ASME Pressure Vessels – Basic Calculations
Here is what you will be able to do when you complete each objective:
1. Calculate the required minimum thickness or the maximum allowable working pressure of piping, tubes, tubes, drums, and headers of steel steel tubing up to and including 125 125 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.
Jurandir Primo, BSME Mechanical Engineer
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1.0 - INTRODUC INTRODUCTION TION:: forest of mandatory mandatory requirement requirements, s, specific specific prohibiti prohibitions, ons, non-mand non-mandatory atory guidance guidance The ASME Code is a forest construction activities, calculations and technical formulae. Every day a student or a professional is looking for a short and timely handbook with practical information and comprehensive calculations the way he can conclude a work without wasting too much his precious time. This is the main subject of this sketch. The ASME Code is divided into eleven sections and each of them, deal with particular areas. However, there are five five sectio sections ns of the ASME Code involved with the day by day to manufacture Pressure Vessels, as described below: The Section I – Power Boilers provides requirements for all methods of construction of power, electric, and miniature boilers; high temperature water boilers used in stationary service; and power boilers used in locomotive, portable, and traction service The Section II – Materials is divided into four parts:
Material Specific Specifications ations - Quality, Quality, Chemical Chemical Composition, Composition, Tensile Tensile Requirements Requirements,, etc. The Part A : Ferrous Material The Part B: Nonferrous Material Specifications. The Part C: Specifications for Welding Rods, Electrodes and Filler Metals. The Part D: Properties for Ferrous Materials. determines the Rules and the procedures for: Radiographic, Radiographic, The Section V - Nondestructive Examination determines Ultrasonic, Liquid Penetrant, Magnetic Particles and Leak Testing. The Section VIII – Pressure Vessels is divided into three divisions: Division 1 - Provides requirements applicable to the design, fabrication, inspection, testing, and certification of pressure vessels operating at either internal or external pressures exceeding 15 psig .
Alternative Rules, Rules, provides requirements requirements applicable applicable to the design, fabrication, fabrication, inspection, inspection, Division 2 - Alternative testing, and certification of pressure vessels operating at internal or external pressures exceeding 15 psig . These rules cover only vessels to be installed in a fixed location for a specific service where operation and maintenance maintenance control control is retained retained during the useful useful life of the the vessel, by the user, who prepares prepares or causes to be prepared the design specifications. Division 3 - Alternative Rules for Construction of High Pressure Vessels, provides requirements applicable to the design, fabrication, inspection, testing, and certification of pressure vessels operating at either internal or external pressures generally above 10,000 psi . The Section IX - Welding and Brazing Qualifications provides requirements such as Procedure Qualifications, the Qualification of Welders, Welding Operators and Brazing Procedures, Welders, Brazers, and Welding and Brazing Operators.
Once any ASME ASME Stamped Pressure Vessel is manufactured, it is checked, tested and approved by the ASME Authorized Inspector, who review all the procedures and all the documentation and sign the Data Report Form before to procedure to Stamp the name plate with the “U” or “UM” symbols, which means that the Pressure Pressure Vessel Vessel fully fully compli complies es with with the ASME Code rules for construction of Pressure Vessels. The National Board of Boiler & Pressure Vessel Inspectors uses the “NB” Symbol as well the “R” Symbol to repair or to alter any previous Stamped Pressure Vessel. 2
Note: Professionals and students should be well versed in conversion practice . Many US customary unit values presented in the ASME codes do not convert directly into metric va lues in the ASME editions.
1.1 - ASME SECTION I - POWER BOILERS The scope of jurisdiction of Section I applies to boilers and boilers external piping. Superheaters, economizers, and other pressure parts connected directly to the boiler, without valves , are considered to be parts of the proper boiler and their construction shall conform to Section I rules. The ASME Section I, paragraphs PG-1, PG-2 covers rules construction of power boilers, electric boilers, miniature boilers, and high temperature water boilers. Materials Steel plates for boilers subject to pressure, whether or not exposed to the fire, shall be in accordance with the specifications in ASME Section I, paragraph PG-6.1. Paragraph PG-9 states that pipes, tubes, and pressure containing parts used in boilers shall be according to the specifications listed in paragraph PG-9.1. Design
The ASME Section I, paragraph PG-16.3 states that the minimum thickness of any boiler plate under pressure shall be 6 mm (1/4”). The minimum thickness of plates where stays may be attached shall be 8 mm (5/16”). When pipe over 125 mm (5.0”) O.D. is used in lieu of plate for the shell of cylindrical components under pressure, its minimum wall thickness shall be 6 mm (1/4”). The ASME Section I, Paragraph PG-21 states that the term maximum allowable working pressure (MAWP) refers to gauge pressure, except when noted in the calculation formula. Cylindrical Components under Internal Pressure.
The formulae in ASME Section I - Paragraph PG-27, are used to determine the minimum required thickness of piping, tubes, drums, and headers, when the maximum allowable working pressure is known. These formulae can also be transposed to determine the maximum allowable working pressure when the minimum required thickness is given . The symbols used in the formulae are found in paragraph PG-27.3 and are defined as follows: C = Minimum allowance for threading and structural stability (mm, inches) (see PG-27.4) D = Outside diameter of cylinder (mm, inches) E = Efficiency of longitudinal welded joints. A fully radiographed major longitudinal butt-welded joint in a cylindrical shell would have a joint efficiency factor ( E ) of 1.0. Non radiographed longitudinal butt-welded joints have a joint efficiency factor ( E ) of 0.7. e = Thickness factor for expanded tube ends (mm, inches) (see PG-27.4) P = Maximum allowable working pressure (MPa, Kg/cm², psi) refers to gauge pressure ) R = Inside radius of cylinder (mm, inches) S = Maximum allowable stress value at the operating temperature (Section II, Part D, Table 1A. See PG-27.4) t = Minimum required thickness (mm, inches) (see PG-27.4) y = Temperature coefficient (see PG-27.4) = 0.4 [ferritic steel less than 480° C (896°F)].
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1.2 - ASME SECTION VIII, DIVISION 1 - PRESSURE VESSELS This Division of Section VIII provides requirements applicable to the design, fabrication, inspection, testing, and certification of pressure vessels operating at either internal or external pressures exceeding 15 psig . It contains mandatory and nonmandatory appendices detailing supplementary design criteria, nondestructive examination and inspection acceptance standards. Rules pertaining to the use of the U, UM and UV Code symbol stamps are also included. Materials
Paragraph UG-4 states that materials subject to stress due to pressure shall be according to the specifications given in Section II, D - Paragraph UG-23 for various materials. Design
The ASME Boiler Code Section I, as well as Section VIII, Division 2 , requires longitudinal and circumferential butt joints to be examined by full radiograph. The Section VIII-1 lists various levels of examination for these joints. Fully radiographed longitudinal butt-welded joint in a cylindrical shell shall have a joint efficiency factor (E ) of 1.0. This factor corresponds to a safety factor (or material quality factor) of 3.5 in the parent metal. Non radiographed longitudinal butt-welded joints shall have a joint efficiency factor ( E ) of 0.7, which corresponds to a safety factor of 0.5 in . plates resulting in an increase of 43% in the thickness of the plates required. ASME Paragraph UG-20: Design temperature
The minimum temperature used in design shall be the lowest temperature that the vessel will experience from any factor, including normal operation, upset condition, or environmental conditions. ASME 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 (MAWP) is known. The symbols used in the formulae are found in paragraph UG-27 (b) and are defined as follows: t = Minimum required thickness (mm, inches) P = Internal design pressure (MPa, Kg/cm², psi) refers to gauge pressure R = Inside radius of shell course under consideration (mm, inches) S = Maximum allowable stress value at the operating temperature (MPa, Kg/cm², psi) E = Efficiency of longitudinal welded joints. A fully radiographed major longitudinal butt-welded joint in a cylindrical shell would have a joint efficiency factor ( E ) of 1.0. Non radiographed longitudinal butt-welded joints have a joint efficiency factor ( E ) of 0.7. In general the factor used is E = 0.85.
Notes: 1. Most relevant mandatory and nonmandatory sections can be found in the ASME Academic Extract (visit www.powerengineering.org ). 4
2. The ASME Academic Extract converts 5 inches to 127mm and ¼ inch is converted to 6.35 mm . The ASME rounds 5 inches to 125 mm and ¼ inch to 6 mm. You are required to use the ASME values and to convert US customary numbers rounding values for calculation purposes.
2.0 - SECTION I – Calculation: The following formulae are found in ASME Section I, paragraph PG-27.2.1. a) Formula for minimum required thickness :
b) Formula for gauge maximum allowable working pressure (MAWP):
2.1 - Example: Section I - Boiler Tube:
Calculate the minimum required wall thickness of a water tube boiler 70 mm (2.75 in) O.D. that is strength welded (e = 0) into place in a boiler. The tube is located in the furnace area of the boiler and has an average wall temperature of 350°C (650°F) . The maximum allowable working pressure is 4000 kPa (580 psi) gauge. Material is carbon steel SA-192. Note: See PG-6 for plate materials and PG-9 for boiler tube materials before starting calculations to check the correct stress table in ASME Section II, Part D if the metal is carbon steel or alloy steel. Solution:
For tubing up to and including 125 mm [5 in] O.D. use equation 1.1 above. P = 4000 kPa = 4.0 MPa [580 psi] D = 70 mm [2.75 in] e = 0 (strength welded) S = 87.8 MPa [~12700 psi] - SA-192 at 350°C [~650°F]). (metric)
(imperial) t= 2.75 x 580 (2 x 12700) + 580
+0.005 x 2.75 + 0
t = 0.075 in t = 1.9 mm = 0.075 in Note: Where the manufacturing process produces only standard plate thickness , so should be used 1/8 in (3.2 mm) minimum. 5
2.2 - Example: Section I – Superheater Tube:
Calculate the maximum allowable working pressure for a 75 mm [2.95 in] O.D . and 4.75 mm [0.19 in] minimum thickness superheater tube connected to a header by strength welding. The average tube temperature is 400°C [~750° F]. The tube material is SA-213-T11. Note: According to table in ASME Section II, Part D, SA-213-T11 alloy steel is 102 MPa (~15000 psi). Solution:
For tubing up to and including 125 mm [~5 in] O.D. Use equation 1.2 above. t = 4.75 mm [0.19 in] D = 75 mm [2.95 in] e = 0 (strength welded.) S = 102 MPa [~15000 psi] - (SA-213-T11 at 400°C ~750° F)
(metric)
(imperial) P = 15000 x [(2 x 0.19) - (0.01 x 2.95) - (2 x 0)] 2.95 – [0.19 – (0.005 x 0.19)] - 0
P = 15000 x
0.38 - 0.0295 2.95 - (0.19 - 0.00095)
P = 1904 psi
3.0 - SECTION VIII – Basic Calculation Formulae: The following formulae (ASME Section VIII-1, paragraph UG-27) are used for calculating wall thickness and design pressure. Paragraph UG-31states that these calculations are also used for tubes and pipes under internal pressure. 3.1 – Formulae for thin Cylindrical Shells: 3.1.1 – Circumferential stress (longitudinal joints):
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3.1.2 –Longitudinal stress (circumferential joints):
3.2 – Formulae for thick Cylindrical Shells
As internal pressures increase higher than 20.6 MPa (~3000 psi), special considerations must be given to the construction of the vessel as specified in paragraph U-1 (d). As the ratio of t/R increases beyond 0.5, a more accurate equation is required to determine the thickness. The formulae for thick walled vessels are listed in ASME Appendix 1, Supplementary Design Formulas 1.1 to 1.3.
a) Where R 0 and R are outside and inside radii, respectively. By substituting R 0 = R + t.
and
b) For longitudinal stress with t > 0.5R or P > 1.25SE
And
Note: Formulae 1.3 to 1.10 are for internal pressure only.
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3.3 – Formula for Spherical Shells: PR t =
2SE + 0.8P
3.4 – Example: Thin Cylindrical Shells when P < 0.385SE, or, 0.385SE > P
A vertical boiler is constructed of SA-515-60 material in accordance with the requirements of Section VIII-1. The inside diameter is 2440 mm [96 in] The internal design pressure is 690 kPa [0.69 Mpa = [~100 psi] at 230°C [446 F°]. The corrosion allowance is 3 mm (0.125 in) The joint welding efficiency is 0.85 Calculate the required thickness of the shell if the allowable stress is 138 MPa [20,000 psi]. Solution: Equation 1.3 - for circumferential stress (longitudinal joints), when P < 0.385SE = 0.69 MPa < 0.385 x 138 MPa x 0.85 = 45.16 MPa Then, 0.69 MPa (P) < 45.16 Mpa (0.385SE), or, ~100 psi < 6550 psi
The inside radius in a corroded condition is equal to - R = 1220 + 3 = 1223 mm [48.15 in]. t=
PR + corrosion allowance (SE - 0.6P)
t=
0.69 x (1220 + 3) (138 x 0.85) - (0.6 x 0.69)
+ 3
t = 10.22 mm = 0.402 in
The calculated thickness must be less than 0.5 R = 0.5 x 1220 mm = 610 mm t = 10.22 mm < 610 mm - therefore, equation 1.3 is acceptable .
3.5 - Example: Thick Cylindrical Shells when P > 0.385SE, or, 0385SE < P
Calculate the required shell thickness of an accumulator with P = 69 MPa [~10,000 psi], R = 45.70 cm [180 in], S = 138 MPa [20000 psi], and E = 1.0. Assume a corrosion allowance of 6.0 mm [0.25 in]. Solution: Thick cylindrical shells when P > 0.385SE = 0.385 x 138 x 1.0 = 53.13 MPa. Then, 69 MPa (P) > 53.13 MPa (0.385SE), or , ~10,000 psi > ~7700 psi (use equation 1.7).
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= 13.34 in 3.5.1 - Example: Thick Cylindrical Shells when P < 0.385SE, or, 0.385SE > P
Calculate the required shell thickness of an accumulator with P = 52.75 MPa [7650 psi], R = 45.7 cm [180 in], S = 138 MPa [20000 psi], and E = 1.0. Assume corrosion allowance = 0. Solution: Thick cylindrical shells when P > 0.385SE = 0.385 x 138 x 1.0 = 53.13 MPa > 52.75 MPa (P).
Then, we can compare equation 1.3 using equation 1.7:
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4.0 – SECTION I - Piping, Drums, and Headers. The following formulae are found in ASME Section I, paragraph PG-27.2.2. The information for piping, drums, or headers may be given with either the inside (R) or outside (D) measurement. a) Using the outside diameter :
b) Using the inside radius:
4.1 - Example: Steam Piping:
Calculate the required minimum thickness of a seamless steam piping which carries steam at a pressure of 6200 kPa [6.2 Mpa = 899 psi] gauge and a temperature of 375°C [707°F]. The piping is 273.1 mm [10.77 in] O.D., (10 inches nominal) plain end; the material is SA-335-P11. Allow a manufacturer's tolerance allowance of 12.5%. Note: Check PG-6 and PG-9 for materials before starting calculations; the information will direct you to the correct stress table in ASME Section II, Part D. The material SA-335-P11 is alloy steel . Note: Plain-end pipe does not have its wall thickness reduced when joining to another pipe. For example, lengths of pipe welded together, rather than being joined by threading, are classed as plain-end pipes. Solution: Use equation 2.1.
P = 6200 kPa = 6.2 MPa [899 psi] D = 273.1 mm [10.77 in] C = 0 S = 104 MPa [15000] - SA-335-P11 alloy steel at 375°C [707°F]) E = 1.0 y = 0.4 (ferritic steel less than 480°C).
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= 0.31 in
Including a manufacturer's allowance of 12.5%. Therefore, 7.95 × 1.125 = 8.94 mm [~0.35 in]. 4.1.1 - Example: Steam Piping - Using the Outside Diameter
Calculate the maximum allowable working pressure for a seamless steel pipe of material SA-209-T1 . The nominal pipe size is 323.9 mm [12.75 in] (~12 in. pipe) with a wall thickness of 11.85 mm [0.46 in]. The operating temperature is 450°C [842°F]. The pipe is plain ended. Assume that the material is austenitic steel. Note: Check PG-6 and PG-9 for materials before starting calculations; the information will direct you to the correct stress table in ASME Section II, Part D. The material SA-209-T1 is alloy steel . Solution: Use equation 2.2.
D = 323.9 mm [12.75 in] t = 11.85 mm [0.46 in] C = 0 S = 101 MPa [~15000 psi] - SA-209-T1 at 450°C [842°F]) E = 1.0 y = 0.4 (austenitic steel at 450°C [842°F])
= ~1104 psi
In cylindrical vessels, the stress set up by the pressure on the longitudinal joints is equal to twice the stress on the circumferential joints. 4.1.2 - Example: Drum - Using the Inside Radius 11
A welded water tube boiler drum of SA-515-60 material is fabricated to an inside radius of 475 mm [18.7 in ] on the tube sheet and 500 mm [19.68 in] on the drum. The plate thickness of the tube sheet and drum are 59.5 mm [2.34 in] and 38 mm [1.49 in] respectively. The longitudinal joint efficiency is 100%, and the ligament efficiencies are 56% horizontal and 30% circumferential. The operating temperature is not to exceed 300°C [572°F]. Determine the maximum allowable working pressure (MAWP) based on: FIGURE 1 Welded Water Tube Boiler Drum
DRUM TUBESHEET
Note: This is a common example of a water tube drum fabricated from two plates of different thickness. The material SA-515-60 is carbon steel plate. Solution:
This example has two parts: a) The drum - consider the drum to be plain with no penetrations. a)Use equation 2.4 (inside radius R). Drum P =
SE (t - C )
R + (1 - y) (t - C )
Where: S = 115 MPa [16600 psi] - SA-515-60 at 300°C [572°F]) E = 1.0 t = 38 mm [1.49 in] C=0 R = 500 mm [19.68 in] (Drum) y = 0.4 (ferritic steeel less than 480° C [896°F])
Drum P =
=
115 x 1.0 (38 - 0) 500 + (1.0 - 0.4) (38 - 0) 4370 500 + 22.8
P = 8.36 MPa - [~1212 psi] (Ans.) 12
Note: In cylindrical vessels, the stress set up by the pressure on the longitudinal joints is equal to twice the stress on the circumferential joints.
b) The tubesheet - consider the drum to have penetrations for boiler tubes. b) Use equation 2.4 (inside radius R).
Where: S = 115 MPa [16600 psi] - SA-515-60 at 300°C [572°F]) E = 0.56 (circumferential stress = 30% and longitudinal stress = 56%; therefore, 0.56 < 2 x 0.30) T = 59.5 mm [2.34 in] C = 0 R = 475 mm [18.7 in] (for the tubesheet). y = 0.4 [ferritic steel less than 480°C (896°F)]
= ~1087 psi Note: The maximum allowable working pressure (MAWP) is based on the lowest number.
5.0 - SECTION I: DISHED HEAD CALCULATIONS 5.1 - Blank, Unstayed Dished Heads
Paragraph PG-29.1 states that the thickness of a blank, unstayed dished head with the pressure on the concave side, when it is a segment of a sphere, shall be calculated by the following formula:
Where: t = minimum thickness of head (mm, in). P = maximum allowable working pressure (MPa,psi). L = radius (mm) to which the head is dished on the concave side. S = maximum allowable working stress (MPa, psi).
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. Where two radii are used, the longer shall be taken as the value of L in the formula.” 13
5.2 – Example: Segment of a Spherical Dished Head
Calculate the thickness of a seamless, blank unstayed dished head having pressure on the concave side. The head has a diameter of 1085 mm [42.7 in] and is a segment of a sphere with a dish radius of 918 mm [36.1 in]. The maximum allowable working pressure is 2500 kPa [2.5 Mpa = 362 psi] and the material is SA-285 A . The temperature does not exceed 250°C [482°F] . State if this thickness meets Code. Solution: Use equation 3.1.
P = 2.5 MPa [362 psi] L = 918 mm [36.1 in] S = 88.9 MPa [12893 psi] - SA-285 A at 250°C [482°F)
= 1.06 in
Therefore, to determine if this head thickness meets the Code , the minimum thickness of the piping must be calculated. Use equation 2.1.
D = 1085 mm [ 42.7 in] y = 0.4 (ferritic steel less than 480°C [896°F]) E = 1.0 (welded)
= 0.59 in
Therefore, a head thickness of 26.89 mm [1.06 in] meets Code requirements. 5.3 - Example: Segment of a Spherical Dished Head with a Flanged-in Manhole
Calculate the thickness of a seamless, unstayed dished head with pressure on the concave side, having a flanged-in manhole 154 mm [6.06 in] x 406 mm [15.98 in] . Diameter head is 1206.5 mm [47.5 in] as a segment of a sphere with a dish radius of 1143 mm [45 in]. The maximum allowable working pressure is 1.55 MPa [225 psi], the material is SA-285-C, and temperature does not exceed 220°C [428°F]. 14
Solution: First thing to check: is the radius of the dish at least 80% of the diameter of the shell?
Therefore, the radius of the dish meets the criteria. Using equation 3.1:
P = 1.55 MPa [225 psi] L = 1143 mm [45 in] S = 108 MPa [~15600 psi] (use 250°C [482°F] since 220°C [428°F] - (use the next higher temperature) .
= 0.672 in This thickness is for a blank head. PG-29.3 requires this thickness to be increased by 15% or 3.0 mm, whichever is greater. As 17.088 × 0.15 = 2.56 mm [0.1 in] is less than 3.0 mm [0.12 in], so the thickness must be increased by 3.0 mm [0.12 in] . Required head thickness = 17.088 + 3.0 = 20.088 mm [0.79 in] (Ans.)
5.4 – Seamless or Full-Hemispherical Head
The following equation applies to drums or headers with a full-hemispherical end . The thickness of a blank, unstayed, full-hemispherical head with the pressure on the concave side shall be calculated by the following formula:
t = minimum thickness of head (mm) [in]. P = maximum allowable working pressure (MPa) [psi]. L = radius to which the head was formed (mm) [in] (measured on the concave side of the head). S = maximum allowable working stress (MPa) [psi]
The above formula shall not be used when the required thickness of the head given by the formula exceeds 35.6% of the inside radius . Instead, use the following formula:
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5.4.1 - Example: Seamless or Full-Hemispherical head
Calculate the minimum required thickness for a blank, unstayed, full-hemispherical head with the pressure on the concave side. The radius to which the head is dished is 190.5 mm [7.5 in]. The maximum allowable working pressure is 6.205 MPa [900 psi], and the head material is SA-285-C. The average temperature of the header is 300oC [572°]. Solution: Use equation 3.2.
P = 6.205 MPa [900 psi] L = 190.5 mm [7.5 in] S = 107 MPa [~15500 psi] - SA-285-C at 300°C).
= 0.22 in
Check if this thickness exceeds 35.6% of the inside radius: 190.5 × 0.356 = 67.8 mm [7.5 x 0.356 = 2.67 in]. It does not exceed 35.6%, therefore the thickness of the head meets Code requirements .
6.0 - SECTION VIII-1: DISHED HEAD CALCULATIONS The Section VIII-1 determines the rules for head configurations including spherical, hemispherical, elliptical or ellipsoidal and torispherical shapes . To know exactly how the shapes are, make some confusing for beginners and even professionals users of ASME VIII. To cast a little light on these subjects we tried to resume below:
Hemispherical Heads
Semi-Elliptical Heads (2:1)
Torispherical Heads
Elliptical or Ellipsoidal Heads (1.9:1)
Flanged and Dished Head
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Flat Dished Heads Non Standard 80-10 Dished Heads
Dished Discs
Toriconical Heads
6.1 - Spherical or Hemispherical Heads
or
Where t < 0.356 R or P < 0.665 SE When the ratio t/R increases beyond 0.356 , use the following equations:
or
Where t > 0.356R or P > 0.665SE 6.1.1 - Example: Spherical or Hemispherical Head
A pressure vessel is built of SA-516-70 material and has an inside diameter of 2440 mm [96 in]. The internal design pressure is 0.690 MPa [100 psi] at 232°C [450°F]. Corrosion allowance is 3 mm [0.118 in], and joint efficiency is 0.85 . What is the required thickness of the hemispherical heads if “S” is 138 MPa [20000 psi]? Solution: Since 0.665SE > P = 78 MPa [11313 psi] > 0.690 MPa [100 psi], use equation 3.4. The inside radius in a corroded condition is equal to - R = 1220 + 3 = 1223 mm [48.15 in].
= 0.26 in
The calculated thickness is less than 0.356R ; therefore, equation 3.3 is acceptable. 17
6.1.2 - Example: Spherical Shell
A spherical pressure vessel with an internal diameter of 3048 mm [120 in] has a head thickness of 25.4 mm [1”]. Determine the design pressure if the allowable stress is 113 MPa[~16400 psi ]. Assume joint efficiency E = 0.85. No corrosion allowance is stated to the design pressure. Solution: Use equation 3.5 since t < 0.356R .
= 463 psi .
The calculated pressure is less than 0.665SE ; therefore, equation 3.4 is acceptable. 6.1.3 - Example: Thick Hemispherical Head
Calculate the required hemispherical head thickness of an accumulator with P = 69 MPa [10007 psi], R = 460 mm [18.1 in], S = 103 MPa [~15000 psi], and E = 1.0. Corrosion allowance is 6 mm (0.25 in]. Solution: As 0.665SE < P = 68.495 MPa [9934 psi]< 69 MPa [10007 psi] , use equation 3.6.
= 6.58 in
This is the minimum thickness for fully corroded state . Total head thickness is 167.3 + 6 mm (corrosion allowance) [6.58 + 0.125 in] = 173.3 mm [6.80 in] (Ans.). 18
6.2 – Elliptical or Ellipsoidal Heads
The commonly used ellipsoidal head has a ratio of base radius to depth of 2:1 (shown in Fig. 2a). The actual shape can be approximated by a spherical radius of 0.9D and a knuckle radius of 0.17D (shown in Fig. 2b.) The required thickness of 2:1 heads with pressure on the concave side is given in paragraph UG-32 (d). Semi-Elliptical or Semi-Ellipsoidal Heads – 2:1
6.2.1 - Example: Elliptical or Ellipsoidal Head (imperial unities):
Dimensions: Inside Diameter of Head (Di) – 18.0 [in] Inside Crown Radius (L) – ( 18.0 x 0.9Di) [in] Inside Knuckle Radius (ri) – ( 18.0 x 0.17Di) [in] Corrosion Allowance - 0.010 corr [in] Straight Skirt Length (h) - 1.500 skirt [in] Material and Conditions:
Material - SA-202 Gr. B – Room Temperature. Internal Pressure - 200 P [psi] Allowable Stress - 20,000 S [psi] 19
Head Longitudinal Joint Efficiency - 0.85 E Radius L - ( 18.0 x 0.9Di) = 16.20 [in] Radius ri - ( 18.0 x 0.17Di) = 3.06 [in] Variable:
L/r = L/ri = 16.2/3.06 = 5.29 [in] a) Required Thickness: t = t = t =
PD
+ corrosion allowance
2SE – 0.2P
200 x 18 2 x 20,000 x 0.85 – 0.2 x 200 0.106 + 0.010 = 0.116 in
b) Maximum Pressure: P =
2 x 20,000 x 0.85 x 0.116 18 + 0.2 (0.116)
P =
216 psi
Note: Ellipsoidal heads designed under K > 1.0 and all torispherical heads made of materials having a specified min. tensile strength > 80 000 psi shall be designed using a value of S = 20 000 psi at room temperature and reduced in proportion to the reduction in max. allowable stress values at temperature for the material as shown in the appropriate table (see UG-23). 6.3 - Torispherical Heads
Shallow heads, commonly referred to as flanged and dished heads (F&D heads), can be built according to paragraph UG-32 (e). A spherical radius L of 1.0.D and a knuckle radius r of 0.06.D , as shown in Fig. 4, approximates the most common F&D heads. FIGURE 4: Torispherical Head
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6.3.1 Torispherical Heads & Flanged and Dished Heads a) ASME Flanged & Dished Head (F&D heads)
The dish radius of an ASME Flanged and Dished Head is 100% of the vessel diameter . And, the knuckle is 6% of the vessel diameter . b) Non Standard 80-10 Flanged & Dished Head
On an 80-10 the outside dish radius equals 80% of the head diameter and the outside corner radius equals 10% of the head diameter.
Why Use An 80-10?
As the inside diameter of high alloy pressure vessel heads increases in size, savings in material cost can be achieved by designing 80-10 Torispherical Heads rather than Standard ASME Torispherical or Ellipsoidal shapes. The 80-10 is typically only 66% the thickness of the ASME Torispherical. The required thickness of an ASME F&D head with r/L=0.06 and L = Di
or
E = joint efficiency factor L = inside spherical radius P = pressure on the concave side of the head S = allowable stress t = thickness of the head
Shallow heads with internal pressure are subjected to a stress reversal at the knuckle. This stress reversal could cause buckling of the shallow head as the ratio D/t increases. 21
6.3.2 - Example: Torispherical Heads (imperial unities)
A drum is to operate at 500°F and 350 psig and to hold 5000 gal. Dished torispherical heads inside radius is equals 78 in . The material is SA285 Grade A . Assume “S” = 11200 psi and “E” = 0.85. Solution: ASME Dished heads with L = Di and r/L = 0.06. Use equation 3.10. t=
0.885PL SE - 0.1P
t=
0.885(350)(78) (11,200)(0.85) – (0.1)(1350)
= 2.54
7.0 – Conical or Toriconical Heads 7.1 – Thickness Calculus: The required thickness of the conical portion (knuckle radius > 6% OD) shall be determined by formula using internal diameter of shell, Di and α ≤ 30º. t =
PD 2.(SE – 0.6P).cos
α
L = Di/(2 cos α) L = Outside spherical or crown radius Di = Internal diameter of conical portion = D -2 r (1- cos α) r = Inside knuckle radius 7.2 – Resume of Pressure Vessels Formulae – ASME Section II, Table 1A:
OBS.: D = Shell/Head Inside Diameter, E = Weld Joint Efficiency (0.7 -1.0), L = Crown Radius, P = Internal Pressure, h = Inside Depth of Head, r = Knuckle Radius, R = Shell/Head Inside Radius, S = Allowable Stress, t = Shell/Head Thickness.
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7.3 – Resume of Maximum Allowable Stress (for the most common materials). a) Carbon and Low Alloy Steels:
b) High Alloy Steels:
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