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Lesson 11 corrosion new2

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Lesson 11 corrosion new2

  1. 1. API 510 Preparatory ClassAPI 510 Preparatory Class Lesson 11Lesson 11 Corrosion CalculationsCorrosion Calculations
  2. 2. A 60 foot tower consisting of four (4) shell courses was found to have varying corrosion rates in each course. Minimum wall thickness readings were taken after 4 years and 6 months of service. All original wall thicknesses included a 1/8" corrosion allowance. The top course's original thickness was .3125". The present thickness is .3000". The second course downward had an original thickness of .375". During the inspection it was found to have a minimum wall thickness of .349". The third course was measured at .440" its original thickness was .500". The bottom course had an original thickness of .625" and measured to be .595". Corrosion Example Problems
  3. 3. Determine the metal loss for the top course, the corrosion rate for the second course, the corrosion allowance remaining in the third course, the retirement date for the bottom course. Corrosion Example Problems
  4. 4. Solution A: TOP COURSE: Metal loss equals the previous thickness minus the present thickness. Previous .3125" Present -.3000" .0125" Metal Loss Corrosion Example Problems
  5. 5. SECOND COURSE: Corrosion rate equals metal loss per given unit of time. Previous .3750“ Present - .3490" .0260" Metal Loss = = .006"/ Per Year Corrosion Example Problems yearsTime Loss 5.4 "0260.
  6. 6. THIRD COURSE: Remaining Corrosion Allowance equals the actual thickness minus the required thickness. Original Thickness .500" Original C. A. -.125" Required Wall Thickness .375" Actual Wall Thickness .440" Required Wall -.375" Remaining C.A. .065“ Corrosion Example Problems
  7. 7. BOTTOM COURSE Remaining life equals the remaining corrosion allowance divided by the corrosion rate. 1. Required Thickness Original Thickness .625" Original C. A. -.125" Required Thickness .500“ Corrosion Example Problems
  8. 8. 2.Remaining Corrosion Allowance Actual Wall Thickness .595“ Required Thickness -.500" Remaining Corr. Allow .095“ 3.Corrosion Rate Original Thickness .625" Present Thickness -.595“ Metal Loss .030" Metal Loss .030“ = Corrosion Rate = .0067"/Yr. Time 4.5 Years Corrosion Example Problems
  9. 9. 4. Remaining Life Corrosion Allowance.095“ = 14.2 Yrs. Remaining Life Corrosion Rate .0067"/Yr. Corrosion Example Problems
  10. 10. Corrosion Just Milling Around In the previous examples we calculated using decimal fractions of an inch such as .006”. However questions on the exam often have the format of Mils or thousandths of and inch. So if you are not familiar with this terminology here are some examples. .1 = 1/10th of an inch .01 = 1/100th of an inch .001 = 1/000th of an inch or 1 Mil.  this is what is used. Take our example above .006” this translates to 6 Mils. .0067” translates to 6.7 Mils.
  11. 11. Class Quiz Just Milling Around 1. Convert the following decimal fractions into Mils. .195” = _____ Mils . 023” = _____ Mils .009” = _____ Mils. .0011” =______Mils. 2. Calculate or previous Remaining Life problem using Mils. Corrosion Allowance.095“ = 14.2 Yrs. Remaining Life Corrosion Rate .0067"/Yr.
  12. 12. Solution Just Milling Around Converting the following decimal fractions into Mils. .195” = 195 Mils .023 = 23 Mils .009 = 9 Mils. .0011 = 1.1 Mils. Corrosion Allowance 95 Mils = 14.2 Yrs. Rem. Life Corrosion Rate 6.7“ Mils/Yr.
  13. 13. Time A year has 12 parts The next issue is how do you handle the odd months in a year. Its all good when it has been an even number of years. However what about 2 years and 5 months? Five months is 5/12th of a year. 5/12 = .416 years rounded to 3 places. Of course 6 months is then 6/12 or .5 years. Example: Based on 4.416 years and a metal loss of 50 Mils we have; Metal Loss 50 Mils = Corrosion Rate = 11.3 Mils/Yr. Time 4.416 Years
  14. 14. Class Quiz Time A vessel shell course was replaced in October of 1980. The initial or, baseline thickness was .505”. The most recent actual thickness measured was .485” in January 1991. What is the corrosion rate, based in Mils per year in January of 1991.
  15. 15. Solution .505(t-initial) - .485(t-actual) = .020” = 20 Mils Oct.1990 – Oct.1980 = 10 years plus November, December of 1990 and January of 1991 so; 10 yrs. 3 Mos. 10 - 3/12th (3/12 = .250) = 10.25 years Therefore; Metal Loss 20 Mils = Corrosion Rate = 1.95 Mils/Yr. Time 10.25 Years
  16. 16. Yes, in the API 510 there are, the Long Term and Short Term. Here is the Long Term. From API 510 6.4 page 6-3 The long term (L.T.) corrosion rate shall be calculated from the following formula: Corrosion rate (L.T.) = Can there be two Corrosion Rates? t(actual)andt(inital)between)time(years t(actual)-t(inital)
  17. 17. The short term (S.T.) corrosion rate shall be calculated from the following formula:   Corrosion rate (S.T.) = Corrosion Example Problems actualtandprevioustbetween(years)time actualtprevioust -
  18. 18. tinitial = the thickness, in inches(millimeters), at the same location as t actual measured at the initial installation or at the commencement of a new corrosion rate environment. tprevious = the thickness, in inches(millimeters), at the same location as t actual measured during a previous inspection. Long-term and short-term corrosion rates should be compared to as part of the data assessment. The authorized inspector, in consultation with a corrosion specialist, shall select the corrosion rate that best reflects the current process. Corrosion Example Problems
  19. 19. Example: A vessel shell had a second set of ultrasonic thickness measurements after 1 year of service, the original baseline wall thickness was 0.500”, and the second set revealed that the shell was now at 0.489”. Five years later a third set of wall readings were taken and the shell was measured to be 0.459”. What value should be used in the Remaining Life calculations? Comparing short term to long term corrosion we find the following values. Normally the most aggressive will be used and in this case it will be the Long Term Corrosion Rate. Corrosion Example Problems
  20. 20. L.T. = = 6.8 Mils/ year S.T. = = 6 Mils/ year The remaining life of the vessel shall be calculated from the following formula: Remaining life (years) = (6.4) Corrosion Example Problems years6 0.4590.500- years5 0.459-0.489 ratecorrosion requiredt-actualt
  21. 21. tactual = the actual minimum thickness, in inches determined at the time of inspection for a given location or component. trequired = the required thickness in inches at the same location or component as the t actual measurement computed by the design formulas (e.g. , pressure and structural) before corrosion allowance and manufacturer’s…. Example: Determine Remaining life of the vessel shell course in the example above. T required thickness of the shell course is 0.388” (known as t minimum). Compare S.T. and L.T. corrosion rates as follows: S.T. rate = 0.006” or 6 Mils a year L.T. rate = 0.0068” or 6.8 Mils a year Corrosion Example Problems
  22. 22. Therefore we will use the most aggressive corrosion rate found to be the Long Term Rate. Remaining life (years) What would be the maximum length of time before the next inspection? ANS: ½ the Remaining Life or 10 years whichever is less. Therefore: 17.79/2 = 8.895 years Corrosion Example Problems Yrs Mils Mils 79.17 8.6 121 "0068.0 "121.0 yeara0.0068" 0.338"-0.459" ===
  23. 23. What would be the maximum length of time before the next inspection? From: API 510 page 6-2 6.4 Internal An On-Stream Inspection The period between internal or on-stream inspections shall not exceed one half the estimated remaining life of the vessel based on corrosion rate or 10 years, whichever is less. In cases where the remaining safe operating life is estimated to be less than 4 years, the inspection interval may be the full remaining safe operating life up to a maximum of 2 years. ANS: ½ the Remaining Life or 10 years whichever is less. Therefore: 17.79/2 = 8.895 years Corrosion Example Problems
  24. 24. (a) (the second a) For a corroded area of considerable size in which the Circumferential Stress Govern, the least thickness along the most critical element of the area may be averaged over a length not exceeding the following: 1. For vessels with the inside diameters less than or equal to 60 inches one half the vessel diameter or 20 inches whichever is less. 2. For vessels with the inside diameters greater than 60 inches one third the vessel diameter or 40 inches whichever is less. Example 5.7 Page 5-3 Corrosion
  25. 25. For a corroded area of considerable size in which the Circumferential Stress Govern, the least thickness along the most critical element of the area may be averaged over a length not exceeding the following: 1. For vessels with the inside diameters less than or equal to 60 inches one half the vessel diameter or 20 inches whichever is less. 2. For vessels with the inside diameters greater than 60 inches one third the vessel diameter or 40 inches whichever is less. Corrosion and Minimum Thickness Evaluation 5.7
  26. 26. When the area contains an opening the distance on either side of the opening within which the thicknesses are averaged shall not extend beyond the limits of the reinforcement as defined in the ASME Code. Corrosion and Minimum Thickness Evaluation See Fig. UG-37 and refer to reinforcement calculations for determining the extra metal in the shell. For Limits: d ( t – tr ) or 2(t + tn ) ( t – tr ) Use the Larger
  27. 27. Widely scattered pits may be ignored as long as the following are true: Corrosion and Minimum Thickness Evaluation 1. The remaining thickness below the pit is greater than one-half the required thickness. (1/2 t required) 1.0” – 0.125” = 0.875” the required thickness Therefore no pit equal to 0.875”/2 = 0.4375” or greater! Pit #3  0.402 < 0.4375” Acceptable.
  28. 28. 2. The total area of the pits does not exceed 7 square inches (45 sq. centimeters) within any 8 inch (20 centimeter) diameter circle.   We will assume the pits to be circles and use the formula Pi x Radius Squared to determine the area of each pit. Pit # One 3.141 x (0.170/2) 2 = 0.085 2 = 0.0226 sq. in. Pit # Two 3.141 x (0.330/2) 2 = 0.165 2 = 0.0855 sq. in. Pit # Three 3.141 x (0.250/2) 2 = 0.125 2 = 0.0490 sq. in. Pit # Four 3.141 x (0.377/2) 2 = 0.1185 2 = 0.1116 sq. in. Total: 0.2678 sq. in. less than 7 sq. inches Acceptable! Corrosion and Minimum Thickness Evaluation
  29. 29. 3. The sum of their dimensions along any straight line within the circle does not exceed 2 inches Straight line for pits 1-3-4 #1. 0.170” #3. 0.250” #4. 0.377” Total 0.797” < 2 inches Acceptable When the surface at a weld with a joint factor other than 1.0 as well as the surfaces remote from the weld is corroded, an independent calculation using the appropriate weld joint factor must be made to determine if thickness at the weld or remote from the weld governs the allowable working pressure. Corrosion and Minimum Thickness Evaluation
  30. 30. For this calculation the surface at a weld includes 1 inch on either side of the weld, (measured from the toe of the weld), or twice the minimum thickness on either side of the weld, whichever is greater. Corrosion and Minimum Thickness Evaluation
  31. 31. When measuring the corroded thickness of ellipsoidal and torispherical heads the governing thickness may be as follows: 1. The thickness of the knuckle region with the head rating calculated by the appropriate head formula. 2. The thickness of the central portion of the dished region in which case the dished region may be considered to be a spherical segment whose allowable pressure is calculated by the Code formula for spherical shells. Corrosion and Minimum Thickness Evaluation
  32. 32. 3. For calculating the spherical portion using the spherical formula the following applies to find the spherical radius L: a. L = the OD of the Torispherical head b. L = the ID of the shell times 90% for 2 to 1 Ellipsoidal heads See 5.7 second e. Corrosion and Minimum Thickness Evaluation
  33. 33. Corrosion Over And Out…

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