ASME Code Calculations

Revised 03/06 to conform with the 2004 ASME ExtractPart A1 CHAPTER 1 ASME Code Calculations: Cylindrical Components Here is what you w i l l be able to do when you complete each objective: 1. Calculate the required minimum thickness or the maximum allowable working pressure of piping, tubes, drums, and headers of ferrous tubing up to and including 125 mm O.D. 2. Calculate the required minimum thickness or the maximum allowable working pressure of ferrous piping, drums, and headers. 3. Calculate the required thickness or maximum allowable working pressure of a seamless, unstayed dished head. 4. Calculate the minimum required thickness or maximum allowable working pressure of unstayed flat heads, covers, and blind flanges. 5. Calculate the acceptability of openings in a cylindrical shell, header, or head. 6. Calculate the compensation required to reinforce an opening in a cylindrical shell, header, or head. Revised 03/06 to conform with the 2004 ASME Extract 1 Chapter 1 • ASME Code Calculations: Cylindrical Components 3 INTRODUCTION As power engineers acquire their second and first class power engineering certification, they find that their roles and areas of responsibility require them to have a more detailed working knowledge of the key engineering codes and standards with which their facility must comply. Power engineers often work on teams or lead teams that are responsible for upgrades within their facilities and/or for making changes to major pressure piping or equipment. Although power engineers are not required to design a boiler or pressure vessel, they often work as team members for equipment design, upgrade, process change, commissioning, operation, or repair. These activities require work to be done in accordance with applicable codes. As well, when you become chief engineer of a facility, you may be called upon to lead teams and give approval for various projects that must comply with specific engineering codes and standards. In the early 1900’s, the American Society of Mechanical Engineers (ASME) appointed various committees to draw up standards for the construction of boilers and pressure vessels together with standards for welding and guidelines for the care of boilers in service. These standards and guidelines have been improved over the years with the improvement in materials and technology. One important component of the standards for pressure vessels is the use of a safety factor. The measured physical properties of a material, including ultimate tensile strength, are divided by a defined safety factor to derive the maximum allowable stress. In this way, allowance is made for limitations in the testing technology, unusual stress concentrations, non-uniform materials, and material flaws. Technological improvements, especially in materials testing, have allowed a reduction in the safety factor to 3.5 in current editions of Section I; this is the same factor used in Sections VIII-1 and VIII-2. Pressures calculated or given in this module refer to gauge pressure unless otherwise indicated. Consult the latest ASME Codes (currently the 2004 Edition)—Section I and Section VIII, Division 1—while studying this module. Figures referenced with a Code section prefix, such as “Fig. PG-32” or “Fig. UG-34,” can be found in the ASME Codes or the Academic Extract and are generally not reproduced here. Note: Material and formulae used in this chapter refer to the 2004 edition of the ASME Codes. Most relevant sections can be found in the 2004 ASME Academic Extract (visit www.powerengineering.ca for more info). Note: Correct units of measure are very important to accurate calculations, and students should be well versed in their use. However, due to the size and complexity of Code calculations, it is common practice to omit the units Conforms with the 2004 ASME Extract • Revised 03/06 4 Revised Second Class Course • Section A1 • SI Units until the final answer is derived. This convention has been used throughout this chapter. Note: It should be noted that many US customary unit values presented in the ASME codes do not convert directly into metric values in the current ASME edition or the 2004 ASME Academic Extract ( i.e. 5 in. converts to 127mm, ASME shows 5 in. (125 mm); ¼ in. coverts to 6.35 mm, ASME shows ¼ in. (6 mm)). You are required to use the ASME values as presented and not to convert US customary numbers to metric. ASME SECTION I - POWER BOILERS Paragraphs PG-1, PG-2: This Code covers rules for construction of power boilers, electric boilers, miniature boilers, and high temperature water boilers. The scope of jurisdiction of Section I applies to the boiler proper and the boiler external piping. Superheaters, economizers, and other pressure parts connected directly to the boiler, without intervening valves, are considered to be parts of the boiler proper and their construction shall conform to Section I rules. Materials Paragraph PG-6 states that steel plates for any part of a boiler subject to pressure, whether or not exposed to the fire or products of combustion, shall be in accordance with specifications listed in paragraph PG-6.1. Paragraph PG-9 states that pipes, tubes, and pressure containing parts used in boilers shall conform to one of the specifications listed in paragraph PG-9.1. Design Paragraph PG-16.3 states that the minimum thickness of any boiler plate under pressure shall be 6 mm. The minimum thickness of plates to which stays may be attached (in other than cylindrical outer shell plates) shall be 8 mm. When pipe over 125 mm O.D. is used in lieu of plate for the shell of cylindrical components under pressure, its minimum wall thickness shall be 6 mm. Paragraph PG-16.4 states that plate material not more than 0.3 mm thinner than the required thickness calculated by Code formula may be used provided the manufacturing process is such that the plate will not be more than 0.3 mm thinner than that specified in the order. Paragraph PG-16.5 states that pipe or tube material shall not be ordered thinner than the required thickness calculated by Code formula. Also, the ordered thickness shall include provisions for manufacturing tolerance. Conforms with the 2004 ASME Extract • Revised 03/06 4.4. Paragraph PG-27 Cylindrical Components Under Internal Pressure The formulae in this section are used to determine the minimum required thickness of piping. This can be seen in that Section VIII-1 applies to small compressed-air receivers sold commercially to the general public as well as to very large pressure vessels used by the petrochemical industry.3 and are defined as follows: C = minimum allowance for threading and structural stability (mm) (see PG-27. These formulae can be transposed to determine the maximum allowable working pressure if the minimum required thickness is given.2. specific prohibitions.4. Table 1A. whichever is lower (the values allowed for E are listed in PG-27. The symbols used in the formulae are found in paragraph PG-27. when the maximum allowable working pressure is known. note 2) t = minimum required thickness (mm) (see PG-27. DIVISION 1 . and the use of good engineering judgement. Conforms with the 2004 ASME Extract • Revised 03/06 . drums. The Code is not a handbook and cannot replace education. The Code contains mandatory requirements.Chapter 1 • ASME Code Calculations: Cylindrical Components 5 Paragraph PG-21 states that the term maximum allowable working pressure (MAWP) refers to gauge pressure. note 3) D = outside diameter of cylinder (mm) E = efficiency of longitudinal welded joints or of ligaments between openings. refers to gauge pressure) R = inside radius of cylinder (mm) S = maximum allowable stress value at the operating temperature of the metal (Section II. note 1) e = thickness factor for expanded tube ends (mm) (see PG-27. and headers. note 4) P = maximum allowable working pressure (MPa). tubes. (see PG-21. See PG-27. experience. Part D. and non-mandatory guidance for pressure vessel construction activities.4. note 7) y = temperature coefficient (see PG-27. except when noted otherwise in the calculation formula of PG-27.4. note 6) ASME SECTION VIII.PRESSURE VESSELS Foreword The Boiler and Pressure Vessel Committee established rules for new construction of pressure vessels that ensure safe and reliable performance.4. or environmental conditions. This results in an increase of 43% in the thickness of the plates required.5 in the parent metal. as well as Section VIII. Use UW53 for ligaments between openings) Conforms with the 2004 ASME Extract • Revised 03/06 = = = = . or the efficiency of ligaments between openings. upset condition. Division 2 (VIII-2). except as otherwise permitted in paragraphs UG-9. including normal operation. This factor corresponds to a safety factor (or material quality factor) of 3. Design ASME Boiler Code Section I. whichever is less (use UW-12 for welded vessels. Paragraph UG-20: Design temperature With pressure vessels.0. UG-10. The minimum temperature used in design shall be the lowest temperature that the vessel will experience from any factor. requires all major longitudinal and circumferential butt joints to be examined by full radiograph.5 in plates. These formulae can be transposed to determine the maximum allowable working pressure if the minimum required thickness is given. which corresponds to a safety factor of 0. or the efficiency of. Nonradiographed longitudinal butt-welded joints have a joint efficiency factor (E) of 0. refers to gauge pressure) inside radius of shell course under consideration (mm) maximum allowable stress value (see UG-23 and the stress limitations specified in UG-24) E = joint efficiency for. UG-11.7. as is the minimum temperature. Paragraph UG-23 (a) lists the tables in Section II. appropriate joint in cylindrical or spherical shells. Section VIII-1 lists various levels of examination for these major joints. UG-15 and the Mandatory Appendices. D for various materials. The symbols used in the formulae are found in paragraph UG-27 (b) and are defined as follows: t P R S minimum required thickness (mm) internal design pressure (MPa) (see UG-21. Paragraph UG-27: Thickness of shells under internal pressure The formulae in this section are used to determine the minimum required thickness of shells when the maximum allowable working pressure is known. the maximum temperature used in the design is important. A fully radiographed major longitudinal butt-welded joint in a cylindrical shell would have a joint efficiency factor (E) of 1.6 Revised Second Class Course • Section A1 • SI Units Materials Paragraph UG-4 states that materials subject to stress due to pressure are to conform to the specifications given in Section II. ( t . the information will direct you to the correct stress table in ASME Section II. tubes.1.1 ) t = PD + 0. Solution For tubing up to and including 125 mm O.e ) ⎥ ⎣ ⎦ 1. (See paragraph PG-27.2 Example 1: boiler tube Calculate the minimum required wall thickness of a watertube boiler tube 70 mm O. that is strength welded into place in a boiler.D. The tube material is carbon steel SA-192.2. The maximum allowable working pressure is 4000 kPa gauge.D.2. SECTION I The following formulae are found in ASME Section I.0.01D . drums and headers of ferrous tubing up to and including 125 mm O.005D + e 2S + P Conforms with the 2004 ASME Extract • Revised 03/06 .2e ⎤ P = S ⎢ ⎥ ⎢ D .Chapter 1 • ASME Code Calculations: Cylindrical Components 7 OBJECTIVE OBJECTIVE 3 1 Calculate the required minimum thickness or the maximum allowable working pressure of piping.1 ⎡ 2t . use equation 1.1. The tube is located in the furnace area of the boiler and has an average wall temperature of 350°C. Part D by indicating if the metal is carbon steel or an alloy steel.005D . Formula for minimum required thickness t = Formula for MAWP PD + 0. paragraph PG-27.0. Note: Check PG-6 for plate materials and PG-9 for boiler tube materials before starting calculations.D.005 D + e 2S + P 1. 9 mm (Ans. the information will direct you to the correct stress table in ASME Section II. and 4. Part D. Note: Check PG-9 for boiler tube materials before starting calculations.56 + 0. SA-213-T11 is alloy steel.D.) ⎡ 2t − 0.35 = 1. Solution For tubing up to and including 125 mm O.8 Revised Second Class Course • Section A1 • SI Units Where P D e S = = = = 4000 kPa = 4. Part D.4. Example 2: superheater tube Calculate the maximum allowable working pressure. Table 1A.) 102 MPa (Section II. The manufacturing process does not produce absolutely uniform wall thickness.) Note: This value is exclusive of the manufacturer’s tolerance allowance (see PG-16. note 4.5% to the minimum thickness calculated.005D − e ) ⎥ ⎣ ⎦ Where t D e S = = = = 4.4. note 4.35 179.75 mm 75 mm 0 (see PG-27.1. Use equation 1. in kPa.6 = 1.8) + 4 280 = + 0. add an allowance of approximately 12. Table 1A.005(70) + 0 2(87. The average tube temperature is 400°C. The tube material is SA-213-T11. for a 75 mm O. Part D. strength welded.0 MPa 70 mm 0 (see PG-27.75 mm minimum thickness superheater tube connected to a header by strength welding. SA-192 at 350°C) t = 4 × 70 + 0. SA-213-T11 at 400°C) Conforms with the 2004 ASME Extract • Revised 03/06 .5).01D − 2e ⎤ P = S⎢ ⎥ ⎢ D − ( t − 0.D.2.2.8 MPa (see Section II. The formula for minimum thickness may be transposed to solve for the maximum allowable working pressure if the tube size and thickness are known. (See paragraph PG-27. strength welded) 87. 0.0 ) ⎣ ⎡ ⎤ 9.( 4.0.64 MPa = 12 640 kPa (Ans.( 0.4 PR (SE . Paragraph UG-31(a) states that these calculations are used for tubes and pipes under internal pressure.625 = 12.( 0.0.6P ) 1. the appropriate value of e is found in paragraph PG-27. Thin Cylindrical Shells (1) Circumferential stress (longitudinal joints) t = Or P = SEt (R + 0. PR t = (2SE + 0.4.75 ) .5 .3 When t < 0.005 × 75 ) .4P) Or 2SEt P = (R .4t ) When t < 0.5 1. note 4.75 70.5 R or P < 1.0.5R or P < 0.Chapter 1 • ASME Code Calculations: Cylindrical Components 9 ⎡ ( 2 × 4.75 = 102 × ⎢ ⎥ ⎢ 75 .75 .25SE Conforms with the 2004 ASME Extract • Revised 03/06 1.385SE (2) Longitudinal stress (circumferential joints) .01 × 75 ) .( 4.375 ) ⎥ ⎣ ⎦ 8. For calculations involving tubes expanded into place.( 2 × 0 ) P = 102 × ⎢ ⎢ 75 .75 .) = 102 × ⎤ ⎥ ⎥ ⎦ The tubes were strength welded in Example 1 and Example 2. paragraph UG-27(c)) are used for calculating wall thickness and design pressure.6 . SECTION VIII The following formulae (found in ASME Section VIII-1.6t ) 1. Conforms with the 2004 ASME Extract • Revised 03/06 .5.1⎟ Where Z = ⎝ ⎠ Where t > 0. a more accurate equation is required to determine the thickness. The corrosion allowance is 3 mm. and joint efficiency is 0.1 to 1.10 Note: Formulae 1. Calculate the required thickness of the shell if the allowable stress is 138 MPa. The formulae for thick walled vessels are listed in Appendix 1.1) Where ⎡(R + t)⎤ Z =⎢ ⎥ ⎣ R ⎦ 2 1.3 to 1.10 are for internal pressure only. respectively.R2 ) ) Where R0 and R are outside and inside radii.1) ⎤ P = SE ⎢ ⎥ ⎢ ( Z + 1) ⎥ ⎣ ⎦ = ⎡ (R + t)⎤ ⎢ ⎥ R ⎦ ⎣ 2 Where Z 1. SE = P R02 + R 2 2 0 ( (R .9 And P = SE ( Z .85.6 MPa. As the ratio of t/R increases beyond 0.5 R or P > 0.P ) 1.385SE ( SE + P ) ( SE .10 Revised Second Class Course • Section A1 • SI Units Thick Cylindrical Shells As internal pressures increase higher than 20.25SE ⎛ 1 ⎞ t = R ⎜ Z 2 .1⎟ Where ⎝ ⎠ ⎛ P ⎞ Z =⎜ ⎟ +1 ⎝ SE ⎠ 1. Example 3: thin shell thickness A vertical boiler is constructed of SA-515-60 material in accordance with the requirements of Section VIII-1. Supplementary Design Formulas 1.8 For longitudinal stress with t > 0.7 And ⎡ ( Z . It has an inside diameter of 2440 mm and an internal design pressure of 690 kPa at 230°C. By substituting R0 = R + t ⎛ 1 ⎞ t = R ⎜ Z 2 .3.5R or P > 1. special considerations must be given to the construction of the vessel as specified in paragraph U-1 (d). 92 mm (Ans.16 MPa.Chapter 1 • ASME Code Calculations: Cylindrical Components 11 Solution The quantity 0. use equation 1.) The calculated thickness is less than 0.385SE = 45.) Conforms with the 2004 ASME Extract • Revised 03/06 . since this is greater than the design pressure P = 690 kPa.92 + 6 = 344.5R.1⎟ ⎝ ⎠ = 463 × 0. (See Section VIII-1.732 ( 457 = 338.13 MPa.85) .92 mm Total including corrosion allowance t = 338. use equation 1.0.1⎟ ⎝ ⎠ (138 × 1) + 69 Z = (138 × 1) − 69 = = t = 207 69 3 Z = SE + P SE .886 = 7.P ⎛ 1 ⎞ + 6 ) ⎜ 3 2 . UG-27. equation 1.) Note R must be in the fully corroded state to determine the minimum thickness.69 ) 843. S = 138 MPa.7.7 cm. since this is less than the design pressure P = 69 MPa.( 0.3. Assume a corrosion allowance of 6 mm. t = = = PR + corrosion allowance ( SE .69 × (1220 + 3) + 3 (138 × 0. ⎛ 1 ⎞ Where t = R ⎜ Z 2 . Solution The quantity 0.385SE = 53.6 P ) 0.6 × 0.3 is acceptable. Example 4: thick shell thickness Calculate the required shell thickness of an accumulator with P = 69 MPa. and E = 1. therefore.22 mm (Ans.87 + 3 116. R = 45.0.22 + 3 = 10. compare the answer using equation 1.6 ( 52. Assume corrosion allowance = 0.1⎟ ⎝ ⎠ (138 × 1) + 52.3. Conforms with the 2004 ASME Extract • Revised 03/06 .7 ⎛ 1 ⎞ Where Z t = R ⎜ Z 2 .75 ) = = 24106.75 × 457 = +0 (138 × 1) .385SE = 53.13 MPa.59 mm (Ans.75 85.6P 52.35 226. Solution The quantity 0.52.75 MPa. t = PR + corrosion allowance SE .0.75 190.3.0.25 = 2.1⎟ ⎝ ⎠ 457 × 0.) This shows that the 'simple to use' equation (1.67 mm (Ans.4958 = SE + P SE . use equation 1. R = 45.75 Z = (138 × 1) .) This example used equation 1.2375 = t = = = 1 ⎛ ⎞ 457 ⎜ 2.2375 2 . since this is greater than the design pressure P = 52.P 226.3) is accurate over a wide range of R/t ratios.75 MPa.12 Revised Second Class Course • Section A1 • SI Units Example 5 Calculate the required shell thickness of an accumulator with P = 52. and E = 1.7 cm.75 106.0. S = 138 MPa. (nominal pipe size of 10 inches) and the material is SA-335-P11. Note: Check PG-6 and PG-9 for materials before starting calculations.C ) 2.3 P = R + (1 . Using the outside diameter t = PD + C 2SE + 2 yP 2.4 Example 6: steam piping Calculate the required minimum thickness of seamless steam piping which carries steam at a pressure of 6200 kPa gauge and a temperature of 375°C.2. or headers may be given with either the inside (R) or outside (D) measurement.2.(1 .1 mm O.y ) P 2. The material SA-335-P11 is alloy steel. In cylindrical vessels.y )( t − C ) SE ( t . drums. drums. Allow a manufacturer's tolerance allowance of 12. Conforms with the 2004 ASME Extract • Revised 03/06 .D.1 P = Using the inside radius t = D . the stress set up by the pressure on the longitudinal joints is equal to twice the stress on the circumferential joints. and headers. SECTION I The following formulae are found in ASME Section I. paragraph PG-27.Chapter 1 • ASME Code Calculations: Cylindrical Components 13 OBJECTIVE OBJECTIVE 3 2 Calculate the required minimum thickness or the maximum allowable working pressure of ferrous piping. the information will direct you to the correct stress table in ASME Section II.C ) 2SE ( t . The information for piping.5%.( 2 y )( t .2 PR + C SE . 273. Part D. The piping is plain end.C ) 2. Part D. Solution Use equation 2. note 3. The nominal pipe size is 323.4 × 6.4.94 mm (Ans.4.) Example 7: steam piping using outside diameter Calculate the maximum allowable working pressure in kPa for a seamless steel pipe of material SA-209-T1. The pipe is plain ended.2 × 273.5%. seamless pipe as per PG-9.14 Revised Second Class Course • Section A1 • SI Units Note: Plain-end pipe does not have its wall thickness reduced when joining to another pipe.1) y = 0. The material SA-209-T1 is alloy steel.) t = Where P D C S PD + C 2 SE + 2 yP 6200 kPa = 6. note 1. are classed as plain-end pipes. For example. Note: Check PG-6 and PG-9 for materials before starting calculations. rather than being joined by threading. Therefore 7.4.1 (See PG-27.1 + 0 2 (104 × 1) + 2 ( 0. lengths of pipe welded together. note 6.2. Table 1A.96 1693. Conforms with the 2004 ASME Extract • Revised 03/06 .22 = 212. pipe) with a wall thickness of 11.95 × 1.2 ) = = = = t = = 1693.125 = 8.4 (see PG-27. The operating temperature is 450°C.2. Assume that the material is austenitic steel.2 MPa 273.96 = 7.9 mm (~12 in.95 mm This value does not include a manufacturer's tolerance allowance of 12. the information will direct you to the correct stress table in ASME Section II. ferritic steel less than 475°C) 6.0 (see PG-27.85 mm. Part D. 4-inch nominal and larger) 104 MPa (see Section II. SA-335-P11 at 375°C) E = 1.1 mm 0 (see PG-27.22 208 + 4. note 6.") t = 11. Formulae and Physical Constants.9 .4 ) × (11.85 . seamless pipe as per PG-9. note 1.9.7 = 314. 4-inch (100 mm) nominal and larger) S = 101 MPa (Section II.( 2 y )( t . The longitudinal joint efficiency is 100%. The plate thickness of the tubesheet and drum are 59.1) y = 0.613 MPa = 7613 kPa (Ans.4.2.0 (see PG-27. "Table of Actual Pipe Dimensions. Part D. Table 1A.85 mm C = 0 (see PG-27.) Example 8: drum using inside radius A welded watertube boiler drum of SA-515-60 material is fabricated to an inside radius of 475 mm on the tubesheet and 500 mm on the drum.4.C ) D = 323.( 2 × 0. and the ligament efficiencies are 56% horizontal and 30% circumferential.85 323.5 mm and 38 mm respectively. austenitic steel at 450°C) P = = 323.42 = 7. The operating temperature is not to exceed 300°C.C ) 2SE ( t . note 3.2.) P = Where D .9 mm (see 2005 Academic Supplement.0 ) 2 (101 × 1) × (11.Chapter 1 • ASME Code Calculations: Cylindrical Components 15 Solution Use equation 2.4 (see PG-27. SA-209-T1 at 450°C) E = 1.0 ) 202 × 11.4. (See PG-27. Determine the maximum allowable working pressure based on: (a) the drum (b) the tubesheet Conforms with the 2004 ASME Extract • Revised 03/06 .85 .9 .48 2393. C ) = 115 MPa (see Section II.2. Greater material thickness is required where the boiler tubes enter the drum than is required for a plain drum.) Drum P = Where S SE (t .C ) R + (1 .4 (see PG-27. SA-515-60 at 300°C) E = 1 (see PG-27. Note: Check PG-6 and PG-9 for materials before starting calculations. (a) Use equation 2.4. ferritic steel less than 480°C) Conforms with the 2004 ASME Extract • Revised 03/06 . (See PG-27. note 6. note 3. Part D. 4-inch (100 mm) nominal and larger) R = 500 mm (for the drum) y = 0. The material SA-515-60 is carbon steel plate. Table 1A.16 Revised Second Class Course • Section A1 • SI Units FIGURE 1 Welded Watertube Boiler Drum DRUM TUBESHEET Note: This is a common example of a watertube drum fabricated from two plates of different thickness. the drum is designed to meet the pressure requirements for each situation.2. note 1) t = 38 mm C = 0 (see PG-27.consider the drum to have penetrations for boiler tubes.4 (inside radius R). b) The tubesheet .consider the drum to be plain with no penetrations.y )(t . Solution This example has two parts: a) The drum . For economy. Part D. the information will direct you to the correct stress table in ASME Section II.4.4. y )(t .8 = 475 + 35.) Tubesheet P = Where S E T C R y SE (t .4. (b) Use equation 2.0) 4370 = 500 + 22.7 = 7. SA-515-60 at 300°C) = 0.0.56 (circumferential stress = 30% and longitudinal stress = 56%. note 6. Table A1.36 MPa (Ans.2.5 .8 = 8. Part D. note 3.4)(38 .Chapter 1 • ASME Code Calculations: Cylindrical Components 17 Drum P = (115 × 1) (38 . SECTION VIII-1 Section VIII-1 does not contain separate formulae for small and large bore cylinders.4 (see PG-27. Conforms with the 2004 ASME Extract • Revised 03/06 .0) 3831.5 MPa (Ans.0.56 )( 59.5 mm = 0 (see PG-27.4)(59. (See PG-27.4 (inside radius R). The formulae given in paragraph UG-27 are used as set out in Objective 1.) (115 Note: The maximum allowable working pressure is based on the lowest number.0 ) 475 + (1 .4.30) = 59.) Note: In cylindrical vessels. ferritic steel less than 480°C) Tubesheet P = × 0.C ) = 115 MPa (see Section II.C ) R + (1 .2.56 < 2 x 0. 4-inch (100 mm) nominal and larger) = 475 mm (for the tubesheet) = 0.0) 500 + (1 .5 . therefore. 0. the stress set up by the pressure on the longitudinal joints is equal to twice the stress on the circumferential joints. Section I: DISHED HEAD CALCULATIONS The paragraphs from PG-29 must be considered when performing calculations on dished heads.” Example 9: the segment of a spherical dished head Calculate the thickness of a seamless. Paragraph PG-29. the longer shall be taken as the value of L in the formula.1 states that the thickness of a blank. blank unstayed dished head having pressure on the concave side. Table 1A). Part D.1 t = minimum thickness of head (mm). measured on the concave side S = maximum allowable working stress (MPa) (see ASME Section II. The head has a diameter of 1085 mm and is a segment of a sphere with a dish radius of 918 mm. Where two radii are used. Paragraph PG-29.2 states: "The radius to which the head is dished shall be not greater than the outside diameter of the flanged portion of the head. when it is a segment of a sphere.8S 3. The metal temperature does not exceed 250°C. unstayed dished head with the pressure on the concave side. Conforms with the 2004 ASME Extract • Revised 03/06 . shall be calculated by the following formula: t Where: = 5PL 4. P = maximum allowable working pressure (MPa). The maximum allowable working pressure is 2500 kPa and the material is SA-285 A. unstayed dished head. L = radius (mm) to which the head is dished. State if this thickness meets Code.18 Revised Second Class Course • Section A1 • SI Units OBJECTIVE 3 Calculate the required thickness or maximum allowable working pressure of a seamless. ) t = Where PD + C 2 SE + 2 yP D = 1085 mm y = 0.1 for segment of a spherical dished head.5 × 1085 2 ( 88.) Note: PG-29.2.086 mm Therefore. SA-285 A at 250°C) t= 5 ( 2.4. Part D.89 mm (Ans.1.4 × 2. Table 1A.5 177.8 + 2 = 15.Chapter 1 • ASME Code Calculations: Cylindrical Components 19 Solution Use equation 3.5 × 918 ) 4. Conforms with the 2004 ASME Extract • Revised 03/06 .9 = 26.9 × 1) + 2 ( 0.89 mm meets Code requirements.) t Where = 5PL 4.2. Use equation 2. the thickness of the shell must be calculated.1 (See paragraph PG-27. the head thickness of 26. shall be of a lesser thickness than that required for a seamless shell of the same diameter. ferritic steel less than 480°C) E = 1 (welded) t = = 2.9 MPa (see ASME Section II. to determine if this head thickness meets Code.5 ) 2712. (See paragraph PG-29.4 (see PG-27.8S P = 2.5 MPa L = 918 mm S = 88. except a full-hemispherical head. note 6." Therefore.8 × 88.6 states “No head. the thickness of which is calculated by this rule. and the metal temperature does not exceed 220oC.8S Conforms with the 2004 ASME Extract • Revised 03/06 . Solution First thing to check: is the radius of the dish at least 80% of the diameter of the shell? (per paragraph PG-29. unstayed dished head with pressure on the concave side. but in no case less than 3. the thickness shall be increased by 15% of the required thickness for a blank head computed by the above formula. (See paragraph PG-29. an increase in thickness over that for a blank head is not required. having a flanged-in manhole 154 mm by 406 mm. has a flanged-in manhole or access opening that exceeds 150 mm in any dimension. Note: This applies to the manhole found on the end of a boiler drum.1. The head has a diameter of 1206. the material is SA285-C.3 states When a head.0 mm additional thickness over a blank head. dished to a segment of a sphere.5 = 0.9473 > 0. the radius of this dish meets the criteria. "Inspection Openings" to see if this manhole size is acceptable.9473 0. Where such a dished head has a flanged opening supported by an attached flue.5) dish radius 1143 = shell diameter 1206.8 Therefore. If more than one manhole is inserted in a head. the minimum distance between the openings shall be not less than one-fourth of the outside diameter of the head. Example 10: the segment of a spherical dished head with a flanged-in manhole Calculate the thickness of a seamless.20 Revised Second Class Course • Section A1 • SI Units Paragraph PG-29. The maximum allowable working pressure is 1550 kPa.) t = 5PL 4.5 mm and is a segment of a sphere with a dish radius of 1143 mm. Note: Check paragraph PG-44. Use equation 3.1. Chapter 1 • ASME Code Calculations: Cylindrical Components 21 Where P = 1.0 mm. Table 1A: use 250°C since 220°C is not listed.088 mm This thickness is for a blank head.7 A blank head of a semi-ellipsoidal form in which half the minor axis or the depth of the head is at least equal to one-quarter of the inside diameter of the head shall be made at least as thick as the required thickness of a seamless shell of the same diameter as provided in PG-27. Therefore 17.2. 2.55 × 1143) 4. flanged-in manhole rule A semi-ellipsoidal head is shown in Fig.5) with a dish radius equal to eight-tenths the diameter of the shell and with the added thickness for the manhole as specified in PG-29. the thickness of the head shall be the same as for a head dished to a segment of a sphere (see PG-29.2. If a flanged-in manhole that meets the Code requirements is placed in an ellipsoidal head.1 and PG-29.3.15 = 2.088 mm (Ans. Part D. Conforms with the 2004 ASME Extract • Revised 03/06 . use the next higher temperature) t = = 5 (1.088 + 3.56 mm This is less than 3.8 (108 ) 17. blank head rule 2.3 requires this thickness to be increased by 15% or 3.088 × 0.0 mm Therefore Required head thickness = 17. so the thickness must be increased by 3. whichever is greater.55 MPa L = 1143 mm S = 108 MPa (see ASME Section II. This rule combines two rules: 1.) Semi-ellipsoidal head Paragraph PG-29. PG-29. therefore.0 mm.0 = 20. 5 mm. full-hemispherical head with the pressure on the concave side shall be calculated by the following formula: t Where = PL 2 S .11: The thickness of a blank.0. and the head material is SA-285-C. Part D). L = radius to which the head was formed (mm) (measured on the concave side of the head). fullhemispherical head with the pressure on the concave side. Maximum allowable working pressure is 6205 kPa. unstayed. Paragraph PG-29.P 3.22 Revised Second Class Course • Section A1 • SI Units FIGURE 2 Semi-ellipsoidal Head h = 1/4 D 1/2 r = 1/4 D h r D L Full-hemispherical head The following rule applies to drums or headers with a full-hemispherical end. 3.2 The above formula shall not be used when the required thickness of the head given by the formula exceeds 35. The average temperature of the header is 300oC.6% of the inside radius. Instead.3 Example 11: full-hemispherical head Calculate the minimum required thickness (mm) for a blank. S = maximum allowable working stress (MPa) (Table A1. use the following formula: ⎛ 1 ⎞ t = L ⎜ Y 3 . unstayed. The radius to which the head is dished is 190. Section II. P = maximum allowable working pressure (MPa).1⎟ ⎝ ⎠ where Y = 2(S + P) 2S . Conforms with the 2004 ASME Extract • Revised 03/06 .2 P t = minimum thickness of head (mm). 2.205 MPa L = 190. and transition sections.5 mm S = 107 MPa (see ASME Section II.) t Where = PL 2S .56 mm (Ans.2 P P = 6.Chapter 1 • ASME Code Calculations: Cylindrical Components 23 Solution Use equation 3.5). This chapter uses only Section VIII1 equations. the thickness of the head shall be the same as for a head dished to a segment of a sphere (see PG-29.05 214 .759 = 5. with a dish radius equal to eight-tenths the diameter of the shell and with the added thickness for the manhole as specified in PG-29. Part D. Section VIII-1 has rules for head configurations including spherical.2 ( 6. Conforms with the 2004 ASME Extract • Revised 03/06 . (See PG-29.12: If a flanged-in manhole that meets the Code requirements (see PG-44) is placed in a full-hemispherical head.6%.) Check if this thickness exceeds 35. heads.11. hemispherical. Table 1A.205 ) 1182.3.0. There are significant differences in the equations due to the different design approaches used. therefore The thickness of the head meets Code requirements.241 1182.5 × 0.0. ellipsoidal.1 and PG-29. SA-285-C at 300oC) t = = 6.05 = 212.1. SECTION VIII-1: DISHED HEAD CALCULATIONS Sections VIII-1 and VIII-2 each contain rules for the design of spherical shells.6% of the inside radius: 190.8 mm It does not exceed 35.5 2 (107 ) .205 × 190. Paragraph PG-29.356 = 67. and torispherical shapes. 665SE For thick shells.0. What is the required thickness of the hemispherical heads if the allowable stress is 138 MPa? Conforms with the 2004 ASME Extract • Revised 03/06 .2t 3. The internal design pressure is 690 kPa at 232°C.2 P 3. As the ratio t/R increases beyond 0.4 P = 2SEt R + 0. where t >0.24 Revised Second Class Course • Section A1 • SI Units Spherical Shells and Hemispherical Heads Paragraph UG-27 (d) gives the required thickness of a thin spherical shell due to internal pressure. use Mandatory Appendix 1 sections 1-3.6 or ⎛ Y -1 ⎞ ⎛ R+t⎞ P = 2SE ⎜ ⎟ where Y = ⎜ ⎟ ⎝Y + 2⎠ ⎝ R ⎠ Where t > 0.356R or P > 0.85.665SE 3 3.356R or P < 0.7 Example 12: hemispherical head A pressure vessel is built of SA-516-70 material and has an inside diameter of 2440 mm.5 Where t < 0. use the following equations ⎛ 1 ⎞ 2 ( SE + P ) t = R ⎜ Y 3 -1⎟ where Y = 2 SE . t = or PR 2SE . and the joint efficiency is 0.356.665SE.356R or P > 0. The corrosion allowance is 3 mm.P ⎝ ⎠ 3. equation 3.4.85 × 25.0.69 ) 843.85.69 × 1223 = +3 2 (138 × 0. Determine the design pressure if the allowable stress is 113 MPa.87 +3 234. therefore.665SE.) The calculated pressure is less than 0.4 is acceptable.2 ( 0.191 MPa (Ans.34 1529.3 is acceptable. Conforms with the 2004 ASME Extract • Revised 03/06 .6 + 3 = 6. since this is greater than the design pressure of 690 kPa.4 ) 4879.46 = 3.4 mm. equation 3.5 since t is less than 0.356R. Use equation 3.2t 2 (113 × 0.) = The calculated thickness is less than 0. Solution As no corrosion allowance is stated the design pressure will act on the given internal diameter.08 = 3. use equation 3. Assume joint efficiency E = 0. P = = = 2 SEt R + 0. Example 13: spherical head A spherical pressure vessel with an internal diameter of 3048 mm has a head thickness of 25.Chapter 1 • ASME Code Calculations: Cylindrical Components 25 Solution The quantity 0.) The inside radius in a corroded condition is equal to R = 1220 + 3 (corrosion allowance) = 1223 mm PR + corrosion allowance t = 2 SE .0.4 ) 1524 + 0.2 P 0.6 mm (Ans.665SE = 78 MPa.2 ( 25.356R. therefore.85) . (See paragraph UG-32 (f). fully corroded state.P ⎝ ⎠ Y = = 2 (103 × 1 + 69 ) 2 (103 × 1) .665SE = 68.69 344 137 = 2.26 Revised Second Class Course • Section A1 • SI Units Example 14: thick hemispherical head Calculate the required hemispherical head thickness of an accumulator with P = 69 MPa. and E = 1.9D and a knuckle radius of 0. Connecting this head to the accumulator shell would require special treatment. use equation 3.51 ⎛ 1 ⎞ t = R ⎜ Y 3 .0.9 Conforms with the 2004 ASME Extract • Revised 03/06 .) The required thickness of 2:1 heads with pressure on the concave side is given in paragraph UG-32 (d).1⎟ ⎝ ⎠ 1 ⎛ ⎞ = 460 + 6 ⎜ 2.).3 mm This is the minimum thickness i.359 ) = 167. 3b. Solution The quantity 0. Ellipsoidal Heads The commonly used ellipsoidal head has a ratio of base radius to depth of 2:1 (shown in Fig.3 mm (Ans.3 + 6 mm (corrosion allowance) = 173. Total head thickness is 167.8 P 3.513 . which is outside of the scope of this module.e. ⎛ 1 ⎞ 2 ( SE + P ) t = R ⎜ Y 3 .0.2t 3.6. 3a). Assume a corrosion allowance of 6 mm.1⎟ where Y = 2 SE .2 P = 2 SEt D + 0. S = 103 MPa. R = 460 mm.495 MPa. t = or PD 2SE .1⎟ ⎝ ⎠ = 466 ( 0. since this is less than the design pressure of 69 MPa.17D (shown in Fig. The actual shape can be approximated by a spherical radius of 0. Torispherical Heads Shallow heads. commonly referred to as flanged and dished heads (F&D heads). can be built according to paragraph UG-32 (e).Chapter 1 • ASME Code Calculations: Cylindrical Components 27 Where D E P S t = = = = = inside base diameter joint efficiency factor pressure on the concave side of the head allowable stress for the material thickness of the head FIGURE 3 Ellipsoidal Head (a) (b) Section VIII-1 does not give any P/S limitations or rules for ellipsoidal heads when the ratio of P/S is large.0D and a knuckle radius r of 0. A spherical radius L of 1. as shown in Fig. approximates the most common F&D heads. 4.06D. Conforms with the 2004 ASME Extract • Revised 03/06 . 1P SEt 0. This stress reversal could cause buckling of the shallow head as the ratio D/t increases.28 Revised Second Class Course • Section A1 • SI Units FIGURE 4 Torispherical Head The required thickness of an F&D head is t = or 0.11 Shallow heads with internal pressure are subjected to a stress reversal at the knuckle.0.10 P = Where E L P S t = = = = = 3.885L + 0. Conforms with the 2004 ASME Extract • Revised 03/06 .885PL SE .1t joint efficiency factor inside spherical radius pressure on the concave side of the head allowable stress thickness of the head 3. it may be a separate component that is attached by bolts or some quick-opening mechanism utilizing a gasket joint attached to a companion flange on the end of the shell.1 When the circular head. or blind flange is attached by bolts (Fig.1 states that the minimum thickness of unstayed flat heads. and blind flanges. AND BLIND FLANGES Flat plates. covers. cover. Alternately. or blind flange is attached by welding t = d CP S 4. Paragraph PG31. covers.3 states formulae for calculating the minimum thickness of flat.2 defines the notations used in this paragraph and in Fig. Bolted flanges are not covered in the scope of this module. cover plates. Paragraph PG-31. (k)) Conforms with the 2004 ASME Extract • Revised 03/06 . UNSTAYED FLAT HEADS. cover. When a flat plate or cover is used as an end closure or head of a pressure vessel. and flanges are used extensively in boilers and pressure vessels. and especially blind flanges are often misunderstood and can be challenging to anyone learning and working on this type of equipment. The concepts of unstayed flat heads. it may be designed as an integral part of the vessel (having been formed with the cylindrical shell) or welded to it. and blind flanges. It is very important for power engineers to have good working knowledge of thickness requirements as this allows them to work safely and provide sound and safe advice. PG-31 (j). PG-31. COVERS. SECTION 1 Paragraph PG-31. When the circular head.Chapter 1 • ASME Code Calculations: Cylindrical Components 29 OBJECTIVE 4 Calculate the minimum required thickness or maximum allowable working pressure of unstayed flat heads. unstayed circular heads. covers. covers. and blind flanges shall conform to the requirements. 3. (j). Note: The formulae used to determine thickness may be transposed to solve for P and find the maximum allowable working pressure for a flat head or cover of known thickness. and 4. obround.1.5 D When the non-circular head. or blind flanges that are square.4 to a maximum of 2.30 Revised Second Class Course • Section A1 • SI Units t = d 1.4. 4.2 Note: W = the total bolt loading and hg = the gasket moment arm.3.3 states two formulae for the required thickness of flat unstayed heads. (k)).3 Where Z is a factor from the ratio of the short and long spans 2.2. rectangular. PG-31 (j).2. Paragraph PG-31. The gasket moment arm is the radial distance from the centre line of the bolts to the line of the gasket reaction force (Fig.9Whg CP + S Sd 3 4.4 Paragraph PG-31. covers. the thickness t shall be calculated for both design conditions (flange sketches j and k) and the greater value used.4d Z = 3. cover. When using equation 4. or blind flange is attached by bolts (Fig. PG31. 4. (k)) t = d 6Whg ZCP + S SLd 2 4. elliptical. or segmental in design and attached by welding. Conforms with the 2004 ASME Extract • Revised 03/06 . t = d ZCP S 4.4 lists the values for C to be used in the formulae 4. Table 1A) As no temperature is given. The shell is seamless.3 to find the value of tr (see paragraph PG-27. C = 0.9 MPa (ASME Section II. the saturation temperature of steam (224°C at 2500 kPa) may be used.2) tr = required minimum thickness of the shell ts = actual thickness of the shell as given Use equation 2. The thickness t is 19 mm.Chapter 1 • ASME Code Calculations: Cylindrical Components 31 Example 15: circular flat head welded to a shell (Illustrated by Fig. Shell’s inside diameter d is 762 mm. PG-31 (e) and Fig. The material for head and shell is SA-285-A. Part D.4. t = PR + C SE . Solution Use equation 4.(1 .33 m (see PG-31. therefore. 5. use the value for 250°C. Maximum allowable working pressure is 2500 kPa.) FIGURE 5 Circular Flat Head Calculate the minimum thickness for the circular head and the depth of the fillet welds required.5 MPa d = 762 mm S = 88.y ) P Conforms with the 2004 ASME Extract • Revised 03/06 .1 t = d CP S Where P = 2.2. Head joint welding meets Code requirements. Fig PG-31 sketch (e). where m is defined as the ratio of tr/ts from paragraph PG-31.2). ) 762 For a welded circular flat head (Fig PG-31 (e)).5 952.5 88. t = = = = d CP S 0.15 mm is required.4 10.4 ) × 2.4 (see PG-27.2 in the formula from PG-31.574 C = 0.4.4) As this value is less than 0. note 1) 0.4. note 3) 2.898 19 0.2.0.0750 57.4 or in equation 4.9 762 × 0.33 × 0.20 × 2.5 × 381 + 0 (88. use 0.15 mm (Ans. a minimum thickness of 57.4.32 Revised Second Class Course • Section A1 • SI Units Where R E y C tr = = = = = = = d/2 = 381 mm 1 (see PG-27.5 87.3 mm Conforms with the 2004 ASME Extract • Revised 03/06 .9 × 1) . ts = 19 mm (given) = 0.574 = 0.31. The depth of each weld would be 0.33 m (from PG .7 ts (see Fig PG-31 (e)).7 × 19 = 13.19 tr ts 10.1.(1 .898 mm Therefore m = = = = 0. note 6) 0 (see PG-27. Table 1A) 610 mm 0.) P = t 2S d 2C 4.6 MPa (see ASME Section II.5 mm 96. Shell and head thickness of 30.Chapter 1 • ASME Code Calculations: Cylindrical Components 33 It is interesting to note that the required minimum shell thickness is 10. SECTION VIII-1 The equations for the design of unstayed plates and covers are found in paragraph UG-34. yet the required minimum thickness of the blank head is approximately 5.858 MPa (Ans.2.898 mm. Material is SA-285-B.13 = 1.33 (depending on the corner details as shown in Fig. unstayed head is 1858 kPa.15 mm. t = d CP SE 4. PG-31 (d)) Use equation 4. sketch (d). PG-31.3.6 = 6102 × 0.5 mm.1 and 0.2 times thicker at 57. (See PG-32. UG-34) P = design pressure S = allowable stress at design temperature Conforms with the 2004 ASME Extract • Revised 03/06 .13 (see Fig.6. Part D. Shell diameter is 610 mm. Example 16: circular flat head maximum allowable working pressure Calculate the maximum allowable working pressure for a circular flat head with the following specifications.) The maximum allowable working pressure for this flat.7 Where d = effective diameter of the flat plate (mm) C = coefficient between 0. Head joint weld meets Code requirements.6 P 30. Operating temperature not to exceed 300°C.52 × 96. Solution t S d C = = = = 30. Head design to Fig. There is a corrosion allowance of 4 mm. E is not a weld efficiency value of the head to shell corner joint.33 d = 610 + 4 = 614 mm (fully corroded state) t = d = 614 × CP + corrosion allowance SE 0.33(1) = 0.76 mm (Ans.33×17 +4 120 ×1 = 136. Example 17: integral flat plate Using the rules of paragraph UG-34.) Conforms with the 2004 ASME Extract • Revised 03/06 .7. The corner detail conforms to Fig. (See Fig UG-34 (b-2)) Where C = 0. and a plate diameter d = 610 mm. There are no butt weld joints within the head.33(m) = 0. UG-34 (b-2) (assume that m = 1). Solution Use equation 4. determine the minimum required thickness of an integral flat plate with an internal pressure P = 17 MPa. an allowable stress S = 120 MPa.34 Revised Second Class Course • Section A1 • SI Units E = butt-welded joint efficiency of the joint within the flat plate t = minimum required thickness of the flat plate The value of E depends on the degree of non-destructive examination performed. only the area replacement method is used. ASME Codes Section I and Section VIII give two methods for examining the acceptability of openings in the pressure boundary for pressure loads only. Curves have been developed to simplify this examination. Reinforcement limits are developed parallel and perpendicular to the shell surface from the opening. or head. This extra material may be provided by increasing the wall thickness of the shell or nozzle or by adding a reinforcement plate around the opening. experience has shown that they are acceptable. called the reinforced opening or area replacement method is used when nearby substitute areas replace the area removed by the opening. areas are substituted when making calculations. Although the methods used are a simplified application of the elastic foundation theory. The second method is the ligament efficiency method. For single openings. The first method. For multiple openings. Conforms with the 2004 ASME Extract • Revised 03/06 .Chapter 1 • ASME Code Calculations: Cylindrical Components 35 OBJECTIVE 5 Calculate the acceptability of openings in a cylindrical shell. Equations have been developed to set the limits for reinforcement. 1946). The stress analysis basis used in the ASME Codes to analyze nozzle reinforcement is called Beams on Elastic Foundation (Hetenyi. Openings through the pressure boundary of a vessel require extra care to keep loading and stresses at acceptable levels. Since stress is related to load and cross-sectional area. Placement and location of the replacement area are very important. either method may be used. This method determines the effectiveness of the material between adjacent openings to carry the stress compared with the area of metal that was there before the openings existed. An examination of the pressure boundary may indicate that extra material is needed near the opening to keep stresses at acceptable levels. header. 6 above. ASME Section I. The area (tr x d) is the cross-sectional area that is removed and has to be compensated for. Paragraph PG-32. Conforms with the 2004 ASME Extract • Revised 03/06 . PG-32.1.5tn + t e t smaller of 2.5tn WL1 tr B d ABCD = Limits of reinforcement d or Rn + t n + t Use larger value d or Rn + t n + t Use larger value C D When an opening is cut into a vessel wall for the attachment of a nozzle with diameter d (as in Fig. Paragraph PG-32.1. paragraph PG-32 "Openings in Shells. providing the diameter of the opening does not exceed that permitted by the chart in Fig. the vessel wall thickness t is usually thicker than the minimum thickness required tr.36 Revised Second Class Course • Section A1 • SI Units Figure 6 Reinforcement Limits tn Rn t rn A smaller of 2. If greater than the cross-sectional area removed. Headers and Heads" contains rules to be applied to maintain the vessel pressure boundary. 6). paragraph UG-40) gives the rules for the “Limits of Metal Available for Compensation. paragraph PG-36 (ASME Section VIII. SECTION I ASME Section I." The limit is shown by box ABCD in Fig.2 provides the rules for openings that do not require reinforcement calculations. 6 within the limit of compensation boundary) may provide adequate compensation.1 states that paragraphs PG-32 to PG-39 shall apply to all openings (except for flanged-in manholes as stated in paragraph PG-29) and to tube holes in a definite pattern that are designed according to paragraph PG-52. the additional material in the shell wall and the additional material in the nozzle wall (the hatched crosssectional area shown in Fig.5t or 2.5t or 2. of the shell is 1000 mm.4 = 25400 mm 2 Conforms with the 2004 ASME Extract • Revised 03/06 .0. ASME Section I rules apply.2.5 MPa 1000 mm 118 MPa 25. allowed without reinforcement. PG-32. The shell material is SA-515-60 and the nozzle material is SA-192. PG-32. the value for the x-axis is calculated from the shell diameter times the shell thickness. The thickness of the nozzle wall is 10 mm. Shell diameter × shell thickness = 1000 × 25. calculate the x-axis value. nozzle located in a cylindrical boiler shell. Example 18: reinforcement of nozzle abutting vessel Determine if reinforcement is required for a 100 mm I.82St 4. All joint efficiencies E = 1.4 mm.4 ) = 0. The point where the x-axis value meets the K value curve is read to the y-axis and gives the maximum diameter of the opening. The O.) Use equation 5.1 Using the chart in Fig. The nozzle abuts the vessel wall and is attached by a fullpenetration weld.4 mm K = PD 1.1. Solution As this is a boiler shell.Chapter 1 • ASME Code Calculations: Cylindrical Components 37 To determine if reinforcement is required.D.825 Using Fig. (See PG-32.82 (118 × 25. K = Where P D S t = = = = PD 1.5 × 1000 = 1. The thickness of the shell wall is 25.1 to calculate the K value. and the design temperature is not to exceed 200°C.D.82St 5. the value K is calculated from the formula K = PD 1.82St 4. The maximum allowable working pressure is 4500 kPa. i. and the design temperature is not to exceed 200°C. or (b) the opening is not larger than 60 mm diameter and the shell thickness is greater than 10 mm. to use reinforcement calculations in paragraph UG-37. not larger than 60 mm in a shell that is thicker than 10 mm. (See UG-36(c)(3).38 Revised Second Class Course • Section A1 • SI Units The intersection of the x-axis value (2540) and the K value curve (0. the nozzle diameter is 60 mm This falls within the second condition. All joint efficiencies E = 1. Therefore. The thickness of the shell wall is 25.) UG-36(c)(3) states that reinforcement is not required if (a)the opening is not larger than 89 mm diameter and the shell is 10 mm thick or less. Conforms with the 2004 ASME Extract • Revised 03/06 . The nozzle abuts the vessel wall and is attached by a fullpenetration weld. nozzle located in a cylindrical shell. The shell material is SA-516-60 and the nozzle is SA-192. unless certain dimensional requirements are met as listed in paragraph UG-36(c)(3). ASME Section VIII-1 rules apply.D. The maximum allowable working pressure is 4500 kPa.). no reinforcement is required (Ans.4 mm.825) give a yaxis value of approximately 134 mm. not subjected to rapid fluctuations. of the shell is 1000 mm.D. and the thickness of the nozzle wall is 10 mm. In this example.0 Solution As this is a not a boiler shell. no additional reinforcement is required (Ans. SECTION VIII-1 Section VIII-1 requires all openings in pressure vessels. Therefore.) for an opening of 100 mm diameter. Example 19: reinforcement of nozzle abutting vessel Determine the reinforcement requirements for a 60 mm I. The O.e. the requirements are illustrated in Fig.Chapter 1 • ASME Code Calculations: Cylindrical Components 39 OBJECTIVE 6 Calculate the compensation required to reinforce an opening in a cylindrical shell. 7.1. For an opening in a shell with a nozzle abutting the shell wall (such as an opening for a safety valve).5t or 2. paragraph PG-33. 7. "Compensation required for openings in shells and formed heads".5tn d ABCD = Limits of reinforcement d or Rn + t n + t Use larger value d or Rn + t n + t Use larger value C D NOZZLE WALL ABUTTING VESSEL WALL Conforms with the 2004 ASME Extract • Revised 03/06 . states the rules for compensation. or head. FIGURE 7 Nozzle Wall Abutting Vessel Wall Dp tn Rn tr n rn A WL1 smaller of WL2 2. Paragraph PG-33. header. shown in Fig.2 states that the total cross-sectional area of compensation required in any given plane for a vessel under internal pressure shall not be less than A as defined in PG-33. SECTION I ASME Section I.5t or 2.5tn + t e t tr B te smaller of 2. The O. and the design Conforms with the 2004 ASME Extract • Revised 03/06 . The nozzle abuts the vessel wall and is attached by a fullpenetration weld.D. nozzle located in a cylindrical boiler shell. if A1 + A2 + A41 +A42 + A5 > A The opening is adequately reinforced. The thickness of the shell wall is 22. Example 20: reinforcement of nozzle abutting vessel Determine the reinforcement requirements for a 100 mm I. The maximum allowable working pressure is 4900 kPa. The nozzle fillet welds are 5 mm wide. Therefore.5 mm and the thickness of the nozzle wall is 8 mm. of the shell is 1000 mm.5tn + te)fr1 where fr1 is the ratio of Snozzle/Sshell The area available from the nozzle to the reinforcement plate welds A41 = (WL1)2 × fr1 where fr1 is the ratio of the lesser of Snozzle or Splate / Sshell The area available from the reinforcement plate to shell weld A42 = (WL2)2fr3 The area available in the reinforcement plate (shown as herring-bone brick hatch) A5 = (Dp – d – 2tn)te/fr3 `Where fr3 is Splate/Sshell (b) (c) (d) (e) (f) If A1 + A2 + A41 > A The opening is adequately reinforced. PG-33 The area in the shell wall thickness available to be used as compensation A1 (shown as the forward sloped hatched areas on either side of the opening) = the larger of d(t – Ftr) or 2(t + tn)(t – Ftr) The area in the nozzle wall thickness available to be used as compensation A2 (shown as the backward sloped hatched area on either side of the nozzle) = the smaller of 2(tn – trn)(2.D. The shell material is SA-516-60 and the nozzle is SA-192.40 Revised Second Class Course • Section A1 • SI Units Where (a) The area to be replaced A (shown as the cross-hatched area) = dtrF where F is taken from the chart Fig. If A1 + A2 + A41 < A The opening is not adequately reinforced. and reinforcing elements (reinforcement plate and welds) must be added and/or the thickness must be increased.5tfr1) or 2(tn – trn)(2. 4 MPa. All joint efficiencies E = 1.) K Where P D S t = PD 1.5 ) = = = = = 1.5 MPa 500 – 22.5 mm K = 4.2 cannot be used.0.4.4. Allowable tensile stress for SA-516-60 is 118 MPa and for SA-192 is 92.Chapter 1 • ASME Code Calculations: Cylindrical Components 41 temperature is not to exceed 200°C.82 (118 × 22.4 (see PG-27. ASME Section I rules apply.5 = 477. note 3) Conforms with the 2004 ASME Extract • Revised 03/06 . Therefore PG-32.2." states that K is limited to a value of 0. PG-32.2) Where P R S E y C = = = = = = 4. Therefore: f r1 = 118 92.4 = 1.1. (See PG-27.1.014 ASME Section I.28 Use equation 2.5 mm 118 MPa 1 0.82St 4.2.9 MPa 1000 mm 118 MPa 22. (See PG-32.3 to determine the minimum required shell thickness (additional thickness may be used towards reinforcement requirements).99. "General Notes. note 6) 0 (see PG-27.9 × 1000 1. Solution As this is a boiler shell. Use equation 5. Fig.1. The reinforcement plate (if required) shall be of SA-192 material and 18 mm thick. 8 mm2 Conforms with the 2004 ASME Extract • Revised 03/06 .4 ) + 4.3375 mm = Therefore tr n = 3.y ) P = Therefore 4.5 × 477.4)4.1) Where P D S e = = = = 4.5 = 56.5 = 20.5 × 116 = + 0. whichever is smaller Y = 2.2.408 ×1 = 2040.408 mm tr = 20. PG-33.58 189.5d + tn + t).5 MPa 100 + (2 x 8) = 116 mm 92.42 Revised Second Class Course • Section A1 • SI Units tr = PR +C SE .3375 mm and tn = 8 mm Limit of compensation parallel to shell surface X = d or X = (0.5 Therefore X = 100 mm Limit perpendicular to the shell surface Y = 2.3.5tn + te). whichever is larger X = 100 or X = (0.3 = 3.25 or Y = (2.(1 . F=1) Ar = 100 ×20.005 (116 ) + 0 2 ( 92.0. PG-33.005D + e 2S + P 4.4.408 mm and t = 22. (See PG-27.5) = 80.1) A = dtrF (where F is taken from the chart Fig.5 522 + 0.5 +0 (118 × 1) .4 MPa 0 (see PG-27.(1 .5 ×100 + 8 + 22.5 mm Use equation 1. note 4) t = PD + 0.1 to determine the minimum required nozzle thickness.5 × 22.5t or Y = (2.5 × 8 + 18) = 38 Therefore Y = 38 mm (a) Reinforcement area required A (according to Fig. 85 mm2 (f) Diameter of the reinforcement pad A5 = ( D p .58 Therefore A42 = 19.5tn + te) in the equation A2 = 2(tn .75 + 100 + 16 92.2 Therefore A1 =209.75 mm ( Ans.58) = 1377. nozzle.d .17 = 414.408) = 209.4 118 = 213.37 + 19.2 × 8) × 18 × ( D p .58 mm2 Area required by reinforcement pad A5 = A – (Ar + A42) A5 = 2040.4/118 = 19.095 = 97.85 Therefore A5 = 1377.85 = ( D p .2 + 414.(643.100 .trn)(Y)fr1 A2 = 2(8 .58 = 643.3375)(38) × 1.2tn )te f r 3 1377.100 .Chapter 1 • ASME Code Calculations: Cylindrical Components 43 (b) Reinforcement area available in the shell (X replaces d in the equation) A1 = X(t – Ftr) A1 = 100(22. and nozzle weld Ar = A1 + A2 + A41 Ar = 209.59 + 19.59 Therefore A2 = 414.2 mm2 Reinforcement area available in the nozzle Y replaces (2.16) = Dp 1377.58 mm2 Total area available from shell.8 .85 14.4/118 = 19.37 mm2 (c) (d) (e) Area provided by the reinforcement plate weld A42 = (WL2)2Fr3 A42 = (5)2 × 92.58 Therefore A41 = 19.5 – 1 x 20.) Conforms with the 2004 ASME Extract • Revised 03/06 .59 mm2 Reinforcement area available in the nozzle weld A41 = (WL1)2fr2 where fr2 = Sn/Ss A41 = (5)2 × 92.3. the limit of compensation parallel to the shell surface X = diameter of the finished opening in corroded condition Or X = radius of the finished opening in corroded condition + shell wall thickness+ nozzle wall thickness Whichever is larger The limit of compensation normal to the shell surface Y = 2. Therefore.5 × nominal shell thickness less the corrosion allowance Or Y = 2. This pad size falls within the limits of compensation.44 Revised Second Class Course • Section A1 • SI Units Thus. SECTION VIII-1 The limits of compensation stated in paragraph UG-40 (b) and (c) are the same used in Section I. except that the vessel shell and nozzle must be treated as being in a corroded condition.5 × nozzle wall thickness + the thickness of the reinforcing plate (te) Whichever is smaller Conforms with the 2004 ASME Extract • Revised 03/06 . a reinforcing pad 213.75 mm diameter and 18 mm thick is required to carry the tensile stress and maintain the vessel pressure boundary. The thickness of the drum is 28. The plate is integrally welded into place as per Fig UG-34(h). All joint efficiencies E = 1. 2. MAWP is 6205 kPa. and the header and head material is SA-204-A. The material used has a stress value of 103 MPa. The tube will be in the furnace area of the boiler and have an average wall temperature of 350°C. The cylinder has an internal diameter of 36 cm. The maximum allowable working pressure is 6000 kPa. Using the rules in Section VIII-1.D. The maximum allowable working pressure is 3500 kPa.D. Conforms with the 2004 ASME Extract • Revised 03/06 . The drum material is SA-516-60.Chapter 1 • ASME Code Calculations: Cylindrical Components 45 CHAPTER QUESTIONS The following questions provide candidates with experience using the ASME Codes. 1.0. full hemispherical head.5 MPa? 5. 3. determine the minimum required thickness of the flat end plate of a rectangular box header 200 mm by 400 mm with an internal pressure of 2500 kPa. and the joint efficiency is 0. and the nozzle material is SA209-T1.D. nozzle located in a cylindrical boiler drum. There is no corrosion allowance and no butt-welded joints in the plate. and E = 1. unstayed. Calculate the required shell thickness for a hydraulic cylinder with a design pressure of 62 000 kPa. S = 142 MPa. and the design temperature is 250°C. The design pressure is 1034 kPa at 200°C. Calculate the thickness of a boiler steam header designed with a seamless. 6. The average temperature of the header is 400°C. The tube material is SA-192. The corrosion allowance is 4 mm. with pressure on the concave side. Assume no corrosion allowance for this cylinder. calculate the reinforcement requirements for a 150 mm I. 4. An air receiver pressure vessel is constructed from SA-204-A with an inside diameter of 1830 mm.575 mm. that is strength welded in place in a boiler drum.85. Using the rules in Section I. The inside radius of the header and the radius to which the head is dished is 304 mm. The nozzle wall thickness is 35 mm. Calculate the minimum required wall thickness of a watertube boiler tube 75 mm O. The nozzle abuts the vessel wall and is attached by a full-penetration weld. of the drum is 780 mm. The I. The header has a flanged-in circular inspection opening 100 mm diameter. The reinforcement plate material (if required) is of SA-515-55 and 10 mm thick. What is the required thickness of the hemispherical heads if the allowable stress is 147.