Pressure Vessel Inspector Certification - By Puspatri

PREPARATION CLASSAPI 510

PRESSURE VESSEL INSPECTOR

CERTIFICATION EXAMINATION

PUSPATRIJOHOR BAHRU

5TH -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’ career- Additional job opportunities & salary increase- Widen employment doors – resume with API certificates- Oil & Gas industry is booming- Middle East offers USD 500 - 700 per day

“ Life is too short to be ordinary”

What Will Be Asked?API Publications

1. API 510, Pressure Vessel Inspection Code

2. API RP 571, Damage Mechanisms Affecting Equipment in Refining Industry

3. 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

Don’t Worry About The ExamThe API Examination- 150 multiple choices with four possible answers- Exam divided into 2 Parts

(a) Open Book – 50 questions for 4 hours durations(b) Closed 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 Exam

1. API Examination Confirmation Letter2. Identification Card3. API & 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.

4. Non-programmable calculator. Make sure enough battery.5. 2B pencil, eraser and other stationery6. Jacket – some classroom is uncomfortably cool7. Earplugs – 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 Code

Sections

I – Rules for Construction of Power Boilers

II – Materials-Ferrous, Nonferrous, Welding Rods, Electrodes

III – Nuclear Power Plant

IV – Rules of Construction of Heating Boilers

V – Nondestructive Examination

VI – Rules for Operation of Heating Boilers

VII- Guidelines for Operations of Power Boilers

VIII- Rules for Construction of Pressure Vessel

IX-Welding & Brazing Qualifications

X-Fiber Reinforced Plastic Pressure Vessel

XI-Rules for In-Service Inspection of Nuclear Power Plant

XII-Rules for Construction and Continued Service of Transport Tank

Settings Rules!Is What The Code Is All About

Important factors during fabrication that affect vessel safety and

reliability seem almost endless.

(a) Design – thickness formulas, welding processes(b) Material – Known physical & chemical properties(c) Fabrication – qualified welding procedure, NDT(d) Pressure Testing – hydro or pneumatic(e) Documentation – nameplate, design calculation data

Scope of the Code

U-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:”(a) Within cope of other division(b) Fired process tubular heaters(c) PV of integral parts/component of pumps, turbines,

compressor(d) Pipe and piping component(e) PV for water under pressure in which Pd<300 psi or Td<2100F(f) Tank hot water in which heat input<200,000 BTU, T<2100F,

capacity <120 gallon(g) PV having internal/external pressure <15 psi; no size limit(h) PV having inside diameter <6 inches(i) PV for human occupancy

Scope of the Code continued…

Vessel 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 vessel

U-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:

(1) How information is organized in the code(2) How to use the organization tools of the code

If you know these 2 secret, every answer can be found with minimum

effort. That’s much better than paging through 700 pages of the

codes!

Organization of CodeSection VIII- Rules for Construction of Pressure Vessel consists

of 2 division;(a)Division 1 – Routine Vessel(b)Division 2 – Alternate Rules for special vessel

Section VIII Division 1 Introduction 3 Subsections

Subsection A – General Subsection B – Fabrication Method Subsection C - Material

2 Appendix Mandatory Non-Mandatory

Organization of Code continued…

Each “Subsection” is further divided into “Parts” Subsection A – General

Part UG – applies to all vessels Subsection B – Fabrication

Part UW – applies to all vessels that are welded Subsection C – Materials

Part UCS – applies to all vessel made of CS or LA steel Part UHT – applies to ferritic vessel that use Heat Treatment

Every “Part” is divided into “ Paragraphs”UG-1, UW-1, UCS-1

Organization of Code continued…

Mandatory AppendicesU-1(b) “address specific subject not covered elsewhere in this

division”Alternate formulaeQC ManualNDE Standards

Non-mandatory AppendicesU-1(b) “provides information and suggested good practices”

Completed 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 Code

Section IX Welding

Section V NDT

Section VIII Pressure Vessel

Section II Materials

Section 1 Boilers

Division 1 Routine

Division 2 Other

Sub A General

Sub B Fabrication

Sub C Materials

Part UF Forge

Part UW Welding

Part UB Brazing

Design

Materials

Fabrication

Inspect

Marking

Exercise 1-2A vessel is welded with Carbon Steel components. Which part in the Code would you find:

(a) Design Requirements : ____________________(b) Hydrotest Pressure : ____________________(c) RT Acceptance Standards : ____________________(d) Nameplate Data : ____________________(e) Limits of Carbon % in Materials : ____________________(f) Material ID Traceability : ____________________(g) Inspection Requirements : ____________________(h) PWHT Requirements : ____________________

Exercise 1-2 answer….

A vessel is welded with Carbon Steel components. Which part in the Code would you find:

(a) Design Requirements : UG, UW, UCS(b) Hydrotest Pressure : UG(c) RT Acceptance Standards : UW(d) Nameplate Data : UG(e) Limits of Carbon % in Materials : UCS(f) Material ID Traceability : UG(g) Inspection Requirements : UG, UW, UCS(h) PWHT Requirements : UW, UCS

Hint: If it’s applicable to all vessel –UG.If it’s only applicable because it’s welded – UWIf it’s based on metallurgy - UCS

Code’s PurposeEstablish Rules for Construction of Pressure Vessel

To 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:(1) User(2) Manufacturer(3) Authorized Inspector(4) Welder(5) NDE Technician

Setting The RulesUser Roles & Responsibilities

U-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 & Responsibilities

U-2(b)(1) “ The manufacturer of any vessel…has 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 ASME• Perform design calculations & develop fabrications drawings• Identify all material used during fabrication• Examine materials before fabrication: thickness, ID, defects• Qualify welding procedures & welders• Perform NDE test and records result• Perform vessel hydro or pneumatic test• Apply the Code Stamp• Prepare Manufacturer’s Data Report

Setting The RulesAuthorized Inspector Roles & Responsibilities

Section 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:”1. Perform required duties2. Authorize the application of the Code Symbol3. Sign the Certificate of Shop Inspection

Setting The Rules continued…

Authorized Inspector Roles & Responsibilities

UG-90(c)(1) “ The inspector shall”• Verify the Manufacturer has a current Certificate of

Authorization• Verify Manufacturer is working to the QC system• Verify design calculations are available• Verify materials meet Codes• Verify weld procedures and welders are qualified• Verify NDE tests have been performed & are acceptable• Perform internal & external inspections• Verify nameplate is attached and has the right markings• Witness the hydrotest• Sign the Manufacturer’s Data Report

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

NDE TechnicianUW-51(a)(2) RT “Qualified and certified in accordance with

employer’s written practice. SNT-TC-1A used as a guidelines”App 12-2 UT “Qualified and certified in accordance with employer’s written practice. SNT-TC-1A used as a guidelines”App 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 method”App 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 Stamp

When 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 Authorization

What 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 symbols…will 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 System

UG-117(e) “ Any Manufacturer shall have and demonstrate a Quality

Control System to establish all Code requirement…will 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 Manufacturer’s Quality Control System tells ASME and the

AI “how things will be done” in the shop to meet the code.

CODE-OLOGYData Reports

UG-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 Vessel

If a vessel Is not covered by U-1 (c), (g), (h) & (i) Is not required to be fully radiographed Does not have a quick actuating device Does not exceed either the following limit

5 ft3 and 250 psi 1.5 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 1-3

1) Full radiography is performed on a vessel shell with a wall thickness

of ½”. 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 ½”. 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 Time

Subsection 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 ft

1 cubic foot of water weighs 62.4 lbs.

12”

12”

12”

1) How much force (weight) is on the bottom of this container?______________

2) How much force on each square inch of the box’s bottom?________ What’s 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 Pressure

1) A deep diving submarine cruising at a depth of 854 feet. What is the static head pressure on this submarine? (external pressure)

2) 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?

3) A 60’ vessel is filled with water. The pressure at the bottom is 210 psi. What is the pressure at the top?

50’

0’

?? psig

0 psig

2a

50’

0’100 psig

?? psig

2b

60’

0’

210 psig

?? psig

3

Exercise 2-1Answer 11) A 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 psig

85

4 f

t

Exercise 2-1Answer 2a2) A vessel is 50’ high. The vessel will be hydrotested

a. When filled with water what is the pressure at the bottom of the vessel?

50’

0’

?? psig

0 psig

2a

Pbtm = 0.433 psi/ft x 50 ft

= 21.7 psig

Exercise 2-1Answer 2b2) A vessel is 50’ high. The vessel will be hydrotested

b. 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

50’

0’100 psig

?? psig

2b

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

60’

0’

210 psig

?? psig

3

Module 2.2Design Pressure

Design Pressure

The pressure used in the design of a vessel component together with

the coincident design metal temperature for the purposes of determining the minimum permissible thickness…static head

shall be added to the design pressure …” App 3-2

Exercise 2-2Design Pressure

A 50 high vessel has a design pressure of 100 psig. The elevations are

shown in the sketch below.

a. The shell should be designed for _________ psigb. The top head should be designed for _________ psigc. The bottom head should be designed for _________ psig

50’

0’

?? psig

100 psig

2’

48’

?? psig

?? psig

Exercise 2-2Design Pressurea. Pshell = Ptop +Psh

= 100 psig + (0.433 psi/ft x 48 ft) = 120.8 psigb. Ptop head = Ptop + Psh

= 100 psig + (0.433 psi/ft x 2 ft)= 100.9 psig

c. Pbtm head = Ptop + Psh= 100 psig + (0.433 psi/ft x 50 ft)= 121.7 psig 50’

0’

?? psig

100 psig

2’

48’

?? psig

?? psig

Note: 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 Division…excluding 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 vessel…and adjusted for any difference in static head…”

How much can he lift?

MAWPVessel & Vessel Part

To 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 link”

12’

34’

A

BPart B: MAWP 336

Part A: MAWP 343

Part A: 343 – 0.433(34) = 328.3 psig

Part B: 336 -0.433(12) = 330.8 psig

Thus, vessel MAWP = 328.3 psig

0’

Exercise 2-3Determining Vessel MAWP

The maximum MAWP this vessel can be rated is ______psig.

0 ft

1 ft

3 ft

24 ft

40 ft

42 ft

Part Part MAWP Static Head Pressure @ TopTop Nozzle 342 psigSide Nozzle 426 psigTop Head 329 psig

Bottom Head 336 psigShell 337 psig

Exercise 2-3Determining Vessel MAWP

The maximum MAWP this vessel can be rated is ______psig.

0 ft

1 ft

3 ft

24 ft

40 ft

42 ft

Part Part MAWP Static Head Pressure @ TopTop Nozzle 342 psig 0.433 341.6Side Nozzle 426 psig 10.4 415.6Top Head 329 psig 1.3 327.7

Bottom Head 336 psig 18.2 317.8Shell 337 psig 17.3 319.7

Design Pressure vs. MAWP

Design Pressure – pressure from the system (process + static head)

MAWP – pressure part of the vessel is “good for”

MAWP

Design Pressure

To meet Code Requirement:

MAWP > Design Pressure

If MAWP < Design Pressure

Exercise 2-3More MAWP & Static Head calculation

1) The 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.

2) A vessel’s 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?

3) 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 ft

1 ft

50 ft

100 ft

Exercise 2-3More MAWP & Static Head calculation

1) The 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 psig

100 psig0 ft

2 ft

34 ft

Exercise 2-3More MAWP & Static Head calculation

2) A vessel’s 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

0 ft

1 ft

50 ft

100 ft

Exercise 2-3More MAWP & Static Head calculation

3) 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.

Ptop = Pgauge – Psh= 136 – (0.433 x 60)= 110.0 psig

0 ft

80 ft

60 ft 136 psig

Module 2.4Calculating Stress

Stress on Welds

Circumferential Weld

Longitudinal Weld

Circumferential Stress affects: ___________ welds

Longitudinal Stress affects: __________ welds

Stress on Welds

Circumferential Weld

Longitudinal Weld

Circumferential Stress affects: Longitudinal welds

Longitudinal Stress affects: Circumferential welds

Calculating Stress

A 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 Stress

A 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 Stress

1) A tension specimen is 0.5” thick and 0.75” wide. It breaks with a load of 21,500 lbs. The failure stress is _______ psi.

2) 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 ½” thick, not including corrosion allowance.What is the actual longitudinal stress on the shell-to-head weld?

Exercise 2.5Calculating Stress

1) A 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 psi

2) 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 ½” 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 2000 4.0 to 1.0 Safety Factor

Post 2000 3.5 to 1.0 Safety Factor

Allowable Stress = Ultimate Tensile Strength / Safety Factor

Exercise 2-6Allowable Stress

A material has an ultimate tensile strength of 70,000 psi at ambient

temperaturea) What is the allowable stress for this material at ambient

conditions if used in a pressure vessel today? b) What is the allowable stress for this material at ambient

conditions if used in a 1977 pressure vessel ?

Exercise 2-6Allowable Stress

A material has an ultimate tensile strength of 70,000 psi at ambient temperaturea) What 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 psi

b) 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 don’t try

-Beverly Sills

Module 3.1Joint Efficiencies

“E” – The Basic

What is Joint Efficiency “E”?What factors that affect “E”?How does Joint Efficiency affect “E”?How is Joint Efficiency determined?

“E” – The Basic

What is Joint Efficiency “E”?- A safety factor for welds- Compensation for possible weld defects

What factors that affect “E”?- Type of Joint, Location of Joint, Amount of RT

How does Joint Efficiency affect “t”?- As ‘E’ decreases, required thickness increases

How is Joint Efficiency determined?- The Code, Section VIII – Table UW-12- (also few exception listed in UW-12)

Exercise 2.7Joint Efficiencies

1) A shell is made with Type 2 joints and spot RT was performed. What is “E”? _________.

2) A shell is made with Type 1 joints and Full RT was performed. What is “E”? _________.

3) A shell is made with Type 3 joints and no RT was performed. What is “E”? _________.

Exercise 2.7Joint Efficiencies

1) A shell is made with Type 2 joints and spot RT was performed. What is “E”? 0.8 .

2) A shell is made with Type 1 joints and Full RT was performed. What is “E”? 1.0.

3) 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 Radiography

Type of Radiography

Full – Generally 100% of welds, some exception

Spot – On RT for each 50 ft of weld

None – Send the RT techs home

Amount specified by Code for some vessels

Based on Service, Thickness or Welding Process (UW-11)

Amount specified by Users/Designer for others

“The 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 Code

Full RT is required by the Code, when:

1. Vessels in Lethal Service – 100% RT Butt Welds

2. Butt Welds > 1-1/2”…- 100% RT

3. Unfired Boilers > 50 psig – 100% RT Butt Welds

4. Butt Welds of nozzles for 1&3 – 100% RT

5. Butt Welds made using Electrogas & Electroslag Process … -100% RT

UW-11(a)

Full RT – Required by User

Full 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 Factors

Describe the amount of RT performed

- RT 1&2: Full Radiography

- RT-3: Spot Radiography

- RT-4: Combo Radiography

The RT Factor is located on Nameplate

They are described in UG-116(e)

Seamless Parts – “E”

For seamless vessel sections or head, where circumferential stresses

govern:

E = 1.0

When the category B and C butt welds are spot RT

And the welds connecting seamless vessel sections or heads are either Type 1 or 2.

E = 0.85

When the butt welds are either not spot RT

Or when the welds connecting seamless vessel sections or heads are

type 3,4,5 or 6.

UW-12(d)

Example – Finding “E”

1) A pressure vessel has butt welds which are single welded. No RT has been performed.

1) “E” for welded shell? _______2) “E” for the seamless head? _______

2) A pressure vessel has butt welds which are double welded. The vessel is stamped as RT-3

1) “E” for welded shell? _______2) “E” for the seamless head? ________

3) A pressure vessel has butt welds which are double welded. The vessel is stamped as RT-2.

1) “E” for welded shell? _______2) “E for the seamless head? _______

4) A pressure vessel has butt welds which are double welded. The vessel is stamped as RT-1

1) “E” for welded shell? _______2) “E” for the seamless head? _______

Example – Finding “E”

1) A pressure vessel has butt welds which are single welded. No RT has been performed.

1) “E” for welded shell? 0.62) “E” for the seamless head? 0.85

2) A pressure vessel has butt welds which are double welded. The vessel is stamped as RT-3

1) “E” for welded shell? 0.852) “E” for the seamless head? 1.00

3) A pressure vessel has butt welds which are double welded. The vessel is stamped as RT-2.

1) “E” for welded shell? 1.002) “E for the seamless head? 1.00

4) A pressure vessel has butt welds which are double welded. The vessel is stamped as RT-1

1) “E” for welded shell? 1.002) “E” for the seamless head? 1.00

Exercise 2-8More Joint Efficiencies

1) A pressure vessel has lap welds which are single welded. The vessel is stamped RT-3

1) “E” for welded shell? _____

2) “E” for the seamless head? ______

2) A pressure vessel has lap welds which are double fillet welded1) “E” for welded shell? ______

2) “E” for the seamless head? ______

3) A pressure vessel has butt welds which are single welded with backing strips.The vessel is stamped as RT-2

1) “E” for welded shell? ______

2) “E” for the welded head? _______

Exercise 2-8More Joint Efficiencies

1) A pressure vessel has lap welds which are single welded. The vessel is stamped RT-3

1) “E” for welded shell? 0.45

2) “E” for the seamless head? 0.85

2) A pressure vessel has lap welds which are double fillet welded1) “E” for welded shell? 0.55

2) “E” for the seamless head? 0.85

3) A pressure vessel has butt welds which are single welded with backing strips.The vessel is stamped as RT-2

1) “E” for welded shell? 0.99

2) “E” for the welded head? 0.99

Module 3.2Tmin Calculations

Successful Calculations• The 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 Calculations

• The formula UG-27(c)(1) – Internal Pressure

– “R” is the inside radius – ½ of diameter– “P” is the Design Pressure: Pressure on the part includes Static

Head

• The “P” formula calc’s the shell MAWP

– The part MAWP – The shell’s “good for” pressure– This is not vessel MAWP

t = PR/SE-0.6P

P = SEt/R + 0.6t

ExampleA vessel has an internal radius of 36”. At the design temperature the material’s 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___________________________________________________________2) The formula: t = PR/SE-0.6P___________________________________________________________3) The plug-in: t = 158 x 36 / (15,000 x 1) – (0.6 x 158)___________________________________________________________4) The solutions with Units t = 0.382”___________________________________________________________

Exercise 2-9Shell Minimum Required Thickness

A vessel shell has an internal radius of 24”. At the design temperatures the material’s 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=___________________________________________________________2) The formula: t = PR/SE-0.6P___________________________________________________________3) The plug-in: t = ___________________________________________________________4) The solutions with Units t = ___________________________________________________________

Exercise 2-9Shell Minimum Required Thickness

A vessel shell has an internal radius of 24”. At the design temperatures the material’s 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___________________________________________________________2) The formula: t = PR/SE-0.6P___________________________________________________________3) The plug-in: t = (250)(24)/(20,000 x 1.0) – (0.6 x 250)___________________________________________________________4) The solutions with Units t = 0.302”___________________________________________________________

Exercise 2-10Shell Minimum Required Thickness

1) A 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 material’s 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 Thickness1) A 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.___________________________________________________________1) The “Givens”: P= ? , R= 48” , S= 12,500 psi, E= 0.85 (Table

UW-12)Pshell = Ptop + Psh = 120 + (0.433x52) = 142.5 psi

___________________________________________________________2) The formula: t = PR/SE-0.6P___________________________________________________________3) The plug-in: t = (142.5)(48)/(12,500 x 0.85) – (0.6 x 142.5)___________________________________________________________4) The solutions with Unitst = 0.649”___________________________________________________________

Exercise 2-10Shell Minimum Required Thickness

2) A horizontal vessel has an internal diameter of 10 ft. The material’s 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

___________________________________________________________2) The formula: t = PR/SE-0.6P___________________________________________________________3) The plug-in: t = (124.33)(60)/(14,000 x 1.00) – (0.6 x 124.33)___________________________________________________________4) The solutions with Unitst = ___________________________________________________________

Rounded Head tmin Calculations

• Ellipsoidal Head UG-32 (d)

• Torispherical Head UG-32 (e)

• Hemispherical Head UG-32 (f)

t = PD/2SE-0.2P

P = 2SEt/D + 0.2t

t = 0.885PL/SE-0.1P

P = SEt/0.885L + 0.1t

t = PL/2SE-0.2P

P = 2SEt/L + 0.2t L = inside radius

L = 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 head’s minimum required thickness?___________________________________________________________• The “Givens”: P= , R= , S= , E=

___________________________________________________________2) The formula: t = PD/2SE-0.2P___________________________________________________________3) The plug-in: t = ___________________________________________________________4) 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 head’s minimum required thickness?___________________________________________________________• The “Givens”: P= 200, R=60” , S= 15,000 psi, E= 1.00 (Table UW-

12(d))___________________________________________________________

2) The formula: t = PD/2SE-0.2P___________________________________________________________3) The plug-in: t = (200)(60)/2(15,000 x 1.00) – (0.2 x 200)___________________________________________________________4) The solutions with Unitst = 0.401”___________________________________________________________

Exercise 2-11Formed Head Minimum Thickness

1) A 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.

2) A horizontal vessel with seamless torispherical heads has an outside diameter of 96 inches. The material’s 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 ___________________________________________________________2) The formula: t = PL/2SE-0.2P

___________________________________________________________3) The plug-in: t = (64.5)(42)/2(8,700 x 0.85) – (0.2 x 64.5)

___________________________________________________________4) 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 material’s 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 ___________________________________________________________2) The formula: t = 0.885PL/SE-0.1P

___________________________________________________________3) The plug-in: t = 0.885(123.46)(96)/(20,000 x 1.00) – (0.1 x 123.46)

___________________________________________________________4) The solutions with Units t = 0.52”

___________________________________________________________

Flat Head tmin Calculations

The 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 “E”E – 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 weld

t = d √CP/SE

Exercise 2-12Flat Head tmin

A 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 tmin

A 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

___________________________________________________________2) The formula: t = d √CP/SE

___________________________________________________________3) The plug-in: t = 60 √C(0.33)(302.165)/(12,500)(1.0)

___________________________________________________________4) The solutions with Unitst = 5.359”

___________________________________________________________

Part MAWPPart MAWP is the pressure a part is “good for”Based on knowing the thickness (don’t 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-0.2P

P = 2SEt/D + 0.2t

Exercise 2-13Let’s 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 + 0.2t

P = SEt/0.885L + 0.1t

P = 2SEt/L + 0.2t

P = SEt/R + 0.6t

Exercise 2-13Let’s 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 Head

P = 2SEt/D + 0.2t

P = SEt/0.885L + 0.1t

P = 2SEt/L + 0.2t

P = 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)

1) Complete all pre-hydrotest work & testing2) Assure vessel & support structure is designed for weight of hydrotest

liquid3) Select hydrotest fluid – any non-hazardous liquid below its boiling point4) Disconnect or “ blind off” appurtenances not to be tested5) Vent of high points to remove possible air-pockets6) Test fluid should be 300F above MDMT7) Pressure gauges must be acceptable range. About 2 times test pressure.

(Acceptable range is 1.5 to 4 times Test Pressure)8) Pressure gauge should be calibrated9) Pressure gauge should be connected directly to vessel. If not visible,

another should be connected near operator.10) Check tightness of test equipment11) Perform pressure test at 1.3 times MAWP corrected for temperature.12) Back pressure down to Test Pressure divided by 1.313) After vessel temperature is below 1200F, perform close visual inspection

of joints and connections.

Hydrotest Procedure

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 psi

Hydrotest 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 psi Sd=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/1.1

Pt = 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 Pressure• Each of the Test Pressure Steps• Inspection Test Pressure

Exercise 3-2Hydrotest Solutiona) Pneumatic 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

c) 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 Brittleness

(a) Material(b) Temperature(c) Thickness(d) Stress Loading(e) Residual 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 material’s toughness is called Charpy impact test.

The Code Controls Toughness

The 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 STEPS

1. 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 MDMT

Example :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 MDMT

Step 1: Initial MDMT: Table UCS-66Material : Curve BInitial MDMT: -70F

Step 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 MDMT

Step 1: Initial MDMT: Table UCS-66

Material : Curve BInitial MDMT: -70F

Example 3-2Calculate MDMT

Step 2: MDMT Reduction: Figure UCS 66-1

Reduction ration: 0.90Reduction: 100FNew MDMT = Initial –

Reduction = -70F - 100F = -170F

Example 3-2Calculate MDMT

Step 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 MDMT

Step 4: Check limits: UCS 66(b)(2)

No restriction as UCS-68 (c) allows

for temperatures below these limits

Exercise 3-3Determine MDMT1) Material is SA-516 Gr. 60. Nominal thickness is 2.0”. Renewal thickness

is 1.750”. Nameplate stamped “HT”

2) Material normalized SA-612. Thickness is 0.750”. Reduction ration is 0.85. Vessel was not PWHT.

3) 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 Reduction 00F [no reduction ratio given assume no

reduction]PWHT Reduction 00F [no reduction as 2” plate required PWHT by

Codes]Final MDMT 260F

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 MDMT3) 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.

Curve BInitial MDMT 31 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 <1 ½” not required PWHT as per

UCS-56]Final MDMT -11 0F

Impact Testing UG-84Impact testing of material is required when minimum operating

temperature is lower than allowed by the UCS-66 MDMT calculations.

(a) Test procedure – SA-370(b) Each set of specimens – 3

specimens(c) Acceptance 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-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 > ____ft-lbs b) To be acceptable each specimen must be > _____ft-lbs

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 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?

1) 18-19-122) 17-12-253) 17-16-174) 18-17-12

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

1) 18-19-12 [Average 16.3, Min value 12]2) 17-12-25 [Average 18, Min value 12]3) 17-16-17 [Average 16.67, Min value 16]4) 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 strength

• Codes 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 Mat’l chart – Fig. CS-2)

Step 6 : Calculate “P” – Max All External Pressure

P =4B/3(Do/t)

D0 = Outside diameter

L = Length between supports (inches)

Factors A & B – numbers from graph

External PressureSample of calculations

A 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 calculations

A 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 = 55.2

Step 2&3: Determine Factor A from figure GFind D0/t curve 55.2

Find 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 calculations

A 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 Pressure

P =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” = 50

Step 2 & 3: Determine Factor A from figure GFind D0/t curve 50

Find intersection with the L/D0 line of 3

At 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 Question

The 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 Question

The 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= 0.635” B=11,600

D0=45” A=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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