API 510 Problems

QUESTIONS 1. A vessel has inside diameter 45 inches , Material is SA-516 Gr.70N , operates at 750 psi and 700 degrees F (maximum) and stress value 17300 psi (MDMT is -30 F). The corrosion allowance is 0.125” and RT-1. One 2” sch. 160 (0.344”) nozzle (SA-106 B) is attached as per UW-16.1c to shell and pad material is SA-516 Gr. 70 , thickness 1.45” . The thickness of the vessel is determine by the last inspection is 1.45". The vessel has been in service for 10 years. The original thickness (measured when installed) was 1.5". Two years previous to the 1.45” measurement the thickness of the vessel was measured to be 1.47". Determine the greatest corrosion rate i.e. short or long term a) 0.0050 inches per year b) 0.0065 inches per year c) 0.0100 inches per year d) 0.0130 inches per year 2. In above question the next planned inspection is scheduled for 7 years. Using the worst corrosion rate (short or long term) determine what pressure the vessel will withstand at the end of its next inspection period ? a) 973.3 psi

b) 700 psi

c) 564 psi

3. Calculate the minimum required thickness in Q-1 a) 1.250 “ b) 1.00” c) 1.45” 4. Calculate the remaining life in Q-1 a) 47 Years b) 45.4 Years 5. Which curve Material is this?

d) 990 psi d) 1.34”

c) 50 Years

a) A b) B c) C 6. What is true about impact test of this vessel

d) 46 Years d) D

a) Required b) not required 7. Which is true for this vessel about minimum impact energy a) 20 J b) 27 J d) 22 J d) 25 J 8. If impact energies of specimens are 27 J , 21 J and 9 J then a) Acceptable b) Not acceptable 9. If impact energies of specimens are 27 J , 19 J and 9 J then a) Acceptable b) Not acceptable 10.The minimum fillet leg size of nozzle with pad and nozzle with shell will be ----------------respectively. a) 0.530” , 0.340” 0.375”

b) 0.241” , 0.375”

c) 0.344”, 0.25”

ANSWERS Inside Diameter: D= 45 in.

d) 0.621”,

MAWP: 750 PSI Stress Value: S=17300 Corrosion Allowance: C= 0.125 in. Radiography :RT=1 1) tr = P * R SE – 0.6P Corroded D= 45-2*.125 =44.75 in. R= 22.375 in. tr = 750 * 22.375 17300*1 – 0.6*750 tr = 0.996 in. t initial = 1.5 in. t trev. = 1.47 in. t act. = 1.45 C.R (Long Term) = (1.5-1.45) /10 = 0.005 in./year C.R (Short Term) = (1.47-1.45) /2 = 0.010 in./year Answer: C 2) P =

t* S * E R + 0.6 * t

t = 1.45 -2*(0.01) * 7 t= 1.31 in. P= 1.31 * 17300 * 1 22.5 + 0.6 * 1.31 P = 973.24 PSI Ans : a 3) Ans: B 4) Remaining Life = (t act. - t req.) / C.R R.L = (1.45-0.996) /0.01 = 45.4 Years 5) Ans: D 6) Ans: A 7) Ans : A 8) Ans : B 9) Ans : B 10) A) t(min) for pad to shell =lesser of ( 0.75” or thickness of thinner part joint by fillet weld) t(min) = Lesser of (0.75 or 1.45 or 1.45) t(min) = 0.75 throat size pad to shell t(min)= ½(0.75)=0.375” Leg size=0.375/0.707=0.530 B) t(min) for pad to nozzle =lesser of ( 0.75” or thickness of thinner part joint by fillet weld) t(min) = Lesser of (0.75 or 1.45 or 0.433)

t(min) = 0.433” throat size pad to nozzle t(min)=lesser of 0.25” or 0.7t(min.)=lesser of 0.25 or 0.7(0.433) =lesser of 0.25 or 0.241=0.241 Leg size=0.241/0.707=0.340”