ASME VIII UGppt
of 208
/208

Author
charliechong 
Category
Engineering

view
1.073 
download
94
Embed Size (px)
description
API 510 ICP Self study note
Transcript of ASME VIII UGppt
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang ASME VIII Div.1 API510 Training for 2013 June.
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang
 Speaker: Fion Zhang 2013/April/15 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang Applicable sections; UG: the G denotes general requirements. UW: the W denotes welding. UCS: the CS denotes carbon steel. UHT: the HT denotes heat treatment. Appendix 1: supplementary design formulae. Appendix 3: definitions.
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang SUBSECTION A GENERAL REQUIREMENTS
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang PART UG GENERAL REQUIREMENTS FOR ALL METHODS OF CONSTRUCTION AND ALL MATERIALS
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang Vessel design features
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang The main ASME VIII design topics required included in the API 510 syllabus are: Internal pressure in shells and heads (clauses UG27 and UG32) External pressure on shells (clause UG28) Nozzle compensation (mainly figure UG37.1) Nozzle weld sizing (mainly figure UW16)
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang About rounding answers. In the ASME Code and for the exam you must round DOWN for pressure allowed. Even if our solution had been 1079.999 we cannot round to 1080, we still round down to 1079 psi. This is the conservative approach taken by the Codes in general and of course is different for the normal rules of rounding. When rounding thickness required we must round UP. The most conservative thing to do. So our example below would round to .230. Even it had been .2291 we would still round up to .230.
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG20 DESIGN TEMPERATURE UG20
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG20 UG20(f) lists an exemption from impact testing for materials that meet All of the following requirements. 1. Material is limited to PNo.1 Gr. No.1 or 2 and the thicknesses don't exceed the following: (a) 1/2 in. for materials listed in Curve A of Fig. UCS66; (b) 1 in for materials from Curve B, C or D of Fig. UCS66; 2. The completed vessel shall be hydrostatically tested 3. Design temperature is no warmer than 650F or colder than  20F. 4. The thermal or mechanical shock loadings are not controlling design. 5. Cyclical loading is not a controlling design requirement.
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG20 1. Material is limited to PNo.1 Gr. No.1 or 2 and the thicknesses don't exceed the following: (a) 1/2 in. for materials listed in Curve A of Fig. UCS66; (b) 1 in for materials from Curve B, C or D of Fig. UCS66; All of the conditions of UG20(f) must be met to take this exemption from impact testing.
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG27 THICKNESS OF SHELLS UNDER INTERNAL PRESSURE. UG27
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG27
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang c) Cylindrical Shells. The minimum thickness or maximum allowable working pressure of cylindrical shells shall be the greater thickness or lesser pressure as given by (1) or (2) below. (1) Circumferential Stress (Longitudinal Joints). When the thickness does not exceed onehalf of the inside radius, or P does not exceed 0.385SE, the following formulas shall apply: (2) Longitudinal Stress (Circumferential Joints). When the thickness does not exceed onehalf of the inside radius, or P does not exceed 1.25SE, the following formulas shall apply: UG27
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang Shell calculations: internal pressure Shell calculations are fairly straightforward and are set out in UG27. Figure below shows the two main stresses existing in a thinwalled vessel shell. Hoop (circumferential) stress This is the stress trying to split the vessel open along its length. Confusingly, this acts on the longitudinal weld seam (if there is one). For the purpose of the API 510 exam this is the governing stress in a shell cylinder. UG27
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang The relevant UG27 equations are: (used when you want to find t) or, rearranging the equation to find P when t is already known: Where: P = maximum design pressure (or MAWP). t = minimum required thickness to resist the stress. S = allowable stress of the material. E = joint efficiency. Ri = the internal radius of the vessel. UG27
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang Remarks: S = allowable stress of the material. This is read from ASME II part D tables or, more commonly, given in the exam question (it has to be as ASME II part D is not in the syllabus). E = joint efficiency. This is a factor (between 0.65 and 1) used to allow for the fact that a welded joint may be weaker than the parent material. It is either read off tables (see UW11 and UW12 later) or given in the exam question. You can think of E as a safety factor if you wish. Ri = the internal radius of the vessel. Unlike some other design codes ASME VIII Div.I prefers to use the internal radius as its reference dimension, perhaps because it is easier to measure. UG27
 ASMEVIIIDiv.1CharlieChong/FionZhang/HeJungang/LiXueliang Figure 9.4 Vessel stresses UG27
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG27
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang A key feature of Ri is that it is the radius in the corroded conditions (i.e. that anticipated at the next scheduled inspection). Dont get confused by this it is just worked out in this way. If a vessel has a current Ri of 10 in and has a corrosion rate (internal) of 0.1 in./years, with the next scheduled inspection in five years, then: Current Ri = 10 in. Ri in 5 years = 10 in. + (5 x 0.1 in) = 10.5 in corroded condition. Hence 10.5 in. is the Ri dimension to use in the UG27 equation. UG27
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang The thickness must not exceed onehalf of the inside radius, i.e. it is not a thick cylinder. The pressure must not exceed 0.385SE, i.e. not be high pressure. In practice this is more than about 4000 psi for most carbon steel vessels. UG27
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG27 The pressure must not exceed 0.385SE, i.e. not be high pressure. In practice this is more than about 4000 psi for most carbon steel vessels. Example: for SA515/Gr. 60 at 700F where S = 14,400 psi. P must not exceed 5544 psi.
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang Shell calculation example The following information is given in the question. Ri = inside radius of 30 in . P = pressure of 250 psi (MAWP). E = 0.85 (type 1 butt weld with spot examination as per UW12). S = 15 800 psi. What minimum shell thickness is necessary to resist the internal MAWP? Using thickness (t) = PR/(SE0.6P) from UG27 Thickness = 250x30 / [15800x0.85 (0.6x250)] t = 0.565 in ANSWER UG27
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang Shell calculation example The following information is given in the question. Ri = inside radius of 30 in. t = 0.625 in. E = 0.85 (type 1 butt weld with spot examination as per UW12). S = 15 800 psi. What is the MAWP? Using pressure (P) = SEt/(R + 0.6t) from UG27 Pressure (P) = 15 800 x 0.85x 0.625 / [30 + (0.6 x 0.625)], MAWP = 276 psi ANSWER. UG27
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG27
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG27 DESIGN INFORMATION Design Pressure = 250 psig. Design Temperature = 700F. Shell and Head Material is SA515 Gr. 60. Corrosion Allowance = 0.125 in. Both Heads are Seamless Shell and Cone Welds are Double welded. Heads are spun and press without welding. Welded and will be Spot Radiographed The Vessel is in All Vapor Service Cylinder Dimensions Shown are Inside Diameters
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UW12 Summary Maximum Weld Joint Efficiency: Joint Type 1~6
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG27 The allowable stress is given in ASME II, as it is not part of API510 examination, the following should be given: S = 14,400 psi for SA515/Gr. 60 at 700F
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG27 If corrosion allowance is specified: (usually not in API510 exam) Rcal for 6 = 36.125 Rcal for 4 =24.125 The Rcal or Dcal used in calculation shall be the vessel RDesign or DDesign plus the corrosion allowance. The required wall thickness shall be pressure thickness + corrosion allowance tp+c c = Corrosion Allowance R design= Designed radius R calculation = R design + c, Radius used for calculation tp = thickness required for internal pressure
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG27 Lower section 2:1 ellipsoidal head Lower section 6ID shell Middle section conical Top section 4ID shell Top hemispherical head Section 1.0 0.85 0.85 0.85 1.0? Required Thickness tp + cEquationE P=250psig, S=14400psi, c=0.125in. Dcal = 72 + 2 x 0.125 = 72.25 in. Rcal = 24 + 0.125 = 24.125 in. Rcal = 24 + 0.125 = 24.125 in. Rcal = 36 + 0.125 = 36.125 in.
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG27 Lower section 2:1 ellipsoidal head Lower section 6ID shell Middle section conical Top section 4ID shell Top hemispherical head Section 1.0 0.85 0.85 0.85 1.0? Required Thickness tp + cEquationE P=250psig, S=14400psi, c=0.125in. Dcal = 72 + 2 x 0.125 = 72.25 in. Rcal = 24 + 0.125 = 24.125 in. Rcal = 24 + 0.125 = 24.125 in. Rcal = 36 + 0.125 = 36.125 in.
 3535 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG32 D = inside diameter of the head skirt; or inside length of the major axis of an ellipsoidal head; or inside diameter of a conical head at the point under consideration, measured perpendicular to the longitudinal axis. Di = inside diameter of the conical portion of a toriconical head at its point of tangency to the knuckle, measured perpendicular to the axis of the cone = D 2r (1 cos ) E = lowest efficiency of any joint in the head; for hemispherical heads this includes head toshell joint; for welded vessels, use the efficiency specified in UW12 L = inside spherical or crown radius. The value of L for ellipsoidal heads shall be obtained from Table UG37. P = internal design pressure (see UG21) r = inside knuckle radius S = maximum allowable stress value in tension as given in the tables referenced in UG23, except as limited in UG24 and (e) below. t = minimum required thickness of head after forming ts = minimum specified thickness of head after forming, in. (mm). ts shall be t = onehalf of the included (apex) angle of the cone at the centerline of the head (see Fig. 14) UG32(a)
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG27 Lower section 2:1 ellipsoidal head Lower section 6ID shell Middle section conical Top section 4ID shell Top hemispherical head Section 1.0 0.85 0.85 0.85 0.85 Required Thickness tp + cEquationE P=250psig, S=14400psi, c=0.125in. Dcal = 72 + 2 x 0.125 = 72.25 in. Rcal = 24 + 0.125 = 24.125 in. Rcal = 24 + 0.125 = 24.125 in. Rcal = 36 + 0.125 = 36.125 in.
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG27 Exercise 2 Required Thickness for Internal Pressure Determine the minimum required thickness for the cylindrical shell and heads of the following pressure vessel: Inside Diameter = 10 6 Design Pressure = 650 psig Design Temperature = 750F Shell & Head Material = SA516 Grade 70 Corrosion Allowance = 0.125 2:1 SemiElliptical heads, seamless 100% radiography of cylindrical shell welds The vessel is in an all vapor service (i.e., no liquid loading) 
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG27 Answers: Inside diameter = 126 + 0.25 = 126.25 in. Ri= 63.125 in. S=14800psi, E=1. Calculations: Shell: tp = (650x63.125)/(148000.6x650) =2.848 in. tshell = tp+c = 2.847+0.125 = 2.973 in.# Head: tp = (650x126.25)/(2x148000.2x650) = 2.785 in. Thead = tp+c=2.785+0.125=2.910 in.#
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG27 Appendix 1 Supplementary Design Formulas 11 THICKNESS OF CYLINDRICAL AND SPHERICAL SHELLS (a) The following formulas, in terms of the outside radius, are equivalent to and may be used instead of those given in UG27 (c) and (d). (1) For cylindrical shells (circumferential stress),
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG27 Example: Given a cylindrical shell with the following variables, solve for the MAWP of the cylinder using both formulas. P = ? , t = 0.500, S = 15,000 psi, E = 1.0, R = 18.0 and Routside = 18.5" Exercise 3
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG27 A cylindrical shell has been found to have a minimum thickness of .353". Its original thickness was .375 with an original inside radius of 12.0. S = 13,800 psi, E = .85 What is its present MAWP ? R = 12.0" + (.375.353) = 12.022 corroded inside radius Ro= 12.0" + 0.375 (orig. t) =12.375 original outside radius Exercise 4
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG27
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG27 You need to consider the hemispherical head joint to shell as category A, but ellipsoidal and torispherical head joint to shell as category B; Do you know why? Why ASME considered the stringent rule for pressure vessel RT test in hemispherical head joint? It is because this joint is more critical, because the thickness obtained from the formula for hemispherical head approximately would be half of the shell thickness; It means if the shell thickness is 1 inch, the hemispherical head thickness would be 0.5 inch.
 44 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG32 Example: For the same pressure, stress and, dimension values will be used for all heads. Lets determine which type of head will be the thickest required and which will be the thinnest allowed. Given: P = 100 psi S = 17500 PSI E = .85 for spot RT of hemispherical head joint to shell E = 1.0 for seamless heads ( Ellipsoidal and Torispherical ) L = 48" for the inside spherical radius for the hemispherical head L = 96" for the inside crown radius of the torispherical head D = 96" inside diameter of the ellipsoidal t = ? Required wall thickness, inches
 45 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG32 t = (100x48)/(2x17500x0.85t = (100x48)/(2x17500x0.850.2x100)0.2x100) t = 0.162t = 0.162 HemisphericalHemispherical t = (0.885x100x96)/(17500x1t = (0.885x100x96)/(17500x10.1x100)0.1x100) T= 0.486T= 0.486 TorisphericalTorispherical t = (100x96) / (2x17500x1t = (100x96) / (2x17500x10.2x100)0.2x100) t = 0.275t = 0.275 EllipsoidalEllipsoidal thicknessthicknessEquationEquationHeadHead
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG27 Spot radiography for ellipsoidal and torispherical heads (Cat. B). Full radiography foe hemispherical head (Cat. A).
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG28 THICKNESS OF SHELLS AND TUBES UNDER EXTERNAL PRESSURE. UG28
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG28
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG28
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG28
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang Overview. The critical pressure that causes buckling is not a simple function of the stress that is produced in the shell, as is true with tensile loads. An allowable stress is not used to design pressure vessels that are subject to elastic instability. Instead, the design is based on the prevention of elastic collapse under the applied external pressure. This applied external pressure is normally 15 psig for full vacuum conditions. The maximum allowable external pressure can be increased by welding circumferential stiffening rings (i.e., stiffeners) around the vessel shell. The addition of stiffening reduces the effective buckling length of the shell, and this length reduction increases the allowable buckling pressure. These stiffener rings may be welded on either the inside or the outside of the shell. UG28
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG28
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG28 Basic Data (example) Temperature = 500F t = 0.530 in. L = 120 in. Do = 10 in. 1. Calculate Do/t 2. Calculate L/Do Find A and B using Chart Fig. G and applicable material chart in Subpart 3 of Section II, Part D. As stated in the API 510 Body of Knowledge, these charts will be provided in the exam body, IF an external calculation is given on the examination. 3. Calculate P
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG28 1. Use common chart and Find A As stated in the API 510 Body of Knowledge, these charts will be provided in the exam body, IF an external calculation is given on the examination.
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG28 2. Select applicable material chart and Find B As stated in the API 510 Body of Knowledge, these charts will be provided in the exam body, IF an external calculation is given on the examination.
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG28 3. Calculate P
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG28, C(1) Cylinders having Do /t values 10: Example #1 The easiest way to understand the UG28 calculations themselves is to look at this worked example. Figure 9.14 shows the parameters for a vessel under external pressure operating at 300oF: . t = thickness of the shell = 0.25 in. . L = distance between stiffeners = 90 in. . Do = shell outside diameter = 180 in. The first step is to calculate the values of the dimensional ratios (L/Do) and (Do/t): L/Do = 90/180 = Do/t = 180 / 0.25 = 720 UG28
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang In a real design situation, these ratios would then be plotted on charts to give values of A and B. In this example, the charts would give values of A = 0.000 15 and B = 2250 (remember that you will generally be given these in an exam question). Pa = 4B/[3 (Do/t)] = 4x 2250/(3x 720) = 4.2 psi Conclusion the vessel is not suitable for full vacuum duty (14.5 psi ). Conclusion the vessel is not suitable for full vacuum duty (14.5 psi ). Pa should be 14.5psi UG28
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG28
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG28, C(1) Cylinders having Do /t values 10: Example #12 Limited data for a vessel are given as: Outside diameter Do = 60 in Length between supports L = 15 feet Factor A = 0.000 18, Factor B = 2500 These are all the data you have. How thick does the vessel wall have to be to be suitable for use under full vacuum? t = 3PaDo/(4B) = 3x14.5x60 / (4x2500) = 0.261in. Select your answer: (a) 1/8 in. (b) in. (c) 3/8 in. (d) others. UG28
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG28, C(1) Cylinders having Do /t values 10: Example #2 Step 1 Assume a value for t and determine the ratios L/Do and Do /t. Example: The cylinder has corroded to a wall thickness of 0.530, its length is 120 and the outside diameter is 10. It operates at 500oF So then; Temp = 500oF t = 0.530 L = 120 Do = 10 Calculate; Do/t = 10/.530 = 18.8 call it 19# (no need to be exact) L/Do = 120/10 = 12# UG28
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG28, C(1) Cylinders having Do /t values 10: Example of Calculation using graphs Normally values A & B are given without using the ASME II graphs UG28
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang Step 2 Enter Fig. G in Subpart 3 of Section II, Part at the value of L/Do determined in Step 1. we must go up the left side of the Fig. G until we reach the value of L/Do of 12. Using the chart we have the following; Do/t = 19 L/Do = 12 UG28
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang Step 3 Move horizontally to the line for the value Do /t determined in Step 1.... Which in our case was 19, but we will round this to 20 since these problems are not meant to be extremely precise. So now we have. From this point of intersection move ertically downward to determine the value of factor A. Do/t = 19 L/Do = 12 UG28
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang Do/t=19,L/Do=12 UG28
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang Step 4 Enter the applicable material chart in Subpart 3 of Section II, Part D for the material under consideration. Move vertically to an intersection with the material/temperature line for the design temperature. Interpolation may be made between lines for intermediate temperatures. To use the next figure we enter at the bottom at the value Factor A = .0028 and then up to our temperature of 500oF. UG28
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang Do/t = 19 L/Do = 12 A=0.0028 UG28
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG28
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG28, C(1) Cylinders having Do /t values 10: Example #3 Problem: A vessel is operating under an external pressure, the operating temperature is 500oF. The outside diameter of the vessel is 40 inches. Its length is 70 inches. The vessels wall is 1.25 inches thick and is of SA51570 plate. Its specified min. yield is 38,000 psi. What is the maximum external pressure allowed? Givens: Temp = 500oF t = 1.25 in. L = 70 in. D0 = 40 in. Determine; Do/t = 40/1.25 = 32 used equation c(1). L/Do = 70/40 = 0.175. Determine value A UG28
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG28
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang Step 4. Using our value of Factor A calculated in Step 3, enter the Factor B (CS2) chart on the bottom. Move vertically to the material temperature line given in the stated problem (in our case 500oF). A=0.0045 UG28
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang Step 5 Then across to find the value of Factor B. We find that Factor B is approximately 13000. Step 6 Using this value of Factor B, calculate the value of the maximum allowable external pressure Pa using the following formula: UG28
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG28 UG28, C(1) Cylinders having Do /t values 10: Example #3 Problem: A vessel is operating under an external pressure, the operating temperature is 500 F. The outside diameter of the vessel is 40 inches. Its length is 70 inches. The vessels wall is 1.25 inches thick and is of SA51570 plate. Its specified min. yield is 38,000 psi. What is the maximum external pressure allowed? Givens: Mtls = SA515 Gr.70. Temp = 500F. t = 1.25 inches. L = 70 inches. Do = 40 inches. Do/t = 40/1.25 = 32 L/Do = 70/40 = 1.75
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG28
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG28
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang UG28
 ASME VIII Div.1 Charlie Chong/ Fion Zhang/ He Jungang / Li Xueliang (2) Cylinders having Do /t values