002 api 510 course puspatri apr 09

API 510: Pressure Vessel Inspection Code for Maintenance Inspection, Rating,

Repair, and Alteration

PUSPATRIPASIR GUDANG, JOHOR27th Apr – 1st May 2009

SECTION 1API Document Practices Question

1. The application of API 510 is restricted to organization that:1) Fabricate or build pressure vessel according to ASME B&PV

Code2) Employ or have access to a authorized inspection agency3) Manufacture or inspect pressure vessel according to NBIC4) Hire or have access to an unsanctioned inspection agency(API 510 - 1.1.2)


1. The application of API 510 is restricted to organization that:

1) Fabricate or build pressure vessel according to ASME B&PV Code

2) Employ or have access to a authorized inspection agency

3) Manufacture or inspect pressure vessel according to NBIC

4) Hire or have access to an unsanctioned inspection agency

(API 510 - 1.1.2)


2. API 510 inspection code is only applicable to pressure vessel used by

the petroleum and chemical industries that:1) Are being fabricated for the petroleum/chemical industries2) Can be fabricated to ASME B&PV Code and inspected by NBIC inspectors3) Will be place in service after fabrication to the ASME B&PV

Code4) Have been placed in service, inspected and/or repaired by an authorized agency(API 510 – 1.1.1)


3) Which of the following is an example of a pressure vessel covered by API 510. (All of the vessels are in-service).

1) PV on an ocean going ship, operates at 100 psig and 100F2) PV in a oil refinery, operates a 5 psig and 70F3) PV in a oil refinery, operates a 100 psig and 200F4) PV in a oil refinery, vol of 4 cu.ft & operates at 70 psig & 70F(API 510 - Appendix A)


3) Which of the following is an example of a pressure vessel covered by API 510.

(All of the vessels are in-service).1) PV on an ocean going ship, operates at 100 psig and 100F2) PV in a oil refinery, operates a 5 psig and 70F3) PV in a oil refinery, operates a 100 psig and 200F4) PV in a oil refinery, vol of 4 cu.ft & operates at 70 psig & 70F(API 510 - Appendix A)


4) If there is a conflict between the ASME Code and API 510 for vessels that have been placed In service, the requirement of:

1. API 510 shall take precedence over the ASME codes2. ASME codes shall take precedence over the API 5103. NBIC shall be used as an arbitration4. The owner/user of pressure vessel shall take precedence

over both codes(API 510 – 1.1.1)


77. A 12 inches reinforced nozzle is being installed in a vessel where the largest

existing reinforced opening is 10 inches. This is considered;1. A repair2. An alteration3. A re-rating4. A maintenance activity(API 510 – 3.2)


84. Which of the following is not an alteration?1) Increasing the diameter of a shell section2) Decreasing the temperature with impact testing of materials

required3) Adding a NPS 12” reinforced nozzle in a vessel that has two NPS 10”

reinforced nozzles4) Adding a NPS 10” reinforced nozzle in a vessel that has two NPS 12”

reinforced nozzles(API 510 – 3.2)


5. One type of Authorized Inspection Agency is:1) An inspection organization that does inspection2) An insurance /inspection agency which does not write

pressure vessel insurance.3) An owner/user of pressure vessel who maintains an

inspection organization only for his own equipment4) An independent third party consultant(API 510 – 3.6)


6.The term required thickness is the thickness:1. Essential for the shell and heads of a vessel2. Without corrosion allowance for each element of a pressure

vessel3. Necessary for the shell of a vessel4. Including the corrosion allowance for the shell of a pressure

vessel(API 510 – 3.55)


7. Lowering of the maximum allowable working pressure or temperature

rating of a vessel or both below the design condition is;1) Not a permissible way to provide for corrosion2) The preferred way to provide for corrosion3) The only way to keep the vessel in service when it is corroded4) A permissible way to provide for additional corrosion(API 510 – 3.56)


8. An owner user is responsible for developing, implementing, executing

and assessing pressure vessel inspection system and inspection procedures that will meet the requirements of API 510. These

system and procedures will be:1)Maintained in engineering procedure document2)Kept as a standard procedure3)Contained in a quality assurance inspection manual4)In hand and available at owner user headquarters(API 510 – 4.2.1)


78. How many years of experience is required when the candidates does not have an education level equivalent to a high school education?

1) 52) 33) 24) 10API 510 – Appendix B.2.1


79. May the API 510 inspector be assisted by non-certified individuals when visual inspections are performed?

1. Yes, when authorized by API2. Yes3. No4. Yes, when properly trained and qualified(API 510 – 4.2.4)


9. Safety precautions are important in pressure vessel inspection because

of the limited access to and the confined space of pressure vessel. Out of

the organization listed, which is the primary one that should be reviewed

and followed.1)ASME2)OSHA3)NFPA4)NBIC(API 510 – 5.3)

SECTION 5API Document Practices

Question

14. If the external of internal covering such as insulation, refractory

protective lining and corrosion resistant linings are in good condition and

there is no reason to suspect that an unsafe condition is behind them.

1) It is not necessary to remove them for inspection2) It is necessary to remove them completely for inspection3) It is necessary to partially remove them for inspection4) It is required remove them completely for inspection on some

set interval(API 510 – 5.2.2.3)


126. Examples of degradation mechanism are:

1) Corrosion2) Cracking3) Creep4) All the above

(API 510 – 5.2.1)


127. It is essential that all RBI assessment be:

1) Reviewed by jurisdictional authorities2) Understood by engineering3) Thoroughly documented4) Defined in terms of API 4467

(API 510 – 5.2.3)


24. When must a pressure test be performed on a pressure vessel?1. When the contractor working on the vessel deems it is necessary2. When the API authorized pressure vessel inspector believes that it is

necessary3. When the safety group of the owner user request it4. When NFPA request it(API 510 – 5.8)


43. A pressure test is normally required after;1. An alteration2. A repair3. A lightning strike4. A unit upsetAPI 510 – 5.8.1


125. When owner-user chooses to conduct RBI assessment it must

includes a systematic evaluation of:

1) Fatigue2) Likelihood of failures3) Associated consequences of failure4) Both 2 & 3

(API 510 – 5.2)


87. How are safety relief devices isolated from the vessel during hydrostatic testing?

1) Test levers are wired down2) Removed or gagged3) Springs are tightened4) None of the above

(API 510 - 5.8.3.2)


85. A pressure vessel whose thickness does not exceed 2 inches has a MDMT of –

20F. Per API 510 what is the minimum test temperature permitted?1. 30 F2. -10 F3. -30 F4. 10 F(API 510 – 5.8.6.2)


18. Under what conditions is an internal field inspection of a newly installed pressure vessel waived?

1)The contractor installing the vessel assures the owner user that the

vessel is satisfactory for its intended service2)A manufacturers data report assuring the vessel is satisfactory

for its intended service is available.3)The owner-user assures the inspector that the vessel is

satisfactory for its intended service.4)The manufacturer orally assures the owner user that the vessel

is satisfactory for its intended service.(API 510 – 6.2.1.2)


128. An RBI assessment may be used to _______the 10 year inspection

limit.

1) Increase or decrease2) Increase3) Decrease4) Limit

(API 510 – 6.3)


19. An above ground pressure vessel shall be given a visual external inspection,

preferably while in operation, at least every _____years or at the same interval as

the internal, whichever is less.1) 22) 33) 54) 10(API 510 – 6.4.1)


20. The period between internal or on-stream inspections for a remaining life of

four years or greater shall not exceed one-half the estimated remaining life of the

vessel based on corrosion rate or _____years whichever is less.• 10• 15• 5• 12(API 510 – 6.5.1.1)


109. You have a corroded area which is 0.900 in thick. The minimum

thickness in calculated to be 1.0 in. The corrosion rate is 0.01 in per

year. Assuming constant corrosion rate how much weld overlay must be

applied to have a 10 year life?

1) 0.10 in.2) 0.20 in.3) 1.3 in.4) Repair not permitted by API 510

API 510 6.3


21. In cases where the safe operating remaining life is estimated to be less than 4

years, the inspection interval may be the full remaining safe operating life up to a

maximum of ______ years.1) 12) 23) 34) 4(API 510 – 6.5.1.1)


22. If both the ownership and the location of a vessel are changed, what must happen before it is reused?

• It must be internally and externally inspected• All the records must be reviewed• It must be thoroughly ultrasonically checked• It must have all the paper work transferred to the new owner(API 510 – 6.2.2)


81. Which of the following conditions must be met to qualify vessel for on-stream inspections when the general corrosion rate is less than 0.005 ipy and has remaining life greater than 10 years?

1) The vessel is strip lined2) The corrosive character of the content has been established for at

least 5 years3) Not more than two questionable conditions were disclosed at the

previous external inspections4) The operating temperature exceed the lower temperatures limits for

the materials creep-rupture range

(API 510-6.5.2)


26. What is the maximum inspection or testing interval of a pressure-relieving device?

1. 15 years2. 12 years3. 10 years4. 20 years(API 510 – 6.6.2.2)


25. Who is authorized to test and repair a pressure relief valve?1) An organization experienced in PRV maintenance2) A valve repair shop3) A certified pressure vessel inspector4) A contractor with valve mechanics(API 510 – 6.6)


11. If a probable corrosion rate cannot be determined from reviewing

data from the same or similar service vessels or estimated form published

data, on-stream determinations shall be made after approximately

_______hours of service by using NDE methods and a corrosion rate is

established.1)5002)10003)50004)10,000(API 510 – 7.1.2)


12. The maximum allowable working pressure for the continued use of a

pressure vessel shall be based on computations determined by using the;

1)Latest edition of the National Boiler Inspection Code or the construction

code to which the vessel was built.2)Latest edition of ASME code or the construction code to which

the vessel was built3)Latest edition of API/ASME code or the construction code to

which the vessel was built4)Latest edition of underwrite laboratories Code or the

construction code to which the vessel was built.(API 510 – 7.3.1)


Question16. For a corroded area of considerable size in which the circumferential

stresses govern, the least thickness along the most critical element of the area

may be averaged over a length not exceeding the following;a. For vessels with inside diameters less than or equal to 60 inches-1) ¼ the vessel diameters or 5 inches whichever is less2) ½ the vessel diameters or 10 inches whichever less3) 1/4 the vessel diameters or 15inches whichever less4) ½ the vessel diameters or 20 inches whichever less(API 510 – 7.4.2.1)

b. For vessels with inside diameters greater than 60 inches-1) 1/3 the vessel diameters or 40 inches whichever is less2) ¼ the vessel diameters or 30 inches whichever less3) 1/3 the vessel diameters or 50 inches whichever less4) 1/4 the vessel diameters or 25 inches whichever less(API 510 – 7.4.2.1)


Question

83. Which of the following widely scattered pit cannot be ignored?

1. The total area of the pits that is deeper than CA does not exceed 7 square inches within an 9 inch diameter circle

2. No pit depth is greater than 1/3 the vessel thickness3. The sum of the pit dimensions along any straight line within the

circle does not exceed 2.5 inches4. The total area of the pits that is deeper than CA does not exceed 7

inch2 within an 8 inch diameter circle(API 510 – 7.4.2.3)


Question17. For corrosion calculations the surface of the weld is considered to be:1. 1 inch on either side of the weld or twice the minimum thickness on

either side of the weld whichever is greater2. 2 inches on either side of the weld or twice the minimum thickness

on either side of the weld whichever is greater3. 4 inches on either side of the weld or twice the minimum thickness

on either side of the weld whichever is greater4. 6 inches on either side of the weld or twice the minimum thickness

on either side of the weld whichever is greater(API 510 – 7.4.5)

People are always blaming their circumstances for what they are. I don’t believe in circumstances. The people who get on in this world are the people who get up and look for the circumstances they want, and if they can’t find them, make them.

-George Bernard Shaw

API 510 Mathematical CalculationsPressure Vessel Inspection Code:-Maintenance Inspection, Rating,

Repair, and Alteration

Section 7.3Maximum Allowable Working Pressure

Determination

Establish using latest ASME Code edition or original Code edition to which vessel constructed

Certain essential details must be known prior to performing computations

In corrosive service the wall thickness used shall be actual thickness determined by inspection minus twice the estimated corrosion loss before next inspection

The “t” in the pressure formula is replaced by the following: t = tactual – 2(Corrosion rate x years to next inspection)

P = SEt/R + 0.6t

Section 7Calculating P when the vessel is corroding

API 510 - During an internal inspection a corroded area on a vessel is discovered with a current thickness of 0.446 inches. You determine that the corrosion rate is 0.004 ipy. Operations would like to have a 6 year run before performing the next internal inspection on this vessel. What pressure is the wasted area “good for” for the end of 6 year. The shell has a 5 inch internal diameter and has joint efficiency of 1.0. The allowable stress of the material at design conditions is 15,000 psi.

ASME VIII-During the inspection of horizontal pressure vessel a corroded area was found on the inside surface, the thickness at the point was measured to be 0.906 inches thick. Checking the ASME data report for the vessel it was found that the shell plate was 1-1/4 inch purchased thickness,48 inch inside diameter of SA-516 Gr. 70 Material, the shell was rolled and welded using type 1 longitudinal seam with full radiography examination. The MAWP is 600 psig at 300 F. All openings in the vessel are fully reinforced, welded connections. S is 20,000 psi. What would be the maximum pressure permitted for this thinned area

(API 510 – 7.3)

Math Problem #1Maximum Allowable Working Pressure Determination

A pressure vessel has an MAWP of 500 psi at 400F and is made of SA 516 Gr.70 material with a stress value of 20,000 psi and has a

measured thickness of 0.875 inches. The efficiency is 1.0, inside radius is 24

inches and there is no corrosion allowance. The observed corrosion rate is

0.01 inches/year and the next planned inspection is 5 years. What will be

the MAWP at the end of 5 years?1) 633 psi2) 500 psi3) 450 psi4) 900 psi

Math Problem #1Maximum Allowable Working Pressure Determination________________________________________________________________________Step 1: List down all “Given” infoMAWP = 500 psi tactual=0.875” R=24” Next inspection = 5 yrsS=20,000 psi E=1.0 CR= 0.01 inch/yearMAWP at the end of next 5 years?________________________________________________________________________Step 2: Formula P = SEt/R+0.6t

t = tactual – 2(CR x next inspection)________________________________________________________________________Step 3: Calculate t = 0.875” – 2(0.01 x 5) = 0.775”________________________________________________________________________Step 4: Plug in value into P = SEt/R+0.6t

=(20,000)(1)(0.775)/24+(0.6)(0.775)=633 psi________________________________________________________________________

Math Problem #2Maximum Allowable Working Pressure DeterminationA pressure vessel has an MAWP of 350 psi at 300F and is made of SA 515

Grade 60material with a stress value of 16,000 psi. The efficiency of the longitudinal

joint is 0.70 and the inside radius is 36 inches with 1.25 inch corrosion allowance. The measured thickness is 0.750 inches and the corrosion rate is 0.02 ipy. The

next planned inspection is in 6 years. What will be the MAWP at the end of six years?

1) 440 psi2) 350 psi3) 602 psi4) 157 psi

Math Problem #2Maximum Allowable Working Pressure Determination__________________________________________________________________________Step 1: List down all “Given” infoMAWP = 350 psi tactual=0.875” R=36” Next inspection = 6 yrsS=16,000 psi E=7.0 CR= 0.02 inch/yearMAWP at the end of 6 years?__________________________________________________________________________Step 2: Formula P = SEt/R+0.6t

t = tactual – 2(CR x next inspection)__________________________________________________________________________Step 3: Calculate t = 0.75” – 2(0.02 x 6) = 0.51”__________________________________________________________________________Step 4: Plug in value into P = SEt/R+0.6t

=(16,000)(0.7)(0.51)/36+(0.6)(0.51) =5,712/36.306 = 157.3 psi

Section 7.4:Fitness For Service Analysis for Corroded

Region• Corrosion may cause uniform loss or may cause a pitted

appearance• Minimum actual thickness and maximum corrosion rate

determined by:– NDE thickness testing and drilled test holes– Measurement through openings– Gauging from uncorroded surfaces

• When the minimum actual thickness or maximum corrosion rate is to be adjusted, one of the following evaluation techniques should be considered

1) Averaging for large areas2) Evaluation of widely scattered pit 3) Evaluation of corrosion on weld surface with joint factor other

than 14) Measuring corroded thickness of ellipsoidal & torispherical head

Section 7.4.2:Evaluation of Locally Thinned Area

Math Problem #3Averaging for Large Areas

An eight foot diameter vessel has a large area of general corrosion (32 inches in length and width) and has thickness readings taken in the circumferential (lettered row) and longitudinal (numbered row) directions as shown below. What would be the least thickness along the most critical element considering circumferential stress applies.

1a-0.298” 2a-0.243” 3a-0.192”4a-0.185” 5a-0.267”

1b-0.264” 2b-0.230” 3b-0.185”4b-0.193” 5b-0.220”

1c-0.223” 2c-0.180” 3c-0.204”4c-0.210” 5c-0.246”

1. 0.206”2. 0.193”3. 0.179”4. 0.195”


An eight foot diameter vessel has a large area of general corrosion (32 inches in length and width) and has thickness readings taken in the circumferential (lettered row) and longitudinal (numbered row) directions as shown below. What would be the least thickness along the most critical element considering circumferential stress applies.

1a-0.298” 2a-0.243” 3a-0.192” 4a-0.185” 5a-0.267”

1b-0.264” 2b-0.230” 3b-0.185” 4b-0.193” 5b-0.220”

1c-0.223” 2c-0.180” 3c-0.204” 4c-0.210” 5c-0.246”

___________________________________________________________________________ 0.2616” 0.2176” 0.1936” 0.196”

0.569”

1. 0.206”2. 0.193”3. 0.179”4. 0.195”

a b c

1

23

4

5


An six foot diameter vessel has a large area of general corrosion (24 inches in length and width) and has thickness readings taken in the circumferential (lettered row) and longitudinal (numbered row) directions as shown below. What would be the least thickness along the most critical element considering circumferential direction.

1a-0.456” 2a-0.443” 3a-0.388”4a-0.443” 5a-0.550”

1b-0.458” 2b-0.423” 3b-0.334”4b-0.457” 5b-0.589”

1c-0.398” 2c-0.399” 3c-0.368”4c-0.447” 5c-0.498”

1. 0.447”2. 0.363”3. 0.191”4. 0.422”


An six foot diameter vessel has a large area of general corrosion (24 inches in length and width) and has thickness readings taken in the circumferential (lettered row) and longitudinal (numbered row) directions as shown below. What would be the least thickness along the most critical element considering circumferential direction.

1a-0.456” 2a-0.443” 3a-0.388” 4a-0.443” 5a-0.550”1b-0.458” 2b-0.423” 3b-0.334” 4b-0.457” 5b-0.589”1c-0.398” 2c-0.399” 3c-0.368” 4c-0.447” 5c-0.498”

1. 0.447”2. 0.363”3. 0.191”4. 0.422”

a b c

1

23

4

5

0.456”

0.452”

0.422”

Section 7.4.3: Evaluation of Pitting

Math Problem #4 Widely Scattered PitA pressure vessel is 48 inches in diameter. There is some widely scattered pitting present. The wall thickness including 0.125 inch corrosion allowance is 0.750 inch. One group is encircled by an eight inch diameter circle that contains the following;

One 1” dia x 0.20” deep pit, three ¾” dis x 0.20” pit and one 1-1/2” dia x 0.15 deep pit. A straight line is drawn through the most pits within the circle and includes the ¾”, 1” and 1-1/2” pits.

a) What is the total area of the its within the 8 inches circle?1)2.5 inch2 2)3.876 inch2 3)1.325 inch2 4)7.25 inch2

b) What would have to be done to the pits in the straight line to make them acceptable?

1) Nothing, they are acceptable as is2) Weld repair the 1.5” dia pt3) Fill all pits with epoxy materials4) None of the above apply

Math Problem #4 Widely Scattered PitA pressure vessel is 48 inches in diameter. There is some widely scattered pitting present. The wall thickness including 0.125 inch corrosion allowance is 0.750 inch. One group is encircled by an eight

inch diameter circle that contains the following;

One 1” dia x 0.20” deep pit, three ¾” dis x 0.20” pit and one 1-1/2” dia x 0.15 deep pit. A straight line is drawn through the most pits within the circle and includes the ¾”, 1” and 1-1/2” pits.

a) What is the total area of the its within the 8 inches circle?1)2.5 inch22)3.876 inch2 3)1.325 inch24)7.25 inch2

b) What would have to be done to the pits in the straight line to make them acceptable?1) Nothing, they are acceptable as is2) Weld repair the 1.5” dia pt3) Fill all pits with epoxy materials4) None of the above apply

D=48”Tnom = 0.75CA=0.125

Pit #1: 1” dia x 0.2” depthPit #2: 0.75” dia x 0.2” depth x (3nos)Pit#3: 1.5” dia x 0.15” depth

a) Total area of the pitsFormula for area: ∏r2Area of Pit#1 + Area of Pit#2 x 3 + Area of Pit#3= ∏(0.5)2 + ∏(0.375)2 x 3 + ∏(0.75)2

= 0.785 + 1.325 + 1.767= 3.878 inches2

b) Sum of dimension of pits within the 8 inches circleDia Pit#1 + Dia Pit#2 + Dia Pit#3= 1 + 0.75 + 1.5= 3.25” which is exceed the limit of 2”. Thus by repair 1.5” pit then the sum of dimension will be 1.75” which <2”

Another ProblemWidely Scattered PitFour scattered pits are found on the shell of a vessel. All the pits are contained

withinan 8 inch diameter circle. Pit no 1 is 0.75 inch diameter and 0.25 inch deep. Pit

no 2 is 0.49 inch diameter and 0.61 inch deep. Pit no 3 is 0.6 inch diameter and 0.51

inch deep. Pit no 4 is 1.5 inch diameter and 0.26 inch deep. The required thickness

is 0.781 inch and the actual thickness is 1 inch.

a) What is the total area of the pits in the 8 inch circle?1) 2.678 inch2 2)3.335 inch2 3)1.002 inch2 4) 6 inch2

b) What pit(s) are unacceptable due to remaining thickness?1) No 1 & 3 2) No 2 3) No 4 4) No 1 & 2

c) What pit(s) are unacceptable because of straight line out tolerance?1) No 1&2 2) No 2&3 3) No 3&4

4) N0 2&4

Another ProblemWidely Scattered PitFour scattered pits are found on the shell of a

vessel. All the pits are contained within an 8 inch diameter circle. Pit no 1 is 0.75 inch diameter and 0.25 inch deep. Pit no 2 is 0.49 inch diameter and 0.61 inch deep. Pit no 3 is 0.6 inch diameter and 0.51 inch deep. Pit no 4 is 1.5 inch diameter and 0.26 inch deep. The required thickness is 0.781 inch and the actual thickness is 1 inch.

a) What is the total area of the pits in the 8 inch circle?1) 2.678 inch22)3.335 inch23)1.002 inch24) 6 inch2

b) What pit(s) are unacceptable due to remaining thickness?1) No 1 & 3 2) No 2 3) No 4 4) No 1 &

2

c) What pit(s) are unacceptable because of straight

line out tolerance?1) No 1&2 2) No 2&3 3) No 3&4 4) N0

2&4

Pit #1 : 0.75” dia x 0.25” depthPit #2 : 0.49” dia x 0.61” depth Pit # 3 : 0.6” dia x 0.51” depth Pit # 4 : 1.5” dia x 0.26” depthtmin = 0.781”tactual = 1”

a) What is the total area of the pits in the 8 inch circle?Formula for area: ∏r2= Area Pit#1 + Area Pit#2 + Area Pit#3 + Area

Pit#4= ∏(0.375)2+ ∏(0.245)2+ ∏(0.3)2+ ∏(0.75)2

= 0.4418 + 0.1886 + 0.2827 + 1.7671= 2.68 inch2

b) What pit(s) are unacceptable due to remaining

thickness?Remaining t below the pit >½(tmin)½ x tmin = ½ x 0.781 = 0.3905”Pit#1: 1” – 0.25” = 0.75”Pit#2: 1” – 0.61” = 0.39”Pit#3: 1” – 0.51” = 0.49”Pit#4: 1” – 0.26” = 0.74”

c) What pit(s) are unacceptable because of straight

line out tolerance?Pit#1 + Pit#2 = 0.75 + 0.49 = 1.24”Pit#2 + Pit#3 = 0.49 + 0.6 = 1.09”Pit#3 + Pit#4 = 0.6 + 1.5 = 2.1” – not

acceptablePit#2 + Pit#4 = 0.49 + 1.5 = 1.99”

Section 7.4.5:Evaluation of Corroded Weld Surface

Math Problem #6Evaluation of Corroded Weld SurfaceA pressure vessel is found to have a corroded area that includes the longitudinal welded seam. An evaluation must be made to determine whether the thickness at the weld or remote from the weld governs. The inside radius is 36 inch which includes 0.125 inch corrosion allowance. The MAWP is 500 psig, S=17500 psi. The nameplate has no extent of radiography shown under the code stamp. The longitudinal seam is

Type 1. What would be the width of the evaluation if the width from toe to

toe was 1 inch?

1) 7 inch2) 2 inch3) 3.022 inch4) 7.044 inch

Math Problem #6Evaluation of Corroded Weld SurfaceA pressure vessel is found to have a

corroded area that includes the longitudinal welded

seam. An evaluation must be made to determine whether the thickness at the weld or remote

from the weld governs. The inside radius is 36

inch which includes 0.125 inch corrosion

allowance. The MAWP is 500 psig, S=17500 psi. The

nameplate has no extent of radiography shown under

the

code stamp. The longitudinal seam is Type 1. What would be the width of the evaluation if

the width from toe to toe was 1 inch?

1) 7 inch2) 2 inch3) 3.022 inch4) 7.044 inch

MAWP = 500 psi, S= 17,500, R=36”, CA=0.125”

E = 0.7 [Type 1 with no RT].Width of area for evaluation of the weld and

area remote from weld?

Section 5.7: The greater of 1” on either side of

weld or 2 x tmin on either side of weld

tmin = PR/SE-0.6P = (500)(36)/(17,500)(0.7) – 0.6(500) = 18,000/11,900 = 1.51”2 x t min = 3.01” which is > 1”

Thus, total area of evaluation= both side of the weld + weld area= (3.01) x 2 + 1= 7.03”

Section 7.4.6:Corroded Areas in Vessel Head

Math Problem#7Measuring Corroded Ellipsoidal Head

#7. There has been a corrosion in the center of ellipsoidal head. The head is a 2:1 elliptical with a D of 60 inches and h of 15 inches. What is the required thickness in the center portion of this head if the MAWP is 300 psi, E=1.0 and the S= 15,600 psi?

a) 0.520 inb) 0.750 inc) 1.050 ind) 0.889 in

Math Problem#7Measuring Corroded Ellipsoidal Head

MAWP = 300 psi, E=1.0, S=15,600 psi,D=60, h=15.What is treq or tmin?As per Section if 5.7 thickness at central portion calculation for elipsoidal or torispherical to use spherical head

formula:t = PR/2SE-0.2P

Calculate RR=kD, find value K from table.For D/2h = 60/2(15) = 2 value k=0.9R = 0.9 (60) =54”

t = (300)(54)/2(15,600)(1)-0.2(300) = 16,200/31,140 = 0.52”

Math Problem#13Measuring Corroded Torispherical

Head#13. A torispherical head is to be repaired by replacement of the

center portion of the head. What is the required thickness for a dishead

repair plate that is 50% of the head diameter. The material is SA 516 Grade

70 and S=18,000, P=300 psi, the crown radius = 96 inch. E=0.65 and a

Type 2 joint will be used.

1) 2.0 in2) 1.233 in3) 0.556 in4) 0.234 in

Math Problem#13Measuring Corroded Torispherical

HeadMAWP = 300 psi, S=18,000, R=96”E=0.65, Type 2 jointWhat is the t min?

As per Section if 5.7 thickness at central

portion calculation for elipsoidal or torispherical to use spherical head formula:t = PL/2SE-0.2P

The radius of torispherical to be used as

radius of spherical segment.

t = (300)(96)/2(18,000)(0.65) – 0.2(300)

= 28,800/23,340 = 1.234”


82. The definition of trequired is:1. Default thickness at the same CML or component in inch

as the t actual measurement. 2. Required thickness at different CML or component in inch

as the t actual measurement. 3. Required thickness at the same CML or component in

inch as the t actual measurement.4. Nominal thickness at the same CML or component in inch

as the t actual measurement.API 510 - 7.2.1


23. A pressure vessel has been in service for 12 years and has shown history of corrosion over its service life. The original thickness was 1.9375” thick and the present thickness is 1.405”. What is the corrosion rate for this vessel?1) 0.266250 ipy2) 0.532500 ipy3) 0.088750 ipy4) 0.044375 ipy(API 510 – 7.1)


28. The following is not normally found in pressure vessel records;

• Manufacturer data reports• Vessel identification numbers• Piping past the first vessel flanges• Relief valve information(API 510 – 7.8.2)

Inspection IntervalsWhy It So Important?

• Selecting the correct inspection interval is a “balancing act”!– Inspection must be conducted to assure the

integrity of the pressure vessel. But, inspections are costly and should be minimized.

• API 510 has developed rules that provide assurance of equipment integrity but at cost effective frequencies.

Inspection Interval Input

C-21210

Next Inspection?

API 510 Interval Rules

Section VIII

“t min”

Equipment Data P, S,E

Inspection Data “Tact/T

prev”

API 510 Requirements

External Inspection• Lesser of;

– 5 years– During Internal Inspection

• Internal Inspection• Lesser of;

– 10 years– ½ Remaining Life– Full life up to 2 years

• If remaining life <4 years

5 years

½ Life

2 years

Full Life

10 years

½ Life

2 years

Full LifeRem

ain

ing L

ife

Rem

ain

ing L

ife

20 yrs

4 yrs

2 yrs

0 yrs

4 yrs

2 yrs

0 yrs

10 yrs

Calculate the Interval

• Calculate the external and Internal Inspection intervals based on the given remaining life

Remaining Life (years) External Inspection (years) Internal Inspection (years)

18

22

2

15

4

8

44

12

1

The Problem

• A vessel head in June 1993 was 0.698 inches thick. Previous thickness in June 1988 at the same area was 0.713 inches. The required thickness of the head is 0.653 inches.

What is the internal inspection

interval?

Calculating the Interval

1. Calculate the Corrosion Rate (CR) CR = tprevious – tactual/time

2. Calculate the Remaining Corrosion Allowance (RCA) RCA = tactual - tminimum

3. Calculate Remaining Life of Vessel (RL) RL = RCA/CR

4. Calculate Inspection Interval (Based on API 510)

The Thickness

Required Thickness

Remaining Corrosion Allowance

Corroded & Gone

t mint a

ctual

t pre

vio

us

Corrosion Rate (CR)

• How do we calculate the corrosion rate?

CR = t previous – tactual period of time

The Problem!CR = 0.713 – 0.698

1998 – 1993 = 0.003 inch/year or ipy

Corrosion Rate – An Exercise

• Calculate the Corrosion Rate

Thickness Data 1TML Jan 2003 Jan 1993

1 0.450" 0.500"

Thickness Data 2TML Jan 2003 Jan 1983

1 0.260" 0.400"

Thickness Data 3TML Nov 2003 Nov 1988

1 0.675" 0.75"

Calculate Corrosion Rate with Odd Months

• Converts months to yearsDate Month

Fractional Year

Decimal Year Total Years

April 1988 4th 4/12 0.33 1988.33Feb 1995 2ndJune 2001 6thNov 1997 11thJan 1978 1stMar 2007 3rd

Thickness DataTML March 2004 Oct 2001N-2 0.275" 0.288"

Thickness DataTML Sept 2003 Apr 1999N-4 0.521" 0.565"

Calculate the CR

Long Term & Short Term CR

Calculate Short Term and Long Term CR and Remaining Life• Step 1 – Calculate Short Term CR

STCR = (tprevious –tcurrent)/time period• Step 2- Calculate Long Term CR

LTCR = (tinitial – tcurrent)/time period• Step 3- Pick the Controlling Corrosion Rate (highest value)• Step 4 – Calculate Remaining Life

RL = (tcurrent – tmin)/ corrosion rate

Inspection Data Recordt minimum Jan 2003 Jan 1998 Jan 1995 Jan 1993

0.380" 0.440 0.465 0.480 0.500

The Problem

• A vessel head in June 1993 was 0.698 inches thick. Previous thickness in June 1988 at the same area was 0.713 inches. The required thickness of the head is 0.653 inches.

What is the internal inspection

interval?

Corrosion Rate (CR)

• How do we calculate the corrosion rate?

CR = t previous – tactual period of time

The Problem!CR = 0.713 – 0.698

1998 – 1993 = 0.003 inch/year or ipy

Remaining Corrosion Allowance

• How do we calculate the Remaining Corrosion Allowance (RCA)

RCA = tactual – tmin

The Problem!RCA = 0.698 -0.653 = 0.045 inches

Remaining Life (RL)

• How do we calculate the Remaining Life (RL)RL = tactual – tmin

corrosion rate= RCA/CR

The Problem!RL = 0.045/0.003RL = 15 yrs

Calculate the Interval

• Calculate the Internal Inspection interval based on the info in the table

Vessel NoActual

Thickness (inches)

Required Thickness (inches)

Remaining CA (inches)

Corrosion Rate (ipy)

Remaining LifeInternal

Inspection Interval (years)

V-100 0.475 0.4 0.008V-101 0.322 0.309 0.003V-102 1.652 0.453 0.012V-103 0.455 0.405 0.015V-104 1.342 1.225 0.002V-105 0.142 0.158 0.01V-106 0.739 0.543 0.011V-107 0.345 0.321 0.004

Test Your Skill!

• Calculate the next inspection date

Inspection Data Recordt minimum May 2003 Oct 1998 May 1995 Jan 1991

0.395" 0.454 0.477 0.492 0.500

Inspection Data Recordt minimum Feb 2003 May 1997 Jun 1992 Nov 1985

0.212" 0.288" 0.296" 0.305" 0.322"

Problem 9

A pressure vessel shell has a measured thickness of 0.678 inches and required thickness of 0.570 inches. The short term CR 0.023 ipy and the long term CR 0.015 ipy. What is the remaining life of the shell?

1) 7.2 yrs2) 6 yrs3) 8.123 yrs4) 4.695 yrs

Problem 10

The bottom head of a vessel has had the following thickness readings taken since being put in service in 1997. 1997=0.75”, 1999=0.72”, 2000=0.65”, 2002=0.623”. The required thickness is 0.553 inches. What is the next inspection interval?

1) 2 yrs2) 2.755 yrs3) 6.2 yrs4) 10 yrs

Problem 11

The following reading were taken for a pressure vessel. New in 1992 – 0.689”,

1994-0.527”, 1996-0.5”, 1998-0.467”.

a) What is the short term corrosion rate?1)0.036 ipy 2) 0.006 ipy 3) 0.016 ipy 4)0.037 ipy

b) Long term corrosion rate1) 0.037 ipy 2)0.0235 ipy 3)0.1 ipy 4)0.001 ipy


29. When repairs and alterations are to be performed on a pressure vessel, all materials and all welding procedures that are to be used must be approved by;

1) The insurance carrier for the company that the pressure vessel belongs to and the owner user of the pressure vessel

2) The owner-user and the contractor performing the repairs or alterations to the pressure vessel.

3) The API authorized pressure vessel inspector and if necessary by engineer experienced in pressure vessel design, fabrication or inspection

4) The original vessel fabricator and the insurance carrier for the company tha owns the pressure vessel

(API 510 – 9th edition 8.12)


30. What type of repairs can an authorized inspector give prior general authorization for?

1) Major repairs that require pressure test2) Alterations that require pressure tests3) Major alteration that requires pressure test4) Limited to routine repairs that will not require pressure test(API 510 – 9th edition 8.11)


110. Who is notified when a repair to an ASME Section VIII Div 2 vessel is

required?

1) Pressure vessel engineer2) API inspector3) Pressure vessel engineer & API inspector4) Manufacturer of the vessel

(API 510 - 8.1.1)


32. Who should be consulted before repairing a crack at discontinuity, where stress concentrations may be serious

1) The operators of the vessel2) The owner-users3) An engineer experienced in the operations of the vessels4) An engineer experienced in pressure vessel design(API 510 – 8.1.5.2.1)


33. All repair and alteration welding shall be in accordance with the applicable requirements of the

• NBIC codes• AWS D1.1 Welding Codes• ASME Code• NFPA Welding Code(API 510 – 9th edition 8.1.6.1)


34. The repair organization should use qualified welders and welding procedures qualified in accordance with the applicable requirements of;

1) Section V of ASME codes2) Section IX of ASME Codes3) AWS D1.1 Welding Codes4) API Standard 1104, Welding(API 510 – 9th edition 8.1.6.2.1)


35. The repair organization shall maintain records of its qualified welding procedures and its welding performance qualifications. These records shall be available to the _______ prior to start of the welding.

1) Operator2) Owner-user3) Welder4) Inspector(API 510 – 9th edition 8.1.6.2.1)


80. When can pre-heat at 300F in lieu of PWHT not be applied in the repairs of a pressure vessel?

1) For P-1 materials2) For P-3 materials3) For P-4 materials4) None of the above(API 510-8.1.6.4.2.2)


36. For alterations or repairs of vessels initially postweld heat treated as a code requirement and constructed of P-1 and P-3 steels listed in the ASME Codes, preheating to not less than ______degrees F may be considered as an alternative to PWHT when impact testing is not required.

1) 2002) 3003) 4004) 500(API 510 – 8.1.6.4.2.2.2)


37. A carbon steel (A 285 Gr C) vessel must be repaired by welding in a flush patch (replacing a corroded area). Impact testing is not required. The vessel is in caustic service and was originally PWHT. Which of the following is correct?

• The repair may be PWHT• The repair may be pre-heated to 300F while welding and PWHT

waived• The repair may be pre-heated to 200F while welding and PWHT

waived• No 1&2 above(API 510 – 8.1.6.4.2.2)


115. How long must the temper bead repair be held at temperature after

completion of the weld repair for an API 510 application?

1) 5 hours2) 1 hour3) 4 hours4) 2 hours

(API 510 – 8.1.6.4.2.3)


130. What is the maximum interpass temperature for a repair weld using the controlled deposition method that was originally PWHT?

1) 400 F2) 600 F3) 300 F4) Not greater than that used in procedure qualifications

(API 510 – 8.1.6.4.2.3 f-4)


38. If local PWHT is approved for a vessel repair ( a complete 360 degree band

around the vessel is not used-only a localized spot), what are the minimum number of thermocouples required around the localized area to monitor the temperatures?

1) 12) 23) 44) 6(API 510 – 9th edition 8.1.6.4.1)


39. When repairing vessels with stainless steel weld overlay and cladding (vessel

constructed of P-3, P-4 or P-5 base materials) the base metal in the area of

repair should be examined for cracking by per ASME Section V. this UT inspection

should be made _______ hours after repairs have been completed for equipment

in hydrogen service, especially for chromium-molybdenum alloys that could be

affected by delayed cracking• 12• 24• 36• 42 (API 510 – 8.1.5.4.4)


76. The pressure vessel engineer must additionally consider which of the following

when welding a P-5 base material to stainless cladding which is in hydrogen

service at elevated temperatures?1) Weld rod diameter2) Effects of temper embrittlement3) Outgassing base metals4) Weld bead size(API 510 – 9th edition 8.1.5.4.3)


116. Which of the following must be considered by the pressure vessel

engineer when repairs are made to stainless steel weld overlay for

equipment in hydrogen service?

1) Hardening of the weld metal due to grinding or arc gouging2) Softening of base metal due to welding3) PWHT to reduce impact properties4) Outgassing base metal

(API 510 - 9th edition 8.1.5.4.3)


116. Which of the following must be considered by the pressure vessel

engineer when repairs are made to stainless steel weld overlay for

equipment in hydrogen service?

1) Hardening of the weld metal [base metal] due to grinding or arc gouging

2) Softening [Hardening] of base metal due to welding3) PWHT to reduce impact properties [reduce hardness]4) Outgassing base metal

(API 510 - 9th edition 8.1.5.4.3)


112.Which NDE method is used to examine P-3, P-4 or P-5 base metal when repairs to stainless steel weld overlay is made?

1) MT2) UT3) RT4) VT

(API 510 – 8.1.5.4.4)


107. Which of the following is a requirement when installing a patch?

1) Always perform pressure test after repair2) Having rounded corners3) Ultrasonically examine adjacent base materials before

welding4) Having rounded corners with a minimum 1 in radius

(API 510 – 8.1.5.1.2.3)


108. Who decides how long a temporary patch may remain in place?

1. Owner user2. Pressure vessel engineer and API inspector3. Pressure vessel engineer4. API inspector

(API 510 – 8.1.5.1.1)


40. When are fillet welded patches (lap patches) allowed?1) They may be only used in vessels with shells 3/8 inch thick or less2) They may be only used if approved by the operators3) They may be only used on low pressure vessels4) They are used for only temporary repairs(API 510 – 8.1.5.1.2)


41. Carbon or alloy steel with a carbon content over _____percent shall not be

welded.1) 0.32) 0.353) 0.44) 0.45(API 510 – 8.1.4)


42. Acceptance criteria for a welded repair or alteration should include NDE

techniques that are in accordance with the;1) Applicable section of the NBIC2) Applicable section of the ASME codes3) Jurisdiction4) Owner user(API 510 – 8.1.7.4)


86. If a filler metal is used that has minimum specified tensile strength that is lower than that of the base metal, which of the following condition must be met?

1. The increased thickness shall have rounded corners2. The repair thickness shall not be more than 60% of the

required base metal thickness excluding corrosion allowance3. The repair shall be made with a minimum of two passes4. The chemistry of the filler metal shall be taken at the

completion of welding

(API 510 – 8.1.5.3)


44. When is re-rating of a pressure vessel considered complete?1) When the pressure vessel engineers approves the re-rating2) When the authorized construction organization attaches the

nameplates to the re-rated vessel3) When API authorized PV inspector oversees the attachment of an

API 510 nameplate or stamping4) When the owner-user accepts the re-rating from the re-rating

organizations

(API 510 – 8.2.2)

Appendix BAPI Document Practices Question

45. An API certified inspector who has not been actively engaged, as such within

the previous three years can be rectified by;• Being employed by a refinery• Being licensed by the jurisdictions• An oral examination• A written examination(API 510 – B3.1)

That’s is all!

002 api 510 course puspatri apr 09

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