PREPARATION CLASS API 510 PRESSURE VESSEL INSPECTOR CERTIFICATION EXAMINATION PUSPATRI JOHOR BAHRU BAHRU 5 TH -9 TH NOVEMBER 2007
Course Outlines DAY 1 - 5 th November 2007 Introduction to API 510 Certification Module 1: ASME Section VIII – Rules For Construction Of Pressure Vessel Module 2: Static Head, MAWP & Stress Calculations Calculations DAY 2 - 6 th November 2007 Module 3: Joint Efficiencies & Internal Pressure Module 4: Pressure Testing, MDMT, Impact Testing Module 5: External Pressure
Course Outlines DAY 1 - 5 th November 2007 Introduction to API 510 Certification Module 1: ASME Section VIII – Rules For Construction Of Pressure Vessel Module 2: Static Head, MAWP & Stress Calculations Calculations DAY 2 - 6 th November 2007 Module 3: Joint Efficiencies & Internal Pressure Module 4: Pressure Testing, MDMT, Impact Testing Module 5: External Pressure
Course Outline DAY 3 – 7 th November 2007 Introduction to API 510 Pressure Vessel Inspection Code Section Section Section Section
1 2 3 4
: Scope : References : Definitions :Owner User Inspection Organization
Section 5 - : Inspection Practices 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8
Preparatory Work Modes of Deterioration and Failure Corrosion Rate Determination Maximum Allowable Working Pressure Determination Defect Inspection Inspection of Parts Corrosion & Minimum Thickness Evaluation Fitness for Service Evaluation
Course Outline DAY 4 – 8 th November 2007 Section 6: Inspection and Testing of Pressure Vessels and Pressure Relieving Devices 6.1 6.2 6.3 6.4 6.5 6.6 6.7
General Risk Based Inspection External Inspection Internal and On Stream Inspection Pressure Test Pressure Relieving Devices Records
Section 7: Repairs, Alterations and Re-rating of Pressure Vessels 7.1 General 7.2 Welding 7.3 Rerating
Course Outline Day 5 – 9 th 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 Vessels 4. API RP 576, Inspection of Pressure Relieving Devices 5. API RP 577, Welding Inspection and Metallurgy
ASME Publications 1. Section V, Nondestructive Examination 2. Section VIII, Division 1, Rules for Constructing Pressure Vessels 3. Section IX, Welding and Brazing Qualifications
Don’t Worry About The Exam The 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 Letter
2.
Identification Card
3.
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 stationery
6.
Jacket – some classroom is uncomfortably cool
7.
Earplugs – you never know who might be seat next to you
Module 1: ASME Section VIII – Rules For Construction Of Pressure Vessel
ASME SECTION VIII RULES FOR CONSTRUCTION OF PRESSURE VESSEL
ASME Boiler & 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 T d<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 Limits U-1(e)1 (a) 1st circumferential weld (b) 1st threaded joint for screwed connection (c) face of 1 st 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 7:: pages of the codes!
Organization of Code Section 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
Subsection B – Fabrication
Part UG – applies to all vessels 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 Appendices U-1(b) “address specific subject not covered elsewhere in this division” Alternate formulae QC Manual NDE Standards Non-mandatory Appendices U-1(b) “provides information and suggested good practices” Completed Sample Problems
Example 1-1 Finding the Right Answer At what base metal temperature is welding not allowed on routine pressure vessel?
Example 1-1 Finding the Right Answer At what base metal temperature is welding not allowed on routine pressure vessel? ASME Code
Section 1 Boilers
Section II Materials
Section VIII Pressure Vessel
Division 1 Routine
Section V NDT
Section IX Welding
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 1-2 A 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 – UW If it’s based on metallurgy - UCS
Code’s Purpose Establish 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 Rules User 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 Rules Manufacturer 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 Rules Authorized Inspector
Roles & Responsibilities
Section VIII requires that all vessels be inspected by a qualified thirdparty 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 duties
2.
Authorize the application of the Code Symbol
3.
Sign the Certificate of Shop Inspection
Setting The Rules
continued…
Authorized Inspector
Roles & Responsibilities
UG-9:(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 & Responsibilities
Welder UW-29(a) “ The welders ..used in welding pressure parts…shall be qualified in accordance with Section IX” NDE Technician UW-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 particle examination 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 particle examination method”
CODE-OLOGY Code 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-OLOGY Certificate 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-OLOGY Quality 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 1:.” 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-OLOGY Data 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-3 For vessel parts – the Form U-2
CODE-OLOGY UM 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 the a. Normal Holding Temperature b. 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 1-3 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? Subsection B Fabrication – Part UW – Inspection UW51(b)2 2) A P-1 material (carbon steel) 2” thick is being PWHT. What is the a. Normal Holding Temperature b. 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? •
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 1:’ of water?
PSHead = 0.433 x liquid height
Exercise 2-1 Static 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 5:’ high. The vessel will be hydrotested a. 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 6:’ vessel is filled with water. The pressure at the bottom is 210 psi. What is the pressure at the top?
:’
0 psig
2a
5:’
100 psig
:’
:’
2b
?? psig
3
?? psig 5:’
?? psig
210 psig 6:’
Exercise 2-1 Answer 1 1)
A deep diving submarine cruising at a depth of 854 feet. What is the static head pressure on this submarine? (external pressure)
t f
845 ft x 0.433 psi/ft = 369.8 psig
4 5 8
Exercise 2-1 Answer 2a 2) A vessel is 5:’ high. The vessel will be hydrotested a.
When filled with water what is the pressure at the bottom of the vessel?
Pbtm = 0.433 psi/ft x 50 ft
:’
0 psig
= 21.7 psig 2a
5:’
?? psig
Exercise 2-1 Answer 2b 2) A vessel is 5:’ 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?
100 psig
:’
Pbtm = Ptop + Psh = 100 + (0.433x50)
2b
= 121.7 psig ?? psig 5:’
Exercise 2-1 Answer 3 3) A 6:” vessel is filled with water. The pressure at the bottom is 210 psi. What is the pressure at the top?
:’
?? psig
Ptop = Pbtm - Psh = 210 - (0.433x60)
3
= 184 psig 210 psig 6:’
Module 2.2 Design 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 2-2 Design 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
_________ psig
b.
The top head should be designed for
_________ psig
c.
The bottom head should be designed for
_________ psig
:’ 2’
100 psig ?? psig
48’
?? psig
5:’
?? psig
Exercise 2-2 Design Pressure a.
Pshell = Ptop +Psh = 100 psig + (0.433 psi/ft x 48 ft) = 120.8 psig
b.
:’
2’
Ptop head = Ptop + Psh
100 psig
?? psig
= 100 psig + (0.433 psi/ft x 2 ft) = 100.9 psig
c.
Pbtm head = Ptop + Psh
48’
?? psig
5:’
?? psig
= 100 psig + (0.433 psi/ft x 50 ft) = 121.7 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 2.3 MAWP Calculations
MAWP Vessel & 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?
MAWP Vessel & Vessel Part To determine Vessel MAWP: Step 1: Determine each part MAWP (based on code formulas) Step 2: For each part subtract appropriate static head Step 3 Pick smallest pressure at top, the “weakest link”
:’ 12’
B
Part A: 343 – 0.433(34) = 328.3 psig Part B: MAWP 336
Part B: 336 -0.433(12) = 330.8 psig Thus, vessel MAWP = 328.3 psig
A
34’
Part A: MAWP 343
Exercise 2-3 Determining 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 Top Nozzle Side Nozzle Top Head Bottom Head Shell
Part MAWP Static Head Pressure @ Top 342 psig 426 psig 329 psig 336 psig 337 psig
Exercise 2-3 Determining 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 Top Nozzle Side Nozzle Top Head Bottom Head Shell
Part MAWP Static Head Pressure @ Top 342 psig 0.433 341.6 426 psig 10.4 415.6 329 psig 1.3 327.7 336 psig 18.2 317.8 337 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”
Design Pressure
MAWP
If MAWP < Design Pressure
To meet Code Requirement: MAWP > Design Pressure
Exercise 2-3 More MAWP & Static Head calculation 1)
The MAWP of a vessel is 1:: 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? 0 ft 1 ft
3)
A 8:’ tall pressure vessel is being hydrotested. A pressure gauge 2:’ up from the bottom reads 136 psig. What is the pressure at 50 ft the top of the vessel.
100 ft
Exercise 2-3 More MAWP & Static Head calculation 1)
The MAWP of a vessel is 1:: 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.
0 ft
Pshell = Ptop + Psh = 100 + 0.433(34)
2 ft
= 114.7 psig
34 ft
100 psig
Exercise 2-3 More 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? 0 ft
Ptop = Pshell – Psh = 87.5 – (0.433 x100)
1 ft
= 44.2 psig 50 ft
100 ft
Exercise 2-3 More MAWP & Static Head calculation 3)
A 8:’ tall pressure vessel is being hydrotested. A pressure gauge 2:’ up from the bottom reads 136 psig. What is the pressure at the top of the vessel. 0 ft
Ptop = Pgauge – Psh = 136 – (0.433 x 60) = 110.0 psig
60 ft
80 ft
136 psig
Module 2.4 Calculating 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 :.25:” thick thic k and 1.:” 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 :.25:” thick thic k and 1.:” wide. The specimen breaks with 12,500 lbs. of load. What is the ultimate tensile strength of this specimen? S = Load/Area S = 12,500/(0.25 x 1.0) S = 50,000 psi
Exercise 2.5 Calculating Stress 1)
A tension specimen specimen is :.5” thick thick and :.75” wide. wide. It breaks with a load of 21,500 lbs. The failure stress is _______ psi.
2)
A vessel is 1:’ 1:’ in diameter and and has pressure pressure of 1:: psi. The force (load) trying trying to launch launch a vessel head head into space is 1,1::,::: lbs. The circumference circumference of the shell is about 4::”. The head and shell are ½” thick, not including corrosion allowance. What is the actual actual longitudinal longitudinal stress on the shell-to-head shell-to-head weld?
Exercise 2.5 Calculating Stress 1)
A tension specimen is :.5” thick and :.75” wide. It breaks with a load of 21,500 lbs. The failure stress is _______ psi. S = Load/Area S = 21,500/(0.5)(0.75) S = 57,333 psi
2)
A vessel is 1:’ in diameter and has pressure of 1:: psi. The force (load) trying to launch a vessel head into space is 1,1::,::: lbs. The circumference of the shell is about 4::”. 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/Area S = 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. Allowable Stress = Ultimate Tensile Strength / 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
Exercise 2-6 Allowable Stress A material has an ultimate tensile strength of 70,000 psi at ambient temperature a)
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 2-6 Allowable Stress A material has an ultimate tensile strength of 70,000 psi at ambient temperature a)
What is the allowable stress for this material at ambient conditions if used in a pressure vessel today? Allowable Stress = UTS/S.F Allowable Stress = 70,000/3.5 Allowable 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.F Allowable 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
Module 3.1 Joint 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 2.7 Joint 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 2.7 Joint Efficiencies 1)
A she shell ll is mad made e wit with h Type Type 2 join joints ts and and spo spott RT RT was was perf perfor orme med. d. What is “E”? 0.8 .
2)
A she shell ll is mad made e wit with h Type Type 1 join joints ts and and Ful Fulll RT RT was was perf perfor orme med. d. What is “E”? 1.0.
3)
A she shell ll is made made with with Type Type 3 joi joint nts s and and no RT was was per perfo form rmed ed.. 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.”
C a te t e g o r y A : Longitudinal welded joints
within the main shell, communicating chambers,2 transitions in diameter, or nozzles; any welded joint within a sphere, within a formed or flat head, or within the side plates3 of a flat-sided flat -sided vessel; circumferential welded joints connecting hemispherical heads to main shells, to transitions in diameters, to nozzles, or to communicating chambers. C a te t e g o r y 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 than hemispherical to main shells, to transitions in diameter, to nozzles, or to communicating chambers. UW-3
C a te g o r y 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. C a te g o r y 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 pre pressu ssure re vesse vessell has has butt butt weld welds s whi which ch are are sing single le weld welded ed.. No RT has has been performed. 1) 2)
2)
A pre pressu ssure re vesse vessell has has butt butt weld welds s whi which ch are are doub double le weld welded ed.. The The vess vessel el is stamped as RT-3 1) 2)
3)
“E” for welded shell? _______ “E” for the seamless seamless head? ________
A pre pressu ssure re vesse vessell has has butt butt weld welds s whi which ch are are doub double le weld welded ed.. The The vess vessel el is stamped as RT-2. 1) 2)
4)
“E” for welded shell? _______ “E” for the seamless head? _______
“E” for welded shell? _______ “E for the seamless head? _______
A pre pressu ssure re vesse vessell has has butt butt weld welds s whi which ch are are doub double le weld welded ed.. The The vess vessel el is stamped as RT-1 1) 2)
“E” for welded shell? _______ “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) 2)
2)
A pressure vessel has butt welds which are double welded. The vessel is stamped as RT-3 1) 2)
3)
“E” for welded shell? 0.85 “E” for the seamless head? 1.00
A pressure vessel has butt welds which are double welded. The vessel is stamped as RT-2. 1) 2)
4)
“E” for welded shell? 0.6 “E” for the seamless head? 0.85
“E” for welded shell? 1.00 “E for the seamless head? 1.00
A pressure vessel has butt welds which are double welded. The vessel is stamped as RT-1 1) 2)
“E” for welded shell? 1.00 “E” for the seamless head? 1.00
Exercise 2-8 More Joint Efficiencies 1)
A pressure vessel has lap welds which are single welded. The vessel is stamped RT-3 1) 2)
2)
A pressure vessel has lap welds which are double fillet welded 1) 2)
3)
“E” for welded shell? _____ “E” for the seamless head? ______ “E” 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-2 1) 2)
“E” for welded shell? ______ “E” for the welded head? _______
Exercise 2-8 More Joint Efficiencies 1)
A pressure vessel has lap welds which are single welded. The vessel is stamped RT-3 1) 2)
2)
A pressure vessel has lap welds which are double fillet welded 1) 2)
3)
“E” for welded shell? 0.45 “E” for the seamless head? 0.85 “E” for welded shell? 0.55 “E” for the seamless head? 0.85
A pressure vessel has butt welds which are single welded with backing strips.The vessel is stamped as RT-2 1) 2)
“E” for welded shell? 0.99 “E” for the welded head? 0.99
Module 3.2 Tmin Calculations
Successful Calculations • The 5 steps to calculating Success 1) Always write the formula. Leave space above the formula for step 2 data 2) Write the “Givens” above formula. Put these in the same order as the formula. 3) Plug-in the values directly below the formula 4) Solve the problem 5) When complete shows the appropriate units e.g. inches, mpy, years
Shell tmin Calculations • The formula UG-27(c)(1) – Internal Pressure t = PR/SE-0.6P – “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 P = SEt/R + 0.6t – The part MAWP – The shell’s “good for” pressure – This is not vessel MAWP
Example A 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 2-9 Shell 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 2-9 Shell 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 2-10 Shell 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 2-10 Shell 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. ___________________________________________________________ 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 Units t = 0.649” ___________________________________________________________
Exercise 2-10 Shell 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 Units t= ___________________________________________________________
Rounded Head tmin Calculations •
Ellipsoidal Head UG-32 (d) t = PD/2SE-0.2P P = 2SEt/D + 0.2t
•
Torispherical Head UG-32 (e) t = 0.885PL/SE-0.1P P = SEt/0.885L + 0.1t
•
L = outside diameter
Hemispherical Head UG-32 (f) t = PL/2SE-0.2P P = 2SEt/L + 0.2t
L = inside radius
Head Calculations Example The 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 Units
t=
___________________________________________________________
Head Calculations Example The 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 Units
t = 0.401”
___________________________________________________________
Exercise 2-11 Formed 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 2-11 Formed Head Minimum Thickness 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. • ___________________________________________________________ 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 Units t = 0.183” ___________________________________________________________
Exercise 2-11 Formed Head Minimum Thickness 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. ___________________________________________________________ 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 heads From UG-34(c)(2) equation (1) t = d √CP/SE
d – generally inside diameter C – 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
Exercise 2-12 Flat 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 2-12 Flat 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 Units t = 5.359” ___________________________________________________________
Part MAWP Part 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 calculations This 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 heads Symbols are the same as used in the tmin formulas t = PD/2SE-0.2P P = 2SEt/D + 0.2t
Exercise 2-13 Let’s Calculate Part MAWP 1) 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”.
P = 2SEt/D + 0.2t
2:1 Ellipsoidal Head: _______ psi Torispherical Head: _______ psi Hemispherical Head: _______ psi Cylinder: ________ psi
P = SEt/0.885L + 0.1t P = 2SEt/L + 0.2t P = SEt/R + 0.6t
Flat Head: ________ psi 2) Which shape is the best for containing pressure? __________ 3) Which shape is the worst for containing pressure? _________
Exercise 2-13 Let’s Calculate Part MAWP 1) 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”.
P = 2SEt/D + 0.2t
2:1 Ellipsoidal Head: 249.6 psig Torispherical Head: 138.8 psig Hemispherical Head: 498.3 psig Cylinder: 247.5 psig
P = SEt/0.885L + 0.1t P = 2SEt/L + 0.2t P = SEt/R + 0.6t
Flat Head: 5.2 psig For info only 2) Which shape is the best for containing pressure? Hemispherical 3) Which shape is the worst for containing pressure? Flat Head
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 Testing Hydro test Requirements in UG-99 The test pressure formula: Pt = 1.3 x MAWP x (St/Sd) Pt = Test Pressure St = Allowable stress at temperature of hydrotest Sd = Allowable stress at design temperature Note: St/Sd always > 1 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 Procedure 1)
Complete all pre-hydrotest work & testing
2)
Assure vessel & support structure is designed for weight of hydrotest liquid
3)
Select hydrotest fluid – any non-hazardous liquid below its boiling point
4)
Disconnect or “ blind off” appurtenances not to be tested
5)
Vent of high points to remove possible air-pockets
6)
Test fluid should be 30 0F above MDMT
7)
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 calibrated
9)
Pressure gauge should be connected directly to vessel. If not visible, another should be connected near operator.
10)
Check tightness of test equipment
11)
Perform pressure test at 1.3 times MAWP corrected for temperature.
12)
Back pressure down to Test Pressure divided by 1.3
13)
After vessel temperature is below 120 0F, perform close visual inspection of joints and connections.
Hydrotest Calculation A vessel is constructed of a P-1 material. The vessel is stamped MAWP is 300 psi at 800 0F. Material allowable stress “S” is: 1000F = 20,000 psi 8000F = 13,500 psi Determine the a) hydrostatic test pressure b) minimum inspection test pressure
Hydrotest Calculation a) Calculate Test Pressure MAWP = 300 psi, St = 20,000 psi, Sd = 13,500 psi Formula: 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 Pressure Pinsp = Pt/1.3 = 577.7/1.3 = 444.4 psi
Exercise 3-1 Hydrotest Solution A vessel is constructed of a P-1 material. The vessel MAWP is 600 psi at 6500F. Allowable stress “S” is @ 100 0F = 17,000 psi, @ 650 0F=17,000 psi. Determine the Hydrostatic Test Pressure and the Minimum Inspection Test Pressure.
Exercise 3-1 Hydrotest Solution A vessel is constructed of a P-1 material. The vessel MAWP is 600 psi at 6500F. Allowable stress “S” is @ 100 0F = 17,000 psi, @ 6500F=17,000 psi. Determine the Hydrostatic Test Pressure and the Minimum Inspection Test Pressure. Calculate Test Pressure MAWP=600 psi, St=17,000 psi Pt= 1.3 (600)(17,000/17,000) = 780 psi Minimum Inspection Test Pressure Pinsp = Pt/1.3 = 780/1.3 = 600 psi
Sd=17,000 psi
Pneumatic Testing Pneumatic Test Requirements UG 100 Safety Issues – Compressed Air The Test Pressure formula
Pt = 1.1 x MAWP x (St/Sd) Pressure increased in steps 0.5Pt – 1st step 0.6Pt – 2nd step 0.7Pt – 3rd step 0.8Pt – 4th step 0.9Pt – 5th step 1.0Pt – at test pressure Pinsp = Pt/1.1
Exercise 3-2 Hydrotest Solution A vessel is constructed of a P-1 material. The vessel MAWP is 100 psi at 7500F. Material allowable stress “S” is: 1000F = 18,000 psi 7500F = 17,000 psi Determine the: •
Pneumatic Test Pressure
•
Each of the Test Pressure Steps
•
Inspection Test Pressure
Exercise 3-2 Hydrotest Solution a)
Pneumatic Test Pressure
Pt = 1.1(MAWP) x (St/Sd) = 1.1x100(18,000/17,000) = 116.5 psi b)
Each of the Test Pressure Steps
0.5Pt – 1st step = 0.5(116.5) = 58.2 psi 0.6Pt – 2nd step = 0.6(116.5) = 69.9 psi 0.7Pt – 3rd step = 0.7(116.5) = 81.5 psi 0.8Pt – 4th step = 0.8(116.5) = 93.2 psi 0.9Pt – 5th step = 0.9(116.5) = 104.8 psi 1.0Pt – at test pressure = 1.0(116.5) = 116.5 c)
Inspection Test Pressure
Pinsp = 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 curve Determine 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 reduction Subtract temperature reduction from initial MDMT 3. UCS-68 ( c ) - Further reduction Has vessel has been PWHT when not required by UCS-56 If yes subtract an addition 30 F from MDMT 4. Check Limits- UCS-66(b)(2) (cannot exceed these limits)
Example 3-2 Calculate 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 3-2 Calculate MDMT Step 1: Initial MDMT: Table UCS-66 Material : Curve B Initial MDMT: -70F Step 2: MDMT Reduction: Figure UCS 66-1 Reduction ration: 0.90 Reduction: 10 0F New MDMT = Initial – Reduction = -70F - 100F = -170F Step 3: PWHT Reduction UCS 68(c) PWHT: Yes Required by Code: No Additional Reduction: 30 0F Final MDMT: Step 2 – PWHT reduction = -17 0F - 300F = -470F Step 4: Check limits: UCS 66(b)(2) No restriction as UCS-68 (c) allows for temperatures below these limits
Example 3-2 Calculate MDMT Step 1: Initial MDMT: Table UCS-66 Material : Curve B Initial MDMT: -70F
Example 3-2 Calculate MDMT
Step 2: MDMT Reduction: Figure UCS 66-1 Reduction ration: 0.90 Reduction: 10 0F New MDMT = Initial – Reduction = -70F - 100F = -170F
Example 3-2 Calculate MDMT Step 3: PWHT Reduction UCS 68(c) PWHT: Yes Required by Code: No Additional Reduction: 30 0F Final MDMT: Step 2 – PWHT reduction = -170F - 300F = -470F
Example 3-2 Calculate MDMT
Step 4: Check limits: UCS 66(b)(2) No restriction as UCS-68 (c) allows for temperatures below these limits
Exercise 3-3 Determine MDMT 1)
Material is SA-516 Gr. 60. Nominal thickness is 2.0”. Renewal thickness
2)
is 1.750”. Nameplate stamped “HT” 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 3-3 Determine MDMT 1) Material is SA-516 Gr. 60. Nominal thickness is 2.0”. Renewal thickness is
1.750”. Nameplate stamped “HT”. Material Curve C Initial MDMT Ratio Reduction PWHT Reduction Final MDMT
260F [SA-516 Gr. 60 as it is not mentioned as normalized] 00F [no reduction ratio given assume no reduction] 00F [no reduction as 2” plate required PWHT by Codes] 260F
Exercise 3-3 Determine MDMT 2) 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 3-3 Determine MDMT 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.
Curve B Initial 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-84 Impact 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 UG84.1 * Average value – from the chart * Minimum value – 2/3 chart
Note: 1 ksi = 1,000 psi
Exercise 3-4 Impact Testing 1) 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 3-4 Impact Testing 1)
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 Testing A 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)
18-19-12
2)
17-12-25
3)
17-16-17
4)
18-17-12
More Exercise Impact Testing A 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-lbs Min 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 Pressure Thickness 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 Pressure Steps for calculations
D0 = Outside diameter P =4B/3(Do/t)
L = Length between supports (inches) Factors A & B – numbers from graph
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
External Pressure Sample 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 Pressure Sample 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/D 0 & 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 G Find D0/t curve 55.2 Find intersection with the L/D0 line of 36.23 At intersection drop line straight down to bottom of graph & read factor A 0.000375
External Pressure Sample 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-2 Find temperature curve (700 0F) Find intersection with Factor A line 0.000375 At 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 3-5 External Pressure A 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 500 0F. Use material chart Fig CS-2. Determine the maximum allowed external pressure.
Exercise 3-5 External Pressure A 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/D 0 & D0/t L/D0 = 30”/10” = 3 D0/t = 10”/0.20” = 50 Step 2 & 3: Determine Factor A from figure G Find 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 3-5 External Pressure A 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-2 Find temperature curve (500 0F) Find intersection with Factor A line =
0.0012 At intersection move horizontally to side of graph and read factor B = 10,500 Step 6: Calculate P – Max All External Pressure P =4B/3(Do/t) = 4(10,500)/3(50) = 280 psi.
Question 12 Sample 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 SA516 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 300 0F.
Question 12 Sample 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 SA516 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 300 0F. t= 0.635” D0=45”
B=11,600 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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