7 - 570 Calcs Summary 6 Pgs

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 304.1.2

Barlow Formula  t = PD ÷2SE (See API 574, Section 11.1.2*) ASME B31.3 Tables A-1 & A-1B for “S” & “E” values respectively API 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  Barlow STEP 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 ) the structural 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 interval API 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 yrs Next 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 thickness as 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 # 1 P = 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-1B D 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 = 19,900 Solution ***Solution: Since calculated value exceeds the given Design Pressure in Example # 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 correct

Easy 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 Time What 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 gaskets dg 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 = 3.250 dg value (gasket pitch diameter) to be used for a blank for Class 600, Size 4 NPS = 5.875 dg 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 with a with a ring joint design (Size 10 NPS Class 600 Flange = dg value) in a piping system with a 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 = C Practice calc using pitch diameters for NPS 2 & 4 given above Remember, add + C last Work (NPS 2):_________________________________________________________________ ______________________________________________________________________= .316 in. Work (NPS 4):_________________________________________________________________ ______________________________________________________________________= .471 in.

API 570 Calcs Summary … Page 5 of 6 Expect Open Book Questions for ASME B31.3 Pressure Test calcs!

SIMPLE CALC IN B31.3 Sect 345.4 PT = 1.5 x given Pressure x Rr factor Note: Rr (ST ÷ S) must always be at least 1, never less than 1.

12. A piping system designed at 500 psi @ 500°F is to be hydrostatically tested at a test temp conducted at 100°F or below. Pipe material is A-106 Gr. B Seamless Sch 80 NPS 6. What is the hydrostatic test pressure? (Compensating for design to test temp difference) A) 750 psig

B) 812 psig

C) 838 psig

D) None of these are correct

Answer = D … Reference: ASME B31.3 Table A-1 & Para. 345.4.2 … Use Formula (API EXAM QUESTIONS WILL NOT GIVE THE INFO ABOVE IN PARENTHESIS) Solution: Determine St & S from A-1 Tables … St = 20,000 & S = 19,900 (St ÷ S = 1.053)

1.5 x 500 (pressure given) x 1.053 (Rr = St/S) = 789.75 = 790 psig Para 345.4.2 (a) requires not less than 1-1/2 times design pressure. 500 x 1.5 = 750 Solution = 754 psi to compensate for the difference between design pressure test conditions!

Practice: Now do this calc assuming 400°F … S = 19.0 ksi (19,000) PT = 1.5 x given Pressure x Rr factor = 1.5 x 500 x (20000 ÷ 19900) = ____ psi

Practice: Now do this calc assuming 600°F … S = 17.9 ksi (17,900) PT = 1.5 x given Pressure x Rr factor = ___ x ___ x (_____÷______) = ____ psi

API 570 Calcs Summary … Page 6 of 6 Impact testing (IT) … MDMT … How it applies to hydrotest temps! MDMT = Lowest temperature where Impact Testing is not required

MINIMUM TEMP °F (Note 6 above) = Letter B  Go to Table 323.2.2A  Use Curve B

This requires a thickness of the material … We return to our Hydrotest question (Page 5) A-106 Gr. B Seamless Sch 80 NPS 6  API 574 Table 1 (below) = 0.432 inches or 10.97 mm A-106 Gr. B Sch 80 = 0.432 in. / 10.97 mm 5TH ROW (ABOVE) IN CURVE B COLUMN Answer = -20°F … any design temp colder requires Impact Tests  Design temp was 400F 12. A piping system designed at 500 psi @ 400°F is to be hydrostatically tested at a test temp conducted at 100°F or below. Pipe material is A-106 Gr. B Seamless Sch 80 NPS 6. What is the MDMT for this piping (where IT is not required by code)? A) -55°F

B) -20°F

C) 15°F

D) None of these are correct

Exercise (Assume same material, but much thicker) Use Tables!!! 1. Assume NPS 24, Sch 160 = 2.344 in/59.54mmRow = 2.3125, MDMT = ____°F / ____°C 2. Assume NPS 24, Sch 80 = 1.219 in/30.96mmRow = 1.1875, MDMT = ____°F / ____°C Use Table 323.2.2A using the two thicknesses above  Use Curve B 1 = 68°F … 2 = 42°F Apply 570 Hydrotest Rule for Brittle Fracture prevention in Section 5.8.3 … What are the minimum test temps for # 1 & 2 above? Solution: Use 2 inch rule per API 570, Para. 5.8.3! Item # 1 MDMT = 68°F … Add 30°F = 98°F = Minimum metal temp during hydrotest)  Item # 2 MDMT = 42°F … Add 10°F = 52°F = Minimum metal temp during hydrotest) 