PREPARATION CLASSAPI 510 PRESSURE VESSEL INSPECTORCERTIFICATION EXAMINATIONPUSPATRIJOHOR BAHRU5TH -9TH NOVEMBER 2007

Course OutlinesDAY 1 - 5th November 2007Introduction to API 510 Certification Module 1: ASME Section VIII Rules For Construction Of Pressure VesselModule 2: Static Head, MAWP & Stress Calculations

DAY 2 - 6th November 2007Module 3: Joint Efficiencies & Internal PressureModule 4: Pressure Testing, MDMT, Impact TestingModule 5: External Pressure

Course OutlineDAY 3 7th November 2007

Introduction to API 510 Pressure Vessel Inspection CodeSection 1 : Scope Section 2 : ReferencesSection 3 : DefinitionsSection 4 :Owner User Inspection Organization

Section 5 - : Inspection Practices 5.1 Preparatory Work5.2 Modes of Deterioration and Failure5.3 Corrosion Rate Determination5.4 Maximum Allowable Working Pressure Determination5.5 Defect Inspection5.6 Inspection of Parts5.7 Corrosion & Minimum Thickness Evaluation5.8 Fitness for Service Evaluation

Course OutlineDAY 4 8th November 2007

Section 6: Inspection and Testing of Pressure Vessels and Pressure Relieving Devices6.1 General6.2 Risk Based Inspection6.3 External Inspection6.4 Internal and On Stream Inspection6.5 Pressure Test6.6 Pressure Relieving Devices6.7 Records

Section 7: Repairs, Alterations and Re-rating of Pressure Vessels7.1 General 7.2 Welding7.3 Rerating

Course OutlineDay 5 9th November 2007

API 572: Inspection of Pressure Vessel (Towers, Drum, Reactors Heat Exchanger & Condensers)API 576: Inspection of Pressure Relieving Devices

Trial Examination

Why Are You Here?Why API Certification?Significant milestone in inspectors careerAdditional job opportunities & salary increaseWiden employment doors resume with API certificatesOil & Gas industry is boomingMiddle East offers USD 500 - 700 per day

Life is too short to be ordinary

What Will Be Asked?API Publications1. API 510, Pressure Vessel Inspection Code2. API RP 571, Damage Mechanisms Affecting Equipment in Refining Industry3. API RP 572, Inspection of Pressure Vessels4. API RP 576, Inspection of Pressure Relieving Devices5. API RP 577, Welding Inspection and Metallurgy

ASME Publications1. Section V, Nondestructive Examination2. Section VIII, Division 1, Rules for Constructing Pressure Vessels3. Section IX, Welding and Brazing Qualifications

Dont Worry About The ExamThe API Examination- 150 multiple choices with four possible answers- Exam divided into 2 PartsOpen Book 50 questions for 4 hours durationsClosed Book 100 questions for 4 hours durations

-The examination handle by Professional Examination Services (PES)-Result approximately 3 months after the examination-API grants three consecutive attempts within 18 months periods- 1st attempt : Applications forms and exam fees USD 800- 2nd attempt : re-scheduling fees is USD 50- 3rd attempt : USD 50 plus updated Employment Reference Form Note: Failed after three attempts - New applications with new applications fees

What To Bring To ExamAPI Examination Confirmation LetterIdentification CardAPI & ASME reference publications. Note: highlighting, underlining, page tabs, written notes on the codes book is acceptable. Loose pages inserted into the codes book is not acceptable.Non-programmable calculator. Make sure enough battery.2B pencil, eraser and other stationeryJacket some classroom is uncomfortably coolEarplugs you never know who might be seat next to you

Module 1: ASME Section VIII Rules For Construction Of Pressure Vessel

ASME SECTION VIIIRULES FOR CONSTRUCTION OF PRESSURE VESSEL

ASMEBoiler & Pressure Vessel CodeSectionsI Rules for Construction of Power BoilersII Materials-Ferrous, Nonferrous, Welding Rods, ElectrodesIII Nuclear Power PlantIV Rules of Construction of Heating BoilersV Nondestructive ExaminationVI Rules for Operation of Heating BoilersVII- Guidelines for Operations of Power BoilersVIII- Rules for Construction of Pressure VesselIX-Welding & Brazing QualificationsX-Fiber Reinforced Plastic Pressure VesselXI-Rules for In-Service Inspection of Nuclear Power PlantXII-Rules for Construction and Continued Service of Transport Tank

Settings Rules!Is What The Code Is All AboutImportant factors during fabrication that affect vessel safety and reliability seem almost endless.

Design thickness formulas, welding processesMaterial Known physical & chemical propertiesFabrication qualified welding procedure, NDTPressure Testing hydro or pneumaticDocumentation nameplate, design calculation data

Scope of the CodeU-1(a) For the scope of the division pressure vessels are containers for the containment of pressure either internal or external.U-1(c)(2) The following classes of vessels are not considered to be within the scope of the division:Within cope of other divisionFired process tubular heatersPV of integral parts/component of pumps, turbines, compressorPipe and piping componentPV for water under pressure in which Pd

Scope of the Code continuedVessel Boundary LimitsU-1(e)1(a) 1st circumferential weld(b) 1st threaded joint for screwed connection(c) face of 1st flange for bolted connection(d) 1st sealing surface for proprietary connection/fitting

U-1(e) 2 Non-pressure part that are welded directly to the vesselU-1(e) 3 Pressure retaining cover i.e. manhole and hand hole covers

Shhhhh!!!!!During API exam you will be asked to find specific information in Section VIII. In order to accomplish this quickly and successfully you must know:

How information is organized in the codeHow to use the organization tools of the code

If you know these 2 secret, every answer can be found with minimum effort. Thats much better than paging through 700 pages of the codes!

Organization of CodeSection VIII- Rules for Construction of Pressure Vessel consists of 2 division;Division 1 Routine VesselDivision 2 Alternate Rules for special vessel

Section VIII Division 1Introduction3 SubsectionsSubsection A GeneralSubsection B Fabrication MethodSubsection C - Material2 AppendixMandatoryNon-Mandatory

Organization of Code continuedEach Subsection is further divided into PartsSubsection A GeneralPart UG applies to all vesselsSubsection B FabricationPart UW applies to all vessels that are weldedSubsection C MaterialsPart UCS applies to all vessel made of CS or LA steelPart UHT applies to ferritic vessel that use Heat Treatment

Every Part is divided into ParagraphsUG-1, UW-1, UCS-1

Organization of Code continuedMandatory AppendicesU-1(b) address specific subject not covered elsewhere in this divisionAlternate formulaeQC ManualNDE Standards

Non-mandatory AppendicesU-1(b) provides information and suggested good practicesCompleted Sample Problems

Example 1-1 Finding the Right AnswerAt what base metal temperature is welding not allowed on routine pressure vessel?

Example 1-1 Finding the Right AnswerAt what base metal temperature is welding not allowed on routine pressure vessel?ASME CodeSection IX WeldingSection V NDTSection VIII Pressure VesselSection II MaterialsSection 1 BoilersDivision 1 RoutineDivision 2 OtherSub A GeneralSub B FabricationSub C MaterialsPart UF ForgePart UW WeldingPart UB Brazing

Exercise 1-2A vessel is welded with Carbon Steel components. Which part in the Code would you find:Design Requirements: ____________________Hydrotest Pressure: ____________________RT Acceptance Standards: ____________________Nameplate Data: ____________________Limits of Carbon % in Materials: ____________________Material ID Traceability: ____________________Inspection Requirements: ____________________PWHT Requirements: ____________________

Exercise 1-2 answer.A vessel is welded with Carbon Steel components. Which part in the Code would you find:Design Requirements: UG, UW, UCSHydrotest Pressure: UGRT Acceptance Standards: UWNameplate Data: UGLimits of Carbon % in Materials: UCSMaterial ID Traceability: UGInspection Requirements: UG, UW, UCSPWHT Requirements: UW, UCS

Hint: If its applicable to all vessel UG.If its only applicable because its welded UWIf its based on metallurgy - UCS

Codes PurposeEstablish Rules for Construction of Pressure VesselTo ensure the construction of pressure vessels will be carried out in a safe and reliable manner.

One way the Code achieves this is by setting requirements for critical work assignment. This will involved:UserManufacturerAuthorized InspectorWelderNDE Technician

Setting The RulesUser Roles & ResponsibilitiesU-2(a) : The user or his designated agent shall establish the requirements for pressure vessel - Specify size and shape- Specify vessel internals; type, spacing- Determine design pressure and temperature- Specify the Corrosion Allowance- Determine whether the PV be classified as Lethal Service- Specify PWHT if needed for process conditions

Setting The RulesManufacturer Roles & ResponsibilitiesU-2(b)(1) The manufacturer of any vesselhas the responsibility of complying with all of the applicable requirements of this Division ..UG-90(b) The Manufacturer shall perform his specified duties.Obtain Certificate of Authorization from ASMEPerform design calculations & develop fabrications drawingsIdentify all material used during fabricationExamine materials before fabrication: thickness, ID, defectsQualify welding procedures & weldersPerform NDE test and records resultPerform vessel hydro or pneumatic testApply the Code StampPrepare Manufacturers Data Report

Setting The RulesAuthorized Inspector Roles & ResponsibilitiesSection VIII requires that all vessels be inspected by a qualified third-party inspector which is employed by An Authorized Inspection Agency.

U-2(e) It is the duty of the Inspector to make all of the inspections specified by the rules of this Division, and of monitoring the quality control and the examinations made by the Manufacturer..

U-2(f) The rules of this Division shall serve as the basis for the inspector to:Perform required dutiesAuthorize the application of the Code SymbolSign the Certificate of Shop Inspection

Setting The Rules continuedAuthorized Inspector Roles & ResponsibilitiesUG-90(c)(1) The inspector shallVerify the Manufacturer has a current Certificate of AuthorizationVerify Manufacturer is working to the QC systemVerify design calculations are availableVerify materials meet CodesVerify weld procedures and welders are qualifiedVerify NDE tests have been performed & are acceptablePerform internal & external inspectionsVerify nameplate is attached and has the right markingsWitness the hydrotestSign the Manufacturers Data Report

Setting The Rules Welders & NDE Techs Roles & ResponsibilitiesWelderUW-29(a) The welders ..used in welding pressure partsshall be qualified in accordance with Section IX

NDE TechnicianUW-51(a)(2) RT Qualified and certified in accordance with employers written practice. SNT-TC-1A used as a guidelinesApp 12-2 UT Qualified and certified in accordance with employers written practice. SNT-TC-1A used as a guidelinesApp 6-2 (a&b) MT He has vision ..read a Jaeger Type No. 2 Chart .. And is competent in the techniques of the magnetic particleexamination methodApp 8-2 (a&b) MT He has vision ..read a Jaeger Type No. 2 Chart .. And is competent in the techniques of the magnetic particleexamination method

CODE-OLOGYCode StampWhen the manufacturer uses the Code Stamps, they are saying We met all the applicable requirement of the Code:

UG-116(a&b) Each pressure vessel shall be marked with official Code U symbol or the official UM Symbol

UG-116 (g) The Manufacturer shall have a valid Certificate of Authorization, and with the acceptance of Inspector shall apply the Code Symbol to the vessel.Note: The Code Symbol shall be applied after hydrostatic or pneumatic test

CODE-OLOGYCertificate of AuthorizationWhat driving license does for the driver, the Certificate of Authorization (CoA) does for the Manufacturer!CoA - authorizes Manufacturer to design & build a Section VII vessel.CoA -authorizes Manufacturer to stamp the vessel with the Code Stamp.

UG-117(a&b) A Certificate of Authorization to use the Code U, UM, & UV symbolswill be granted by the Society Each applicant must agree that each Certificate of Authorization and each Code Symbol Stamp are at all times the property of the Society

CODE-OLOGYQuality Control SystemUG-117(e) Any Manufacturer shall have and demonstrate a Quality Control System to establish all Code requirementwill be met. The Quality Control System shall be in accordance with Appendix 10.

The Code tells the Manufacturer what must be done when building a vessel. The Manufacturers Quality Control System tells ASME and the AI how things will be done in the shop to meet the code.

CODE-OLOGYData ReportsUG-120(a) A Data Report shall be filled out on Form U-1..by the Manufacturer and shall be signed by the Manufacturer and Inspector for each pressure vessel marked with the Code U symbol.

For UM vessel the Form U-3For vessel parts the Form U-2

CODE-OLOGYUM Vessel Mini VesselIf a vesselIs not covered by U-1 (c), (g), (h) & (i)Is not required to be fully radiographedDoes not have a quick actuating deviceDoes not exceed either the following limit5 ft3 and 250 psi1.5 ft3 and 600 psi

U-1(j) The vessel may be exempted from inspection by Inspectors Vessel fabricated .. With this rule shall be marked with the UM symbol

Exercise 1-31) Full radiography is performed on a vessel shell with a wall thickness of . What is the maximum allowed length for a slag inclusion?

2) A P-1 material (carbon steel) 2 thick is being PWHT. What is thea. Normal Holding Temperatureb. Minimum Holding Time

3) A relief device is required on a vessel so that the pressure does not rise more than _____% or ______ psi above MAWP (whichever greater)

Exercise 1-31) Full radiography is performed on a vessel shell with a wall thickness of . What is the maximum allowed length for a slag inclusion?Subsection B Fabrication Part UW Inspection UW51(b)2

2) A P-1 material (carbon steel) 2 thick is being PWHT. What is thea. Normal Holding Temperatureb. Minimum Holding TimeSubsection C Materials Part UCS Design UCS56

3) A relief device is required on a vessel so that the pressure does not rise more than _____% or ______ psi above MAWP (whichever greater)Subsection A General Part UG Pressure Relief Devices UG125(c)

I have learned that success is to be measured not so much by the position that one has reached in life as by the obstacles which he has overcome while trying to succeed. -Booker T. Washington

Module 2: Static Head, MAWP & Internal Pressure

Module 2.1 Static Head

What is Static Head?The weight of liquid applied a force (lbs) & a pressure (psi). The higher the liquid height, the greater the pressure. The pressure resulting from liquid height is called, Static Head!Box is 1 cubic ft1 cubic foot of water weighs 62.4 lbs.How much force (weight) is on the bottom of this container?______________How much force on each square inch of the boxs bottom?________ Whats this pressure? _______

Static Head Factor?Water, 1 foot high will exert 0.433 psi at the bottom of the container?

62.4 lbs/144 sq inch = 0.433 psi per foot of water

What is the pressure at the bottom of 10 of water?

PSHead = 0.433 x liquid height

Exercise 2-1Static Head PressureA deep diving submarine cruising at a depth of 854 feet. What is the static head pressure on this submarine? (external pressure)A vessel is 50 high. The vessel will be hydrotesteda. When filled with water what is the pressure at the bottom of the vessel?b. When the hydrotest pressure at the top of the vessel 100 psi, what is total pressure at the bottom?A 60 vessel is filled with water. The pressure at the bottom is 210 psi. What is the pressure at the top?

Exercise 2-1Answer 1A deep diving submarine cruising at a depth of 854 feet. What is the static head pressure on this submarine? (external pressure)845 ft x 0.433 psi/ft = 369.8 psig854 ft

Exercise 2-1Answer 2a2) A vessel is 50 high. The vessel will be hydrotestedWhen filled with water what is the pressure at the bottom of the vessel?Pbtm = 0.433 psi/ft x 50 ft = 21.7 psig

Exercise 2-1Answer 2b2) A vessel is 50 high. The vessel will be hydrotestedb. When the hydrotest pressure at the top of the vessel 100 psi, what is total pressure at the bottom?Pbtm = Ptop + Psh = 100 + (0.433x50) = 121.7 psig

Exercise 2-1Answer 33) A 60 vessel is filled with water. The pressure at the bottom is 210 psi. What is the pressure at the top?Ptop = Pbtm - Psh = 210 - (0.433x60) = 184 psig

Module 2.2Design Pressure

Design PressureThe pressure used in the design of a vessel component together with the coincident design metal temperature for the purposes of determining the minimum permissible thicknessstatic head shall be added to the design pressure App 3-2

Exercise 2-2Design PressureA 50 high vessel has a design pressure of 100 psig. The elevations are shown in the sketch below.

The shell should be designed for _________ psigThe top head should be designed for _________ psigThe bottom head should be designed for_________ psig

Exercise 2-2Design PressurePshell = Ptop +Psh = 100 psig + (0.433 psi/ft x 48 ft) = 120.8 psigPtop head = Ptop + Psh= 100 psig + (0.433 psi/ft x 2 ft)= 100.9 psigPbtm head= Ptop + Psh= 100 psig + (0.433 psi/ft x 50 ft)= 121.7 psigNote: Each component should be designed for the highest pressure it will see at conditions. The highest pressure is at the bottom of the part

Module 2.3MAWP Calculations

MAWPVessel & Vessel Part

UG98(b) The maximum MAWP for a vessel part is the maximum pressure including static head(based) upon rules and formulae in this Divisionexcluding any metal thickness specified as corrosion allowance

UG98(a) The maximum MAWP for a vessel is the maximum pressure permissible at the top of the vessel in its normal operating position...

It is the least of the values found for maximum allowable working pressure for any of the part of the vesseland adjusted for any difference in static headHow much can he lift?

MAWPVessel & Vessel PartTo determine Vessel MAWP:Step 1: Determine each part MAWP (based on code formulas)Step 2: For each part subtract appropriate static headStep 3: Pick smallest pressure at top, the weakest link1234ABPart B: MAWP 336Part A: MAWP 343Part A: 343 0.433(34) = 328.3 psigPart B: 336 -0.433(12) = 330.8 psigThus, vessel MAWP = 328.3 psig0

Exercise 2-3Determining Vessel MAWPThe maximum MAWP this vessel can be rated is ______psig.0 ft1 ft3 ft24 ft40 ft42 ft

Exercise 2-3Determining Vessel MAWPThe maximum MAWP this vessel can be rated is ______psig.0 ft1 ft3 ft24 ft40 ft42 ft

Design Pressure vs. MAWPDesign Pressure pressure from the system (process + static head)MAWP pressure part of the vessel is good forMAWPDesign PressureTo meet Code Requirement:MAWP > Design PressureIf MAWP < Design Pressure

Exercise 2-3More MAWP & Static Head calculationThe MAWP of a vessel is 100 psig. Each head depth is 2 and the cylindrical portion of the shell is 32. The shell should be designed for a pressure of ______ psig.A vessels MAWP is limited by the lower shell portion. This shell part has a MAWP of 87.5 psig. What is the maximum allowed MAWP for this vessel?

A 80 tall pressure vessel is being hydrotested. A pressure gauge 20 up from the bottom reads 136 psig. What is the pressure at the top of the vessel.0 ft1 ft50 ft100 ft

Exercise 2-3More MAWP & Static Head calculationThe MAWP of a vessel is 100 psig. Each head depth is 2 and the cylindrical portion of the shell is 32. The shell should be designed for a pressure of ______ psig.

Pshell = Ptop + Psh = 100 + 0.433(34) = 114.7 psig100 psig0 ft2 ft 34 ft

Exercise 2-3More MAWP & Static Head calculationA vessels MAWP is limited by the lower shell portion. This shell part has a MAWP of 87.5 psig. What is the maximum allowed MAWP for this vessel?

Ptop = Pshell Psh= 87.5 (0.433 x100)= 44.2 psig

Exercise 2-3More MAWP & Static Head calculationA 80 tall pressure vessel is being hydrotested. A pressure gauge 20 up from the bottom reads 136 psig. What is the pressure at the top of the vessel.

Ptop = Pgauge Psh= 136 (0.433 x 60)= 110.0 psig0 ft80 ft60 ft136 psig

Module 2.4Calculating Stress

Stress on WeldsCircumferential WeldLongitudinal WeldCircumferential Stress affects: ___________ weldsLongitudinal Stress affects: __________ welds

Stress on WeldsCircumferential WeldLongitudinal WeldCircumferential Stress affects: Longitudinal weldsLongitudinal Stress affects: Circumferential welds

Calculating StressA weld specimen that is 0.250 thick and 1.0 wide. The specimen breaks with 12,500 lbs. of load. What is the ultimate tensile strength of this specimen?

Calculating StressA weld specimen that is 0.250 thick and 1.0 wide. The specimen breaks with 12,500 lbs. of load. What is the ultimate tensile strength of this specimen?

S = Load/AreaS = 12,500/(0.25 x 1.0)S = 50,000 psi

Exercise 2.5Calculating StressA tension specimen is 0.5 thick and 0.75 wide. It breaks with a load of 21,500 lbs. The failure stress is _______ psi.A vessel is 10 in diameter and has pressure of 100 psi. The force (load) trying to launch a vessel head into space is 1,100,000 lbs. The circumference of the shell is about 400. The head and shell are thick, not including corrosion allowance.What is the actual longitudinal stress on the shell-to-head weld?

Exercise 2.5Calculating StressA tension specimen is 0.5 thick and 0.75 wide. It breaks with a load of 21,500 lbs. The failure stress is _______ psi.S = Load/AreaS = 21,500/(0.5)(0.75)S = 57,333 psi2)A vessel is 10 in diameter and has pressure of 100 psi. The force (load) trying to launch a vessel head into space is 1,100,000 lbs. The circumference of the shell is about 400. The head and shell are thick, not including corrosion allowance.What is the actual longitudinal stress on the shell-to-head weld?S = Load/AreaS = 1,100,000/(400)(0.5)S = 5,500 psi

What is Allowable Stress?Stress level that the designer is allowed to used.The allowable stress is generally determined by dividing the Ultimate Tensile Strength by Codes safety Factor.

The Allowable Stresses for Section VIII Pressure Vessel are provided in the B&PV Code Section II.

The Safety Factor for Section VIII vessel is:

Pre 20004.0 to 1.0 Safety FactorPost 20003.5 to 1.0 Safety FactorAllowable Stress = Ultimate Tensile Strength / Safety Factor

Exercise 2-6Allowable StressA material has an ultimate tensile strength of 70,000 psi at ambient temperatureWhat is the allowable stress for this material at ambient conditions if used in a pressure vessel today? What is the allowable stress for this material at ambient conditions if used in a 1977 pressure vessel ?

Exercise 2-6Allowable StressA material has an ultimate tensile strength of 70,000 psi at ambient temperatureWhat is the allowable stress for this material at ambient conditions if used in a pressure vessel today? Allowable Stress = UTS/S.FAllowable Stress = 70,000/3.5Allowable Stress = 20,000 psib)What is the allowable stress for this material at ambient conditions if used in a 1977 pressure vessel ?Allowable Stress = UTS/S.FAllowable Stress = 70,000/4.0 Allowable Stress = 17,500 psi

You may be disappointed if you fail, but you are doomed if you dont try -Beverly Sills

Module 3.1Joint Efficiencies

E The BasicWhat is Joint Efficiency E?What factors that affect E?How does Joint Efficiency affect E?How is Joint Efficiency determined?

E The BasicWhat is Joint Efficiency E?- A safety factor for welds- Compensation for possible weld defectsWhat factors that affect E?- Type of Joint, Location of Joint, Amount of RTHow does Joint Efficiency affect t?- As E decreases, required thickness increasesHow is Joint Efficiency determined?- The Code, Section VIII Table UW-12- (also few exception listed in UW-12)

Exercise 2.7Joint EfficienciesA shell is made with Type 2 joints and spot RT was performed. What is E? _________. A shell is made with Type 1 joints and Full RT was performed. What is E? _________. A shell is made with Type 3 joints and no RT was performed. What is E? _________.

Exercise 2.7Joint EfficienciesA shell is made with Type 2 joints and spot RT was performed. What is E? 0.8 . A shell is made with Type 1 joints and Full RT was performed. What is E? 1.0. A shell is made with Type 3 joints and no RT was performed. What is E? 0.6.

Weld Joint Categories The term Category as used herein defines the location of a joint in a vessel, but not the type of joint.

Category A: Longitudinal welded joints within the main shell, communicating chambers,2 transitions in diameter, or nozzles; any welded joint within asphere, within a formed or flat head, or within the side plates3 of a flat-sided vessel; circumferential welded joints connecting hemispherical heads to main shells, to transitions in diameters, to nozzles, or to communicating chambers.

Category B: Circumferential welded joints within the main shell, communicating chambers, nozzles, or transitions in diameter including joints between the transition and a cylinder at either the large or small end; circumferential welded joints connecting formed heads other than hemispherical to main shells, to transitions in diameter, to nozzles, or to communicating chambers. UW-3

Category C: Welded joints connecting flanges, Van Stone laps, tubesheets, or flat heads to main shell, to formed heads, to transitions in diameter, to nozzles, or to communicating chambers,any welded joint connecting one side plate to another side plate of a flat sided vessel.

Category D: Welded joints connecting communicating chambers or nozzles to main shells, to spheres, to transitions in diameter, to heads, or to flat-sided vessels, and those joints connecting nozzles to Communicating chambers (for nozzles at the small end of a transition in diameter, see Category B).

Degree of RadiographyType of RadiographyFull Generally 100% of welds, some exceptionSpot On RT for each 50 ft of weldNone Send the RT techs home

Amount specified by Code for some vesselsBased on Service, Thickness or Welding Process (UW-11)

Amount specified by Users/Designer for othersThe User or his designated agent shall establish the type of joint and the degree of examination when the rules of this Division do not mandate specific requirements (UW-12)

Full RT Required by CodeFull RT is required by the Code, when:Vessels in Lethal Service 100% RT Butt WeldsButt Welds > 1-1/2- 100% RTUnfired Boilers > 50 psig 100% RT Butt WeldsButt Welds of nozzles for 1&3 100% RTButt Welds made using Electrogas & Electroslag Process -100% RT

UW-11(a)

Full RT Required by UserFull RT can be specified by the User:- If full RT is not required by the Code- Selected to increase E and lower t- Full RT does not mean 100% RT- The following RT must be performed- Category A welds -100% RT- Category A&B welds Type 1 or 2- Category B welds Spot RT UW-11(a)(5)

The RT FactorsDescribe the amount of RT performed- RT 1&2: Full Radiography- RT-3: Spot Radiography- RT-4: Combo Radiography

The RT Factor is located on NameplateThey are described in UG-116(e)

Seamless Parts EFor seamless vessel sections or head, where circumferential stresses govern:E = 1.0When the category B and C butt welds are spot RTAnd the welds connecting seamless vessel sections or heads are either Type 1 or 2.

E = 0.85When the butt welds are either not spot RTOr when the welds connecting seamless vessel sections or heads are type 3,4,5 or 6.UW-12(d)

Example Finding EA pressure vessel has butt welds which are single welded. No RT has been performed.E for welded shell? _______E for the seamless head? _______A pressure vessel has butt welds which are double welded. The vessel is stamped as RT-3E for welded shell? _______E for the seamless head? ________A pressure vessel has butt welds which are double welded. The vessel is stamped as RT-2.E for welded shell? _______E for the seamless head? _______A pressure vessel has butt welds which are double welded. The vessel is stamped as RT-1E for welded shell? _______E for the seamless head? _______

Example Finding EA pressure vessel has butt welds which are single welded. No RT has been performed.E for welded shell? 0.6E for the seamless head? 0.85A pressure vessel has butt welds which are double welded. The vessel is stamped as RT-3E for welded shell? 0.85E for the seamless head? 1.00A pressure vessel has butt welds which are double welded. The vessel is stamped as RT-2.E for welded shell? 1.00E for the seamless head? 1.00A pressure vessel has butt welds which are double welded. The vessel is stamped as RT-1E for welded shell? 1.00E for the seamless head? 1.00

Exercise 2-8More Joint EfficienciesA pressure vessel has lap welds which are single welded. The vessel is stamped RT-3E for welded shell? _____E for the seamless head? ______A pressure vessel has lap welds which are double fillet weldedE for welded shell? ______E for the seamless head? ______A pressure vessel has butt welds which are single welded with backing strips.The vessel is stamped as RT-2E for welded shell? ______E for the welded head? _______

Exercise 2-8More Joint EfficienciesA pressure vessel has lap welds which are single welded. The vessel is stamped RT-3E for welded shell? 0.45E for the seamless head? 0.85A pressure vessel has lap welds which are double fillet weldedE for welded shell? 0.55E for the seamless head? 0.85A pressure vessel has butt welds which are single welded with backing strips.The vessel is stamped as RT-2E for welded shell? 0.99E for the welded head? 0.99

Module 3.2Tmin Calculations

Successful CalculationsThe 5 steps to calculating Success1) Always write the formula. Leave space above the formula for step 2 data2) Write the Givens above formula. Put these in the same order as the formula.3) Plug-in the values directly below the formula4) Solve the problem5) When complete shows the appropriate units e.g. inches, mpy, years

Shell tmin CalculationsThe formula UG-27(c)(1) Internal Pressure

R is the inside radius of diameterP is the Design Pressure: Pressure on the part includes Static Head

The P formula calcs the shell MAWP

The part MAWP The shells good for pressureThis is not vessel MAWPt = PR/SE-0.6PP = SEt/R + 0.6t

ExampleA vessel has an internal radius of 36. At the design temperature the materials allowable stress is 15,000 psi. The pressure on the shell is 158 psi (static head included). The joint efficiency is 1.0. Determine the minimum required thickness___________________________________________________________The Givens: P=158 psi, R= 36, S=15,000 psi, E=1.0___________________________________________________________The formula: t = PR/SE-0.6P___________________________________________________________The plug-in: t = 158 x 36 / (15,000 x 1) (0.6 x 158)___________________________________________________________The solutions with Unitst = 0.382___________________________________________________________

Exercise 2-9Shell Minimum Required ThicknessA vessel shell has an internal radius of 24. At the design temperatures the materials allowable stress is 20,000 psi. The pressure on the shell is 250 psi (static head is included). The joint efficiency is 1.0. Determine the minimum required thickness?___________________________________________________________The Givens: P= , R= , S=, E=___________________________________________________________The formula: t = PR/SE-0.6P___________________________________________________________The plug-in: t = ___________________________________________________________The solutions with Unitst = ___________________________________________________________

Exercise 2-9Shell Minimum Required ThicknessA vessel shell has an internal radius of 24. At the design temperatures the materials allowable stress is 20,000 psi. The pressure on the shell is 250 psi (static head is included). The joint efficiency is 1.0. Determine the minimum required thickness?___________________________________________________________The Givens: P= 250 psi , R= 24 , S= 20,000 psi, E= 1.0___________________________________________________________The formula: t = PR/SE-0.6P___________________________________________________________The plug-in: t = (250)(24)/(20,000 x 1.0) (0.6 x 250)___________________________________________________________The solutions with Unitst = 0.302___________________________________________________________

Exercise 2-10Shell Minimum Required ThicknessA vertical vessel has an internal radius of 48. The material allowable stress is 12,500 psi. The MAWP of the vessel is 120 psi. The welds are double-welded and the nameplate says RT-3. The top of this shell section is 4 ft from the top of the vessel and the bottom of this shell section is 52 ft from the top of the vessel. Determine the minimum required thickness.

2) A horizontal vessel has an internal diameter of 10 ft. The materials allowable stress is 14,000 psi. The MAWP of the vessel is 120 psi. The welds are all Type 1 and full RT was performed. Determine the minimum required thickness.

Exercise 2-10Shell Minimum Required ThicknessA vertical vessel has an internal radius of 48. The material allowable stress is 12,500 psi. The MAWP of the vessel is 120 psi. The welds are double-welded and the nameplate says RT-3. The top of this shell section is 4 ft from the top of the vessel and the bottom of this shell section is 52 ft from the top of the vessel. Determine the minimum required thickness.___________________________________________________________The Givens: P= ? , R= 48 , S= 12,500 psi, E= 0.85 (Table UW-12)Pshell = Ptop + Psh = 120 + (0.433x52) = 142.5 psi___________________________________________________________The formula: t = PR/SE-0.6P___________________________________________________________The plug-in: t = (142.5)(48)/(12,500 x 0.85) (0.6 x 142.5)___________________________________________________________The solutions with Unitst = 0.649___________________________________________________________

Exercise 2-10Shell Minimum Required Thickness2) A horizontal vessel has an internal diameter of 10 ft. The materials allowable stress is 14,000 psi. The MAWP of the vessel is 120 psi. The welds are all Type 1 and full RT was performed. Determine the minimum required thickness.___________________________________________________________The Givens: P= ? , R= 5=60 , S= 14,000 psi, E= 1.00 (Table UW-12)Pshell = Ptop + Psh = 120 + (0.433x10) = 124.33 psi___________________________________________________________The formula: t = PR/SE-0.6P___________________________________________________________The plug-in: t = (124.33)(60)/(14,000 x 1.00) (0.6 x 124.33)___________________________________________________________The solutions with Unitst = ___________________________________________________________

Rounded Head tmin CalculationsEllipsoidal Head UG-32 (d)

Torispherical Head UG-32 (e)

Hemispherical Head UG-32 (f)

t = PD/2SE-0.2PP = 2SEt/D + 0.2tt = 0.885PL/SE-0.1PP = SEt/0.885L + 0.1tt = PL/2SE-0.2PP = 2SEt/L + 0.2tL = inside radiusL = outside diameter

Head Calculations ExampleThe design pressure (with static head) on a 2:1 seamless elliptical head is 200 psi. The vessel ID is 60. The allowable stress is 15,000 psi. Welds are Type 1 with Spot RT. Find heads minimum required thickness?___________________________________________________________The Givens: P= , R= , S= , E= ___________________________________________________________The formula: t = PD/2SE-0.2P___________________________________________________________The plug-in: t = ___________________________________________________________The solutions with Unitst = ___________________________________________________________

Head Calculations ExampleThe design pressure (with static head) on a 2:1 seamless elliptical head is 200 psi. The vessel ID is 60. The allowable stress is 15,000 psi. Welds are Type 1 with Spot RT. Find heads minimum required thickness?___________________________________________________________The Givens: P= 200, R=60 , S= 15,000 psi, E= 1.00 (Table UW-12(d))___________________________________________________________The formula: t = PD/2SE-0.2P___________________________________________________________The plug-in: t = (200)(60)/2(15,000 x 1.00) (0.2 x 200)___________________________________________________________The solutions with Unitst = 0.401___________________________________________________________

Exercise 2-11Formed Head Minimum ThicknessA vertical vessel has an internal diameter of 84. The heads are hemispherical and made in segments. The head material has n allowable stress of 8,700 psi at the design temperature. The MAWP of the vessel is 48 psi. The overall vessel height is 38. The welds are double welded and Spot RT was performed. Determined the minimum required thickness for the bottom head.A horizontal vessel with seamless torispherical heads has an outside diameter of 96 inches. The materials allowable stress is 20,000 psi. The MAWP of the vessel is 120 psi. The joint efficiency is 1.0. Determine the minimum required thickness for the heads.

Exercise 2-11Formed Head Minimum ThicknessA vertical vessel has an internal diameter of 84. The heads are hemispherical and made in segments. The head material has n allowable stress of 8,700 psi at the design temperature. The MAWP of the vessel is 48 psi. The overall vessel height is 38. The welds are double welded and Spot RT was performed. Determined the minimum required thickness for the bottom head. ___________________________________________________________1)The Givens: P= ?, L=R =42 , S= 8,700 psi, E= 0.85 (Table UW-12)Pshell = Ptop + Psh = 48 + (0.433 x 38) = 64.5 psi ___________________________________________________________The formula: t = PL/2SE-0.2P___________________________________________________________The plug-in: t = (64.5)(42)/2(8,700 x 0.85) (0.2 x 64.5)___________________________________________________________The solutions with Unitst = 0.183___________________________________________________________

Exercise 2-11Formed Head Minimum ThicknessA horizontal vessel with seamless torispherical heads has an outside diameter of 96 inches. The materials allowable stress is 20,000 psi. The MAWP of the vessel is 120 psi. The joint efficiency is 1.0. Determine the minimum required thickness for the heads. ___________________________________________________________1)The Givens: P= ?, L=OD =96=8 , S= 20,000 psi, E= 1.0 (Seamless)Pshell = Ptop + Psh = 120 + (0.433 x 8) = 123.46 psi ___________________________________________________________The formula: t = 0.885PL/SE-0.1P___________________________________________________________The plug-in: t = 0.885(123.46)(96)/(20,000 x 1.00) (0.1 x 123.46)___________________________________________________________The solutions with Unitst = 0.52___________________________________________________________

Flat Head tmin CalculationsThe code allows for many different types of flat head designs. See figure UG-34 for illustrations.You are responsible for welded flat heads, not bolted headsFrom UG-34(c)(2) equation (1)

d generally inside diameterC a factor based on head design concept similar to EE joint efficiency normally 1.0. Only needs to be determined if the flat head is made with multiple plates. This does not apply to head-to-shell weldt = d CP/SE

Exercise 2-12Flat Head tminA flat circular head is made from seamless A-285 Grade B plate with a corner design illustrated in Figure UG-34 (e). The allowable stress is 12,500 psi. The vessel MAWP is 300 psi. The horizontal vessel is stamped RT-3. Assume m=1.0. The vessel inside diameter is 60.

t = d CP/SE

Exercise 2-12Flat Head tminA flat circular head is made from seamless A-285 Grade B plate with a corner design illustrated in Figure UG-34 (e). The allowable stress is 12,500 psi. The vessel MAWP is 300 psi. The horizontal vessel is stamped RT-3. Assume m=1.0. The vessel inside diameter is 60. ___________________________________________________________1)The Givens: P= ?, D = 60 = 5, C= 0.33m = 0.33 x 1 =0.33 , S= 12,500 psi, E= 1.0 (Seamless)Pshell = Ptop + Psh = 300 + (0.433 x 5) = 302.165 psi ___________________________________________________________The formula: t = d CP/SE ___________________________________________________________The plug-in: t = 60 C(0.33)(302.165)/(12,500)(1.0)___________________________________________________________The solutions with Unitst = 5.359___________________________________________________________

Part MAWPPart MAWP is the pressure a part is good forBased on knowing the thickness (dont include the CA)The a typically used in re-rate calculationsThis is not vessel MAWP. Vessel MAWP is based on the weakest link after subtracting Static Head.

Part MAWP formulas are given in the same paragraphs as the tmin formulas.No P formula for flat headsSymbols are the same as used in the tmin formulast = PD/2SE-0.2PP = 2SEt/D + 0.2t

Exercise 2-13Lets Calculate Part MAWP1) The thickness of each part is 0.5. The allowable stress ofte materials is 15,000 psi. The joint efficiency is 1.0. The inside diameter is 60. Calculate the maximum pressure each part is good for.2:1 Ellipsoidal Head: _______ psiTorispherical Head: _______ psiHemispherical Head: _______ psiCylinder: ________ psiFlat Head: ________ psi

2) Which shape is the best for containing pressure? __________3) Which shape is the worst for containing pressure? _________P = 2SEt/D + 0.2tP = SEt/0.885L + 0.1tP = 2SEt/L + 0.2tP = SEt/R + 0.6t

Exercise 2-13Lets Calculate Part MAWP1) The thickness of each part is 0.5. The allowable stress ofte materials is 15,000 psi. The joint efficiency is 1.0. The inside diameter is 60. Calculate the maximum pressure each part is good for.2:1 Ellipsoidal Head: 249.6 psigTorispherical Head: 138.8 psigHemispherical Head: 498.3 psigCylinder: 247.5 psigFlat Head: 5.2 psig For info only

2) Which shape is the best for containing pressure? Hemispherical3) Which shape is the worst for containing pressure? Flat HeadP = 2SEt/D + 0.2tP = SEt/0.885L + 0.1tP = 2SEt/L + 0.2tP = SEt/R + 0.6t

Destiny is no matter of chance. It is a matter of choice. It is not a thing to be waited for, it is thing to be achieved -William Jennings Bryan

Module 4: Pressure Testing, MDMT, Impact Testing

Hydrostatic TestingHydro test Requirements in UG-99The test pressure formula:

Pt = Test PressureSt = Allowable stress at temperature of hydrotestSd = Allowable stress at design temperatureNote: St/Sd always > 1

Pt = 1.3 x MAWP x (St/Sd)What is (???)x MAWP x (St/Sd)Corrected for Temperature As the temperature increases, materials get weaker. Since vessel designed for hot temperatures are hydro tested at ambient conditions (where materials are stronger) the test pressure needs to be compensated (increased)

Hydrotest ProcedureComplete all pre-hydrotest work & testingAssure vessel & support structure is designed for weight of hydrotest liquidSelect hydrotest fluid any non-hazardous liquid below its boiling pointDisconnect or blind off appurtenances not to be testedVent of high points to remove possible air-pocketsTest fluid should be 300F above MDMTPressure gauges must be acceptable range. About 2 times test pressure. (Acceptable range is 1.5 to 4 times Test Pressure)Pressure gauge should be calibratedPressure gauge should be connected directly to vessel. If not visible, another should be connected near operator.Check tightness of test equipmentPerform pressure test at 1.3 times MAWP corrected for temperature.Back pressure down to Test Pressure divided by 1.3After vessel temperature is below 1200F, perform close visual inspection of joints and connections.

Hydrotest CalculationA vessel is constructed of a P-1 material. The vessel is stamped MAWP is 300 psi at 8000F. Material allowable stress S is:1000F = 20,000 psi8000F = 13,500 psi

Determine the a) hydrostatic test pressure b) minimum inspection test pressure

Hydrotest Calculationa) Calculate Test Pressure MAWP = 300 psi, St = 20,000 psi, Sd = 13,500 psiFormula: Pt = 1.3 (MAWP) x (St/Sd) = 1.3 (300) x (20,000)/(13,500) = 390 x 1.481 = 577.7 psiHydrotest pressure is 577.7 psi at the top of vessel

b) Calculate the min Inspection Test PressurePinsp = Pt/1.3 = 577.7/1.3 = 444.4 psi

Exercise 3-1Hydrotest SolutionA vessel is constructed of a P-1 material. The vessel MAWP is 600 psi at 6500F. Allowable stress S is @ 1000F = 17,000 psi, @ 6500F=17,000 psi. Determine the Hydrostatic Test Pressure and the Minimum Inspection Test Pressure.

Exercise 3-1Hydrotest SolutionA vessel is constructed of a P-1 material. The vessel MAWP is 600 psi at 6500F. Allowable stress S is @ 1000F = 17,000 psi, @ 6500F=17,000 psi. Determine the Hydrostatic Test Pressure and the Minimum Inspection Test Pressure.

Calculate Test PressureMAWP=600 psi, St=17,000 psiSd=17,000 psiPt= 1.3 (600)(17,000/17,000) = 780 psi

Minimum Inspection Test PressurePinsp = Pt/1.3= 780/1.3= 600 psi

Pneumatic TestingPneumatic Test Requirements UG 100Safety Issues Compressed AirThe Test Pressure formula

Pressure increased in steps0.5Pt 1st step0.6Pt 2nd step0.7Pt 3rd step0.8Pt 4th step0.9Pt 5th step1.0Pt at test pressurePinsp = Pt/1.1Pt = 1.1 x MAWP x (St/Sd)

Exercise 3-2Hydrotest SolutionA vessel is constructed of a P-1 material. The vessel MAWP is 100 psi at 7500F. Material allowable stress S is:1000F = 18,000 psi7500F = 17,000 psi

Determine the:Pneumatic Test PressureEach of the Test Pressure StepsInspection Test Pressure

Exercise 3-2Hydrotest SolutionPneumatic Test PressurePt = 1.1(MAWP) x (St/Sd) = 1.1x100(18,000/17,000) = 116.5 psib)Each of the Test Pressure Steps0.5Pt 1st step = 0.5(116.5) = 58.2 psi0.6Pt 2nd step = 0.6(116.5) = 69.9 psi0.7Pt 3rd step = 0.7(116.5) = 81.5 psi0.8Pt 4th step = 0.8(116.5) = 93.2 psi0.9Pt 5th step = 0.9(116.5) = 104.8 psi1.0Pt at test pressure = 1.0(116.5) = 116.5

Inspection Test PressurePinsp = Pt/1.1 = 105.9 psi

Minimum Design Metal Temperature (MDMT)

Why MDMT?- The code is very concerned about the lower operating temperature. It is so important that it is one of the few pieces of information that is required on the nameplate. UG-116(a)(4)

- The reason for all the concern? Generally as the temperature of a material is lowered the material become brittle

- A Brittle Fracture can be instantaneous and thus Catastrophic. This must be avoided!

Factors Affecting BrittlenessMaterialTemperatureThicknessStress LoadingResidual Stress

The material property that is opposite brittleness is called Toughness.Toughness describes the ability of a material to absorb an impact and is measured in ft-lb. One tests to determine a materials toughness is called Charpy impact test.

The Code Controls ToughnessThe Code control brittle fracture by:1. Controls Material Selection only certain material can be used in a pressure vessel.2. Provides method to calculate a vessels allowable MDMT.3. Specifies impact testing for materials that operate below the temperature limits determined in item 1 & 2.

Our goal: Determine the lowest allowable MDMT without impact testing.

MDMT 4 EASY STEPS1. Table UCS-66 or Fig. UCS-66 Initial MDMT (Factor thickness, temperature, material type)Determine Material curveDetermine Initial MDMT from table base on nominal thickness and material type A,B,C,D

2. Fig. UCS-66.1 MDMT reduction (factor-maximum operation stress compare to max. material stress loading.Ratio will be given determine temperature reductionSubtract temperature reduction from initial MDMT

3. UCS-68 ( c ) - Further reductionHas vessel has been PWHT when not required by UCS-56If yes subtract an addition 30 F from MDMT

4. Check Limits- UCS-66(b)(2) (cannot exceed these limits)

Example 3-2Calculate MDMTExample :A horizontal vessel is made from SA 516 gr 70 plates that are not normalized. The vessel is rated at 250 psig at 700 F. The wall thickness is 0.500 and has a corrosion allowance of 0.100. The nameplate is stamped RT-3 and HT.

Find : The lowest possible MDMT for this vessel. Reduction ratio is 0.90.

Example 3-2Calculate MDMTStep 1: Initial MDMT: Table UCS-66Material : Curve BInitial MDMT: -70FStep 2: MDMT Reduction: Figure UCS 66-1Reduction ration: 0.90Reduction: 100FNew MDMT = Initial Reduction = -70F - 100F = -170F Step 3: PWHT Reduction UCS 68(c)PWHT: YesRequired by Code: NoAdditional Reduction: 300F Final MDMT: Step 2 PWHT reduction = -170F - 300F = -470F Step 4: Check limits: UCS 66(b)(2)No restriction as UCS-68 (c) allows for temperatures below these limits

Example 3-2Calculate MDMTStep 1: Initial MDMT: Table UCS-66Material : Curve BInitial MDMT: -70F

Example 3-2Calculate MDMTStep 2: MDMT Reduction: Figure UCS 66-1Reduction ration: 0.90Reduction: 100FNew MDMT = Initial Reduction = -70F - 100F = -170F

Example 3-2Calculate MDMTStep 3: PWHT Reduction UCS 68(c)PWHT: YesRequired by Code: NoAdditional Reduction: 300F Final MDMT: Step 2 PWHT reduction = -170F - 300F = -470F

Example 3-2Calculate MDMTStep 4: Check limits: UCS 66(b)(2)No restriction as UCS-68 (c) allows for temperatures below these limits

Exercise 3-3Determine MDMTMaterial is SA-516 Gr. 60. Nominal thickness is 2.0. Renewal thickness is 1.750. Nameplate stamped HTMaterial normalized SA-612. Thickness is 0.750. Reduction ration is 0.85. Vessel was not PWHT.Material SA-516 Gr. 70, material retirement thickness 0.875. New thickness 1.0. Vessel is PWHT for environmental cracking. The reduction ratio is 0.88.

Exercise 3-3Determine MDMT1) Material is SA-516 Gr. 60. Nominal thickness is 2.0. Renewal thickness is 1.750. Nameplate stamped HT.

Material Curve CInitial MDMT 260F [SA-516 Gr. 60 as it is not mentioned as normalized]Ratio Reduction00F [no reduction ratio given assume no reduction]PWHT Reduction00F [no reduction as 2 plate required PWHT by Codes]Final MDMT260F

Exercise 3-3Determine MDMT2) Material normalized SA-612. Thickness is 0.750. Reduction ration is 0.85. vessel was not PWHT.

Material Curve D for SA-612 normalized.Initial MDMT-420F [Either figure UCS-66 or Table UCS-66 for tabular values]Ratio reduction-150F [from figure UCS-66.1]PWHT reduction-00F [as no PWHT carried out]Final MDMT-570F

UCS-66(b)(2) limits the MDMT to -550F otherwise impact testing is required.

Exercise 3-3Determine MDMTMaterial SA-516 Gr. 70, material retirement thickness 0.875. New thickness 1.0. Vessel is PWHT for environmental cracking. The reduction ratio is 0.88.

Curve BInitial MDMT31 0F [Either figure UCS-66 or Table UCS-66 for tabular values]Ratio reduction-12 0F [from figure UCS-66.1]PWHT reduction-30 0F [P-1 material

Impact Testing UG-84Impact testing of material is required when minimum operating temperature is lower than allowed by the UCS-66 MDMT calculations.

Test procedure SA-370Each set of specimens 3 specimensAcceptance criteria Figure UG-84.1 * Average value from the chart * Minimum value 2/3 chart

Note: 1 ksi = 1,000 psi

Exercise 3-4Impact Testing1) Impact testing is performed on a 3 thick plate that has yield strength of 55,000 psi. To be acceptable, the average for the set must be at or above ________ ft-lbs2) Impact testing is performed on a 1 thick plate that has yield strength of 45,000 psi.a) To be acceptable, the average for the set must be > ____ft-lbs b) To be acceptable each specimen must be > _____ft-lbs

Exercise 3-4Impact TestingImpact testing is performed on a 3 thick plate that has yield strength of 55,000 psi. To be acceptable, the average for the set must be at or above 30 ft-lbs

2) Impact testing is performed on a 1 thick plate that has yield strength of 45,000 psi.a) To be acceptable, the average for the set must be > 15 ft-lbs b) To be acceptable each specimen must be > 10 ft-lbs

More Exercise Impact TestingA welding procedure requires impact testing for a thickness range 3/16 2. The specimen is 1 having 45 ksi yield strength. What is the minimum acceptable impact test values for the three specimens?

18-19-1217-12-2517-16-1718-17-12

More Exercise Impact TestingA welding procedure requires impact testing for a thickness range 3/16 2. The specimen is 1 having 45 ksi yield strength. What is the minimum acceptable impact test values for the three specimens?

From figure UG-84.1 find value required for average of 3 specimen using 2 the thickest range.Average = 17 ft-lbsMin value = 2/3 (17) = 11.3

18-19-12 [Average 16.3, Min value 12]17-12-25 [Average 18, Min value 12]17-16-17 [Average 16.67, Min value 16]18-17-12 [Average 15.67, Min value 12]

External PressureThickness of shells and tubes under external pressure (UG-28)Shells or tubes under external pressure are required to resist collapse by buckling. Methods for calculating minimum thickness are primarily based on factors influencing stiffness rather than material strengthCodes provides a series of charts in section II Part-D to eliminates tedious calculation.Shells of pressure vessel that fails the external pressure design may be stiffened using stiffening rings.

External PressureSteps for calculations

Step 1 : Calculate L/Do & Do/t

Step 2 & 3 : Determine Factor A (from Fig. G graph)

Step 4/5 : Determine Factor B (from Matl chart Fig. CS-2)

Step 6 : Calculate P Max All External Pressure

P =4B/3(Do/t)

D0 = Outside diameterL = Length between supports (inches)Factors A & B numbers from graph

External PressureSample of calculationsA tube has an outside diameter of 6.625. The distance between supports is 20. The wall thickness is 0.120. Tube material is SA 516 Gr. 70. The tube is rated for 125 psi at 700 0F. Determine the maximum allowed external pressure.

External PressureSample of calculationsA tube has an outside diameter of 6.625. The distance between supports is 20. The wall thickness is 0.120. Tube material is SA 516 Gr. 70. The tube is rated for 125 psi at 700 0F. Determine the maximum allowed external pressure.Step 1: Calculate L/D0 & D0/t L/D0 = (12x20)/6.625 = 36.23 D0/t = 6.625/0.120 = 55.2Step 2&3: Determine Factor A from figure GFind D0/t curve 55.2Find intersection with the L/D0 line of 36.23At intersection drop line straight down to bottom of graph & read factor A 0.000375

External PressureSample of calculationsA tube has an outside diameter of 6.625. The distance between supports is 20. The wall thickness is 0.120. Tube material is SA 516 Gr. 70. The tube is rated for 125 psi at 700 0F. Determine the maximum allowed external pressure.

Steps 4&5: Determine factor B from Fig CS-2Find temperature curve (700 0F)Find intersection with Factor A line 0.000375At intersection move horizontally to side of graph and read factor B = 4500

Step 6: Calculate P Max All External PressureP =4B/3(Do/t) = 4(4500)/3(55.2) = 108.7 psi.

Exercise 3-5External PressureA tube has length of 30 and outside diameter of 10. The nominal thickness is 0.375 and the renewal thickness is 0.20. The design temperature is 5000F. Use material chart Fig CS-2.Determine the maximum allowed external pressure.

Exercise 3-5External PressureA tube has length of 30 and outside diameter of 10. The nominal thickness is 0.375 and the renewal thickness is 0.20. The design temperature is 5000F. Use material chart Fig CS-2.Determine the maximum allowed external pressure.Step 1: Calculate L/D0 & D0/t L/D0 = 30/10 = 3 D0/t = 10/0.20 = 50Step 2 & 3: Determine Factor A from figure GFind D0/t curve 50Find intersection with the L/D0 line of 3At intersection drop line straight down to bottom of graph & read factor A = 0.0012

Exercise 3-5External PressureA tube has length of 30 and outside diameter of 10. The nominal thickness is 0.375 and the renewal thickness is 0.20. The design temperature is 5000F. Use material chart Fig CS-2.Determine the maximum allowed external pressure.

Steps 4&5: Determine factor B from Fig CS-2Find temperature curve (500 0F)Find intersection with Factor A line = 0.0012At intersection move horizontally to side of graph and read factor B = 10,500

Step 6: Calculate P Max All External PressureP =4B/3(Do/t) = 4(10,500)/3(50) = 280 psi.

Question 12Sample of API QuestionThe inner wall of a jacketed vessel is 0.635 wall, the cylinder is 45 outside diameter, the unsupported length is 120 and is made of SA-516 Gr.70 material. Factor A is 0.0008 and Factor B is 11,600. What is the maximum pressure permitted on the inner wall of the jacket with temperature rating of 3000F.

Question 12Sample of API QuestionThe inner wall of a jacketed vessel is 0.635 wall, the cylinder is 45 outside diameter, the unsupported length is 120 and is made of SA-516 Gr.70 material. Factor A is 0.0008 and Factor B is 11,600. What is the maximum pressure permitted on the inner wall of the jacket with temperature rating of 3000F.

t= 0.635B=11,600D0=45A=0.0008

P =4B/3(Do/t) = 4(11,600)/3(45/0.635) = 218 psi

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To

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