUT Figure E6.5- Level Sreening Curve Cirumferential Extent The point shown is between the TSF = 0.8 / TSF = 0.9 sreening urve whih is above the urve for TSF =.384, so that the irumferential extent is ACCEPTABLE 6-43

147 e) STEP 5 Summarize and alulate the ombined RSF and MAWP The RSF for the ombined LTA and pitting damage is RSF RSF RSF pit lta = = = RSF RSF = = 0.74 omb pit lta ( )( ) The region of pitting and loal thin area fails the Level Assessment proedure. Determine the redued MAWP of the damaged omponent RSF omb MAWPr = min MAWP, MAWP RSFa 0.74 = min 3.80,3.80 = psi

148 6.6 Example Problem 6 Inspetion of a pressure vessel during a sheduled turnaround has deteted widespread pitting on both ID and OD surfaes. The vessel and inspetion data are shown below. The vessel was onstruted to the ASME B&PV Code, Setion VIII Division 97. Determine if the vessel is aeptable for ontinued operation at the urrent MAWP and temperature. Vessel Data Material = SA 56 Grade 70 Year 97 Design Conditions = F Inside Diameter = 96 in Wall Thikness = in Uniform Metal Loss = 0.0 in Future Corrosion Allow. = 0.0 in Allowable Stress = 7500 psi Weld Joint Effiieny = 0.85 Distane to Nearest Disontinuity = 96 in Supplemental Loads = 0( Negligible ) The inspetion data taken from the ID and OD surfaes is shown in Table E6.6-. Depths of the pitting are measured from eah respetive surfae. 6-45

149 Table E6.6- Inspetion Data Pit Couple, k P, in θ,deg dik,, in wik,, in djk,, in wjk,, in k k ID Surfae OD Surfae

150 Sine the pitting is on both sides of the omponent a Level assessment per is required. a) STEP - Determine the following parameters: D, DO, FCA, LOSS, RSF a, and either rd LOSS D = 96 in DO = 96 + ( )( ) = 98 in FCA = 0.0 in LOSS = 0.0 in RSFa = 0.9 t =.0 in nom Additional Variables Sa = 7500 psi P= 300 psi EL = 0.85 E = 0.85 R = = 48 in L = 30 in msd t or t nom and b) STEP - Determine the wall thikness to be used in the assessment using Equation (6.) or Equation (6.), as appliable. t = t LOSS FCA= =.0 in nom ) STEP 3 - Determine the pit-ouple sample for the assessment (see ), and the following parameters for eah pit-ouple. In addition, determine the orientation of the pit-ouple measured from the diretion of the σ stress omponent, θ k (see Figure 6.) For the first Pit-Couple. All results for the other pit-ouples are shown in the embedded tables. θ = 0 P = 3.5 in d, = 0.3 in w, = 0. in d = 0.4 in w = 0.3 in,, d) STEP 4 - Determine the depth of eah pit in all pit-ouples, w ik and the average pit depth, w avg w jk (See Figure 6.b) and ompute, onsidering all readings. In the following equations the subsript represents a alulation for pit-couple. The remaining alulations are performed in an embedded matrix w + w wavg = = = 0.65 in 6-47

151 e) STEP 5 - Calulate the omponents of the membrane stress field σ and σ (see Figure 6.). Membrane stress equations for shell omponents are inluded in Annex A. t sl = 0.0 in P R σ = = = psi E t 0.85 P R σ = 0.4 = psi EL t t = sl ( )( 0.85) 0.0 f) STEP 6 - Determine the MAWP for the omponent using the thikness from STEP. See Annex A paragraph A.5 MAWP MAWP ( 7500)( 0.85)( ) psi t ( )( ) ( ) ( )( )( )( ) ( ) ( )( ) SEt = = = R C a L SE a L t tsl = = R 0.4 t t sl = 65 psi The MAWP is the lowest of the longitudinal and irumferential MAWPs [ ] MAWP = min , 65 = psi g) STEP 7 - For pit-ouple, alulate the Remaining Strength Fator. Calulations are shown for the first pit-ouple. Use subpart for multiple layer analysis. Multiple Layer Analysis This analysis is used to aount for pitting on both sides of the omponent (see Figure 6.5). In this analysis, E avg, k, is alulated for eah pit-ouple using Equations (6.7) through (6.3). The value of E avg, k is then used along with the thikness of all layers that the pit-ouple penetrates to alulate a value of RSFk for the pit-ouple. The seletion of the number of layers, N, is based on the depth of pits on both sides of the omponent. The omponent thikness is divided into layers based on the pitting damage (see Figure 6.5), and the RSF k is omputed using Equation (6.4) onsidering all layers ontaining the pit-ouple. Eah layer thikness, t L, is determined by the depth of the deeper of the two pits in the pit-ouple that establishes the layer. For layers where a pit-ouple does not penetrate the layer, and E in Equation (6.4) equals.0. The MAWP used with this the solid layer for all pit-ouples, avg, k expression should be based on t. If the pitting damage is overlapped from both surfaes (Figure 6.5), it is not aeptable per Level. A Level 3 assessment or the reommendations provided in paragraph an be used. RSF k N t L = ( Eavg, k) L L= t (6.4) 6-48

152 d μ ρ ρ API 579-/ASME FFS- 009 Fitness-For-Servie Example Problem Manual d + d in P d P 3.5 σ psi avg = = = avg avg = = = = = = μavg 0.9 σ 8400 = = = μavg ( os [ α] sin [ α] ) 3sin [ α] ρ ρ psi ρ + ψ = + 4 ( sin [ α] + sin [ α] ) ρ 4 ( os ( 0) sin ( )( 0) )( ) 3sin ( )( 0) ( )( ) + ψ = + = 3.63(0) psi 4 ( sin ( 0) + sin ( )( 0) )( ) max [,, ] [ ] Φ = μ ρ ρ ρ ρ avg = 0.9 max , , = psi E avg Φ = = 3.63( 0) min, min, Ψ

153 Determine the maximum pit depth of eah pit-ouple. Table Maximum Pit Depth for eah Pit-Couple ID Surfae OD Surfae Pit-Couple w max Pit-Couple w max Based on reviewing the maximum pit depth data for all pit-ouples, layers of the following thiknesses are required for the evaluation (Refer to Figure 6.5). Layer # (from ID) Table Layers Thikness (in)

154 Compute the RSF for the first pit-ouple using equation 6.4 N t L RSF = ( Eavg, k ) L L= t 0.7( 0.9) ( 0.9) ( 0.9) ( 0.9) + 0.0( 0.9) ( ) + RSF = L= 0.05( ) ( ) ( ) + 0.0( ) + 0.( ) RSF = h) STEP 8 - Repeat STEP 7 for all pit-ouples, n, reorded at the time of the inspetion. Determine the average value of the Remaining Strength Fators, RSF k, found in STEP 7 and designate this value as RSF pit for the region of pitting. Results are shown in Table E Table E6.6-5 Pit-Couple Calulations Pit Couple, k w avg, k avg, k ( in) d ( in) ρ μ k ρk avg, k ( psi) ( psi) ψ k ( psi) Φ k ( psi) E avg, k RSF k E E E E E E E E E E E E E E E E E E E E

155 RSF RSF pit pit = n n k = 0 = RSFk = k = RSF k i) STEP 9 - Evaluate results based on the type of pitting damage (see Figure 6.): Widespread Pitting For widespread pitting that ours over a signifiant region of the omponent, if RSFpit RSFa, then the pitting damage is aeptable for operation at the MAWP determined in STEP 6. If RSFpit < RSFa, then the region of pitting damage is aeptable for operation at MAWP r, where MAWPr is omputed using the equations in Part, paragraph.4... The MAWP from STEP 6 shall be used in this alulation. In this ase RSFpit < RSFa, thus the redued MAWP an be determined from equation. as: RSFpit MAWPr = MAWP RSF a MAWPr = = psi 0.9 j) STEP 0 - Chek the reommended limitations on the individual pit dimensions ) Pit Diameter If the following equation is not satisfied for an individual pit, then the pit should be evaluated as a loal thin area using the assessment methods of Part 5. The size of the loal thin area is the pit diameter and the remaining thikness ratio is defined below. This hek is required for larger pits to ensure that a loal ligament failure at the base of the pit does not our. d Q D t The value of Q shall be determined using Part 4, Table 4.5 and is a funtion of the remaining thikness ratio, R t, for eah pit as given by either of the following equations where w ik, is the depth of the pit under evaluation. R t + FCA w i, k t = t For the first pit of the first pit-ouple R Q ( ) ( 96)( ) t = = Q = (.3) = 0.9 Dt Is diameter less than allowable? D Q Dt =.5690 = in Yes

156 ) Pit Depth The following limit on the remaining thikness ratio is reommended to prevent a loal failure haraterized by pinhole type leakage. The riterion is expressed in terms of the remaining thikness ratio as follows: R 0.0 t The alulations are summarized in Table E6.6-6 for all Pit-Couples Pit Couple, k R ti,,k Table E6.6-6 Pit-Couple Sizing Calulations Q Q Dt Single Pit Dia. Ok? R t, j Q,k Q Dt Single Pit Dia Ok? R Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes Ok Yes Yes SUMMARY The pitting fails the level assessment; the redued MAWP is 303 psi t 6-53

157 THIS PAGE INTENTIONALLY LEFT BLANK 6-54

158 PART 7 ASSESSMENT OF HYDROGEN BLISTERS AND HYDROGEN DAMAGE ASSOCIATED WITH HIC AND SOHIC EXAMPLE PROBLEMS 7. Example Problem Example Problem Example Problem Example Problem HIC damage has been disovered on a ylindrial pressure vessel. Both subsurfae and surfae breaking HIC damage are present. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division. Determine if the vessel is aeptable for ontinued operation fully pressurized at 50 F. Vessel Data Material = SA-56 Grade70 Year 984 Design Conditions = F Inside Diameter = 96 in Fabriated Thikness =.5 in FCA = 0.5 in Weld Joint Effiieny = 0.85 PWHT = Yes Inspetion Data A shemati of a pressure vessel ontaining the HIC damage is shown in Figure E7.-. The inspetion data for the HIC damage is in Table E

159 Shell HIC Area HIC Area a Nozzle HIC Area b HIC Area 3 HIC Area 4 Figure E7.- - HIC Damage 7-

160 Table E7.- Size, Loation, Condition, and Spaing for HIC Damage Enter the data obtained from a field inspetion on this form. Inspetion Date: Equipment Identifiation: Equipment Type: X Pressure Vessel Storage Tank Piping Component Component Type & Loation: t nom :.5 LOSS : FCA : 0.5 t rd :.5 Data Required for Level and Level Assessment HIC Identifiation HIC Area HIC Area a HIC Area b HIC Area 3 HIC Area 4 Diameter s () Dimension () Edge-To-Edge Spaing To Nearest HIC or Blister L H () Minimum Measured Thikness to Internal Surfae tmm ID (3) Minimum Measured Thikness to External Surfae tmm OD (3) Minimum Measured Thikness ; Total of Both Sides t mm (3) Spaing To Nearest Weld Joint L W () Spaing To Nearest Major Strutural Disontinuity Depth of HIC damage L msd w H Notes:. The HIC-to-HIC spaing may affet the size of the HIC damage to be used in the evaluation (see paragraph i.).. See Figure See Figure

161 Perform a Level Assessment per paragraph 7.4. on HIC Area a) STEP Determine the wall thikness to be used in the assessment. t = trd FCA t = =.5 in b) STEP Determine the information in paragraph D = ID+ FCA= 96.5 in tmm ID tmm OD = 0. in = 0.55 in wh = in ( L = 0.0 in) (8t = 9.0 in) H ) STEP 3 - Satisfy the following requirements, then proeed to STEP 4. Otherwise, the Level Assessment is not satisfied. ) The planar dimensions of the HIC damage satisfy Equations (7.7) and (7.8). ( = 3.0 ) ( 0.6 = 6.43 ) ( = 5.0 ) ( 0.6 = 6.43 ) s in D t in True in D t in True ) The through-thikness extent of the damage satisfies Equation (7.9). t wh = in min = in, 0.5 in False 3 ( ) 3) HIC Damage must not be surfae breaking in aordane with paragraph h ( tmm ID = 0. in ) ( 0.t = 0.5 in) False t = 0.55 in 0.t = 0.5 in True ( ) ( ) mm OD 4) Distane between the edge of the HIC damage and the nearest weld seam satisfies the following equation. ( Lw = 0.0 in) > max ( t =.5 in ),.0 in True 5) Distane from edge of HIC damage to the nearest major strutural disontinuity satisfies following the equation. ( msd = 50.0 ) (.8 = 8.73 ) L in D t in True 6) Further hydrogen harging of the metal has been stopped. False The Level Assessment Criteria are not satisfied. A Level assessment must be performed. Perform a Level Assessment per paragraph on HIC Damage Area. a) STEP See Level, STEP b) STEP See Level, STEP ) STEP 3 See Level, STEP 3, item 4) 7-4

162 d) STEP 4 See Level, STEP 3, item 5) e) STEP 5 See Level, STEP 3, item 3) damage is lassified as surfae breaking, therefore wh = wh + min[ tmm ID, tmm OD ] w = min[0., 0.55] = = in H f) STEP 6 Determine the MAWP of the omponent per Annex A, paragraph A.. E is set to.0 beause the damage is in the base metal. ( 7500)(.0)(.5) ( ) SEt MAWP = 403 psi R+ 0.6t = = g) STEP 7 Calulate the RSF based on surfae breaking HIC damage.85( 3.0) ( 96.5)(.5).85s λ = = = Dt M =.03 Part 5, Table5. t ( 0.575)( 0.8) wh D H t.5 RSF = = = w ( 0.575)( 0.8 H D H ) M t t.03.5 h) STEP 8 Evaluate the longitudinal extent of the flaw. Sine ( = 0.980) ( = 0.9) RSF RSF True a then the longitudinal extent of the HIC damage satisfies the LTA portion of the assessment. with the MAWP from STEP 6. MAWP = MAWP = 403 psi but the MAWP is limited by MAWP adjusted for r weld joint effiieny of 0.85=343 psi i) STEP 9 Evaluate the irumferential extent of the HIC damage as an LTA using the proedures in Part 5, paragraph See Example Problem 5.3 STEP 0 for the omplete proedure. The depth used in this analysis is given by ( )( ) d = w D = = 0.46 in. HIC H H Per the results of the LTA analysis of the irumferential extent of the HIC damage, MAWP = 506 psi. Sine MAWP from STEP 9 is greater than PDesign = 300 psi then the irumferential extent of the flaw is aeptable. j) STEP 0 Determine whether a frature assessment is required. This is the ase if any of the following are true. ) The equipment will remain in hydrogen servie. True ) The HIC damage is surfae breaking. True t 3) ( wh = in ) min = in, 0.5 in 3 True 7-5

163 k) STEP Evaluate the HIC as a rak like flaw in aordane with the proedures in Part 9. An example of this proedure and the assoiated alulations is provided in the Part 9 example problems. The parameters used in the rak like flaw assessment are speified below. ) Flaw Size two rak like flaw assessments must be performed, one for the irumferential extent of the HIC damage and one for the longitudinal extent. The rak dimensions are as follows. i) Cirumferential rak a = wh = in = = 5.0 in ii) Longitudinal rak a = w = in H = s = 3.0 in ) Frature Toughness If hydrogen harging of the steel has not been halted by means of a barrier oating, overlay, or proess hange, the lower bound arrest frature toughness as speified in Annex F must be used in the assessment. l) STEP Confirm that further HIC damage has been either prevented or is limited to a known or verifiable rate based on one of the methods provided. The Level Assessment Criteria are satisfied. ( MAWP = 403psi) > ( PDesign = 300 psi) but the MAWP is limited by MAWP adjusted for weld joint effiieny of 0.85=343 psi The equipment is fit for ontinued operation at design stress and temperature pending the outome of a frature assessment following proedures listed in Part 9 and the outome of the assessments of other damaged areas. Perform a Level Assessment per paragraph 7.4. on HIC Area a. a) STEP Determine the wall thikness to be used in the assessment. t =.5 in b) STEP Determine the information in paragraph tmm ID tmm OD = 0.0in = in L =.0 in 9.0 in False ( ) H HIC Area a is within 8t of HIC Area b, therefore; aording to the proedures desribed in Part 4, Figure 4-7, the two areas are ombined for analysis. s = 3.0 in = 5.0 in Repeating STEP satisfies the requirements and the Level Assessment an ontinue. ) STEP 3 - Satisfy the following requirements, then proeed to STEP 4. Otherwise, the Level Assessment is not satisfied. ) The planar dimensions of the HIC damage satisfy Equations (7.7) and (7.8). 7-6

164 ( s = 3.0 in) 6.43 in True ( = 5.0 in) 6.43 in True ) The through-thikness extent of the damage satisfies Equation (7.9). ( w = 0.65 in) in False H 3) HIC Damage must not be surfae breaking in aordane with paragraph h ( ) tmm ID = 0.0 in 0.5 in False ( t = in) 0.5 in True mm OD 4) Distane between the edge of the HIC damage and the nearest weld seam satisfies the following equation. ( L =.5 in) >.5 in False w 5) Distane from edge of HIC damage to the nearest major strutural disontinuity satisfies following the equation. ( ) L = 50.0 in 8.73 in True msd 6) Further hydrogen harging of the metal has been stopped. False Therefore, Level Assessment riteria are not satisfied. Sine item 4) is also required for a Level Assessment, a Level 3 analysis must be onduted. Perform a Level Assessment per paragraph 7.4. on HIC Area 3. a) STEP Determine the wall thikness to be used in the assessment. t =.5 in b) STEP Determine the information in paragraph tmm ID tmm OD = 0. in = 0.65 in ( L = 8.5 in) 9.0 in True H ) STEP 3 - Satisfy the following requirements, then proeed to STEP 4. Otherwise, the Level Assessment is not satisfied. ) The planar dimensions of the HIC damage satisfy Equations (7.7) and (7.8). ( s = 3.0 in) 6.43 in True ( = 5.5 in) 6.43 in True ) The through-thikness extent of the damage satisfies Equation (7.9). ( w = 0.75 in) in True H 3) HIC Damage must not be surfae breaking in aordane with paragraph h ( t = 0.0 in) 0.5 in False mm ID ( t = 0.65 in) 0.5 in True mm OD 7-7

165 4) Distane between the edge of the HIC damage and the nearest weld seam satisfies the following equation. ( ) L = 9.0 in >.5 in True w 5) Distane from edge of HIC damage to the nearest major strutural disontinuity satisfies following the equation. ( ) L =.0 in 8.73 in False msd 6) Further hydrogen harging of the metal has been stopped. False Therefore, Level Assessment riteria are not satisfied. Sine item 5) is also required for a Level Assessment, a Level 3 analysis must be onduted. Perform a Level Assessment per paragraph on HIC Damage Area 4. a) STEP Determine the wall thikness to be used in the assessment. t =.5 in b) STEP Determine the information in paragraph tmm ID tmm OD = 0.5 in = in L = 3.0 in 9.0 in True ( ) H ) STEP 3 - Satisfy the following requirements, then proeed to STEP 4. Otherwise, the Level Assessment is not satisfied. ) The planar dimensions of the HIC damage satisfy Equations (7.7) and (7.8). ( ) ( ) s = 7.0 in 6.43 in False = 4.5 in 6.43 in True ) The through-thikness extent of the damage satisfies Equation (7.9). ( ) w = 0.3 in in True H 3) HIC Damage must not be surfae breaking in aordane with paragraph h ( tmm ID = 0.5 in) 0.5 in True t = in 0.5 in True ( ) mm OD 4) Distane between the edge of the HIC damage and the nearest weld seam satisfies the following equation. ( ) L = 8.0 in >.5 in True w 5) Distane from edge of HIC damage to the nearest major strutural disontinuity satisfies following the equation. ( ) L = 50.0 in 8.73 in True msd 6) Further hydrogen harging of the metal has been stopped. False 7-8

166 The Level Assessment riteria are not satisfied; therefore, a Level Assessment per paragraph must be onduted on HIC Damage Area 4. Perform a Level Assessment per paragraph on HIC Damage Area 4. a) STEP See Level, STEP b) STEP See Level, STEP ) STEP 3 See Level, STEP 3, item 4) d) STEP 4 See Level, STEP 3, item 5) e) STEP 5 See Level, STEP 3, item 3) damage is lassified as sub surfae f) STEP 6 Determine the MAWP of the omponent per Annex A, paragraph A. ( 7500)(.0)(.5) ( ) SEt MAWP = 403 psi R+ 0.6t = = g) STEP 7 Calulate the RSF based on the subsurfae HIC damage. The minimum longitudinal distane to the nearest region of HIC damage is inhes. LHs.0 LR = min,8t = min,8(.5) = 9.0 in ( ) ( 0.3)( 0.8) wh D H LR + s 9 ( ) ( 7) t +.5 RSF = = = 0.94 L + s R h) STEP 8 Evaluate the longitudinal extent of the flaw. Sine ( = 0.940) ( = 0.9) RSF RSF True a then the longitudinal extent of the HIC damage satisfies the LTA portion of the assessment. with the MAWP from STEP 6. MAWP = MAWP = 403 psi but the MAWP is limited by MAWP adjusted for r weld joint effiieny of 0.85=343 psi i) STEP 9 Evaluate the irumferential extent of the HIC damage as an LTA using the proedures in Part 5, paragraph See Example Problem 5.3 STEP 0 for the omplete proedure. The depth used in this analysis is given by ( )( ) d = w D = = 0.4 in. HIC H H Per the results of the LTA analysis of the irumferential extent of the HIC damage, MAWP = 487 psi. Sine MAWP from STEP 9 is greater than PDesign = 300 psi then the irumferential extent of the flaw is aeptable. j) STEP 0 Determine whether a frature assessment is required. This is the ase if any of the following are true. ) The equipment will remain in hydrogen servie. True ) The HIC damage is surfae breaking. False w = 0.3 in in False 3) ( ) H 7-9

167 k) STEP Evaluate the HIC as a rak like flaw in aordane with the proedures in Part 9. An example of this proedure and the assoiated alulations is provided in the Part 9 example problems. The parameters used in the rak like flaw assessment are speified below. ) Flaw Size two rak like flaw assessments must be performed, one for the irumferential extent of the HIC damage and one for the longitudinal extent. The rak dimensions are as follows. i) Cirumferential rak a= wh = 0.3in = = 4.5 in d = 0.5 in ii) Longitudinal rak a= w = 0.3in H = s = 7.0 in d = 0.5 in ) Frature Toughness If hydrogen harging of the steel has not been halted by means of a barrier oating, overlay, or proess hange, the lower bound arrest frature toughness as speified on Annex F must be used in the assessment. l) STEP Confirm that further HIC damage has been either prevented or is limited to a known or verifiable rate based on one of the methods provided. The Level Assessment Criteria are satisfied. ( MAWP = 403psi) > ( PDesign = 300 psi) but the MAWP is limited by MAWP adjusted for weld joint effiieny of 0.85=343 psi The equipment is fit for ontinued operation at design stress and temperature pending the outome of a frature assessment following proedures listed in Part 9. and the outome of the assessments of other damaged areas. 7-0

168 7. Example Problem A ylindrial vessel with both internal and external blisters is shown below. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division. Determine if the vessel is suitable for ontinued operation at the urrent MAWP and temperature using the Level Assessment riteria and Level Assessment riteria if neessary. Vessel Data Material = SA 56 Grade70 Year 980 Design Conditions = 50 F Inside Diameter = 96 in Nominal Wall Thikness =.4 in Future Corrosion Allowane = 0.5 in LOSS = 0.0 in Allowable Stress = 7,500 psi Weld Joint Effiieny = 0.85 Inspetion Data The pressure vessel setion ontaining the blisters is shown below. The inspetion data for the blisters is shown in the following table. G D Pressure Vessel Shell with Blisters E F A C B H External Blister Internal Blister Figure E7.- Blister Damage 7-

169 Table E7.- Size, Loation, Condition, and Spaing for Blisters Enter the data obtained from a field inspetion on this form. Inspetion Date: Equipment Identifiation: Equipment Type: X Pressure Vessel Storage Tank Piping Component Component Type & Loation: Data Required for Level and Level Assessment Blister Identifiation A B C D E Diameter s, in () Dimension, in () Edge-To-Edge Spaing To Nearest Blister LB, in () Bulge Diretion (inside/ outside) outside outside inside inside Inside Blister Projetion Bp, in Minimum Measured Thikness t mm, in Craking At Periphery (Yes/No) No No No No No Crown Craking or Venting (Yes/No) () Length Of Crown Crak or Diameter of Vent Hole, Spaing To Nearest Weld Joint L in (3) W, s in () Crak Vent Vent Crak Vent Spaing To Nearest Major Strutural Disontinuity Lmsd, in Notes:. The blister-to-blister spaing may affet the size of the blister to be used in the evaluation (see paragraph a & b). If the blister has rown raks, enter the length of the rak, see dimension s in Figure 7.6. If the blister has a vent hole, indiate as suh with the diameter of the hole (see Figure 7.7). 3. See Figure

170 Table E7.- Size, Loation, Condition, and Spaing for Blisters Enter the data obtained from a field inspetion on this form. Inspetion Date: Equipment Identifiation: Equipment Type: X Pressure Vessel Storage Tank Piping Component Component Type & Loation: Data Required for Level and Level Assessment Blister Identifiation F G H Diameter s, in () 4 Dimension, in () 8 8 Edge-To-Edge Spaing To Nearest Blister LB, in () Bulge Diretion (inside/ outside) 6 8 inside outside outside Blister Projetion Bp, in Minimum Measured Thikness t mm, in Craking At Periphery (Yes/No) No Yes(inward) No Crown Craking or Venting (Yes/No) () Length Of Crown Crak or Diameter of Vent Hole, Spaing To Nearest Weld Joint L in (3) W, s in () No Crak Vent Spaing To Nearest Major Strutural Disontinuity Lmsd, in Notes:. The blister-to-blister spaing may affet the size of the blister to be used in the evaluation (see paragraph a & b). If the blister has rown raks, enter the length of the rak, see dimension s in Figure 7.6. If the blister has a vent hole, indiate as suh with the diameter of the hole (see Figure 7.7). 3. See Figure

171 Perform a Level Assessment per paragraph on Blister A a) STEP Determine the wall thikness to be used in the assessment. t = trd FCA t = =.05in b) STEP Determine the information in paragraph ( ) D = ID + FCA = = 96.5 in The blister-to-blister spaing hek satisfies "sx " ( L = 8.0 in) ( t =.03in) b s = 0.0in = 8.0in box riteria ) STEP 3 Chek the blister aeptane riteria, blisters are aeptable without repair if all of the following are satisfied: ) The blister diameter and venting requirements meet one of the following: ( [ ] ) ( max [, ] = 0.0 ) ( 0.6 = 5.93 ) max s, = 0.0 in.0 in False s in D t in and is vented False ) The minimum measured undamaged thikness measured from the non-bulged surfae satisfies the follow. ( = 0.70 ) ( 0.5 = ) t in t in True mm 3) The blister projetion satisfies the following. ( p =.5 ) ( 0. min [, ] = 0.8 ) B in s in False 4) The blister has no periphery raks. True 5) The distane between the edge of the blister and the nearest weld seam satisfies Equation (7.0). ( = 0.0 ) max[ =.03,.0 ] L in t in in True w 6) The distane from the blister edge to the nearest major strutural disontinuity satisfies Equation (7.). ( msd = 5.0 ) (.8 = 7.79 ) L in D t in True Therefore, Level Assessment riteria are not satisfied. 7-4

172 Perform a Level Assessment per paragraph on Blister A. a) STEP See Level, STEP b) STEP See Level, STEP ) STEP 3 See Level, STEP 3, item 6) d) STEP 4 The blister has no periphery raks. Proeed to STEP 6. e) STEP 6 The blister has a rown rak. Proeed to STEP 9. f) STEP 9 Evaluate the blister as an equivalent loal thin area using the proedures in Part 5. ) STEP 9. Determine the remaining thikness ratio and the longitudinal flaw length parameter. λ. D= 96 in tmm = 0.70 in FCA = 0.5 in s = 0 in Lmsd = 5 in t =.05in Design Pr essure = 50 psig RSFa = 0.90 tmm FCA Rt = = = t.05 D= 96 + ( FCA) = 96 + (0.5) = 96.5in.85s.85 0 λ = = =.3 Dt ( ) ( ) ) STEP 9. Chek the limiting flaw size riteria. ( Rt = ) 0.0 ( mm = = ) 0.0 ( ) ( ) True t FCA in in True ( ) L = 5in.8 Dt = = 7.79 in True msd 3) STEP 9.3 Determine the MAWP for the omponent (see A.3.4). Note that E =.0 sine the LTA is remote from weld seams (see paragraph A..5.b) of Annex A) D 96.5 R = = = 48.5 in C SEt 7500(.0)(.05 ) MAWP = = = psi R+ 0.6t (.05) L SE ( t tsl) 7500(.0)( ) MAWP = = = psi R 0.4 t t ( ) sl ( ) [ ] C L MAWP = min MAWP, MAWP = min 364 psi,744 psi = 364 psi 7-5

173 4) STEP 9.4 Evaluate the longitudinal extent of the flaw. From Part 5 Figure 5.6 with λ =.3 R t = aeptable. Using Table 5. and equation (5.): M t =.36, the longitudinal extent of the flaw is not R t RSF = = = < a = 0.9 ( Rt ) ( ) M t.36 RSF.8448 MAWPr = MAWP = 364 = 34 psig RSFa.9 ( RSF ) The LTA is aeptable at a MAWP r of 34 psig. g) STEP 0 Evaluate irumferential extent of the flaw. ) STEP 0. From the irumferential CTP, = 8 in.85( 8 ) ( 96.5)(.05).85 λc = = =.06 Dt ) STEP 0. Chek the following onditions ( λ = ).06 9 D 96.5 = = t = = =.0 ( RSF ) ( EL ) ( E ) C 3) STEP 0.3 Calulate tensile strength fator, True True True True True E 4 3E C L 4 3 TSF = + = + =.8 RSF E L From Part 5, Figure 5.8 with aeptable. From Table 5.4, R t _min = 0. ( Rt = ) > ( Rt_min = 0.) λ C =.06 R t = , the irumferential extent of the flaw is 7-6

174 The irumferential extent of the flaw is aeptable. h) STEP See Level, STEP 3, item 5) i) STEP An in-servie monitoring program should be developed to monitor potential blister growth. The Level Assessment Criteria are satisfied. ( MAWP = 34 psi) > ( PDesign = 50 psi) but the MAWP is limited by MAWP adjusted for weld joint effiieny of 0.85=30 psi The equipment is fit for ontinued operation. Perform a Level Assessment per paragraph on Blister B a) STEP Determine the wall thikness to be used in the assessment. t =.05 in b) STEP Determine the information in paragraph The blister-to-blister spaing hek satisfies "sx " box riteria ( ) L = 8.0 in.03 in b s = 5.5 in = 5.0 in ) STEP 3 Chek the blister aeptane riteria, blisters are aeptable without repair if all of the following are satisfied: ) The blister diameter and venting requirements meet one of the following: ( [ ] ) ( [ ] ) max s, = 5.5 in.0 in False max s, = 5.5 in 5.93 in and is vented True ) The minimum measured undamaged thikness measured from the non-bulged surfae satisfies the follow. ( ) t = 0.80in in True mm 3) The blister projetion satisfies the following. ( p = 0.3 ) ( 0. min [, ] = 0.5 ) B in s in True 4) The blister has no periphery raks. 5) The distane between the edge of the blister and the nearest weld seam satisfies Equation (7.0). ( ) L = 5.0 in.03 in True w 6) The distane from the blister edge to the nearest major strutural disontinuity satisfies Equation (7.). ( ) L = 0.0in 7.79 in True msd Therefore, Level Assessment riteria are satisfied. 7-7

175 Perform a Level Assessment per paragraph on Blister C a) STEP Determine the wall thikness to be used in the assessment. t =.05 in b) STEP Determine the information in paragraph The blister-to-blister spaing hek satisfies "sx " box riteria ( ) L = in.03 in b s = 5.0 in = 5.0 in ) STEP 3 Chek the blister aeptane riteria, blisters are aeptable without repair if all of the following are satisfied: ) The blister diameter and venting requirements meet one of the following: ( [ ] ) ( [ ] ) min s, = 5.0 in.0 in False min s, = 5.0 in 5.93 in and is vented True ) The minimum measured undamaged thikness measured from the non-bulged surfae satisfies the follow. ( ) t = 0.80 in in True mm 3) The blister projetion satisfies the following. ( p = 0.4 ) ( 0. min [, ] = 0.5 ) B in s in True 4) The blister has no periphery raks. True 5) The distane between the edge of the blister and the nearest weld seam satisfies Equation (7.0). ( ) L = 5.0 in.03 in True w 6) The distane from the blister edge to the nearest major strutural disontinuity satisfies Equation (7.). ( ) L = 0.0 in 7.79 in True msd Therefore, Level Assessment riteria are satisfied. Perform a Level Assessment per paragraph on Blister D a) STEP Determine the wall thikness to be used in the assessment. t =.05 in b) STEP Determine the information in paragraph The blister-to-blister spaing hek satisfies "sx " box riteria ( ) L = 0.0 in.03 in b 7-8

176 s =.0 in = 0.0 in ) STEP 3 Chek the blister aeptane riteria, blisters are aeptable without repair if all of the following are satisfied: ) The blister diameter and venting requirements meet one of the following: ( [ ] ) ( [ ] ) max s, =.0 in.0 in False max s, =.0 in 5.93 in and is vented False ) The minimum measured undamaged thikness measured from the non-bulged surfae satisfies the follow. ( ) t = 0.60 in in True mm 3) The blister projetion satisfies the following. ( p = 0.8 ) ( 0. min [, ] =.0 ) B in s in True 4) The blister has no periphery raks. True 5) The distane between the edge of the blister and the nearest weld seam satisfies Equation (7.0). ( ) L = 8.0 in.03 in True w 6) The distane from the blister edge to the nearest major strutural disontinuity satisfies Equation (7.). ( ) L = 30.0 in 7.79 in True msd Therefore, Level Assessment riteria are not satisfied. 7-9

177 Perform a Level Assessment per paragraph on Blister D. a) STEP See Level, STEP b) STEP See Level, STEP ) STEP 3 See Level, STEP 3, item 6) d) STEP 4 The blister has no periphery raks. Proeed to STEP 6. e) STEP 6 The blister has a rown rak. Proeed to STEP 9. f) STEP 9 Evaluate the blister as an equivalent loal thin area using the proedures in Part 5. Perform a Level Assessment per paragraph for STEP 9 ) STEP 9. Determine the CTP (Critial Thikness Profiles) (see paragraph ) D= 96 in t =.05 in FCA = 0.5 in tmm = 0.60 in Lmsd = 5 in s = in Design Pr essure = 50 psig RSFa = 0.90 ) STEP 9. Determine the remaining thikness ratio and the longitudinal flaw length parameter. λ. tmm 0.6 Rt = = = 0.59 t.05 D= 96 + ( FCA) = 96 + (0.5) = 96.5 in ( ) ( ).85s.85 λ = = =.56 Dt ) STEP 9.3 Chek the limiting flaw size riteria. ( Rt = 0.59) 0.0 ( mm = 0.6 ) 0.0 ( ) ( ) True t in in True ( ) L = 30in.8 Dt = = 7.4 in True msd 4) STEP 9.4 Determine the MAWP for the omponent (see A.3.4). Note that E =.0 sine the LTA is remote from weld seams (see paragraph A..5.b) of Annex A) D 96.5 R = = = 48.5 in C SEt 7500(.0)(.05 ) MAWP = = = psi R+ 0.6t (.05) L SE ( t tsl) 7500(.0)( ) MAWP = = = psi R 0.4 t t ( ) sl ( ) [ ] C L MAWP = min MAWP, MAWP = min 364 psi,744 psi = 364 psi 7-0

178 5) STEP 9.5 Evaluate the longitudinal extent of the flaw. From Part 5 Figure 5.6 with λ =.56 R t = 0.59 aeptable. Using Table 5. and equation (5.): M t =.4, the longitudinal extent of the flaw is not R 0.59 t RSF = = = 0.83 < a = 0.9 ( Rt ) ( 0.59) M t.4 Sine the alulated RSF in Part, paragraph.4... < RSFa, a r RSF 0.83 MAWPr = MAWP = 364 = 336 psig RSFa 0.9 ( RSF ) MAWP must be alulated using the equations The LTA is aeptable at a MAWPr of 336 psig. g) STEP 0 Evaluate irumferential extent of the flaw. ) STEP 0. From the irumferential CTP, = 8 in.85( 0) ( 96.5)(.05).85 λc = = =.3 Dt ) STEP 0. Chek the following onditions ( λ = ).3 9 D 96.5 = = t = = =.0 ( RSF ) ( EL ) ( E ) C 3) STEP 0.3 Calulate tensile strength fator, True True True True True E 4 3E C L 4 3 TSF = + = + =.05 RSF E L 0.83 From Part 5, Figure 5.8 with aeptable. From Table 5.4, R t _min = 0. λ C =.3 R t = 0.59, the irumferential extent of the flaw is 7-

179 ( Rt = 0.59) > ( Rt_min = 0.) The irumferential extent of the flaw is aeptable. h) STEP See Level, STEP 3, item 5) i) STEP An in-servie monitoring program should be developed to monitor potential blister growth. The Level Assessment Criteria are satisfied. ( MAWPr = 336 psi) > ( PDesign = 50 psi) but the MAWP r is limited by MAWP adjusted for weld joint effiieny of 0.85=30 psi The equipment is fit for ontinued operation. Perform a Level Assessment per paragraph on Blister E a) STEP Determine the wall thikness to be used in the assessment. t =.05 in b) STEP Determine the information in paragraph The blister-to-blister spaing hek satisfies "sx " box riteria ( ) L = 6.0 in.03 in b s =.0 in = 4.0 in ) STEP 3 Chek the blister aeptane riteria, blisters are aeptable without repair if all of the following are satisfied: ) The blister diameter and venting requirements meet one of the following: ( [ ] ) ( [ ] ) max s, = 4.0 in.0 in False max s, = 4.0 in 5.93 in and is vented True ) The minimum measured undamaged thikness measured from the non-bulged surfae satisfies the follow. ( ) t = 0.9 in in True mm 3) The blister projetion satisfies the following. ( p = 0. ) ( 0. min [, ] = 0. ) B in s in True 4) The blister has no periphery raks. True 5) The distane between the edge of the blister and the nearest weld seam satisfies Equation (7.0). ( ) L = 0.0 in.03 in True w 6) The distane from the blister edge to the nearest major strutural disontinuity satisfies Equation (7.). 7-

180 ( ) L = 40.0 in 7.79 in True msd Therefore, Level Assessment riteria are satisfied. Perform a Level Assessment per paragraph on Blister F a) STEP Determine the wall thikness to be used in the assessment. t =.05 in b) STEP Determine the information in paragraph The blister-to-blister spaing hek satisfies "sx " box riteria ( ) L = 6.0 in.03 in b s =.0 in =.0 in ) STEP 3 Chek the blister aeptane riteria, blisters are aeptable without repair if all of the following are satisfied: ) The blister diameter and venting requirements meet one of the following: ( [ ] ) ( [ ] ) max s, =.0 in.0 in True max s, =.0 in 5.93 in and is vented False ) The minimum measured undamaged thikness measured from the non-bulged surfae satisfies the follow. ( ) t = 0.60 in in True mm 3) The blister projetion satisfies the following. ( p = 0. ) ( 0. min [, ] = 0. ) B in s in True 4) The blister has no periphery raks. True 5) The distane between the edge of the blister and the nearest weld seam satisfies Equation (7.0). ( ) L = 6.0 in.03 in True w 6) The distane from the blister edge to the nearest major strutural disontinuity satisfies Equation (7.). ( ) L = 40.0 in 7.79 in True msd Therefore, Level Assessment riteria are satisfied. Perform a Level Assessment per paragraph on Blister G a) STEP Determine the wall thikness to be used in the assessment. t =.05 in b) STEP Determine the information in paragraph The blister-to-blister spaing hek satisfies "sx " box riteria 7-3

181 ( ) L =.0 in.03 in b s =.0 in s = 8.0 in ) STEP 3 Chek the blister aeptane riteria, blisters are aeptable without repair if all of the following are satisfied: ) The blister diameter and venting requirements meet one of the following: ( [ ] ) ( [ ] ) max s, =.0 in.0 in False max s, =.0 in 5.93 in and is vented False ) The minimum measured undamaged thikness measured from the non-bulged surfae satisfies the follow. ( ) t = 0.6 in in True mm 3) The blister projetion satisfies the following. ( p = 0.3 ) ( 0. min [, ] = 0.8 ) B in s in True 4) The blister has no periphery raks. False 5) The distane between the edge of the blister and the nearest weld seam satisfies Equation (7.0). ( ) L = 3.0 in.03 in True w 6) The distane from the blister edge to the nearest major strutural disontinuity satisfies Equation (7.). ( ) L = 4.0 in 7.79 in True msd Therefore, Level Assessment riteria are not satisfied. 7-4

182 Perform a Level Assessment per paragraph on Blister G a) STEP See Level, STEP b) STEP See Level, STEP ) STEP 3 See Level, STEP 3, item 6) d) STEP 4 Inspetion information gathered indiates periphery raking inward from an external blister; therefore, a Level Assessment annot be performed. Level Assessment riteria are not satisfied; therefore, a Level 3 Assessment onsisting of a detailed stress analysis must be onduted. Perform a Level Assessment per paragraph on Blister H a) STEP Determine the wall thikness to be used in the assessment. t =.05 in b) STEP Determine the information in paragraph The blister-to-blister spaing hek satisfies "sx " box riteria ( ) L = 8.0 in.03 in b s = 4.0 in s = 8.0 in ) STEP 3 Chek the blister aeptane riteria, blisters are aeptable without repair if all of the following are satisfied: ) The blister diameter and venting requirements meet one of the following: ( [ ] ) ( [ ] ) max s, = 4.0 in.0 in False max s, = 4.0 in 5.93 in and is vented False ) The minimum measured undamaged thikness measured from the non-bulged satisfies the follow. ( ) t = 0.55 in in True mm 3) The blister projetion satisfies the following. ( p =.5 ) ( 0. min [, ] =.8 ) B in s in True 4) The blister has no periphery raks. True 5) The distane between the edge of the blister and the nearest weld seam satisfies Equation (7.0). ( ) L = 6.0 in.03 in True w 6) The distane from the blister edge to the nearest major strutural disontinuity satisfies Equation (7.). ( ) L = 5.0 in 7.79 in True msd Therefore, Level Assessment riteria are not satisfied. A Level Assessment is required. 7-5

183 --``,,`,,,,`,,,,` API 579-/ASME FFS- 009 Fitness-For-Servie Example Problem Manual Perform a Level Assessment per paragraph on Blister H a) STEP See Level, STEP b) STEP See Level, STEP ) STEP 3 See Level, STEP 3, item 6) d) STEP 4 The blister has no periphery raks. Proeed to STEP 6. e) STEP 6 The blister does not have a rown rak. Proeed to STEP 7. f) STEP 7 The blister projetion riteria is satisfied. See Level, STEP 3, item 3). Proeed to STEP 8. g) STEP 8 The blister is vented. Proeed to STEP 0. h) STEP 0 See Level, STEP 3, item 5) i) STEP An in-servie monitoring program should be developed to monitor potential blister growth. Therefore, Level Assessment riteria are satisfied. 7-6

184 7.3 Example Problem 3 An AUT (Automated UT) inspetion was performed on a pressure vessel in hydrogen harging servie. Two areas with a varying degree of HIC damage were identified by AUT. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division, 989 Edition. The plant has deided to use weld overlay to stop future hydrogen harging of the steel. Determine if the vessel is fit for ontinued operation. Vessel Data Material = SA-56 Grade70 Year 980 Design Conditions = 48 F Inside Diameter = 74 in Nominal Thikness =.0 in Measured Uniform Thikness = 0.9 in FCA = 0.0 in Weld Joint Effiieny = 0.85 Supplemental Loads = 0( negligible ) Frature Evaluation Temp = 00 F Inspetion Data Data on HIC damaged areas with inreasing severity are given below. These HIC areas are at least 50 in from one another. Eah of the HIC area is loated a minimum distane of 30 in away from the nearest strutural disontinuity and 0 in away from a weld. HIC Area Longitudinal Length = 6 in Cirumferential Length = 7 in HIC to ID surfae = 0.5 in HIC to OD surfae = 0.35 in HIC Area Longitudinal Length = in Cirumferential Length = 0 in HIC to ID surfae = 0.5 in HIC to OD surfae = 0.35 in 7-7

185 Perform a Level Assessment per paragraph HIC Area a) STEP Determine the wall thikness to be used in the assessment. t = t FCA= 0.9 in rd ( L = 50.0 in) ( 8t = 7. in) H Therefore, s = 6.0 in = 7.0 in b) STEP Determine the information in paragraph HIC spaing to the nearest HIC or blister LH = 50.0 in HIC spaing to weld joints Lw = 0.0 in HIC spaing to major strutural disontinuities Lmsd = 30.0 in Minimum remaining wall thikness of undamaged metal, internal side, t mm ID = 0.5 in Minimum remaining wall thikness of undamaged metal, external side, t mm OD = 0.35 in HIC through-wall extent of damage w = t t t = 0.3 in H mm ID mm OD ) STEP 3 Chek all the onditions listed below. ) The planar dimensions of the HIC damage LOSS = t t = 0. in nom rd D = ID+ ( LOSS + FCA) = 74. in ( = 6.0 ) ( 0.6 = 7.5 ) ( = 7.0 ) ( 0.6 = 7.5 ) s in D t in True in D t in True ) The through-thikness extent of the damage t wh = 0.3 in min, 0.5 in = 0.3 in True 3 ( ) 7-8

186 3) The HIC damage is not surfae breaking ( tmm ID = 0.5 in) ( 0.t = 0.8 in) True t = 0.35 in 0.t = 0.8 in True ( ) ( ) mm OD 4) The distane between the edge of the HIC damage and the nearest weld seam L = 0.0 in > max t,.0in =.8 in True ( w ) ( [ ] ) 5) The distane from the edge of the HIC damage to the nearest major strutural disontinuity ( msd = 30.0 ) (.8 =.54 ) L in D t in True 6) Further HIC damage has been prevented Weld overlay will be applied True The Level Assessment Criteria are Satisfied. Perform a Level Assessment per paragraph HIC Area a) STEP Determine the wall thikness to be used in the assessment. t = 0.9 in ( ) L = 50.0 in 7. in H Therefore, s =.0 in = 0.0 in b) STEP Determine the information in paragraph HIC spaing to the nearest HIC or blister LH = 50.0 in HIC spaing to weld joints Lw = 0.0 in HIC spaing to major strutural disontinuities Lmsd = 30.0 in Minimum remaining wall thikness of undamaged metal, internal side, t mm ID = 0.5 in Minimum remaining wall thikness of undamaged metal, external side, t mm OD = 0.35 in HIC through-wall extent of damage w = t t t = 0.3 in H mm ID mm OD 7-9

187 ) STEP 3 Chek all the onditions listed below. ) The planar dimensions of the HIC damage ( ) ( ) s =.0 in 7.5 in False = 0.0 in 7.5 in False ) The through-thikness extent of the damage ( ) w = 0.3 in 0.3 in True H 3) The HIC damage is not surfae breaking ( tmm ID = 0.5 in) 0.8 in True t = 0.35 in 0.8 in True ( ) mm OD 4) The distane between the edge of the HIC damage and the nearest weld seam ( ) L = 0.0 in >.8 in True w 5) The distane from the edge of the HIC damage to the nearest major strutural disontinuity ( ) L = 30.0 in.54 in True msd 6) Further HIC damage has been prevented Weld overlay will be applied HIC Area is not aeptable per the Part 7 Level Assessment Criteria. True Perform a Level Assessment per paragraph HIC Area a) STEP See Level, STEP b) STEP See Level, STEP ) STEP 3 See Level, STEP 3, item 4) d) STEP 4 See Level, STEP 3, item 5) e) STEP 5 See Level, STEP 3, item 3) damage is lassified as sub surfae f) STEP 6 Determine the MAWP of the omponent per Annex A, paragraph A. ( 7500)( 0.9) ( ) SEt MAWP = 80 psi R+ 0.6t = = g) STEP 7 Calulate the RSF based on the sub surfae HIC damage. The minimum longitudinal distane to the nearest region of HIC damage is inhes. LHs 50.0 LR = min,8t = min,8( 0.9) = 7. in ( ) ( 0.3)( 0.8) wh D H LR + s 7. ( ) ( ) t RSF = = = L + s 7. + R 7-30

188 h) STEP 8 Evaluate the longitudinal extent of the flaw. Sine ( = ) ( = 0.9) RSF RSF False a RSF MAWPr = MAWP = 80 = psi RSFa 0.9 then the longitudinal extent of the HIC damage satisfies the LTA portion of the assessment. with the MAWP from STEP 8. r i) STEP 9 Evaluate the irumferential extent of the HIC damage as an LTA using the proedures in Part 5, paragraph See Example Problem 5.3 STEP 0 for the omplete proedure. The depth used in this analysis is given by ( )( ) d = w D = = 0.4 in. HIC H H Per the results of the LTA analysis of the irumferential extent of the HIC damage, MAWP = 0 psi. Sine MAWP from STEP 9 is greater than PDesign = 48 psi then the irumferential extent of the flaw is aeptable. j) STEP 0 Determine whether a frature assessment is required. This is the ase if any of the following are true. ) The equipment will remain in hydrogen servie (overlay applied). False ) The HIC damage is surfae breaking. False w = 0.3 in > 0.3 in False 3) ( ) H A rak like flaw assessment does not need to be performed. Proeed to STEP. k) STEP Confirm that further HIC damage has been either prevented or is limited to a known or verifiable rate based on one of the methods provided. The Level Assessment Criteria are satisfied. ( MAWPr = 75 psi) ( PDesign = 48 psi) but the MAWPr is limited by MAWP adjusted for weld joint effiieny of 0.85=53 psi The equipment is fit for ontinued operation. 7-3

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190 PART 8 ASSESSMENT OF WELD MISALIGNMENT AND SHELL DISTORTIONS EXAMPLE PROBLEMS 8. Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem A NPS 36 long seam welded pipe is to be used on a refinery projet. Inspetion of the pipe indiates peaking at the long seam weld. The pipe was onstruted to ASME B3.3. Determine if the pipe is suitable for servie. Pipe Data Material = ASTM A-69 Grade - 4Cr Year 990 Pipe Outside Diameter = 36 in Wall Thikness = 0.5 in Design Pressure = 35 psig Design Temperature = 800 F Joint Effiieny = 00 % Future Corrosion Allowane = 0.05 in Inspetion Data Peaking distortion δ = 0.3 in Perform a Level Assessment per Part 8 paragraph Limitations for weld peaking misalignment are not speified in ASME B3.3 (see Part 8 Table 8.4). Typially, the rules for out-of-roundness are applied to this type of misalignment. ( ) ( ) ( ) Dmax Dmin = = 0.3 in 0.0D = 0.36 in True The Level Assessment is Satisfied if the Out-Of-Roundness Criterion is Applied 8-

191 Perform A Level Assessment per Part 8 paragraph a) STEP Identify the omponent and weld misalignment type (see Part 8 Table 8.0) and determine the following variables as appliable (see Figures 8., 8.3, and 8.4) The weld misalignment is peaking whih ours on a longitudinal weld seam. The following data is required for the assessment: D = 36 in LOSS = 0.0 in o tnom = 0.5 in FCA = 0.05 in P = 35 psig δ = 0.3in Ey 6 = 5.(0 ) psi S = 6,800 a psi ν = 0.3 H = 3.0 b) STEP - Determine the wall thikness to be used in the assessment t = t LOSS FCA = = 0.45 in nom ) STEP 3 Determine the membrane stress based on the urrent design pressure (see Annex A, Equation (A.90) ). MA= 0.0 in Y B3 = 0.4 ( ) σ m = 0.4 =, 474 psi.0 ( )( ) d) STEP 4 Calulate the ratio of the indued bending stress to the applied membrane stress using the equations in Part 8 Table 8.0 based on loal peaking R = = in S p 3 {( ν ) PR } ( ) { }( )( ) { 5.5 ( 0) }( 0.45) f (7.775 = = =.88 Et δ 0.3 = = R (7.775) y S p =.88 From Figure 8.3, with δ C f 0.83, and = R ( )( ) ( ) ( 0.45) R = 0.83 = 3.43 lja b lj lja b b b R = R + R = R =.0 bs 8-

192 e) STEP 5 Determine the remaining strength fators set H = 3.0 (the indued bending stress is evaluated as a seondary stress) ( 3.0)( 6,800) (, 474)( ) + ( 0.0) + (.0) RSF = min,.0 = 0.9 ( ) f) STEP 6 Evaluate the results. ( = 0.9) ( = 0.90) RSF RSF True The Level Assessment Criterion is Satisfied. a 8-3

193 8. Example Problem Determine if the pipe in the Example Problem an operate for,000 yles at 35 psig. Perform A Level Assessment Fatigue Analysis per Part 8 Paragraph Paragraph permits a Level assessment as long as the geometri flaw satisfies the requirements of the Part 8 paragraph The results of Example Problem shows that this restrition is met by the flaw sine the Level riterion for the assessment of the weld misalignment was satisfied. Additional Pipe Data Material Yield Strength = 5,00 psi@800 F a) STEP Determine the nature of the loading, the assoiated membrane stress and the number of operating yles. The loading onsists of pressure loading. From Example Problem the irumferential membrane stress is σ m =, 474 psi. The desired number of operating yles is 000. b) STEP Determine the ratio of the indued bending stress to the applied membrane stress, R b. lj b R = 0.0 sine enterline offset is not present lja b R = 3.43 or b R = 0.0 sine neither general or arbitrary out-of-roundness is present. lj lja or b b b R = R + R + R = = 3.43 b ) STEP 3 Using the loading history and membrane stress from STEP and the R b from STEP, alulate the stress range for the fatigue analysis using Table 8.. From Table 8., for a ylinder with a longitudinal weld joint with weld misalignment: Δ σ m = σ m = 474 psi lj lja or ( b b ) ( )( ) Δ σ = σ R + R + R =, = 4, 756 psi b m b lj lja or ( b b )( ) Δ S = σ + R + R + R K P m b f Sine we apply the fatigue strength redution fator when using Equation (B.30) below, we will set f.0 Δ S K = in the equation for P lj lja or ( b b )( ) ( )( )( ) Δ S = σ + R + R + R.0 =, = 55, 60 psi P m b d) STEP 4 Compute the number of allowed yles using the stress range determined in STEP 3. Table 8. referenes Annex B, paragraph B.5. Paragraph B.5 provides three methods for determining the permissible number of yles: ) Elasti Stress Analysis and Equivalent Strength in aordane with paragraph B.5.3 ) Elasti-Plasti Stress Analysis and Equivalent Strain in aordane with paragraph B.5.4 3) Elasti Stress Analysis and Strutural Stress in aordane with paragraph B

194 Sine an elasti-plasti stress analysis has not been onduted, the permitted number of yles will be determined using Methods and 3. In both ases the stresses onsidered onsist of those due to pressure loading, stresses from supplementary loads and thermal gradients are onsidered negligible. Method : For a fatigue assessment using an elasti stress analysis and equivalent stresses, STEPS through 3 in paragraph B.5.3. are similar to STEPS through 3 in paragraph with the exeption that the elasti stress range is alulated from the stress tensors and that the stress state from both mehanial and thermal loading are onsidered. For this example problem the stress range due to thermal loading is onsidered negligible and the mehanial loading onsists of internal pressure. Thus the stress range is given by STEP 3 and is 55,60 psi. STEP 4 Determine the effetive alternating stress from Equation (B.30), modified to ignore yli thermal stress, (i.e., Δ S LT = 0.0 ): Kf Ke ΔSP Salt = K f is a fatigue strength redution fator determined from Table B.0 based on type of weld and the quality level determined from Table B.. The quality level in Table B. is based on the type of inspetion performed on the weld. For the pipe material, the speifiation alled for full volumetri and full visual examination, but neither MT nor PT were performed on the weld. Thus from Table B. the quality level is 4. The weld being assessed is a full penetration weld. For a full penetration weld inspeted to quality level 4, Table B.0 stipulated a weld fatigue redution fator of K f =.0. The fator Ke is a fatigue penalty fator that may be determined from Equations (B.3) to (B.33) depending on the value of the stress range ΔSP ompared to the permitted primary plus seondary stress range, S PS. The value of S PS is the larger of three times the allowable stress at temperature or two times the material yield strength at the average temperature during a stress yle. The allowable stress at temperature, S a, equals 6,800 psi and the yield strength for the A-69 Grade -/4Cr material, S y, equals 5,00 psi at 800 F and 35,000 at ambient temperature. The average yield stress during the yle is thus 30,00 psi. ( )( ) ( )( ) SPS = max 3.0 Sa, Sy = max 3 6,800, 30,00 = 60, 00 psi Compare the value of ΔSP to S PS : ( Δ = 55, 60) ( = 60, 00) S S True P PS Therefore from Equation (B.3) K e = S alt ( )( )( 55, 60) Kf Ke ΔSP = = = 55, 60 psi STEP 5 Determine the permitted number of yles, N, for the alternating stress omputed in STEP 4 and the smooth bar fatigue urves as provided in Annex F, paragraph F.6... For temperatures not in the reep range, the permitted number of yles is given by Equation (F.4) and Equation (F.5): ( 0) X E N T = EFC 8-5

195 where X = Salt Salt Salt Salt Salt Cus Cus Cus Cus Cus Salt Salt Salt Salt S C alt + + C4 + C6 + C8 + C0 Cus Cus Cus Cus Cus C C C C C C C are given in Table F.3 for low allow steels where σ 80 The values of the oeffiients i ksi. Examining Table F.3, the values of C6 through C all equal zero. Substituting the values for C through 5 ( ) 3 E = ksi, FC C, S = 55.6 ksi, alt us UTS C =, 5.( 0) 3 E = ksi, and T + ( )( ) ( ) + ( )( ) ( ) 4 ( )( ) ( ) ( )( ) ( ) X = = and ( ) ( ) N = ( 0) =,80 yles Method 3: a) STEP Determine the load history for the omponent, onsidering all signifiant operating loads. The load applied to the pipe onsists of internal pressure, P, of 35 psig. b) STEP For the weld joint subjet to fatigue evaluation determine the individual number of stressstrain yles. The desired number of yles, N, is,000. ) STEP 3 Determine the elastially alulated membrane and bending stress normal to the hypothetial rak plane at the start and end of the yle. Using this data alulate the membrane and bending stress ranges between the time of maximum and minimum stress for the yle. From Example Problem the maximum membrane stress for the yle ours at a pressure of 35 psig, and the minimum membrane stress for the yle ours at zero pressure. Similarly, the maximum bending stress for the yle ours at a pressure of 35 psig, and the minimum bending stress for the yle ours at zero pressure. The values of the two stress ranges given by Equations (B.46) through (B.50) are: m e n e Δ σm = σm σm = =.474 ksi ( R )( ) ( )( ) m e n e m e n e b b b b m m Δ σ = σ σ = σ σ = = ksi m e m e n e n e ( ) ( ) ( ) ( ) σmax = max σ + σ, σ + σ = max , = 55.6 m b m b ksi m e m e n e n e ( ) ( ) ( ) ( ) σmin = min σ + σ, σ + σ = min , = 0 σ mean e m b m b ksi e σmax + σmin = = = 7.63 ksi 8-6

196 d) STEP 4 Determine the elastially alulated strutural stress range, Equation (B.5) e e e Δ σ =Δ σm +Δ σb = = 55.6 ksi e Δ σ, for the yle using e Δ ε, from the elastially alulated e) STEP 5 Determine the elastially alulated strutural strain, e strutural stress range, Δ σ, using Equation (B.5) and the elasti modulus for the material at the average temperature of 435 F, e e Δσ 55.6 Δ ε = = = E 4 ya ( ) ( ) 4 e and the values of the stress range, Δ σ, and strain range, Δ ε, by orreting Δσ and hysteresis stress-strain loop by solving Equations (B.53) and (B.54) simultaneously, where e e ( ) ( ) ( ) ( ) 4 Δσ Δ ε =Δσ Δ ε = = Δσ Δσ Δ ε = + Eya Kss n ss e Δ ε for Kss and nss are determined from Table F.8 in Annex F for the average temperature during the yle. The losest material to the ASTM A69 Grade 4 is the Cr--Mo-/4V material. The average value of Kss is given by, K ss ( )( ) ( ) Kss + Kss = = = The average value of n ss is given by, n ss ( )( ) ( ) nss + nss = = = 0.30 Substituting these values into the Equations (B.54) and (B.54) for simultaneously gives, Δ σ = ksi ( ) 3 Δ ε = Modify the value of Δσ for low-yle fatigue using Equation (B.55), ( ) ( ) 3 ( ( ) ) 4 E ya Δ σ = ε ksi Δ = = ν 0.3 Δσ and Δ ε, and solving them --``,,`,,,,`,,,,`,,``,`````- 8-7

197 f) STEP 6 Compute the equivalent strutural stress range Δ Sess using Equation (B.56) where the input parameters are as follows: Δ σ = tess R I b ksi = 0.65 in sine the omponent thikness, t = in Δσ b = = = Δ σ + Δ σ m b m R 0.7 ss b Rb Rb 0.78Rb ( ) ( ) ( ) ( ) = = =.757 min 0 R = σ = = 0 σ 55.6 max f =.0, sine R = 0 0 (see Equations (B.63) and (B.64)) M m = 3.6 ss Δσ Δ S = = = ksi ess ( ) m ss mss m ss ess M t I f g) STEP 7 Determine the permitted number of yles, N, using the value of Δ Sess from STEP 6 and the welded omponent fatigue urves in Annex F. The welded omponent fatigue urves are represented in Annex F by Equation (F.8): where f fmt C N = fe ΔSess h f =.0, sine no work has been done to improve the fatigue of the pipe longitudinal weld f = 4.0, sine the proess fluid is onsidered mildly aggressive E f MT ( ) ( ) EACS.94 0 = = =.484 E 4 T From Table F.9, for a lower 99% predition interval ( 3σ ), the values of C and h for low alloy steel are, C = 88.3 h = h ( )( ) f fmt C N = = = 3,505 yles fe ΔSess

198 h) STEP 8 Evaluate the omponent by omparing the number of permitted yles to the number of desired yles: Method : N =,80,000 True Method 3: N = 3,505,000 True The Level assessment for fatigue is satisfied by both Method and Method 3 8-9

199 8.3 Example Problem 3 An existing pressure vessel is being repaired during a shutdown. After field PWHT, inspetion of the vessel indiates that out-of-roundness along the length of the ylindrial setion of the vessel has ourred. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division. Determine if the vessel is suitable for servie. Vessel Data Material = SA 56 Grade 70 Year 998 Design Conditions = F Wall Thikness =.875 in Inside Diameter, D in = 0 in Joint Effiieny = 00 % FCA = 0.5 in Inspetion Data D max = 0.5 in D min = 9.4 in Based on other measurements, the deformed shape signifiantly deviates from the perfet oval shape. Perform A Level Assessment per Part 8 paragraph 8.4. ( ) ( ) ( ) ( ) ( )( ) Dmax Dmin = =. in 0.0D = =. in True The Level Assessment Criterion Is Satisfied Perform A Level Assessment per Part 8 paragraph a) STEP Determine the following variables based on the type of out-of-roundness. R = D = 0 / = 60 in LOSS = 0.0 in in tnom in =.875 in FCA = 0.5 in P = 500 psig θ = 0 (hosen beause this is the loation of the longitudinal weld seam) Ey ( ) 6 = 6. 0 psig S = 7,500 a psi ν = 0.3 H = 3.0 Dmax = 0.5 in Dmin = 9.4 in C s = 0. (the deformed shape signifiantly deviates from a perfet oval) b) STEP Determine the wall thikness to be used in the assessment. t = t LOSS FCA= =.75 in nom ) STEP 3 Determine the irumferential membrane stress based on the urrent design pressure (see Annex A). ( ) ( ) ( + ) ( ) σ m = = 7,479 psi.0.75 f 8-0

200 d) STEP 4 Determine the ratio of the indued irumferential bending stress to the irumferential membrane stress from Equation (8.): or Rb (.5)( ) ( os( (. )( 0. ))) = abs = ( 0.)( 500) ( ( 0.3) ) (.75) ( 0 ).75 e) STEP 5 Determine the remaining strength fator using Equation (8.): ( 3.0)( 7,500) ( 7479)( ) RSF = min,.0 = min[.8855,.0] =.0 f) STEP 6 Evaluate the results. If RSF RSFa, the out-of-roundness is aeptable per Level ; otherwise, refer to Part 8 paragraph ( =.0) ( = 0.90) RSF RSF True The Level Assessment Criterion Is Satisfied a 8-

201 8.4 Example Problem 4 On further inspetion of the vessel in Example Problem Number 3, the out-of-roundness was relassified as weld misalignment on one of the longitudinal seams. The weld misalignment is ategorized as enterline offset and loal peaking. Determine if the vessel is suitable for operation, and the maximum allowable working pressure. Inspetion Data Dmax = 0.5 in Dmin = 9.4 in Based on additional field measurements, the deformed shape signifiantly deviates from the perfet oval shape. The enterline offset and loal peaking were measured to be: e = 0.5 in ( enterlineoffset) δ = 0.60 in ( peaking) Perform A Level Assessment per Part 8 paragraph a) STEP The omponent is a ylindrial shell with enterline offset and peaking (angular) weld misalignment. The variables neessary to perform a Level Assessment were determined as part of Example Problem 3. b) STEP The wall thikness to use in the assessment was determined in Example Problem 3, STEP, and equals.75 in ) STEP 3 Determine the irumferential membrane stress based on the urrent design pressure (see Annex A) from Example Problem 3: σ = 7, 479 psi m d) STEP 4 Calulate the ratio of the indued bending stress to the applied irumferential membrane stress for weld misalignment using Part 8, paragraph lj R b for enterline offset misalignment, (see Table 8.9 for the equation to alulate S p and Table 8.0 lj for the equation to alulate R b ): S C C R p 3 {( ) PR } ( ) ( ) ( )( ) ( )( ) ( 6.( 0) )(.75) y ( ) ν = = =.98 Et = 3.839( 0) lj b 4 ( ) ( ) ( ) 6 3 ( ) = (0) (0) = ( 0) (.98) = C 0.47 = = = C

202 lja R b for peaking misalignment (see Table 8.0) S p =.98 δ 0.60 = = 0.00 R 6 From Figure 8.3, with ( ) ( ) ( ) R = 0.87 =.79 lja b S p =.98 δ C f 0.87, and = 0.00 R R b for enterline offset misalignment and peaking weld misalignment is: lja lj b b b R = R + R = =.4 e) STEP 5 Determine the remaining strength fator. ( 3.0)( 7500) ( 7479)( +.4) RSF = min,.0 = min[ 0.97,.0] = 0.93 f) STEP 6 Evaluate the results. ( = 0.97) ( = 0.90) RSF RSF True The Level Assessment Criterion is Satisfied. Thus, from Equation (.3), MAWPr MAWP 500 a = = psig 8-3

203 8.5 Example Problem 5 A vertial, ylindrial pressure vessel subjeted to an upset ondition has been inspeted and found to have deformed to an out-of-round shape along the length of about a third of the vessel. From the measurements, it appears that the deformation an be lassified as arbitrary out-of-roundness. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division. Determine if the vessel is suitable for servie. Vessel Data Material = SA-54 Grade B Year 004 Design Conditions = F t nom =.5 in D = 8 in Joint Effiieny = 00 % ( both irumferential and longituidinal ) FCA = 0.5 in LOSS = 0.0 in Allowable Stress = 3,00 psi Inspetion Data As part of the inspetion, twenty-four measurements of the distane to the vessel inside surfae were made from a point near the enter of the vessel. The distane from the point to the equally spaed loations at the vessel inside surfae, taken at fifteen degree inrements are shown in Table E8.5-. Table E8.5- also shows the diameter obtained by adding the radii of two points on opposite sides of the vessel, and the perent the diameter varies from the design diameter. Table E8.5- Measured Distanes To Vessel Inside Surfae Point Angle (degrees) Radius (in) Point Angle (degrees) Radius (in) Diameter (in) % Out-Of-Round

204 Perform a Level Assessment Table 8.3 shows that for a ylindrial shell under internal pressure, the value of Dmax Dmin shall not exeed one perent of the design diameter. For this to be satisfied, the absolute value of the algebrai differene between two values in the last olumn should not exeed.0. The maximum absolute algebrai differene between any two values ours between the 0-80 diameter and the diameter and is equal to [(.88) (-.8)] = 4.6. Therefore, the vessel out-of-roundness does not satisfy the Level assessment riterion. Perform a Level Assessment Classify the shell deformation as arbitrary out-of-roundness. a) STEP Using the measured radii from the inspetion results alulate the Fourier series oeffiients that represent the shape of the ylindrial shell using the method provided in Part 8, Table 8.. Before we an alulate the Fourier oeffiients we must apply Equations (8.5) through (8.9) to orret the measurements so that they aount for the differene between the enter of the vessel and the point from whih the measurements were made. This may be done using a spreadsheet or a omputer program written for this purpose, as shown in sub-steps. through 6. below: ) Apply Equation (8.7), inside radius, R m M M i = R m, equals 4 in. = Ri, to the twenty-four measured radii, to determine the mean ) Determine the values of A and B, the oeffiients of the seond terms in the Fourier series for the osine and sine funtions respetively. For A apply Equation (8.8) ( i ) M π An = Rios n M i= M and for B Equation (8.9) ( i ) M π Bn = Risin n M i= M with the value of n equal to. Doing this gives A = in and B = in 3) Use the values of A and B in Equation (8.5) ( i ) π( i ) π Ri = Ri Aos Bsin M M to determine the radii of the twenty-four measured points adjusted for the true enter of the vessel R. i 4) Determine the value of the orretion to eah of the twenty-four measured radii, ε i, from Equation (8.6) ε = R R. i i m 5) Using the previously determined values, alulate the adjusted radius at eah of the twenty-four loations using Equation (8.4) R( θ ) = R + A osθ + B sinθ + ε m 8-5

205 The values of i, θ i, in Table E8.5-. R i, ε i, and R( θ ) at eah of the twenty-four measured loations are shown Table E8.5- Corretions To Measured Radii i θ i (degrees) R i (in) ε i (in) R( θ ) (in) ) Using the values shown in the last olumn of this table (i.e., R( θ )) as a new value for Equations (8.8) and (8.9) determine the values of A and n R i in B n for n = to 4. This may be aomplished with either a spreadsheet or omputer program that implements the pseudo-ode of Table 8.. Beause there are 4 measurement points, there an only be twenty-six values total for both A n and B n, where A0 equals times the mean radius Rm and B 0 = 0 (i.e., A to A and B to B ). The values of the twenty-six oeffiients for the thirteen terms of the Fourier series are shown in Table E

206 Table E8.5-3 Fourier Coeffiients To Calulate The True Shape Of The Vessel Shell Index Fourier Term A n B n E E-06 b) STEP Determine the wall thikness to be used in the assessment, t, using Equation (8.0) or (8.) as appliable. t = t LOSS FCA= =.375 in nom ) STEP 3 Determine the irumferential membrane stress using the thikness from STEP (see Annex A). C P R σ m = = =,54 psi E t d) STEP 4 Determine the ratio of the indued irumferential bending stress to the irumferential membrane stress at the irumferential position (denoted by the angleθ ) of interest using Equation (8.3). where and R k or b n D ( θ ) = ( A os( nθ) + B sin( nθ) ) N 6 n n = t + k ( n ) n= n PR 3 Et = 3 y D ( ν ). 8-7

207 Calulate D, and n D ( ν ) k as a funtion of n, ( 7,50,000)(.375) ( ) 3 Et y = = = 6,64, ( ) ( ) ( )( ) 3 PR kn ( n) = = = n D n 6, 64,89.8 n ( )( ) ( ) 3 3 in-lb Substitute the appropriate values oft, R or b ( θ ) A n, 4 6 ( Anos( nθ) + Bnsin( nθ) ) = n= + ( n ) R θ. or Determine the maximum value of ( ) b Bn and kn into Equation (8.3) for eah value of θ. or The maximum value ours at θ = 300 degrees, where R max =.6607 and the minimum value ours at θ = 0 degrees where R or min =.3377 b e) STEP 5 Determine the remaining strength fator RSF using Equation (8.) and the value of or or ( ( max ) ( )) ( ) R = max abs R, abs R min = max.6607,.3377 =.6607, b b b R =.0, bs H = 3, and f σ = 0.0 psi (supplemental loads are negligible) ms H fsa RSF = min,.0 σm( + Rb) + σms( + Rbs) f) STEP 6 Evaluate the results, ( 3)( 3, 00) (,54)( ) + ( 0)(.0.0) [ ] = min,.0 = min.807,.0 =.0 ( =.0) ( = 0.90) RSF RSF True a ( 300)(.0)(.375) ( ) ( ) SEt MAWP = 77.5 psi 65 psi True R+ 0.6t = = The vessel satisfies the Level riteria. b 8-8

208 8.6 Example Problem 6 A pressure vessel has experiened general shell distortion in the ring stiffened ylindrial setion. The vessel is subjet to both internal pressure and external pressure. In addition, the stiffening rings provided for vauum servie have also been distorted. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division. Determine if the vessel is suitable for ontinued servie. Vessel Data Shell and Head Material = SA 85 C FBX Year 965 Design Pressure (top) = 5 psig (bottom) = 50.7 psig (entire vessel) = -5 psig (external) Design Temperature = 650 F Inside Diameter of Cylinder = 80 in Cone Height = 63 in Cylinder Tangent-Tangent = 938 in Cone Small Inside Diameter = 7 in Shell Wall Thikness = in Cone Wall Thikness = 0.65 in Hemispherial Head Thikness = in Future Corrosion Allowane = 0.5 in Allowable Stress = 3,750 psi Yield Stress = 3 ksi Tensile Stress = 55 ksi Joint Effiieny = 0.85 Inspetion Data The vessel was 360 degrees sanned by laser along its length from tangent line to tangent line. The inspetion revealed that the largest radial deformation was.5 in inwards and 5.5 in outward. Figure E8.6- shows the mapped laser san data on the north side (top plots) and south side (bottom plots) of the model. Figure E8.6-3 shows the shell distortion built into a finite element model. Perform a Level Assessment Table 8.3 shows that for a ylindrial shell under internal pressure, the value of Dmax Dmin shall not exeed one perent of the design diameter. For the ylindrial shell this would be.8 in. Assuming that the ylinder opposite the maximum inward or outward deformation is a true ylinder, the diametrial deviation would equal the maximum radial deformation. The maximum radial deformation of 5.5 in exeeds the permitted one perent deviation. Therefore, the Level assessment riteria are not satisfied. 8-9

209 Perform a Level Assessment Based on the measured shell deformation, the shell distortion was lassified as general shell distortion. Therefore it was deided to forgo a Level assessment and ondut a Level 3 assessment. Perform a Level 3 Assessment A Level 3 FFS assessment was onduted in aordane with Annex B. The assessment used a threedimensional shell finite element model as shown in Figure E Four proedures were followed: ) A Limit Load analysis in aordane with paragraph B..3 using elasti-perfetly plasti material behavior and linear geometry ) A hek of loal strain riteria in aordane with paragraph B.3.3 using a model with elastiplasti material properties that inluded strain hardening. 3) An elasti bukling analysis in aordane with paragraph B.4 to determine the strutural stability of the deformed shell 4) A hek of the fatigue requirements in aordane with paragraph B

210 C L t=0.500" t = " 80" I.R. C x 0.7 " 90" C x " 40" C x 0.7 C x " t = " 0" 40" W x 6 0" W x 6 t = " 80" 40" Liquid Level 70" from Tangent Line t = 0.78" 80" W x 40 40" 40" W x " Cone t=.03" 55.00" Cone t= 0.750" 40" I.R. 3" 8" " Cone 3 t= 0.406" 6.000" x 0.750" Plate.65" Nozzle t=0.750" Figure E8.6- Vessel Drawing 8-

211 Figure E Mapped Laser San Data 8-

212 Figure E Dimensional Shell Finite Element Model 8-3

213 Limit Load Analysis An elasti-perfetly plasti limit load stress analysis was performed on the model of the damaged ylinder. In the model and analysis it was assumed that the damaged stiffening rings would be replaed or repaired to adequately reinfore the shell. The loads applied to the model inluded the vessel weight, stati head from the ontents, and the internal pressure. These loads were inreased in aordane with Table B.3 by a fator of.35. This fator is.5 RSF a where the allowable remaining strength fator RSF a is taken equal to 0.9. In the model the applied loads were: Weight of 57,00 lb, based on a shipping weight of 383,000 lb. A 34.7 psig (.35 x 5.7 psig) hydrostati load applied beginning at an elevation of 94 inhes from the bottom tangent line A psig (.35 x 5 psig) onstant pressure, applied to the model in addition to the stati head The Limit Load finite element analysis onverged to a solution indiating that the deformed shell was adequate for the imposed loading. Loal Strain Criteria An elasti-plasti analysis was performed using the loads stipulated for loal strain riteria in Table B.4. The loal strain riteria require that the loads be fatored by.7 times the allowable remaining strength fator, giving a fatored load of.53. Thus the fatored loads were: Weight of 586,000 lb, based on a shipping weight of 383,000 lb. A 39.4 psig (.53 x 5.7 psig) hydrostati load applied beginning at an elevation of 94 in from the bottom tangent line A 38.3 psig (.53 x 5 psig) onstant pressure, applied to the model in addition to the stati head The equivalent maximum plasti strain was determined and shown to beε peq = From Equation (B.6) the permitted strain from fabriation and applied loading is given by, ε ε exp α σ + σ + σ 3 3 sl 3 L = Lu + m σ e From Table B.6, σ ys 3 ε Lu = m = 0.6 = 0.6 = σ uts 55 α =. sl From the elasti-plasti analysis, the values of the prinipal stresses and equivalent stress at the point of evaluation were: σ = 3.34 ksi σ = 6.73 ksi σ 3 = 6.8 ksi σ = 4. ksi e Substituting into the equation for ε Lu resulted in ε L = exp = ( 4.) 3 The old forming strain was alulated from the radius of urvature, ρ, and the thikness, t,as 8-4

214 ε f = t ρ = 90 = Chek the riteria of Equation (B.7). ε peq + εf εl ( ) = True The loal strain riterion of Equation (B.7) is met. Bukling Analysis Sine the pressure vessel is subjet to external pressure, it was neessary to determine the deformed shells stability. For this purpose a linear elasti bukling analysis in aordane with Paragraph B.4 was onduted to determine the ritial eigenvalue bukling modes and assoiated bukling pressures for the vessel. The bukling analysis was aomplished in two steps. The first step onsisted of a preload that inluded the vessel weight of 383,000 lb, applied as a body load, along with an initial external pressure of - psig. This first step produed displaements in the vessel that formed the basis for linear perturbation eigenvalue bukling analysis. In the seond step a perturbation external pressure of - psig was applied and a linear eigenvalue bukling analysis that sought the first three bukling modes and eigenvalues was onduted. In the linear perturbation analysis the finite element program sales the perturbation load by multipliers that produe a solution to the eigenvalue problem (i.e., the eigenvalues). The ritial bukling loads were then obtained by adding the preload pressure of - psig to the perturbation load saled by the eigenvalue. The first bukling mode was identified as the ritial mode and its eigenvalue plus the initial - psig as the ritial bukling pressure. The first three bukling modes are shown in Figure E The ritial bukling pressure for the deformed shell geometry was alulated as -6.9 psig. For bifuration bukling performed using an elasti stress analysis without geometri non-linearities, a apaity redution fator Φ = / β B r shall be used to determine the permissible external load. The permissible external pressure is the ritial bukling pressure divided by the apaity redution fator. For unstiffened and ring stiffened ylinders, Therefore, β = r Φ = / 0.80 =.5. B Using this fator, the permissible external pressure is Pext = 6.9 /.5 = 6.75 > 5 psig external design pressure Sine the permissible external pressure P ext exeeds the design external pressure of 5 psig, the deformed ylindrial shell is adequate for ontinued servie. 8-5

215 Mode : 7.7 psig Mode 3: 9. psig Mode : 6.9 psig Figure E Lowest Three Bukling Modes Fatigue Assessment To determine whether fatigue was a onern a fatigue sreening in aordane with B.5..4, Fatigue Analysis Sreening Method B was performed. Only pressure loads were onsidered. The smooth bar fatigue urves in Annex F were used for this purpose. Thermal stresses were onsidered to be neglible. a) STEP Determine the number of full range pressure yles for the vessel The pressure vessel is filled with atalyst one a week and emptied one every three weeks. During filling operations, the vessel experienes its -5 psig external pressure and while it is emptied, the pressure at the top of the vessel is 5 psig. Thus the pressure vessel experienes 5 vauum pressure yles and 7 internal pressure yles per year for a total of 69 pressure yles per year. b) STEP Determine the fatigue sreening fators, C and C based on the type of onstrution in aordane with Table B.9. Table B.9 shows that for omponents with a flaw as haraterized by Part 8, the values of C andc are given by: 3 3 C = = = 3.33 RSFa 0.90 C = = =. RSF 0.90 a ) STEP 3 Based on the number of yles determined in STEP, and the allowable stress of the material S ompare the number of full range yles to the number of permitted yles: m ( ) ( ) ( ) ( )( ) NΔ FP N CS m = N , 750 = N 45,

216 N ( 45,833) API 579-/ASME FFS- 009 Fitness-For-Servie Example Problem Manual may be determined from Equation (F.4) and (F.5) or from logarithmi interpolation using the data from Table (F.) adjusted for the modulus of elastiity at the assessment temperature of 650 F to the fatigue urve modulus at 700 F. For arbon steel the oeffiients of Equation (F.5) for an alternating stress between 3 ksi and 580 ksi are: ( ) ( ) 4 5 ( ) C ( ) = = = C C C C = = Substituting into Equation (F.5) and Equation (.4) respetively Dividing N by ( ) + ( ( 0) )( ) + ( ( 0) )( ) 4 + ( ( 0) )( ) + ( ( 0) )( ) 5 X = = ( ) ( ) ,708 6 N = = FP yles N Δ FP we an determine the number of years that the vessel may be used, N 5, 708 = = 8.7 years N Δ 69 Based on this and the other assessment riteria, the vessel satisfies the Level 3 riteria and may be put bak in servie. 8-7

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218 PART 9 ASSESSMENT OF CRACK-LIKE FLAWS EXAMPLE PROBLEMS 9. Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem Example Problem A rak-like flaw has been found on a ylindrial shell of a pressure vessel during a sheduled turnaround. The vessel and inspetion data are provided below. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division, 00 Edition. Determine if the vessel is aeptable for ontinued operation using a Level Assessment. Vessel Data Material = SA-56 Grade 70 Year 00 Design Conditions = F Inside Diameter = 96 in Fabriated Thikness =.5 in Uniform Metal Loss = 0.0 in FCA = 0.5 in Weld Joint Effiieny =.0 PWHT = Yes, Original Fabriation Requirement Operating Conditions The vessel is not fully pressurized until the temperature is 00 o F. Below this temperature, the startup pressure remains under 40 psig. At shutdown, the pressure is dereased to 40 psig before letting the temperature drop below 00 F. Inspetion Data The flaw is loated in a longitudinal weld seam on the inside surfae of a ylindrial vessel. The flaw is parallel to the weld joint. The longitudinal seam is a double V-groove weld. The depth of the flaw was established by UT; however, many different values were obtained during the inspetion with a maximum value of 0.5 in being reported. The flaw length was established by MT and is. in. The distane of the rak-like flaw to the nearest strutural disontinuity is 60 in. Perform a Level Assessment per paragraph First, hek that the onditions to perform a Level Assessment are satisfied Geometry: Component is a flat plate, ylinder or sphere: (ylinder) True 9-

219 Cylinder with R/ t 5 (t being the urrent thikness) t = tnom LOSS =.5 0.0=.5 in R/ t = 48/.5 = R= D/= 96/= 48in Wall thikness at the loation of the flaw is less than.5 in: (.5 in.50 in) True < True Flaw of surfae or through-thikness type with a maximum rak-length of 8 in: Surfae rak with length equal to. in Cylindrial shell: flaw oriented in the axial or irumferential diretion: (longitudinal = axial) With a distane to the nearest strutural disontinuity greater than or equal to.8 Dt Lmsd = 60 in Lmsd.8.8 Dt =.8 (96)(.5) = 8.99 in Loads: Pressure produing only a membrane stress field Membrane stress within the design limits of original onstrution ode Welded joint is single or double V: (double V-groove weld) Material: Carbon Steel (P, Group or ) with S 5 ksi, σ 40 ksi and σ 70 ksi (From ASME Setion II, Part D, SA-56 Grade 70 is a arbon steel, P, Group, With S = 0 ksi, σ = 38 ksi and 70 ksi ys uts ys Dt uts True True True True True True σ = ) True Frature toughness greater than or equal to the lower-bound K IC in Annex F Carbon steel not degraded beause of environmental damage True a) STEP Determine the temperature to be used in the assessment based on operating and design onditions The primary membrane tensile stress σ due to startup or shutdown pressure (40 psig), alulated per formula in paragraph A.3.4 of Annex A is less than 8 ksi. Per Part 3 paragraph 3.., a brittle frature assessment is not needed for these pressures. The temperature used in the assessment will be the minimum temperature for whih the pressure is above 40 psig. Therefore T = 00 o F b) STEP Determine the length and depth of the rak-like flaw from inspetion data. a = 0.5 in =.0in ) STEP 3 Determine the figure to be used in the assessment The flaw is loated in a longitudinal weld seam in a ylindrial vessel and is parallel to the weld joint; therefore Figure 9.3 will be used. d) STEP 4 Determine the sreening urve. The maximum flaw depth reported from UT measurements is 0.5 in. The urrent omponent thikness is t = =.5 in whih is greater than in ; therefore, the maximum permissible flaw depth for an assessment with ¼-t sreening urve is 0.5 in. Based on NDE results, this is the maximum flaw depth reported. The flaw is in a weldment and the vessel was subjet to PWHT at the time of onstrution. Based on the above, the ¼-t (solid line) Curve B of Figure 9.3 will be used. C m 9-

220 e) STEP 5 Determine the Referene Temperature T ref is established using Table 9.. Inputs for this table are the exemption urve as per Table 3. in Part 3 and the minimum speified yield strength at ambient temperature based on the original onstrution ode. SA-56 Grade 70 is a Curve B Carbon Steel with σ = 38 ksi, therefore: ys Curve B Carbon Steel Tref = 43 F σ ys = 38 ksi f) STEP 6 Determine the maximum permissible rak-flaw length using Figure 9.3 (see STEP 3). ( ref ) T T + 00 = ( ) = 57 F = 8.00 in ¼-t Curve B of Figure 9.3 g) STEP 7 Evaluate Results. Sine ( 8.00 in ) (.0 in ) Sreening Curve = > =, the flaw is aeptable. Measured The Level Assessment Criteria are Satisfied. The vessel is aeptable for ontinued operation. 9-3

221 9. Example Problem A rak-like flaw was found on a spherial pressure vessel that was onstruted to the ASME B&PV Code, Setion VIII, Division, 998 Edition. The vessel and inspetion data are provided below. Determine if the vessel is aeptable for ontinued operation using a Level Assessment. Vessel Data Material = SA-56 Grade 70 Year 998 Design Conditions =.0 MPa (0 350 C Operating Conditions =.5 MPa (5 300 C Inside Diameter =.4 m Fabriated Thikness = 30 mm Uniform Metal Loss =.5 mm FCA = 3 mm Weld Joint Effiieny =.0 PWHT = Yes, Original Fabriation Requirement Operating Conditions At startup the vessel is warmed up to 30 C prior to pressurizing. At shutdown, the vessel is depressurized before letting the temperature drop below 30 C. Inspetion Data The flaw is loated in a irumferential weld seam on the inside surfae of a spherial vessel. The flaw is perpendiular to the weld joint. The seam is a single V-groove weld. The maximum measured depth of the flaw using UT is 0 mm. A flaw length of 30 mm is established by MT. The distane of the rak-like flaw to the nearest strutural disontinuity is 500 mm. Perform a Level Assessment per paragraph First, hek that the onditions to perform a Level Assessment are satisfied Geometry: Component is a flat plate, ylinder or sphere: (sphere) True Sphere with R/ t 5 (t is the urrent thikness) t = tnom LOSS = = 7.50 mm R/ t = 00 / 7.50 = R= D/ = 400 / = 00 mm Wall thikness at the loation of the flaw is less than 38 mm: ( 7.50 mm mm) Flaw of surfae or through-thikness type with a maximum rak-length of 00 mm: Surfae rak with length equal to 30 mm Spherial shell: flaw oriented in the axial or irumferential diretion: Perpendiular to irumferential weld = axial diretion True < True True True 9-4

222 With a distane to the nearest strutural disontinuity greater than or equal to.8 Dt Lmsd = 500 mm Lmsd.8.8 Dt =.8 (400)(7.50 ) = mm Loads: Pressure produes only a membrane stress field Membrane stress from operation is within the design limits of original onstrution ode Welded joint is single or double V: (single V-groove weld) Material: Carbon Steel (P, Group or ) with S 7 MPa, σ 76 MPa and σ 483 MPa (From ASME Setion II, Part D, SA-56 Grade 70 is a arbon steel, P, Group, With S = 38 MPa, σ = 60 MPa and σ = 485 MPa ys uts ys uts Dt True True True True Frature toughness greater than or equal to the lower-bound K IC in Annex F Carbon steel not degraded beause of environmental damage True a) STEP Determine the temperature to be used in the assessment based on operating and design onditions Based on the operating onditions: T = 30 C b) STEP Determine the length and depth of the rak-like flaw from inspetion data. a = 0.0 mm = 30.0mm ) STEP 3 Determine the figure to be used in the assessment The flaw is loated at a irumferential weld seam of a spherial vessel and is perpendiular to the joint; therefore Figure 9.8M will be used. d) STEP 4 Determine the sreening urve. The maximum flaw depth reported from UT measurements is 0.0 mm. The urrent omponent thikness is 7.5 mm whih is greater than 5.0 mm; therefore, the maximum permissible flaw depth for an assessment with ¼-t sreening urve is 6.0 mm. Sine the maximum flaw depth is 0.0 mm, then the -t sreening urves are to be used. The flaw is in a weldment and the vessel was subjet to PWHT at the time of onstrution. Based on the above, the -t (dashed line) Curve B of Figure 9.8M will be used. e) STEP 5 Determine the Referene Temperature T ref is established using Table 9.M. Inputs for this table are the exemption urve as per Table 3. in Part 3 and the minimum speified yield strength at ambient temperature based on the original onstrution ode. SA-56 Grade 70 is a Curve B Carbon Steel with σ = 60 MPa, therefore: ys Curve B Carbon Steel Tref = 6 C σ ys = 60 MPa 9-5

223 f) STEP 6 Determine the maximum permissible rak-flaw length using Figure 9.8M (see STEP 3). ( ref ) T T + 56 = ( ) = 80 C = 37.5 mm t Curve B of Figure 9.8M g) STEP 7 Evaluate Results. Sine the maximum permissible flaw length from the sreening urve of 37.5 mm is greater than the measured flaw length of 30.0 mm, the flaw is aeptable. The Level Assessment Criteria are Satisfied. The vessel is aeptable for ontinued operation. --``,,`,,,,`,,,,`,,``,`````-`- 9-6

224 9.3 Example Problem 3 A rak-like flaw has been found on a ylindrial shell of a pressure vessel during a shedule turnaround. The vessel and inspetion data are provided below. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division, 00 Edition. In order to determine if the vessel is aeptable for ontinued operation using a Level or Level Assessment, the flaw length used in the assessment must be omputed. Vessel Data Material = SA-56 Grade 70 Year 00 Design Conditions = Inside Diameter = 96 in Fabriated Thikness =.5 in Uniform Metal Loss = 0.0 in FCA = 0.5 in Weld Joint Effiieny =.0 PWHT = Yes, Original Fabriation Requirement Operating Conditions The vessel is not fully pressurized until the temperature is 00 F. Below this temperature, the startup pressure remains under 40 psig. At shutdown, the pressure is dereased to 40 psig before letting the temperature drop below 00 F. Inspetion Data The flaw is loated primarily in a longitudinal weld seam on the inside surfae of a ylindrial vessel. The flaw is perpendiular to the inside surfae and oriented at 30 with respet to the horizontal seam weld joint. The longitudinal seam is a double V-groove weld. The depth of the flaw was established by UT with a maximum value of 0.5 in being reported. The flaw length was established by MT and is.8 in. The distane of the rak-like flaw to the nearest strutural disontinuity is about 60 in. Before performing a Level or Level Assessment per paragraph 9.4.., the equivalent flaw length onto the prinipal plane needs to be omputed first. Compute the equivalent flaw length parallel to the seam weld. a) The prinipal stresses are the hoop stress due to pressure ( σ ) and the axial stress due to the end effet ( σ ). Both of them are positive and σ > σ. This leads to a biaxiality ratio B F B σ σ = = 0.50 b) From Equation (9.), for the plane of the flaw projeted onto the plane normal to the hoop stress, σ : m [ α ] ( B) sin[ α] os[ α] = os + + B sin In the above equation, the dimension orresponds to the half flaw length to be used in alulations and m is the measured half length for the flaw oriented at an angle α from the σ plane Thus in this ase, m =.8 / = 0.59 in α = 30 [ α ] 9-7

225 m [ ] ( 0.5) sin[ 30] os[ 30] [ ] = + + = os 30 (0.5) sin For =.8in, =.0865 in m The equivalent flaw length, parallel to the seam weld, to be taken into aount in a Level or a Level Assessment is rounded to.0 in. 9-8

226 9.4 Example Problem 4 A rak-like flaw has been found in the longitudinal seam on the inside surfae of a ylindrial pressure vessel during a sheduled turnaround. The vessel and inspetion data are provided below. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division, 00 Edition. In order to determine if the vessel is aeptable for ontinued operation using a Level or Level Assessment, the flaw depth used in the assessment must be omputed. Vessel Data Material = SA-56 Grade 70 Year 00 Design Conditions = F Inside Diameter = 0.0 in Fabriated Thikness =.0 in Uniform Metal Loss = 0.0 FCA = 0.0 in Weld Joint Effiieny =.0 PWHT = No Inspetion Data The flaw is loated in a longitudinal weld seam on the inside surfae of the vessel. The longitudinal seam is a double V-groove weld with bevel angle of 5 degrees. The depth of the flaw was established by UT; onsistent readings were noted and a final value for the flaw depth was established at 0.7 in. The flaw length was established by MT and is 3. in. The distane of the rak-like flaw to the nearest strutural disontinuity is 30 in. Before performing a Level or Level Assessment per paragraph 9.4.., sine the flaw is not normal to the surfae (due to a lak of fusion, the flaw is oriented parallel to the bevel angle as shown in Figure 9.4, the flaw depth dimension, a, must be omputed first. Compute the flaw depth to be used in the assessment. a) STEP Projet the flaw onto a plane that is normal to the plate surfae, designate this flaw depth as a. m am = 0.7 in b) STEP Determine W using the Equations (9.6) and (9.7) in whih the angle,θ, expressed in degrees and defined in Figure 9.4, is the bevel angle of the weld (5 in this ase) W = max[ W Theta,.0] W Theta ( ) θ + ( ) θ ( ) θ ( ) θ ( ) θ ( ) θ + ( ) θ ( ) + ( ) + ( ) ( ) ( ) ( ) + ( ) = = =

227 ) STEP 3 Multiply a m by W to obtain the dimension a, whih is used in flaw alulations. ( )( ) a= a W = = in m The flaw depth to be taken into aount in a Level or a Level Assessment is rounded to 0.0 in. 9-0

228 9.5 Example Problem 5 A rak-like flaw has been found in the longitudinal seam on the inside surfae of a ylindrial pressure vessel during a sheduled turnaround. The vessel and inspetion data are provided below. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division, 998 Edition. Determine if the vessel is aeptable for ontinued operation. Vessel Data Material = SA-56 Grade 70 Year 998 Design Conditions = F Inside Diameter = 0.0 in Fabriated Thikness =.0 in Uniform Metal Loss = 0.0 in FCA = 0.0 in Weld Joint Effiieny =.0 PWHT = No Operating Conditions The vessel is not fully pressurized until the temperature is 40 o F. Below this temperature, the startup pressure remains under 00 psig. At shutdown, the pressure is dereased to 00 psig before letting the temperature drop below 40 F. Inspetion Data The flaw is loated in the HAZ of a longitudinal weld seam on the inside surfae of the vessel. The longitudinal seam is a double V-groove weld. The flaw is parallel to the weld seam. The depth of the flaw was established by UT; onsistent readings were noted and a final value for the flaw depth was established at 0.0 in. The flaw length was established by MT and is 3. in. The distane of the rak-like flaw to the nearest strutural disontinuity is 30 in. Perform a Level Assessment per paragraph First, hek that the onditions to perform a Level Assessment are satisfied: See Example 9. a) STEP Determine the temperature to be used in the assessment based on operating and design C onditions The primary membrane tensile stress σ m due to startup or shutdown pressure (00 psig), alulated per formula in paragraph A.3.4 of Annex A is less than 8 ksi. Per Part 3 paragraph 3.., a brittle frature assessment is not needed for these pressures. The temperature used in the assessment will be the minimum temperature for whih the pressure is above 00 psig. Therefore T = 40 F b) STEP Determine the length and depth of the rak-like flaw from inspetion data. a = 0.0 in = 3.0in ) STEP 3 Determine the figure to be used in the assessment The flaw is loated in a longitudinal weld seam in a ylindrial vessel and is parallel to the weld joint; therefore Figure 9.3 will be used. 9-

229 d) STEP 4 Determine the sreening urve. The maximum flaw depth reported from UT measurements is 0.0 in. The urrent omponent thikness is in ; therefore, the maximum permissible flaw depth for an assessment with ¼-t sreening urve is (. 00) / 4 = 0. 5 in. Based on NDE results, the maximum flaw depth reported is 0.0 in The flaw is in a weldment and the vessel was not subjet to PWHT at the time of onstrution. Based on the above, the ¼-t (solid line) Curve C of Figure 9.3 will be used. e) STEP 5 Determine the Referene Temperature T ref is established using Table 9.. Inputs for this table are the exemption urve as per Table 3. in Part 3 and the minimum speified yield strength at ambient temperature based on the original onstrution ode. SA-56 Grade 70 is a Curve B Carbon Steel with σ = 38 ksi, therefore: ys Curve B Carbon Steel Tref = 43 F σ ys = 38 ksi f) STEP 6 Determine the maximum permissible rak-flaw length using Figure 9.3 (see STEP 3). ( ref ) T T + 00 = ( ) = 97 F 0. in ¼-t Curve C of Figure 9.3 g) STEP 7 Evaluate Results. Sine, ( = 0. in) < ( = 3.0 in) the flaw is not aeptable. Sreening Curve Measured The Level Assessment Criteria are Not Satisfied 9-

230 Perform a Level Assessment per paragraph a) STEP Evaluate operating onditions and determine the pressure, temperature and supplemental loading ombinations to be evaluated There are no signifiant supplemental loads, pressure is the only signifiant load. T = 40 F P = 00 psig b) STEP Determine the stress distribution at the loation of the flaw - The primary stress distribution is based on the applied loads. ) Primary Stress The flaw is loated away from all major strutural disontinuities. Therefore, the primary stress at the weld joint perpendiular to the rak fae is a membrane hoop stress. From Annex C, Table C., the flaw geometry, omponent geometry, and loading ondition orrespond to KCSCLE and RCSCLE, Cylinder - Surfae Crak, Longitudinal Diretion - Semi-Elliptial Shape, Internal Pressure. The stress intensity fator solution for KCSCLE is provided in Annex C, paragraph C.5.0. The referene stress solution for RCSCLE is provided in Annex D, paragraph D.5.0. Ri = D/ = 0.00 / = in Ro = Ri + t = = 6.00 in t/ R =.00 / = i The membrane and bending omponents of the primary stress for the alulation of the referene stress are given by Equations (D.47) and (D.48): P = PR / t = (00) (60.00) / (.00) = psi m i 3 PR o t t t.5.8 o i i i i Pb = + R R R R R (00) (6.00) = + = (6.00) (60.00) 3 (0.067).5(0.067).8(0.067) The bending omponent is less than % of the membrane omponent and, therefore, will be negleted. The alulations will be performed with: C Pm = σ m = 000 psi P = 0 psi b ) Seondary Stress Thermal gradients do not exist in the vessel at the loation of the flaw, and the flaw is loated away from all major strutural disontinuities. Therefore, there are no seondary stresses. 3) Residual Stress The flaw is loated at a weldment in a vessel that was not subjet to PWHT at the time of fabriation. From Annex E, paragraph E.3.. r σ ys = σ ys + 0 = = 48 ksi The flaw is loated at the limit between the weld seam and the base metal. The residual stress field used in the assessment an be based on the surfae distribution or the through-thikness distribution. The more onservative stress distribution is hosen (see Example 9.6 or 9.7 for an assessment using a less onservative residual stress field based on the through-thikness distribution).the residual stress is alulated from Annex E, paragraph E.4.4..a with y=w / psi 9-3

231 It has been verified that the rak-like flaw was in the vessel during a field hydrotest previously performed as part of a rerate. Therefore the residual stress may be redued. The irumferential membrane stress during hydrotest is alulated: S750F = 4800 psi@ 750 F SRT = 0000 psi@ Ambient C S RT 0.0 σm, t =.3 σm = (.3) ( 000) = psi S750F 4.8 The perentage of yield strength reahed during hydrotest is: T p σ m, t.08 = (00) = ( 00) = % r σ ys Sine T p < 75 %, then the redution fator on the residual stress is R r =.0. Therefore: r Q = σ. R = (48000)(.0) = psi Q m ys r b = 0 psi ) STEP 3 Determine the material properties; yield strength, tensile strength and frature toughness. Material properties for the plate ontaining the flaw are not available; therefore, the speified minimum speified yield and tensile strength are used. Based on the material speifiation and grade, the material frature toughness is established using the lower-bound urve in Annex F, paragraph F σ ys = 38 ksi σ uts = 70 ksi T = 43 F ( see STEP 5 of the Level Assessment) ref ( ref ) KIC = exp 0.0 T T 00 + = exp 0.0 ( ) = ksi in d) STEP 4 Determine the rak-like flaw dimensions from inspetion data. a = 0.0 in = 3.0in e) STEP 5 Modify the primary stress, material frature toughness, and flaw size using Partial Safety Fators. Based on a risk assessment, it was deided that the most appropriate probability of failure to 3 use in the FFS assessment would be p f = 0. The mean frature toughness to speified minimum yield strength ratio, R ky, is required to determine the Partial Safety Fators. Using the information in mean Notes 5 and 6 of Table 9.3 (Note that sigma = is used in alulating the Kmat K IC ratio per Table F. of Annex F): 9-4

232 Δ T = T T = = 3 0 F K K mean mat IC ref sigma=.0 = = ( ) mean K =.686 K =.686 (5.763) = ksi in R mat ky IC mean Kmat = = =.598 σ 38.0 ys From Table 9.3, with ( R.598) ( R.9) ky in = > =, the Partial Safety Fators are: ( a = 0.0 in) 0.0 in PSFs =.50 COVs = 0.0 PSFk =.00 R =.9 PSFa =.00 The primary stress, pressure on the rak fae, frature toughness, and flaw size are fatored by the Partial Safety Fators as follows: P = P PSF = (000)(.5) = 8000 psi m m s P = P PSF = (0)(.5) = 0 psi b b s p= p PSF = (00)(.5) = 300 psig s K = K / PSF = (85.870) / (.0) = ksi in mat mat k a= a PSF = (0.0)(.0) = 0.0 in a Note: The frature toughness data is the lower-bound estimate in Annex F. Therefore, per Table 9.3 Note 6, the Partial Safety Fator on frature toughness is applied on K. f) STEP 6 Compute the referene stress for the primary stress. The referene stress solution for RCSCLE is provided in Annex D, paragraph D.5.0. a = 0.0 in = 3.0 / =.60 in ( ) ( ) (.60) λa = = = R a 60 (0.0) i λa λ a M t( λa) = λa +.533(0 ) λa (0.8397) (0.8397) = = (0.8397) +.533(0 )(0.8397) NS M s = = =.059 a a t t Mt( λa) mean mat 9-5

233 a 0.0 α = t =.00 = 0.3 t σ P ref { ( ) 9 ( α) } gp + gp + M P = 3( α) 0.5 b b s m { } (0) + 9 (.0659)(8000)( 0.3) = = ( 0.3) In the above formula M s = M (as reommended). g) STEP 7 Compute the Load Ratio ( L ) or absissa of the FAD. σ L ys P r = psi NS s P σ ref = = = σ ys P r psi P h) STEP 8 Compute K - The stress intensity fator for KCSCLE is provided in Annex C, paragraph C.5.0. Note that beause the applied loading is a membrane stress, only the data required to evaluate the G 0 influene oeffiient is needed to ompute the stress intensity fator. The flaw ratios and parameters to determine the G 0 influene oeffiient from Annex C Table C. are: A0,0 = t A,0 = = = R 60 i A,0 = a 0. = = 0.5 A3,0 = a 0. A4, = = = 0. t.0 A5,0 = A 6,0 = The influene oeffiients required for the assessment are: At the base of the flaw ϕ = 90 o π ϕ π : ϕ = 90 = rad β = = = G0 =.9838 π π ϕ At the edge of the flaw 0 o ϕ = 0 = 0 rad β = = 0 = 0 G = The stress intensity fators are: ϕ = : ( ) ( ) a 0. Q = = = At the base of the flaw ϕ = 90 o : π π 9-6

234 K P I pr G π a (300)(6) (.9838) π (0.0) = = = ksi in R R Q (6) (60).0474 o 0 o i At the edge of the flaw ϕ = 0 o : K P I pr G π a (300)(6) (0.4498) π (0.0) = = = ksi in R R Q (6) (60).0474 o 0 o i i) STEP 9 Compute the referene stress for seondary stresses. Note that SR σ ref used in this alulation is r based on the residual stress ( σ ) from STEP. From Annex C, Table C., the flaw geometry, omponent geometry, and loading ondition orrespond to KCSCLE and RCSCLE, Cylinder - Surfae Crak, Longitudinal Diretion - Semi-Elliptial Shape, Through-Wall Fourth Order Polynomial Stress Distribution. The referene stress solution for RCSCLE is provided in Annex D, paragraph D.5. whih referenes paragraph D.5.0. Details regarding the alulation of the referene stress are provided in STEP 6. σ SR ref { ( ) 9 ( α) } gq + gq + M Q = 3( α) 0.5 b b s m { } 0 + (0) + 9 (.059)(48000)( 0.3) = = psi 3( 0.3) SR j) STEP 0 Compute K. The stress intensity fator solution for KCSCLE is provided in Annex C, paragraph C.5.. Details regarding the alulation of oeffiients Q and G 0 used in the formula for the stress intensity fator are provided in STEP 8. The stress intensity fators are: At the base of the flaw ϕ = 90 o : SR π a π (0.) KI = G0σ 0 = (.9838)(48.0) = ksi in Q.0474 At the edge of the flaw ϕ = 0 o : SR π a π (0.) KI = G0σ 0 = (0.4498)(48.0) = ksi in Q.0474 k) STEP Compute the plastiity interation fator, with 0.5 P L r from STEP 7 L SR r SR σ ref = = =.959 σ ys 9-7

235 ψ and ϕ are alulated from Tables 9.3 and 9.5 respetively P L r = ψ = SR L.959 ϕ = r = and, Φ ψ =.0 + =.0 + =.448 Φ ϕ < = , then Φ =.0 and Φ =.448 SR Sine ( L r ) o l) STEP Determine toughness ratio or ordinate of the FAD assessment point. At the base of the flaw At the edge of the flaw ϕ = 90 o : ϕ = 0 o : m) STEP 3 Evaluate the results. ) STEP 3. Determine the ut-off for the K K r r ( ) P SR K (.448) I +ΦK + I = = = K mat ( ) P SR K (.448) I +ΦK + I = = = 0.74 K mat P Lr -axis of the FAD Sine the hardening harateristis of the material are not known, the following value an be used (see Figure 9.0, Note ): L = P r(max).0 ) STEP 3. Plot the assessment point on the FAD shown in Figure o P At the base of the flaw ϕ = : ( L, K ) = (0.486,0.789) 0 o r r the point is inside the FAD (see Figure E9.5-) P At the edge of the flaw ϕ = : ( L, K ) = (0.486,0.74) r r the point is inside the FAD (see Figure E9.5-) Note: Equation (9.33) under Figure 9.0 gives the maximum allowable { } P P ( ) ( ) 6 r,max = 0.4 r exp 0.65 r K L L { } ( ) ( ) 6 = exp = 0.96 P K for L = : r r 9-8

236 Kr Lr Assessment Curve Base of the Flaw Edge of the Flaw Figure E FAD with Assessment Points The Level Assessment Criteria are Satisfied. The vessel is aeptable for ontinued operation. Note: Should the Level riteria not be satisfied, then the assessment ould be repeated with a less onservative residual stress field based on the through-thikness distribution (see Example 9.6 or 9.7) 9-9

237 9.6 Example Problem 6 A rak-like flaw has been found in the irumferential seam on the outside surfae of a pipe during a sheduled turnaround. The pipe and inspetion data are provided below. The piping system was onstruted to the ASME B3.3 Code, 003 Edition. Determine if the pipe is aeptable for ontinued operation. Pipe Data Material = SA-06 Grade B Year 003 Design Conditions = 3.0 MPa (30 50 C Fluid Density = 0.8 Pipe Outside Diameter = 508 mm ( NPS 0) Pipe Thikness = 9.53 mm ( Shedule 0) Uniform Metal Loss = 0.0 mm FCA = 0.0 mm Weld Joint Effiieny =.0 PWHT = No Operating Conditions The piping system is not fully pressurized until the temperature is 0 C. Below this temperature, the startup pressure remains under.0 MPa (0 bar). At shutdown, the pressure is dereased to.0 MPa (0 bar) before letting the temperature drop below 0 C. Inspetion Data The flaw is loated in a irumferential weld seam on the outside surfae of the pipe. The seam is a single V- groove weld. The flaw is parallel to the weld seam. The depth of the flaw was established by UT; onsistent readings were noted and a final value for the flaw depth was established at 3.0 mm. The flaw length is suh that the flaw may be onsidered as a 360 degree rak. The rak-like flaw is situated midway between supports, the distane of whih is 0.5 m. Perform a Level Assessment per paragraph First, hek that the onditions to perform a Level Assessment are satisfied Geometry: Component is a flat plate, ylinder or sphere: (ylinder) True Cylinder with R/ t 5 (t being the urrent thikness) t = 9.53 mm R/ t = / 9.53 = R= D/ t = (508 / ) 9.53 = mm Wall thikness at the loation of the flaw is less than 38mm : (9.53 mm 38 mm Flaw of surfae or through-thikness type with a maximum rak-length of 00 mm Surfae rak with length equal to ( mm) True < ) True 508π = mm False Cylindrial shell: flaw oriented in the axial or irumferential diretion: (longitudinal = axial) True 9-0

238 Loads: Membrane stress field produed by pressure only Considering the loation of the flaw, a global bending moment shall be taken into aount. Membrane stress within the design limits of original onstrution ode Welded joint is single or double V: (single V-groove weld) Material: Carbon Steel (P, Group or ) with S 7 MPa, σ 76 MPa and σ 483 (From ASME Setion II, Part D, SA-06 Grade B is a arbon steel, P, Group, With S = 8 MPa, σ = 40 MPa and 45 MPa ys uts ys uts MPa False True True σ = ) True Frature toughness greater than or equal to the lower-bound K IC in Annex F (Carbon steel not degraded beause of environmental damage) True The Level Assessment Criteria are Not Satisfied Perform a Level Assessment per paragraph a) STEP Evaluate operating onditions and determine the pressure, temperature and supplemental loading ombinations to be evaluated: Due to the loation of the flaw, a global bending moment shall be onsidered. The pipe setion is onsidered as simply supported at both ends C The irumferential primary membrane tensile stress σ m due to startup or shutdown pressure (.0 MPa) alulated per formula in paragraph A.3.4 of Annex A are less than 55 MPa. L The longitudinal primary membrane tensile stress σ m due to startup or shutdown pressure and to the global bending moment, alulated per formula in paragraph A.3.4 of Annex A is less than 55 MPa too. Per Part 3 paragraph 3.., a brittle frature assessment is not needed for these loads. Therefore, the temperature used in the assessment will be the minimum temperature for whih the pressure is above.0 MPa. T = 0 C P = 3.0 MPa M = 36.8 (0) N mm ( ) 6 b) STEP Determine the stress distribution at the loation of the flaw - The primary stress distribution is based on the applied loads. ) Primary Stress The flaw is loated away from all major strutural disontinuities. From Annex C, Table C., the flaw geometry, omponent geometry, and loading ondition orrespond to KCSCCL and RCSCCL, Cylinder - Surfae Crak, Cirumferential Diretion Degrees, Pressure with a Net Setion Axial Fore and Bending Moment. The stress intensity fator solution for KCSCCL is provided in Annex C, paragraph C.5.7. The referene stress solution for RCSCCL is provided in Annex D, paragraph D.5.7. Ro = D/ = / = mm R = R t = = mm i o { } M / π ( R R ) = (36.8) (0) / π (54.00) (44.47) = N/ mm o i The membrane and bending omponents of the primary stress for the alulation of the stress intensity fator are (with F = 0 ): 9-

239 σ σ pr M ( ) P i m = + R 4 4 o + Ri ( Ro Ri ) π ( Ro Ri ) (3.0) (44.47) = + + (54.00) (44.47) = MPa P b 4 4 o i π ( Ro Ri ) { (0.0984) ( ) } M = ( R R ) = (0.0984) ( ) = MPa The membrane and bending omponents of the primary stress for the alulation of the referene stress are (with F = 0 ): P P m bg pri (3.0) (44.47) = = = MPa ( Ro Ri ) (54.00) (44.47) M R = = (0.0984) = MPa π o 4 4 ( Ro Ri ) ) Seondary Stress Thermal gradients do not exist in the pipe at the loation of the flaw, and the flaw is loated away from all major strutural disontinuities. Therefore, there are no seondary stresses. 3) Residual Stress The flaw is loated at a girth in a pipe that was not subjet to PWHT at the time of fabriation. From Annex E, paragraph E.3.. r σ ys = σ ys + 69 = = 309 MPa The flaw is loated at the limit between the weld seam and the base metal. The weld is a single V- groove. The through-thikness residual stress field is alulated from Annex E, paragraph E.4...b The basi parameters used in the equation representing the through-thikness residual stress field are σ and σ. r m r σ m = r b 0.30 r σ b is a funtion of the mean radius to thikness ratio and of the heat input of the welding proess. It has been established that the first pass was a GTAW one and that all subsequent passes were SMAW ones. Sine the rak is on the opposite side of the root, the seleted heat input orresponds to the SMAW passes reorded as q = 500 J/ mm The parameters in the r σ b equations are: q 500 Q J mm between and t (9.53) ˆ 3 = = (0.744) =.896 / (.5 5.0) ˆ r R= = = 6.57 t 9.53 < 30.0 Rˆ = 30.0 ln Rˆ =

240 Leading to: ˆ ln R 7.599(0) Qˆ (0) ( ln ˆ ) (0) ˆ R + Q 3 r ˆ ˆ 3 σ (0) ln (0) ( ln ˆ b = Q R R ) 5 ˆ3 4 ˆ.05665(0) Q (0) Q ln Rˆ (0) ˆ ( ln ˆ ) Q R (3.40) 7.599(0) (.896) (0) (3.40) (0) (.896) 3 3 = (0) (.896) (3.40) (0) (3.40) (0) (.896) (0) (.896) (3.40) (0) (.896)(3.40) = 0.96 s = K σ σ = = r r r o b m r r si = 0.5 so = (0.5) (0.6704) = r r 5 so + s i C = artan artan.03 r r = = π so s i π With K =. orresponding to residual stresses perpendiular to the weld There is no indiation that the rak-like flaw already existed in the pipe during the last field hydrotest performed as part of a rerate. Therefore the residual stress may not be redued: R r =.0 The values of the residual stress with respet to the depth ς = x / t ( ς = 0 on the inside surfae and ς = on the outside surfae) together with the intermediate oeffiients in paragraph E.3.4.a r r r { σm + σb ζ } σ ysrr and A B σ R. are given in Table E9.6- where the olumn "Linear" orresponds to ( ) the olumn "Auto-Eqlb" orresponds to the self-equilibrating part of the stress { } r ys r 9-3

241 Table E Through-Thikness Distribution of Residual Stress per Annex E R ς A B D E σ ( ς ) Linear Auto-Eqlb MPa Through-Thikness Distribution of Residual Stress per Appendix E Linear Auto-Eqlb s_r (Tau) ID ===> OD Figure E Through-Thikness Distribution of Residual Stress per Annex E In order to alulate the stress intensity fator and the referene stress, the through-thikness distribution will be represented by a polynomial funtion. From Annex C, Table C., the flaw geometry, omponent geometry, and loading ondition orrespond to KCSCCL and RCSCCL, Cylinder - Surfae Crak, Cirumferential Diretion Degrees, Through-Wall Fourth Order Polynomial Stress Distribution. The stress intensity fator solution for KCSCCL is provided in Annex C, paragraph C.5.8. The referene stress solution for RCSCCL is provided in Annex D, paragraph D

242 A best-fit 4 th order polynomial is determined by generating a graph of the through-thikness distribution versus ς in a Mirosoft Exel spreadsheet and adding a trend urve to it. Sine the flaw is on the outside surfae, the variable ς of the polynomial is set to 0 on the outside surfae and on R the inside surfae. Values of σ ( ς ) are input data with a step Δ ζ = The residual stress for the alulation of the stress intensity fator is written as: 3 4 R x x x x σ ( x) = σ0 + σ + σ + σ3 + σ4 t t t t with: σ 0 = σ = σ = σ = σ = Figure E9.6- shows the through-wall residual stress distribution as determined per Annex E (s_r) and the best-fit polynomial urves of different degrees (deg to 4). This figure validates the use of a 4 th order polynomial for the representation of the residual stress. MPa 350 Polynomial Regressions s_r deg 4 deg 3 deg deg ID ===> OD Figure E Through-Thikness Distribution with Polynomial Trend Curves The membrane and bending omponents of the residual stress for the alulation of the referene stress may be based on: i) The linear part of the through-thikness distribution given in Annex E: r r Q = σ σ R = (0.30) (309.00) (.0) = MPa m m ys r r r Q = σ σ R = ( 0.96) (309.00) (.0) = MPa b b ys r 9-5 or

243 ii) Q m Q b Equivalent membrane and bending stresses for the 4 th order polynomial stress distribution used for the stress intensity fator alulation, as desribed in Annex D, paragraph D...3: σ σ σ 3 σ4 = σ = ( 3.584) = MPa σ σ 9 σ 3 6 σ4 = ( ) 6 ( ) = = 7.96 MPa 0 5 The methods give similar values. Sine the seond set ( Q, Q ) is based on an approximate solution, the first set ( Q, Q ) will be retained for the alulation of the referene stress m b ) STEP 3 Determine the material properties; yield strength, tensile strength and frature toughness. Material properties for the pipe ontaining the flaw are not available; therefore, the speified minimum speified yield and tensile strengths are used. σ σ ys uts = 40 MPa = 45 MPa Based on the material speifiation and grade, the material frature toughness is established using the lower-bound urve in Annex F, paragraph F.4.4. Determine the Referene Temperature T ref is established using Table 9.. Inputs for this table are the exemption urve as per Table 3. in Part 3 and the minimum speified yield strength at ambient temperature based on the original onstrution ode. ASTM A06 Grade B is a Curve B Carbon Steel with σ = 40 MPa, therefore: ys m b Curve B Carbon Steel Tref = 0 C σ ys = 40 MPa This leads to ( ref ) KIC = exp T T 56 + = exp ( ) = MPa m d) STEP 4 Determine the rak-like flaw dimension from inspetion data. a = 3.0 mm e) STEP 5 Modify the primary stress, material frature toughness, and flaw size using Partial Safety Fators. Based on a risk assessment, it was deided that the most appropriate probability of failure to 3 use in the FFS assessment would be p f = 0. The mean frature toughness to speified minimum yield strength ratio, R ky, is required to determine the Partial Safety Fators. Using the information in mean Notes 5 and 6 of Table 9.3 (Note that sigma = is used in alulating the Kmat K IC ratio per Table F. of Annex F): 9-6

244 Δ T = T T = 0 0 = 0 C = 8 F K K mean mat IC ref sigma=.0 = = mean K =.5996 K =.5996 ( ) =.474 MPa m mat IC mean Kmat.474 Rky = Cu = (6.75) =.945 in σ 40.0 ys From Table 9.3, with ( R =.945) > ( R =.4), the Partial Safety Fators are: ky ( a = 3.0 mm) < 5mm PSFs =.50 COVs = 0.0 PSFk =.00 R =.4 PSFa =.00 The primary stress, frature toughness, and flaw size are fatored by the Partial Safety Fators as follows: Pm = Pm PSFs = ( )(.5) = MPa P = P PSF = (0.583)(.5) = MPa bg bg s P m m s P b b s P σ = σ PSF = (57.534)(.5) = MPa P σ = σ PSF = (0.378)(.5) = MPa Kmat = Kmat / PSFk = (.474) / (.0) =.474 MPa m a= a PSF = (3.0)(.0) = 3.0 mm a f) STEP 6 Compute the referene stress for the primary stress. The referene stress solution for RCSCCL is provided in Annex D, paragraph D.5.7. R a = = 5.0 mm o = {( ) ( ) }/{ ( ) ( ) } = (56.650) {(54.0) (44.47) }/ { (5.0) (44.47) } = MPa = (3 π /6) {( ) ( ) }/ { ( ) ( ) ( ) } = (30.374) (3 π /6) {(54.0) (44.47) }/ { (54.0) (5. 0) (44.47) } N P R R R a R r m o i o i M P R R R R a R σ r bg o i o o i = MPa = ( M /) + ( N ) + ( M /) P ref r r r 0.5 { } { } = + + = 0.5 ( / ) (83.53) ( / ) (see Note STEP 5 of Example 9.5) Note: Usually the same primary and seondary stresses are used for the alulation of the stress intensity fator and for the alulation of the referene stress. This is not true for the type of rak under evaluation. MPa 9-7

245 P P P The alulation of σ ref from σ m and σ b instead of P m and P bg, based on the formula given for a through-wall fourth order polynomial stress distribution (paragraph D.5.8) would have led to a very P onservative value of the referene stress ( σ = MPa ) due to the fats that σ represents ref the mean stress on the ross setion at the loation of maximum global bending stress and that the distribution of this global bending stress would not have been taken into aount. g) STEP 7 Compute the Load Ratio ( L ) or absissa of the FAD. σ L ys P r = 40.0 MPa P r P σ ref = = = σ 40.0 ys P h) STEP 8 Compute K - The stress intensity fator solution for KCSCCL is provided in Annex C, paragraph C.5.7. This solution is based on a through-wall first order polynomial stress distribution in whih the onstant oeffiients are: P P σ = σ + σ = = MPa 0 P σ = σ = (0.567) =.345 MPa m b b The parameters used to determine the G 0 and G influene oeffiients from Annex C Table C. are: t R a t i 9.53 = = = = The influene oeffiients required for the assessment are alulated by interpolation between values given in Table C for raks on the outside surfae (see Table E9.6-): P m Table E Influene Coeffiients used in the Assessment t/r i a/t G 0 G G G 3 G Note that oeffiients G 0 and G only are used to alulate SR to alulate K I in STEP 0. P K I. Coeffiients G, G 3 and G 4 will be needed 9-8

246 Sine the rak lies on the outside surfae, the rak fae pressure is nil ( p = 0 ). The stress intensity fator is: P a KI = G0{ σ0 + p} + Gσ π a t K K P I P I { } = ( ) ( ) (.345) (0.348) π (3.0) = MPa mm = 3.57 MPa m i) STEP 9 Compute the referene stress for seondary stresses. The referene stress solution for RCSCCL is provided in Annex D, paragraph D.5.8. a 3.0 α = = = t 9.53 t 9.53 τ = = = R o τ + ατ (0.0375) (0.0375) Z = α τ = = = σ SR ref { ( ) 9 ( α) } Q + Q + Z Q = 3( α) 0.5 b b m { } ( ) + ( ) + 9 (.4507)(9.7000) ( 0.348) = 3( 0.348) = MPa 0.5 SR j) STEP 0 Compute K. The stress intensity fator solution for KCSCCL is provided in Annex C, paragraph C SR a a a a KI = G0{ σ 0 + p} + G σ + G σ + G3 σ3 + G4 σ4 π a t t t t SR KI SR KI ( ) (3.584) + ( ) ( ) (0.348) = + π = 3 + ( ) ( ) (0.348) 4 + (0.4894) ( ) (0.348) (0.6036) ( ) (0.348) (3.0) = MPa m MPa mm 9-9

247 k) STEP Compute the plastiity interation fator, with P L r from STEP 7 L SR r SR σ ref = = = σ 40.0 ys ψ and ϕ are alulated from Tables 9.3 and 9.5 respetively P L r = ψ = SR L ϕ = r = and, Φ ψ =.0 + =.0 + =.0709 Φ ϕ < = , then Φ =.0 and Φ =.0709 SR Sine ( L r ) l) STEP Determine toughness ratio or ordinate of the FAD assessment point. o K r P SR KI +Φ KI (.0709)(3.6009) = = = K.474 mat m) STEP 3 Evaluate the results. P Lr -axis of the FAD Sine the hardening harateristis of the material are not known, the following value an be used (see Figure 9.0, Note ): ) STEP 3. Determine the ut-off for the L = ( ).0 r max ) STEP 3. Plot the assessment point on the FAD shown in Figure 9.0. P ( L, K ) = (0.399,0.50) r r The point is inside the FAD (see Figure E9.6-3) Note: Equation (9.33) under Figure 9.0 gives the maximum allowable { } P P ( ) ( ) 6 r,max = 0.4 r exp 0.65 r K L L { } ( ) ( ) 6 = exp = P K for L = : r r 9-30

248 . Kr Lr Assessment Curve Assessment Point Figure E FAD with the Assessment Point The Level Assessment Criteria are Satisfied. The pipe is aeptable for ontinued operation. 9-3

249 9.7 Example Problem 7 A rak-like flaw has been found in the irumferential seam on the outside surfae of a pipe during a sheduled turnaround. The pipe and inspetion data are provided below. The piping system was onstruted to the ASME B3.3 Code, 003 Edition. Determine if the pipe is aeptable for ontinued operation. Pipe Data: Idential to those of Example 9.6 Operating Conditions: Idential to those of Example 9.6 Inspetion Data The flaw is loated in a irumferential weld seam on the outside surfae of the pipe on its lower part. The seam is a single V-groove weld. The flaw is parallel to the weld seam. The depth of the flaw was established by UT; onsistent readings were noted and a final value for the flaw depth was established at 4.0 mm. The flaw length was established by MT and is 5.0 mm. The rak-like flaw is situated midway between supports, the distane of whih is 0.5 m. Perform a Level Assessment per paragraph at the deepest point and at surfae points of the rak (minimum required for a semi-elliptial surfae rak) and at points at 45 degrees on the rak front of the flaw. a) STEP Evaluate operating onditions and determine the pressure, temperature and supplemental loading ombinations to be evaluated: See Example 9.6 T = 0 C P = 3.0 MPa M x 36.8 (0) M = 0 N mm y 6 = N mm b) STEP Determine the stress distribution at the loation of the flaw - The primary stress distribution is based on the applied loads. ) Primary Stress The flaw is loated away from all major strutural disontinuities. From Annex C, Table C., the flaw geometry, omponent geometry, and loading ondition orrespond to KCSCCE and RCSCCE, Cylinder - Surfae Crak, Cirumferential Diretion - Semi-Elliptial Shape, Through-Wall Fourth Order Polynomial Stress Distribution with a Net Setion Bending Stress. The stress intensity fator solution for KCSCCE is provided in Annex C, paragraph C.5.4. The referene stress solution for RCSCCE is provided in Annex D, paragraph D.5.4. Ro = D/ = 508 / = mm Ri = Ro t = = mm R = ( R + R) / = ( ) / = mm m o i { } M / π ( R R ) = 36.8 (0) / π (54.00) (44.47) = N/ mm x o i The primary stress for the alulation of the stress intensity fator is written as: 3 4 P P P x P x P x P x P P x = 0 + p σ ( ) ( σ ) σ σ σ σ ( σ σ ) t t t t 9-3

250 With: p P 0 P 5 { } σ = PR / ( R R ) = (3.0) (44.47) / (54.00) (44.47) = MPa i o i σ = MR π R R = = MPa x o / ( o i ) 4(54.00) (0.0984) = σ = σ = σ = σ = σ = 0 P P P P P The membrane and bending omponents of the primary stress for the alulation of the referene stress are: P Pm = σ 0 = MPa P = 0 MPa b ) Seondary Stress: See Example 9.6 3) Residual Stress: See Example 9.6 The residual stress for the alulation of the stress intensity fator is written as: 3 4 R R R x R x R x R x ( x) = σ σ σ σ σ σ t t t t With: R R σ 0 = σ R σ 3 = = σ = σ = The membrane and bending omponents of the residual stress for the alulation of the referene stress are: R R 4 r m ys r r Q = σ m σ R = MPa r b ys r r Q = σ b σ R = MPa ) STEP 3 Determine the material properties; yield strength, tensile strength and frature toughness. See Example 9.6 σ σ ys uts KIC = 40 MPa = 45 MPa = MPa m d) STEP 4 Determine the rak-like flaw dimension from inspetion data. a = 4.0 mm = 5.0 mm 9-33

251 e) STEP 5 Modify the primary stress, material frature toughness, and flaw size using Partial Safety Fators. See Example 9.6 Pm = Pm PSFs Pb = Pb PSFs = ( ) (.5) = MPa = (0) (.5) = 0 MPa σ = σ PSFs = ( ) (.5) = MPa σ = PSF = (0.583) (.5) = MPa P P 0 0 P P 5 σ5 s 6 6 M = M PSFs = 36.8 (0) (.5) = 55. (0) N mm Kmat = Kmat / PSFk = (.474) / (.0) =.474 MPa m a= a PSF = (4.0) (.0) = 4.0 mm a f) STEP 6 Compute the referene stress for the primary stress. The referene stress solution for RCSCCE is provided in Annex D, paragraph D.5.4. x= a/ t = (4.00) / (9.53) = τ = t/ Ro = (9.53) / (54.0) = θ = π / 4 Ro = π (7.5) / 4(54.0) = α = ( a/ t) / ( + t/ ) = (4.00) / (9.53) / + (9.53) / (7.50) = Intermediate oeffiients for { } { [ ]} D.5.3 σ ref are: ( τ)( τ + xτ) + ( τ + xτ) { + ( τ)( τ) } ( )( ( ) (0.4973)( )) ( (0.4973)( )) + ( )( ) A= x + + = (0.4973) = 0.05 { } ψ = aros sin = aros 0.05sin(0.039) =.5660 [ A θ ] [ ] τ + ατ Z = π / ψ xθ τ ( ) + (0.4973)( ) = π / (.5660) (0.4973)(0.039) =.006 Leading to: σ D.5.3 ref { ( ) 9 ( α) } P + P + Z P = 3( α) 0.5 b b m { } 0.5 (0) + (0) + 9 (.006)(56.650) ( 0.848) = = MPa 3 ( 0.848) --``,,`,,,,`,,,,`,,``,`````-`-`,,`,,`,`,,` 9-34

252 Intermediate oeffiient for the bending moment part of the referene stress is: π θ a P m π β = = =.954 π t σ ys π Therefore: θ + β π Leading to: σ P ref = M σ a + Rt m sinβ sinθ t D.5.3 ref 6 55.(0) = + (56.96) 4.00 (49.35) (9.53) sin(.954) sin(0.039) 9.53 = (5.88) + (56.96 ) = MPa g) STEP 7 Compute the Load Ratio ( L ) or absissa of the FAD. σ L ys P r = 40.0 MPa P σ ref = = = σ 40.0 ys P r P h) STEP 8 Compute K - The stress intensity fator solution for KCSCCE is provided in Annex C, paragraph C.5.4. The influene oeffiients required for the assessment are alulated by interpolation between values given in Table C5 for raks on the outside surfae with: a/ t = (4.00) / (9.53) = a/ = (4.00) / (7.50) = t / R = (9.53) / (44.47) = i They are given in the Table E9.7- exept oeffiients for 6 Influene oeffiients G i are alulated by: P R G sine σ 6 = 0 and 6 G = A + A β + A β + A β + A β + A β + A β i 0, i, i, i 3, i 4, i 5, i 6, i With β = φ / π σ do not exist. At the deepest point of the flaw: φ = π / β = : At the surfae points of the flaw: φ = 0 β = 0 : G G G 0 5 G G G 0 5 =.907 = =.697 =.047 = =

253 Table E Coeffiients A i, j used to alulate influene oeffiients G 0, G and G 5 t/ri a/ a/t A 0 A A A 3 A 4 A 5 A 6 G i G G G G G G G G G G G G G G G G G G G G G G G G G G G G G G G G G G G G G G G G G G G G G

254 At 45 degrees on the rak front of the flaw: φ = π /4 β = 0.5 : G G G 0 5 =.09 = 0.50 =.073 Influene oeffiients G, G 3 and G 4 will be needed for the stress intensity fator due to residual stresses. They are alulated per paragraphs C.4. and C.4.3 of Annex C for the surfae points and the deepest point of the rak a 4.0 Q = = = At the deepest point of the flaw: π / Q = π / (.589) = ( π )( ) { } M = / Q 3G G 4.8 = (3.605) 3(0.7396) (.907) 4.8 =.0937 M 0 = ( π )( ) { } M = 3 / Q G G +.6 = 3(3.605) (.907) (0.7396) +.6 = Q / π = (.589) / π = Q 6 M 6 M M 3 G = π (3.000).500 = (0.5548) = Q 3 M 3 M M 3 G3 = π (3.000).500 = (0.5548) = Q 56 M 56 M M 3 G4 = π (3.000).500 = (0.5548) = At the surfae points of the flaw: 3 π / Q = 3 π /.589 = ( π )( ) { } ( π )( ) { } ( π )( ) { } N = 3 / Q G 5G 8 = (7.6473) (.047) 5(0.868) 8 = N = 3 / Q 3G G + 5 = 5(7.6473) 3(.047) (0.868) + 5 = N = 3 / Q 3G 0G 8 = (7.6473) 3(.047) 0(0.868) 8 =

255 G G G Q / π =.589 / π = Q 4 N 4N N3 π (0.8057) 4( 3.443) (.6374) = (0.393) = = Q 4 N 4 N N3 π (0.8057) 4( 3.443) (.6374) = (0.393) = = Q 4 N 4 N N π (0.8057) 4( 3.443) (.6374) = (0.393) = = The influene oeffiients G, G 3 and G 4 for points at 45 degrees on the rak front of the flaw are alulated per paragraph C.4.4 of Annex C with = /4 φ z = sinφ = sin ( π / 4) = Q / π =.589 / π = δ = + z = =.3066 π / Q = π /.589 =.549 ω = z = = 0.54 η = (/ z) = (/ 0.707) = M M π { 050G 05 G (3 7 z) } { 35 70z+ 35z + 89δz + 6δz } π + + = 4 Q ( zz ) δ ( zz ) = ( π / Q)( / ) 4( / ) = { 0G 90G z} π + = + Q ( + z+ 9 z ) η { 8 + 4z 5z + 44δ z } ( + z+ 9 z ) η = ( π / Q)( / 6.49) + ( / 6.49) = G = z+ 576z 864 z + ( M ) δ z 3.5 = (7.809) + ( M )(0.5493) = G M z z z z.5 3 = 3 (45η+ 54η + 7η 35ω + 44 η ) = M ( ) = ( ) G = ( Q / π ) G + G / 945 = δ 9-38

256 G = z+ z z 64 z + ( M ) δ z = (480.74) + ( M)(0.3884) = G3 = M3 (385η+ 440ηz+ 58ηz + 704ηz 3465ωz ηz ) = M ( ) = ( ) G = ( Q / π ) G + G /3860 = G = z+ 338z + 584z z 3380 z + ( M ) δ z = ( ) + ( M)(0.747) = G4 = M3 (89η+ 909ηz+ 040ηz + 48ηz + 664ηz 9009ωz ηz ) = M 3 (80.893) = G = ( Q / π )( G + G )/ = The stress intensity fators are: At the deepest point of the flaw: [ ] [ ] P KI = G0σ0 + G5σ5 π a/ Q = (.907)(56.650) + (.697)(30.374) π (4.0) /.589 = MPa mm = MPa m At the surfae points of the flaw: [ ] [ ] P KI = G0σ0 + G5σ5 π a/ Q = (.047)(56.650) + (.05)(30.374) π (4.0) /.589 = MPa mm = 8.67 MPa m At 45 degrees on the rak front of the flaw: [ ] [ ] P KI = G0σ0 + G5σ5 π a/ Q = (.09)(56.650) + (.073)(30.374) π (4.0) /.589 = 7.0 MPa mm = MPa m SR i) STEP 9 Compute the referene stress for seondary stresses. Note that σ ref used in this alulation is based on the membrane and bending omponents of the residual stress ( Q m and Q b ) from STEP. SR Details regarding the alulation of the referene stress are provided in STEP 6 with σ ref restrited to its D.5.3 part. σ SR ref { ( ) 9 ( α) } Q + Q + Z Q = 3( α) 0.5 b b m { } 0.5 ( ) + ( ) + 9 (.006)(9.7000) ( 0.848) = = MPa 3 ( 0.848) j) STEP 0 Compute K STEP 8. SR. Details regarding the alulation of the stress intensity fator are provided in 3 4 SR a a a a KI = G0σ0 + Gσ + Gσ + G3σ3 + G4σ4 π a/ Q t t t t 9-39

257 The stress intensity fators are: At the deepest point of the flaw: SR KI 4.0 (.907)(3.584) + (0.7396)( ) = + (0.579)( ) + (0.485)( ) π (4.0) / (0.444)( ) 9.53 = MPa mm = MPa m At the surfae points of the flaw: SR KI 4.0 (.047)(3.584) + (0.868)( ) = + (0.0739)( ) + (0.0388)( ) π (4.0) / (0.037)( ) 9.53 = MPa mm = 6.34 MPa m At 45 degrees on the rak front of the flaw: SR KI 4.0 (.09)(3.584) + (0.50)( ) = + (0.3094)( ) + (0.993)( ) π (4.0) / (0.339)( ) 9.53 = 0.49 MPa mm = MPa m k) STEP Compute the plastiity interation fator, with L SR r SR σ ref = = = σ 40.0 ys ψ and ϕ are alulated from Tables 9.3 and 9.5 respetively P L r from STEP 7 P L r = ψ = Φ ψ SR =.0 + =.0 + =.065 L ϕ = 0.30 r = Φ0 ϕ < = , then Φ =.0 and Φ =.065 SR Sine ( L r ) o 9-40

258 l) STEP Determine toughness ratio or ordinate of the FAD assessment point. At the deepest point of the flaw: K r P SR KI +Φ KI (.065)( ) = = = 0.7 K.474 mat At the surfae points of the flaw: K r P SR KI +Φ KI (.065)(6.34) = = = 0.68 K.474 mat At 45 degrees on the rak front: m) STEP 3 Evaluate the results. ) STEP 3. Determine the ut-off for the K r P SR KI +Φ KI (.065)( 6.393) = = = K.474 mat L -axis of the FAD See Example L P (max) =.0 ) STEP 3. Plot the assessment point on the FAD shown in Figure 9.0. P r P At the deepest point of the flaw: ( L, K ) = (0.34,0.7) ; the point is inside the FAD r r P At the surfae points of the flaw: ( L, K ) = (0.34,0.7) ; the point is inside the FAD r r P At 45 degrees on the rak front: ( L, K ) = (0.34,0.38) ; the point is inside the FAD r r r. Kr Lr Assessment Curve Surfae Points Deepest Point 45 Degrees Point Figure E FAD with Assessment Points The Level Assessment Criteria are Satisfied. The pipe is aeptable for ontinued operation. 9-4

259 9.8 Example Problem 8 A rak-like flaw has been found in the longitudinal seam on the inside surfae of a ylindrial pressure vessel during a sheduled turnaround. The vessel and inspetion data are provided below. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division, 998 Edition. Determine if the vessel is aeptable for ontinued operation. Vessel Data: Idential to those of Example 9.5 exept the operating temperature whih is 00 F The fluid inside the vessel is non aggressive. Operating Conditions The vessel is not fully pressurized until the temperature is 40 o F. Below this temperature, the startup pressure remains under 00 psig. At shutdown, the pressure is dereased to 00 psig before letting the temperature drop below 40 F. In servie, the vessel is subjet to yli loading between no pressure and full pressure. Inspetion Data: Idential to those of Example 9.5 Subritial rak growth by fatigue is verified: Perform a Level 3 Assessment per paragraph to determine the remaining life of the vessel The methodology is the following: a) Perform an assessment of the rak at maximum loading b) If the assessment point is inside the FAD: ) Calulate the stress intensity fators at minimum loading ( K min ) and at maximum loading ( K max ), for the surfae points and for the deepest point ) Calulate the variation of stress intensity fators between minimum and maximum loading ( Δ K = K K ) max min 3) If Δ K is greater than the threshold then alulate the size inrement on eah dimension ( Δ a and Δ ) by applying the fatigue propagation law with a given number of yles, update the dimensions of the rak ( a+δ a and +Δ ) and the overall number of yles before repeating the whole proedure. 4) If Δ K is smaller than the threshold then the rak stops to propagate and the proedure ends ) If the assessment point is outside the FAD, then the flaw is not aeptable and the proedure ends. d) The remaining life is the overall number of yles before the FAD boundary is reahed The overall proedure requires a Level 3 assessment. However, the aeptability of any urrent flaw an be based on a Level Assessment. a) STEP Evaluate operating onditions and determine the pressure, temperature and supplemental loading ombinations to be evaluated: See Example 9.5 T = 40 F P = 00 psig b) STEP Determine the stress distribution at the loation of the flaw at maximum and at minimum loadings - The primary stress distribution at maximum loading is based on the applied loads. The primary stress distribution at maximum loading orresponds to the full pressure. The primary stress is nil at minimum loading --``,,`,,,,`,,,,`,,``,`` 9-4

260 ) Primary Stress: See Example 9.5 C Pm,max = Pm = σ m = 000 psi Pm,min = 0 psi P = P = P = 0 psi b,max m,min b ) Seondary Stress: See Example 9.5: No seondary stresses. 3) Residual Stress: This stress is onstant throughout the life of the vessel: See Example 9.5 r Qm,max = Qm,min = Qm = σ = psi Q = Q = 0 psi b,max b,min ) STEP 3 Determine the material data For yield strength, tensile strength and frature toughness, see Example 9.5 σ ys = 38 ksi σ uts = 70 ksi T = 43 F ( see Step 4 of the Level Assessment of Example 9.5) ref ( ref ) KIC = exp 0.0 T T 00 + = exp 0.0 ( ) = ksi in The rak lies in a ferriti steel in a non aggressive environment. The fatigue rak growth law used in the assessment is given in Annex F paragraph F.5.3..a, with a threshold given in paragraph F.5.3..d. da ( in / yle ) 8.6 (0) ( ) dn 0 K ksi in 3.0 = Δ for Δ K > ( Δ Kth =.8 ksi in) The number of yles for eah inrement must be small enough so as to onsider the stress intensity fators as onstant for the whole inrement. This will be heked during the iterations. The retained value is: Δ N = 00 yles Determine the ut-off for the L = P r(max).0 L -axis of the FAD See Example STEP 3. P r d) STEP 4 Determine the rak-like flaw dimensions from inspetion data. a = 0.0 in = 3.0 in =.60 in e) STEP 5 Modify the primary stress, pressure on the rak fae, material frature toughness, and flaw size using Partial Safety Fators. See Example 9.5 Pm,max = Pm,max PSFs = (000)(.5) = 8000 psi Pm,min = Pm,min PSFs = (0)(.5) = 0 psi P = P = P PSF = (0)(.5) = 0 psi b,max b,min b s 9-43

261 p = p PSF = (00)(.5) = 300 psig,max,max s p = p PSF = (0)(.5) = 0 psig,min,min s K = K / PSF = (85.870) / (.0) = ksi in mat mat k a= a PSF = (0.0)(.0) = 0.0 in a ASSESSMENT OF THE CRACK Note: Sine the proedure for growing raks is very similar but not idential to the proedure for non-growing raks, the differenes are emphasized by using bold haraters. f) STEP 6 Compute the referene stress for the primary stress at maximum loading. See Example 9.5 The P NS bending stress is nil, therefore the referene stress formula may be written as:.88.88(.60) λa = = = Ra 60 (0.) i 0.5 σ = M P ref s m,max λa λ a M t( λa) = λa +.533(0 ) λa (0.8397) (0.8397) = 6 4 = (0.8397) +.533(0 )(0.8397) NS M s = = =.059 a a t t M ( ) t λa P NS σ = M P = (.059)(8000) = psi ref s m,max g) STEP 7 Compute the Load Ratio ( L ) or absissa of the FAD assessment point at maximum loading. σ L ys P r = 38 ksi P σ ref = = = σ ys P r P h) STEP 8 Compute K at maximum loading and at minimum loading for the apex of the flaw. See Example 9.5 The influene oeffiients required for the assessment are: At the deepest point of the flaw ϕ = 90 o G0 = A0,0 =.984 At the surfae points of the flaw The stress intensity fators are: ϕ = 0 o G 6 = A = i,0 i=

262 At the deepest point of the flaw pr G π a (300)(6) (.984) π (0.0) K = K = = = ksi in (6) (60).0474 K P P o 0 I,max I Ro Ri Q P I,min = 0 ksi in At the surfae points of the flaw pr G π a (300)(6) (0.4498) π (0.0) K = K = = = ksi in (6) (60).0474 K P P o 0 I,max I Ro Ri Q P I,min = 0 ksi in i) STEP 9 Compute the referene stress for seondary stresses at maximum loading. See Example 9.5 SR NS There is no bending omponent, therefore: σ = M Q,max = (.059)(48000) = psi ref s m SR j) STEP 0 Compute K at maximum loading and at minimum loading for the apex of the flaw. See SR Example 9.5 Sine the seondary stresses are nil, K is based only on the residual stresses whih are onstant with respet to time. The stress intensity fators are: At the deepest point of the flaw: SR SR SR π a π (0.) KI,max = KI,min = KI = G0σ 0 = (.984)(48.0) = ksi in Q.0474 At the surfae points of the flaw: SR SR SR π a π (0.) KI,max = KI,min = KI = G0σ 0 = (0.4498)(48.0) = ksi in Q.0474 k) STEP Compute the plastiity interation fator at maximum loading. See Example 9.5 L SR r SR σ ref = = =.959 σ ys P L r = ψ = Φ ψ SR = + = + = L.959 ϕ = Φ0 ϕ r = 0 < = , then Φ =.0 and Φ =.448 SR Sine ( L r ) o l) STEP Determine toughness ratio (K r ) or ordinate of the FAD assessment point at maximum loading. At the deepest point of the flaw: At the surfae points of the flaw: m) STEP 3 Evaluate the results at maximum loading. K K r r P SR KI +Φ KI (.448) = = = K mat P SR KI +Φ KI (.448) = = = 0.74 K mat --``,,`,,,,`,,,,`,,``,````` 9-45

263 Determine the maximum allowable Chek that P K for L = : r P P ( ) ( ) r { } 6 r,max = 0.4 r exp 0.65 r K L L K r { } ( ) ( ) 6 = exp = 0.96 P P K and that Lr Lr(max) r,max P At the deepest point of the flaw: ( L, K ) = (0.4859,0.789) ; the point is inside the FAD r r P At the surfae points of the flaw: ( L, K ) = (0.4859,0.74) ; the point is inside the FAD r r Both deepest point and surfae points are aeptable, the propagation of the flaw by fatigue is then evaluated. FATIGUE CRACK GROWTH n) STEP 4 Calulate the stress intensity fators at maximum loading and at minimum loading and their variation At the deepest point of the flaw: K = K + K = = ksi in P SR max I,max I,max K = K + K = = ksi in P SR min I,min I,min Δ K = K K = = ksi in max min At the surfae points of the flaw: K = K + K = =.383 ksi in P SR max I,max I,max K = K + K = = ksi in P SR min I,min I,min Δ K = K K = = ksi in max o) STEP 5 Chek that the rak is propagating min At the deepest point of the flaw: ( Δ K = ksi in ) ( Δ K.8 ) th = ksi in At the surfae points of the flaw: ( Δ K = ksi in ) ( Δ Kth =.8 ksi in ) The rak is propagating in the through thikness diretion and in the surfae diretion p) STEP 6 Calulate the size inrements in through thikness diretion ( Δ a ) and in surfae diretion ( Δ ) for the number of yles in STEP 3. At the deepest point of the flaw: Δ = Δ Δ = = a ( N) 8.6(0) ( K) (00) 8.6(0) (7.035) 4.54 (0) in At the surfae points of the flaw: Δ = Δ Δ = = ( N) 8.6(0) ( K) (00) 8.6(0) (5.994).786 (0) in 9-46

264 Table E Fatigue Crak Propagation - Dimensions and Referene Stress Parameters STEP Inrement N a a/ a/ t λ a M t NS M s 9-47

265 Table E Fatigue Crak Propagation -Referene Stresses and Plastiity Interation Fators STEP Inrement σ P ref L P r σ SR ref SR L ϕ ψ r Φ ``,,`,,,,`,,,,`,,``,` 9-48

266 Table E Fatigue Crak Propagation - Parameters and Stress Intensity Fators STEP G deep G0 surf K P I deep K P I surf K SR I deep Inrement Q K SR I surf 9-49

267 Table E Fatigue Crak Propagation - Evaluation of Result and Propagation STEP Chek Inr. deep r r r Δ a Δ Inrement Size deep surf E E-05 N.A N.A E E E-04.83E E-04.86E E E E E E E E-04.88E E E E-04.9E E-04.95E E E E E E E E E E-04.00E E-04.06E E-04.03E E E E E E E E E E-04.3E E-04.30E E-04.46E E-04.64E E-04.8E E-04.98E E-04.6E E-04.34E E-04.43E E E E-04.87E E E E E E E E E E E E E E E E E E E E E E E E E E E E E

268 q) STEP 7 Chek that the number of yles is small enough for the stress intensity fators to be onsidered as onstant during the inrement. ) Chek on rak dimensions: Δa 0.5 % ( a)? 4.54 (0) (0. )? 4.54 (0) 0.0 (0)? True in in Δ in in % ( a)?.786 (0) (0. )?.786 (0) 0.0 (0)? True ) Chek on stress intensity fators: (This does not apply to the st inrement) Δ K (at inrement k+) - Δ K (at inrement k) % Δ K (at inrement k)? For the seond inrement (see Tables E9.8- to E9.8-4 with detailed results of all inrements): At the deepest point of the flaw: ( = 0.045) ( % = 0.703) True At the surfae points of the flaw: ( = 0.046) ( % = 0.059) True r) STEP 8 Inrement rak dimensions and total number of yles 4 ak+ = ak + ( Δ a) k = (0) = in 5 k+ = k + ( Δ ) k = (0) = in N = N + ( Δ N) = = 00 yles k+ k s) Repeat STEPs 6 to 8 until the FAD boundary is reahed. The detailed results obtained with Mirosoft Exel are given in Tables E9.8- to E The hek on stress intensity fators is written as: (.0) Δ K (at inrement k) - Δ K (at inrement k+) 0 Note: Due to rak size inrements used in the rak propagation analysis, double preision is needed to ensure auray Table E9.8-4 shows that the assessment point is outside the FAD at inrement number 4. Therefore the allowable number of yles is given by inrement number 3, leading to: The remaining life of the vessel orresponds to 00 yles between no pressure and full pressure. An additional safety fator on this number of yles is reommended for atual vessel operation. 9-5

269 9.9 Example Problem 9 A rak-like flaw has been found in the base metal of a vessel. In order to take advantage of the atual properties of the material it is deided to perform a Level 3 Method B Assessment. Determine the material-speifi FAD used in the Level 3 Method B Assessment per paragraph for the material of the vessel a) STEP Obtain the engineering stress-strain urve data for the material of the vessel and determine the 0.% offset yield strength, tensile strength and modulus of elastiity. These data are obtained from test. The engineering stresses ( σ e ) and engineering strains ( ε e ) are smoothened. They are given in olumns and 3 of Table E9.9- and are represented by the urve in Figure E9.9-. Not all neessary values are output by the test. Missing values are obtained by interpolation; they are printed in bold haraters. The other material properties are: - 0.% offset yield strength: σ = 33.9 ksi - tensile strength: σ = modulus of elastiity: E = 9350 ys uts y ksi ksi b) STEP Convert the engineering stress-strain urve into a true stress strain urve as shown in Annex F, paragraph F..3.. The true stresses ( σ t ) and true strains ( ε t ) are given in olumns 4 and 5 of Table E9.9-. For σ / σ = 0.9: e ys σ t = ( + εe) σe = ( ) (30.50) = ksi ε = ln + = ln = = % t [ ε ] [ ] ) STEP 3 Determine the material-speifi FAD - K r (0.0000) =.000 For σ / σ = 0.9: e ys L = σ / σ = / = K r t ys r e y εref Lr σ ys 9350 ( ) (0.904) (33.90) E = + = + = Lr σ ys Ey ε ref (0.904) (33.90) (9350) ( ) σ σ ratios are given in olumns 6 and 7 of Table E9.9-. Column 8 gives the The values for the other e / ys value of K r for the Level FAD as given in Figure 9.0. The resulting FAD is shown in Figure E9.9-. σ σ = : K = 0.4( L ) exp 0.65( L ) For / 0.9 e ys { } 6 { } = 0.4( 0.904) exp 0.65( 0.904) = r,max r r 9-5

270 stress (ksi) Engineering Stress-Strain Curve strain (%) Figure E Engineering Stress-Strain Curve of the Material of the Vessel.00 Kr Failure Assessment Diagram Kr Method B Kr Figure Lr Figure E Material Dependent of the FAD used in the Level 3 Method B Assessment of the Vessel 9-53

271 Table E Stress-Strain Curves and Failure Assessment Diagram Parameters σ e / σ ys σ e (ksi) ε e (%) σ t (ksi) ε (%) L σ / σ t K r = r t ys Method B K r Figure

272 9.0 Example Problem 0 A rak-like flaw has been found in a forged nozzle of a ylindrial pressure vessel on its inside surfae during a sheduled turnaround. The vessel and inspetion data are provided below. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division, 00 Edition. Determine if the vessel is aeptable for ontinued operation. Vessel Data Material = SA-8 Grade F304 Year 00 Design Conditions = 6.0 MPa (60 0 C Shell Mean Diameter = 000 mm Shell Fabriated Thikness = 5 mm Nozzle Mean Diameter = 500 mm Nozzle Fabriated Thikness = 0 mm Angle between Shell and Nozzle = 90 degrees Fillet Radius between Shell and Nozzle = 0 mm (outside surfae) Fillet Radius between Shell and Nozzle = 5 mm (inside surfae) Uniform Metal Loss = 0.0 mm Future Corrosion Allowane = 0.0 mm Weld Joint Effiieny =.0 PWHT = No Inspetion Data The flaw is a orner rak loated in the longitudinal plane of the nozzle. Its shape is quarter-elliptial. Its dimensions were established by MT leading to a small axis of 0 mm on the shell side and 0 mm on the nozzle side with a enter at the intersetion of the inside surfaes of the shell and nozzle without fillet radius (see Figure E9.0-). The distane of the rak-like flaw to the nearest weld seam is large enough to neglet the residual stresses due to welding. It is deided to perform a Level 3 Method C Assessment per paragraph In 3D, the values of K I and σ ref vary along the rak front. It is deided to divide the rak front into 3 parts of equal length and to perform assessments at division points and at points lose to the shell and nozzle surfaes. The first assessment is performed at the division point on the shell side (see Figure E9.0-4.a) Level 3 Method C involves elasti-plasti Finite Element analyses. The omputations are performed with ANSYS, following the rules of Annex B paragraphs B.7.3 and B.7.4. The units are mm for the lengths and MPa for the stresses and pressures. Sine a rak is to be meshed in a large 3D struture in an elasti-plasti analysis, the mesh refinement around the rak front and the size of the load steps are first validated on a D model with a similar rak size. The material is the one of the nozzle. The speifi ommands to be added for the omputation of the J-integral are validated at the same time. A through-wall rak 40 mm long is onsidered at the enter of a plate 400 mm wide and 600 mm long. Due to the symmetries only one quarter of the plate is modeled with nil normal displaements on the surfaes of symmetry. The maximum tension applied on the plate is hosen as twie the yield strength. This phase shows that the elements near the rak tip on the lip side are so distorted at high loading that it is neessary to use P the "large displaement" option of the elements in order to obtain results near the ut-off for the L r of the FAD. 9-55

273 C L O = 500 mm t = 0 mm 0.5 mm Nozzle r = 0 mm r = 5 mm 3000 mm Shell t = 5 mm O = 000 mm C L (a) Overall Geometry of Nozzle 0 mm 0 mm (b) Detail of the raked region Figure E Craked Nozzle a) STEP - Categorize loads as primary and seondary. The only load is the pressure whih is onsidered as primary. There is no other mehanial or thermal load, therefore there is no seondary load b) STEP - Construt an elasti-plasti finite element model. 9-56

274 Due to the symmetries, only one quarter of the nozzle is modeled with SOLID86 elements. This is a 0-node brik element that degenerates into prism, pyramid or tetrahedron by merging nodes (see Figure E9.0-). The length of modeled shell is 3000 mm, the length of nozzle is 500 mm. The rak front itself is embedded in a torial region in order to ensure a regular spider mesh around it (see Figure E9.0-3). The size of the triangular faes next to the rak front is 0.5 mm. I M Y X T U Q P B L W N Z S J V R O A K M Y I U Q X N Z T J V R O,P,W A,B K,L,S (a) Cube (b) Prism M,N,O,P,U,V,W,X M,N,O,P,U,V,W,X I Y B T Q L A Z S R J K I Y Q T Z J A,B R K,L,S () Pyramid (d) Tetrahedron Figure E D 0-Node Solid Elements used for Elasti-Plasti Analysis A plane perpendiular to the rak front is also defined in order to enter the nodes that will be used by the software to desribe the ontour on whih the J-integral will be performed (see Figure E9.0-4), the first of these nodes being on the rak lip. 9-57

275 B A (a) Overall Mesh (b) Mesh in the Viinity of the Crak-Front Figure E D Mesh of the Craked Nozzle --``,,`,,,,`,,,,`,,``,`````-`-`,,`,,`,`,,`- 9-58

276 F (a) Assessment Node and Loal Axis of Crak Extension E A B D C (b) Nodes Defining the Contour for the J-Integral Projeted on the Shell Surfae Figure E Nodes used in Frature Mehanis Computations 9-59

277 For the boundary onditions, zero normal displaements are applied on surfaes of symmetry exept on the rak fae. The nodes on the line from point A to point B (see Figure E9.0-3) are assigned a nil displaement in the z-diretion (nozzle axis) where the resultant of the fores is equal to zero. For the loading, the applied pressure is a little bit above the upper limit that must be reahed by the elastiplasti analysis. This pressure orresponds to a membrane irumferential stress equal to the yield strength (05 MPa for SA-8 Grade F304) in the shell, alulated from the MAWP formula in paragraph A.3.4.a of Annex A. This leads to P = 0.0 MPa. It is thus deided to speify a pressure equal to.0 MPa on the internal surfaes of the shell and of the nozzle, and on the fae of the rak End effets are applied on the radial surfaes of the shell (tension = 4.07 MPa) and of the nozzle (tension = 69. MPa) alulated as Pr /( r r ). i o i For the material, the stress-strain relationship is based on the MPC Model as desribed in paragraph F..3. of Annex F for stainless steels with the basi properties E y, ν, σ ys and σ uts from ASME Setion II Part D. These data, expressed as a funtion of engineering stress and strain, are given in Table E9.0-: - Young modulus E y = MPa - Poisson ratio ν = Yield strength σ ys = 05 MPa - Tensile strength σ uts = 55 MPa - The stress-strain urve above the yield strength is pieewise linear. The inputs are the stress versus the plasti strain. Table E Data for the non-linear part of the stress-strain urve ε p σ - MPa ) STEP 3 - The elasti-plasti analysis is performed with an inreasing load. The size of eah step is regulated by the software. The default value is 0.833% of the presribed loading in order to generate load steps equal to 0. MPa; the maximum and minimal values are respetively % and 0. % of the presribed loading. At eah step the software outputs (olumns of Table E9.0-): the perentage of the total load reahed the orresponding pressure ui the J-integral alulated by formula J = σij εij dy σij nj ds Γ x on the ontour Γ defined by the user with the loal axes x and y suh that x lies in the rak plane and is normal to the rak-front toward the inside of the material and y is perpendiular to the rak lips (opening diretion) toward the inside of the material if only one side of the rak is modelled due to symmetry onsiderations d) STEP 4 - Calulate K defined as J K = J E ν based on the J integral alulated in STEP 3. J. y /( ) 9-60

278 Table E Frature Mehanis Parameters used to Define the FAD Curve % Total Load Pressure MPa J-Integral MPa mm K J MPa mm P K I MPa mm e) STEP 5 - Infer the elasti solution P K I for eah step. From olumns and 4 of Table E9.0- the urve K f ( pressure) figure, or from the values of ( K / ) P K I J = P. The values are given in olumn 5 of Table E9.0-. f) STEP 6 - Compute the vertial oordinate of the FAD, J K r = is drawn (Figure E9.0-5). In this pressure, the linear part of the urve is identified, leading to r K = K K J - Elasti-Plasti Equivalent Stress Intensity Fator - is represented by the Solid Line in Figure E P K I - Elasti Stress Intensity Fator - is represented by the Dotted Line in Figure E The values of Kr are given in olumn 6 of Table E9.0-. K P I J L r --``,,`,,,,`,,,,`,,``,`````-`-`,,`,,`,`,,` 9-6

279 K P Figure E Stress Intensity Fators versus Pressure g) STEP 7 - Compute the horizontal oordinate of the FAD, = / Calulate the value of K orresponding to L = r r L σ σ r ref ys ( σ ) Kr( Lr = ) = + ( 0.00 Ey) / σ ys + + ( 0.00 Ey) / ys 0.5 (0.00) (95000) (0.00) (95000) = = This orresponds to P = MPa = % max loading in Table E9.0- enabling to situate the value L r = in olumn 7 of Table E9.0- and then dedue the other values of this olumn. h) STEP 8 - Plot the FAD urve K r versus L r - See Figure E Kr Lr Figure E FAD with Assessment Point 9-6

280 i) STEP 9 - Compute L r for the operating load The applied primary stress at L r = is alulated per Annex A paragraph A.3.4 in the shell: P PL ( r = ) R (8.477) σ ( Lr = ) = = = 70.4 MPa E t (.0) 5 Leading to a referene stress geometry fator F P = σ / σ ( L = ) = (05.00) / (70.4) =.03 ref ys r The MAWP is 6.0 MPa. It generates an applied primary stress in the shell equal to P P R (6.0) σ = = = 0.6 MPa E t (.0) 5 and a referene stress equal to σ ref P = F. σ = (.03) (0.6) = 45. MPa ref Therefore L = σ / σ = (45.) / (05.0) = r ref ys Note: Sine there is only one primary load, L r ould also have been alulated as the ratio between the MAWP and the pressure orresponding to L r = i.e L r = (6.0) / (8.477) = j) STEP 0 - Compute the elasti P K for the operating load using formula in STEP 5 K = P P I I K = (350.4) (6.0) = 0. MPa mm = MPa m P I k) STEP - Compute the toughness ratio The toughness of this material, austeniti stainless steel, is taken from paragraph F.4.8. in Annex F. A onservative value equal to 3 MPa m is seleted (value for weld material) Therefore K = K P / K = (66.49) / (3.0) = r I mat l) STEP - Plot the assessment point ( L, K ) = (0.708, 0.504) on the FAD of STEP 8 - See Figure r r E9.0-6 m) STEP 3 - Evaluate the result The assessment point lies in the Aeptable Region of the FAD. A very onservative value of toughness is taken into aount and other data are known with suffiient auray, therefore The Level 3 Method C Assessment Criteria are Satisfied for the Assessment Point The assessment is repeated for other points on the rak-front. For point E of Figure E9.0-4(a) it is not possible to define a plane perpendiular to the rak-front beause this plane would be tangent to the shell interior surfae. Therefore the assessment at the apex of the small axis is performed at the next point on the rak-front. The same omment applies to point F (long axis) of Figure E9.0-4(a) on the nozzle side. 9-63

281 THIS PAGE INTENTIONALLY LEFT BLANK 9-64

282 PART 0 ASSESSMENT OF COMPONENTS OPERATING IN THE CREEP RANGE EXAMPLE PROBLEMS 0. Example Problem Example Problem Example Problem Example Problem Example Problem A liquid knok-out vessel that is part of a pressure relief system typially operates at temperatures below the reep range. During a reent upset ondition, high temperature liquid was relieved in the vessel for a period of time, subjeting the vessel to temperatures in the reep range. Details regarding the vessel and the upset ondition are given below. The shell ontains a weld seam whih was exposed to the exursion onditions. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division. Estimate the level of reep damage sustained by the vessel shell during the upset ondition. Vessel Data Material = SA 56Grade60 Year 998 Design Conditions = F Inside Diameter = 60 in Fabriated Thikness = in Future Corrosion Allowane (FCA) = 0.0 in Weld Joint Effiieny =.0 Unsupported Length = 44in Cylindrial Shell with : Elliptial Heads ( R ell = ) PWHT = Yes, Original Fabriation Requirement Temperature Exursion Data Exursion Pressure: = 8.6 psig Exursion Temperature: = 950 F Exursion Duration: = 0 hours Inspetion Data There are no visual signs of damage to the vessel, no bulging, metal loss, or exessive sale was noted. UT thikness readings indiated light general metal loss within the original orrosion allowane. Looking through the inspetion reords, this is the first operational exursion into the reep range for this omponent. 0-

283 Perform a Level Assessment for the omponent in reep servie per paragraph Eah omponent of the vessel must be analyzed separately. In this example, the ylindrial shell is analyzed first, followed by the elliptial heads. Nozzles and supplemental loadings are ignored for the purposes of this example. Level Assessment for the ylindrial shell a) STEP Determine the maximum operating temperature, pressure, and servie time the omponent was exposed to. Sine the omponent ontains a weld seam exposed to the exursion onditions, 5 F shall be added to the maximum operating temperature. Tmax = = 975 F Pmax = 8.6 psig time = 0 hours b) STEP Determine the operating stress of the omponent for the operating ondition defined in STEP using Annex A. The omputed nominal stress shall inlude the effets of servie-indued wall thinning. Definition of ommon variables: ( FCA) ( ) ID R = = = 30. in t = t FCA= = 0.75 in nom Supplemental loadings are not onsidered in this example. Cylindrial shell irumferential membrane stress (A.) C P R σ m = E t C σ m = σ = 909 psi C m Cylindrial shell longitudinal membrane stress (A.7) L P R σ m = 0.4 E t L σ m = L m ( ) σ = 4,504 psi max max max C L ( m m) σ = max σ, σ σ σ ( ) = max 909, 4504 = 909 psi 0-

284 ) STEP 3 Determine the material of onstrution for the omponent and find the figure with the sreening and damage urves to be used for the Level assessment from Figures 0.3 through 0.5. The ylindrial shell is onstruted of SA-56 Grade 60, arbon steel; therefore, Figure 0.3 shall be used for the analysis. d) STEP 4 Determine the maximum permissible time for operation based on the sreening urve obtained from STEP 3, the nominal stress from STEP, and the assessment temperature from STEP. If the time determined from the sreening urve exeeds the servie time for the omponent from STEP, then the omponent is aeptable per the Level Assessment proedure. From Figure 0.3, the aeptable reep life of the ylindrial shell at 0 ksi and 975 F is over 5 hours. Sine the omponent was only exposed to these onditions for 0 hours, and has no history of prior temperature exursions on reord, the omponent is fit for servie without further evaluation. However, it is important to note the temperature exursion in the vessel s files so that future analyses an aurately take into aount all past temperature exursions. Level Assessment for the : elliptial heads a) STEP Determine the maximum operating temperature, pressure, and servie time the omponent was exposed to. Sine the omponent ontains a weld seam exposed to the exursion onditions, 5 F shall be added to the maximum operating temperature. T max = = 975 F Pmax = 8.6 psig time = 0 hours b) STEP Determine the operating stress of the omponent for the operating ondition defined in STEP using Annex A. The omputed nominal stress shall inlude the effets of servie-indued wall thinning. Definition of ommon variables: ( ) ( ) D= ID+ FCA = = 60. in t = t FCA= = 0.75 in nom Elliptial head membrane stress (A.33 and A.3) ( ( ) ) K =.0 + R ell 6 K = K =.0 ( ) ( ) P DK σ m = + 0. = + 0. = 9049 psi ( E) t (.0) 0.75 ) STEP 3 Determine the material of onstrution for the omponent and find the figure with the sreening and damage urves to be used for the Level assessment from Figures 0.3 through 0.5. The heads are onstruted of SA-56 Grade 60, arbon steel; therefore, Figure 0.3 shall be used for the analysis. d) STEP 4 Determine the maximum permissible time for operation based on the sreening urve obtained from STEP 3, the nominal stress from STEP, and the assessment temperature from STEP. If the time determined from the sreening urve exeeds the servie time for the omponent from STEP, then the omponent is aeptable per the Level Assessment proedure. 0-3

285 From Figure 0.3, the aeptable reep life of the ylindrial shell at 0 ksi and 975 F is over 5 hours. Sine the omponent was only exposed to these onditions for 0 hours, and has no history of prior temperature exursions on reord, the omponent is fit for servie without further evaluation. However, it is important to note the temperature exursion in the vessel s files so that future analyses an aurately take into aount all past temperature exursions. 0-4

286 0. Example Problem A fired rude heater experiened a temperature exursion for a short duration. The refinery needs to know how muh additional damage ourred to the tubes to understand how the exursion impats the remaining tube life. This information will be used to help determine if the heater will need to be re-tubed at an upoming sheduled turn-around, or if the tubes are likely to last for another run. Evaluate the tube remaining life (typial past history plus known temperature exursion) and determine if they are fit for servie for another run. Heater Tube Data Material = SA 335 Grade P Year 998 Typial Conditions (j=) = 0 F Outside Diameter = 8.65 in Fabriated Thikness = 0.3 in Future Corrosion Allowane (FCA) = 0.0 in Weld Joint Effiieny =.0 Unsupported Length = 44 in Cylindrial Shell Past Operating Time (j=) = 3400 hours Past Operating Time (j=) = 336 hours Future Expeted Time (j=) = hours Temperature Exursion Data Exursion Pressure: = 0 psig Exursion Temperature: = 0 F Exursion Duration: = 336 hours Inspetion Data There are no visual signs of damage to the tube, no bulging, metal loss, or exessive sale was noted. UT thikness readings indiated light general metal loss within the original orrosion allowane. Looking through the inspetion reords, this is the first operational exursion into the reep range for this omponent. There are no weld seems in the fire box. Perform a multiple ondition Level Assessment for the omponent in reep servie per paragraph Eah omponent of the vessel must be analyzed separately. In this example, the tube bends are loated outside the firebox, so only the ylindrial portion of the tubes will be analyzed. For the purposes of this example, assume the tubes are adequately supported and that irumferential pressure stress is the limiting design ondition. 0-5

287 Level Assessment for the heater tube a) STEP Determine the maximum operating temperature, pressure, and servie time the omponent was exposed to. The omponent does not ontain any weld seams exposed to the exursion onditions; therefore, it is not neessary to add the 5 F to the maximum operating temperature. The supersript j, indiates either for the typial operating onditions (design) or for the temperature exursion. T max max = total max max = total P time T P time = 5 F 0 psig = = 7500 hours = 0 F 0 psig = 336 hours b) STEP Determine the nominal stress of the omponent for eah of the operating onditions defined in STEP using Annex A. The omputed nominal stress shall inlude the effets of servie-indued wall thinning. Definition of ommon variables: OD 8.65 R = tnom + FCA= = in t = t FCA= = 0.75 in nom Supplemental loadings are not onsidered in this example. Cylindrial shell irumferential membrane stress (A.) C P R σ m = E t C σ m = σ = 3995 psi C m Cylindrial shell longitudinal membrane stress (A.7) L P R σ m = 0.4 E t L σ m = L m ( ) σ = 893 psi max max max C L ( m m) σ = max σ, σ σ σ ( ) = max 3995,893 = 3995 psi ) STEP 3 Determine the material of onstrution for the omponent and find the figure with the damage urves to be used for the Level assessment from Figures 0.3 through 0.5. The ylindrial shell is onstruted of SA-335 P,.5Cr-.0Mo annealed steel; therefore, Figure 0.9 shall be used for the analysis. 0-6

288 d) STEP 4 Determine the reep damage rate, R and assoiated reep damage, j j D for eah of the j operating onditions defined in STEP using the damage urve obtained from STEP 3, the nominal stress from STEP, and the assessment temperature from STEP. The reep damage for eah operating ondition, j, an be omputed using Equation (0.6) where the servie exposure time is determined from STEP. D = R t ( ) j j j se The reep damage rate, j = ) are: R =.75 0 / Hr D R t D D 6 = ( se) 6 = = j R and the assoiated reep damage, ( ) The reep damage rate, R j and the assoiated reep damage, ondition (j = ) are: 5 R = / Hr D = R ( tse) 5 D = ( 336) D = 0.08 j D for the typial operating ondition ( j D for the temperature exursion e) STEP 5 Determine the reep damage for the total number of operating onditions, J, using Equation (0.7). D total J = D j= j The reep damage for the total number of operating onditions, J, is determined as follows: D D total total J j= j j D j= ( ) total D = D + D total D = + D total = = = D ( ) f) STEP 6 If the total reep damage determined from STEP 5 satisfies Equation (0.8), then the omponent is aeptable per the Level Assessment proedure. Otherwise, the omponent is not aeptable and the requirements of paragraph shall be followed. total D 0.5 total In this ase, the total reep damage determined in STEP 5, D = exeeds the allowable per Equation (0.8). Therefore, the Level assessment riteria are not satisfied. This same problem is examined further with a Level assessment in Example Problem

289 0.3 Example Problem 3 A fired rude heater experiened a temperature exursion for a short duration. The refinery needs to know how muh additional damage ourred to the tubes to understand how the exursion impats the remaining tube life. This information will be used to help determine if the heater will need to be re-tubed at an upoming sheduled turn-around, or if the tubes are likely to last for another run. The tubes have already failed a Level assessment (Example Problem ). Evaluate the remaining life of the tubes, using the Level assessment proedures, and determine if they are fit for servie for another run. Heater Tube Data Material = SA 335 Grade P Year 998 Typial Conditions (j=) = 0 F Outside Diameter = 8.65 in Fabriated Thikness = 0.3 in Future Corrosion Allowane (FCA) = 0.0 in Weld Joint Effiieny =.0 Unsupported Length = 44 in Cylindrial Shell Past Operating Time (j=) = 3400 hours Past Operating Time (j=) = 336 hours Future Expeted Time (j=) = hours Temperature Exursion Data Exursion Pressure: = 0 psig Exursion Temperature: = 0 F Exursion Duration: = 336 hours Inspetion Data There are no visual signs of damage to the tube, no bulging, metal loss, or exessive sale was noted. UT thikness readings indiated light general metal loss within the original orrosion allowane. Looking through the inspetion reords, this is the first operational exursion into the reep range for this omponent. There are no irumferential weld seams in the fire box. Perform a multiple ondition Level Assessment for the omponent in reep servie per paragraph Eah omponent of the vessel must be analyzed separately. In this example, the tube bends are loated outside the firebox, so only the ylindrial portion of the tubes will be analyzed. For the purposes of this example, assume the tubes are adequately supported and that irumferential pressure stress is the limiting design ondition. 0-8

290 Level Assessment for the heater tube a) STEP Determine a load history based on past operation and future planned operation. The load history for this example inludes three operating onditions as listed below: Table E0.3- Past (m = ) Exursion (m= ) Future (m = 3) Design Pressure (P) 0 psig 0 psig 0 psig Design Temperature (T) 5 F 0 F 5 F Servie Time (hours) b) STEP For the urrent operating yle m, determine the total yle time, m t, and divide the yle into a number of time inrements, n t as shown in Figure 0.7. Define N as the total number of time inrements in operating yle m. For this illustration, N is set to even though the ondition for eah sub-yle is the same. In general, N should be set to math any hange in pressure, temperature, or tube thikness. Eah of the operating yles in the load history is split into its respetive sub-inrements below: Table E0.3- Operating Cyle Past (m = ) Exursion (m = ) Future (m = 3) Sub-Inrement n = n = n = n = n = n = Design Pressure (P) Servie Time (hours) ) STEP 3 Determine the assessment temperature, n T, for the time inrement n t. Table E0.3-3 Operating Cyle Past (m = ) Exursion (m = ) Future (m = 3) Sub-Inrement n = n = n = n = n = n = Design Pressure (P) Servie Time (hours) Design Temperature (T)

291 d) STEP 4 Determine the stress omponents, n σ ij, for the time inrement n t. First, the tube dimensions are heked to insure the tubes are onsidered to be thin-walled per the definition given in paragraph OD > 6 tnom 8.65 > > 6 Sine the thin-walled riterion is met, the mean diameter stress equations per Table 0. are appliable. For this example a fully-orroded thikness is used for simpliity. A more realisti approah is to alulate the stress as a funtion of the thikness aording to the past and predited orrosion rates. An example of this alulation is worked out below for the first sub-inrement of the first operating yle. The subsequent inrements are alulated similarly. n n PD ( mean ) σ = σmean = ( torr ) POD ( tnom + FCA) σ = ( tnom FCA) 0( ) σ = (0.3 0.) σ = 3974 psi n n σ = 0.5( σmean) σ = 0.5( σ) σ = 0.5(3974) σ = 987 psi n σ 3 = 0.0 psi n n σe = 0.866( σmean) σe = 0.866( σ) σ e = 0.866(3974) σ = 344 psi e 0-0

292 Eah of the stress omponents are inluded in the table shown below: Table E0.3-4 Operating Cyle Past (m = ) Exursion (m = ) Future (m = 3) Sub-Inrement n = n = n = n = n = n = Design Pressure (P) Servie Time (hours) Design Temperature (T) n σ xx = n σ (psi) n σ yy = n σ (psi) n σ zz = n σ 3 (psi) n τ xy (psi) n σ e (psi) e) STEP 5 Determine if the omponent has adequate protetion against plasti ollapse. Sine the primary load referene stress, n σ p ref, is less than 75% of the minimum yield strength, the plasti ollapse riteria are satisfied. The stress in the omponent is onstant in this example, therefore the results below are valid for all operating yles and sub-inrements. n n n σ σ σ p ref p ref p ref = ( 9( )) P + P + P n n n b b L 3 ( ( )) = 3 = 3974 psi ( ) ( ) σ ys 5 = 985 psi σ ys 0 = 5034 psi n p σref min 0.75 σ ys 5,0.75 σ ys 0 n p σ ref min , psi 76 psi ( ( )) ( ) ( ) ( ) ( ) n n n f) STEP 6 Determine the prinipal stresses, σ, σ, σ 3 and the effetive stress, n σ e. Thin-walled tubes experiene a bi-axial stress state and the shear stress is zero; therefore, the prinipal n n n n n n stresses are given by the stress omponents alulated in STEP 4 ( σ xx = σ, σ yy = σ, σ zz = σ 3 ). The table given at the end of STEP 4 inludes the prinipal stresses. g) STEP 7 Determine the remaining life at the stress level n σ e and temperature n T for time inrement n t by utilizing reep rupture data for the material and designate this value as n L. All stresses are in ksi and all temperatures are in F, the orresponding time to rupture is in hours. 0-

293 Material onstants for the Omega method reep remaining life alulation, see Annex F, Table F.30 for.5cr-mo annealed. A o =.86 A = 5005 A = 5436 A3 = 500 A = B o =.85 B = 705 B = 436 B3 = 0.0 B = For a ylinder or one α Ω =. The MPC Projet Omega parameter is defined as β Ω =. An example 3 alulation for the remaining life at the stress level n σ and temperature n T for time inrement n t is shown below. For this example, the adjustment fators for reep dutility set to 0.0 n S σ l S S l l = log 0 ( e) ( ) = log = e d Δ Ω and reep strain sr Δ Ω are sr 3 log ε = 0 o ( A +Δ ) + A AS AS AS o Ω n l 3 l 4 l T 3 log ε = 0 ( ) ( ) + 500( ) ( o ) log ε = o 8 ε =.99 0 / Hr o 3 log0 Ω= d ( Bo +Δ Ω ) + B BSl B3Sl B4S n l 460 T log0 Ω= ( ) ( 0.537) 0.0( 0.537) 0.0( 0.537) log Ω= Ω= ``,,`,,,,`,,,,`,,``,`````-`-`,,`, 0-

294 nbn = A AS 3 l 3A n + + 4Sl T n n BN BN = 5436 ( 500)( 0.537) 3( 3400)( 0.537) = ( BN) ( ) Ω n = max Ω n, 3.0 Ω n = max , 3.0 Ω = n δ δ δ Ω Ω Ω n n n σ+ σ + σ 3 = βω.0 n σ e = = 0.44 Ω =Ω + α n δω m n Ω BN Ω = + m ( ) Ω = 9.43 n m L = ε Ω L = o m ( ) L = hours The remaining life for eah other inrement is alulated similarly. 0-3

295 h) STEP 8 Repeat STEP 3 through STEP 7 for eah time inrement n t in the mth operating yle to determine the rupture time, n L, for eah inrement. The results for eah time period are inluded in the table below. Table E0.3-5 Operating Cyle Past (m = ) Exursion (m = ) Future (m = 3) Sub-Inrement n = n = n = n = n = n = Design Pressure (P) Servie Time (hours) Design Temperature (T) n σ xx = n σ (psi) n σ yy = n σ (psi) n σ zz = n σ 3 (psi) n τ xy (psi) n σ e (psi) Remaining Life = n L (hrs) i) STEP 9 Compute the aumulated reep damage for all points in the mth yle using Equation (0.5). m D D D = = n n= N n n n= t L n t L t t = + = + = L L j) STEP 0 Repeat STEP through STEP 9 for eah of the operating yles defined in STEP. The results for eah operating yle are inluded in the table below. Table E0.3-6 Operating Cyle Past (m = ) Exursion (m = ) Future (m = 3) Sub-Inrement n = n = n = n = n = n = Design Pressure (P) Servie Time (hours) Design Temperature (T) n σ xx = n σ (psi) n σ yy = n σ (psi) n σ zz = n σ 3 (psi) n τ xy (psi) n σ e (psi) Remaining Life = n L (hrs) Damage = m D

296 k) STEP Compute the total reep damage for all yles of operation. M total m allow = m= D D D 3 total m allow = m= D D D total 3 allow D = D + D + D D = total D = l) STEP The reep damage predition is omplete for this loation in the omponent. Follow the requirements of Part 0 to determine the reommended ations. total allow For this example, sine the total damage, D = 0.494, is less than the allowable damage, D = 0.80, the omponent is aeptable for ontinued operation, inluding a future run of five years (operating ondition m = 3). The remaining life for operation ould be determined by repeating this exerise and total allow determining the time when D = D. Larson Miller Parameter Approah g) Alternative STEP 7 - Determine the remaining life at the stress level using the Larson-Miller parameter data per Annex F, Table F.3. For SA335Grade P material (.5 Cr-Mo) Parameters Table E0.3-7 Minimum Larson-Miller Parameter - LMP m Average Larson-Miller Parameter - LMP a A E E+0 A E E-0 A E E+0 A E E-0 A E E+0 A E E-0 A E E-0 C LMP Where Larson-Miller parameter is given by σ in ksi LMP A + A σ + Aσ + Aσ.5 o 4 6 ma, =.5 + A σ + A3σ + A5σ Rupture Life L is evaluated using Equation (0.) to (0.4) log n 0 [ L] n LMP( Seff ) n ( T + 460) 000 = C LMP 0-5

297 Where n J S J S n S n n n 3 s s S n eff = σ e exp 0.4 Ss = σ + σ + σ n n n ( σ σ σ3 ) 0.5 = ( ) = 596 eff = + + J = + + = 596 = 344 exp 0.4 = Calulate the rupture life using the minimum Larson-Miller parameter data LMP log 0 A + A S + A S + A S =.5 o eff 4 eff 6 eff min.5 + A Seff + AS 3 eff + AS 5 eff = ( ) ( ) ( ) ( ) ( ) ( ) = LMPmin L = CLMP = 0 = ( T + 460) ( ) L = 0 = 76,850 hours t = 65, 700 hours Life Fration used for first sub-inrement = = ,850 Similarly it an be shown that for the other 5 sub-inrements, the life frations are: 0.375, , , 0.38, 0.38, therefore [ ] D = =.06 > 0.80 total Therefore, the omponent is not aeptable per Level analysis using the minimum Larson-Miller parameter data 0-6

298 Calulate the rupture life using the average Larson-Miller parameter data. LMP log 0 avg A + A S + A S + A S = + A S + AS + AS = o eff 4 eff 6 eff.5 eff 3 eff 5 eff ( ) ( ) ( ).5 ( ) ( ) ( ) = L= = = ( T + 460) ( ) 000 LMPmin CLMP L = 0 = 506, 70 hours t = 65, 700 hours Life Fration used for first sub-inrement = = , 70 Similarly it an be shown that for the other 5 sub-inrements, the life frations are: 0.97, 0.034, 0.034, 0.043, 0.043, therefore [ ] D = = < 0.80 total Therefore, the omponent is aeptable per Level analysis using the average Larson-Miller parameter data Comparison with API 530 Method If the same data were to be analyzed using the API 530 method, the Huddleston uniaxial stress Seff = ksi is replaed by the mean diameter hoop stress σ mean = ksi in Equation 0.. Sine S eff is 0.94 times σ mean, the orresponding life frations onsumed using API 530 beome higher. a) Using minimum LMP data: Life frations onsumed are: , , 0.048, 0.048, 0.689, Total life fration D total =.448 total total ompared with D =.063 using S eff and D = using Omega data with both adjustment fators for reep strain and reep dutility set to zero. b) Using average LMP data: Life frations onsumed are: 0.685, 0.685, 0.07, 0.07, 0.056, Total life fration total total total D = ompared with D = using S eff and D = using Omega data with both adjustment fators for reep strain and reep dutility set to zero. ) Analysis using Atual Corroded Tube Wall Thikness Assuming aurate and reliable historial tube wall orrosion rates are available, atual tube wall thikness an be used to obtain a more aurate estimate of rupture life and life fration. For this example, assuming the tubes were orroding at inh per year from the inside surfae, the load history orresponding to the tube dimension during eah of the operating yles and sub-inrements an be derived. Rupture life and damage results based on various methods are summarized in Table E Note that with this approah, all umulative damages are below

299 Table E0.3-8 Operating Cyle Past (m = ) Exursion (m = ) Future (m = 3) Sub-Inrement n = n = n = n = n = n = Servie Time in Hours L_hours 65,700 65, ,900,900 Operating Pressure, psig P_avg Tube Wall Temperature, F T_avg ID Corrosion Rate, 0.00 inh / year Beginning Tube OD, inh Do_begin Ending Tube OD, inh Do_end Beginning Tube ID, inh Di_begin Ending Tube ID, inh Di_end Beginning Tube Thikness, inh t_begin Ending Tube Thikness, inh t_end Average Outside Diameter, inh Do_avg Average Inside Diameter, inh Di_avg Average Tube Wall, inh t_avg Omega Method Prinipal Stress, psi σ Prinipal Stress, psi σ Prinipal Stress 3, psi σ Effetive Stress, psi σe Ω Rupture Life, hrs LΩ 879,35 633,665 3,839 3,85 48,08 4,347 Ω Life Used (This Period) L / LΩ Damage (Cumulative) Σ (L / LΩ ) LMP Using Huddleston Unaxial Stress Approah J = (σ + σ + σ3) J S S = (σ + σ + σ 3 ) 0.5 S S Huddleston Uniaxial Stress S EFF LMP min Minimum LMP at S EFF (S EFF ) Rupture Life, hours L EFF 75,60 46,094 7,60 7,607 63,03 0,58 Life Used (This Period) L / L EFF Damage (Cumulative) Σ (L / L EFF ) LMP avg Average LMP at S 5EFF (S EFF ) Rupture Life, hours L EFF,579,34,0,363 8,775 8, , ,34 Life Used (This Period) L / L EFF Damage (Cumulative) Σ (L / L EFF ) API STD 530 Approah API 530 Mean Diameter Stress, psi σ mean API 530 Minimum LMP at LMPmin σ mean (σ mean ) Rupture Life, hours L ,463 37,5 5,80 5,800 95,805 49,8 Life Used (This Period) L / L Damage (Cumulative) Σ (L / L 530 ) API 530 Average LMP at LMP avg σ mean (σ mean ) Rupture Life, hours L 530,308, 80,47 5,03 5,000 55, ,088 Life Used (This Period) L / L Damage (Cumulative) Σ (L / L 530 ) Therefore the heater tubes now pass Level analysis regardless of the approah used. 0-8

300 0.4 Example Problem 4 In August 006, a longitudinal rak-like flaw has been found in the base metal on the outside surfae of a ylindrial pressure vessel during a sheduled turnaround. The vessel and inspetion data are provided below. The vessel was onstruted to the ASME B&PV Code, Setion VIII, Division, 998 Edition. Determine how long is the vessel aeptable for ontinued operation. Vessel Data Material = SA 40 Grade 36 Year 998 Design Conditions = 500 F Inside Diameter = 60 in Fabriated Thikness =.5 in Future Corrosion Allowane (FCA) = 0.0 in Weld Joint Effiieny =.0 Operating Conditions The vessel is not fully pressurized until the temperature is 40 o F. Below this temperature, the startup pressure remains under 300 psig. At shutdown, the pressure is dereased to 300 psig before letting the temperature drop below 40 F. The equipment operates 8 days a month. At the end of the 8 th day, it is shutdown. It is restarted at the beginning of the next month. Every year in August, the plant is shutdown for a month for omplete inspetion. Exept at start-up and at shutdown the operating onditions remain onstant. The vessel was put into servie on September 999 and is due to operate until July 09. Inspetion Data The flaw is loated in the base metal on the outside surfae of the vessel away from any weld seam. The flaw is oriented in the longitudinal diretion. The depth of the flaw was established by UT; onsistent readings where noted and a final value for the flaw depth was established at 0.30 in. The flaw length was established by MT and is 4.8 in. The distane of the rak-like flaw to the nearest strutural disontinuity is 50 in. During the preeding inspetion (August 005) the same area was inspeted and no rak was disovered. Perform a Level 3 Assessment per paragraph to determine the remaining life. The overall assessment is split into 3 parts: ) From the date at whih the vessel was put into servie (September 999) up to the last inspetion when it is established that there was no rak (August 005), the damage in the material ahead of the rak is alulated on the unraked omponent ) From the date of the last inspetion when it is established that there was no rak (August 005) up to the present inspetion when a rak has been disovered (August 006), the damage in the material ahead of the rak is alulated on the raked omponent without any rak growth 3) From the date of the present inspetion (August 006) up to the sheduled end of operations (July 09), the rak is propagating. The damage in the material ahead of the rak is alulated on the raked omponent taking rak growth into aount. a) STEP Determine a load history based on past and future planned operation The plant has been operated as stated in the original operating onditions and no hange is foreseen. In steady state onditions the only load is the pressure, the thermal loading is onsidered as negligible. The start-up and shutdown operations reate yli loadings with pressure variation and transient thermal loadings. The temperature inrease/derease rate is suh that the orresponding stress field is very small ompared to the stress field due to pressure. The yli loads our one a month and are negleted. Therefore the load taken into aount in the assessment is: P 500 psig = at T = 00 F during 67 hours/month or 739 hours/year. b) STEP Determine the material properties: yield strength, tensile strength and frature toughness. Yield and tensile strengths are needed at operating temperature and at the temperature of the FAD assessment. Frature toughness is needed only at the temperature of the FAD assessment. 0-9

301 C The primary membrane tensile stress σ m due to startup or shutdown pressure (300 psig), alulated per formula in paragraph A.3.4 of Annex A is less than 8 ksi. Per Part 3 paragraph 3.., a brittle frature assessment is not needed for these pressures. The temperature used in the FAD assessment will be the minimum temperature for whih the pressure is above 300 psig. Therefore TFAD = 40 F At this temperature, the yield and tensile strengths of SA-40 Grade 36 per ASME Setion II Part D are: σ σ ys, FAD uts, FAD = 30 ksi = 75 ksi SA-40 Grade 36 is an austeniti stainless steel. The toughness of this material is taken from paragraph F.4.8. in Annex F. The flaw lies in the base metal, therefore: KIC = 00 ksi in Sine no safety oeffiients are embedded in the Failure Assessment Diagram, a Partial Safety Fator is applied on the toughness: K = K / PSF mat IC K PSF =.5 K = (00.0) / (.5) = 33.3 ksi in 33.0 ksi in K mat At operating temperature, the yield and tensile strengths are alulated from their nominal value and Tables F. and F.4 in Annex F. In Table F.3 PYS=4 for Type 36 materials; in Table F.5 PUS=5 for Type 36 materials; leading to: σ σ σ σ σ ys ys ys uts uts rt = σ exp ys C + CT + C T + CT + C T + C T = (.59384) (0) + ( ) (0) (00) + ( ) (0) (00) (30) exp ( ) (0) 0 (00) 3 ( ) (0) 3 (00) 4 ( ) (0) ( = 5.37 ksi rt = σ exp uts C + CT + C T + CT + C T + C T = ( ) (0) + (-.455) (0) (00) + ( ) (0) (00) + 00) 3 6 (75) exp ( ) (0) 9 (00) 3 ( ) (0) ( ) + ( ) (0) (00) σ uts = 5.53 ksi + 5 The Young's Modulus is also needed at operating temperature. From ASME Setion II Part D Table TM for Group G (SA-40 Grade 36 is a 6Cr-Ni-Mo material): 6 Ey =.0 (0) psi = 000 ksi ) STEP 3 PAST - DAMAGE PRIOR TO CRACKING ) STEP 3A Determine the past damage in the material ahead of the rak prior to raking, D b The yli loadings are onsidered as negligible, then the reep damage prior to raking is alulated using the proedure in paragraph The operating onditions are onstant: M b = i) STEP 3A. Determine a load history based on past and future planned operation - See STEP 0-0

302 ii) STEP 3A. For the urrent operating yle, determine the total yle time, and divide the yle into a number of time inrements. Sine the operating onditions are onstant, no inrements are needed (see above) iii) STEP 3A.3 Determine the assessment temperature - See STEP : T = 00 F iv) STEP 3A.4 Determine the stress omponents σ ij - The prinipal stresses are alulated per Table 0.. tomp = tnom FCA= =.5 in tsl = 0.0 in Lf =.0 ( ylinder) Dmean = ID+ tnom + FCA= = 6.5 in PDmean Lf σ = σ = tomp = (500.0) (6.5) (.0) / (.5) = 050 psi PDmean Lf σ = σm = = (500.0) (6.5) (.0) / 4 (.5 0.0) = 55 psi 4( tomp tsl ) σ = σ = 0 psi 3 z v) STEP 3A.5 Determine if the omponent has adequate protetion against plasti ollapse - The stress omponents are determined from an elasti analysis: p 0.5 σ = P + ( P + 9 P ) /3 P { } ref b b L L p = σ and P = 0 σ = P = 050 psi b ref L p The riterion ( σ = 050 psi) ( σ = 5370 psi) for austeniti stainless steel is satisfied ref ys vi) STEP 3A.6 Determine the effetive stress σ e from prinipal stresses in STEP 3A σe = ( ) ( 3) ( 3) σ σ + σ σ + σ σ 0.5 σ e = (050 55) + (050 0) + (55 0) = 8877 psi vii) STEP 3A.7 Determine the remaining life L at stress level σ e = ksi and temperature T = 00 F by utilizing reep rupture data for SA-40 Grade 36 - The MPC projet Omega Data in Annex F Table F.30 are used. The parameters A 0 to A 4, and B 0 to B 4 for Type 36 materials in Table F.30 are: A0 = -8.9 B0 = A = B = A = B = A3 = 84.3 B3 = A 4 = 00. B4 = From Annex F the additional data are: 0-

303 αω =.0 ( ylinder) β = 0.33 Ω d Δ = 0.3 ( dutile behaviour austeniti stainless steel) Ω sr Δ = 0.5 ( bottom of satter band onservative value) Ω Leading to: S l [ σ ] [ ] = log = log = e 0 nbn = A + A3 Sl + 3 A4 Sl T nbn = + + = ( ) (84.3) (0.9483) 3 (00.) (0.9483) δ Ω σ + σ + σ β.0 (0.33).0 = = Ω = σ e d 3 log0 [ Ω ] = ( B0 +Δ Ω ) + B B Sl B3Sl B4 Sl 460 T (704.76) + (-60.0) (0.9483) + log0 [ Ω ] = ( ) = (3949.5) (0.9483) + (400.0) (0.9483).35 Ω = max [( Ω n ),3.0] = max (0 [ ] n BN 7.775),3.0 = max ( ),3.0 = 3.38 Ω =Ω + = (3.38 ) + (.0) (7.775) = m δω n α n + Ω BN log = A +Δ + A + A S + AS + A S T sr [ ε ] ( Ω ) 3 0 o 0 l 3 l 4 l (5790.0) + ( ) (0.9483) + log 0 [ ε ] = ( ) o + = (84.3) (0.9483) + (00.) (0.9483) ε = = o (0) -9 L = hours -9 ε o Ω = 9.33 (0) (40.58) = m Between September 999 and August 005 the operating time is t = 6 years = hours op 0-

304 The damage is then: D top = = = L total viii) STEPs 3A.8 to 3A. N.A: No inrement - D = D = ix) STEP 3A. The past damage in the material ahead of the rak prior to raking is then: D b total = D = allow This damage is aeptable: ( D = ) ( D = 0.80) b ) STEP 3 PAST - DAMAGE AFTER CRACKING WITHOUT CRACK GROWTH ) STEP 3B Determine the past damage in the material ahead of the rak after raking, The yli loadings are onsidered as negligible, then the past reep damage after raking is alulated using the proedure in paragraph 0.5. but based on the referene stress of the raked omponent The operating onditions are onstant: M = 0 a i) STEP 3B. Determine a load history based on past and future planned operation - See STEP ii) STEP 3B. For the urrent operating yle, determine the total yle time, and divide the yle into a number of time inrements. Sine the operating onditions are onstant, no inrements are needed (see above) iii) STEP 3B.3 Determine the assessment temperature - See STEP : T = 00 F iv) STEP 3B.4 Determine the stress omponents σ ij in the unraked onditions through the wall of the omponent ontaining the rak-like flaw - See STEP 3A.4 σ = σ = 050 psi σ = σm = 55 psi σ = σ = 0 psi 3 z v) STEP 3B.5 Determine if the omponent has adequate protetion against plasti ollapse STEP 3B.5. Determine the rak-like flaw dimensions - From inspetion data in August 006: a = 0.30 in = 4.80in STEP 3B.5. Determine the referene stress σ ref using Annex D The primary stress omponents are: P σ 050 psi m = = and P = 0 From Annex C, Table C., the flaw geometry, omponent geometry, and loading ondition orrespond to RCSCLE. The referene stress solution for RCSCLE is provided in Annex D, paragraph D.5.. t = t FCA= =.50 in nom R = ID/ = 30.0 in i The bending stress is nil, therefore the referene stress formula may be written as: σ = M P NS ref s m b 0 D a 0-3

305 .88.88(.40) λa = = =.454 Ra i ( 30.0)( 0.30) ( λa) ( λ ) a M t( λa) = 6 4 = ( λa) +.533(0 )( λa) M σ NS s (.454) (.454) 6 4 = (.454) +.533(0 )(.454) = = =.057 a a t ( ) t Mt λa NS = M P = (.057)(050) = 0835 psi ref s m STEP 3B.5.3 Evaluate the protetion against plasti ollapse based on the referene stress The riterion ( σ = 0835 psi) ( σ = 5370 psi) for austeniti stainless steel is satisfied ref ys vi) STEP 3B.6 Determine the effetive stress σ e from prinipal stresses in STEP 3B σe = ( ) ( 3) ( 3) σ σ + σ σ + σ σ 0.5 σ e = (050 55) + (050 0) + (55 0) = 8877 psi vii) STEP 3B.7 - Determine the damage in the material ahead of the rak - See STEP 3A.7 with Sl = log 0 σ ref

306 S l 0 [ ] = log =.035 nbn = A + A3 Sl + 3 A4 Sl T n BN = ( ) + (84.3) (.035) + 3 (00.) (.035) = δ σ + σ + σ β (0.33) Ω = Ω = = σ e d 3 log0 [ Ω ] = ( B0 +Δ Ω ) + B B Sl B3Sl B4 Sl 460 T (704.76) + (-60.0) (.035) + log0 [ Ω ] = ( ) = (3949.5) (.035) + (400.0) (.035) [ n ] Ω n = Ω =.5 max ( BN ),3.0 max ( ),3.0 = Ω =Ω + α n = (5.947 ) + (.0) (7.394) = 3.94 m δω n Ω BN log = A +Δ + A + A S + AS + A S T sr [ ε ] ( Ω ) 3 0 o 0 l 3 l 4 l (5790.0) + ( ) (.035) + log 0 [ ε o ] = ( ) = (84.3) (.035) + (00.) (.035) ε o = 0 = 4.3 (0) L 0 a = = = hours -8 ε Ω 4.3 (0) (3.94) o m Between August 005 and August 006 the operating time is t = year = 739 hours op The damage is then: D t 739 = = = 7.487(0) L op 0 a 3 viii) STEPs 3B.8 to 3B. N.A: No inrement - D total = D = (0) ix) STEP 3B. The past damage in the material ahead of the rak after raking is then: D = D = (0) 0 total 3 a 3 0-5

307 Chek that the overall damage is aeptable: 0-3 allow The riterion ( D + D = (0) = 0.048) ( D = 0.80) is satisfied b a d) STEP 4 FUTURE - DAMAGE AFTER CRACKING WITH CRACK GROWTH Note: Due to rak size inrements used in the rak propagation analysis, double preision is needed to ensure auray Initialize initial flaw dimension sizes, starting time and damage after raking: i= 0 i= a = a = 0.30 in i= 0 i= = =.40 in i= 0 t = 0.0 i= D = D = (0) a a e) STEP 5 For the urrent operating yle, the total yle time, m t, is set to the sheduled remaining operating life (3 years between August 006 and August 09). This yle is divided into time periods of month (67 hours) f) STEP 6 Determine the temperature, and ompute the stress omponents in the unraked onditions i= i= through the wall of the omponent ontaining the rak-like flaw at time t - ( t = 67 hours ) See STEPs 3B.3 and 3B.4: T = 00 F σ = σ = 050 psi σ = σm = 55 psi σ = σ = 0 psi The effetive stress is then: σ = e z psi σ g) STEP 7 Determine the referene stress = ref at time = t using Annex D - The primary stress omponents are: Pm = σ = 050 psi and P b = 0 From Annex C, Table C., the flaw geometry, omponent geometry, and loading ondition orrespond to RCSCLE. The referene stress solution for RCSCLE is provided in Annex D, paragraph D.5.. i= a= a= 0.30 in i= = =.40 in t =.50 in R = 30.0 in i The bending stress is nil, therefore the referene stress formula may be written as: i= σ = M P NS ref s m λ =.454 See STEP 3 B.5. a M ( λ ) =.370 See STEP 3 B.5. M i= t NS s σ a =.057 See STEP 3 B.5. NS = M P = (.057)(050) = 0835 psi ref s m Chek that the omponent has adequate protetion against plasti ollapse: The riterion i= ( σref 0835 psi) ( σ ys 5370 psi) = = for austeniti stainless steel is satisfied 0-6

308 h) STEP 8 Perform a FAD assessment at time t i= using proedures in Part 9 ) STEP 8. Evaluate operating onditions and determine the pressure, temperature and supplemental loading ombinations to be evaluated - See STEPs and ) STEP 8. Determine the stress distribution at the loation of the flaw Primary Stress: See STEP 7 - P 050 m = psi and P = 0 Seondary Stress: Thermal loading is onsidered negligible (See STEP ), and the flaw is loated away from all major strutural disontinuities. Therefore, there are no seondary stresses. Residual Stress: The flaw is in the base metal far away from any weldment. Therefore, there are no residual stresses. 3) STEP 8.3 Determine the material properties; yield strength, tensile strength and frature toughness - See STEP σ σ ys, FAD uts, FAD KIC = 30 ksi = 75 ksi = 00 ksi in P Determine the ut-off for the Lr -axis of the FAD per Figure 9.0 of Part 9 SA-40 Grade 36 is an austeniti stainless steel: L P,max =.8 r 4) STEP 8.4 Determine the rak-like flaw dimensions - See STEP 4. 5) STEP 8.5 Modify the primary stress, material frature toughness, and flaw size using Partial Safety Fators - See STEP - Partial Safety Fator is applied on toughness only K = K / PSF mat IC K PSF =.5 K = (00) / (.5) = 33.3 ksi in 33.0 ksi in K mat 6) STEP 8.6 Compute the referene stress for the primary stress - See STEP 7 σ P i ref = = σref = 0835 psi 7) STEP 8.7 Compute the Load Ratio ( L ) or absissa of the FAD L P r P σ ref 0835 = = = 0.36 σ ys, FAD P r P 8) STEP 8.8 Compute K - From Annex C, Table C., the flaw geometry, omponent geometry, and loading ondition orrespond to KCSCLE. The stress intensity fator for KCSCLE is provided in Annex C, paragraph C.5.. The applied loading is a membrane stress ( σ 0 = P m, σ = σ = σ3 = σ4 = 0 ), therefore only the data required to evaluate the G 0 influene oeffiient is needed to ompute the stress intensity fator. The flaw ratios and parameters to determine the G 0 influene oeffiient from Annex C Table C.3 are: b 0-7

309 A0,0 = t A,0 = = = R 30.0 i A,0 =.7667 a 0.30 = = 0.5 A3,0 = a 0.30 A4, = = = 0. t.50 A5,0 = A 6,0 =.48 The influene oeffiients required for the assessment are: G = A + A β + A β + A β + A β + A β + A β ,0,0,0 3,0 4,0 5,0 6,0 - at the deepest point of the flaw ϕ = 90 o : π ϕ π ϕ = 90 = rad β = = = G0 =. π π ϕ = : - at the surfae points of the flaw 0 o ϕ ϕ = 0 = 0 rad β = = 0 = 0 G = 0.45 ( ) ( ) 0 The stress intensity fators are: π π a 0.30 Q = = = at the deepest point of the flaw ( ϕ = 90 o ): P π a π (0.30) KI = G0σ 0 = (.)(0.50 ksi) =.7855 ksi in Q at the surfae points of the flaw ( ϕ = 0 o ): P π a π (0.30) KI = G0σ 0 = (0.45)(0.50 ksi) = ksi in Q ) STEP 8.9 Compute the referene stress for seondary stresses - σ SR = 0 SR SR 0) STEP 8.0 Compute K for seondary stresses - KI = 0 ksi in SR ) STEP 8. Compute the plastiity interation fator - Sine 0 I ref psi K = then Φ=.0 ) STEP 8. Determine toughness ratio or ordinate of the FAD assessment points. K r K = P I +Φ K mat K SR I At the deepest point of the flaw ( At the surfae points of the flaw ( ϕ = 90 o ): ϕ = 0 o ): K r K r ( ) (.0) 0.0 = = (.0)(0.0) = =

310 3) STEP 8.3 Evaluate the results. Determine the maximum allowable P K for L = 0.36 : r 6 P { ( ) } P Kr,max = 0.4( Lr) exp 0.65 Lr = Chek that ( ) ( ) r { } exp = L P P r ( Lr,max =.80) and that Kr Kr,max ( = 0.98) P At the deepest point of the flaw: ( L, K ) = (0.36,0.089) ; the point is inside the FAD r r P At the surfae points of the flaw: ( L, K ) = (0.36,0.030) ; the point is inside the FAD Both deepest point and surfae points are aeptable i) STEP 9 Determine the damage in the material ahead of the rak growth at time See STEP 3A.7 with S l 0 S [ ] i l = log 0 = σ ref = log =.035 r r i= t. nbn = A + A3 Sl + 3 A4 Sl T n BN = ( ) + (84.3) (.035) + 3 (00.) (.035) = δ Ω σ + σ + σ β (0.33) = Ω = = σ e d 3 log0 [ Ω ] = ( B0 +Δ Ω ) + B B Sl B3Sl B4 Sl 460 T (704.76) + (-60.0) (.035) + log0 [ Ω ] = ( ) = (3949.5) (.035) + (400.0) (.035) [ n ] Ω n = Ω =.5 max ( BN ),3.0 max ( ),3.0 = m δω n α n + Ω BN Ω =Ω + = (5.947 ) + (.0) (7.394) =

311 log = A +Δ + A + A S + AS + A S T sr [ ε ] ( Ω ) 3 0 o 0 l 3 l 4 l (5790.0) + ( ) (.035) + log 0 [ ε o ] = ( ) = + (84.3) (.035) + (00.) (.035) o = 0 = 4.3 (0) ε L i= a = = = hours -8 ε Ω 4.3 (0) (3.94) o m Leading to: D = D + ( t t) / L = (0) + (67 / ) = 8.68 (0) i = i = 0 i = i = 0 i = 3-3 a a a Chek that the overall damage is aeptable: i= -3 allow The riterion ( D + D = (0) = ) ( D = 0.80) is satisfied b a i ε ref i= ref = i= j) STEP 0 Determine the referene strain rate at time t - The damage alulations are based on the MPC Projet Omega Creep Data, therefore: ε = ε of STEP 9: = 4.3 (0) i= -8 ε ref i= k) STEP Determine the stress intensity fator at the deepest point and at surfae points at time t - P These stress intensity fators are based on those alulated in STEP 8.8 ( K I for primary stresses) and SR STEP 8.0 ( K I for seondary stresses) - at the deepest point of the flaw: K = K ( ϕ = 90 ) + K ( ϕ = 90 ) = =.7855 ksi in 90 P SR I I I - at the surfae points of the flaw: K = K ( ϕ = 0 ) + K ( ϕ = 0 ) = = ksi in 0 P SR I I I l) STEP Determine the rak driving fore at the deepest point and at surfae points at time C t i= i= ( KI a ) 90 i= i= ( KI a ) i= (, ) -8 ε *90 ref 4.3 (0) (.7855) -7 = = = (0) i= Db D a σ ref relax (, ) 0.9(.7855) = = n E C *90 ( BN + ) y ( ) (000) 5.56 o 4 (0) nbn nbn *90 t relax t = C + = + = i= -7 = 30 C (0).399 (0) t 67 t i= 0-30

312 C t i= i= ( KI a ) 0 i= i= ( KI a ) i= (, ) -8 ε *0 ref 4.3 (0) (4.037) -8 = = = 6.59 (0) i= Db D a σ ref relax (, ) 0.9 (4.037) = = ( n + ) E C ( ) (000) 6.59 (0) BN y *0-8 = 30 C 6.59 (0).64 (0) t 67 nbn nbn *0 t relax t = C + = + = i= Note: It may be notied that t 90 = t 0. No differene will be made between them in Table E0.4-5 relax giving the rak growth parameters and propagation for eah inrement. relax m) STEP 3 Calulate the rak growth rate at time H =Ω = =.5 / 500 (0) / ( ) t i= μ = n / ( n + ) = / ( ) = i= BN BN - Based on the MPC projet Omega da 90 μ = H ( Ct ) = (0.0668) (.399 (0) ) =.860 (0) in/ hour dt i= d 0 μ = H ( Ct ) = (0.0668) (.64 (0) ) =.88 (0) in/ hour dt i= n) STEP 4 Calulate the time step for integration at time t i= i= da d -7 Δ t = ( Cintg t ) / max, = (.50) / (.860 (0) ) = 4030 hours dt dt Sine Δ t is greater than the time period of month (67 hours), o) STEP 5 Update the flaw dimensions and the aumulated time i= i= i= i= -7-4 i= i= i= -8-5 i= i= i= Δ t is set to the time period da a= a+ Δt dt a= (.860 (0) ) (67) = (0) = in i= d = + Δt dt =.40 + (.88 (0) ) (67) = (0) = in t = t+δt t = (67) + (67) = 344 hours p) STEP 6 Repeat STEPs 5 to 5 until the sheduled remaining operating life The detailed results obtained with Mirosoft Exel are given in Tables E0.4- to E Up to July 009, the results are given at the end of eah month. Between July 009 and July 09, the results are given at the end of eah operating year, but the alulations are still performed with time periods equal to month. q) STEP 7 N.A - The yli loads are negleted - No fatigue rak growth is taken into aount r) STEP 8 There is only yle in the histogram - The rak growth alulations are omplete 0-3

313 s) STEP 9 Evaluate the rak growth results Tables E0.4- to E0.4-5 show that the propagation of the rak due to reep is very slow. At the time of sheduled end of operating time, the depth of the flaw has inreased less than 0 %. The remaining life of the vessel is greater than the sheduled operating life A sensitivity analysis on the pressure shows that: With a Partial Safety Coeffiient of., leading to a pressure of 600 psig and a primary stress of.3 ksi lose to maximum allowable stress value per ASME Setion II Part D Table A, of.4 ksi at T=00 F, the values of the main parameters at the end of the sheduled operating time are: a = in Db + Da = =.4049 in Ct = (0) 0 7 σ ref = 3.03 ksi Ct = (0) 90 7 K = 5.6 ksi in da / dt = 3.90 (0) in/ hour 8 0 K = ksi in d / dt = (0) in/ hour With a Partial Safety Coeffiient of.367, the remaining life is equal to the sheduled operating life limited by a damage of The values of the main parameters at the end of the sheduled operating time are then: a = 0.47 in Db + Da = =.4 in Ct = (0) 0 6 σ ref = 4.94 ksi Ct = (0) 90 6 K = 0.4 ksi in da / dt = 3.7 (0) in/ hour 7 0 K = ksi in d / dt = (0) in/ hour With a Partial Safety Coeffiient of.5, the remaining life goes up to July 0 limited by a damage of The values of the main parameters at this date are then: a = in Db + Da = =.48 in Ct =.089 (0) 0 5 σ ref = 6.37 ksi Ct =.73 (0) 90 6 K =.70 ksi in da / dt = (0) in/ hour 6 0 K = ksi in d / dt =.085 (0) in/ hour 0-3

314 Table E Creep Crak Growth - Times, Dimensions and Referene Stress Parameters STEP Inr Date i t a a/ a/ t λ a M t NS M s P σ ref Sep Ot Nov De Jan Feb Mar Apr May Jun Jul Sep Ot Nov De Jan Feb Mar Apr May Jun Jul Sep Ot Nov De Jan Feb Mar Apr May Jun Jul Jul Jul Jul Jul Jul Jul Jul Jul Jul Jul

315 Table E Creep Crak Growth - Failure Assessment Parameters of the Crak STEP Inr P L r Q G0 deep G0 surf P K I deep P K I surf Kr deep Kr surf Kr allow

316 Table E Creep Crak Growth - Damage and Stain Rate Parameters STEP Inr S l n [ ] BN Ω log0 Ω n Ω log [ ] m 0 ε = ε ε o o ref i L a E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E

317 Table E Creep Crak Growth - Crak Growth Rate Parameters STEP 9 9 Inr D + D i D a b a 90 K I deep 0 K I surf 6.807E E E E-08.64E E E-07.08E E-08.69E E E E E-08.E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E-08 8.E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E-08 *90 C deep 90 C t deep *0 C surf 0 C t surf 0-36

318 Table E Creep Crak Growth - Crak Growth Parameters and Propagation STEP Inr t relax H μ da / dt d / dt Δ t Δ a Δ E-07.88E E+04.50E E E-07.46E E E E E-07.03E E E E E-07.88E E E-05.65E E E E E-05.08E E E E E-05.67E E-07.69E E E-05.36E E E E E-05.3E E-07.67E E E E E E E E E E E E E E E E E E E E E E E E E-07.54E E E E E E E E-05.08E E-07.50E E E-05.0E E-08.5E E E-05.06E E E E E-05.0E E E E E E E-08.49E E E-05.00E E E E E E E-08.48E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E-08.45E E E E E E E E E E E E E E E E E E E E-08.44E E E E E E E E E E E E E E E E E E E E-08.48E E E E E E E E E E E E E-05.04E E-08.54E E E-05.04E

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320 PART ASSESSMENT OF FIRE DAMAGE EXAMPLE PROBLEMS. Example Problem Example Problem Example Problem Example Problem A 75 feet high by 60 in inside diameter by 5/8 in thik wall low arbon steel insulated distillation olumn is subjet to fire damage. The vessel is not stress relieved and the weather barrier is galvanized arbon steel. The vessel is used to distill light fuel oil/gasoline type produts during normal operation. Perform a Level Assessment per paragraph.4. a) Observations After Fire: ) The galvanize oating on the weather barrier is disolored but there is no indiation of molten zin running down the barrier. ) The aluminum onduit on the side of the vessel is intat. 3) The alkyd oating on the vessel under the insulation is not disolored or blistered. 4) Light bulbs on the vessel struture are not distorted. b) Conlusions: ) The temperature experiened by the vessel an be determined from the following onditions: ) Sine the oating is only disolored, the surfae temperature during the fire never reahed 790 F, in aordane with Table.3. 3) The temperature is below 5 F at whih point aluminum melts (see Table.4). 4) The temperature of the vessel below the insulation did not exeed 300 F at whih alkyd oatings disolor (see Table.). 5) Light bulbs distort and melt at 950 F (see Table.4) The above observations indiate that the Heat Exposure Zone for the vessel is IV or below (see Table.). a) Further Ation: ) Leak hek the vessel omponents and onsider replaing the disolored areas of the weather barrier. ) Doument in the inspetion files that the vessel was assessed for fire damage and that the vessel s pressure envelope was not affeted by the fire exposure. b) Instrutional Comment: The objetive of a Level I assessment is to doument the observations that led to the onlusion that the pressure ontaining omponent was not degraded by the fire exposure. The Level Assessment Criteria are Satisfied. -

321 . Example Problem A horizontal vessel with an inside diameter of 50 in, a thikness of 9/6 in, and a length of 35 feet is subjet to fire damage. The distane from the tangent point to the enterline of the saddles is 3 feet and the height of the head is 37.5 in (i.e. the head is a : elliptial head). The vessel is fabriated from SA-56 Grade 70 arbon steel, is not insulated, and is oated on the exterior with an epoxy phenoli system. The vessel is not stress relieved and ontains a heavy diesel type produt during normal operation. The vessel design onditions are 80 psig at 650 F, and the weld joint effiieny is A future orrosion allowane of /6 in is required for operation. Perform a Level Assessment per paragraph.4. a) Observations After Fire: ) The aluminum onduit next to the vessel has melted. ) The vessel is not sagged based on a visual inspetion. 3) Iron oxide sale has spalled off of the side of the vessel faing the fuel soure of the fire. 4) Paint disoloration of the vessel surfae on the opposite side of the fuel soure is visible. 5) An internal inspetion of the vessel indiates no damage. 6) Thikness readings indiate a 0.03 in metal loss (attributed to past operation). 7) The formation of oke like produts was not observed from the heating of the fuel oil inside of the vessel. b) Conlusions: ) The aluminum onduit next to vessel has melted; therefore, the surfae temperature of the vessel during the fire ould have been in exess 5 F in aordane with the information in Table.4. ) The Heat Exposure Zone for the vessel is possibly greater than Zone IV, in Table.. The Level Assessment Criteria are Not Satisfied. Inspetion Results ) Sagging of the vessel from the horizontal has not ourred based on atual field measurements. ) Loalized shell distortions have not been found based on an internal inspetion of the vessel. 3) Hardness test are performed to determine the ondition of the shell material. Perform a Level Assessment per paragraph.4.3 a) STEP Hardness Test Results: ) Vessel Hot Side 3 HB, orresponds to a tensile strength of 65 ksi ) Vessel Cool Side 5 HB, orresponds to a tensile strength of 75 ksi b) STEP Determine an allowable stress for the vessel based on the material strength for the Hot Side ht S S uts at S min afd =., at 50 C ism S aa { S } = min.,7. 5 = 6, psi ) STEP 3 Determine the MAWP for the shell setion in the Hot Side. Cirumferential Stress (Annex A): t = t FCA LOSS = = in R = R + FCA + LOSS = = in S E t = = = C MAWP 86.4 [ R t ] [ ] psig -

322 Longitudinal Stress at mid-span loation (Annex A): Q= 97,350 lb L= 35 ft = 40 in H = 37.5 in A= 36 in Rm = 75.8 in { ( 75.8) ( 37.5) } + ( 97,350 )( 40 ) ( 40) 4( 36) tsl = = 0.05 in 4( 6, 50) π ( 0.85)( 75.8 ) 4( 37.5) ( 40) + 3 ( 40 ) L ( 6, 50)( 0.85)( ) MAWP = = 55 psig ( ) ( ) [ ] MAWP = min 86 psig, 55 psig = 86 psig d) STEP 4 Evaluate omponent for other forms of damage ) Iron oxide sale was removed from the vessel wall, subsequent inspetion of the surfae indiates no visible damage (e.g. loal thinning, blisters, shell distortions, and rak-like flaws). ) Field metallographi examination indiates that the steel mirostruture representative of the Hot Side of the vessel has notieable grain oarsening ompared to the Cool Side of the vessel where the oating is only disolored; an evaluation for MAT and CET requirements in aordane with Part 3 is reommended. e) STEP 5 Evaluate the potential for reep damage An evaluation for reep damage in aordane with Part 0 is reommended. The Level Assessment Criteria are Satisfied Pending the Outome of the Brittle Frature Assessment per Part 3. -3

323 .3 Example Problem 3 The depropanizer tower of a HF alkylation unit was subjet to fire damage. The tower was onstruted to ASME setion VIII, Division. The tower omprises two ylindrial setions, the upper one 8 feet-6 in inside diameter, 49 feet-3 in tall, and.5 in thik; the lower one 0 feet inside diameter, 3 feet-3 in tall, and - 5/6 in thik. Both ylindrial setions are joined with a onial setion feet tall and 3/8 in thik. The material of onstrution of the tower was arbon steel SA- Grade B (minimum speified ultimate tensile strength of 70,000 psi) and was given a orrosion allowane of /8 in, and the weld joint effiieny was.00. The design pressure was 38 psig, the design temperature was 400 F. The tower was post weld heat treated (PWHT). The tower had 7 ourses or ans and the onial setion; the ourses were numbered through 7 from top to bottom. A future orrosion allowane of /6 in is required for operation. Perform a Level Assessment per paragraph.4. a) Observations After Fire: ) The fire was produed by a large leak at a bottom nozzle in ourse 5. The flames expanded upward and overed the whole tower, reahing a height at least twie as tall as the tower. ) Two pipes setions around the tower ruptured at a height near the transition onial setion, reating flames that diretly impinged on the tower. Other pipe setions around the tower that did not rupture were found severely distorted and badly deformed. 3) The steel struture around the tower was badly distorted and the platforms in some plaes simply ollapsed by their own weight as they softened exessively during the fire. 4) Most of the insulation was missing, stripped off or burned, remaining partially only in some plaes, mainly on the lower ourses. Aluminum sheet melted and even rok wool was found molten in some plaes. 5) The oating in some plaes was missing or burned too. In some areas where the oating remained it was disolored and damaged exept in the lower tower ourses, where it looked intat in some plaes. 6) The tower appeared tilted and in some plaes bulged, but it was diffiult to be 00% sure without arrying out any measurement b) Conlusions: ).The tower was exposed to a Heat Exposure Zone of at least V or higher and may have suffered dimensional hanges. The Level Assessment Criteria are Not Satisfied. Inspetion Results ) The vessel appeared tilted and bulged. ) Loalized shell distortions have been found based on an external and internal inspetion of the vessel. 3) Hardness tests needed to be performed to determine the ondition of the shell material. 4) Thikness readings in the shell setion indiate a 0.0 in metal loss. The onial setion showed thikness values below the minimum required thikness. The metal loss was attributed to past operation. -4

324 Perform A Level Assessment per paragraph.4.3 The depropanizer tower was fabriated from arbon steel. Therefore, the assessment proedure established in paragraph was followed a) STEP Heat Exposure Zones, In-situ Field Metallography, Hardness Test, and Dimensional Cheks ) Defining Heat Exposure Zones V and VI: i) A Heat Exposure Zone V was established based on finding molten aluminum. Aording to Table.4, ommerially pure aluminum melts at 5 F (655 C). So, in this fire damage assessment, the fire Zone V involved heat exposure up to 5 F (655 C), rather than greater than 800 up to 350 F ( C) range used in Table.. ii) Heat Exposure Zone VI was established based on in-situ field metallography repliation. Figure E.3- shows the plaes where in-situ metallography repliation was done. Evidene of austenitizing temperatures was found, implying temperatures in the order of F ( C), greatly exeeding the limit of 350 F (73 C) for Fire Zone V. Steel within this fire Zone VI often had a thik oxide sale on their surfae. So, in this fire damage assessment, the fire Zone VI involved heat exposure higher than 580 F (860 C), greater than 350 F (73 C) limit used in Table.. iii) Heat Exposure Zones involving overheating temperatures between 5 F (655 C) and 580 F (860 C) ould not be established. iv) Figures E.3- and E.3-3 show elevation views of the Heat Exposure Zones V and VI, as defined here for this partiular ase. Figure E.3-4 shows a plan view with different gray tones representing different elevations, rather than different Heat Exposure Zones. So the blak is also within the Heat Exposure Zones V but on the floor level, then the dark gray is an intermediate level, and the lighter grey is the upper level. ) Hardness Tests: i) Hardness tests were first performed with the portable hardness tester Telebrineller (using a omparative arbon steel bar and hitting with a hammer to produe an indentation on the shell and on the bar). It was taken along vertial lines overing the omplete height of the tower on the North, South, East, and West sides, Figure E.3-5. ii) A Brinell hardness number (HBN) of 43 was taken as the aeptable limit, below whih the steel was delared too soft. Aording to Table F- of Annex F, it orresponds to an ultimate tensile strength of 7 ksi. The minimum speified ultimate tensile strength for this steel was 70 ksi. A maximum allowable hardness was taken as 00 HBN, as per unit material speifiation. Only two values were below 43 HBN, on the East side, 3 HBN in ourse and 4 HBN in ourse 7. iii) Vikers hardness tests were also performed on the plaes where in-situ metallography was done. A portable digital equipment was used whih applies 5 kgf load on a diamond pyramidal indenter. The result is given as Vikers hardness numbers. The results are summarized in Table E.. The Vikers hardness values were onverted to HBN. Values lower than 43 HBN were found for ourses, 3, 4, 5, the onial setion, and ourse 6. These plaes orresponded to those inside the Heat Exposure Zone VI, as defined earlier for this partiular fire damage assessment. 3) Dimensional Cheks: i) Thikness measurements were taken on the same plae where hardness was measured. There were thikness values below the retirement thikness in the onial setion. There was not any value below the retirement thikness for the ylindrial parts on the North and South sides. There were two values below the retirement thikness for the ylindrial parts on the East side and one on the West side. These were isolated values though. ii) The proedure used to derive tower profiles onsisted of measuring the distane from the plumb bob wire to the tower wall outside surfae every foot while the loation of every girth weld was also reorded. Due to limited aess and time onstraint, this was done on vertial lines to over the omplete irumferene of the tower, as indiated in Figure E.3-6, and the whole tower height. The atual loation of eah line was defined by hoosing a plae without any obstale, nozzle, of steel struture interfering with the plumb bob wire. In some -5

325 ase it was neessary to ut old steel struture or platform to allow the plumb bob wire falling freely to the bottom of the tower, as per Figure E.3-4. iii) Eah distane from the shell wall to the plumb bob wire was multiplied by - to onvert it into a negative value and then added to the largest distane measured on that partiular vertial line. If the rude data were plotted as they were, the presene of inlinations ould be notied. Figure E.3-7 shows a typial example, where there were 3 inlinations noted. The proedure used for orreting the data was essentially the same as shown in Figure E.3-8, but more omplex. To measure the inlinations, the data were separated into three distint groups and for eah one a straight line was fitted by the least square method. As seen, the fator R^ for eah inlination was lose to unity (0.979, 0.90, and 0.970, respetively), and were therefore taken to be real inlinations. iv) In the above example, on this side of the tower, the distane between the plumb bob stream and the tower wall should be in if they both were parallel and straight. By ompensating due to the inlinations, the profile ould be plotted, representing the atual profile on the tower surfae on this line 7, Figure E.3-8. Notie in this partiular example that the maximum bulge ourred in ourse 5 and it was 3.9 m (.535 in) in height. The inlinations are given by the slope of eah fitted straight line. Flange N-8 was the one that leaked and started the fire. v) The proedure used allowed opying the atual surfae profile of the tower at eah partiular line (there were suh vertial lines). The deviations from ideal profile are not neessarily symmetrial with regard to the tower axis. To alulate the perentage of deformation, the tower radius is used, rather than the diameter. Aording to onstrution drawings, the outside radius for the upper ylindrial setion was 5.5 in (3.4 m). For the lower ylindrial setion, it was 6.33 in (55.7 m). On this partiular profile, Figure E.3-9, deviations exeeding % were deteted in the ourses,, 3, 5, 6, 7, 8, 9, 0,,, 3, and 4. The maximum allowable is % aording to onstrution ode Setion VIII of the ASME ode, Divisions and. Eleven additional profiles were plotted to obtain the perentage of deformation for eah one. b) STEP Determine an allowable stress for the upper ylindrial part of the depropanizer tower, using Equation., and based on the material strength and hardness testing, ht S S uts at S min afd =., at 000 C ism S aa ) STEP 3 Determine the MAWP for the shell setion. Cirumferential Stress (Annex A): { S } = min.,7. 5 = 5, psi t = t FCA LOSS = =.045 in R = R + FCA + LOSS = = in ( )( ) ( ) SEt = = = [ R + 0.6t] C MAWP C MAWP = 30 < DESIGN = 38 psig P psig The Level Assessment Criteria are Not Satisfied for the upper ylindrial part of the depropanizer tower. Conlusions of the Level Assessment: ) It was not feasible to define Heat Exposure Zones with temperatures between 5 F (655 C) and 580 F (860 C), simply beause this ould not be determined. Therefore, the desription of Heat Exposure Zones as per Table., using 350 F (73 C) as the limit, ould not be followed. New definitions had to be reated, based on the melting point for pure aluminum 5 F (655 C) and autenitizing temperature for this arbon steel, approximately 580 F (860 C). psig -6

326 ) It was deided to replae the steel that fell into this partiular Heat Exposure Zone VI for this fire damage assessment, whih was the upper ylindrial setion, the onial setion, ourses 6 and 7. Course, most of ourse, and the top head were in this partiular Heat Exposure Zone V, involving exposure to temperatures below 5 F (655 C). The deision to replae was based on the fat that ourses, 3, 4, 5, the onial setion, ourses 6, 7 were partially austenitized, whih meant an extreme heat exposure to temperatures higher than 580 F (860 C). 3) At this very high exposure temperature, the steel softened exessively and distorted loally. The tower did not ollapse beause this heating did not our around the whole irumferene but only in one half of the irumferene. The hardness at those overheated ourses did not omply with the minimum speified, as derived from the speified minimum ultimate tensile strength. Suh as uneven heating to suh high temperatures is believed to be harmful sine it may have introdued high residual stresses. That would have required applying a stress relieving treatment to the omplete affeted areas, not just the welds. 4) The onial setion was already showing thikness values below the retirement one. So the opportunity to replae it this time was taken to restore it to aeptable ondition, even though this thinning did not have anything to do with fire damage. Further Ations: ) All piping showing deformation, distortion, sagging were identified, marked with red paint, and replaed. ) All eletrial motors and pumps in the fire zone were taken to the shop for serviing, and replaing if found neessary. 3) Eletrial ables and instruments in zones beyond the fire zone V were all replaed. Damage in these omponents of the unit was determined by maintenane and instrumentation personnel, based on visual inspetion. All damaged steel struture, platforms, ladders, and steel stairs were replaed. This was also determined visually and was based on distortion, sagging, ollapse or severe deformation. 4) All pressure vessels and towers in the Heat Exposure Zone V were assessed for fire damage and with the exeption of the upper part of the depropanizer tower, the rest of the vessels, heat exhangers and towers were found fit for further servie without the need to arry out any major repair. They were all thoroughly inspeted, leaned and painted again. Insulation and fire proofing were replaed. 5) All bolting and gaskets in the Heat Exposure Zones V and VI were removed and replaed by new ones. 6) The Level assessment riteria were not satisfied for the lower ylindrial part of the depropanizer tower. Condut a Level 3 assessment in order to determine if replaement is required. -7

327 Perform A Level 3 Assessment per paragraph.4.4 The assessment proedure established in paragraph.4.4 was followed a) STEP Stress Analysis for Shell Distortions ) An assessment of the shell distortion was ompleted per Part 8 paragraph The surfae ontour of the lower part of the tower was derived from the available profiles, as illustrated in Figure E.3-0, with the objetive to measure the bulge size. Several bulges beame evident, the more relevant ones with the following dimensions: 36 in wide x. in of protuberane, ourses 4, 5, and 6, line 3; 34 in wide and. in of protuberane, ourses 9 through 5, line 7; and 38 in wide and.6 in of protuberane, ourses 3, 4 and 5, line 0. ) Out of roundness measurements were also taken in the lower ylindrial portion of the tower, Figure E.3-. The largest deviation (Dmax-Dmin) was in the order of 4.9 in. The nominal diameter was 0 in, the aeptable value of out of roundness should not be larger that %, that is,. in. 3) Three models were used for finite element stress analyses of the bulged areas. Details of the FE analysis are not part of the sope of this example. Part 8 example problems inludes an example of an FE analysis of a bulged vessel. b) STEP Testing and Metallographi Evaluation on Material Samples ) The strength of the reusable steel in the lower portion of the tower was assumed to omply with the minimum speified ultimate tensile strength of 70 ksi but this was estimated based on hardness alone. Testing on material samples was performed to determine the atual strength. ) The opportunity was taken to extrat the material samples from the parts of the tower that were disarded. The steel in ourse was assumed to represent the one used in the original onstrution of the tower. This was verified again and onfirmed by metallography examination. The steel mirostruture in ourse looked very similar to the steel mirostruture in ourses 6 and 7 that were not affeted by the fire, as suggested by the fat that they still had the oating intat on their surfae. 3) A onern that arose in Table E. was weld softening. Hardness testing was again done but this time using a Vikers benh testing mahine, instead of a portable one used in the field. The results are summarized in Table E.. All findings previously made were onfirmed. The weld steel softened in Heat Exposure Zone VI, those that remained in the Heat Exposure Zone V or less, retained aeptable hardness values. The hardness tests done in the field shown in Table E. was found to produe very similar values than those obtained in the laboratory. 4) Metallographi examination of softened welds onfirmed that they were normalized by the fire in the Heat Exposure Zone VI. This implied temperatures in exess of 580 F (860 C). The HAZ disappeared in these normalized welds, but the welds ould still be reognized. Metallographi examination of welds in Heat Exposure Zone V showed the original weld, HAZ, and base metal mirostruture. This was expeted sine heating up to about 5 F (655 C) is very similar to the heating and temperature used in a PWHT, whih is known not to produe any visible and distinguishable mirostrutural hanges that ould beome evident in the optial mirosope, exept for a possible spheriodization of the ementite plates within the pearlite olonies. 5) Tensile testing results from the welds are summarized in Table E.3. The data were analyzed statistially. The 95% onfidene limits were estimated for the mean values. The differene of strength between the weld normalized by the fire and those not affeted by it, was found to be signifiant. For the normalized welds the ultimate tensile strength was ksi, for the unaffeted weld was ksi. 6) The differene of yield strength between the weld normalized by the fire and those not affeted by it, was found to be signifiant too. For the normalized welds the yield strength was ksi, for the unaffeted weld was ksi. Obviously, the fire softened the welds. The differene regarding the perentage of elongation for the welds was not signifiant. 7) The tensile testing results for base metal are shown in Table E.4. In this ase the differene found between the base metal normalized by the fire and not affeted by it, was not signifiant. For the normalized base metal the ultimate tensile strength was ksi, for the unaffeted base metal was ksi. For the normalized base metal the yield strength was ksi, for -8

328 the unaffeted base metal was ksi. The fire did not affet the strength of the base metal, it affeted only the welds. Conlusions of the Level 3 Assessment: ) Stress analyses were onduted to assess whether the existing distortions and bulges in the lower part of the tower were aeptable or not. They were found to be below the ritial distortion size. The lower part of the tower ould therefore be reused in the ondition it was found after the fire. ) Unaeptable weld softening was found earlier with field portable hardness testing. The tests were repeated but using atual samples extrated from the tower and measuring in a benh hardness testing mahine in the laboratory. It was onfirmed that the fire in the Heat Exposure Zone VI did soften the welds below the aeptable limit for a speified minimum tensile strength of 70 ksi. 3) Tensile testing results for the welds showed that indeed the ultimate tensile strength of the normalized welds was lower than that of not affeted welds. However, in all the ases it still omplied with the speified minimum of 70 ksi. This was assessed statistially. 4) Tensile testing results for the base metal showed that the ultimate tensile strength and the yield strength of the steel were not greatly modified by the fire. There was a large variation of properties between samples from the fire-normalized base metal as well as the base metal not affeted by the fire. However, in neither ase, the ultimate tensile strength fell below the speified minimum of 70 ksi. The steel also omplied with the minimum speified perentage of elongation. 5) The steel did not omply with the speified minimum yield strength of 38 ksi. However, the designed of the tower was based on the speified minimum ultimate tensile strength. Sine this non ompliane with the speified yield strength was found also for the steel that was not affeted by the fire, it was assumed to have been present sine installation of the tower in 96. 6) Based on these results the lower part of the tower that was going to be reused was therefore not rejeted beause of this old non ompliane. Therefore, The Level 3 Assessment Criteria are Satisfied for the lower ylindrial part of the depropanizer tower Further Ations: ) A new upper portion of the tower was fabriated out of arbon steel SA-56 Grade 70, inluding the top head, ourses through 7, and the onial setion. The new welds were PWHT. The lower portion of the tower, ourses 8 through 7 were reused in the onditions they were found after the fire. ) All the required doumentation for the fire damage assessment as well as the FFS were produed and kept in the respetive equipment files. Instrutional Comments: ) There are no speifi rules to define the onditions and amount of measurements and dimensional heks required to omplete the assessment of a vessel distorted during a fire. However, it is reommended to omplete as many measurements and dimensional heks as possible within the physial limitations and time onstraint after the fire. ) Details of the finite element stress analyses of bulged and other distorted areas are not part of the sope of Part Example Problems. Refer to Part 8 Example Problems for details of assessment proedures and aeptane riteria. 3) Hardness testing results should not lead diretly to a rejetion but rather to the deision to ondut a proper tensile test to determine atual properties, instead of relying on hardness values alone. 4) In-situ metallography results need proper interpretation. In this ase the plates of arbon steel SA- Grade B were not normalized originally. The initial mirostruture was oarse and exhibited a similar appearane than the oarse grained mirostruture of the HAZ immediately adjaent to the fusion line of a weld. This oarse steel mirostruture omprises grain boundary allotriomorph ferrite, loated at what were the prior austenite grain boundaries, and Widmanstatten ferrite plates at these former austeniti grain boundaries and within these grains too. In between these ferrite phases, the struture was dark ething, most likely omprising fine pearlite or even a mixture of pearlite and bainite. Under otherwise normal ferriti-pearliti steel mirostruture, the appearane of suh a oarse steel mirostruture would indiate exessive austenitizing temperatures, in the order of 000 F or higher, ausing the austenite grain struture to grow larger. In arbon steel oming from manufaturing, it usually implies that the steel was delivered in the as-hot rolled or ashot forged ondition. Steel hot working is usually arried out at about 000 F or higher. However, -9

329 not all hot worked arbon steel omes with suh a oarse mirostruture sine this depends of the ontrol of the finishing temperature, rather than the normal hot work temperature. 5) For short term heating and in partiular PWHT arbon steels, there will not be any notieable mirostrutural hanges in the steel, unless the heating temperature reahes or exeeds the subritial temperature of about 350 F (73 C), at whih the pearlite would transform to austenite, but the ferrite would remain unaffeted. 6) The amount of ferrite that survives dereases as the heating temperature within this region of partial transformation inreases, until reahing 580 F (860 C), the point of omplete transformation of all prior ferrite into austenite. Upon ooling, this newly formed austenite transforms bak into a ferrite and ementite (Fe 3 C) mirostruture that depends on the ooling rate but whih would leaves distint metallographi evidene indiating this level of overheating. 7) Below the subritial temperature of about 350 F (73 C), there are usually not enough mirostrutural hanges to reognize that the steel has been heated below this limit during the fire. Sine the desription of Heat Exposure Zone V implies temperatures between 800 F and 350 F (47-73 C), in-situ metallography may not be useful to detet overheating to temperature below 350 F (73 C). If the original ferrite-pearlite mirostruture is normal, it might be possible to determine if the steel was heated beyond 350 F (73 C) and if so, the Heat Exposure Zone V as desribed in Part ould be defined via metallography. 8) Depending on the original mirostruture of the arbon steel, it is sometimes possible to detet some degree of spheriodization of the ementite plates within the pearlite olonies and this would indiate overheating between about F ( C). The problem is that a PWHT arried about at about 50 F (60 C) may also produe a similar effet. So if this mirostrutural feature is seen, it would not be possible to asertain whether it was during the PWHT or the fire. 9) Grain growth is a phenomenon that strongly depends on the amount of prior old working that the metal exhibits. If there is some amount of old working in the arbon steel, rerystallization and grain growth ould start ourring at temperatures as low as 50 F (60 C) or even lower. However, arbon steel in pressure vessels and piping is never in the old worked state but either hot-worked, normalized, quenh and tempered or annealed. Under these irumstanes, grain growth does not our below the subritial temperature for partial transformation 350 F (73 C), not matter how prolonged is the heating. 0) If the steel is heated in the fire to temperatures exeeding 350 F (73 C), there would be partial transformation of the steel involving the disappearane of all the pearlite olonies to austenite. Any single pearlite olony will give rise to more than one austenite grains sine the transformation proess is governed by nuleation and growth and there are many nuleation sites in eah ferriteementite interphase of a pearlite olony. By prolonging the time at whih the steel is heated, one austenite grain an grow at the expense of others. However, even if eventually one single austenite grain manages to oupy the whole spae of a former pearlite olony, upon ooling it will have to transform bak to ferrite and pearlite again. Sine this transformation proess is also governed by nuleation and growth, it would be highly unlikely that one single austenite grain will nuleate only one single ferrite grain beause this would wrongly suppose that a former pearlite olony an transform into a single ferrite grain. It would be equally unlikely that one single austenite grain will nuleate one single pearlite olony. The transformation ours by nuleating first some ferrite grains at existing austenite grain boundaries until eventually the remaining arbon enrihed austenite in between these newly formed ferrite grains transforms into pearlite. This would result in more than one ferrite grains and more than one pearlite olony. The proess is illustrated in Figure E.3-. ) Figure E.3- also shows a orrelation with respet to the hot yield strength to demonstrate that above the subritial temperature A (350 F), overheated arbon steel greatly looses its strength. Notie the representation of the spheriodization of ementite plates within the pearlite olonies when heating just below this subritial temperature. The original ondition was taken to be an ideal pearlite olony with parallel alternate plates of ferrite and ementite. Also notie that at a temperature just above the subritial level, the pearlite olonies transform to austenite upon heating and then the austenite transform into a finer ferrite-pearlite mirostruture upon ooling. ) At 350 F (73 C), all the pearlite olonies must transform and as the heating temperature exeeds and departs from350 F (73 C), more austenite will form, also by a nuleation and growth proess starting at existing ferrite grain boundaries. Even if every ferrite boundary -0

330 nuleates a single austenite grain and even if eventually one single austenite grain manages to oupy the whole spae of a former ferrite grain, upon ooling, this single austenite grain will give rise to more than one ferrite grain and more than one pearlite olony. The final mirostruture might thus look finer or similar than the original beause of this double transformation proess, driven by nuleation and growth, that will multiply the amount of grains and pearlite olonies. This is why normalizing to temperatures of 580 F (860 C) or slightly above, will either ause grain refinement or will ause no hanges in the final ferrite-pearlite grain size. So even though this is undoubtedly a very high overheating, it may easily result in grain refinement or not grain growth at all. This is against the ommon belief of expeting grain growth in most ases of overheated arbon steels. 3) Grain growth may indeed be observed but this requires overheating in exess of 800 F (000 C) or higher. While the steel is this hot it would then omprise only austenite grains, but when it is metallographially examined, it would be ferrite and Fe 3 C with a morphology and distribution that depends on the level of overheating and the ooling rate. At this high temperature the austenite grain may grow larger rather fast, the higher the overheating temperature the faster it grows, up to a point where a oarse steel mirostruture omprising grain boundary allotriomorph ferrite and Widmanstatten ferrite plates may result in a matrix of fine pearlite or even bainite, after it ools down. In this ase suh a high overheating may result in a very distint mirostrutural hange as ompared with the previous one, this rather than just grain growth of a normal ferrite-pearlite mirostruture. -

331 Table E. - Hardness Values (HBN) Obtained In The Same Plaes Where In-Situ Metallography Were Performed Replia Side Plae Weld HAZ Base Metal R North Course and R South Course and R3 East Course R0 South Course and R8 South Course 3 and R9 North Course 4 and R5 Southeast Course 4 and R6 North Cone-Course R7 South Cone-Course R6 North Course R4 South Course 7 and R7 North Course 9 and R5 South Course 0 and R0 West Course 6 and Table E. - Weld Hardness Values (HBN) Obtained On Atual Material Samples Removed From The Tower Side Weld Plae One Side Another Side West upper Longitudinal Course West middle Longitudinal Course West lower Longitudinal Course Southeast Cirumferential Courses 4 & North Cirumferential Courses 4 & West upper Longitudinal Course West middle Longitudinal Course West lower Longitudinal Course Note: For eah ell, the value is the average out of six Vikers hardness tests, onverted into HBN. -

332 Table E.3 - Tensile Test Results Obtained On Atual Weld Material Samples Removed From The Tower Side Fire Zone Weld VI West NO Longitudinal, Course North NO Cirumferential Courses 4 and 5 Southwest YES Cirumferential Courses 4 and 5 West NO Longitudinal Course 7 West YES Longitudinal Course 7 Ultimate Tensile Strength, ksi Yield Strength, ksi % Elongation Table E.4 - Tensile Test Results Obtained On Atual Base Material Samples Removed From The Tower Ultimate Tensile Side Fire Zone Course Strength, ksi VI Southwest YES Southwest YES West YES North NO North NO North NO South NO West NO Yield Strength, ksi % Elongation

333 NORTH Mirostruture not affeted Mirostruture not affeted Mirostruture Normalized Mirostruture not affeted R R9 R6 R R R R R3 R8 R5 R7 R0 R4 Mirostruture not affeted Mirostruture Normalized Mirostruture Normalized Mirostruture Normalized Mirostruture Normalized ZONE VI Mirostruture not affeted 0 R5 Mirostruture not affeted Coating still intat in these lower ourses 6 7 R0 Mirostruture not affeted SKIRT Figure E.3- - Plaes Where In-Situ Field Metallography Was Performed. Heat Exposure Zone VI Was Defined Based On Metallographi Findings -4

334 North 3 ZONE VI Burst Pipe Northwest 4 5 Burst Pipe South Side 78'-6" (54,4 m) ZONE V Leaking Flange Southeast 7 Skirt Figure E.3- - Elevation View Showing Heat Exposure Zones V and VI, Looking Toward The East -5

335 West 3 4 ZONE VI Bursst Pipe Northwest side Burst Pipe South Side 8 ZONE V 9 0 ZONE V 78'-6" (54,4 m) 3 Leaking Flange Southeast side Skirt Pumps Figure E Elevation View Showing Heat Exposure Zones V and VI, Looking Toward The North -6

336 NORTH Level of this drum Upper level of towers Floor Level Figure E Plan View Showing Heat Exposure Zone V -7

337 (a) (b) () (d) Figure E Telebrineller Hardness On Sides (a) North, (b) South, () East, And (d) West. There Are 4 Measurements Per Eah Course, Making A Total Of 68 Per Line -8