510 math cal

API 510 Mathematical CalculationsPressure Vessel Inspection Code:-Maintenance Inspection, Rating, Maintenance Inspection, Rating,

Repair, and Alteration

PUSPATRI5th – 9th October 2007

Section 5.4Maximum Allowable Working Pressure Determination§ Establish using latest ASME Code edition or original Code edition to which

vessel constructed§ Certain essential details must be known prior to performing computations§ In corrosive service the wall thickness used shall be actual thickness

determined by inspection minus twice the estimated corrosion loss before next inspection

§ The “t” in the pressure formula is replaced by the following:t = tactual – 2(Corrosion rate x years to next inspection)

P = SEt/R + 0.6t

Math Problem #1Maximum Allowable Working Pressure Determination

A pressure vessel has an MAWP of 500 psi at 400F and is made of SA 516 Gr.70 material with a stress value of 20,000 psi and has a measured thickness of 0.875 inches. The efficiency is 1.0, inside radius is 24 inches and there is no corrosion allowance. The observed corrosion rate is 0.01 inches/year and the next planned inspection is 5 years. What will be the MAWP at the end of 5 years?1) 633 psi1) 633 psi2) 500 psi3) 450 psi4) 900 psi

Math Problem #1Maximum Allowable Working Pressure Determination________________________________________________________________________Step 1: List down all “Given” infoMAWP = 500 psi tactual=0.875” R=24” Next inspection = 5 yrsS=20,000 psi E=1.0 CR= 0.01 inch/yearMAWP at the end of next 5 years?________________________________________________________________________Step 2: Formula P = SEt/R+0.6t

t = tactual – 2(CR x next inspection)________________________________________________________________________________________________________________________________________________Step 3: Calculate t = 0.875” – 2(0.01 x 5) = 0.775”________________________________________________________________________Step 4: Plug in value into P = SEt/R+0.6t

=(20,000)(1)(0.775)/24+(0.6)(0.775)=633 psi________________________________________________________________________

Math Problem #2Maximum Allowable Working Pressure DeterminationA pressure vessel has an MAWP of 350 psi at 300F and is made of SA 515 Grade 60material with a stress value of 16,000 psi. The efficiency of the longitudinal joint is 0.70 and the inside radius is 36 inches with 1.25 inch corrosion allowance. The measured thickness is 0.750 inches and the corrosion rate is 0.02 ipy. The next planned inspection is in 6 years. What will be the MAWP at the end of six years?

1) 440 psi2) 350 psi3) 602 psi4) 157 psi4) 157 psi

Math Problem #2Maximum Allowable Working Pressure Determination__________________________________________________________________________Step 1: List down all “Given” infoMAWP = 350 psi tactual=0.875” R=36” Next inspection = 6 yrsS=16,000 psi E=7.0 CR= 0.02 inch/yearMAWP at the end of 6 years?__________________________________________________________________________Step 2: Formula P = SEt/R+0.6t

t = tactual – 2(CR x next inspection)__________________________________________________________________________Step 3: Calculate t = 0.75” – 2(0.02 x 6) = 0.51”Step 3: Calculate t = 0.75” – 2(0.02 x 6) = 0.51”__________________________________________________________________________Step 4: Plug in value into P = SEt/R+0.6t

=(16,000)(0.7)(0.51)/36+(0.6)(0.51)=5,712/36.306= 157.3 psi

Section 5.7Corrosion and Minimum Thickness Evaluation

• Corrosion may cause uniform loss or may cause a pitted appearance• Minimum actual thickness and maximum corrosion rate determined by:

– NDE thickness testing and drilled test holes– Measurement through openings– Gauging from uncorroded surfaces

• When the minimum actual thickness or maximum corrosion rate is to be • When the minimum actual thickness or maximum corrosion rate is to be adjusted, one of the following evaluation techniques should be considered

1) Averaging for large areas2) Evaluation of widely scattered pit 3) Evaluation of corrosion on weld surface with joint factor other than 14) Measuring corroded thickness of ellipsoidal & torispherical head

Section 5.7Averaging for Large Areas

Math Problem #3Averaging for Large Areas

An eight foot diameter vessel has a large area of general corrosion (32 inches in length and width) and has thickness readings taken in the circumferential (lettered row) and longitudinal (numbered row) directions as shown below. What would be the least thickness along the most critical element considering circumferential stress applies.

1a-0.298” 2a-0.243” 3a-0.192” 4a-0.185” 5a-0.267”1b-0.264” 2b-0.230” 3b-0.185” 4b-0.193” 5b-0.220”1b-0.264” 2b-0.230” 3b-0.185” 4b-0.193” 5b-0.220”1c-0.223” 2c-0.180” 3c-0.204” 4c-0.210” 5c-0.246”

1. 0.206”2. 0.193”3. 0.179”4. 0.195”


An eight foot diameter vessel has a large area of general corrosion (32 inches in length and width) and has thickness readings taken in the circumferential (lettered row) and longitudinal (numbered row) directions as shown below. What would be the least thickness along the most critical element considering circumferential stress applies.

1a-0.298” 2a-0.243” 3a-0.192” 4a-0.185” 5a-0.267”1b-0.264” 2b-0.230” 3b-0.185” 4b-0.193” 5b-0.220”1b-0.264” 2b-0.230” 3b-0.185” 4b-0.193” 5b-0.220”1c-0.223” 2c-0.180” 3c-0.204” 4c-0.210” 5c-0.246”___________________________________________________________________________ 0.2616” 0.2176” 0.1936” 0.196” 0.569”

1. 0.206”2. 0.193”3. 0.179”4. 0.195”

a b c

1

23

4

5


An six foot diameter vessel has a large area of general corrosion (24 inches in length and width) and has thickness readings taken in the circumferential (lettered row) and longitudinal (numbered row) directions as shown below. What would be the least thickness along the most critical element considering circumferential direction.

1a-0.456” 2a-0.443” 3a-0.388” 4a-0.443” 5a-0.550”1b-0.458” 2b-0.423” 3b-0.334” 4b-0.457” 5b-0.589”1b-0.458” 2b-0.423” 3b-0.334” 4b-0.457” 5b-0.589”1c-0.398” 2c-0.399” 3c-0.368” 4c-0.447” 5c-0.498”

1. 0.447”2. 0.363”3. 0.191”4. 0.422”


An six foot diameter vessel has a large area of general corrosion (24 inches in length and width) and has thickness readings taken in the circumferential (lettered row) and longitudinal (numbered row) directions as shown below. What would be the least thickness along the most critical element considering circumferential direction.

1a-0.456” 2a-0.443” 3a-0.388” 4a-0.443” 5a-0.550”1b-0.458” 2b-0.423” 3b-0.334” 4b-0.457” 5b-0.589”

0.456”

0.452”1b-0.458” 2b-0.423” 3b-0.334” 4b-0.457” 5b-0.589”1c-0.398” 2c-0.399” 3c-0.368” 4c-0.447” 5c-0.498”

1. 0.447”2. 0.363”3. 0.191”4. 0.422”

a b c

1

23

4

5

0.452”

0.422”

Section 5.7Widely Scattered Pit

Math Problem #4Widely Scattered Pit

A pressure vessel is 48 inches in diameter. There is some widely scattered pitting present. The wall thickness including 0.125 inch corrosion allowance is 0.750 inch. One group is encircled by an eight inch diameter circle that contains the following;

One 1” dia x 0.20” deep pit, three ¾” dis x 0.20” pit and one 1-1/2” dia x 0.15 deep pit. A straight line is drawn through the most pits within the circle and includes the ¾”, 1” and 1-1/2” pits.and includes the ¾”, 1” and 1-1/2” pits.

a) What is the total area of the its within the 8 inches circle?1)2.5 inch2 2)3.876 inch2 3)1.325 inch2 4)7.25 inch2

b) What would have to be done to the pits in the straight line to make them acceptable?

1) Nothing, they are acceptable as is2) Weld repair the 1.5” dia pt3) Fill all pits with epoxy materials4) None of the above apply

Math Problem #4Widely Scattered Pit

A pressure vessel is 48 inches in diameter. There is some widely scattered pitting present. The wall thickness including 0.125 inch corrosion allowance is 0.750 inch. One group is encircled by an eight inch diameter circle that contains the following;

One 1” dia x 0.20” deep pit, three ¾” dis x 0.20” pit and one 1-1/2” dia x 0.15 deep pit. A straight line is drawn through the most pits within the circle and includes the ¾”, 1” and 1-1/2” pits.

D=48”Tnom = 0.75CA=0.125

Pit #1: 1” dia x 0.2” depthPit #2: 0.75” dia x 0.2” depth x (3nos)Pit#3: 1.5” dia x 0.15” depth

a) Total area of the pitsFormula for area: ∏r2and includes the ¾”, 1” and 1-1/2” pits.

a) What is the total area of the its within the 8 inches circle?1)2.5 inch22)3.876 inch2 3)1.325 inch24)7.25 inch2

b) What would have to be done to the pits in the straight line to make them acceptable?1) Nothing, they are acceptable as is2) Weld repair the 1.5” dia pt3) Fill all pits with epoxy materials4) None of the above apply

Formula for area: ∏r2Area of Pit#1 + Area of Pit#2 x 3 + Area of Pit#3= ∏(0.5)2 + ∏(0.375)2 x 3 + ∏(0.75)2

= 0.785 + 1.325 + 1.767= 3.878 inches2

b) Sum of dimension of pits within the 8 inches circleDia Pit#1 + Dia Pit#2 + Dia Pit#3= 1 + 0.75 + 1.5= 3.25” which is exceed the limit of 2”. Thus by repair 1.5” pit then the sum of dimension will be 1.75” which <2”

Another ProblemWidely Scattered PitFour scattered pits are found on the shell of a vessel. All the pits are contained withinan 8 inch diameter circle. Pit no 1 is 0.75 inch diameter and 0.25 inch deep. Pit no 2 is 0.49 inch diameter and 0.61 inch deep. Pit no 3 is 0.6 inch diameter and 0.51 inch deep. Pit no 4 is 1.5 inch diameter and 0.16 inch deep. The required thickness is 0.781 inch and the actual thickness is 1 inch.

a) What is the total area of the pits in the 8 inch circle?1) 2.678 inch2 2)3.335 inch2 3)1.002 inch2 4) 6 inch21) 2.678 inch2 2)3.335 inch2 3)1.002 inch2 4) 6 inch2

b) What pit(s) are unacceptable due to remaining thickness?1) No 1 & 3 2) No 2 3) No 4 4) No 1 & 2

c) What pit(s) are unacceptable because of straight line out tolerance?1) No 1&2 2) No 2&3 3) No 3&4 4) N0 2&4

Another ProblemWidely Scattered PitFour scattered pits are found on the shell of a vessel. All the pits are contained within an 8 inch diameter circle. Pit no 1 is 0.75 inch diameter and 0.25 inch deep. Pit no 2 is 0.49 inch diameter and 0.61 inch deep. Pit no 3 is 0.6 inch diameter and 0.51 inch deep. Pit no 4 is 1.5 inch diameter and 0.16 inch deep. The required thickness is 0.781 inch and the actual thickness is 1 inch.

a) What is the total area of the pits in the 8 inch

Pit #1 : 0.75” dia x 0.25” depthPit #2 : 0.49” dia x 0.61” depth Pit # 3 : 0.6” dia x 0.51” depth Pit # 4 : 1.5” dia x 0.16” depthtmin = 0.781”tactual = 1”

a) What is the total area of the pits in the 8 inch circle?Formula for area: ∏r2= Area Pit#1 + Area Pit#2 + Area Pit#3 + Area Pit#4= ∏(0.375)2+ ∏(0.245)2+ ∏(0.3)2+ ∏(0.75)2

= 0.4418 + 0.1886 + 0.2827 + 1.7671= 2.68 inch2

b) What pit(s) are unacceptable due to remaining thickness?a) What is the total area of the pits in the 8 inch

circle?1) 2.678 inch22)3.335 inch23)1.002 inch24) 6 inch2

b) What pit(s) are unacceptable due to remaining thickness?1) No 1 & 3 2) No 2 3) No 4 4) No 1 & 2

c) What pit(s) are unacceptable because of straight line out tolerance?1) No 1&2 2) No 2&3 3) No 3&4 4) N0 2&4

thickness?Remaining t below the pit >½(tmin)½ x tmin = ½ x 0.781 = 0.3905”Pit#1: 1” – 0.25” = 0.75”Pit#2: 1” – 0.61” = 0.39”Pit#3: 1” – 0.51” = 0.49”Pit#4: 1” – 0.16” = 0.84”

c) What pit(s) are unacceptable because of straight line out tolerance?Pit#1 + Pit#2 = 0.75 + 0.49 = 1.24”Pit#2 + Pit#3 = 0.49 + 0.6 = 1.09”Pit#3 + Pit#4 = 0.6 + 1.5 = 2.1” – not acceptablePit#2 + Pit#4 = 0.49 + 1.5 = 1.99”

Section 5.7Evaluation of Corroded Weld Surface

Math Problem #6Evaluation of Corroded Weld SurfaceA pressure vessel is found to have a corroded area that includes the longitudinal welded seam. An evaluation must be made to determine whether the thickness at the weld or remote from the weld governs. The inside radius is 36 inch which includes 0.125 inch corrosion allowance. The MAWP is 500 psig, S=17500 psi. The nameplate has no extent of radiography shown under the code stamp. The longitudinal seam is Type 1. radiography shown under the code stamp. The longitudinal seam is Type 1. What would be the width of the evaluation if the width from toe to toe was 1 inch?

1) 7 inch2) 2 inch3) 3.022 inch4) 7.044 inch

Math Problem #6Evaluation of Corroded Weld SurfaceA pressure vessel is found to have a corroded area that includes the longitudinal welded seam. An evaluation must be made to determine whether the thickness at the weld or remote from the weld governs. The inside radius is 36 inch which includes 0.125 inch corrosion allowance. The MAWP is 500 psig, S=17500 psi. The nameplate has no extent of radiography shown under the

code stamp. The longitudinal seam is Type 1.What would be the width of the evaluation if the

MAWP = 500 psi, S= 17,500, R=36”, CA=0.125”E = 0.7 [Type 1 with no RT].Width of area for evaluation of the weld and area remote from weld?

Section 5.7: The greater of 1” on either side of weld or 2 x tmin on either side of weld

tmin = PR/SE-0.6PWhat would be the width of the evaluation if the width from toe to toe was 1 inch?

1) 7 inch2) 2 inch3) 3.022 inch4) 7.044 inch

tmin = PR/SE-0.6P= (500)(36)/(17,500)(0.7) – 0.6(500)= 18,000/11,900= 1.51”

2 x t min = 3.01” which is > 1”

Thus, total area of evaluation= both side of the weld + weld area= (3.01) x 2 + 1= 7.03”

Section 5.7 Measuring Corroded Thickness of Ellipsoidal &

Torispherical Head

Math Problem#7Measuring Corroded Ellipsoidal Head

#7. There has been a corrosion in the center of ellipsoidal head. The head is a 2:1 elliptical with a D of 60 inches and h of 15 inches. What is the required thickness in the center portion of this head if the MAWP is 300 psi, E=1.0 and the S= 15,600 psi?

a) 0.520 inb) 0.750 inb) 0.750 inc) 1.050 ind) 0.889 in

Math Problem#7Measuring Corroded Ellipsoidal Head

MAWP = 300 psi, E=1.0, S=15,600 psi,D=60, h=15.What is treq or tmin?As per Section if 5.7 thickness at central portion calculation for elipsoidal or torispherical to use spherical head formula:t = PR/2SE-0.2Pt = PR/2SE-0.2P

Calculate RR=kD, find value K from table.For D/2h = 60/2(15) = 2 value k=0.9R = 0.9 (60) =54”

t = (300)(54)/2(15,600)(1)-0.2(300)= 16,200/31,140= 0.52”

Math Problem#13Measuring Corroded Torispherical Head

#13. A torispherical head is to be repaired by replacement of the center portion of the head. What is the required thickness for a dishead repair plate that is 50% of the head diameter. The material is SA 516 Grade 70 and S=18,000, P=300 psi, the crown radius = 96 inch. E=0.65 and a Type 2 joint will be used.

1) 2.0 in1) 2.0 in2) 1.233 in3) 0.556 in4) 0.234 in

Math Problem#13Measuring Corroded Torispherical Head

MAWP = 300 psi, S=18,000, R=96”E=0.65, Type 2 jointWhat is the t min?

As per Section if 5.7 thickness at central portion calculation for elipsoidal or torispherical to use spherical head formula:formula:t = PR/2SE-0.2P

The radius of torispherical to be used as radius of spherical segment.

t = (300)(96)/2(18,000)(0.65) – 0.2(300)= 28,800/23,340= 1.234”

Section 5Calculating P when the vessel is corroding

API 510 - During an internal inspection a corroded area on a vessel is discovered with a current thickness of 0.446 inches. You determine that the corrosion rate is 0.004 ipy. Operations would like to have a 6 year run before performing the next internal inspection on this vessel. What pressure is the wasted area “good for” for the end of 6 year. The shell has a 5 inch internal diameter and has joint efficiency of 1.0. The allowable stress of the material at design conditions is 15,000 psi.

ASME VIII-During the inspection of horizontal pressure vessel a corroded area was found on ASME VIII-During the inspection of horizontal pressure vessel a corroded area was found on the inside surface, the thickness at the point was measured to be 0.906 inches thick. Checking the ASME data report for the vessel it was found that the shell plate was 1-1/4 inch purchased thickness,48 inch inside diameter of SA-516 Gr. 70 Material, the shell was rolled and welded using type 1 longitudinal seam with full radiography examination. The MAWP is 600 psig at 300 F. All openings in the vessel are fully reinforced, welded connections. S is 20,000 psi. What would be the maximum pressure permitted for this thinned area

API 5.4