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Revised 03/06 to conform with the 2004 ASM
art A1
CH A PT ER 1
ASME Code Calculations: Cylindrical Components
Here is what what you w i l l be able to do do when you complete complete each each objective: objective:
1. Calculate the required minimum thickness or the maximum allowable wor 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 wor pressure of ferrous piping, drums, and headers.
3. Calculate the required thickness or maximum allowable working pressure pressure seamless, unstayed dished head.
working 4. Calculate the minimum required thickness allowable Signor up maximum to vote on this title pressure of unstayed flat heads, covers, and blind flanges.
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5. Calculate the acceptability of openings in a cylindrical shell, header, or he
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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 changes to major 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 Sign up to vote on this title same factor used in Sections VIII-1 and VIII-2.
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Pressures calculated or given in this module refer to gauge pressure unless otherwise indicated.
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Revised Second Class Course • Section A1 • SI Units
until the final answer is derived. This convention has be throughout this chapter.
N ote: It should be noted that many US customary customary unit values values p in the ASME codes do not convert directly into metric v the current ASME edition or the 2004 ASME Academic E i.e. 5 in. converts to 127mm, ASME shows 5 in. (125 mm cover co verts ts to 6.35 mm, ASME ASME shows sho ws ¼ in. in . (6 mm)). You ar to use the ASME values as presented and not to con customary numbers to metric.
ASME SECTION I - POWER BOILERS
Paragraphs PG-1, PG-2: This Code covers rules for construction o boilers, electric boilers, miniature boilers, and high temperature water The scope of jurisdiction of Section I applies to the boiler proper and th external piping. Superheaters, economizers, and other pressure parts co directly to the boiler, without intervening valves, are considered to be 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 su pressure, whether or not exposed to the fire or products of combustion, in accordance with specifications listed in paragraph PG-6.1. Paragra states that pipes, tubes, and pressure containing parts used in boil conform to one of the specifications listed in paragraph PG-9.1. Sign up to vote on this title
Design
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Paragraph PG-16.3 states that the minimum thickness of any boiler pla
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Chapter 1 • ASME Code Calculations: Cylindrical Components
5
Paragraph PG-21 states that the term maximum allowable working pressure (MAWP) refers to gauge pressure, except when noted otherwise in the calculation formula of PG-27.2. Paragraph PG-27 Cylindrical Components Under Internal Pressure
The formulae in this section 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 be transposed to determine the maximum allowable working pressure if 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 = D = E =
e P
= =
R S
= =
t y
= =
minimum allowance for threading and structural stability (mm) (see PG-27.4, note 3) outside diameter of cylinder (mm) efficiency of longitudinal welded joints or of ligaments between openings, whichever is lower You're (the values allowed for E are listed in Reading a Preview PG-27.4, note 1) access with(see a free trial. thickness factor for expandedUnlock tube full ends (mm) PG-27.4, note 4) maximum allowable working pressure (MPa). (see PG-21, refers to gauge pressure) Download With Free Trial inside radius of cylinder (mm) maximum allowable stress value at the operating temperature of the metal (Section II, Part D, Table 1A. See PG-27.4, note 2) minimum required thickness (mm) (see PG-27.4, note 7) temperature coefficient (see PG-27.4, note 6) Sign up to vote on this title
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ASME SECTION VIII, DIVISION 1 - PRESSURE VESSELS
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Revised Second Class Course • Section A1 • SI Units
Materials
Paragraph UG-4 states that materials subject to stress due to pressur conform to the specifications given in Section II, except as otherwise p in paragraphs UG-9, UG-10, UG-11, UG-15 and the Mandatory App Paragraph UG-23 (a) lists the tables in Section II, D for various materials
Design
ASME Boiler Code Section I, as well as Section VIII, Division 2 requires all major longitudinal and circumferential butt joints to be exam full radiograph. Section VIII-1 lists various levels of examination for the joints. A fully radiographed major longitudinal butt-welded joint in a cy shell would have a joint efficiency factor ( E ) of 1.0. This factor corres a safety factor (or material quality factor) of 3.5 in the parent met radiographed longitudinal butt-welded joints have a joint efficiency facto 0.7, which corresponds to a safety factor of 0.5 in plates. This resu increase of 43% in the thickness of the plates required.
Paragraph UG-20: Design temperature You're Reading a Preview With pressure vessels, the maximum temperature used in the design is im Unlock full access with a free trial. as is the minimum temperature.
With Free The minimumDownload temperature used inTrial design shall be the lowest tempera the vessel will experience from any factor, including normal operatio condition, or environmental conditions. Paragraph UG-27: Thickness of shells under internal pressure
to vote on this title The formulae in this section are Sign usedupto determine the minimum useful pressure is Useful Notworking thickness of shells when the maximum allowable These formulae can be transposed to determine the maximum allowable pressure if the minimum required thickness is given.
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Chapter 1 • ASME Code Calculations: Cylindrical Components
7
OBJECTIVE O B BJ J ECTIVE 3 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.
SECTION I
The following formulae are found in ASME Section I, paragraph PG-27.2.1. Formula for minimum required thickness t
PD
=
2S + P
Formula for MAWP
P = S
+
1.1
0.005 D + e
You're Reading a Preview
⎡ ⎢ ⎢⎣
Unlock full ⎤ access with a free trial. 2t - 0.01D - 2e D-
⎥
( t - 0.005 D - e) ⎥⎦ Download With Free Trial
1.2
Example 1: boiler tube Calculate the minimum required wall thickness of a watertube boiler tube 70 mm O.D. that is strength welded 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. The maximum allowable working pressure is 4000 kPa gauge. TheSign tube up material to vote onis this title carbon steel SA-192. Useful Not useful
Note :
Check PG-6 for plate materials and PG-9 for boiler tube materials before starting calculations; the information will direct you to the
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Asme Section Viii Div-1,2,3
Revised Second Class Course •
Pressure Vessel Design - Guides
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Section A1
•
SI Units
Where P D e S
= = = =
4000 kPa = 4.0 MPa 70 mm 0 (see PG-27.4, note 4, strength welded) 87.8 MPa (see Section II, Part D, Table 1A, SA-192 at 350°C) t
= =
4 × 70 2(87.8) + 4
+ 0.005(70) + 0
280
+ 0.35 179.6 = 1.56 + 0.35 = 1.9 mm (Ans.)
Note :
This value is exclusive of the manufacturer’s tolerance allow PG-16.5). The manufacturing process does not produce a uniform wall thickness; add an allowance of approximately 1 the minimum thickness calculated.
You're Reading a Preview The formula for minimum thickness may be transposed to solve maximum allowable working pressure if the tube size and thickness are k Unlock full access with a free trial.
Example 2: superheater tubeFree Trial Download With Calculate the maximum allowable working pressure, in kPa, for a 75 m and 4.75 mm minimum thickness superheater tube connected to a h strength welding. The average tube temperature is 400°C. The tube m SA-213-T11. Note :
Solution
Check PG-9 for boiler tube startingcalculat Signmaterials up to votebefore on this title information will direct you toUseful stress table in ASME useful the correct Not II, Part D. SA-213-T11 is alloy steel.
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Chapter 1 • ASME Code Calculations: Cylindrical Components
P
⎡ ( 2 × 4.75) - ( 0.01 × 75) - ( 2 × 0 ) ⎢⎣ 75 - ( 4.75 - ( 0.005 × 75) - 0)
= 102 × ⎢
⎡
9
⎤ ⎥ ⎥⎦
⎤ ⎥ ⎢⎣ 75 - ( 4.75 - 0.375) ⎦⎥
= 102 × ⎢ = 102 ×
9.5 - 0.75
8.75
70.625 = 12.64 MPa = 12 640 kPa (Ans.)
The tubes were strength welded in Example 1 and Example 2. For calculations involving tubes expanded into place, the appropriate value of e is found in paragraph PG-27.4, note 4.
SECTION VIII The following formulae (found in ASME Section VIII-1, paragraph UG-27(c)) You're Reading a Preview are used for calculating wall thickness and design pressure. Paragraph UG-31(a) with a freeinternal trial. states that these calculations are used for Unlock tubes full andaccess pipes under pressure.
Thin Cylindrical Shells
Download With Free Trial
(1) Circumferential stress (longitudinal joints) t
=
PR
(SE - 0.6P )
Or P
=
SEt
( R + 0.6t )
1.3 Sign up to vote on this title
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Revised Second Class Course • Section A1 • SI Units
Thick Cylindrical Shells
As internal pressures increase higher than 20.6 MPa, special considerati be given to the construction of the vessel as specified in paragraph U-1 the ratio of t/ R increases beyond 0.5, a more accurate equation is req determine the thickness. The formulae for thick walled vessels are Appendix 1, Supplementary Design Formulas 1.1 to 1.3.
SE =
( ( R
P R0 + R 2
2 0
2
)
2 - R )
Where R0 and R are outside and inside radii, respectively. By substituting R0 = R + t
⎛ 12 ⎞ t = R ⎜ Z - 1⎟ ⎝ ⎠
Where
Z =
( SE + P ) ( SE - P )
Where t > 0.5 R or P > 0.385SE And
You're Reading a Preview Unlock full access with a free trial.
⎡ ( -Z1) ⎤ P = SE ⎢ ⎥ Where Download With 1) ⎥⎦ Trial + Free ⎢⎣ ( Z
Z
⎡ ( = ⎢ ⎣
For longitudinal stress with t > 0.5R or P > 1.25SE
⎛ 12 ⎞ ⎛ P ⎞ t = R ⎜ Z - 1⎟ Where Z = ⎜ ⎟ +1 Sign up to vote on this ⎝ SE ⎠ title ⎝ ⎠ Useful Not useful And
+R R
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Chapter 1 • ASME Code Calculations: Cylindrical Components
11
Solution The quantity 0.385SE = 45.16 MPa; since this is greater than the design pressure P = 690 kPa, use equation 1.3. (See Section VIII-1, UG-27.) Note R must be in the fully corroded state to determine the minimum thickness. t =
= =
PR
( SE
- 0.6P )
0.69 × (1220 + 3)
(138 ×
0.85) -
843.87
+ 3
116.886 = 7.22 + 3
=
+ corrosion allowance
10.22 mm
( 0.6 ×
0.69)
+ 3
(Ans.)
The calculated thickness is less than 0.5 R ; therefore, equation 1.3 is acceptable. Example 4: thick shell thickness Calculate the required shell thickness of an accumulator with P = 69 MPa, You're Reading a Preview R = 45.7 cm, S = 138 MPa, and E = 1.0. Assume a corrosion allowance of 6 mm. Unlock full access with a free trial. Solution Download With Free Trial The quantity 0.385SE = 53.13 MPa; since this is less than the design pressure P = 69 MPa, use equation 1.7.
⎛ 12 ⎞ t = R⎜Z - 1⎟ Where ⎝ ⎠ (138 × 1) + 69 Z = (138 × 1) − 69 207
Z
=
SE + P SE - P
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Revised Second Class Course • Section A1 • SI Units
Example 5 Calculate the required shell thickness of an accumulator with P = 52 R = 45.7 cm, S = 138 MPa, and E = 1.0. Assume corrosion allowance =
Solution The quantity 0.385SE = 53.13 MPa; since this is greater than the design P = 52.75 MPa, use equation 1.3. PR
t =
+ corrosion allowance SE - 0.6P 52.75 × 457 +0 138 1 0.6 52.75 × ( ) ( )
=
24106.75
=
106.35 226.67 mm
=
(Ans.)
This example used equation 1.3; compare the answer using equation 1.7
⎛
1
⎞
You're t =Reading R ⎜ Z 2 a Preview - 1⎟ Where Z
⎝
⎠ (138 × 1) + 52.75 Z = - 52.75 × 1) Trial (138 Free Download With
=
SE + P SE - P
Unlock full access with a free trial.
= = t =
= =
190.75 85.25 2.2375 1 ⎛ ⎞ 2 457 ⎜ 2.2375Sign -up1to⎟ vote on this title ⎝ ⎠ Useful Not useful 457 × 0.4958
226.59 mm
(Ans.)
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Chapter 1 • ASME Code Calculations: Cylindrical Components
13
OBJECTIVE O B BJ ECTIVE 3 2 Calculate the required minimum thickness or the maximum allowable working pressure of ferrous piping, drums, and headers.
In cylindrical vessels, the stress set up by the pressure on the longitudinal joints is equal to twice the stress on the circumferential joints.
SECTION I 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. You're Reading a Preview Using the outside diameter t =
P =
Unlock full access with a free trial.
PD
+ C With Free Trial Download 2SE + 2 yP 2SE ( t - C ) D-
( 2 y)( t -
2.1
2.2
C)
Sign up to vote on this title
Using the inside radius t =
PR SE - (1 - y ) P
+ C
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Revised Second Class Course • Section A1 • SI Units
Note: Plain-end pipe does not have its wall thickness reduced when jo another pipe. For example, lengths of pipe welded together, ra being joined by threading, are classed as plain-end pipes. Solution Use equation 2.1 (See PG -27.2.2.) t =
PD
2SE + 2 yP
+ C
Where P D C S
= = = =
E = y =
6200 kPa = 6.2 MPa 273.1 mm 0 (see PG-27.4, note 3, 4-inch nominal and larger 104 MPa (see Section II, Part D, Table 1A, SA-335-P11 at 375°C) 1.0 (see PG-27.4, note 1, seamless pipe as per PG 0.4 (see PG-27.4, note 6, ferritic steel less than 47
You're Reading a Preview 6.2 × 273.1 t = + 0 Unlock full access 2 with a free trial. 104 1 2 0.4 6.2 × + × ( ) ( ) 1693.22 Download With = Free Trial 208 + 4.96 1693.22
=
212.96 = 7.95 mm
Sign up totolerance vote on this title This value does not include a manufacturer's allowance of 12.5
Therefore
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Chapter 1 • ASME Code Calculations: Cylindrical Components
15
Solution Use equation 2.2. (See PG-27.2.) P =
2SE ( t - C ) D-
( 2 y)( t
- C)
Where D = t = C = S
=
E = y =
323.9 mm (see 2005 Academic Supplement, Formulae and Physical Constants, "Table of Actual Pipe Dimensions.") 11.85 mm 0 (see PG-27.4, note 3, 4-inch (100 mm) nominal and larger) 101 MPa (Section II, Part D, Table 1A, SA-209-T1 at 450°C) 1.0 (see PG-27.4, note 1, seamless pipe as per PG-9.1) 0.4 (see PG-27.4, note 6, austenitic steel at 450°C) P =
=
2 (101 × 1) × (11.85 - 0 ) 323.9 - ( 2 × 0.4 ) × (11.85 - 0 ) You're Reading a Preview 202 × 11.85
Unlock full access with a free trial. 323.9 - 9.48 2393.7 = Download With Free Trial 314.42 = 7.613 MPa
=
7613 kPa
(Ans.)
Example 8: drum using inside radius Sign uptotoan vote on this title A welded watertube boiler drum of SA-515-60 material is fabricated inside UsefulTheplate Not useful radius of 475 mm on the tubesheet and 500 mm on the drum. thickness of the tubesheet and drum are 59.5 mm and 38 mm respectively. The longitudinal joint efficiency is 100%, and the ligament efficiencies are 56%
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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 f plates of different thickness. Greater material thickness is require the boiler tubes enter the drum than is required for a plain dr economy, the drum is designed to meet the pressure requirem each situation. You're Reading a Preview Note: Check PG-6 and PG-9 for materials before starting calculat Unlock full access withyou a free information will direct totrial. the correct stress table in ASME S Part D. The material SA-515-60 is carbon steel plate. Download With Free Trial Solution This example has two parts: a) The drum - consider the drum to be plain with no penetrations. b) The tubesheet - consider the drum to have penetrations for boiler tu
(a)
Sign up(See to vote on this title Use equation 2.4 (inside radius R). PG-27.2.2.)
Drum P =
Useful
Not useful
SE (t - C ) R + (1 y)( t - C)
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Chapter 1 • ASME Code Calculations: Cylindrical Components
Drum P = =
17
(115 × 1) (38 - 0) 500 + (1 - 0.4)(38 - 0) 4370
500 + 22.8 = 8.36 MPa (Ans.) 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) Use equation 2.4 (inside radius R). (See PG-27.2.2.) Tubesheet P =
SE (t - C ) R + (1 - y)( t - C)
Where S
=
E = T = C = R y
= =
115 MPa (see Section II, Part D, Table A1, SA-515-60 at 300°C) 0.56 (circumferential stress = 30% and longitudinal stress = 56%; therefore,You're 0.56
(115 ×
0.56 ) ( 59.5 - 0 )
475 + (1 - 0.4)(59.5 - 0) 3831.8
475 + 35.7 = 7.5 MPa (Ans.)
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Revised Second Class Course • Section A1 • SI Units
OBJECTIVE 3
Calculate the required thickness or maximum allowable working pressure of a seamless, unstayed dished head.
Section I: DISHED HEAD CALCULATIONS
The paragraphs from PG-29 must be considered when performing cal on dished heads.
Paragraph PG-29.1 states that the thickness of a blank, unstayed dish with the pressure on the concave side, when it is a segment of a sphere, calculated by the following formula: 5PL t = Reading a Preview You're 4.8S Unlock full access with a free trial.
Where:
t = minimum thickness Download With Free Trial of head (mm). P = maximum allowable working pressure (MPa). L = radius (mm) to which the head is dished, measure concave side S = maximum allowable working stress (MPa) (see Section II, Part D, Table 1A). Sign up to vote on this title
Paragraph PG-29.2 states: "The radiusUseful Nothead usefulis dished sha to which the greater than the outside diameter of the flanged portion of the head. W radii are used, the longer shall be taken as the value of L in the formula.”
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Chapter 1 • ASME Code Calculations: Cylindrical Components
19
Solution Use equation 3.1. (See paragraph PG-29.1 for segment of a spherical dished head.) t =
5PL 4.8S
Where P L S
= = =
2.5 MPa 918 mm 88.9 MPa (see ASME Section II, Part D, Table 1A, SA-285 A at 250°C) t =
5 ( 2.5 × 918 )
4.8 × 88.9 = 26.89 mm (Ans.)
Note: PG-29.6 states “No head, except a full-hemispherical head, shall be of a lesser thickness than that required for a seamless shell of the same diameter." You're Reading a Preview Therefore, to determine if this head thickness meets Code, the thickness of the Unlock full access with a free trial. shell must be calculated.
Use equation 2.1 (See paragraph PG-27.2.2.) Download With Free Trial t =
Where D = y = E =
PD
2SE + 2 yP
+ C
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1085 mm Useful than Not useful 0.4 (see PG-27.4, note 6, ferritic steel less 480°C) 1 (welded)
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Revised Second Class Course • Section A1 • SI Units
Paragraph PG-29.3 states When a head, dished to a segment of a sphere, has a flanged-in manhole or access opening that exceeds 150 mm in any dimension, the thickness shall be increased by 15% of the required thickness for a blank head computed by the above formula, but in no case less than 3.0 mm additional thickness over a blank head. Where such a dished head has a flanged opening supported by an attached flue, an increase in thickness over that for a blank head is not required. If more than one manhole is inserted in a head, the thickness of which is calculated by this rule, the minimum distance between the openings shall be not less than one-fourth of the outside diameter of the head. Note: This applies to the manhole found on the end of a boiler drum.
Example 10: the segment of a spherical dished head with a fla manhole You're Reading a Preview Calculate the thickness of a seamless, unstayed dished head with pressu concave side, having a flanged-in manhole 154 mm by 406 mm. The he Unlock full access with a free trial. diameter of 1206.5 mm and is a segment of a sphere with a dish radius mm. The maximum allowable working pressure is 1550 kPa, the materi With Freedoes Trialnot exceed 220 oC. 285-C, and theDownload metal temperature Note: Check paragraph PG-44, "Inspection Openings" to see if this size is acceptable.
Solution to vote on this titleof the diamet First thing to check: is the radius ofSign the up dish at least 80% shell? (per paragraph PG-29.5) Useful Not useful
dish radius
1143
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Chapter 1 • ASME Code Calculations: Cylindrical Components
21
Where P L S
= = =
1.55 MPa 1143 mm 108 MPa (see ASME Section II, Part D, Table 1A: use 250°C since 220°C is not listed; therefore, use the next higher temperature) t =
=
5 (1.55 × 1143) 4.8 (108 ) 17.088 mm
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. Therefore 17.088 × 0.15 = 2.56 mm This is less than 3.0 mm, so the thickness must be increased by 3.0 mm
You're Reading a Preview
Therefore
Unlock full access with a free trial.
Required head thickness = 17.088 + 3.0 mm (Ans.) = 20.088 With Download Free Trial
Semi-ellipsoidal head Paragraph PG-29.7 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 Signthickness up to vote of on athis title diameter of the head shall be made at least as thick as the required Useful If Not useful seamless shell of the same diameter as provided in PG-27.2.2. a flanged-in manhole that meets the Code requirements is placed in an ellipsoidal head, the thickness of the head shall be the same as for a head dished to a segment of a
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Revised Second Class Course • Section A1 • SI Units
FIGURE 2
1/2 r = 1/4 D
h = 1/4 D
Semi-ellipsoidal Head
r
h
D
L
Full-hemispherical head The following rule applies to drums or headers with a full-hemispherical
Paragraph PG-29.11: The thickness of a blank, unstayed, full-hemispher with the pressure on the concave side shall be calculated by the f You're Reading a Preview formula: Unlock full access with a free trial.
t =
PL
2S - 0.2 P Free Trial Download With
Where t P L
= = =
S
=
minimum thickness of head (mm). maximum allowable working pressure (MPa). radius to which the head was formed (mm) (measured onSign theup concave side oftitle the head). to vote on this maximum allowable useful(MPa) Usefulworking Notstress (Table A1, Section II, Part D).
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Chapter 1 • ASME Code Calculations: Cylindrical Components
23
Solution Use equation 3.2. (See PG-29.11.) t =
PL
2S - 0.2 P
Where P L S
= = =
t =
= = =
6.205 MPa 190.5 mm 107 MPa (see ASME Section II, Part D , Table 1A, SA-285-C at 300oC) 6.205 × 190.5 2 (107 ) - 0.2 ( 6.205 ) 1182.05 214 - 1.241 1182.05 212.759 5.56 mm
(Ans.)
You're Reading a Preview Check if this thickness exceeds 35.6% of the inside radius:
Unlock full access with a free trial.
190.5 × 0.356 = 67.8 mm Download With Free Trial It does not exceed 35.6%, therefore The thickness of the head meets Code requirements. Paragraph PG-29.12: If a flanged-in manhole that meets the Code requirements up to voteshall on this title (see PG-44) is placed in a full-hemispherical head, the thicknessSign of the head Not useful and be the same as for a head dished to a segment of a sphere (seeUseful PG-29.1 PG-29.5), 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-29.3.
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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 she internal pressure. PR
t =
2SE - 0.2 P
or P
2SEt
=
R + 0.2t
Where t < 0.356 R or P < 0.665SE
For thick shells, where t > 0.356R or P > 0.665SE, use Mandatory Appen sections 1-3. As the ratio t/You're R increases beyond 0.356, use the following equations Reading a Preview Unlock full access ⎛ 1 with ⎞ a free trial.
t = R ⎜ Y 3 -1 ⎟
⎝
⎠
where Y =
Download With Free Trial
2 ( SE + P ) 2SE - P
or
⎛ Y -1 ⎞ ⎛ R + t ⎞ P = 2 SE ⎜ where Y = ⎜ ⎟ ⎟ ⎝Y + 2⎠ ⎝ R ⎠
3
Sign up to vote on this title
Useful Where t > 0.356R or P > 0.665SE Example 12: hemispherical head
Not useful
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Chapter 1 • ASME Code Calculations: Cylindrical Components
25
Solution The quantity 0.665SE = 78 MPa; since this is greater than the design pressure of 690 kPa, use equation 3.4. (See paragraph UG-32 (f).) The inside radius in a corroded condition is equal to R = 1220 + 3 (corrosion allowance)
= 1223 mm PR
t =
+ corrosion allowance 2 SE - 0.2 P 0.69 × 1223 +3 = 2 (138 × 0.85 ) - 0.2 ( 0.69 )
=
843.87
234.46 = 3.6 + 3
=
+3
6.6 mm
(Ans.)
The calculated thickness is less than 0.356R; therefore, equation 3.3 is acceptable. Example 13: spherical head You're Reading a Preview A spherical pressure vessel with an internal diameter of 3048 mm has a head Unlock pressure full access with a free trial. thickness of 25.4 mm. Determine the design if the allowable stress is 113 MPa. Assume joint efficiency E = 0.85.
Download With Free Trial Solution As no corrosion allowance is stated the design pressure will act on the given internal diameter. Use equation 3.5 since t is less than 0.356R. P =
=
Sign up to vote on this title
2SEt
R + 0.2t
2 (113 × 0.85 × 25.4 )
(
)
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Revised Second Class Course • Section A1 • SI Units
Example 14: thick hemispherical head Calculate the required hemispherical head thickness of an accumulator w P = 69 MPa, R = 460 mm, S = 103 MPa, and E = 1.0. Assume a corrosion allowance of 6 mm.
Solution The quantity 0.665SE = 68.495 MPa; since this is less than the design pr 69 MPa, use equation 3.6.
⎛ 13 ⎞ 2 ( SE + P ) t = R ⎜ Y - 1⎟ where Y = 2SE - P ⎝ ⎠ Y =
=
2 (103 × 1 + 69) 2 (103 × 1) - 69 344
137 = 2.51
⎛ 3 ⎞ You're Reading a Preview t = R Y -1 1
⎜ ⎟ ⎝ ⎠ Unlock full access with a free trial. 1 ⎛ ⎞ = 460 + 6 ⎜ 2.513 - 1⎟ Download With Free Trial ⎝ ⎠ = 466 ( 0.359 )
= 167.3 mm This is the minimum thickness i.e. fully corroded state. Total head thickness is 167.3 + 6 mm (corrosion allowance) = 173.3 mm Sign up to vote on this title
Connecting this head to the accumulator shell would require special tr Not useful Useful which is outside of the scope of this module.
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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)
You're Reading a Preview Unlock full access with a free trial.
Download With Free Trial
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(b)
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Revised Second Class Course • Section A1 • SI Units
FIGURE 4 Torispherical Head
The required thickness of an F&D head is t =
or
0.885PL SE - 0.1P
You're Reading a Preview SEt P = Unlock full access with L a+free t 0.885 0.1trial.
WhereDownload With Free Trial 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 Sign up to vote on this title
Shallow heads with internal pressure are subjected to a stress revers Useful Not useful knuckle. This stress reversal could cause buckling of the shallow head as D/ t increases.
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Chapter 1 • ASME Code Calculations: Cylindrical Components
29
OBJECTIVE 4 Calculate the minimum required thickness or maximum allowable working pressure of unstayed flat heads, covers, and blind flanges.
UNSTAYED FLAT HEADS, COVERS, AND BLIND FLANGES Flat plates, covers, and flanges are used extensively in boilers and pressure vessels. When a flat plate or cover is used as an end closure or head of a pressure vessel, it may be designed as an integral part of the vessel (having been formed with the cylindrical shell) or welded to it. Alternately, 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. Bolted flanges are not covered in the scope of this module. You're Reading a Preview The concepts of unstayed flat heads, covers, and especially blind flanges are Unlock full access with a free trial. often misunderstood and can be challenging to anyone learning and working on this type of equipment. It is very important for power engineers to have good Download Freethem Trialto work safely working knowledge of thickness requirements as thisWith allows and provide sound and safe advice.
SECTION 1 Sign up to vote on this title
Useful
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Paragraph PG-31.1 states that the minimum thickness of unstayed flat heads, cover plates, and blind flanges shall conform to the requirements. Paragraph PG-
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Revised Second Class Course • Section A1 • SI Units
t
=
d
CP S
+
1.9Whg 3
Sd
Note: W = the total bolt loading and hg = the gasket moment arm. T moment arm is the radial distance from the centre line of the bo line of the gasket reaction force (Fig. PG-31 (j), (k)).
When using equation 4.2, the thickness t shall be calculated for bot conditions (flange sketches j and k) and the greater value used.
Note: The formulae used to determine thickness may be transposed to P and find the maximum allowable working pressure for a flat cover of known thickness.
Paragraph PG-31.3.3 states two formulae for the required thickness unstayed heads, covers, or blind flanges that are square, rectangular, obround, or segmental in design and attached by welding. t
ZCP
d = You're Reading a Preview S
Unlock full access with a free trial.
Where Z is a factor from the ratio of the short and long spans 2.4d Download With Z = 3.4 - Free Trial to a maximum of 2.5 D
When the non-circular head, cover, or blind flange is attached by bolts (F 31. (j), (k))
t
= d
ZCP S
+
6Whg Sign up to vote on this title SLd 2 Useful
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Chapter 1 • ASME Code Calculations: Cylindrical Components
31
Example 15: circular flat head welded to a shell (Illustrated by Fig. PG-31 (e) and Fig. 5.) FIGURE 5
Circular Flat H
Calculate the minimum thickness for the circular head and the depth of the fillet welds required. The material for head and shell is SA-285-A. The shell is seamless. The thickness t is 19 mm. Maximum allowable working pressure is 2500 kPa. Shell’s inside diameter d is 762 mm. Head joint welding meets Code requirements.
You're Reading a Preview
Solution Use equation 4.1
t
Unlock full access with a free trial.
= d
CP
Download With Free Trial
S
Where P = d = S =
2.5 MPa 762 mm Sign1A) up to vote on this title 88.9 MPa (ASME Section II, Part D, Table As no temperature is given, the saturation of useful Useful Not temperature steam (224°C at 2500 kPa) may be used; therefore, use the value for 250°C.
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Revised Second Class Course • Section A1 • SI Units
Where R E y C t r
= = = =
d/ 2 = 381 mm 1 (see PG-27.4, note 1) 0.4 (see PG-27.4, note 6) 0 (see PG-27.4, note 3)
= = =
2.5 × 381
(88.9 × 1)
-
(1
- 0.4 ) × 2.5
+ 0
952.5 87.4 10.898 mm
Therefore t r
m =
t s
10.898
=
19 0.574
=
You're Reading a Preview C = 0.33 m (from PG - 31.4) Unlock full access with a free trial. = 0.33 × 0.574 = 0.19
Download With Free Trial
As this value is less than 0.2, use 0.2 in the formula from PG-31.4 or in equation 4.1. t
= d =
CP
762
S
Sign up to vote on this title Useful 0.20 ×2.5
88.9
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Chapter 1 • ASME Code Calculations: Cylindrical Components
33
It is interesting to note that the required minimum shell thickness is 10.898 mm, yet the required minimum thickness of the blank head is approximately 5.2 times thicker at 57.15 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. Head design to Fig. PG-31, sketch (d). Shell and head thickness of 30.5 mm. Material is SA-285-B. Head joint weld meets Code requirements. Shell diameter is 610 mm. Operating temperature not to exceed 300°C. Solution t S d C
= = = =
30.5 mm 96.6 MPa (see ASME Section II, Part D, Table 1A) 610 mm 0.13 (see Fig. PG-31 (d))
Use equation 4.6. (See PG -32.3.2.) 2
P =
P
=
t S d 2C
30.52 × 96.6
You're Reading a Preview
4.6
Unlock full access with a free trial.
6102 × 0.13 Download With Free Trial = 1.858 MPa (Ans.)
The maximum allowable working pressure for this flat, unstayed head is 1858 kPa. Sign up to vote on this title
SECTION VIII-1
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Revised Second Class Course • Section A1 • SI Units
E = t
=
butt-welded joint efficiency of the joint within th plate minimum required thickness of the flat plate
The value of E depends on the degree of non-destructive exa performed. E is not a weld efficiency value of the head to shell corner jo
Example 17: integral flat plate Using the rules of paragraph UG-34, determine the minimum required of an integral flat plate with an internal pressure P = 17 MPa, an allowa S = 120 MPa, and a plate diameter d = 610 mm. There are no butt we within the head. There is a corrosion allowance of 4 mm. The corn conforms to Fig. UG-34 (b-2) (assume that m = 1). Solution Use equation 4.7. (See Fig UG-34 (b-2))
Where C = 0.33(m) = 0.33(1) = 0.33 d = 610 + 4 = 614 mm (fully corroded state) You're Reading a Preview CP Unlock a free trial. allowance t =fulld access with + corrosion SE
Download With Free 0.33× 17 Trial = 614× +4 120× 1 = 136.76 mm (Ans.)
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Chapter 1 • ASME Code Calculations: Cylindrical Components
35
OBJECTIVE 5 Calculate the acceptability of openings in a cylindrical shell, header, or head.
Openings through the pressure boundary of a vessel require extra care to keep loading and stresses at acceptable levels. An examination of the pressure boundary may indicate that extra material is needed near the opening to keep stresses at acceptable levels. 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. The stress analysis basis used in the ASME Codes to analyze nozzle reinforcement is called Beams on E lastic Foundation (Hetenyi, 1946). Although the methods used are a simplified application of the elastic foundation theory, experience has shown that they are acceptable. You're Reading a Preview ASME Codes Section I and Section VIII give two methods for examining the Unlock full access with a free trial. acceptability of openings in the pressure boundary for pressure loads only. The first method, called the reinforced opening or area replacement method is With Free used when nearby substitute areas replaceDownload the area removed byTrial the opening. The second method is the ligament efficiency method. 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. Curves have been developed to simplify this examination. For single openings, only the area replacement method is used. For multiple openings, either method Sign up to vote on this title may be used.
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Since stress is related to load and cross-sectional area, areas are substituted when making calculations. Placement and location of the replacement area are very
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Revised Second Class Course • Section A1 • SI Units
Figure 6 Reinforcement Limits
tn
Rn t rn
A smaller of 2.5t or 2.5tn + t e
WL1 tr
t smaller of 2.5t or 2.5tn
C
d
ABCD = Limits of reinforcem d or Rn + t n + t Use larger value
d or Rn + t n + t Use larger value
You're Reading a Preview
Unlock full access with a free trial. When an opening is cut into a vessel wall for the attachment of a noz diameter d (as in Fig. 6), the vessel wall thickness t is usually thicker Download Witht r Free minimum thickness required . TheTrial area (t r x d ) is the cross-sectional ar removed and has to be compensated for. ASME Section I, paragrap (ASME Section VIII, paragraph UG-40) gives the rules for the “Limits Available for Compensation." The limit is shown by box ABCD in Fig. 6
If greater than the cross-sectional area removed, the additional materi Sign up to vote on this title shell wall and the additional material in the nozzle wall (the hatche Useful useful sectional area shown in Fig. 6 within limit of Not compensation bound the provide adequate compensation.
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Chapter 1 • ASME Code Calculations: Cylindrical Components
37
To determine if reinforcement is required, the value K is calculated from the formula K
PD
=
5.1
1.82St
Using the chart in Fig. PG-32, the value for the x-axis is calculated from the shell diameter times the shell thickness. 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, allowed without reinforcement. Example 18: reinforcement of nozzle abutting vessel Determine if reinforcement is required for a 100 mm I.D. nozzle located in a cylindrical boiler shell. The nozzle abuts the vessel wall and is attached by a fullpenetration weld. The O.D. of the shell is 1000 mm. The thickness of the shell wall is 25.4 mm. The thickness of the nozzle wall is 10 mm. The shell material is SA-515-60 and the nozzle material is SA-192. The maximum allowable working pressure is 4500 kPa, and the design temperature is not to exceed 200°C. All joint efficiencies E = 1.0.
You're Reading a Preview Solution As this is a boiler shell, ASME Section Unlock I rulesfullapply. access (See with a PG-32.1.2.) free trial. Use equation 5.1 to calculate the K value. K =
PD
Download With Free Trial
1.82St
Where P D S t
= = = =
4.5 MPa 1000 mm 118 MPa 25.4 mm
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Revised Second Class Course • Section A1 • SI Units
The intersection of the x-axis value (2540) and the K value curve (0.825) axis value of approximately 134 mm.
Therefore, no additional reinforcement is required (Ans.) for an open 100 mm diameter.
SECTION VIII-1
Section VIII-1 requires all openings in pressure vessels, not subjected fluctuations, to use reinforcement calculations in paragraph UG-37 certain dimensional requirements are met as listed in paragraph UG-36(c
Example 19: reinforcement of nozzle abutting vessel Determine the reinforcement requirements for a 60 mm I.D. nozzle loc cylindrical shell. The nozzle abuts the vessel wall and is attached b penetration weld. The O.D. of the shell is 1000 mm. The thickness of wall is 25.4 mm, and the thickness of the nozzle wall is 10 mm. The shell is SA-516-60You're and the nozzlea Preview is SA-192. The maximum allowable Reading pressure is 4500 kPa, and the design temperature is not to exceed 20 Unlock full1.0 access with a free trial. joint efficiencies E = Solution Download With Free Trial As this is a not a boiler shell, ASME Section VIII-1 rules apply. (See UG-36(c)(3).)
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 less; or (b) the opening is not larger than 60to mm diameter and the shell th Sign up vote on this title is greater than 10 mm. Useful Not useful In this example, the nozzle diameter is 60 mm
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Asme Section Viii Div-1,2,3
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Chapter 1 • ASME Code Calculations: Cylindrical Components
39
OBJECTIVE 6 Calculate the compensation required to reinforce an opening in a cylindrical shell, header, or head.
SECTION I ASME Section I, paragraph PG-33, "Compensation required for openings in shells and formed heads", states the rules for compensation. Paragraph PG-33.2 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-33.1, shown in Fig. 7. For an opening in a shell with a nozzle abutting the shell wall (such as an opening for a safety valve), the requirements are illustrated in Fig. 7. You're Reading a Preview
Nozzle Wall Vessel Wall
Unlock full access with a free trial.
tn
Dp Download With Free Trial Rn t rrnn
B
A WL1 smaller of WL2 2.5t or 2.5tn + t e t
FIGURE 7
te tr
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Revised Second Class Course • Section A1 • SI Units
Where (a)
The area to be replaced A (shown as the cross-hatched area) = dt rF where F is taken from the chart Fig. PG-33
(b)
The area in the shell wall thickness available to be used as compensation A 1 (shown as the forward sloped hatched areas on either side of the = the larger of d (t – Ft r) or 2(t + t n)(t – Ft r)
(c)
The area in the nozzle wall thickness available to be used as compensation A 2 (shown as the backward sloped hatched area on either side of the = the smaller of 2(t n – t rn)(2.5tf r1) or 2(t n – t rn)(2.5t n where f r1 is the ratio of S nozzle / S shell
(d)
The area available from the nozzle to the reinforcement plate we = (W L 1)2 × f r1 where f r1 is the ratio of the lesser of S nozzle or You're Reading a Preview The area available from the reinforcement plate to shell weld A Unlock full access with = (W L 2)2 f r3 a free trial.
(e)
(f)
Download Free Trial The area availableWith in the reinforcement plate (shown as herring-b brick hatch) A 5 = ( D p – d – 2t n)t / e f r3 `Where f r3 is S plate / S shell
If A 1 + A 2 + A 41 > A The opening is adequately reinforced.Sign up to vote on this title
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If A 1 + A 2 + A 41 < A The opening is not adequately reinforced, and reinforcing elements
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Chapter 1 • ASME Code Calculations: Cylindrical Components
41
temperature is not to exceed 200°C. All joint efficiencies E = 1.0. The reinforcement plate (if required) shall be of SA-192 material and 18 mm thick. Solution As this is a boiler shell, ASME Section I rules apply. Use equation 5.1. (See PG-32.1.2.) K
=
PD
1.82 St
Where P D S t
= = = =
4.9 MPa 1000 mm 118 MPa 22.5 mm K =
4.9 × 1000 1.82 (118 × 22.5)
= 1.014
You're Reading a Preview ASME Section I, Fig. PG-32, "General Notes," states that K is limited to a value Unlock full access with a free trial. of 0.99. Therefore PG-32.1.2 cannot be used. Allowable tensile stress for SA-516-60 is 118 MPa and for Free SA-192 is 92.4 MPa. Download With Trial Therefore: f r1 =
118
92.4 = 1.28
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Useful Not useful Use equation 2.3 to determine the minimum required shell thickness (additional thickness may be used towards reinforcement requirements). (See PG-27.2.2)
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Revised Second Class Course • Section A1 • SI Units
t r =
=
PR SE - (1 - y ) P
+ C
4.5 × 477.5
(118 × 1) - (1 - 0.4)4.5
+0
= 20.408 mm Therefore t r = 20.408 mm and t = 22.5 mm
Use equation 1.1 to determine the minimum required nozzle thickness. (See PG-27.2.1) Where P D S e
= = = =
4.5 MPa 100 + (2 x 8) = 116 mm 92.4 MPa 0 (see PG-27.4, note 4)
PD t = a Preview + 0.005D + e You're Reading 2S + P Unlock full access with 4.5 a free × trial. 116 = + 0.005 (116) + 0 2 ( 92.4 ) + 4.5 Download With Free Trial 522 = + 0.58 189.3 = 3.3375 mm
Therefore t r n = 3.3375 mm andSign t n =up8 to mm vote on this title
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Limit of compensation parallel to shell surface X = d or X = (0.5d + t n + t ), whichever is larger
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Chapter 1 • ASME Code Calculations: Cylindrical Components
(b)
Reinforcement area available in the shell ( X replaces d in the equation) A 1 = X (t – Ft r) A 1 = 100(22.5 – 1 x 20.408) = 209.2 Therefore A 1 = 209.2 mm2
(c)
Reinforcement area available in the nozzle Y replaces (2.5t n + t e) in the equation A 2 = 2(t n - t rn)(Y ) f r1 A 2 = 2(8 - 3.3375)(38) × 1.17 = 414.59 Therefore A 2 = 414.59 mm2
(d)
Reinforcement area available in the nozzle weld A 41 = (W L 1)2 f r2 where f r2 = S n / S s A 41 = (5)2 × 92.4/ 118 = 19.58 Therefore A 41 = 19.58 mm2
43
You're Reading Preview Total area available from shell, nozzle, and nozzle aweld A r = A 1 + A 2 + A 41 Unlock full access with2a free trial. A r = 209.2 + 414.59 + 19.58 = 643.37 mm (e)
Download With Free Trial Area provided by the reinforcement plate weld 2 A 42 = (W L 2) F r3 A 42 = (5)2 × 92.4/ 118 = 19.58 Therefore A 42 = 19.58 mm2 Area required by reinforcement pad A 5 = A – ( A r + A 42) A 5 = 2040.8 - (643.37 + 19.58) = 1377.85 Therefore
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Revised Second Class Course • Section A1 • SI Units
Thus, a reinforcing pad 213.75 mm diameter and 18 mm thick is req carry the tensile stress and maintain the vessel pressure boundary. This falls within the limits of compensation.
SECTION VIII-1
The limits of compensation stated in paragraph UG-40 (b) and (c) are used in Section I, except that the vessel shell and nozzle must be treated in a corroded condition. Therefore, the limit of compensation parallel to the shell surface
X = diameter of the finished opening in corroded condit
Or
X = radius of the finished opening in corroded conditio wall thickness+ nozzle wall thickness
Whichever is larger You're Reading a Preview The limit of compensation normal to the shell surface Unlock full access with a free trial.
Or
Y = 2.5 × nominal shell thickness less the corrosion allow Download With Free Trial Y = 2.5 × nozzle wall thickness + the thickness of the reinforcing plate (t e)
Whichever is smaller Sign up to vote on this title
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