Day - 2 api

API 510 IMPORTANT TIPS

UG

UG 4 – UG 15 Pressure Part Material Testing UG 10 – code for the material not fully identified UG 16 – UG 20 Pressure Part Design

UG-20 Design Temperature UG-22 Loading UG-25 Corrosion

UG-27 Thickness of Shells Under Internal Pressure UG-28 Thickness of Shells and Tubes (External Pressure) UG-32 Formulas and Rules for Using Formed Heads UG-34 Un stayed Flat Heads and Covers (Circular) UG-36 Openings in Vessels- 3/8 ‘ and 2 3/8 ‘ formula UG-37 Reinforcement of Openings

UG-40 Limits of Reinforcement UG-41 Requirements for Strength of Reinforcement

UG-42 Reinforcement of Multiple Openings UG-77 Material Identification

UG 78 – Repair of effects in the Material UG -80 Permissible out of roundness UG 81 – tolerance for formed Heads

UG 84 Impact force, Ave impact strength, Temperature reduction below MDMT Charpy Impact Test Requirements

UG-93 Inspection of Materials

UG-98 Maximum Allowable Working Pressure UG-99 Hydrostatic Test Pressure and Procedure UG-100 Pneumatic Test Pressure and Procedure

UG-102 Test Gages UG-116 Name Plate Markings UG-119 Name Plates UG-120 Data Reports

UW

UW-3 Weld Categories

UW-11 RT and UT Examinations

UW-12 Maximum Allowable Joint Efficiencies Table

UW-16 Weld Size Determination Minimum Requirements for Attachment Welds at

Openings UW 33 Misalignment tolerance , UW 35 Re- inforcement

UW-40 Procedures for Postweld Heat Treatment UW 50 NDT

UW-51 RT Examination of Welded Joints UW-52 Spot Examination of Welded Joints UW 53 UT

UCS

UCS-56 Requirements for Post weld Heat Treatment UCS-66 Materials- MDMT , Exemption from impact testing UCS-67 Impact Testing of Welding Procedures

UCS-68 Design

Exercises UG 20 –UCS 66 – 67

Tips

Film width – Minimum 5 times the thickness of the thinner plate Film length minimum 6 inch

Film length should be = 3 times the length of opening

RT not required for Cat b & Cat C , if it is below NPS 10 or 1 1/8 inch thick or both

Pressure Tolerance of the Pressure relieving device is 3 % 1. Minimum Thickness For Cylindrical Shell.

P x R t =

P = Mawp

R = Inner Radius S = Stress of the Mass E = Efficiency

T = Thickness of the shell – Minimum

2. Minimum Thickness of Ellipsoid Head

T = PD / (2SE-0.2P ) , Where P = MAWP, D = internal dia , in all these formulae check which is dia or which is radius, inner or outer and whether corrosion allowance has to be included , if so how much

3. Minimum thickness for torispherical Head

T = 0.885 P L / ( SE- 0.1P ) , where L is inside spherical or Crown radius 4. Minimum thickness for Hemispherical Head =

T = PL /( 2SE- 0.2P )

5. ALLOPWABLE External Pressure

Pa = 4B/ 3 ( D0/t) , or Pa = 2AE / ( 3 D0 /t), where Constant A & B values will be given, and E is the modulus of elasticity.

6. Hydrostatic test pressure = 1.3 x MAWP x Stress Ratio 7. Hydrostatic Inspection pressure = Test pressure/ 1.3 8. Pneumatic test pressure = 1.1 x MAWP x Stress Ratio

9. Pneumatic inspection pressure = Test Pressure /1.1, and increase in terms of six steps from 50 % by 10 %.

10. For MDMT use UCS 66 table., for High toughness ie for lower brittlity, use curve D of UCS 66

11. To find out joint efficiency use UW 12

12. For PWHT use UCS 56, PWHT is a must for P no 3 group no3 for all thickness,

13. Write what’s is P no, A no, S no ,F no .and Which table we have to locate 14. Reinforcement = Dia x Thickness

16. Crown Dia

Flat Dish end = o.8 x dia of shell Ellipsoid dish= 0.9 x Dia of shell

Torispherical Head = 1 X dia of the D Shell

Knuckle radius = 6 % crown radius + Thickness of the Head 17. Thickness of the Head

Hemispherical= 0.5 x t Ellipse = 1 x t

Torrispherical= 1.77 t

18. Crown radius is different from crown portion. For crown portion use Flat Disc formulae ie 0.8 x dia, and for crown portion thickness use Hemispherical thickness head formulae.

19. Charpy Impact – Full size specimen 10 mm x 10 mm, Sub size specimen is 10m x 6.7 mm

20. Minimum 3 specimen should be taken. Find the average. If any one is having the value less than the 2/3 of the average, do the retest of 3 more specimen 21. For the Materials having specified minimum tensile strength of 95 Ksi, use

minimum lateral expansion opposite to the notch

22. For reduction in MDMT based on thickness (table) & impact test temperature differential etc refer UG 84 tables

23. When heat treatment is not performed by the manufacturer, fabricator or people under his control has to do and place letter T followed by G

24. For plates MTR ie Material Test Report is a must

25. VT is done at inspection pressure ie test pressure/1.3 for hydrostatic and for test pressure /1.1 f Pneumatic test

26. For Hydrostatic test Small liquid relief valve set to 1.33 times the test pressure is recommended. It may get warmed up during the test when the personnel is absent

27. Marking on the pressure Vassal :

28. U – Pressure vessal, UM- Mimniature Pressure vessal, UV for PRV, and UD for rupture Disc

Full pressure vessel U1 or U1A Pressure Vessel Part U2 or U2A

W – arc weld or gas weld P – Pressure weld

B =-braze

RES – resistance weld L – Lethal service

UB – unfired Steam Boiler DF—directly fired

UB SEC VIII Div 1, DF is SEC VIII Div 2 1- means single chamber

A – means shop fabricated

29. Partial data reports should have the PART mentioned in the sketch. This does not apply to Hand hole cover and man hole cover

30. Certificate of authorization for U,UV,UD is valid for 3 years. For UM it is valid for one year but reviews for second and third year are done by Authorized inspection agency

31. Name plate thickness shall not be less than .5 mm . Characters Height should not be less than 4mm and on a minimum raised level of 0.1 mm

32. The manufacturers certificate of compliance shall be maintained by the manufacturer for 5 years

33. The spring in the PRV shall not deviate by + or – 5% than the marking on the valve.. The set Pressure tolerance hall not exceed 2 Psi up to 70 Psi and 3 % above 70 Psi. Pressure tolerance of PRV device is 3 %

34. RV size should be atleast NPS 1//2

35. The set pressure of PRV shall not exceed 110 % MAWP or 3 psi 36. Rupture disc is a non re closing PRV

37. Acceptance Standards of defects as per Sec VIII Appendix RT APX 4

UT APX 12 MT APX 6 PT APX 8

RT Ace

Rounded Indication

For t > 2 inch Relevant Indication is . 1/16 inch

Maximum size of rounded indication is ¼ t or 5/32 inch which ever is small For thickness . 2 inch, the maximum size is 3/8 inch

Sum of dia of aligned rounded indication should be less than t in thee length of 12t RT Linear Indication For Thk up to 19 mm ¼ inch For Thk 19 mm to 57 mm, it is 1/3 t For Thk over 57 mm , it is 19 mm MT / PT ACE

Relevant . 1/16 inch dia Linear is l > 3w

IT is acceptable , if it is free from a) relevant linear indication

b) relevant rounded indication greater than 3/16 inch

c) Four or more rounded indication gap of less than 1/16 inch

UT : Imperfection having response greater than 20 % has to be investigated Cracks, LP and LF are not allowed

Other imperfection

For Thk up to 19 mm ¼ inch For Thk 19 mm to 57 mm, it is 1/3 t For Thk over 57 mm , it is 19 mm

Manufacturer should make and retain thee UT report until the Manufacturers data report is signed by the inspector

Tolerance for the formed Heads.: The inners surface of Torri spherical, Hemispherical or ellipsoidal head shall be between 5/8 % D and 1 1/4 % D , where D is ID

Pipe Wall thikness 12.5%

General mill tolerance = .01 “ or 6 % thickness which ever is smaller

Ovality Tolerance , 1 % nominal dia + 2 % ID of the opening if the cross section passes through the 1 ID of the opening

Allignment taper : If difference in thickness is ¼ t or 1/8 inch which ever iis less ie within the allowable tolerance use 3;1 taper

Reinforcement Pad not required when the size finished opening does not exceed a) 3 ½ inch for thickness equal to less than 3/8 inch

b) 2 3/8 inch for thickness greater than 3/8 inch

For Calculating the average length for the corrosion over the large area

For the Pipes ID Less than 60 inch , Choose D /2 or 20 inch which ever is less For ID more than 60 inch choose D/3 or 40 inch which ever is less

39. When thickness of CS weld or P no 1 exceeds 16 mm , then PWHT is required

40. seamless weld joints efficiency is 1. for full radiography

E=1 for spot radiography also when 11a- 5b conditions are met

E = 0.85, when spot radiography conditions for UW 11a -5b are not met

44. P1 Gr 1, No Special Restriction NO PHT 45. P3, Gr 3, PWHT is Mandatory

46. If thickness is . 4 inch, and MDMT is less than 120 deg F then impact testing is needed

47. Impact test exemption is : Upper Limit 120 Deg F, Lower limit is -55 deg f for curve and -20 deg F for curve B

48. PWHT is for UCS 56 or UHT 56

49. Filler metal containing more than .06 % of Vanadium shall not be used for the weldments Subjected to PWHT

50. Max weld reinforcement shall not exceed 10 % plate thickness or 1/8 inch which ever is less

51. MTR is to be prepared by the original manufacturer only. Supplier can generate only the certificate of compliance

52. Hydrostatic test has to be conducted at least 30 deg F above MDMT 53. In spot radiography rounded indication is not concerned

54. Leak tightness of the PPRV is checked at the water head 0.5 inch at 90 % set pressure

55. For Short circuit transfer ( GMAW) RT is not required 56. HOT TAPPING

To minimize burn through, the first weld pass to equipment or piping less than 1/4 in. (6.4 mm) thick should be made with a 3/32 in. (2.4 mm) or smaller diameter welding electrode to limit heat input. Subsequent passes should be made with a 1/8 in. (3.2 mm) diameter electrode, or smaller if the metal thickness does not exceed 1/2 in. (12.7 mm).

Minimum base thickness requirements shall be stated in the written documentation for the job. A minimum base metal thickness of 3/16 in. (4.8 mm) is recommended for most applications of welding and hot tapping.

The actual minimum thickness is a function of the thickness required for strength, plus a safety factor, usually 3/32 in. (2.4 mm), to prevent burn through.

57. For Pre heat WPS temp can go down maximum by 100 Deg F than the PQR 58. But For interpass tem this can increase maximum by 100 deg F than PQR

59. For P1,P2 ,P3,P5 materials Lower transformation temp is 1333 Deg F and upper transformation tem is 1600 Deg F

CASE STUDY 2

THICKNESS CALCULATIONS

Corr./Erro. Allow = 2.5 mm (0.1 in) Allowable Stress = 19400 psi

Length (T-T) = 315 in (8000mm) Radiography = Full / Spot

Calculate required thickness, Design thickness and Nominal thickness for shell and heads considering both cases of radiography.

683 x 602.5 19400 x 0.85 – 0.6x683

683 x 1205 2 x 19400 – 0.2 x 683

SOLUTION:

AA) Shell Thickness

t =

= = 21.7……….for E=1

Required thickness = 21.7+2.5 = 24.2 mm. Nominal thickness= 26.0 mm. By similar calculations for E=0.85

t = + 2.5 = 25.5+2.5=28.0mm (E=0.85), Say, 28.0 mm.

BB) Dished Head :

1) Hemspherical t = = = 10.64 mm

(E=1)

Adding C.A, Required thk. = 13.14, Nomonal thk= 14.0 mm

With E = 0.85, t = 12.75mm . Adding C.A , Thk = 15.25. Say, 16.0 mm. 2) 2:1 Ellip. Head With E =1.0, t = = PR SE – 0.6P 683 x 602.5 19400 – 0.6 x 683 P x R 2SE – 0.2P _{2 x 19400 x 1 –0.2x683}683 x 602.5 PD 2SE – 0.2P

0.885 x 683 x 1205 19400 – 0.1 x 683 0.885 x 683 x 1205 19400 x 0.85 – 0.1 x 683 By similar calculations for E=0.85,

t = == 25.5 mm (E=0.85)

Adding C.A. = 28.0 mm. Say, 28 mm plate

3) Torispherical head t = + C L= Dia = 1205 mm. = = 37.5 +2.5 = 40.0 mm Choose thk=40 mm t = = 44.35 + 2.5 = 46.8 mm Say, 47.0 mm. thk COMPARISON OF THICKNESS Hemisph. Item Shell E=1.0 Shell E=0.85 Torisph. E=1.0 Torisph. E=0.85 2:1 Ellip E=1.0 2:1Ellip

E=0.85 E=1.0 E=0.85 1) Calculated _Thickness 21.7 25.5 37.5 44.35 21.28 25.5 10.64 12.75 2) Thk + C.A 25.0 28.0 40.0 48.0 24.0 28.0 14.0 16.0 683 x 1205 2 x 19400 x 0.85 –0.2x683 0.885 x P x L SE – 0.1 P

CASE STUDY 3

MAWP ANALYSIS

DATA:

Following data is presented for a tall process column Vessel ID (actual) = 8’-0”. (96”)

Vessel height (Tan – Tan) = 60’ Types of Dished head = 2:1 Ellip. Corrosion Allowance = Nil

Vessel MAWP (Stamped) = 180 psi Safe stress for material = 20,000 psi Vessel thk = 0.55” (14.0 mm )

Vessel was fully radiographed.

Due to erosion, three patches were observed at locations A,B,C located at 5’, 54’ and 57’ from top-tan line. The thicknesses measured at locations A,B,C were 0.45” (11.4mm), 0.47”(12.0mm), 0.50”(12.7mm).

Calculate:

1. Vessel part MAWPs at A,B and C 2. Hydrostatic Pressure at A,B and C 3. Total Pressure at A, B, C

4. Identify Unsafe patch (es), if any

SOLUTION:

1 Vessel part MAWP = S E t R+0.6t MAWP at A = 20000 x 0.45 = 186.4 psi 48 +0.6x0.45 MAWP at B = 2 20000 x 0.47 = 194.7 psi 48 + 0.6x0.47 MAWP at C = 20000 x 0.5 = 207.0 psi 48 + 0.6x0.5

2 Hydrostatic pressure at any point = h x 0.433 psi (h = height from top in ft.)

Head depth = D/4 = 8/4 = 2’

At A, h = 5’ + 2’ = 7’, Hyd. Pr. = 7’x0.433=3.03psi At B, h = 54’ + 2’ = 56’ Hyd. Pr. = 56’ x 0.433 = 24.25 psi At C, h = 57’ + 2’ = 59’ Hyd. Pr. = 59’ x 0.433 = 25.55 psi 3 Total pressure at any point = Vessel MAWP + Hyd. Pressure

At A, 180 + 3.03 = 183.03 psi At B, 180 + 24.25 = 204.25 psi At C, 180 + 25.55 = 205.55 psi

4 Since Total pressure at B is > part MAWP at B, it is unsafe. 5 For safety, total pressure at B < part MAWP at B

Take total pressure at B = MAWP = 194.7 psi Deducting hydraulic pressure at B = 24.25 psi Safe MAWP for vessel = 170.45

CASE STUDY 4

IMPACT TESTING

DATA:

Verify whether impact testing shall be specified for plates required for following 4 vessels. All vessels are to be Hydrotested after completion and no shock loading is expected during operation.

Vessel A: MOC = SA516 Gr 70 ( Normalised )

MDMT = ( -5°F), Thk = 2.0”

Vessel B: MOC = SA515 Gr 60

MDMT = ( -15°F), Thk = 1”

Vessel C: MOC = SA285 Gr C

MDMT = ( +40°F), Thk = 0.75”

Vessel D: MOC = SA516 Gr 60

MDMT = ( -30°F), Thk = 1”

SOLUTION:

Vessel A: - Refer fig UCS – 66

- The material is represented by curve D MDMT – Thk combination falls almost on curve but not certain. - Go to Table UCS-66,

- Exemption temp >=(-4°F) - Impact testing is required

Vessel B: - From UCS – 66, for curve B, Impact testing is required

- Ref UG – 20 (f), material is PNo. 1 Group 1, Thk <= 1” and MDMT > - 20°F

- Impact testing is exempted

Vessel C: - From UCS-66, for curve A, Impact tecting is required

- Ref UG – 20 ( f ) since thk > 0.5” - Impact testing is required

Vessel D: - From UCS – 66, for curve C Impact testing is required (

transition temp from table= -3 deg.F)

- Since MDMT < - 20° F, Impact testing is not exempt per UG – 20 (f)

- But since PWHT is carried out as service requirement ( and not code requirement, Since thk < 1.25”.),

- UCS-68 (c) is applicable.

Exemption temp = -3 – 30 = - 33°F - Impact testing not required

CASE STUDY 5

PRESSURE TESTING

DATA:

Vessel Design Pressure = 180 psig, Vessel MAWP = 200 psig

Vessel Design temp = 500˚F, Safe stress at Design temp = 16000 psig Vessel MDMT = 40˚ F, Safe stress (from -20˚ F to 200˚ F) =17600 psig Answer the following:

1 What is the recommended minimum test temp. for hydrostatic test ? 2 What is mandatory minimum test temp for Pneumatic test

3 What is Hydrostatic Test pressure?

4 What is minimum inspection pressure for Hydrostatic Testing? 5 What is Pneumatic test pressure ?

6 What is inspection pressure for pneumatic test?

7 What is the first stage of pressurisation for pneumatic test?

8 What is incremental pressure for second stage of pressurization for pneumatic test ?

9 What will be total pressure at the end of 3rd stage?

10 Following 5 pressure gauge ranges are available. Which of the two you will choose for Hydrostatic test?

0–400psi, 0–600 psi, 0–750 psi, 0–1000psi, 0–1200psi

11 For Pneumatic testing which of the two gauges from the given five you find suitable?

12 If ambient temp. is 125˚F, the hydrostatic test inspection shall be carried out at : a. 125˚F

1 The recommended minimum test temp. for hydrostatic test Test temp > MDMT + 30˚F , i.e 70˚F

2 Mandatory minimum test temp for Pneumatic test Test temp > MDMT + 30˚F , i.e 70˚F

3 Hydrostatic Test pressure

Test pressure = 1.3 x MAWP x Stress ratio = 1.3 x 200 x 17600 = 286 psig

16000

4 Minimum inspection pressure for Hydrostatic Testing Insp. Pressure = Test pressure = 286 = 220 psig 1.3 1.3

5 Pneumatic test pressure

Test pressure = 1.1 x MAWP x stress ratio = 1.1 x 200 x 176 = 242 psig

160 6 Inspection pressure for pneumatic test

Inspection Pressure = Test pressure = 242 = 220 psig 1.1 1.1

7 First stage of pressurisation for pneumatic test First stage = 50 % of test pressure = 121 psig

8 Incremental pressure for second stage of pressurization for pneumatic test 10% of test pressure = 24.2 psig

9 Total pressure at the end of 3rd stage 121 + (2 x 24.2) = 169.4 psig

10 The two gauges for Hydrostatic test

Lower limit of range 0 – 1.5x286 = 0 – 429 psi Upper limit of range 0 – 4 x 286 = 0 – 1144 psi Preffered range = 0 – 2x286 = 0 – 572 psi Choose 0 – 600psi and 0 – 750psi

11 For Pneumatic testing the two gauges from the given five Lower limit = 0 – 1.5 x 242 = 0 – 363 psi

Upper limit = 0 – 4 x 242 = 0 – 968 psi Preferred range = 0 – 2 x 242 = 0 – 484 psi Choose 0-400 psi and 0-600 psi

NOZZLE WELDS

Fig. A – This configuration is represented by code Fig. (g) in ASME code.

1. The radius at top 2 mm R shall be 3.2 mm min. 2. The 70° angle shall be 60° max.

3. Nozzle to shell fillet weld :

Min. throat required = tc = smaller of 6mm or 0.7 t min

= 6mm or 0.7 (smaller of 19, 22, 26) = 6mm or 0.7x 19 i.e. 6 mm Change throat size to 6 mm

4. Min. inner corner radius (r1) = smaller of ¼ t or 19mm (r1) = smaller of ¼ x 28 or 19 mm

Change (r1 ) = 7 mm min.

Fig. B – This configuration is represented by code Fig (l) in ASME code

1. Shell to nozzle outer fillet throat t1 >= smaller of 6 mm or 0.7 tmin

>= 6mm or 0.7 (smaller of 19, 20, 40) Min. size of fillet weld = 6 .= 8.57 say… 9 mm

0.7

Change 6mm to 9mm (size of weld).

2. Shell to nozzle weld (t2) shall be 6 mm – ok

3. t1 + t2 >= 1 ¼ ( t min ) >= 1 ¼ ( smaller of 19,20,40) >= 23.75

t1 = 9 x 0.7 = 6.3 therefore t2 >= 23.75 – 6.3

t2 >= 17.45 say 18 mm.

4. Inside corner fillet weld throat ( tc ) >= smaller of 6mm or 0.7 tmin Min. tc = 6mm

Fillet weld size = 6 = 8.5 0.7

CASE STUDY 7

NOZZLE REINFORCEMENT ANALYSIS

DATA:

A new vessel has the following data:

Shell thickness provided (t)= 20mm , corr. Allow=Nil. Design shell thickness for seamless shell (tr)= 18.5mm Nozzle thickness provided (tn)=12mm

Nozzle thickness for seamless pipe (trn)= 6.5mm

Finished nozzle opening (d)=160mm, Nozzle type=set on

Answer the following:

1) What is the total area required to be compensated (Ar) 2) What is reinforcement limit(X) parallel to vessel wall 3) What is reinforcement limit(Y) normal to vessel wall 4) What is extra area available in shell ( As)

5) What is extra area available in Nozzle

6) What is total area (Av) available inherant to the vessel (Av) 7) Is the additional reinforcement pad required for the nozzle? 8) What shall be the area ( Ap) to be provided by pad

9) What is the Max. pad O.D. possible?

SOLUTION:

1) Total area required to be compensated (Ar)

Ar = d x tr = 160 x 18.5 = 2960 mm2

2) Reinforcement limit (X) parallel to vessel wall

X = 2d X = 2x160 . X = 320 mm

3) Reinforcement limit(Y) normal to vessel wall

Y = 2.5t or 2.5tn use smaller value = 2.5 x 20 or 2.5x12 , Y=30.0mm

4) Extra area available in shell ( As)

As = d(t-tr) = 160 ( 20 – 18.5) = 240 mm2

5) Extra area available in Nozzle

An= Y ( tn-trn)x2 = 30 x (12 - 6.5)x2 = 330 mm2

6) Total area (Av) available inherant to the vessel (Av)

Av= (As + An) = 240 + 330 = 570 mm2

7) Is the additional reinforcement pad required for the nozzle?

Since Av < Ar , R.F. pad is required

8) Area ( Ap) to be provided by pad

Ap = Ar – Av = 2960 – 570 = 2390 mm2

9) Max. pad O.D. possible

Max. pad O.D. = X = 320 mm

10) If pad OD = 320 mm, pad thickness (tp)

tp = Ap = 2390 . (pad OD – pad ID) ( 320 – 184 ) 160 +12+12= 184

So

ASME VIII AND API 510 SAMPLE CALCULATIONS

1. A horizontal deaerator has been in-service for approximately 10 years. An onstream inspection shows that the vessel shell thickness is .275” (uniform) and that the heads have both pitted reducing the thickness at the crown radius to 0.300. The MDR for this vessel reflects that the shell is made from SA 414 GR E plate with Type #2 longitudinal joints in all three courses with an I.D of 66”. The heads are made from SA 285 GR C material, and are full hemispherical, with weld seams (Type 1) with an I.D. of 66” The nameplate stamping shows that the original MAWP is 280 psig @ 550ºF and complies with the rules for spot radiography (RT-3) with no static head considered can this vessel be allowed to continue to operate at this pressure and temperature? If it should be reduced. What is the MAWP that can safely be applied to this vessel?

(Shell S = 16,200. Head S= 13,800)

a. No – allowable pressure should be reduced to approximately 212 psig b. No – allowable pressure should be reduced to approximately 107 psig c. Yes – vessel is acceptable for operation at 280 psig

d. Yes – vessel is acceptable for operation at 280 psig, if impact tests are conducted.

2. A tubular heat exchanger is constructed with a flat, unstayed, seamless circular head, welded to the shell with inside and outside fillet welds as shown in Fig.UG-34, Sketch (F) (C=0.20). The measured thickness of the head is 1”, and is corroding approximately 1/32” (uniform) every year. The thickness of the shell has not corroded, and an on-stream inspection shows the shell to be 0.375” thick. There are Type #1 joints in the shell, with full RT, and a vessel I.D of 30”. The fillet welds are in good condition, and are measured at .375” on both the inside and outside welds. The diameter of the head is 30”, and the vessel is stamped for an MAWP of 90 psig @ 500ºF. The head is constructed of SA-516 GR 70 material, and the shell is constructed of SA 285 GR C material. Assuming that the corrosion rate of 1/32” per year will continue, how may more years may this vessel be allowed to operate within the principles of the ASME Code?

3. A new pressure vessel has been received from a manufacturer with the following information made available to the inspector about the shell: MAWP 500 psig @ 780ºF

MDMT 10ºF 200 psig Spot RT 60” ID

Hydro pressure 750 psig @ 70ºF

Material: SA 387 GR 21. CL 1 Thickness. 0.350” (P# 5 material Stress = 14000)

Type 1 Category A welds Vertical height: 140 feet No impact tests performed No heat treatment performed Material not normalized

From the above given information. How many individual Code violations can you as the inspector find as reason for not accepting this replacement part?

a. No violations – This part meets all Code requirements b. 3 Code violations

c. 5 Code violations d. 10 Code violations

4. A fractionating tower is 14’ ID x 21’ long, bend line to bend line, and is fitted with fractionating trays. The tower is designed for an external design pressure of 15 psig @ 700ºF. The tower is constructed of SA-285 GR C carbon steel, yield strength 30,000psi, and the design length is 39” between the fractionating trays, which are adding support to the vessel. Does this construction comply with ASME VIII requirements (assuming a designed thickness of ½”)?

a. Yes, meets Code requirements

b. No – does not meet code – pressure should be increased to 30 psig c. No – does not meet code – pressure should be decreased to 10 psig d. No – does not meet code – thickness should be increased to 3.6”

5. An ASME – stamped pressure vessel has been altered and now requires a hydrostatic pressure test to be applied. The vessel is 175’ tall and has a pressure gauge at the top of the vessel and another gauge 25’ up from the bottom of the vessel for the inspector to look at. The MAWP is 125 psig. The ratio of design to material test stress = 1. What pressure should be shown on the gauge at the 25’ level to meet API 510 requirements?

a. Approximately 275 psig b. Approximately 185 psig c. Approximately 125 psig d. Approximately 228 psig

6. An existing carbon steel pressure is stamped for lethal vapor service, and has an elliptical 2:1 head. The head is measured at 60.25” I.D in the corroded condition. The head, when new was 1.375” thick and 60” I.D. The stress value is 13,800, the MAWP is 300 psig, and the head is attached to the shell with a Type 1 Category B weld. Assuming a corrosion rate of 1/8” per year. Answer the following questions:

y What are the radiography requirements for the head-to-shell joint?

y Does the head in its corroded condition, meet ASME Code

requirements?

y If the answer to B is yes, how many more years can the vessel

operate within the parameters of ASME Code requirements?

a. Full. Yes. 4.75 years b. Spot. Yes. 8.95 years c. Partial. No. 10.65 years d. None of the above

7. A torispherical head is connected to a seamless vessel with a single welded butt joint with backing. The seam has been welded by a single welder, and is spot radiographed per UW11(a)(5)(b)

8. A 60”I.D pressure vessel will require a fillet welded with a (temporary) patch plate. The patch and the shell are both SA 515-60 material (S=15,000). The patch is .375” thick and the vessel is .622” thick with no corrosion allowance. The vessel has Type 1 Category A welds, and is stamped for RT-2, 200 psig @ 500˚F, and an MDMT of -15˚F. From the information given will this repair require the use of a welding procedure that has been impact testing?

a. yes impact tests are required on the welding procedure b. no. impact tests are not required on the welding procedure

c. yes, impact tests are required on both the base metal and welding procedure

d. no, impact tests are only required on the base metal

9. A nozzle is installed in a vessel shell, as illustrated in Fig. UW-16.1(i), using two equal size fillet welds. The minimum shell thickness is 3/4 inch and the nozzle wall is 7/16 inch minimum thickness. Using equal leg fillet welds, what is the leg dimension of the welds rounded up to the next larger 1/16 inch?

a. 7/16” b. 3/16” c. 9/16” d. 11/16

10. A vertical vessel is to be rerated to a new maximum Allowable Working Pressure based on calculations of the vessel parts. The top of the vessel is located at an elevation of 75 feet. The following calculation values (P) have been determined by the Engineer (elevations are given to the bottom of the items being considered, (static head of water equals 0.433 psi per vertical foot):

1) top head. Elevation 72.5 feet, P-351.3 psi 2) top shell section, elevation 65 feet, P-352.6 psi 3) manway connection. Elevation 50 feet, P = 360 psi 4) Reducer section. Elevation 30 feet, P = 372.5 psi 5) bottom head. elevation 6 feet. P = 425 psi

What is the maximum value of MAWP which can be applied to this vessel? a. 450 psig b. 360 psig c. 395 psig d. 348 psig

11. During the inspection of a horizontal pressure vessel. A torispherical head is measured and found to have the following dimensions. Thickness equals 1.25 inches. Inside diameter of skirt = 48 inches. The distance from the bottom of the head to the top of the vessel is 5 ft 6 in. The weight of the water equals 0.433 psi/ft. Form the vessel data report S = 15000 psi. and ‘RT-2’ has been met. At what Maximum Allowable Working Pressure can this head be used with no corrosion allowance?

a. 490 psig b. 390 psig c. 416 psig d. 426 psig

12. A lap patch is to be installed on a pressure vessel built to ASME code. Section VIII, Div. 1 as part of a repair of the vessel. The patch is made of SA-515 Gr. 70 material (P-no. 1). 1-1/8 inch thickness without normalization. The owner’s engineer has determined the ratio of the allowable stress to the actual stress to be 1.0. The vessel nameplate lists the MDMT as 50ºF with “HT” for the heat treatment, therefore, the patch will be voluntarily heat treated. Will the patch plate require impact testing?

a. Yes b. No

c. No. if welding procedure is impact tested d. None of the above

13. A pressure vessel cylindrical shell is measured and found to be 1.36 inches thickness at its thinnest point. The inside radius was measured at 28.625 inches. Plane records provide the following information:

1) The vessel has been in service for 4 years

2) the original vessel thickness was 1.4375 inches minimum

3) the allowable stress of the vessel material is 17500 psi at design temperature

4) the weld seam efficiency is 1

5) the maximum allowable working pressure is 745psi with a static head of water equal to 5psi

ƒ Based on the above, how much material thickness is available as

remaining corrosion allowance?

ƒ What is the remaining life of the vessel? a. 111”/10.61 years

b. 250”/3.6 years c. 101”/5.31 years d. 202”/4.1 years

14. A vessel’s cylindrical shell has corroded down to 25”in thickness. The cylinder is 40º od with an unsupported length of 10’ Design temperature is 300ºF and the material yield strength is 30,000psi. What is the allowable external pressure allowed on this vessel?

a. 38psi(approximately) b. 45psi(approximately) c. 12psi(approximately) d. 23psi(approximately)

15. During the inspection of an existing pressure vessel you find it necessary to determine the weld seam efficiency of several joints on a vessel. The vessel nameplate shows RT-4. The joints type and degree of RT we read from ASME data reports for the vessel. What are the joint efficiencies for the following?

Type Category RT joint

Efficiency

1. Type1 Cat A spot __________

2. Type2 Cat B Full RT __________

3. Type3 Cat C Full RT __________

a. 1.85 2.60 3.90 b. 1.90 2.90 3.100 c. 1.100 2.1 00 3.100 d. 1.85 2.80 3.80

16. A vessel cylindrical shell is measured today and found to be 1.0625” at the thinnest point. The inside radius is 24” Plant records provide the following:

1) Vessel has been in service 64 years. 2) Original t was 1.1875” min.

3) SV=15000 at design 4) Efficiency = 85

5) MAWP = 500psi with a static head of water equal to 6psi

6) Previous(last) inspection was completed 8 years ago and the wall thickness was 1.087

• Based on the above information, how much material t is available as

remaining corrosion allowance?

• What is the remaining life of the vessel? a. .065”/40.6 years

b. .0868”/29.93 years c. .001”/32 years d. .862”/15.6 years

17. A flat unstayed circular head with a diameter of 14” is operating at 350psi at 500ºF. The SV = 17500 with an efficiency of 1.0 the C factor = .33. Can this head continue in service its present state or would a repair be necessary, if the present thickness is 1.25”?

a. No, head must be replaced b. No, head must be repaired

c. Yes, head can continue in service

d. No, head thickness must be 2.5” to be acceptable

18. A vessel owner is to repair a pressure vessel by replacing one of the vessels seamless ellipsoidal heads with a duplicate head, but welded to the shell. The original vessel name plate is stamped “W” “Rt-2” and ”HT”

• What type or types welded joints may be used in the repair? • What Radiographic Testing of the joint is required?

a. Type 1/full RT b. Type 2/spot RT c. Type 3/full RT d. Type 1or 2/spot RT

19. A vertical pressure vessel in water service with type 1 Category “A” Long seam welds is 10’ seam/seam is made from ½” thick SA516 GR70 material (S = 17.500) is stamped for an MAWP of 100 psig @ 650ºF, and is also stamped as “RT-3” (satisfies spot radiography rules) with an I D of 60”. What is the actual minimum thickness of this vessel including hydrostatic head

a. .211” b. .250” c. .350” d. .360”

20. The heads on the vessel in #19 are 2.1 elliptical heads are seamless and are made from the same material, same diameter, same thickness and are welded to the vessel with Category “B” Type 1 circumferential welds. What is the minimum thickness of the bottom head if the extra radiograph required by UW-11(a)(5)(b) is taken on head-to-shell weld? (Remember static head)

a. .250” b. .200” c. .179” d. .105”

21. Assuming the same parameters for the above pressure vessel in #19 but the heads are seamless hemispherical heads with a 30” spherical radius attached with a category “A” Type 1 full penetration weld, what is the minimum thickness of the bottomhead?

a. .250 b. .220” c. .179 d. .105”

22. An 8 feet I.D horizontal pressure vessel with Type 1 weld joints is constructed totally of SA285 GR C (S = 12,100) plate with two courses (one circumferential seam joining two cylinders.) The original thickness is .375” uncorroded (new and cold) and the vessel is stamped for full radiography (RT-1). The MAWP is 50 psig @750ºF. The heads are torispherical, 6% knuckle, 96.75” O.D skirt. And were .375” thick also when new. A onstream, inspections shows the vessel has corroded evenly over the head and shell with a uniform ¼” external corrosion. What MAWP can this vessel be operated at, assuming no static head?

a. 25 psig b. 31 psig c. 17 psig d. 50 psig

23. What is the required thickness of a seamless flat, unstayed circular head with a diameter (or short span) of 24” an internal design pressure of 250 psig @ 650ºF with material of SA105 (S=17.500)? Attachment is as shown in Fig UG-34(A), and the inside corner radius is not less than three the required head thickness

a. 1.1” b. 1.9” c. 2.3” d. 1.66”

24. Given the parameters of the above flat head in #23, assume the head is not circular but elliptical with the same short span and a long span of 36”. What is the required thickness of this head?

(NOTE: This question is not supposed to be in the test but a similar question has been asked previously)

a. 1.9” b. 1.5” c. 2.1” d. 1.66”

25. A 60” I.D 1” thick pressure vessel constructed of SA442 GR60 material is stamped RT-3 and is also stamped for an MAWP of 70 psig @ 650ºF. A nozzle is located in the shell and doesn’t pass through a welded joint. Details of the attachment are as follows.

Nozzle material – SA106 GR B Nozzle I.D – 16”

Nozzle thickness – 375”

Nozzle attached to shell by penetration weld into shell and a cover fillet weld on the outside of the shell only. Fillet weld leg lengths are ½”х ½”. Attachment detail is as shown in Fig UW 16 1 sketch (C).

Does this construction need a repad? Assuming Fr, F and E=1.0 and tR = .890” and tRS = .290”

a. No b. Yes

c. Not enough information given d. None of the above

26. What are parallel and perpendicular (or normal) limits of reinforcement for the nozzle in #25 above?

a. Parallel – 16” Normal – 2.5” b. Parallel – 9.375” Normal – .9375” c. Parallel – 9.375” Normal – 2.5” d. Parallel – 16” Normal – .9375”

27. A pressure vessel has a new 18” ID manway installed in the shell, with a configuration similar to Fig UW = 16(a-1). The shell thickness is .350”, the manway is .280” thick and the repad is .375”thick. The cover weld attaching the pad to the shell is .300”in size and the cover weld attaching the pad nozzle is .300” in size. The nozzle is SA 516-70 rolled and welded plate (17.500 stress) fully RT’d and the vessel is also SA 516-70 (fully RT’d. the vessel is 50” ID and is constructed for 200 psig @ 500ºF. The od of the repad is 24” and the ID of the hole in the pad is 19”. The repad is also SA 516-70 material. Is this manway property reinforced?

(All Fr. E and F = 1.0, tR = .287” and tRs = .103”)

a. Yes b. No

c. Not enough information given d. None of the above

28. A vertical one course pressure vessel in vapor service is 12” tall is made of .300” nominal wall seamless pipe, SA 106 Gr B (S = 15,000). Design pressure is 250 psig 500ºF. The outside radius of the shell is 18”. The vessel is stamped RT=3 (spot RT) attached to the shell are two seamless torispherical heads made from SA516 Gr 70 plate (S = 17,500). The inside crown radius of the heads is also 18”. The heads are also .300” thick. What is the MAWP of this vessel based on the shell and heads?

29. A 20’ tall pressure vessel is stamped for 1000 psig MAWP @ 900ºF. The hydrostatic test is to be applied at 70ºF. Materials are SA516 GR70 and SA240 type 302 S.S. plate. What is the minimum hydrostatic test pressure that should be applied at the vessel to satisfy ASME Code requirements? (stress value for SA240 Type 304 is 14,700 at 900º and 18,800 at 70º)

a. 4038 psig b. 1500 psig c. 2000 psig d. 1659 66

30. what is the maximum allowable external pressure allowed on the

following pressure vessel: OD = 24”

Material = SA106 GR C (yield strength = 40,000psi) Nominal thickness = .500”

Total length between lines of support = 48” Design temperature = 500ºF

a. 327 psig b. 390 psig c. 456 psig d. 512 psig

31. A stationary vessel is made from SA516 GR70 plate that has been normalized. The MDMT is 30ºF @ 470psig. The actual material thickness is 3.0” thick and the vessel id is 48” and the joint efficiency is 1.0 Does this material require impact testing?

a. Yes b. No

c. Not enough information d. None of the above

32. Assume the same vessel as in #31 above except that the plate has not been normalized. Does this material require impact tests? If so. What FT LBS of energy would be required for this plate?

a. No impacts required

b. Yes impacts required, 25ft.lbs for 3 specimens(avg.) c. Yes impacts required, 17ft.lbs for 3 specimens(avg.) d. Yes impacts required, 35ft.lbs for 1 specimens(minimum)

33. A pressure vessel has been inspected and found to be thinned over a 20” long area parallel with the long seam. Thickness readings in this area are .275”, .279” .280” and .295”. Original thickness is .375”. The vessel is now 11 years old. MAWP is 80 psig @ 100ºF. 24” ID and material stress is 16,800. Joint efficiency is .85

1. What is the minimum shell thickness?

2. What is the longest dimension that can be corrosion averaged per API 510?

3. What is the internal or onstream inspection interval for the vessel based on the above?

a. 1) .067” 2) 12” 3) 10 years b. 1) .100” 2) 6” 3) 10 years c. 1) .500” 2) 2.44” 3) 10 years d. 1) .050” 2) 2.14” 3) 10 years

34. An existing pressure vessel material thickness is measured at .500” on an inspection. 4 years later this same thickness is measured at .250” at the same location. Required thickness (by calculation) shows that the vessel must be .125” thick to withstand the given pressure per API 510, and from this information, what is the:

a. Metal loss = b. Corrosion rate = c. Corrosion allowance = d. Remaining life = e. Inspection interval =

35. A pressure vessel made of SA 285 GR B (12,100 = stress) material has been in service 10 years it has a measured shell thickness of ½” at the thinnest section. If this vessel is to be operated with a stamping that indicates an internal MAWP of 300 psig@ 700 Deg F.RT-2, Type 1 joints, and an ID of 80”. What will the minimum thickness of the shell be to support this pressure?

36. What is the minimum thickness required for a pressure vessel that is stamped with a 600 psig @ 500º MAWP is 70” OD complies with the rules for spot radiography has Type 2 joints is made from SA 515 GR 60 (S=15,000) material and is 25’ high in water service?

a. 1.950” b. 1.074” c. 1.560” d. 1.746”

37. A pressure vessel head is thinned at the knuckle radius to 25C” thick. The head is attached to the vessel with a Type 1 joint that is fully radiographed and operated at 600ºF. The head is a 2:1 elliptical head with an ID of 45” and is made from SA 285 GR C (S= 12 000) material. What MAWP can be operated on this head with no static head considered?

a. 175 psi b. 153 psi c. 190 psi d. 142 psi

38. A pressure vessel shell is 80” ID. 375” thick and the heads are

torispherical (6% knuckle radius) and are also 80” ID and .375” thick. Both shell and heads are made from SA 36 plate (14.500 stress), and the stress complies with spot radiography. The heads are spliced (welded) and comply with spot radiography. Assuming all joints are Type 1. What is the MAWP allowed on this vessel based on the heads assuming a 500” temperature and vapor pressure only?

a. 70 psi b. 85 psi c. 64 psi d. 110 psi

39. If a vessel is built from SA 106 GR B (S = 15 000) seamless pipe is 375” nominal wall thickness and has one circumferential weld joint and is 24” ID, what is the MAWP allowed if the temperature is 500ºF and the vessel is stamped “RT-2”?

a. 403 psi b. 425 psi c. 387 psi d. 415 psi

40. An 8” nozzle on a vessel is replaced with an identical nozzle with an attachment similar to UW 16.1 Sketch C. If the nozzle thickness is .500” and the vessel shell thickness is 1”. What is the minimum size of throat and leg dimensions for the attachment fillet welds?

a. .170” throat/.250” leg b. .750” throat/1.00” leg c. .250” throat/.353” leg d. .250” throat/.250” leg

41. A vessel nozzle has corroded around the attachments fillet welds, reducing them to a 125 throat thickness. With a nozzle wall thickness of .350” and a shell thickness of .500” and assuming a joint configuration in compliance with UW 16.1 Sketch (i) will this condition meet ASME Code?

a. Yes b. No

c. Fillet welds not required for this nozzle d. Not enough information given

42. An 8” nozzle in a pressure vessel if to be replaced with a 10” ID SA 106 B nozzle that is .280” thick. The vessel is 75” thick and is stamped for an MAWP of 350 psig @ 600ºF. The vessel ID is 60” and the vessel complies with the rules for spot RT (Type #1 joints). The installation is similar to UW–16 1 Sketch (c) with a 750” throat fillet weld. Does this nozzle require a reinforcing pad? The S.V for the shell is 15,000 psi. The required thickness of the shell is 490” and the required thickness of the nozzle is 160” (All E.F. FR= 1.0)

a. Yes b. No

c. Not enough information given

43. A 50’ high Amine Tower has been altered and requires a hydrostatic test. The MAWP is 350 psig @ 750ºF. The vessel material are SA 516 GR 70, SA 285 GR A, and SA 53 GR B (seamless) pipe. If the test is to be conducted to ASME VIII requirements. What is the minimum hydrostatic pressure required on the bottom head if the test will be conducted at 70ºF?

a. 455 psi b. 546 psi c. 518 psi d. 670 psi

44. A vessel is to pneumatically tested @ 70ºF per the Code. The MAWP is 200 psig @ 700 Deg F. The materials are SA 240 Type 304 stainless steel and SA 515 GR 65. What is the minimum pneumatic pressure required on this vessel? The S.V. for SA 240 Type 304 @ 700º F is 16,800 psi, and 18,000 psi @ 70ºF.

a. 231 psi b. 220 psi c. 300 psi d. 425 psi

45. A circular flat head is seamless and is 20” diameter and is attached similar to Figure UG 34, (b-1). If the MAWP of the vessel is 300 psig @ 500 Deg F and the material is SA 105 (S = 17,500), what is the minimum required thickness of this head?

a. .894” b. .970” c. .900” d. 1.07”

46. A circular flat head is 30” in diameter and is attached to the shell with a weld similar to Fig. UG 34.(h). The head is splice-welded (seamed) with a Type 1 joint and has been spot radiographed. The heat is made from SA 515 GR 60 (S = 15,000) What is the minimum thickness required on this head. Assuming a temperature of 650ºF and an MAWP of 375 psig?

a. 1.677” b. 2.09” c. 2.955” d. 3.650”

47. A vessel is constructed for external pressure and is supported at 7’ intervals. The OD is 48” the thickness is .500” and the temperature is 600ºF. What is the approximate maximum external pressure allowed on this vessel? The material yield strength is 28 000psi

a. 111 psi b. 160 psi c. 181 psi d. 195 psi

48. A vessel is made from SA 662 GR A material SA 182 GR 21 normalized and tempered material and SA 516 GR 70 material. All materials are .375” nominal thickness and the vessel is made for a design temperature of -30ºF. Which materials, if any will require impact testing?

a. All materials

b. Only the SA 662 and SA 182 materials c. Only the SA 182 and SA 516-07

d. Only the SA 516-70

49. A 30” ID vessel is fully radiographed, has Type 1 joints, is .500” thick and is stamped for an MAWP of 100 psig @ 300 Deg F; with a corrosion allowance of 1/16”, and a minimum temperature of -40ºF. If the material is SA 516 GR 70 (not normalized), does this vessel require impact testing? (A reduction stress ratio of 1.0 will be used, per UCS 66 1).

a. Yes, requires impact testing

b. No. does not require impact testing c. Exempted from impacts per UG-20(f) d. Not enough information provided

50. A vessel is checked during an internal inspection and is found to be .753 inches thick. 5 years later the vessel is shown to be .500” thick. With a minimum thickness required of 350”.

a. Metal Loss = b. Corrosion rate = c. Corrosion allowance = d. Remaining service life =

51. An 80” ID vessel is fully radiographed, is 1” thick and is made from SA 516 GR 70 (S = 17,500) material with Type #1 joints with an MAWP of 150 psig @ 600ºF. If this vessel corrodes at an even rate of 1/8” per year, how many years may the vessel operate within the principals of the ASME Code?

a. 5 24 years b. 2.62 years c. 10 48 years d. 3 15 years

52. A pressure vessel is 175’ tall and is stamped with an MAWP of 150 psig. What is the minimum hydrostatic test pressure that should be shown on a pressure gauge that is placed 25’ up from the bottom of the vessel, assuming the ratio of design stress to test stress is 1.0, and all other rules of ASME have been met?

a. 225 psi b. 235 psi c. 250 psi d. 260 psi

53. An elliptical head (2:1 ratio) is attached to an existing pressure vessel. The heat has internally corroded around the skirt and is measured at 1/8” uniform corrosion. The original inside diameter of the head was 60”, and the MAWP of the vessel is 150 psig @ 650ºF allowable stress value is 17.500. With a stamping of RT-2 applied to the vessel using a Type 2 weld what is the minimum thickness required for this head?

a. .208” b. 258” c. 312” d. 335”

54. A seamless ellipsoidal head is attached to pressure vessel using a single “Vee” grove weld with a backing strip. If spot radiography per RT-2 is conducted on this vessel, determine the following and the applicable ASME Code paragraph?

a. Head Efficiency _________________ Para. _____________________ b. Joint Category __________________ Para. _____________________ c. Joint Type ______________________ Para. _____________________

55. A pressure vessel has the following measurements (averaged) at the below locations on one year. The same readings are taken 5 years later at the same locations. With a minimum thickness of .125” at all location determining the remaining life of each component:

Top Head Bottom Head Shell # 1 Shell # 2 Nozzle # 1 Nozzle # 2 1st _{year .350 .300 .285 .275 .265 .250}

5th _{year .300 .270 .270 .200 .150 .230}

Remaining

Life ________ ________ ________ ________ ________ ________

56. A vessel is stamped for 400 psig design pressure and is currently measured to be .788” thick. The shell material stress value is 16,800, and the joint efficiency is .85. The i.d. of the vessel is 47.5”. If the corrosion rate is known to be .072” per year, and the next inspection is scheduled for 6 years from the current inspection, per API 510 this vessel:

a. May continue to be operated for 6 years at the current design pressure b. Should be reduced in pressure or inspection interval

c. Should be allowed to operate at 550 psi d. Should be immediately removed from service

57. A 60 KSI tensile strength weld metal is used to repair a 75 KSI tensile strength base metal. Total base metal thickness is .390” and the depth of the repair is .195” What is the required total thickness of this weld deposit, per API 510?

a. .195” b. .390’ c. .520” d. .243”

ASME VIII AND API 510 SAMPLE CALCULATIONS

ANSWERS

2. t = ? d = 30 P = 90 S = 17,500 E = 1.0 C = .20 t = 30 (0.20 x 90 / 17,500)1/2 t = .962 1 – .962 = .38CA / 03125CR = 1.21 years ANSWER : D 3.

1. Pressure not to Code for thickness 2. Impacts required

3. Hydro pressure insufficient 4. Heat treatment required 5. Full Radiography required

ANSWER: C 4. L = 39 A = 0.001 Dρ = 169” t = .5 L/Dρ = .230 Dρ/t = 338 Yield = 30,000 PA = (4 x 8000) / (3 x 338) PA = 32,000 / 1014 PA = 31. 55 psig

5. 175’ tall with a gauge at 25’ = 150 ft of head pressure acting on gauge MAWP = 125 DS/MS = 1.0 1.3 X MAWP X (1.0) + (H.H.) = 1.3 X 125 + (.433 X 150) = 162.5 + 64.95 = 227.45 PSIG ANSWER ANSWER: D

6. A FULL, Per UW – 11(a)(1)

B. T = 1.25”

P = 300 (No H.H) S = 13,800 E = 1.0 D = 60.25

From UG-32 t = PD / (2SE-0.2P) t = (300 x 60.25) / (2 x 13,800 x 1 – 0.2 x 300) t = (18,075) / (27,600-60)

t = 18,075 / 27,540 = 0.656 REQUIRED

ANSWER B – YES 0.656 REQU < 1.25 ACTUAL

CA = RL , R.L= (1.25- 0.656)/).125 = 0.54/0.125= 4.75 CR

ANSWER: A

7. Type 2. Category B weld with a joint efficiency of or 1.0 per UW- 11(A)(5)(b) ANSWER: B

8. S = 15,000 E = 1.0 R = 30 t = .622” P = 200 t = (200 x 30) / (15,000 x 1– 0.6 x 200) t = 6,000 / 14,880 t = 0.403

From Fig UCS – 66 – Curve B @ 0.622 = 5ºF

From Fig UCS – 66 1 – (0.403 x 1.0) / (0.622 – 0) = 0.64 Allowable reduction = 35ºF

Allowable = +5 -35ºF = -30ºF, which is lower than -15ºF

ANSWER: B No, impact tests not required.

9. UW – 16.1 Sketch (I):

tmn =0.4375” 7 tmn = 0.306”

1.1/4t =0 .546

From UW -16.1 Sketch (I) Welds must be: (a) t1_+t2 _{= 1.1/4 t}

mn

(b) t1 or t2 not less than smaller of ¼” or 7 tmn Step 1 0.546/2 =0. 273 +0. 273 = 1 1/4 tmn “a” is satisfied

10.

a) 351.3 – (2.5 x0.433) =350.218

b) 352.6- ( 10 x0.433) = 348.27

c) 360 – 10.825 = 349.175

d). Reduce elev 30’ @ 372.5 -( 75 – 30 ) 0.433 =372.5 – 17.32 = 355.18 e). Bottom head elev 6’ @ 425 psi = 75’-6’ = 69’ x .433 = 29.877

425 – 29.877 = 395.123

Question. What is the max. value of MAWP which can be applied to this vessel?

11. Torispherical head-given:

t = 1.25

Skirt i d = 48 L = skirt OD = 50.5

S = 15,000

E = 1.0 (UW – 11 (A)(5)(b) has been met) H.H = 5.5 x .433 = 2.381 psig

From UG-32

P = SEt / (885L – 0.1t)

P = (15,000 x 1 x 1.25) / (.885 x 50.5 + .1 x 1.25) P = 18,750 / (44.69 + .125)

P = 418.3 psig – 2.3 psig (H.H.) = 416 pig

ANSWER:C

12. SA 515 Gr 70 not normalized 11/8” thick

Allowable stress ratio = 1, MDMT -10°F a) UG-20 (f) – Not exempt

b) UCS 66 (Figure) General notes – curve A material

c) UCS 66 (Figure) – Requires impacts @ 75° for 1.25” material d) UCS 66 (b) Figure – Allows reduction of 0 for 1.0 ratio

e) UCS 68 (c) - 30° reduction allowed for voluntary PWHT, 75°F – 30°F = 45°F<50°F therefore, vessel is exempt from impact testing

ANSWER: B Per UCS 66 (a) and (b) this material will not require

13. Measured t = 1.36” Original t = 1.4375” Inside Radius = 28.625” 4 years in service S = 17,500 E = 1.0 P = 745 psi H.H = 5 psig t = PR / (SE-0.6P) t = (750 x 28.625) / [(17,500 x l) – (0.6 x 750)] t = 21,468.75 / 17,050 t = 1.259

A. 1.36 – 1.259 = 0.101” Corrosion Allowance - ANSWER

B. 14375 – 1.36 = [(.07 metal loss) / (4 years)] = .019 per year corrosion rate

B 0.101 / .019 = 5.13 years Remaining Life - ANSWER

14. t = 25” O d = 40” (D.O) L = 10’ or 120” Da /t = 160 L / Da = 30 Factor A = .0002 Factor B = 2,800 From UG-2B(C): Step1: ƒ Da /t = 40/.25 = 160 (use Path (1)) ƒ L/D = 120/40 = 3.0

Step 2 and 3 • Fig G determine “A”

ƒ Enter Chart at 3.0 – over to 160 Intersects at approximately .0002 = Factor “A”

Step 4 and 5

ƒ Using Figure CS-2 (provided @ test location):

Enter bottom @ .0002 up to 300° line – read right – approximately 2,800 = Factor B

Step 6:

ƒ PA = 4B / 3(DO / t)

ƒ PA = (4 x 2800) / (3(160)) ƒ PA = 11200 / 480

NOTE: Step 7 is n/a for this problem Step 8:

ƒ PA = 23.333 psi allowed

15. • RT – 4 stamping. • degree of RT from Data Report

• Efficiencies obtained from Table UW – 12

A Type 1 Cat A Spot RT Efficiency 0 .85 B Type 3 Cat B No RT Efficiency 0 .60 C Type 2 Cat C Full RT Efficiency 0 .90

ANSWER: 1: 0.85 joint Efficiency

ANSWER: 2: 0.60 joint Efficiency

ANSWER: 3: 0.90 Joint Efficiency

ANSWER: A 16. 1.062” = t (corroded), 1.1875” original 24” = R 15,000 = S .85 = E (6+) 500 = P+H H = 506 A. From UG-27 ƒ t = PR/SE- 6P ƒ (t = 506 x 24) / [(15,000 x 85) – (.6 x 506)] ƒ 12,144 / (12,750 – 303.6) ƒ 12,144 / 12446.4 t = .9757” required thickness 1.0625” (present “t”) – .9757 (required ”t”) = .0868”

ANSWER: - A: 0 .0868” Remaining Material for corrosion Allowance B. Remaining Life = Corrosion Allowance / Corrosion Rate (from API 510)

Corrosion Rate = 1.087 (8 years ago) – 1.0625 (present thickness) 1.087 – 1.0625 = .0245 in 8 years

.0245/8 = .003 per year

.0868 / .003 = 28.933 Remaining Years

ANSWER: B – Remaining Life = 28.933 years

17. From UG – 34 (c)(2) Editor’s note: Æ t = 1.25” Questionable Values d = 14 Provided! Æ P = 350 Æ C = 0. 33 S = 17,500 E = 1 t = d √(CP/SE) t = 14 √0.(33 x 350) / (17.500 x 1) t = 14 √(115.5 / 17,500) t = 14 √0. 0066 t = 14 √X .081240384 1.13” < 1.25” t = 1.13

ANSWER: C. Yes this Head can continue to operate at the pressure shown

18. From UG-116: HT = Heat Treated W = Welded

RT2 = Full (per UW 11(A)(5)(b)

ANSWER A – Type 1 or Type 2 Per UW 12(d)

ANSWER B – One spot Radiograph in accordance with UW – 52 for each

seam. This Radiograph would be required over and above any other RT requirements

Para UW 11(a)(5)(b)

19. t = .5 P= 100+ ( 10 x 0.433) =100 +4.33 HH = 104.33 From UG-27 E=0. 85 t = PR S= 17,500 SE - 0.6P ID= 60” IR= 30 t = [(104.33 x 30)] / [(17.500 x 85) – (0.6 x 104.33)] = [(3,129.9) / (14,875 – 62.598)] = (3,129.9) / (14812.402) = 0.211 ANSWER: A 20. t = .5 P= 100 + 4.33 HH = 104.33 From UG-32 E= 1.0 t = PD S= 17,500 2SE - 0.2P ID= 60” IR= 30 t = (104.33 x0. 60) / [(2 x 17,500 x 1.0) – (.2 x 104.33)] = (6,259.8) / (35,000 - 20,866) = (6,259.8) / (34,979.134) = 0.179 ANSWER: C 21. t =. 5 P= 100 + 4.33 HH = 104.33 From UG-32(f) E= .85 t = PL S= 17,500 2SE - 0.2P L= 30“ ID= 60” t = (104.33 x 30) / [(2 x 17.500 x .85) – (.2 x 104.33)] = (3,129.9) / (29,750- 20.866) = (3,129.9) / (29,729.134) = 0 .105” ANSWER: D

22. Shell Heads t = .375 – .250 = .125 t = 375” P = ? ID = 96” S= 12,100 IR = 48” L = ICR = OD of skirt = 96.75” S = 12,100 E = 1.0 P = ? E = 10

Shell: From UG – 27 Heads: From UG-32

Shell: P = 12,100 x 1 x .125 Heads: P = 12.100 x 1 x .125 . 48 + .6 x .125 .885 x 96.75 + .1 x .125 P = (1512.5)/(48.075) P = 1512.5/85.635 P = 31.46 psig P = 17.662 Answer = 17.662 psig ANSWER: C 23. t = ?

D =24 From UG-34 t = d (CP-SE)1/2

P = 250 t = 24 x (.17 X 250) / (17,500 X 1) 1/2 S = 17,500 t = 24 x (42.5) / (17,500) 1/2 E = 10 t = 24 x .0024285 C = 17 t = 24 X .04592 t = 1.182“ Answer ANSWER: A

24. t = ? From UG-34 d =24 t = d√ZCP/SE D =36 But z = 3.4 – [(2.4X24) / 36] S =17,500 z = 3.4 – 1.6 E = 1.0 z = 1.8 Z =1.8 t = 24 [(1.8X.17X250) / (17,500X1)] C = 17 t = 24 (76.5 / 17,500) P = 250 t = 24 X .0661 t = 1.586 ANSWER ANSWER:B 25. d =16 tr = 0.890 trn =0.290 tn = 0.375 t = 1.0 From UG-36 A = 16 x .890 x 1 + 0 A = 14.24 ← A1 =16 X (1 - 890) - 0 A1 = 1.76 ← or A1 = 2(1.375)(.11) = .302 A2 = 5(.085) x 1 = .425 or A2 = 5(.085) x .375 = .159 ← A3 = 0 A41 = 52 = .25 A43 = 0 A1 = 1.76 +A2 = .159 +A3 = 0 +A41 = .25 +A43 = 0 _________

2.169 < 14.24 yes need repad

26. Parallel =

16’ or 8 + .375” + 1” Larger value – use 16” Perpendicular =

2.5 x 1 or 2.5 x .375 + 0

Use smaller value – use .9375

ANSWER: D

27. 1 Set Nomenclature and compute “tr” and “trn” d= 18 tr=.287 trn=.103 tn=.280“ t=.350 f=1.0 E1=1.0 fr1 = fr4 = 1.0 te=.375 Dp = 24 tr = (200x25) / (47500x1 - .6x200) tr = (5000) /(17,380) tr = .287 trn = (200x9) / (17.500x1 – .6x200) trn = 1800/17380 trn = .103 2. Compute”A”: A = 18 X .287 X 1 + 0 A = 5.166 3. Compute A1: A1 = 18 (1x .350 – 1 X .287) - 0 A1 = 1.13 OR

4. Compute A2: A2 = 5(.280 - .103) x 1 x .350 A2 = .309 OR A2 = 2(.280-.103) x (2.5x.280 +.375) x1 =.885 x 1.075

A2 = .380, USE SMALLER VALUE

5. compute A3: A3 = 0 6. compute A41 = .3002 = 09 7. compute A42 = .3002 = 09 8. Compute A43 = 0 9. Compute A5 = (24 - 18 – 2 x .280) X .375 / 1 A5 = 5.44 x .375 A5 = 2.04

10. Add values and compare to “A”

A = 5.166 A1 = 1.13 A2 =.309 A3 = 0 A41 = .09 A42 = .09 A5 =2.04 TOTAL = 3.659 < 5.166

28. From Appendix 1 From UG-32 Shell Head OR = 18” S = 17,500 S= 15.000 E = .85 E =1 L = 18 T = .300 X .875 = .2675 t = .300 P = (15.000x1x.0.2625) P = (17,000x.85x.300) (18 – .4x.2625) (.885x18+.1x.300) P = 3937.5/17.895 P = 4462.5/15.96 P = 220 psig P = 279.60 Answer: 220 psig ANSWER: C

29. UG – 99 Stress value of SA516 70 @ 70° = 17,500 = 2.6

@ 900° = 6.500

OR

Stress value of SA 240 type 304 18,800 @ 70° = 1.27 use lowest

14,700 @ 900° value

1.3 x 1000 x 1.27 = 1,651

+ HH 8.66

1,659.99 psig on bottom head

30. From UG – 28 DO = 24” t = 500 L = 48” Dot = 48 L/Do = 2 Temp = 500°F Yield = 40,000

From Fig G = .002 (Factor A) From Fig CS-2 = 11,800 (Factor B) From UG-28 = Pa = 4B . 3 (Do/t) = (4 x 11,800) / (3 x 48) = 47,200 / 144 = 372.77 Pa = 327.77 psig ANSWER: A 31. SA 516 GR 70 is Curve “D” material

Curve D 3.0” thick material is good for +10° per UGS-66(figure) +10°F < 30ºF, which is vessel rating

Answer No, does not require impact tests

32. SA 516 70 not normalized = Curve B material Curve B @ 3” thick = 78° Yes, requires impacts

From Fig UCS-66 1 t R = (470x24) / (17.500x1-.6x470) t R = .655 (.655x1)/(3-0) = 0.218 Reduce 105° from UCS-66 1 No impacts required.

ANSWER: A

33. t = (80 X 12) / (16,800 X .85 – .6 X 80) 1 .067

t = 960 / (14,280 - 48) 2 12”

t = .067 3 10 YEARS

Corrosion Rate = .008 per year Avg = .283 - .067 = .216 Corrosion Allowance = .216

Remaining Life = 0.216/0.008= 27 years – default to 10 years per API 510

ANSWER: A

34.

A. Metal Loss = .250

B. Corrosion Rate = .0625 per year C. Corrosion Allowance = 125 D. Remaining Life = 2 Yrs

E. Inspection Interval = 2 Yrs per API 510

35. From UG-27 t =? P = 300 R = 40 E = 1 S = 12,100 t = (300x40) / (12.100x1 – .6x300) t = 12,000 / 12.100 – 180 t = 12.000 / 11,920

36. From Appendix 1 P = 600 +HH HH = 25 X .433 = 10.825 P = 610.825 E = 80 Ra = 35 S = 15.000 t = (610.825x35) / (15,000x80 –0. 4x610.825) t = (21378.875) / (12,000 - 244.53) t = 21378.875/12244.33 t = 1.746” ANSWER: D 37. From UG – 32 P = ? t = .250 D = 45 S = 13,800 E = 1.0 P = (2x13,800x1x.250) / (45+.2x.250) P = 6900 / 45.05 P = 153.16 psig ANSWER: B 38. From UG - 32 P = ? t = .375“ L = 81.5 S = 14,500 E = .85 P = (14,000x.85x1x.375) / (.885x81.5+.1x.375) P = 4621.875 / (72.127 + .0375) P = 64 PSI ANSWER: C

39. From UG – 27 P = ? t =0. 375 X 0.875 = 0.328 S = 15 000 E = 1.0 R = 12 “ P = 15,000x1x0.328 / 12+0.6x0.328 P = 4920 / 12.196 P = 403 psi ANSWER: A 40. From UW – 16 (b) tc = ¼” or .7tmin (smaller)

tmin = 3/4” or .500” / 1.00” smaller – use .500 0.7 x .500 = .350” or .250” – use .250” throat 0.250 x 1.414 = .353” leg

ANSWER: C

41. From Fig UW -16.1 Sketch (l) t1 = .125 t2 = .125 t1 + t2 = .250 tmin = 3 / 4” or .350 / .500” (smaller) tmin = .350” 1.25 X 350” = 0.4375” required 0 .250” actual

t1 or t2 not less than smaller of .250” or (.7 X .350) = .245” Æ .125<.245

42. From UG – 37 d = 10 tr = .490 t = .750 tn = .280 trn = .160 A = 10 X .490 = 4.9 At = 10 (.750 - .490) = 2.6 or 2(.750 + .280)(.75 - 49) =.53 A2 = 5(.280 - .160) .750 = .45 or 5(.280 - .160).280 = .168 A3 = 0 A41 = (.75 X 1.414)2 = 1.124 A43 = 0 2.6 + .168 + 1.124 = 3.892” 4.9” required > 3.892” actual Reinforcement is required ANSWER: A

43. From UG – 99 P = 350 HH = 21.65 St/SD = 1.09 1.3 X MAWP X (St/SD) + H.H. = H.H. = 50 X .433 = 21.65 SA 516 70 @ 750°F = 14,800 @ 70°F = 17,500 17.500 / 14.800 = 1.18 SA 585 A @ 750°F = 10,300 @ 70°F = 11,300 11,300 / 10,300 = 1.09 SA53 B @ 750°F = 13,000 @ 70°F = 15,000 15,000 / 13,000 = 1.15 Use lowest ratio - use 1.09

1.3 x 350 x 1.09 + 21.65 = 517.6 or 518 psig