Questions & Answers

Formulas to be memorise :-

a) Min thickness / design pressure of shells (cylinders) = UG 27 – IR , APP1 –OR = Thickness of the shells under internal pressure - circumferential stress ( longitudinal joints ) when the thickness does not exceed one half of the inside radius, or P does not exceed 0.385 SE –

T = PR / SE - 0.6P or P = SEt / R + 0.6t

b) Min thickness / design pressure of shells (cylinders) = UG 27 – IR , APP1 –OR Thickness of the shells under internal pressure - longitudinal stress ( circumferential joints ) when the thickness does not exceed one half of the inside radius, or P does not exceed 1.25 SE –

T = PR / 2SE +0 .4P or P = SEt / R - 0.4t

c) Min thickness / design pressure of formed heads = UG 32 (d)(e)(f) = Thickness of the shells under internal pressure - spherical shells - when the thickness of the shell of a wholly spherical vessel ( sphere )/ 2:1 elliptical does not exceed 0.356R, or P does not exceed 0.665 SE –

hemispherical ( sphere )/ 2:1 elliptical T = PD or PL / 2SE -0 .2P or P = 2SEt / D or L + 0.2t

Thickness of torispherical head t = 0.885PL / SE – 0.1 P + C

d) Minimum Thickness of flat unstayed circular heads From UG 34 t= d square root of CP/SE

e) MAWP of cyliderical shell circumferential with uniform external corrosion P=Set / Ro – 0.4 t

f) The weld joint efficiencies - UW 11, UW 12 table UW12RT 1 – full 1.0 or 0.90RT 2 - full on cat A spot on cat B (UW -11(a)(5)(b) - 1.0 or 0.9RT 3 - spot – 1 – 50 foot weld - 1 for each welder 85 or 80RT 4 - combination of aboveNo RT - 0.70 or 0.65Seamless - 0.85 if UW 11(a)(5)(b) is not met

1.0 if UW 11(a)(5)(b) is met

g) The value of factor C of flat circular heads = C = 0.33 x m , m = Tr / Ts

h) Hydrostatic testing pressure = 1.3 x MAWP x stress ratio

i) MAWP = hydrostatic head x 0.433 & depth of head = ¼ ID +flange

j) Pneumatic test pressure = 1.1 x MAWP x stress ratio

k) Metal temperature during PT = 30 deg F above MDMT for more than 2”= 10 deg F above MDMT for 2” or less and max 120 deg F

l) Evaluation of locally thinned areas - ID less than or equal to 60” = ½ of vessel diameter or 20” whichever is lessID greater than 60” = 1/3 of vessel diameter or 40” whichever is less

m) Corroded areas of ( 2 : 1 ) ellipsoidal head , crown radius = 0.9 D

n) ellipsodical head spherical (crown) radius Spherical radius – K1D i.e D/2h = K1 and for many ellipsodical heads d/2h = 2.0, Then from table 7-1 – 2.0 of d/2h equals 0.9 of K1

o) For torisph head, for crown portion, use hemisphere formula using R = D

p) Corroded areas of ( 2 : 1 ) ellipsoidal head, Crown portion lies in circle of diameter equal to 0.8D

q) Torispherical head Crown radius h = D = 60” & knuckle radius = 6% of crown radius

r) Internal inspection = ½ RL or 10 yrs whichever is lessExternal inspection = does not exceed lesser of 5 yrs or the required internal inspection

s) area of a circle = pie r square , Tensile strength = load / area or UTS x width x thk

t) external pressure on cylinders = UG 2B and external pressure parts =

thickness of shells and tubes under external pressure L / Do & Do / t and maximum allowable external working pressure Pa = 4B / 3 ( Do / t )

u) for coil technique ( longitudinal magnetization technique ) - a. if L/D ratios less than 4 but not less than 2 = ampere turns = 45000 / (L/D) b. if L/D ratios equal to or greater than 4 = ampere turns = 35000 / (L/D)+2

v) prod technique – a. Current shall be minimum 100 amp / inch to maximum 125 amp / inch for sections ¾” thick or

greater. b. For less than ¾” thick, the current shall be 90 amp / inch to 110 amp/inch prod spacing.

w) circular magnetization techniquea. Current shall be 300 amp / inch to 800 amp / inch of outer diameter

x) pit may not exceed ½ the part’s t min

y) Nozzle weld sizes UW – 16 & fig UW – 16Tc = not less than the smaller of ¼” or 0.7 t minTmin =smaller of ¾” or thickness of partsMin. throat required = t1,t2= smaller of 6mm or 0.7 t min

Min. inner corner radius (r1) = smaller of ¼ t or 19mmLeg = 1.414 x throat & throat = 0.707 x leg

z) The permissible ovality of vessel = Cross section passes through within 1 ID of the opening measured from the centre of the opening = 1% of nominal dia + 2 % of ID of opening

aa) Reinforcement = d ( t – tr ) or 2 ( t + tn ) ( t – tr ) use larger

bb) Widely scattered pits may be ignored the remaining thickness below the pit is greater than one-half the required thickness. (1/2 t required)

cc) The total area of the pits does not exceed 7 square inches (45 sq. centimeters) within any 8 inch (20 centimeter) diameter circle.

dd) The sum of their dimensions along any straight line within the circle does not exceed 2 inches

ee) Hydrostatic head = UG 99= 0.433 psi x each (1) foot of height

ff) Head depth (dish) = UG 32= ¼ x D for elliptical heads, ½ x D for hemispherical heads

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1) An existing 30” tall pressure vessel is stamped “RT-2” and is constructed of seamless 2:1 elliptical head s, shell material and head material is SA-515-60. All joints are type 2. Inside diameter of the vessel is 66”. The

vessel has corroded in the shell and heads approximately 1/8” and the original thickness of the both the heads and the shell was 5/8” according to the manufacturer data report. Stress is 14,760 psi. MAWP is 300 psig @ 670 deg F. the minimum thickness of the shell, considering hydrostatic head is __________?

a) 0.788” b) 0.701” c) 0.650” d) 0.500”

MAWP – 300 psig @ 670 deg F + HH = 312.99, ts = 0.500th = 0.500, id = 66, ir = 33, SV = 14,760650 = 15,000 50=600 15000E = RT2 = 0.90 700=14,400 1=12 x 20 = 15,000 – 240 = 14,760T = PR / SE - .6P= 312.99 x 33 / 14,760 x 0.9 - .6 x 312.99 = 10,328.67 / 13284 – 187.794= 10,328.67 / 13,096.20 =0.788” required Answer – A

2) The required thickness of the bottom head on the above ( Question no – 1 )vessel will be

a) 0.788” b) 0.701” c) 0.650” d) 0.500”

t = PD / 2SE - .2P = t = 312.99 x 66 / 2 x 14,760 x 1.0 – 2 x 312.99= 20657.34 / 29520 – 62.598 = 20657.34 / 29457.402 = 0.701Answer – B

3) During the internal inspection of a vertical vessel a minimum thickness reading of 1.270” was found on a shell course. The data report reflects an original thickness of 1.50”, and allowable stress at 600 degrees F (315 deg C) of 14,500 psi, full radiography, and a MAWP of 650 psi. The MAWP was based on a retirement diameter of 49.02 inches after the corrosion allowance was expended. Inspection records indicate a L.T. corrosion rate of 0.025 inches a year. If the corrosion rate does not change what is the maximum time before the next on stream or internal inspection? a) 5.6 yearsb) 10 yearsc) 90 daysd) 2.815 years

Ans = D ( API 510 7.1.1)Corrosion rate = 0.025” per year P=650, S = 14,500 psi, E=1.0 full RT, R=49.02 / 2 = 24.51”Calculate min shell t ( from UG 27 ) = PR / SE – 0.6 P = 650 x 24.51 / (14,500 x1.0 ) – (0.6 x650) = 15931.5/14110= 1.1290928 Present thickness = 1.2700000 retirement thickness = 1.1290928Remaining C.A = 0.1409072Remaining life = C.A / C.R = 0.1409072 / 0.025000 = 5.63 yearsNext inspection shall not exceed one half of the estimated remaining corrosion rate life of the vessel or 10 years, whichever is less Therefore : 5.63 / 2 = 2.815 years

4) A fifteen year old vessel made of SA-203. Gr.B was measured as having a thickness of 0.645”. The inside diameter is 42” and the MAWP is 450 psi at 650 deg F. the vessel did comply with UW-11(a) (5) (b). Material stress is 17500 psi. The previous reading on the material had a thickness of 0.68” during the first 10 year interval. Data report indicated that the material had a nominal thickness of 0.75”. (Take E=1).The corrosion rate is (in)a) 0.007b) 0.07c) 0.035d) 0.0035

Ans - CR = 0.68 – 0.645 = 0.035 /5 = 0.007 = Ans = A

5) For Q4, The required minimum thickness is (in)a) 0.548

b) 0.645c) 0.610d) 0.788

Ans –ATk = 0.645” after 15 yrs, D=42”, MAWP=450, S = 17,500 psi, tk after 10 yrs = 0.68”, E=1.0 , nominal tk=0.75” R=42 / 2 = 21”Calculate min shell t ( from UG 27 ) = PR / SE – 0.6 P T = 450 x 21 / (17500x1) – (0.6x450) = 0.548 = ans = A

6) For Q4, The remaining corrosion allowance is (in)a) 0.097b) 0.00350c) 0.035d) 0.0097

Ans - CA = 0.645 - 0.548 = 0.097 = ans =A

7) For Q4, The remaining service life in years isa) 5b) 2.5c) 13.8d) 6.9

RL = 0.097 / 0.007 = 13.85 yrs = ans = C

8) Vessel 60” ID, a nozzle opening of finished opening = 8 inches was present in thinned out patch on inside surface and lying close to a reinforcing nozzle with RF pad OD = 8”. What length will be taken for corrosion averaging is taken for corrosion analysis

a) 8 inches b) 16 inches c) 20 inches d) 30 inches

Ans - B

9) The minimum nominal thickness of new cylindrical shell with design pr = 555 psi, inside diameter = 58”, S = 20000 psi and E = 0.85 with corrosion allowance of 1/8” ( C.A for expected life of 20 years ) will be a) 1.0 inchb) 0.75 inchc) 1.125 inchd) 1.25 inch

T = PR / SE – 0.6 P= 555 x 29.125 ( R = 29.125” ( half of diameter ) / 20000 x 0.85 – 0.6 x 555 = 16164 / 16667=0.967” + 0.125 = 1.09” = Ans = C

10) After 10 years of service, extensive corrosion was observed on the shell over a patch lying entirely beyond 6 inches from weld for the vessel in Q133 above. Thinning as measured on this area was 3/16”, while on weld area; the corrosion was within expected limits. As on today, is the vessel is:

a) Safe for design conditionsb) unsafe for design conditionsc) safe only if welds are fully radiographed and okd) safe only if welds are UT checked and ok

Thickness away from weld ( required thk ) t = PR / SE – 0.6P take E=1T = 555 x 30 / 20000-333 = 0.8465”Available shell thickness = 1.125 – 0.1875 ( 3/16” = 0.1875)

=0.9375 > 0.8465 ( safe ) = ans = A

11) A 12 year old vessel made of SA-516-70 material has a current thickness of 0.955”. The inside diameter is 58” and MAWP of the vessel is 500 psi at 650 deg F. what is the minimum thickness required to operate the vessel in lethal service at the working pressure? The material stress at 650 deg F is 17,500 psi

a) 0.825” b) 0.955” c) 0.843” d) 0.765”

Ans -

12) What is the required thickness of a 60 inch ID cylindrical shell if: it has an allowable stress at design of 17500 psi. The vessel category A and D type 1 joints are fully radiographed. All category B joints are type 1 also and have been spot radiographed per UW-11(a) (5) (b). The vessel MAWP must be 350 psi at 450 deg F. the shell will see 11 psi of static head at its bottom.a) 0.6072” b) 0.7388” c) 0.7159” d) 0.6266”

ans=D ( UG-27 and API 510 body of knowledge)t=? from UG-27 =t = PR / SE-0.6 P P=350+11 = 361 psi S=17500 psi E=1.0 per UW 12(a) R=60/2=30

t=361 x 30 / (17500x1.0)-(0.6x361) = 0.6266”

13) What is the minimum hydrostatic pressure for a pressure vessel given the followingShell material is SA-240T304 S=10200 psi@900 deg F S=18800 psi @ 100 deg FT=1/2” CA=0 design temp=900 deg F MDMT = 50 deg FCategory “A” type 1 joint RT 1 diameter = 38”OD not for lethal service

a) 271.2 b) 499.8 c) 406.8 d) 749.7

Ans -

14) You are designing a vessel to contain water, with multiple course vertical vessel using rolled and welded plates without RT on all joints. The design parameters are Service = water tank Head and shell material = SA-516-70 MAWP = 250 psiTemp = 400 deg F ID = 24” type 1 welds stress = 17,500 piShell height=24 feet 2:1 seamless ellipsoidal headsWhat is the minimum material thickness you can use for the bottom shell course?

a) 0.2587” b) 0.2479” c) 0.2126” d) 0.1988”

Ans -

15) What is the minimum material thickness you can use for the upper head?

a) 0.2020” b) 0.2104” c) 0.3141” d) 0.2288”

Ans -

16) What is the minimum material thickness you can use for the lower head?

a) 0.2020” b) 0.2104” c) 0.3141” d) 0.2288”

Ans -

17) A vessel has undergone welded repairs and will require a hydrostatic test. This vessel is horizontal and has an ID of 6 feet. The vessel operates at 150 psi and the data report reflects an E of 0.85 for the vessel based on RT-3. The limiting thickness of the vessel is its shell coarse. The vessel is made of SA-515-gr 70 material with an allowable stress at temperature of 15500 psi. Its allowable stress at test temperature 17500 psi. What is the required thickness of this vessel and what shall the pressure be as read from a gage on the bottom of the

vessel during the test. The upper most part of the vessel is 6 feet above the centreline of the vessel’s horizontal shell. The static head of water is 0.433 psi per vertical foot.a) T = 0.3621 and gage pressure will be 169.3 psib) T = 0.875 and gage pressure will be 260.624 psic) T = 0.4234 and gage pressure will be 224.057 psid) T = 0.4234 and gage pressure will be 254.032 psi

ans=C (UG-99 and API 510 body of knowledge)t=? P=150+3.897=153.897 psi S=15500 psi design S=17500 psi testE=0.85 R=36” HH=0.3897 psiHydro pressure 1.3 x 17500 /15500 x 150 = 220.16 psi at the top gauge. Bottom gauge reads 220.16 +3.897 =224.057 psi during test. Thickness required from UG 27=t=PR/SE-0.6P=153.897 x 36 / (15500 x 0.85)-(0.6x153.897)=5540.292 / 13082.6618=0.4234Answers=t min 0.4234” pressure at bottom during test is 224.057 psi

18) A pressure vessel was manufactured and ready for hydro test. The operating pressure is 90 psi. The MAWP is 100 psi. the hydro test pressure for this vessel is

a) 130 psi b) 117 psi c) 150 psi d) 135 psi

Ans = a ( UG – 99)1.3 x MAWP x stress ratio = 1.3 x 90 x 100/90 = 130 psi

19) choose correct hydrostatic test pressure from given options for a vessel with following data: design pr = 200 psi, design temp = 675 deg F, MOC = SA 537 cl 1, allow stress ( ambient) = 18000psiallow stress (675 deg F)=15000 psi

a) 350 psig b) 260 psig c) 312 psig d) none

Ans = c ( UG – 99)1.3 x MAWP x stress ratio = 1.3 x 200 x 18000/ 15000 = 312 psi

20) A hydrostatic test is conducted on a vessel that is 60” tall. A pressure gauge is located 35” up from the bottom of the vessel. If the stress ration is 1.0 and the MAWP of the vessel is 100 psi, what approximate pressure should the gauge read to meet ASME VIII requirements?

a) 141 psig b) 145 psig c) 130 psig d) 186 psig

Ans - stress ratio = 1.0 Hydro pressure at top = 1.3 x MAWP x stress ratio =1.3x100x1.0=130 psigHydrostatic head at 35” elevation = 60” – 35” = 25 25 x 0.433 = 10.825 psigPressure at gauge level during test = 130 + 10.825 psig = 140.825 psig = Ans = A

21) A hydrostatic test is being performed on a vertical vessel with an elevation of 48 feet. The MAWP of the vessel in its normal operating position is 225 psi. All RVs has been removed and the vessel is blinded off from all other components. The gauge will be installed at an elevation of 22.5 feet. What shall this gauge pressure be during the hydrostatic test at the 22.5’ elevation? The material design stress allowed is 16,800 psi and the stress allowed at test temperature of 70 degree F (21 deg C) is 17,500 psi. H.H is 433 psi per vertical foot.a) 362.6 psib) 351.0 psic) 339.9 psid) 315.7 psi

Ans - D - UG 99MAWP - 225 psi stress at test - 17,500 psi stress by design - 16,800 psiGauge pressure during test at the 22.5’elevation=?Least ratio of stresses = 17,500 / 16,800 = 1.0416666Hydrostatic test pressure at the top = 1.3 x 1.04166 x 225 = 304.6870 pressure at top

Calculating H.H present at the 22.5’ elevation = 48.0 – 22.5 = 25.5 25.5 x 0.433psi = 11.0415 psiPressure at gauge level during test = 304.6870 + 11.0415 = 315.7285 psi

22) Choose correct hydrostatic pressure from given option for a vessel with the following data:Design pressure = 360 psiDesign temp = 675 deg FMOC = SA 537 cl.1Allow stress (ambient) = 18600 psiAllow stress (675 deg F) =16900 psi

a) 540 psigb) 515 psigc) 450 psigd) None of above

Ans - Hydrostatic test pressure = 1.3 x design pr x stress ratio ( ambient/deg F)= 1.3 x 360 x 18600 / 16900 = 515 psig = ans = B

23) The pressure vessel has design pressure = 300 psi. It’s safe stress values at ambient and design temperatures are 17100 psi and 16300 psi respectively. The hydrostatic test pressure and inspection pressure for above will be:a) 450 psi and 346 psib) 409 psi and 315 psic) 415 psi and 300 psid) None of the above

Ans - Hydrostatic test pressure = 1.3 x design pr x stress ratio ( ambient/deg F)= 1.3 x 300 x 17100 / 16300 = 409.141 psiInspection pressure = hydrostatic pressure / 1.3 = 409 / 1.3 = 314.65 psi = Ans = B ( check C? )

24) A pressure vessel has design pressure = 300 psi. It’s safe stress values at ambient and design temperatures are 19800 psi and 18000 psi respectively. The minimum hydrostatic test pressure and inspection pressure respectively for above will be :a) 450 psi, 390 psib) 390 psi, 330 psic) 429 psi, 330 psid) None of the above

Ans - C = UG – 99 (b) and (g)Hydrostatic test pressure = 1.3 x design pr x stress ratio ( ambient/deg F)= 1.3 x 300 x 19800 / 18000 = 429 psi

Inspection pressure = hydrostatic pressure / 1.3 = 429 / 1.3 = 330 psi = Ans = C

25) What is the gauge reading when performing a hydrostatic test on a vessel having an MAWP of 75 psi at 800 F. The material stress at 800 F is 12,800 psi and the material stress at hydro test temp is 17,500 psi. The gauge is located at the bottom of 50 ft vessel positioned vertically during hydro testing

a) 1538 psi b) 1560 psi c) 1025 psi d) 1047 psi

Ans -

26) A vessel is to undergo a hydrostatic test after a repair. The vessel’s stress allowable at the test temperature is 15000 psi and its stress allowable at its operating temperature is 13800 psi. The vessel’s MAWP at its design temperature is 125 psi. What will be the required test pressure at the top of the vessel?a) 203.8 psi b) 169.6 psi c)176.6 psi d) 143.7 psi

Ans - C ( UG-99)

Stress at test = 15000 psi design stress = 13800 psi MAWP = 125 psiTest pressure = 1.3 x 15000/13800 x 125 = 176.63 psi

27) What is the minimum hydrostatic pressure for a pressure vessel given the followingShell material is SA-240T304 S=10200 psi@900 deg F S=18800 psi @ 100 deg FT=1/2” CA=0 design temp=900 deg F MDMT = 50 deg FCategory “A” type 1 joint RT 1 diameter = 38”OD not for lethal service

b) 271.2 b) 499.8 c) 406.8 d) 749.7

Ans - ref UG 99 B

28) A vessel is undergoing a hydrostatic test after repair. The vessel is marked with an MAWP of 100 psig at 750 degree F ( 399 deg C). the materials allowable stress at operating temperature is 15,000 psi and its allowable stress at test temperature is 17,500 psi. the visual inspection shall take place ata) 100 psib) No less than the MAWP of the vesselc) 116.66 psid) 106.66 psi

Ans - C ( UG -99(b))1.3 x MAWP x least ratio of stressesRatio of stress = 17,500 / 15,000 = 1.166661.3 x 1.166 x 100 = 151.58”Inspection not less test press / 1.3 per UG 99 = 151.58 / 1.3 = 116.6 psi

29) A vessel is to be pneumatically tested to a pressure of 172.7 psi and this pressure has been adjusted for a stress ratio of 1.1666. the pressure which the vessel should be inspected at will be;a) 104.6 psib) 125.6 psic) 157.0 psid) 97.1 psi

Ans - c1.1 x 157 = 172.7UG 100 (d) - visual inspection at no loss than test pressure / 1.1 pressure = 172.7 / 1.1 = 157 psi

30) Choose correct pneumatic test pressure for a vessel with the following data:Design pressure = 160 psiDesign temp = 675 deg FAllow stress (ambient) = 20000 psiAllow stress (675 deg F) =19200 psi

a) 176.0 psigb) 215.6 psigc) 183.3 psigd) None of above

Ans - 1.1 x 160 x 20000/ 19200 = C ASME VIII, UG – 100 (b)

31) A 250 psi vessel (design pressure) is to be pneumatic tested in accordance with the original ASME Sec VIII div 1 rules. The material is SA-516-70 and has a design stress of 12000 psi and a stress of 17500 psi at ambient test temperature.

What is the pneumatic test pressure required?

a) 455 psi b) 401 psi c) 375 psi d) 313 psi

Ans - B ( ASME VIII, UG – 100 (B))Pneumatic test pressure = 1.1 x 250 x 17500 / 12000 = 401.041 psi

32) What is the minimum temperature at which a fabricated vessel can be tested?

a) 10 degrees above operation tempb) 10 degrees above MDMTc) 30 degrees above MDMT d) 30 degrees above zero

Ans - C ( ASME VIII, UG – 100 (C))

33) What is the first test pressure reading of the vessel gauge when pneumatically tested?

a) 401 psi b) 227 psi c) 187 psi d) 201 psi

Ans - D ( ASME VIII, UG – 100 (D))

34) What is the pressure reading when inspection is performed?

a) 401 psi b) 412 psi c) 364 psi d) 318 psi

Ans - C ( ASME VIII, UG – 100 (D))

35) A 250 psi vessel 100ft tall is to be pneumatic tested in accordance with the original ASME Sec VIII div 1 rules. The material is SA-516-70 and has a design stress of 12000 psi and a stress of 17500 psi at ambient test temperature.

What is the pneumatic test pressure required?

b) 455 psi b) 546 psi c) 375 psi d) 313 psi

Ans - ( ASME VIII, UG – 100 (B)) check

36) What is the minimum temperature at which a fabricated vessel can be tested?

a) 10 degrees above operation tempb) 10 degrees above MDMTc) 30 degrees above MDMT d) 30 degrees above zero

Ans - check thickness

37) What is the first test pressure reading of the vessel gauge when pneumatically tested?

a) 313 psi b) 227 psi c) 187 psi d) 273 psi

Ans - check

38) What is the pressure reading on the gauge after the THIRD step in testing the vessel?

a) 344 psi b) 412 psi c) 225 psi d) 318 psi

Ans - check

39) A pressure vessel is scheduled for pneumatic testing using the methodology of ASME code, Section VIII Div 1.with the MAWP = 635 psig at 375 deg F. The vessel is constructed of SA-516-65 material and is neither enabled for lined. The test metal temperature will be a minimum of 50 Deg F, per the owner specifications.

How many pressurization steps are needed to achieve the minimum required test pressure, and at what minimum pressure will the visual examination be performed?a) 6 steps : 635 psigb) 6 steps : 793.75 psigc) 9 steps : 635 psig d) 9 steps : 793.75 psig

Ans - A

40) A 250 psi ( design pressure) vessel is to be pneumatic tested in acc with original ASME section VIII div 1 rules. The material is SA-516-70 and has a design stress of 12000 psi and stress of 17500 psi at ambient test pressure. What is the pneumatic test pressure required?a) 455 psib) 401 psic) 375 psid) 313 psi

Ans - B ( SEC VIII – UG 100 B )Pneumatic test pressure = 1.1 x 250 x 17500 / 12000 = 401.041 psi = B

41) Following four new seamless (2:1) ellipsoidal heads are available for use under following design conditions. Choose the correct one.Design pr = 332 psi, Head I.D = 60” and S = 20000 psi with no C.Aa) Design thk = 0.7382”. provided thk after forming = 0.75”b) Design thk = 0.9788”. provided thk after forming = 1.0”c) Design thk = 0.4985”. provided thk after forming = 0.5”d) Design thk = 0.7377”. provided thk after forming = 0.75”

Thickness of new ellipsoidal head (2:1) is given by ASME formula, t = PD / 2SE – 0.2 P= 332 x 60 / 2 x 20000 x 1 – 0.2 x 332 = 19920 / 39933.6 = 0.4985”Provided thk of 0.5” is OK. Correct ans – C

42) Head in Q15 ( ellip head – B) value of crown whose radius can be taken as:a) 72”b) 60”c) 48”d) 54”

Ans - For ( 2 : 1 ) ellip head, for API analysis crown radius = 0.9 D = 54” Ans = D

43) For head B, the crown portion would lie within similar circle of diametera) 60”b) 30”c) 54”d) 48”

Ans - Crown portion lies in circle of diameter equal to 0.8D = 0.8 x 60 = 48” Ans = D

44) A vessel with 60 inches ID is to be provided std. torispherical heads. The heads shall be formed with crown radius and min. knuckle radius as follows ( head thickness = 1.2 inch):a) 60 inch, 4 inchb) 60 inch, 3.6 inchc) 25 inch, 3 inchd) None of the above

Ans - Crown radius h = D = 60” & knuckle radius = 6% of crown radius = 3.6 inch = Ans B ( UG32-j)

45) An ellipsoidal head has experienced corrosion at a radius 31 inches above the centre of the head. The head is attached to a shell that is 8” in diameter using a type 2 weld that has been fully radiographed. This corroded area may be considered within the spherical portion because the greatest distance allowed for this calculation from the heads centre is:a) 76.8 inches b) 38.4 inches c) 80.0 inches d) 40.0 inches

Ans - b ( API 510 – 7.4.6.1(b))80% of the shells diameter 0.80 x 96 = 76.8” Radius = 76.8 / 2 = 38.4 since 31” is with in the 38.4” radius it may consider within the spherical portion

46) An ellipsodical head has an internal diameter of 76 inches and depth of 21 inches ( including a straight face dimension of 2 inch ) . what should its spherical (crown) radius be?

a) 76.9 inches b) 15.6 inches c) 68.4 inches d) all the above

Ans - c ( API 510 – 7.4.6.3 )Spherical radius – K1D i.e D/2h = K1 and for many ellipsodical heads d/2h = 2.0Then from table 7-1 – 2.0 of d/2h equals 0.9 of K1= 0.9 x 76 = 68.4 inches

47) The crown portion for torisph. Head (ID 40 inch) may be considered as the portion lying entirely within a circle whose centre will be same as head centre and diameter will be

a) 36” b) 40” c) 32” d) 200”

Ans - API 510 – 7.4.6.1(b)

48) For 2:1 ellip head ( ID = 75 inches ), the crown portion would lie within similar circle of diameter

a) 60” b) 30” c) 54” d) 48”

Ans - API 510 – 7.4.6.1(b)

49) A vessel with seamless 2:1 elliptical head 60 inch ID, 400 psi design pressure, CA = 0, S = 20000 psig was inspected. Available thickness of only 5/8” was observed on the crown portion. The knucle thickness was found to be adequate. Your assessment is

a) Head thickness in crown portion is still ok for operationb) thickness is adequatec) depends on opinion of third party inspector d) API 510 inspector

Ans - A check

50) A 2 to 1 ellipsoidal head has experienced corrosion in the central portion of the head and will require calculations for continued service. Using the following information, what is to be used as the diameter of the spherical segment when evaluating the corrosion? The vessel‘s shell diameter is 96 inches.a) 96 inches b) 76.8 inches c) 80 inches d) 38.4 inches

Ans - b ( API 510 – 7.4.6.1(b) = 80 % of the shell’s diameter 0.80 x 96 = 76.8”

51) Following four new seamless standard torispherical heads are available for use under following design conditions. Design pr = 235 psi, Head I.D = 72” and S = 20000 psi with no C.A. Required minimum thickness of crown portion for analysis as per API 510 for head A shall be a) 0.3695”b) 0.4075”c) 0.4235”d) 0.4825”

For torisph head, for crown portion, use hemisphere formula using R = DT = PR / 2SE – 0.2P use R=D=72”, S = 20000 PSI, E = 1, P = 235 = 235 x 72 / 2 SE – 0.2 P = 16920 / 39933.6 = 0.4235” = Ans = C

52) Following four new seamless (2:1) ellipsoidal heads are available for use under following design conditions. Choose the correct one. Design pr = 332 psi, Head I.D = 60” and S = 20000 psi with no C.A. Required minimum thickness of crown portion for analysis as per API 510 for head B shall be a) 0.4490”b) 0.4262”c) 0.3652”d) 0.4863”

Again use hemispherical head formula for crown part: t = PR / 2SE -0.2PFor hemispherical head, use R = 0.9 D, S = 20000 psi, E = 1, P =332= 332 x 0.9 x 60 / 2 SE – 0.2P = 332 x 54 / 39933.6 = 0.4489 = Ans = A

53) A cylindrical shell has a seamless hemispherical head on one end and a seamless ellipsoidal head on the other end. The elliptical head is attached by a type 2 weld meeting the spot radiography requirements of UW-11(a) (5) (b). The shell has a single longitudinal joint. All category A butt welds are type 1 with full radiography applied. The vessel will require an MAWP of 50 psi. The material of the shell and head is SA-515-gr.60. The allowable stress at the operating temperature is 15000 psi. The heads inside diameter is 10’-0”. What is the required thickness of the 2 to 1 ellipsoidal head?a) 0.250 b) 0.235 c) 0.200 d) 0.1003

Ans = C (UG-32-d)P=50 S=15000 E=1.0 D=10’ x 12 = 120” t=?T= PD / 2SE – 0.2 P = 50 x 120 / (2x15000x1.0)-(0.2x50) = 0.200”

54) A vessel was designed to the 2001 edition of the ASME section VIII div 1 code. The ID of the head is 48 inches. The design pressure is 500 psig and the design temperature is 300 F. the vessel seams have been radiographically inspected per UW-51. The vessel material is SA-516-70. The corrosion allowance is 0.125 inch. The minimum required thickness of the 2:1 elliptical head per the code of construction isa) 0.940 inch b) 0.730 inch c) 0.610 inch d) 0.820 inch

Ans - B ( ASME Sec VIII UG 32 (d) checkT=PD / 2SE – 0.2P = 0.125 +(500)(48)/((2)(1)-(0.2)(500))=0.727

55) A hemispherical head formed from solid plate is 48.0 inches in inside diameter this head is attached to a seamless shell and has not had radiography on the category A type 1 weld that attaches the head to the shell. The vessel is horizontal and operates at 500 psi water pressure with an allowable stress on the head’s material of 15000 psi. The head’s thickness required is?a) 0.5741” b) 0.577” c) 1.1432” d) 0.2356”

ans = B ( UG-32)P= 500 + hydrostatic head H.H=4feet x 0.433=1.732 psiP=501.732 psi S=15000 L=48/2=24 t=? E=0.70 for category A weld no RTFrom UG 32 t=PL / 2E – 0.2 P = 501.732 x 24 / (2x15000x0.70)-(0.2x501.732)=0.576”

56) A pressure vessel has been in service for 12 years and has shown a history of corrosion over its service life. The original thickness was 1.9375” thick and the present thickness is 1.405”. What is the corrosion rate for this vessel?

a) 0.266250 in / yr b) 0.532500 in/yr c) 0.088750 in/yr d)0.044375 in/yr

Ans – d i.e. CR = (1.9375”-1.405”)/12 = 0.044375”

57) During an internal inspection a vessel’s shell was measured and found to have a remaining wall thickness of 0.486 inches. It was last inspected 4 years ago. The retirement thickness of the vessel shell is 0.475 inches and its previous thickness was 0.500 inches. Based on this data, per API 510 what is the present corrosion allowance of the vessel?a) 0.025”b) 0.110”c) 0.011”d) 0.014”

Ans - C ( API 510 body of knowledge )C.A = present – previous = 0.486 – 0.475 = 0.011”

58) Calculate the remaining life and the external inspection interval of a vessel the following data: actual thickness = 0.955 inch, minimum thickness required = 0.755 inch, thickness at previous inspection 5 year prior to present inspection = 1.025 inch

a) 20yrs & 10yrs b) 14yrs & 5yrs c) 28yrs & 10 yrs d)none

Ans - dCR = (1.025 – 0.955)/5 = 0.014RL = (0.955 – 0.755)/ CR = 14.28 yrsExternal inspection = lesser of 5 yrs or ½ RL = 5yrs

59) A vessel was inspected six years ago and had shell thickness =0.870”. as on today the thickness as reported by filed inspection is 0.786”. What was the corrosion rate over last six years

a) 10mpy b) 12 mpy c) 14mpy d) none

Ans - cCR = (0.870 – 0.786) / 6 = 0.014 i.e. 14 mpy

60) If the vessel in above question has minimum permissible thickness = 0.618” for the present design conditions. What will be estimated remaining life of the vessel if the corrosion rate remains same?

a) 10yrs b) 12 yrs c) 14yrs d) none

Ans - bRL = (0.786 – 0.618)/0.014 = 12 yrs

61) For a vessel in Q18 & Q19 find interval for next thickness inspection

a) 10 b) 6 c) 8 d) 5

Ans - b i.e. ½ RL

62) For a vessel which is planned for internal inspection after 8 years what should be the minimum remaining corrosion allowance in the vessel if corrosion rate is 150 microns / year

a) 2.0 mm b) 2.4 mm c) 1.2 mm d) none

Ans - RL = x / 150 microns = 8 yrs i.e. x = 150 microns x 8 yrs = 1.2 mm Corrosion rate = RL x 2 i.e. 1.2 x 2 = 2.4 mm Ans = b

63) Calculate the remaining life of a vessel given the following data: actual thickness = 0.955 inch, minimum thickness required = 0.787 inch, thickness at previous inspection 5 year prior to present inspection = 1.015 inch

a) 20yrs b) 14yrs c) 28yrs d)none

Ans - CR = 1.015 – 0.955 / 5 yrs = 0.012” per year RL = 0.955 – 0.787 / 0.012 = 14 yrs Ans = b

64) A pressure vessel has been in service for 12 years and has a history of corrosion over its service life. The original thickness was 1.9375 inches. The current thickness is 1.405 inches. What is the corrosion rate of this vessel?

a) 0.044 in/ yr b) 0.089 in/yr c) 0.266 in/yr d) 0.532 in /yr

Ans - CR = 1.9375 – 1.405 / 12 =0.044 in / year = a

65) A 15 year old vessel was measure to have a general average thickness of 0.645”. A previous inspection reading taken 10 years ago was 0,895” and the original data report of the vessel indicated that the thickness was 1.00”. To determine the remaining life of the vessel which corrosion rate would you as the inspector use?

a) 0.0210” b) 0.0180” c) 0.0236” d) 0,0250”

Ans - CR = 1.00 – 0.645 /15 yrs = 0.0236” = C

66) An existing pressure vessel has been inspected on a given year and found to be 0.750” thick in the shell and heads. The next inspection is performed 3 years later, and the thickness of the shell and heads is shown to be 0.560” (uniform). Calculations show that the vessel required thickness is 0.450”. From this information determine the following information, per API 510

a) Total metal loss b) corrosion rate c)CA d) remaining life e) inspection interval

Ans -

a) Total metal loss - 0.750” – 0.560” = 0.190”b) Corrosion rate - 0.190 / 3 = 0.0633” / yearc) CA - 0.560” – 0.450” = 0.110”d) Remaining life - 0.110 / 0.0633 = 0.86 check for correct anse) Inspection interval - check for correct ans

67) A fifteen year old vessel made of SA-203,Gr B was measured as having a thickness of 0.645”.The inside diameter is 42” and the MAWP is 450 psi at 650 Deg F. The backing ring was left in place and the vessel did comply with UW-11(a) (5) (b). Material stress is 17,500 psi. The previous reading on the material had a thickness of 0.68” during the first 10 year interval. Data report indicated that the material had a nominal thickness of 0.75”. The corrosion rate is

a) 0.007 b) 0.07 c) 0.035 d) 0.0035

Ans -CR = T initial – T actual / years i.e. 0.68” – 0.645” / 5 = 0.007” per year = Ans =a

68) As per above Q36, The required minimum thickness is

a) 0.548” b) 0.645” c) 0.610” d) 0.788”

Ans - check material SA 203 Gr B not correct

69) As per above Q36, The remaining service life in years is

a) 5 b) 2.5 c) 13.8 d) 6.9

Ans - first do Q37

70) During the internal inspection of a vertical vessel a minimum thickness reading of 1.270” was found on a shell course. The data report reflects an original thickness of 1.50”, and allowable stress at 600 degrees of 14,500 psi, full radiography, and a MAWP of 650 psi. The MAWP was based on a retirement diameter of 49.02 inches after the corrosion allowance was expended. This yields a retirement thickness of 1.129 inches. Inspection records indicate a corrosion rate of 0.025 inches a year. If the corrosion rate does not change what is the maximum time before the next on stream or internal inspection? a) 5 yearsb) 5.6 yearsc) 10 yearsd) 2.8 years

Ans - d1.270 – 1.129 / 0.025 = 5.64 yrs remaining lifeNext inspection no more than ½ remaining life or 10 years use the lesser5.64/2 = 2.82 years

71) A pressure vessel has been measured to have a minimum wall thickness of 0.235 inch after 5 years of service. The vessel’s original wall thickness was 0.250. Its minimum allowable wall thickness is 0.195 inch. How long until the next required inspection as per API 510?a) 13.3 yearsb) 10 yearsc) 5 yearsd) 6 and 2/3 years

Ans - d API 510 ( 7.1.1 )Wall loss = T initial – t actual = 0.250”-0.235”=0.015”Corrosion rate = 0.015 / 5 yrs = 0.003”/year = 3 mils / yearRem C.A = t actual – t required = 0.235-0.195 = 0.040 or 40 mils Remaining life = C.A / C.R = 0.040 / 0.003 = 13.33 yearsNext inspection = one half estimated C.R. life or ten years whichever less13.333 / 2 = 6.66 = 6 – 2/3 years

72) A vessel was placed in service on 1/15/60 with an original wall thickness of 0.875”. It lost wall at an average rate of 0.005” per year until 1/15/75. Service conditions have changed and this inspection, 1/15/96 the vessel has a present wall thickness of 0.630”. The previous thickness 5 year ago was 0.789”. If the present rate of corrosion continues and the retirement thickness is 0.525”, what will be the retirement year of the vessel?a) First quarter of 1997b) Second quarter of 1999c) Second quarter of 2001d) last quarter of 1999

ans -bMetal loss calculation : previous – actual = metal loss0.789”-0.630” = 0.159”Short term corrosion rate = 0.159 / 5 yrs = 0.0318”Previous – actual = remaining C.A = 0.630” – 0.525” = 0.105”Remaining life = remaining C.A / short term corrosion rate = 0.105” / 0.0318” = 3.3 years R.LPresent date + 3.3 yrs = second quarter 1999

73) What is the remaining life of a vessel corroding at a rate of 0.017 inch per year when the last inspection measured 0.638 inch and the minimum required thickness is 0.513 inch.a) 4 yearsb) 10 years

c) 6.3 yearsd) 7.3 years

Ans - dT actual – t required = 0.638” – 0.513 = 0.125”Remaining corrosion allowance = 0.125”Remaining life = t actual – t required / corrosion rate = 0.125 / 0.017 =7.35 yrs

74) For a vessel find estimated remaining life and next planned external inspection if actual thickness measured during inspection is 0.45” and minimum required thickness is 0.37”. Estimated corrosion rate is 10 mpy.a) 10 years, 5 yearsb) 8 years, 5 yearsc) 8 years, 4 yearsd) 10 years, 10 years

Actual tk = 0.45” min tk reqd= 0.37” CR = 10 mpyRL = 0.45-0.37 / 10 = 0.08 / 10 mpy = 8 yrsInternal inspection = ½ RL or 10 yrs whichever is less = 4 yrsExternal inspection = does not exceed lesser of 5 yrs or the required internal inspection = 4 yrsAns = 8 yrs of RL and 4 yrs of next planned internal inspection = C

75) In above question, if estimated corrosion rate is 20 mpy. The next planned inspection and external inspection shall not be later than a) 8 years, 4 yearsb) 4 years, 4 yearsc) 2 years, 2 yearsd) 8 years, 5 years

Ans = CRL = 0.45-0.37 / 10 = 0.08 / 20 mpy = 4 yrsInternal inspection = ½ RL or 10 yrs whichever is less = 2 yrsExternal inspection = does not exceed lesser of 5 yrs or the required internal inspection = 2 yrsAns = 2 yrs of internal and 2 yrs of next planned external inspection = C

76) For a vessel remaining life is estimated as 24 years from now, next planned internal and external inspection shall not be later thana) 12 years, 5 yearsb) 10 years, 10 yearsc) 10 years, 5 yearsd) None of the above

Ans = C RL = 24 yrsInternal inspection = ½ RL or 10 yrs whichever is less = 10 yrsExternal inspection = does not exceed lesser of 5 yrs or the required internal inspection=5 yrs Ans = 10 yrs of internal and 5 yrs of next planned external inspection = C

77) For a vessel remaining life is estimated as 3 years from now, next planned internal and external inspection may be planned ________ years from nowa) 1.5 years, 3 yearsb) 2 years, 2 yearsc) 1.5 years, 1.5 yearsd) 2 and 3 above

Ans = D RL = 3 yrsInternal inspection = ½ RL or 10 yrs whichever is less = 1.5 yrsExternal inspection = does not exceed lesser of 5 yrs or the required internal inspection = 1.5 yrsAns = D

78) For a vessel remaining life is estimated as 2 years from now, next planned internal and external inspection shall not be later thana) 1 years,1 yearsb) 2 years, 2 yearsc) 1 years, 2 yearsd) any of the above

Ans =AInternal inspection = ½ RL or 10 yrs whichever is less = 1 yrExternal inspection = does not exceed lesser of 5 yrs or the required internal inspection = 1 yrAns = 1 yr of internal and 1 yr of next planned external inspection = A

79) for a non-continuous corrosive service with remaining life 25 years, external inspection shall be performed not later thana) 10 yearsb) 12,5 yearsc) 5 yearsd) None of above

Ans = CInternal inspection = ½ RL or 10 yrs whichever is less = 10 yrsExternal inspection = does not exceed lesser of 5 yrs or the required internal inspection = 5 yrAns = 10 yrs of internal and 5 yrs of next planned external inspection = C

80) A vessel which was inspected six years ago and had shell thickness = 0.870”. As on today the thickness as reported by filed inspection is 0.786”. What was the corrosion rate over last six years?a) 10 mpyb) 12 mpyc) 14 mpyd) None of the above

Ans = CShell tk = 0.870”@ 6 yrs ago now tk =0.786”CR = 0.870 – 0.786 / 6 = 0.084” / 6 = 84 mpy / 6 = 14 mpy = Ans = c

81) If the vessel in above question has minimum permissible thickness = 0.618” for the present design condition. What will be estimated remaining life of the vessel if corrosion rate remains same?a) 10 yrsb) 12 yrsc) 14 yrsd) None of the above

Ans = BPermissible tk = 0.618” now tk = 0.786” RL = 0.786-0.618 / 14 = 12 yrs = Ans = B

82) Calculate the remaining life and external inspection of a vessel given the following date: actual thickness – 0,955 inch, minimum thickness required – 0.759 inch, thickness at previous inspection 5 years prior to present inspection – 1.025 inch.a) 20 years, 10 yearsb) 14 years, 5 yearsc) 28 years, 10 yearsd) None of the above

Ans - BCR= 1.025 – 0.955 / 5 = 0.014”RL = 0.955 – 0.759 / 0.014 = 14 yrsInternal inspection = ½ RL or 10 yrs whichever is less = 7 yrs

External inspection = does not exceed lesser of 5 yrs or the required internal inspection = 5 yrAns = 14 yrs of RL and 5 yrs of next planned external inspection = B

83) For a vessel remaining life is estimated as 22 years from now, next planned internal and external inspection shall not be later thana) 11 years, 5 yearsb) 10 years, 5 yearsc) 5 years, 5 yearsd) None of the above

Ans = B RL = 22 yrsInternal inspection = ½ RL or 10 yrs whichever is less = 10 yrsExternal inspection = does not exceed lesser of 5 yrs or the required internal inspection = 5 yrAns = 10 yrs of internal and 5 yrs of next planned external inspection = B

84) An existing vessel is inspected on a given year and found to be 0.500” thick in the shell and heads. The next inspection is performed 3 years later, and the thickness of the shell and heads is shown to be 0.310” (uniform). The vessel minimum thickness is calculated to be 0.200". from this information, the remaining life (RL) and internal inspection interval (I.I)on this vessel is approximatelya) 5 years RL, 10 years I.I b) 7 years RL, 15 years I.I c) 1.73 years RL, 1.73 years I.I d) 10 years RL, 10 years I.I

Ans - permissible tk – 0.500” next inspection – 3 years later and present tk = 0.310”, min tk reqd = 0.200”CR = 0.500 – 0.310 / 3 = 0.0633”RL = 0.310-0.200 / 0.0633 = 1.73 yearsInternal inspection = ½ RL or 10 yrs whichever is less or whenever the remaining life is less than 4 years, the inspection interval may be the full remaining life upto maximum of 2 years. Hence 1.73 yrs I.IAns = 1.73 yrs of RL and 1.73 yrs of next planned internal inspection = C

85) A vessel shell course was replaced on September 1st of 1982. The initial thickness was 0.378” on sep 1st of 1982. The second reading for thickness was on sep 1st of 1987 and was 0.364”. the last thickness measured was 0.345” in sep 1st of 1992. Based on the most aggressive corrosion what is the value in mils per year to be used in calculating the remaining life of the vessel?a) 3.8 mils / yr b) 5.15 mils/yr c) 4.77 mils/yr d) 3.3 mils/yr

Ans - A ( API 510 – 7.1.1.1)Time in years long term = sep 1st 1982 to sep 1st 1992 = 10 yearsCorrosion rate (L.T) = t initial – t actual / time = 0.378 – 0.345 /10 = 0.0033 – 3.3 mils/yrTime in years short term = sep 1st 1987 to sep 1st 1992 = 5 yearsCorrosion rate (S.T) = t previous – t actual / time = 0.364 – 0.345 /5 = 0.0038 – 3.8 mils/yrAns – the most aggressive corrosion is short term – 3.8 mils /yr

86) A vessel was built to the 1980 edition of section VIII Div 1 and put into service that year. Shell material is SA516-70 with an internal diameter of 120 inches. All welded seams were radiographed. The design pressure is 140 psig and design temperature is 650 deg F. origin nal shell thickness is 5/8 in with 1/8 in corrosion allowance. An inspection in dec 1999 showed a wall thickness was 0.610 in. in dec 2002, the inspected wall thickness was 0.585 in. what is the long term corrosion rate?a) 0.0018 in/yr b) 0.018 in / yr c) 0.015 in / yr d) 0.0015 in / yr

Ans - A ( API 510 6.4)LTCR = ( 0.625 – 0.585)/22 = 0.0018 in /yr

87) A vessel put into service in december 1980. The vessel had the original shell thickness of 0.625 inches. An inspection in December 1990 measured a wall thickness of 0.610 inches. In December 2002, the inspected wall thickness was 0.585 in. what is the long term corrosion rate?a) 0.0015 in /yr b) 0.018 in / yr c) 0.0018 in / yr d) 0.015 in /yr

Ans - C ( API 510 – 6.4)LTCR = (0.625-0.585)/22 = 0.0018 in / yr

88) A vessel is ultrasonically inspected when it is new and cold in June 1987 and found to be 0.800” thick in the shell. The next inspection is done in June 1995 and the vessel found to be 0.650” thick. If the minimum thickness by calculation is 0.510”, when (approximately) should the next internal or on stream inspection be scheduled per API 510?a) Feb 1999b) Nov 2010c) Jun 2005d) Oct 2002

Ans - AJune 1987 - 0.800”June 1995 - 0.650” ( after 8 years )Min tk - 0.510”CR - 0.800 – 0.650 = 0.150”/8= 0.01875”RL - 0.650 – 0.510 / 0.01875 = 7.46 yearsInternal inspection = ½ RL or 10 yrs whichever is less i.e. 7.46 / 2 = 3.73 yrs = 44.76 yrs = feb 1999

89) During an internal inspection the following minimum thickness readings were taken on a vertical vessel. Top head present = 0.235”, shell coarse present = 0.225”, bottom head present = 0.195”.this vessel has been in service for 6.5 years and all components had an original thickness of 0.250 inch. The engineer calculated a minimum thickness of the top head to be 0.145” and the bottom head minimum thickness to be 0.155” and the shell’s minimum thickness to be 0.205”. Based on the readings what is the corrosion rate for the shell element?a) 0.025”/year b) 0.0023”/year c) 0.020”/year d) 0.00384’ / year

Ans - D (API 510 – 7.1.1.1)0.250 – 0.225 / 6.5 = 0.0038461

90) For a vessel which is planned for internal inspection after 8 years. What should be the minimum remaining allowance in the vessel if corrosion rate is 150 mpya) 2.0 mmb) 2.4 mmc) 1.2 mmd) None of the above

Ans - D – check

91) A round weld tensile specimen is tested and found to fail in the base metal at a load of 14.876 lb. the specimen was measured before the test to have a diameter of 0.503 inches. The ultimate tensile strength of this specimen is: ( hint : the area of circle is equal to 3.141 (pi) times its radius squared)a) 74.878 psi b) 18.829 psi c) 235.185 psi d) 74.380 psi

Ans - A (IX art I QW-153)load = 14876 lbs diameter=0.503” radius=0.503/2 = 0.2515 area of a circle = pie r square= 3.141 x (0.2515)2=3.141x0.2515x0.2515=0.19867”Finally compute the tensile strength = load / area = 14876 / 0.19867 = 74877.938 psi

92) A 3-1/2 inch thick plate weld procedure test coupon must be tensile tested. This will require the cutting of the test coupon into smaller pieces. The tensile testing machine is capable of pulling up to 1” thick specimens. How many pieces will be required to for a complete test per section IX?a) 6 b) 8 c) 2 d) 12

Ans - B ( IX art I QW-151.1(c))thickness of coupon=3.5” 3.5 /4 = 0.875” ( 4 pieces for each tensile specimen)2 tensile tests are always required therefore; 2 x 4 = 8 pieces

93) You have received the following tension test results for a welding procedure. What is the ultimate unit stress for these two specimens?Specimen width thickness area ultimate total load lb ultimate unit stress1 0.750 0.410 0.308 22600 ?2 0.748 0.409 0.306 25800 ?

a) 72380 & 83314 b) 72377 & 83296 c) 73377 & 84314 d) 74377 & 83314

Ans - C (Sec ix – QW-152)Specimen 1 = 22600/0.308 = 73376 psi specimen 2 = 25800 / 0.306 = 84313 psi

94) The ultimate load on a 0.510” diameter tensile specimen is 15 ksi. The ultimate tensile strength on this specimen is approximately :

a) 73,529 psi b) 90,200 psi c) 85,100 psi d) 18,359 psi

Ans -

95) Two plate tensile test specimens have been tested and found to be acceptable. The characteristics of each specimen are as follows : specimen 1 : width =0.752”, tk =0.875”,UTS=78,524 psispecimen 2 : width =0.702”, tk =0.852”,UTS=77,654 psiWhat is the ultimate load for each specimen that was reported on the lab report?

a) 51,668 & 46,445 b) 67,453 & 56,443 c) 78,524 & 77,654 d)none

Ans - a UTS x width x thk

96) A pressure vessel has been designed for external pressure using SA-516-60 material. The total length of support is 8 ft. The MAWP ( external) stamped on the vessel is 250 psig@500 deg F. The OD is 48” can this vessel be safety operated with a minimum shell thickness of 0.250”? (yield strength = 28ksi)a) Yes. Vessel is acceptableb) No. vessel is not acceptablec) Not enough information given to form an answerd) Material cannot be used for external pressure without full RT.

Ans - SA – 516-60 yield = 32,000 ( see ASME Sec II part D – page 10 for YS of SA 516 Gr 60 )L=8’ or 96” P=250 Do=48’ t=0.250” Do/t=48/0.250=192Do/t=192 L/Do=2 =96/48=2A=0.00025 b=3400Pa= 4 x 3400 / 3 (192) = 13,600 / 576 =23.611No. this vessel cannot be operated at 250 psig. Answer = B

97) A pressure vessel has been designed using SA516-60 material (32,000 yields). Total length of support is 8 ft. The stamped MAWP (external) is 150 psig@500 deg F. The OD is 48”. Can this vessel be safely operated if the thickness has corroded down to 0.250”

a) Yes. This vessel is safe down to 0.250”

b) No. this vessel is not safe for 150 psigc) Not enough information provided to solve the problemd) Yes. This vessel could actually be operated at 250 psig

Ans - copied from Q52. Check for correct answer SA – 516-60 yield strength = 32,000 ( see ASME Sec II part D – page 10 for YS of SA 516 Gr 60 )L=8’ or 96” P=150 Do=48’ t=0.250” Do/t=48/0.250=192Do/t=192 L/Do=2 =96/48=2A=0.00025 b=3400Pa= 4 x 3400 / 3 (192) = 13,600 / 576 =23.611 No.this vessel cannot be operated at 250 psig. Answer = B

98) A small exchanger has experienced internal corrosion of its tubes. The tubes are required to withstand an external pressure of 350 psi. The tubes have and OD of 2.0 inch and are 60 inches in length with a wall thickness of 0.025 inch. What would be the approximate factor a value for this combination of length diameter and thickness of tube?a) 0.0056 b) 0.00055 c) 0.057 d) 0.10

Ans - B (UG-28)Calculate Do / t and L/Do Do/t = 2.0 / 0.025 = 80 L/Do = 60/2.0 = 3.0Factor A = 0.00055

99) A pressure vessel (MOC 516 gr 70) has external jacket spread over 72 inches length of the shell. ( jacket pressure = 180 psi, jacket temperature = 350 deg F) the vessel OD = 36 inches, shell thickness=0.5 inch, CA=nil. Decide which of the following are applicable in the present casea) Shell thickness is okb) Shell thickness is not okc) Data are not adequated) Acceptance of shell thickness depends on opinion of vessel engineer

Ans - L / Do = 72 / 36 = 2.0 and Do / t = 36 / 0.5 = 72From fig G of ASME II part D , A=0.002From fig CS-2, B=11500Allowable ext pressure = 4B / 3 (Do/t) = 4 x 11500 / 3 x 72 = 212.9 psi > 180 psi =ok= Ans =A

100) A 30” ID 0.280” thick circular man way has been welded to an existing pressure vessel with a full penetration weld into the shell. The shell is made of SA-516-70 material is 0.625” thick and is 72”ID. The manhole is SA-106-Cand extends past the vessel wall. 3” on the outside of the vessel. The man way is flushed to the ID of the vessel. MAWP is 250 psig@600 deg F. The man way does not pass through a welded joint and has a ½” x ½” leg fillet weld around the nozzle on the top of the full penetration weld. Radiography is RT-1. The required (calculated) thickness of the shell is 0.518” and the required (calculated) thickness of the nozzle 0.216”. From this information, this nozzle is

a) Is adequately reinforced without a repadb) Does not need to be calculated for additional reinforcementc) Will need reinforcing pad to be acceptabled) Not enough information to form an answer

Ans - A = 30 x 0.518 x 1 +0 = 15.54A1=30( 1x0.625-1*0.518)-0=3.21A1=2(0.625+0.280)(1x0.625-1x0.518+0)=1.81 x 0.107 =0.1936A2=5(0.280-0.216)1x0.625=0.20A2=5(0.280-0.216)1x0.280=0.0896A3=0A41=0.5 square x 1 = 0.25A43=03.21+0.0896+.25=3.549 opening need pad

101) Two nozzles are to be added to an existing pressure vessel. One nozzle is 3.5 in OD and the other is 2 in OD. How close may these nozzles be placed in the vessel wall without their areas of reinforcement overlapping?a) 2.5 inchesb) 3 inchesc) 5.5 inchesd) 2 inches

Ans - C ( UG 42 a ) page 49Minimum distance without overlapping areas of reinforcement per UG – 42 not less than two times their average diameter3.5” nozzle 1 + 2.0” nozzle 2 = 5.5” /2 = 2.75” average diameter2 x 2.75” = 5.5” answer

102) For a certain nozzle to shell joint ( no CA reqd) following data presented.Shell thickness provided -=0.42 inch, nozzle ID = 3.2 inch, pad size = 6.4 in OD, thk of pad = 0.42

a) Nozzle is adequately reinforcedb) Reinforcement is inadequatec) Data is inadequate - nozzle thickness must be givend) Data is inadequate - design shell thickness must be given

Ans - A

103) 6” nozzle, vessel thickness = 0.500, required t = 0.416. Nozzle thickness = 0.3, diameter D = 5.71. Calculate the reinforcement area?

Ans -

104) A steel bar is 15” long x 3” in diameter. What are the proper ampere/turns to be applied when using the coil technique?

a) 4375 b) 5000 c) 35000 d) 2000

Ans - for coil technique ( longitudinal magnetization technique ) - if L/D ratios less than 4 but not less than 2 = ampere turns = 45000 / (L/D)- if L/D ratios equal to or greater than 4 = ampere turns = 35000 / (L/D)+2- 35000 / ( 15/3 ) +2 = 5000 ampere turns = Ans = B

105) On 1” thick piece that will be MT tested by prods, the spacing between the prods will be 5”. The correct amperage range for this spacing would be

a) 1000 – 2000 amps b) 150-300 amps c)100-125 amps d) 500-625 amps

Ans - prod technique - Current shall be minimum 100 amp / inch to maximum 125 amp / inch for sections ¾” thick or

greater. For less than ¾” thick, the current shall be 90 amp / inch to 110 amp/inch prod spacing.

- 100 to 125 amp /inch x 5” = 500 to 625 amps = ans = d

106) A round steel bar is 14” long x 7”diameter. What is the proper amperage to be applied when using the circular direct contact technique?

a) 4900-6300 b) 2100-5600 c) 5 turns d) 1500-1600

Ans - circular magnetization technique

- Current shall be 300 amp / inch to 800 amp / inch of outer diameter- OD = 7” i.e. 300 x 7 to 800 x 7 = 2100 to 5600 = ans = B

107) If 6000 amp-turns are required to magnetize a part, how many amps are required for a 5-turn coil?

a) 1000b) 6000c) 1200d) 100

Ans - C

108) A vertical vessel has been determined to need de-rating to a lower MAWP. The new MAWP is based on the engineer’s calculation for pressure allowed on the top shell section listed below. The vessel has an elevation of 55 feet to its very top. The following pressure is calculated to the bottom of the element listed. Top shell section, elevation 40.3 feet, P=458.3 psi. What is the MAWP of this vessel based on this calculation?a) 439.6 psib) 440.8 psic) 451.9 psid) 457.6 psi

Ans - C ( UG 99 & body of knowledge)Total elevation = 55 feet , elevation of interest = 40.3 feet ( bottom of top shell coarse)55.0-40.3 = 14.7 feet of H.H14.7’ x 0.433 psi = 6.365 psi, MAWP of vessel shell coarse at its bottom – H.H present= 458.3 psi – 6.365 psi = 451.935 psi Do not exceed 451.9 at the top or pressure will exceed 458.3 at the 40.3’ elevation

109) A vessel is constructed of rolled and welded SA-516 gr 70 steel plate. It is 96 inches in diameter with a communicating chamber, which extends 18 inches below this horizontal vessel. This vessel is in water service and operates at working pressure of 150 psi. What will be the gauge pressure at working pressure as read on a gauge mounted on the bottom of the vessel’s communicating chamber?a) 150 psib) 154.11 psic) 153.46 psid) 199.36 psi

Ans - B ( UG 99 – API 510 body of knowledge )H.H= 8 + 1.5 = 9.5’ x 0.433 = 4.1135 psiPressure at the bottom = Working pressure 150 + hydrostatic head 4.1135 = 154.1135 psi

110) A column was hydro tested having total height of 45 feet. The top pressure gauge is reading 45 psig. The bottom pressure gauge will read approximately

a) 22.5 psig b) 67.5 psig c) 45 psig d) none of the above

Ans = = 1.5 top pr = bottom pr = 67.5 psig

111) Following four new seamless standard torispherical heads are available for use under following design conditions. Choose the correct one.Design pr = 235 psi, Head I.D = 72” and S = 20000 psi with no C.Aa) Design thk = 0.9882”. provided thk after forming = 1.0”b) Design thk = 0.4886”. provided thk after forming = 0.5”c) Design thk = 0.6184”. provided thk after forming = 0.625”d) Design thk = 0.7496”. provided thk after forming = 0.75”

Ans - D

Thickness of new torispherical head is given by ASME formula, t = 0.885PL / SE – 0.1 PL= crown radius therefore L = diameter of head, E = 1 ( seamless )= 0.885 x 235 x 72 / 20000 – 0.1 x 235 = 14974.2 / 19976.5 = 0.7496 mmProvided thk of 0.75 is OK. Correct ans – d

112) Head in Q62 ( torisph head – A) is to be analysed according to API 510, after placing the vessel in service for 10 years. For the purpose of API 510 analysis, value of crown whose radius can be taken as:a) 72”b) 60”c) 48”d) 54”

Ans - Torisph head D = 72” For API 510 analysis crown radius = diameter = 72” = Ans = A

113) The crown portion for head A may be considered as the portion lying entirely within a circle whose centre will be same as head centre and diameter will bea) 68.5”b) 36.4”c) 57.6”d) 54.0”

Ans - Crown portion lies in circle of diameter equal to 0.8D = 0.8 x 72 = 57.6” Ans = C

114) The standard seamless , ASME F &D head (torispherical) is used for the following conditions:Material of construction = SA515 gr 60Design pressure =250 psigDesign temp =580 deg FInside crown radius =72 inchesMinimum available thk =1.14 inchesCategory B welds are spot radiographed. Corrosion allowance = 0.125 inchesAllowable stress at design temp = 16400 psigYour assessment isa) The head does not meet code requirementsb) The head meets the code requirementsc) Suitability would depend on decision of pressure vessel inspectord) Data are not adequate

Ans - Ref ASME VIII, UG – 32 (e), t = 0.885 PL / SE – 0.1P +CE=1 (as head is seamless and category B welds are spot radiographed)= ( 0.885 x 250 x 72.125 / 16400 x 1 – 0.1 x 250 ) + 0.125 = (15957.6 / 16375)+0.125=0.974”+ 0.125 = 1.0995”Available thickness = 1.14” = ok = Ans = B

115) A circular flat head at the top of a vessel is measured at 1.25 inches thickness during inspection. The flat head has a diameter of 14 inches. The vessel’s data is as follows: MAWP is 500 psi C = 0.33 S=17,500 psi E=1.0. What is required thickness of this part?a) 1.35940 inchb) 1.24563 inchc) 0.958633 inchd) 1.75200 inch

Ans - A ( sec VIII UG – 34 )P = 500 psi s=17500 E=1.0 C=0.33 d=14”From UG 34 t= d square root of CP/SET= 14 square root of 0.33 x 500 / 17500x1.0 = 14 square root of 165/17,500=1.3594056”

116) A flat unstayed circular head with a diameter of 14” is operating at 350 psi@500 deg F. The stress value is 17,500 psi with an efficiency of 1.0 and a C factor of 0.33. Can this head be operated with the corroded thickness of 1, 25?

a) Yes. Can continue to operateb) No. cannot be operatedc) C factor is incorrect for designd) Circular heads cannot be use at 1.25”

Ans -AP = 300 psi s=17500 E=1.0 C=0.33 d=14”From UG 34 t= d square root of CP/SET= 14 square root of 0.33 x 300 / 17500x1.0 = 14 square root of 99/17,500=1.0529”

117) A flat circular head of 24” ID is attached by inside and outside fillet weld as shown in UG – 34 sketch f. the head is made from SA 516 gr 70 material and the MAWP is 200 psig@500g. assuming the fillet weld comply with the code and a C factor of 0.20. what is the minimum thickness required for this head?

Ans -P = 200 psi s=17500 E=1.0 C=0.20 d=24”From UG 34 t= d square root of CP/SET= 24 square root of 0.20 x 200 / 17500x1.0 = 14 square root of 40/17,500=1.14”

118) A circular flat head is to be attached to a cylindrical shell as depicted in fig.UG-34(f). The value of the thickness required for the shell has been computed to be 0.358” and the actual thickness is 0.500”. The value of factor C in this calculation will be?a) 0.46089b) 0.20c) 0.23628d) 0.33

Ans - C Fig UG – 34 ( f)C = 0.33 x m , m = Tr / Ts = 0.358 / 0.500 = 0.716”C = 0.33 x 0.716 = 0.23628

119) A vessel has pitting in a small area; the area will fit in an 8 inch diameter circle with its centre at the deepest pit. The shell course has a minimum thickness of 0.740”. the pits within the circle are more than 2 inches apart in any straight line with the following depths;#1.385”#2.301”#3.235”#4.287”, this area cannot be ignored because;a) The pits are too much in numberb) The pits are too close togetherc) The square inches of the pits exceed 7 square inchesd) One of the pits is two deep to allow averaging

Ans -Pit depth of deepest pit = 0.385 Shell coarse minimum t = 0.740”0.740 / 2 = 0.370” < 0.385”

120) A cluster of scattered of pits was observed on exterior of vessel shell. The description of pits is as follows: nominal vessel thickness = 5/8”, CA = 1/8”, total area of pits = 5.8 sq.inch,Pit of largest depth : 0.5 inch dia, 0.20 deepAll pits could be enclosed in a circle of 8 inch dia and maximum cumulative length is 1.75 inch within circle.a) Pits may be ignored as per API510b) Pits are unsafe as the total area of the pits is beyond limitsc) The cluster is unsafe as depth of deepest pit is beyond limits

d) The cluster is unsafe, they are exceeding cumulative length criteria

Ans - AArea within 8” diameter circle = 5.8 sq.in (ok)Length along straight line in circle = 1.75 (ok)Depth 5/8” = 1/8” 0.25”(ok) Ans = A

121) If in above Q69, depth of deepest pit was 0.15”, cumulative length was 2.3 inch ( other data remaining same ) your analysis will be:a) The cluster is still safeb) Cluster is unsafe as depth is still beyond limitc) Unsafe as cumulative length is beyond limitd) 2 and 3

Ans - Length 2.3 inch > 2 inch----------(not ok) = Ans = C

122) If total area of cluster is 8 sq.in other data remain same as Q136, what is your evaluation?a) Cluster is unsafe due to increased cluster areab) Cluster is unsafe due to pit depth still being unsafec) Cluster is unsafe due to 1 and 2d) Cluster is safe

Ans - Area 8 sq.in > 7 sq.inch-----------(not ok)=Ans = A

123) A vessel shell has scattered corrosive pits caused by hydrochloric acid. The greatest pit depth found excluding corrosion allowance in 0.333 inch deep in the shell. Which has minimum required thickness of 0.521 inch? The sum of pit’s areas within an 8 inch diameter circle does not exceed 7 sq.inches. this pit must be repaired because the limit of the pit depth in this situation is:a) 0.1302” b) 0.2605” c) 0.3000” d) 0.3250”

Ans - b ( API 510 – 7.4.3 (a))shell’s t min = 0.521” pit depth = 0.333” the rule says that the pit may not exceed ½ the part’s t min = 0.521 / = 0.2605” the pit must be repaired

124) While inspecting a torispherical head it was discovered to have some severe corrosion and must be evaluated for continued service. This corrosion occurs at a radius 33.256” from the exact centre of the head. The shell diameter of the vessel the head is attached to is 7’-0”. To evaluate this head for continued service what is the greatest distance from the centre that the corrosion can be and still be allowed to use the formula for spherical shells given in section VIII Div.1 of the ASME code?a) 67.2”b) 48.0”c) 33.6”d) 84.0”

Ans - CAPI510 – 5.7 (h) - diameter = 80% of shell’s diameter= 0.80 x 84” ( 7’ x 12 ) = 67.2Greatest distance must be with a radius = 67.2 / 2 = 33.6 radius

125) The minimum throat dimension for a fillet weld as shown fig UW-16(c) for a 1” thick nozzle joined to a 2” thick head isa) 0.530”b) 0.750”c) 0.250”d) 1.00”

Ans - C

Tc = not less than the smaller of ¼” or 0.7 t minT min = ¾” = 0.7 x 750 = 0.525” Lesser of 0.250 or 0.525 = 0.250”

126) A cylindrical shell has been discovered to have uniform external corrosion. The shells original thickness was 7/8 inch; it is presently 0.745 inch in thickness. The original OD of the shell was 30 inches. The vessel operates at 650 deg F with a stress allowable on the material of 15000 psi. 100 % RT was performed on the vessel. All joints are types 1. What is the vessel’s present MAWP?a) 760 psi b) 772.13 psi c) 766.88 psi d) 382.10 psi

Ans - C ( UG-27 & UG-116)from appendix 1 P=Set / Ro – 0.4 t wall loss= 0.875 – 0.745 = 0.130” outside dia = 30.0”Ro=30.0/2 = 15.0-0.130 =14.87 corroded outside radius = 14.87”Calculate pressure allowed t=0.745 S=1500 psi Ro=14.87” E=1.0 P=?P=15000 x 1.0 x 0.745” / 14.87 – (0.4 x 0.745) = 766.88 psi

127) While inspecting a vessel you discovered the cylindrical shell has corroded externally to a present thickness of 0.498” in thickness at its thinnest point. The vessel’s corroded outside diameter is 95.82” and its material has an allowable stress of 13,800 psi at 750 deg F. the vessel shell is seamless and it has met the spot radiography requirements listed in UW-12(d). What will be the MAWP of the vessel?a) 71.87 psib) 143.5 psic) 144.0 psid) 83.45 psi

Ans - CAppendix 3, section VIII P = S E t / Ro – 0.4 tT = 0.498” E=1.0 Ro = 95.82 / 2 = 47.91” S = 13,800 psiP = 13,800 x 1.0 x 0.498 / 47.91 – ( 0.4 x 498) = 6872.4 / 47.7108 = 144.04

128) For a certain pressure vessel with spot radiography, and all welds type 1, OR ( Name plate of a certain pressure vessel has name plate stamping as “ RT 3 “ for the radiography, all Weds are type 1 welds. )

Vessel diameter = 72 inch IDMOC =SA 515 gr 70Design temp = 600 deg FAllowable stress at design temp = 20000 psiIf shell thk = 5/8” and CA = 1/8”, the MAWP for the shell will be

a) 234 psib) 168 psic) 256 psid) None of the above

Ans - ARef UG – 27 c (1) =MAWP = Set / R +0.6t ( E=0.85 from table UW-12)And t=5/8”-1/8” = ½” after deducting corrosion allowanceMAWP=20000 x 0.85 x 0.5 / 36.125 + 0.6 x 0.5 Note R in corroded condition = 36.125”=233.36 psi = Ans = A

129) A vessel was built to the 1980 edition of section VIII division 1 and put into service in 1980. Shell material is SA516-70 with an internal diameter of 120 inches. All welded seams were fully radiographed. The design pressure is 140 psig and design temperature is 650 deg F. original shell thickness is 5/8 in with 1/8 in corrosion allowance. In December 2002, the inspected wall thickness was 0.585 in. what is the maximum allowable working pressure to which the vessel can be re-rated per figure 7.1 of API 510?a) 193 psig b) 165 psig c) 207 psig d) 171 psig

Ans - A ( ASME Section VIII UG 27 (c) (1)

from UG 27 P=SEt / R + 0.6 t i.e P= (20000)(1)(0.585)/(60+(0.6)(0.585)) = 193 psigAssuming no corrosion allowance and a lower design temperature

130) A shell course is being replaced with the new course being 60 inches in inside diameter and 7/8 inches thick. The vessel coarse material is SA-515 gr 60 plate at a design temperature of 750 deg F the stress allowed is 13000 psi. the vessel joints are all type 2 and the vessel is stamped RT-3. What is the MAWP of this shell coarse?a) 298.11 psi b) 316.74 psi c) 150.35 psi d) 335.38 psi

Ans - A (UG-27 & UG-116)from UG 27 P=SEt / R + 0.6 t S=13000 E=0.80 t=0.875” R=60”/2=30”P=13000 x 0.80 x0.875 / 30 + (0.6x0.875) = 9100 / 30.525 = 298.11 psi

131) A sphere constructed of SA-516-70 has an ID of 80 feet. The joint efficiency of the welded joints is 0.85. the operating temperature will not exceed 200 deg F. if the minimum wall thickness is 2.5 inches. What is the maximum allowable working pressure of the sphere if it was designed to the 1998 edition of the ASME BPV code section VIII division 1, 2000 addenda?a) 155 psig b) 208 psig c) 88 psig d) 177 psig

Ans - A (ASME Section VIII UG 27 (d)P = 2Set / R + 0.2 t i.e P= 2(17500)(0.85)(2.5)/(480+0.2(2.5))=155 psig

132) The inside surfaces of a vessel’s shell have corroded down to a minimum thickness of 0.369”. The original thickness was 0.500”. The vessel’s new and cold inside diameter was 96 inches. What will be the internal dimension used to calculate the vessel’s MAWP?a) 96.262”b) 48.0”c) 96.0”d) 48.131”

Ans - D96/2 = 48” radius = 48” + ( 0.500 – 0.369) = 48.131”

133) A pressure vessel of overall height 80’ has its part MAWP at various parts as detailed in the sketch. What is the vessel MAWP ( hydrostatic head 1’ = 0.433 psi)?a) 419 psib) 420 psic) 430 psid) 426 psi

Ans - AFor location at 4 feet from top = 426 – 0.433 x 4’ = 424.265 psiFor location at 32 feet from top = 440 – 0.433 x 32’ = 426.144 psiFor location at 54 feet from top = 442.5 – 0.433 x 54’ = 419.118 psiFor location at 80 feet from top = 454.64 – 0.433 x 80’ = 420 psiMAWP is the lowest of above = 419.118 psi

134) Vessel MAWP is 200 psi . what will be the total pressure at a point 20 feet from top , if vessel is completely filled with liquid ( specific gravity = 1)a) 220.00 psib) 208.66 psic) 204.33 psid) 191.34 psi

Ans - for location 20 feet from top = 200 -0.433x20’ = 208.66 psi = Ans = B

135) A standard ellipsoidal head with a 1-1/2” flange is attached to the bottom of the vertical vessel. The heads inside diameter is 56 inches. The vessel has a total elevation of 100 feet. The elevation at this head to shell joint is 5 feet. What amount of hydrostatic head must be considered for use in this head’s required thickness calculation?a) 41.135 psi b) 41.909 psi c) 41 psi d) 41.69 psi

Ans - D (41.135 + 0.559(hydro hd. In head depth))=41.69depth of head = ¼ ID +flange56/4 +1-1/2”=14+1-1/2=15-1/2” hydrostatic pressure on head=95 feet + 1.2916 feet=96.2916 x 0.433 = 41.69 psi

136) For the given configuration total area to be compensated will be _________.data with usual notation. D=100mm, t=12mm, tn=6mm, tr=10.5mm, trn=5mma) 920 mm sq b) 870 mm sq c) 1020 mm sq d) 1050 mm sq

Ans- D ( ASME Sec VIII UG 37, fig UG 37.1 see A )

137) Two nozzles are to be added to an existing pressure vessel’s shell which has a thickness of ¼” (0.250”). One nozzle’s finished opening is 2.5 in ID and the other is 2 in OD. How close may these nozzles’s centres be placed in the vessel wall without requiring reinforcement calculations?a) 2.5 inchesb) 3 inchesc) 4.5 inchesd) 2 inches

Ans - C ( UG-42(a))Minimum spacing for the centres of nozzles must be not less than the sum of their diameters apart.2.5 + 2.0 = 4.5” therefore nozzles must be 4.5” a part center to center

138) An additional 4 inch nozzle was added to a pressure vessel and is similar to the one in fig UW-16.1(i). The minimum thickness of the nozzle is 0.377 inch and the shell minimum thickness is 0.750 inch. What is the decimal value of the minimum throat required for each of two equal leg size fillet welds on this attachment?a) 0.235” b) 0.267” c) 0.250” d) 0.750”

Ans - C (UW-16)UW-16 / fig UW-16.1(i). t min = 0.377”T1 + t2 >=1-1/4 t min = t1 + t2 >= 1.25 x 0.377 = 0.47125 ( total throat required)T1 or t2 = 0.47125 / 2 = 0.2356” ( split between two equal sized fillet welds)By the rule : t1 or t2 not less than the smaller of ¼ in or 0.707 x t min 0.707 x 0.377 = 0.2665 = 0.267”Ans : 0.250” no individual throat can fall below this value per this ruleThe t1 + t2 >= rule is still satisfied ( not less than 0.47125) since 0.250 +0.250 = 0.500”

139) A pressure vessel with nominal diameter 1200 mm was inspected for ovality of shell. An opening of 200 mm ID exists. The permissible ovality at the vessel cross section which is 150 mm from centre of the opening is :a) 12 mmb) 16 mmc) 14 mmd) 18 mm

Ans - Ans =B = UG 80 (a)2 page 62Cross section passes through within 1 ID of the opening measured from the centre of the opening = 1% of nominal dia + 2 % of ID of opening = 12 mm + 4 mm = 16 mm

140) A vessel shell is 65” I.D and is made from material with a stress value of 16200 psi. The vessel has category A type 1 butt welds and is stamped “RT-2”. If the thickness has corroded down to 0.285” thick. What MAWP can be allowed on this vessel?a) 400 psigb) 190 psigc) 250 psigd) 139 psig

Ans -

141) During an alteration, a 16” circular man way is attached to an unstayed vessel head by a method similar to that shown in fig UW 16.1(c) of Sec VIII. The head is SA-516-60 (S=15,000 psi), 48” ID and 1-1/8 plate thickness. Working pressure is 500 psi. The circular man way is SA-335-P12 (s=15,000 psi). ¾” wall thickness and does not extends to the inside surface of the head. Calculate the size of reinforcement window.

Ans -

142) Using the above Q, if the required thickness “tr” of the head to resist the 500 psi pressure is 7/8”, what is the area “A” required for compensation?

a) 13 sq.in b) 14 sq.in c) 15 sq.in d) 16 sq.in

Ans -

143) Using the above Q, the area of excess thickness in the head “A1” is

a) 4 sq.in b) 3.8 sq.in c) 4.5 sq.in d) 5.0 sq.in

Ans -

144) A 10” ID pressure vessel 2” thick made of SA-516-70 material, minimum thickness is 1.390” was designed for 400 deg F. A 24” ID SA-516-70 nozzle ¾” thick is to be installed during a shutdown. It will be installed in accordance with fig UW-16.1(a-1). The attaching fillet welds will have ¾” legs. A reinforcing pad 6” diameter by 1” thickness will be added. What is the size of the window?

Ans -

145) What is the MAWP for a pressure vessel where shell measurement thickness is 3/8” thick, SA-516-70(normalised). Corrosion allowance is 1/16”, ID is 42”, and the proposed MDMT is -20 deg F@MAWP? Category “A” joint is type 2 weld. Material stress is 17,500 psi

Ans -

146) Are impact test and PWHT is required for the above conditions?

Ans -

147) A vessel shell is 66” ID and is made from material with a stress value of 16,200 psi. The vessel has category A type 1 butt welds and is stamped RT-2. If the thickness has corroded down to 0.285”thick, what MAWP can be allowed on this vessel?

a) 400 psig b) 190 psig c) 250 psig d) 139 psig

Ans -

148) A standard seamless torispherical head has a skirt OD of 56”and a corroded thickness of 1.27”. The material stress value is 16,100 psi and the head is attached to the bottom of a vertical vessel that is 20” tall full of water. If the vessel is stamped RT-2, what approximate MAWP would be allowed on this vessel?

a) 600 psig b) 402 psig c) 704 psig d) 302 psig

Ans -

149) A forged cylindrical vessel having an outside diameter of 16” was measured to have an actual shell thickness 0f 0.312”. Both ends of the circular forged flat head are ½ inch. Using fig UG 34(b-1) what is the maximum allowable pressure for both the heads? S=17,500 psi

a) 104 psi b) 118 psi c) 123 psi d) 135 psi

Ans -

150) For a vertical with column internal dia =48” and height (T-T )=80ft, the vessel MAWP is 60psi. the vessel part MAWP for bottom dished head (2:1 ellip type ) will be

a) 94.6 psi b) 82.8 psi c) 95.5 psi d)none of above

Ans - C id = 48” = 48/12 = 4 ft & r = 2 fth = 82 ft.' head = 82 x 0.433 = 35.506 psi (BOK) & MAWP = 60 + 35.506 = 95.506 psi

151) For a vertical with column internal dia =48” and height (T-T )=98 ft, thehydrostatic head for bottom dished head (2:1 ellip type ) will be

b) 40.6 psi b) 42.4 psi c) 43.3 psi d)none of above

Ans - C id = 48” = 48/12 = 4 ft & r = 2 fth = 100 ft.' head = 100 x 0.433 = 43.3 psi (BOK)

152) A horizontal vessel is provided with 2:1 elliptical heads on both ends. The tan – tan shell length = 120 in. vessel dia = 40 in. What will be overall length of the vessel?a) 150 inches b) 140 inches c) 160 inches d) None of above

Ans -

153) A vessel is ultrasonically inspected when it is new and cold in June 1987 and found to be 0.800” thick in the shell. The next inspection is done in June 1995 and the vessel found to be 0.650” thick. If the minimum thickness by calculation is 0.510”, when (approximately) should the next internal or on stream inspection be scheduled per API 510?a) Feb 1999b) Nov 2010c) Jun 2005d) Oct 2002

Ans -

154) A flat unstayed circular head with a diameter of 14” is operating at 350 psi @ 500 deg F. the stress value is 17500 psi with an efficiency of 1.0 and a C factor od 0.33. Can this head be operated with a corroded thickness of 1.25”?

a) Yes. Can continue to operateb) No cannot be operatedc) C factor is incorrect for designd) Circular heads cannot be use at 1.25”

Ans -

155) Shell dia = 50” the pitch for the above reading is 4” x 4”. The required thickness for the shell is 2 (circumferential stress governed). What is the minimum thickness to be considered after averaging for calculation of MAWP?a) 1.322”b) 1.392”c) 1.316”d) 1.426”

Ans -

156) A 12” dia pipe, ½” wall is to be installed as a nozzle on the shell. With corrosion allowance of 1/16”. What is the minimum thickness “t” allowed to be used for calculating the MAWP of the pipe?

a) 0.500” b) 0.375” c) 0.437” d) 0.475”

Ans -

157) The minimum thicknesses available for a seamless torispherical head for the following data are given in options. Select suitable head thickness for a new construction of vessel. P = 200 psi, S = 17,500 psi, ID = 96 inches no CA

a) 1 inch thk b) 1.25 inch thk c) 0.75 inch thk d) 1.125 inch thk

Ans - A

158) For a given configuration, correct minimum weld dimension (x) required as per ASME Sec VIII div 1 shall bea) 10mm b) 12 mm c) 14mm d) none

Ans - B ( ASME SEC VIII – UW 16 fig (P))

159) A pressure vessel (MOC 516 Gr 70) has external jacket spread over 72 inches length of the shell ( jacket pressure =200 psi, jacket temp =350 deg F) the vessel OD = 36 inches, shell thickness = 0.5 inch, CA = nil . Decide which of the following are applicable in the present case.

a) Shell thickness is okb) Shell thickness is not okc) Data not adequate d) acceptance of shell thickness depends vessel engineer

Ans - A

160) A 15” OD vessel 3/8” thick is attached to a forged circular flat head as shown in fig UG 34(b-1) of Sec VIII. What is the minimum thickness required using the forged material with a stress value of 11,500 psi for the 150 psi vessel?

a) 0.635” b) 0.529” c) 0.565” d) 0.420”

Ans -

161)

Nozzle reinforcement :-

Total area required to be compensated Ar = d x tr mm2

Reinforcement limit (X) parallel to vessel wall X = 2d mm

Reinforcement limit(Y) normal to vessel wall Y = 2.5t or 2.5tn use smaller value

Extra area available in shell As = d(t-tr) mm2

Extra area available in Nozzle An= Y ( tn-trn)x2 mm2

Total area (Av) available inherant to the vessel Av= (As + An) mm2

Area ( Ap) to be provided by pad Ap = Ar – Av mm2

If pad OD = 320 mm, pad thickness (tp) tp = Ap ( pad OD – pad ID )