7 - 570 Calcs Summary 6 Pgs

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570 calculations
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  API 570 Calcs Summary … Page 1 of 6    Expect Closed Book & Open Book for Corrosion Rate & RL calcs! RL*, ST & LT    Round 3, Questions 26  –  40! See note for t required*Reference = API 570, Section 7.1.1 (Formulas 1, 2, and 3)   *Note: RL calcs use t required. This is for basis for API 570 formulas!   t required = Barlow OR Structural Minimum (using the greater value)   t required (minimum required thickness) calcs *DO NOT USE ASME B31.3 CALCULATION 3A IN 304.1.2Barlow Formula  t = PD ÷2SE (See API 574, Section 11.1.2*)ASME B31.3 Tables A-1 & A-1B for “S” & “E” values respectivelyAPI 574 Table 1 gives you value “D”   … This is the OD (Col. 3 & 4)  Structural Minimum  (See API 574, Section 11.1.3 & Table 6)API 574 Minimum Required Thickness (t required) Examples GO TO API 574 NOW  See 11.1.4 Minimum Required Thickness Minimum required thickness is the greater value of the pressure design thickness or the structural minimum thickness.   STEP 1 Calculate pressure design thickness per rating code  BarlowSTEP 2 Determine structural minimum thickness (given on exam)  574 Table 6 STEP 3 Select minimum required thickness … This is the larger of the pressure design thickness or structural minimum thickness determined in Step 1 and Step 2.   Examples # 1 & # 2  Note these use the B31.3 calc  Use the Barlow instead Reason to use Barlow always : It’s much quicker & answers are nearly the same     API 574 Example 1 (Barlow): t = PD ÷ 2SE … PD = 100 x 2.375 = 237. 5, 2SE = 2 (20000 x 1) Final solution = 237.5 ÷ 40,000 = .0059375 (or 5.9375 to the -03), this = 6 mils or .006    Structural minimum from Table 6 = .070 mils … Therefore, t required = .070 inches   Work Example # 2 (Barlow)  8400 ÷ 40,000 = .210 (Only 2 mils away from .208  ) thestructural minimum = .110  Therefore, minimum required thickness or t required = .210  API 570 Calcs Summary … Page 2 of  6 MAWP calc used to determine inspection interval MAWP CALC is P = 2SEt ÷ D    API 570 Sec 7.2 + Table 4 Example Tricky, solve “t” first, t = 2 x estimated corrosion loss over intervalAPI 570, Table 4 example # 1. Observed corrosion rate = .010/year* Treat mils like ryals (in your mind) … This is 10 mils* = .01* or .010  Next planned inspection = 5 years, so we lose 50 mils (10 x 5) in 5 yrsNext is tricky! We must apply the API 570 Para. 7.2 code rule here! When the MAWP is recalculated, wall thickness used in these computations shall be the actual thicknessas determined by inspection minus twice the estimated corrosion loss  before date of next inspection   The expected 50 mils to be lost over 5 years now becomes 100 mils t = .320 (thickness determined from insp) minus - .100 = .220 or 220 mils …    The rest is very simple after that … P = 2SEt ÷ D … Example # 1P = 2 (   19,900**) (1.0) (.220) ÷ D = 8756 ÷ 16 = 547.25 = 547 psi***S & E values are found in B31.3 Appendix A, Tables A-1 & A-1BD value = OD = in the API 574 Table 1 (CS) in the Columns 3 & 4 **Note: 2008 Code Value for S = 20,000 and 2010 Code value for S = 19,900Solution***Solution: Since calculated value exceeds the given Design Pressure inExample # 1 of Table 4 (500 psi)  Then a five year interval is acceptable      Do example 2 Table # 4 (7 year interval) on your own  t = .180 = 448 psi      Math: t = .320  –  (2 x .070) = .180, and 2 (19,900) (1) (.180) ÷ 16 = 447.75   Calc results in a value less than 500 psi, therefore 7 yrs interval = too long      Extra practice  6 yr. interval  It’s just under 500 psi = too long     SHOW MATH 6 YR INTERVAL: ____________________________________  ______________________________________________________= ________ psi  API 570 Calcs Summary … Page 3 of 6   For Dt calcs, expect both Closed Book & Open Book Questions! Square Root of Dt = lap patch spacing using “toe to toe” value  API 570, Sect. 8.1.4.1 Temporary Repairs … D * = API 574 Table 1   *Note: Value for D = Inside Diameter in Columns 7 & 8 (NOT OD) 14 NPS STD Sch = 13.250 What is required spacing between 2 lap patches (minimum required patch thickness = .21 in .),when the pipe is 14 NPS, ASTM A106, Grade B, STD sch. (.375 in), Design 600 psig/100°F?A) 1.67 inches B) 2.78 inches C) 3.33 inches D) None of these are correctEasy Solution: Dt = 13.250 x .210 (given) = 2.7825 in.    Hit square root key* = 1.668 in.      *Most common error     forget to hit square root key … API knows this … Answer = A    Practice TimeWhat is required spacing between 2 lap patches (minimum required patch thickness = .35 in .),when the pipe is 14 NPS, ASTM A106, Grade B, STD sch. (.375 in), Design 600 psig/100°F?   A) 2.15 inches B) 4.31 inches C) 4.64 inches D) None of these are correct Show Math: __________________________________________________________________  Easy Solution: Dt = 13.250 x .35 (given) = ______  in.    Hit square root key* = ______  in.   What is required spacing between 2 lap patches (minimum required patch thickness = .49 in .),when the pipe is 14 NPS, ASTM A106, Grade B, STD sch. (.375 in), Design 600 psig/100°F?A) 1.67 inches   B) 2.55 inches C) 6.49 inches D) None of these are correct Show Math: __________________________________________________________________  Expected question on a closed Book Exam … Tricky or easy      What formula is to be used for the spacing of temporary repairs using lap patches? A)   2 B) + c C) ÷ 2 D)  API 570 Calcs Summary … Page 4 of 6 Expect Open Book Questions for the ASME B31.3 Blank calc! Blank = Blind Flange (Flat Cover Plate)   … W = 1 for API exams  Use ASME B31.3 Formula 15 (On Page 28)   dg value = will be given on exam for RF Spiral wound gasketsdg value = P (Pitch diameter) in ASME B16.5, Table 5 (p156) dg value (gasket pitch diameter) to be used for a blank for Class 600, Size 2 NPS =   3.250dg value (gasket pitch diameter) to be used for a blank for Class 600, Size 4 NPS = 5.875dg value (gasket pitch diameter) to be used for a blank for Class 600, Size 10 NPS = _____ What is the minimum required thickness of a blank made of ASTM A516 Gr. 70 Plate witha with a ring joint design (Size 10 NPS Class 600 Flange = dg value) in a piping system witha design pressure of 400 psi @ 400°F & Corrosion Allowance of .125 inches? A) .625 inches B) .750 inches C) .875 inches D) .940 inches Apply formula 15 of ASME B31.3 … dg = 12.75    Step 1 … Solve 3P & 16SE … 3P = 400 x 3 = 1200 … 16 SE = 16 x 21,600 (E=1) = 345,600   Step 2 … (3P ÷ 16SE) or 1200 ÷ 345,600 = .0034722 … Remember “c” is saved for last!!!   Step 3 … The square root of .0034722 = .058925, and dg = 12.75 x .058925 = .7513 = .751 in.   Step 4 … Add CA last … Solution is .751 inches + .125 inches = .876 inches = Answer = CPractice calc using pitch diameters for NPS 2 & 4 given above  Remember, add + C lastWork (NPS 2):_________________________________________________________________  ______________________________________________________________________= .316 in.Work (NPS 4):_________________________________________________________________  ______________________________________________________________________= .471 in.
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