Understanding api icp653 reading 4 worksheet 01a

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Petroleum Tank

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LNG Tank

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Fion Zhang Fi Zh att Shanghai Sh h i 27th December 2015

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

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

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ATTENTION: Only the following sections / mechanisms from RP 571 are included on the exam: Section 3, Definitions P 4.2.7 Brittle Par. B i l Fracture F 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 4 3 9 Soil Corrosion 4.3.10 Caustic Corrosion 4.5.1 Chloride Stress Corrosion Cracking (Cl-SCC) 4.5.3 Caustic Stress Corrosion Cracking (Caustic Embrittlement) 5.1.1.10 Sour Water Corrosion (Acidic) 5.1.1.11 Sulfuric Acid Corrosion

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API Recommended Practice 575, Inspection of Atmospheric and LowPressure Storage Tanks, Third Edition, April 2014 API Recommended Practice 577 – Welding Inspection and Metallurgy, Second Edition 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.

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

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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) ii. Section IX, Welding and Brazing Qualifications (Welding Only)

See end of this study note for API Official BOK

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http://independent.academia.edu/CharlieChong1 http://www.yumpu.com/zh/browse/user/charliechong http://issuu.com/charlieccchong

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http://greekhouseoffonts.com/

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The Magical Book of Tank Inspection ICP

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闭门练功

W k h t 作业 Worksheet 作

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Question1: An Aboveground Storage Tank is to be constructed using ASTM A 516 gr 60 plates. The plates are normalized. The plates were tested for impact value and the following were the test results results. The test results are presented for 32 mm 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 ft-lb, lb 14ft 14ft-lb lb and 14ft 14ftb respectively Your assessment is: Chose the correct answer: a. Plate A is okay. B is not. b Plate B is okay b. okay. A is not not. c. Both A and B are okay. d. Both A and B are not okay.

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Material A: ASTM A 516 gr 60 plates t=32mm Material B: ASTM A 516 gr 60 plates t=30mm API60 Data: 4.2.2 ASTM Specifications Plates 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 (1 5 in in.)) (insert plates and flanges to a maximum thickness of 100 mm [4 in.]).

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

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

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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 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 value value.

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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 2. ft-lb, 14ft-lb and 14ftlb respectively Your assessment is: Longitudinal Specimen: average= 15.333 ft-lb, OK Transverse Specimen: average average=13 13.333 333 ft ft-lb lb, OK Chose the correct answer: a. Plate Pl t A iis okay. k B iis not. t b. Plate B is okay. A is not. y c. Both A and B are okay. d. Both A and B are not okay.

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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 2. ft-lb, 14ft-lb and 14ftlb respectively Your assessment is: Longitudinal Specimen: average= 14.666 ft-lb, Not OK Transverse Specimen: average average=13 13.333 333 ft ft-lb lb, OK Chose the correct answer: a. Plate Pl t A iis okay. k B iis not. t b. Plate B is okay. A is not. y c. Both A and B are okay. d. Both A and B are not okay.

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Question 2: The following data are presented for an Aboveground Storage Tank: 1) Code of construction: API 650 2) Tank diameter = 60ft 3) Tank height = 48ft; width of plates used = 8ft g = 48ft 4)) Maximum fill height 5) Product specific gravity = 0.92 6) Material of construction; ASTM A 516 gr 60 7) Corrosion allowance = 1/8 in Determine: a. Required thickness for bottom course (first course) b. Required thicknesses for courses 2 to 6

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5.6 Shell Design 5.6.1 General 5.6.1.1 The required shell thickness shall be the greater of the design shell thickness including any corrosion allowance thickness, allowance, or the hydrostatic test shell thickness, but the shell thickness shall not be less than the following:

The required minimum shell thickness ≼6mm

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

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The shell thickness calculation Course

Height

Calculation, t=

t for Sd 21300

t+c

t for St 24000

Shell Thickness mm

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

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Question 3: Is the tank safe until next inspection? p Data: Code of Constr. API 650 – Basic. Year of Constr. 1984 1) Tank diameter = 100 feet 2) Tank height = Maximum liquid height = 56 feet 3) Course width = 8 feet p g gravity y = 0.98 4)) Product specific 5) Corrosion rate = 0.1 mm / year 6) Thickness provided = 20mm (new) 7) Material = ASTM-A-36 ASTM A 36 8) 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:

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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 thickness, in inches inches, in an area of corrosion corrosion, exclusive of pits pits.

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

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Worksheet: The taverage La = 18.2mm Lb = 18.1mm 18 1mm Lc = 17mm Ld = 18mm Use least of L, used taverage = 17mm A36 API653 Table4.1, A36, Table4.1 Sd=24900psi 24900psi, St=27400psi 27400psi

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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 = 16mm Remaining thickness of least thickness t2 = 15.25 – 10(0.1) = 14.25mm Answer: t10> tmin, t2>0.6tmin, OK.

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Question 4: Case Study 4: Minimum Thickness of Bottom Tank Evaluation (Question) An aboveground g storage g tank p provided with sketch p plate type yp bottom construction was observed to be having following thickness readings on bottom plate by UT measurement. (The tank does not have leak detection system on bottom.) bottom )  Bottom thickness provided when tank was new (excluding corrosion allowance = 10 mm 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 / year year.

Calculate when the next inspection interval should be due.

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MRT = (Minimum of RTbc or RTipp) – Or (StPr + UPr) 2.54 = (5.5)- Or (0.1+ 0.15) 0.25x Or = 5.5- 2.54 Or = (5.5-2.54)/0.25 (5 5-2 54)/0 25 years Or = 11.84 year Note: the corrosion allowance allowance= 5.5 5.5- 2.54 = 2.96mm

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6.4.2.1 Initial Internal Inspection Interval The 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 inspection 6.4.2.1.1 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.

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Or = 11.84 year Safeguard = No The interval from initial service date until the first internal inspection shall not exceed 10 years years. (6.4.2.1.1) (6 4 2 1 1) Used = 10 years if safeguard is none.

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Optional Exercise: For tank with Safeguard

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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 iv.

650, Annex II. The next inspection interval will be: 10 years (initial) + 5 years (fiberglass lining) + 10 years (release prevention barrier) = 25 years.

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

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Question 5: Q1. Find the minimum acceptable shell plate thickness of the first course of an AST. tmin = Minimum acceptable thickness ?

D = Nominal diameter of tank, in feet 120 g of tank in feet 48 H = Height G = Specific gravity of contents 1 S = Maximum allowable stress 21000 E = Joint efficiency of tank .85 85 Answer: tmin = 2.6(H-1)DG/SE = 2.6(47)120•1/(21000x.85) = 0.822�

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Q2. Find the corrosion rate of an AST given the following information.

Original thickness = .875” Measured thickness = .625 Years since last thickness measurement = 11 Answer: 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 is 0.436”?

Answer: T = (.625-.436)/ ( 625 436)/ 0.0227 0 0227 = 8.3years 83 Q Shell settlement evaluation measurements are to be taken on a 146 feet Q4. diameter AST. Given the following formula, how many elevation measurements would be required?

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

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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 12.5.2 5 2 Initial Settlement Survey When 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 number b off elevation l ti measurementt points, i t N, N uniformly if l di distributed t ib t d around d th the circumference. An initial settlement survey, prior to the first hydrostatic test, provides 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 q number of settlement measurement p points,, but no 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 ft ft.

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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: 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 inches. inches Answer: .4375 x sin45 = 0.31” Q7. Using the following information, what is the maximum vertical length (also g ), in inches,, over which hoop p stresses are assumed to known as critical length), “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” 563” Answer: ( ) = 32.37” L = 3.7√(136x.563) Note: need not exceed 40” Charlie Chong/ Fion Zhang

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: Areinforce= ODopening x .875 = 6.75 x .875 = 5.91n.2

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

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API650: 5.7.2 Reinforcement and Welding 5.7.2.1 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, 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 cross sectional area of the reinforcement shall be measured vertically, vertically coincident with the diameter of the opening.

5.7.2.2 The only shell openings that may utilize welds having less than full penetration t ti through th h th the shell h ll are th those th thatt d do nott require i reinforcement i f t and d those that utilize a thickened insert plate as shown in Figure 5.7b and Figure p g listed in Table 3 of the Data Sheet that are 5.8. However,, anyy openings marked “yes” under “Full Penetration on Openings” shall utilize welds that fully penetrate the shell and the reinforcement, if used.

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

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Q9. The maximum ultrasonic thickness measurement interval shall be five years when the corrosion rate is not known. What is the interval based on a known corrosion rate? The tank information is as follows: Initial thickness of the tank was 1.0625 1 0625”, the actual thickness is .815 815”, the calculated required thickness is .473”, the tank has been in service for 16 years with no measurements taken since it was installed.

Calculation: Interval base for known corrosion rate is 15 years or RCN/2N Corrosion rate= (1.0625-0,815)/16 = 0.0155”/year Remaining life= RCN/2N = (0.815-0.473)/(2x 0.0155)= 11 years Answer: y T= 11 years

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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 =.602 Answer: t = 2.6(H-1)DG/SE, S= 2.6(28)78/ S 2 6(28)78/ (.602x.85) ( 602x 85) = 11097psi. t’ = 2.6(H)DG/SE, S 2.6(29)78/ S= 2 6(29)78/ (.602x.85) ( 602 85) = 11493psi.

11493 i Comment: which is which?

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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 = 1.125 Answer: tt’ = 2.6(H)DG/SE, 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 =.9375 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’ 2 6(H)DG/Et’ = 28404psi

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

In USC units: C = 0.14D2H where C is the capacity of tank tank, 42-gal 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

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

Maximum allowable hydrostatic stress (St)= 21350 Diameter of the tank = 110 feet Hydrostatic test shell thickness (tmin) is .625�

Shell joint efficiency = 70% Answer: t = 2.6HtD/StE Ht = StEt/ (2.6D) = 32.66’

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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 0 35”

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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 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 0.88Y 88Y or 0 0.472T 472T for all other courses 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: ( ) = 2.6(55.5)135x0.87/ ( ) ((22400)) = 0.757” t = 2.6(H-1)DG/SE

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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 3.7.2.1 Workout: 5 7 2 1 Openings in tank shells larger than required to accommodate a NPS 2 5.7.2.1 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, conditions the required thickness may be used in lieu of the nominal plate thickness. The crosssectional area of the reinforcement shall be measured vertically, coincident with ith the th diameter di t off th the opening. i Answer: A= 14 1/8 x 1.0625 =15.01in.2

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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 Bulges B.3.3 Measure the bulge or depression. The permissible bulge or depression is given by the following equation (see Note).

BB = 0.37R where BB is maximum height of bulge or depth of local depression depression, in inches; R is radius of inscribed circle in bulged area or local depression, in feet.

Answer: BB = 0.37R 0 37R = 0 0.37(10.8)= 37(10 8) 4�

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

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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 10.5.1.2 W k t Workout: Circumference= πD= 175π= 550ft, p g = 550/18 = 30.54’ Spacing Answer: Spacing less than 32’ OK Quoted from: Figure g B.1- Measurements of Shell Settlement ((External)) NOTE 1 There must be at least eight settlement points. The maximum spacing of the settlement points is 32 ft around the circumference.

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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 provided 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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