Understanding api icp653 reading 4 worksheet 01a

Charlie Chong/ Fion Zhang


Petroleum Tank


LNG Tank


Fi Zh t Sh h iFion Zhang at Shanghai27th December 2015


API 653 Exam Administration -- PublicationsAPI 653 Exam Administration Publications Effectivity Sheet FOR: November 2015, March 2016 and July 20162016 and July 2016 Listed below are the effective editions of the publications required for this exam for the date(s) shown above.

d d h ff dAPI Recommended Practice 571, Damage Mechanisms Affecting Fixed Equipment in the Refining Industry, Second Edition, April 2011


ATTENTION: Only the following sections / mechanisms from RP 571 are included on the exam:

Section 3, Definitions P B i l FPar. 4.2.7 Brittle Fracture 4.2.16 Mechanical Fatigue 4.3.2 Atmospheric Corrosion 4.3.3 Corrosion Under insulation (CUI) 4.3.8 Microbiologically Induced Corrosion (MIC) 4.3.9 Soil Corrosion 4 3 94.3.10 Caustic Corrosion 4.5.1 Chloride Stress Corrosion Cracking (Cl-SCC) 4.5.3 Caustic Stress Corrosion Cracking (Caustic Embrittlement)4.5.3 Caustic Stress Corrosion Cracking (Caustic Embrittlement) 5.1.1.10 Sour Water Corrosion (Acidic) 5.1.1.11 Sulfuric Acid Corrosion


• API Recommended Practice 575, Inspection of Atmospheric and Low-Pressure Storage Tanks, Third Edition, April 2014

• API Recommended Practice 577 – Welding Inspection and Metallurgy, Second Edition December 2013Second Edition, December 2013

• API Standard 650, Welded Tanks for Oil Storage, Twelfth Edition, March 2013 with Addendum 1 (September 2014), Errata 1 (July 2013), and Errata 2 (December 2014).

• API Recommended Practice 651, Cathodic Protection of Aboveground Petroleum Storage Tanks, Fourth Edition, September 2014.Petroleum Storage Tanks, Fourth Edition, September 2014.

• API Recommended Practice 652, Lining of Aboveground Petroleum Storage Tank Bottoms, Fourth Edition, September 2014 API St d d 6 T k I ti R i Alt ti d• API Standard 653, Tank Inspection, Repair, Alteration, and Reconstruction, Fifth Edition, November 2014.


• American Society of Mechanical Engineers (ASME), Boiler and Pressure Vessel Code, 2013 Edition i. ASME Section V, Nondestructive Examination, Articles 1, 2, 6, 7 and

23 (section SE-797 only)23 (section SE-797 only) ii. Section IX, Welding and Brazing Qualifications (Welding Only)

See end of this study note for API Official BOK


http://independent.academia.edu/CharlieChong1http://www.yumpu.com/zh/browse/user/charliechonghttp://issuu.com/charlieccchong


http://greekhouseoffonts.com/Charlie Chong/ Fion Zhang


The Magical Book of Tank Inspection ICP



闭门练功


W k h tW k h t 作作Worksheet Worksheet 作业作业


Question1:An Aboveground Storage Tank is to be constructed using ASTM A 516 gr 60plates. The plates are normalized. The plates were tested for impact value and the following were the test results The test results are presented for 32and the following were the test results. The test results are presented for 32mm thk plates (Lot A) and 30 mm thk plates (Lot B).

A) Longitudinal Specimen: Three specimens failed at 13 ft-lb, 15 ft-lb and 18 ft-lb respectively

B) Transverse Specimen: Three specimens failed at 12 ft lb 14ft lb and 14ftB) Transverse Specimen: Three specimens failed at 12 ft-lb, 14ft-lb and 14ft-b respectively Your assessment is:

Chose the correct answer:a. Plate A is okay. B is not.b Plate B is okay A is notb. Plate B is okay. A is not.c. Both A and B are okay.d. Both A and B are not okay.


Material A: ASTM A 516 gr 60 plates t=32mmMaterial B: ASTM A 516 gr 60 plates t=30mm

API60 Data:API60 Data:4.2.2 ASTM SpecificationsPlates that conform to the following ASTM specifications are acceptable as long as the plates are within the stated limitations.e) ASTM A516M Grades 380, 415, 450, 485/A516, Grades 55, 60, 65, and 70, for plates to a maximum thickness of 40 mm (1 5 in ) (insert plates andfor plates to a maximum thickness of 40 mm (1.5 in.) (insert plates and flanges to a maximum thickness of 100 mm [4 in.]).


Optional Information:Group IIIA: Figure 4.1a—Minimum Permissible Design Metal Temperature for Materials Used in Tank Shells without Impact Testing (SI)


Table 4.5b—Minimum Impact Test Requirements for Plates (USC) (See Note)


4.2.9.3 An impact test shall be performed on three specimens taken from a single test coupon or test location. The average value of the specimens (with no more than one specimen value being less than the specified minimum value) shall comply with the specifiedbeing less than the specified minimum value) shall comply with the specified minimum value. If more than one value is less than the specified minimum value, or if one value is less than two-thirds the specified minimum value, three additional specimens shall be tested, and each of these must have a value greater than or equal to the specified minimum valueor equal to the specified minimum value.


For:1. Longitudinal Specimen: Three specimens failed at 13 ft-lb, 15 ft-lb and 18

ft-lb respectively 2 Transverse Specimen: Three specimens failed at 12 ft-lb 14ft-lb and 14ft-2. Transverse Specimen: Three specimens failed at 12 ft-lb, 14ft-lb and 14ft-

lb respectively

Your assessment is:Longitudinal Specimen: average= 15.333 ft-lb, OKTransverse Specimen: average=13 333 ft-lb OKTransverse Specimen: average 13.333 ft lb, OK

Chose the correct answer:Pl t A i k B i ta. Plate A is okay. B is not.

b. Plate B is okay. A is not.c. Both A and B are okay.yd. Both A and B are not okay.


For:1. Longitudinal Specimen: Three specimens failed at 13 ft-lb, 15 ft-lb and 16

ft-lb respectively 2 Transverse Specimen: Three specimens failed at 12 ft-lb 14ft-lb and 14ft-2. Transverse Specimen: Three specimens failed at 12 ft-lb, 14ft-lb and 14ft-

lb respectively

Your assessment is:Longitudinal Specimen: average= 14.666 ft-lb, Not OKTransverse Specimen: average=13 333 ft-lb OKTransverse Specimen: average 13.333 ft lb, OK

Chose the correct answer:Pl t A i k B i ta. Plate A is okay. B is not.

b. Plate B is okay. A is not.c. Both A and B are okay.yd. Both A and B are not okay.


Question 2:The following data are presented for an Aboveground Storage Tank:1) Code of construction: API 6502) Tank diameter = 60ft2) Tank diameter = 60ft3) Tank height = 48ft; width of plates used = 8ft4) Maximum fill height = 48ft) g5) Product specific gravity = 0.926) Material of construction; ASTM A 516 gr 607) Corrosion allowance = 1/8 in7) Corrosion allowance = 1/8 in

Determine:a. Required thickness for bottom course (first course)b. Required thicknesses for courses 2 to 6


5.6 Shell Design5.6.1 General5.6.1.1 The required shell thickness shall be the greater of the design shell thickness including any corrosion allowance or the hydrostatic test shellthickness, including any corrosion allowance, or the hydrostatic test shell thickness, but the shell thickness shall not be less than the following:

The required minimum shell thickness ≥6mm


Worksheet: Code API650:2013The EQ: t = 2.6(H-1)DG/SEE assumed =1S derived from table 5 2b: S =21300psi S =24000psiS derived from table 5.2b: Sd=21300psi, St=24000psi


The shell thickness calculation

Course Height Calculation, t= t for Sd

21300t + c t for St

24000Shell Thicknessmm

1st 8 2.6( 8-1)DG/SE 0.0472 0.1722 0.0419 4.373 use 6mm

2nd 16 2.6(16-1)DG/SE 0.1011 0.2261 5.742 used 6mm

3rd 24 2.6(24-1)DG/SE 0.1550 0.28 7.112 used 8mm

4th 32 2.6(32-1)DG/SE 0.2089 0.334 8.48 used 9mm(10mm)

5th 40 2.6(40-1)DG/SE 0.2629 0.388 9.85 used 10mm

Bottom 48 2.6(48-1)DG/SE 0.3168 0.442 11.22 used 12mm


Question 3: Is the tank safe until next inspection?pData: Code of Constr. API 650 – Basic. Year of Constr. 19841) Tank diameter = 100 feet2) Tank height = Maximum liquid height = 56 feet2) Tank height = Maximum liquid height = 56 feet3) Course width = 8 feet4) Product specific gravity = 0.98) p g y5) Corrosion rate = 0.1 mm / year6) Thickness provided = 20mm (new)7) Material = ASTM A 367) Material = ASTM-A-368) After 20 years, a patch was observed on the second shell course, at 46

feet from the top (i.e. in 2nd course). Least thickness in the area of corrosion was 15.25mm (0.6”). The next interval for inspection is 10 years.

9) Readings in vertical planes a, b, c, d over the length L were:


Optional Information:

L = 3.7√(Dt2) = 3.7 √(100x15.25/25.4) = 28.7”

Where:L is the maximum vertical length, in inches, over which hoop stresses are assumed to “average out” around local discontinuities;D is the tank diameter, in feet;t2 is the least thickness in inches in an area of corrosion exclusive of pitst2 is the least thickness, in inches, in an area of corrosion, exclusive of pits.

NOTE The actual vertical length of the corroded area may exceed L.


Worksheet:The taverageLa = 18.2mmL = 18 1mmLb = 18.1mmLc = 17mmLd = 18mmUse least of L, used taverage = 17mm

A36 API653 Table4.1 Sd=24900psi St=27400psiA36, API653 Table4.1, Sd 24900psi, St 27400psi


t’min = 2.6(H)DG/SE = 2.6(46)100 x 0.98/ (24900) = 0.471”tmin = t’min+ c = 11.96mm (for hydraulic integrity)(c, corrosion allowance was not given thus assumed=0)

Remaining thickness of existing tank at end of 10 years t10= 17 – 10x0.1mm = 16mmRemaining thickness of least thickness t2 = 15.25 – 10(0.1) = 14.25mm

Answer:Answer: t10> tmin, t2>0.6tmin, OK.


Question 4:Case Study 4: Minimum Thickness of Bottom Tank Evaluation (Question)

An aboveground storage tank provided with sketch plate type bottomg g p p ypconstruction was observed to be having following thickness readings on bottom plate by UT measurement. (The tank does not have leak detection system on bottom )system on bottom.)

Bottom thickness provided when tank was new (excluding corrosion allowance = 10 mmallowance = 10 mm.

Maximum depth of corrosion pits from inside of tank on the tank bottom = 3.5 mm.

Maximum corrosion from bottom side (metal loss) = 4.5 mm. Rate of corrosion on topside of bottom = 100 microns / year. Rate of

corrosion on bottom side = 150 microns / yearcorrosion on bottom side 150 microns / year.

Calculate when the next inspection interval should be due.


MRT = (Minimum of RTbc or RTip) – Or (StPr + UPr)p2.54 = (5.5)- Or (0.1+ 0.15)0.25x Or = 5.5- 2.54O = (5 5-2 54)/0 25 yearsOr = (5.5-2.54)/0.25 yearsOr = 11.84 yearNote: the corrosion allowance= 5.5- 2.54 = 2.96mmNote: the corrosion allowance 5.5 2.54 2.96mm


6.4.2.1 Initial Internal Inspection IntervalThe initial internal inspection intervals for newly constructed tanks and/or refurbished tanks shall be established either per 6.4.2.1.1 or 6.4.2.1.2.6 4 2 1 1 The interval from initial service date until the first internal inspection6.4.2.1.1 The interval from initial service date until the first internal inspection shall not exceed 10 years unless a tank has one or more of the leak prevention, detection, corrosion mitigation, or containment safeguards listed in Table 6.1. The initial internal inspection date shall be based on incremental credits for the additional safeguards in Table 6.1 which are cumulative.


Or = 11.84 yearSafeguard = NoThe interval from initial service date until the first internal inspection shall not exceed 10 years (6 4 2 1 1)exceed 10 years. (6.4.2.1.1)

Used = 10 years if safeguard is none.


Optional Exercise: For tank with Safeguard


If following were assumed:Fiberglass-reinforced lining of the product-side of the tank bottom installed per API RP 652.iv Release prevention barrier installed per API Std 650 Annex Iiv. Release prevention barrier installed per API Std 650, Annex I.

The next inspection interval will be:

10 years (initial) + 5 years (fiberglass lining) + 10 years (release prevention barrier) = 25 years.barrier) 25 years.

Note: the initial is 10 years ≠ Or(11.84 year)


Question 5:Q1. Find the minimum acceptable shell plate thickness of the first course of an AST.tmin = Minimum acceptable thickness ?tmin = Minimum acceptable thickness ?D = Nominal diameter of tank, in feet 120H = Height of tank in feet 48gG = Specific gravity of contents 1S = Maximum allowable stress 21000E = Joint efficiency of tank 85E = Joint efficiency of tank .85

Answer:tmin = 2.6(H-1)DG/SE = 2.6(47)120•1/(21000x.85) = 0.822”


Q2. Find the corrosion rate of an AST given the following information.Original thickness = .875”Measured thickness = .625Years since last thickness measurement = 11Years since last thickness measurement = 11Answer:Corrosion rate= (.875-.625)/11 = 0.0227”/year

Q3. Using the corrosion rate determined in problem number 2, what would the remaining life expectancy of this tank be if the minimum required thickness isremaining life expectancy of this tank be if the minimum required thickness is 0.436”?Answer:T ( 625 436)/ 0 0227 8 3T = (.625-.436)/ 0.0227 = 8.3years

Q4. Shell settlement evaluation measurements are to be taken on a 146 feet Qdiameter AST. Given the following formula, how many elevation measurements would be required?Answer:Answer:N = D/10 = 14.6 used 16 (even number)


Q4. Shell settlement evaluation measurements are to be taken on a 146 feet diameter AST. Given the following formula, how many elevation measurements would be required?

N = D/10 = 14.6 used 16 (even number)

API653: 12 5 2 Initial Settlement SurveyAPI653: 12.5.2 Initial Settlement SurveyWhen a settlement survey is required in accordance with 12.5.1, the tank settlement shall initially be surveyed with the tank empty, using an even

b f l ti t i t N if l di t ib t d d thnumber of elevation measurement points, N, uniformly distributed around the circumference. An initial settlement survey, prior to the first hydrostatic test, provides baseline readings for future settlement evaluation. In the absence ofprovides baseline readings for future settlement evaluation. In the absence of this initial survey, the tank shall be assumed to be initially level.

N is the minimum required number of settlement measurement points, but no q p ,less than eight. All values of N shall be rounded to the next higher even whole number. The maximum spacing between settlement measurement points shall be 32 ftshall be 32 ft.


Q5. Using the information in number four, does this number meet the requirements of API-653, which does not permit the measurement points to exceed 32 feet?Answer:Answer:Circumference= πD= 146π = 459’Distance between each point = 459/16 = 28.7’ OK

Q6. Find the throat of a right angle fillet weld with a weld leg length of .4375 inchesinches.Answer:.4375 x sin45 = 0.31”

Q7. Using the following information, what is the maximum vertical length (also known as critical length), in inches, over which hoop stresses are assumed to g ), , p“average out” around local discontinuities? Diameter of the tank is 136 feet, the least thickness in an area of corrosion, exclusive of pits, is determined to be 563”be .563” Answer:L = 3.7√(136x.563) = 32.37”


( )Note: need not exceed 40”

Q8. A 6 inch nozzle is to be placed in the tank shell of an AST. The shell course where the nozzle will be installed is .875” thick. What is the required area of reinforcement?Answer:Answer:Areinforce= ODopening x .875 = 6.75 x .875 = 5.91n.2


Figure 5.8- Shell Nozzles (see Tables 5.6a, 5.6b, 5.7a, 5.7b, 5.8a, and 5.8b)


API650: 5.7.2 Reinforcement and Welding5.7.2.1 Openings in tank shells larger than required to accommodate a NPS 2 flanged or threaded nozzle shall be reinforced. The minimum cross-sectional area of the required reinforcement shall not be less than the product of thearea of the required reinforcement shall not be less than the product of the vertical diameter of the hole cut in the shell and the nominal plate thickness,Area= ODopening x tshell but when calculations are made for the maximum required thickness considering all design and hydrostatic test load conditions, the required thickness may be used in lieu of the nominal plate thickness. The cross-sectional area of the reinforcement shall be measured verticallycross sectional area of the reinforcement shall be measured vertically, coincident with the diameter of the opening.5.7.2.2 The only shell openings that may utilize welds having less than full

t ti th h th h ll th th t d t i i f t dpenetration through the shell are those that do not require reinforcement and those that utilize a thickened insert plate as shown in Figure 5.7b and Figure5.8. However, any openings listed in Table 3 of the Data Sheet that are , y p gmarked “yes” under “Full Penetration on Openings” shall utilize welds that fully penetrate the shell and the reinforcement, if used.


API650: Table 5.6b—Dimensions for Shell Nozzles (USC)


Q9. The maximum ultrasonic thickness measurement interval shall be five years when the corrosion rate is not known. What is the interval based on aknown corrosion rate? The tank information is as follows: Initial thickness ofthe tank was 1 0625” the actual thickness is 815” the calculated requiredthe tank was 1.0625 , the actual thickness is .815 , the calculated requiredthickness is .473”, the tank has been in service for 16 years with nomeasurements taken since it was installed.

Calculation:Interval base for known corrosion rate is 15 years or RCN/2NInterval base for known corrosion rate is 15 years or RCN/2NCorrosion rate= (1.0625-0,815)/16 = 0.0155”/yearRemaining life= RCN/2N = (0.815-0.473)/(2x 0.0155)= 11 years

Answer:T= 11 yearsy


Q10. What is the circumference of an AST that is 96 feet in diameter?Answer:πD = π96= 302’

Q11. Find “S” when tmin, D, H, G, and E are known where: D=78, H=29, G = 1, E=.85, and tmin =.602Answer:t = 2.6(H-1)DG/SE, S= 2 6(28)78/ ( 602x 85) = 11097psi.S 2.6(28)78/ (.602x.85) 11097psi.

t’ = 2.6(H)DG/SE, S 2 6(29)78/ ( 602 85) 11493 iS= 2.6(29)78/ (.602x.85) = 11493psi.

Comment: which is which?


Q12. Find E when D, H, G, S, G, and tmin are known where: D =128, H=60, S = 22,300 G = .87, and tmin = 1.125Answer:t’ = 2 6(H)DG/SEt = 2.6(H)DG/SE,E = 0.692.

Q13. Find “D” when H, E, S, G, and tmin are known where: H=45, E=.90, S=22500, G= .7 and tmin =.9375Answer:Answer:t’ = 2.6(H)DG/SE,D = 231’

Q14. Find “St” when E, Ht, D, G, and tmin are know where: E = .80, Ht = 54, D= 70, G = 1 and tmin =.4325,Answer:t’ = 2.6(H)DG/SE,St = 2 6(H)DG/Et’ = 28404psiSt = 2.6(H)DG/Et’ = 28404psi


Q15. Find “C” when D and H are know where D = 60, H = 54Workout:A.1.5 Typical sizes, capacities, and shell-plate thicknesses are listed in Tables A 1a through A 4b for a design in accordance with A 4 (joint efficiencyTables A.1a through A.4b for a design in accordance with A.4 (joint efficiency = 0.85; design specific gravity = 1.0; and corrosion allowance = 0). In USC units:

2C = 0.14D2H where C is the capacity of tank 42-gal barrels;C is the capacity of tank, 42 gal barrels; D is the diameter of tank, in ft (see A.4.1); H is the height of tank, in ft (see A.4.1).

Answer:C= 0.14x 602 x 54 = 27216 barrels


Q16. What is the required hydrostatic test height (Ht) of an AST given the following data?

Maximum allowable hydrostatic stress (St)= 21350Maximum allowable hydrostatic stress (St)= 21350Diameter of the tank = 110 feetHydrostatic test shell thickness (tmin) is .625”Shell joint efficiency = 70%

Answer:Answer:t = 2.6HtD/StE Ht = StEt/ (2.6D) = 32.66’


Q17. What is the area of a tank bottom that is 121 feet in diameter?Answer:Area= π (112/2)2 = 9852ft.2

Q18. A reinforcing plate is installed around a nozzle installed in the first course of an AST. What is the throat dimension of the ½” leg fillet weld around the periphery of the reinforcing plate.Answer:Throat= ½ x sin45 = 0 35”Throat ½ x sin45 0.35


Q19. What is the minimum acceptable thickness of the bottom course of an existing AST that is 135 feet in diameter, maximum fill height is 56.5 feet? The product stored in the tank has a specific gravity of .87 and all shell butt- welds were 100% radiographically examined The yield strength of the material iswere 100% radiographically examined. The yield strength of the material is 28,000 and the tensile strength is 60,000. Worksheet:API653: 4.3.3.1- S is the maximum allowable stress in pound force per square inch (lbf/in.2); use the smaller of 0.80Y or 0.429T for bottom and second course; use the smaller of 0 88Y or 0 472T for all other coursessecond course; use the smaller of 0.88Y or 0.472T for all other courses. Allowable shell stresses are shown Table 4.1 for materials listed in the current and previous editions of API 12C and API 650;

■ Sd= 0.8Y=22400psi■ Sd= 0.429T= 25740psi

Answer:t = 2.6(H-1)DG/SE = 2.6(55.5)135x0.87/ (22400) = 0.757”( ) ( ) ( )


Q20. What is the required area of reinforcement for a 14 NPS nozzle installed in a shell course that is 1.0625” thick. Use API 650 Paragraph 3.7.2.1Workout:5 7 2 1 Openings in tank shells larger than required to accommodate a NPS 25.7.2.1 Openings in tank shells larger than required to accommodate a NPS 2 flanged or threaded nozzle shall be reinforced. The minimum cross-sectional area of the required reinforcement shall not be less than the product of the vertical diameter of the hole cut in the shell and the nominal plate thickness,but when calculations are made for the maximum required thickness considering all design and hydrostatic test load conditions the requiredconsidering all design and hydrostatic test load conditions, the required thickness may be used in lieu of the nominal plate thickness. The cross-sectional area of the reinforcement shall be measured vertically, coincident

ith th di t f th iwith the diameter of the opening.Answer:A= 14 1/8 x 1.0625 =15.01in.2



Q21. A bulge is noted in a tank bottom, what is the maximum height of the bulge permitted if the radius of the inscribed circle is 10.8 feet? Workout:B 3 3 Internal Bottom Settlements or BulgesB.3.3 Internal Bottom Settlements or BulgesMeasure the bulge or depression. The permissible bulge or depression is given by the following equation (see Note).BB = 0.37RwhereBB is maximum height of bulge or depth of local depression in inches;BB is maximum height of bulge or depth of local depression, in inches;R is radius of inscribed circle in bulged area or local depression, in feet.Answer:B 0 37R 0 37(10 8) 4”BB = 0.37R = 0.37(10.8)= 4”


Figure B.9- Localized Bottom Depressions or Bulges Remote from Shell


Q22. If the throat of a fillet weld is .309, what is the leg length of the weld?Answer:L = .309/ sin45 = 0.437”

Q23. An AST is 175 feet in diameter. How many tank settlement survey points are required? Use API-653 paragraph 10.5.1.2.Answer:N=175/10 = 17.5 used 18 (even number)

Q24. Do the number of settlement measurement locations and maximum spacing comply with API –653? Use API –653 Paragraph 10.5.1.2W k tWorkout:Circumference= πD= 175π= 550ft,Spacing = 550/18 = 30.54’ p gAnswer: Spacing less than 32’ OK

Quoted from: Figure B.1- Measurements of Shell Settlement (External)g ( )NOTE 1 There must be at least eight settlement points. The maximum spacing of the settlement points is 32 ft around the circumference.


Q25. What is the design liquid level of an AST that is 140 feet in diameter, thickness of the bottom or first course is .765”, stress of the material is 21300. The specific gravity of the product to be stored is 1 and no corrosion allowance is needed nor providedallowance is needed nor provided.Answer:tmin= 2.6(H-1)DG/SE + c(H-1)= .765x 21300/(2.6x 140) = 44.765’, H=45.765’






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Understanding api icp653 reading 4 worksheet 01a