asme viii div1 2010-2011a

ASME VIII division 1

AutoPIPE Vessel (Microprotol) procal V33.1.0.11 1 prodia2 V33.1.0.11 Bentley Systems, Inc.

ASME VIII div.1 verification document

22 August 2013 11 July 1,2010 – 2011a AutoPIPE Vessel 33.1.0.11 No changes

05 Mar 2013 10 July 1,2010 – 2011a Microprotol 33.0.7.6 Feb 2013 Same as Rev 9 including Bentley Inc.

24 Oct 2012 9 July 1,2010 – 2011a Microprotol 33.0.7.5 Oct 2012 Including conical head (L-6.4)

16 July 2012 8 July 1,2010 – 2011a Microprotol 33.0.7.2 July 2012 UG-45 checking (L7-8)

20 June 2012 7 July 1,2010 – 2011a Microprotol 33.0.7.1 May 2012

23 April 2012 6 July 1,2010 – 2011a Microprotol 33.0.7.0 Mar 2012 Editing modification

15 Feb 2012 5 July 1,2010 – 2011a Microprotol 33.0.6.0 Jan 2012 Code release

17 Feb 2011 4 July 1,2010 Microprotol 33.0.1.0 Feb 2011 Code release + misc developpements

02 Nov 2009 3 July 1,2009 Microprotol 32.8.8.2 Oct 2009 Add expansion joint calc. +upd. code

25 Mar 2009 2 July 1,2008 Microprotol 32.8 April 2009 Check + add examples

15 Dec 2008 1 July 1,2008 Microprotol 32.8 Nov 2008 Check + new material data base

15 May 2008 0 July 1,2007 Microprotol 32.7 April 2008 Initial issue

Date Rev ASME VIII div. 1 Edit. Software release Comments



Table of contents

Introduction 3

Appendix L-2 [Thickness Calculation for shells under internal pressure with supplemental loadings ] 4

L-2.3.1 – L-2.3.2 4 Appendix L-3 [Vessels under External Pressure ] 5 L-3.1 5 L-3.2 6 L-3.3 without stiffening ring 7 L-3.3 with stiffening rings 12

Appendix L-5 [Design of circumferential stiffening ring and attachment weld for a cylindrical shell under External

Pressure] 15

L-5.1 (using stiffening ring of 2 in. x 3.75 in.) 16

Appendix L-6 [Required thickness for formed heads with pressure on the convex side] 18

L-6.1 19 L-6.2 [t = 0.5 in.] 20 L-6.2 [t = 0.5625 in.] 21 L-6.3 22 L-6.4 [t = 0.75 in.] 23 L-6.4 [t = 0.563 in.] 24

Appendix L-7 [Openings and reinforcements] 25

L-7.1 26 L-7.2 (without reinforcing plate) 29 L-7.2 (with reinforcing plate of 3 in. wide x 0.375 in. thick 31 L-7.2 (with reinforcing plate of 3.125 in. wide and 0.375 in. thick) 33 L-7.4 35 L-7.5 38 L-7.6 41 L-7.7 (nozzle thickness = 1/2 in.) 43 L-7.7 (nozzle thickness = 7/8 in.) 45 L-7.8 + UG-45 47

Part UHX 50

UHX-20.1.1 Tubesheet Integral with Shell and Channel 51 UHX-20.1.2 Tubesheet gasketed with Shell and Channel 56 UHX-20.1.4 Tubesheet gasketed with Shell and Integral with Channel extended as a flange 60

UHX-20.2.1 Tubesheet integral with Shell and gasketed with Channel extended as a flange 63 UHX-20.2.3 Fixed Tubesheet Exchanger , Configuration a 70 UHX-20.3.1 Stationary Tubesheet Gasketed With Shell and Channel; Floating Tubesheet Gasketed, Not extended as a

Flange 78

Other calculations 81

Welding Neck Flange ( Taylor Forge Bulletin) 82 ASME VIII Div.1 2010 App. 26 Unreinforced Bellows 85



Introduction

This document is a part of the AutoPIPE Vessel ® validation procedure intended to demonstrate the application of the

ASME VIII div 1 Code through the examples provided in the various parts of Code.

AutoPIPE Vessel ® input data files used are printed front of the paragraph reference :

Appendix L-2 :

Paragraph L-2.3.1 – L-2.3.2 : L-2.3.emvd

Appendix L-3 :

Paragraph L-3.1 : L-3 1.emvd


Paragraph L-3.3 : L-3.3 without stiffener.emvd

L-3 3.emvd

Appendix L-5 :


Appendix L-6 :


Paragraph L-6 2 L-6 2 1.emvd for L-6 2 [t=0.5 in.]

L-6 2 2.emvd for L-6 2 [t=0.525 in.]

Paragraph L-6 3 : L-6 3.emvd

Paragraph L-6 4 : L-6 4 1.emvd for L-6 4 [t=0.75 in.]

L-6 4 2.emvd for L-6 4 [t=0.563 in.]

Appendix L-7 :


Paragraph L-7 2 without reinforcing plate: L-7 2 without reinf.emvd

Paragraph L-7 2 with reinforcing plate : L-7 2 with reinf.emvd 3 in x 0.375 in

L-7 2 with reinf incr.emvd 3.125 in x 0.375 in




Paragraph L-7.7 : L-7 7-1.emvd (thk = ½ in.)

L-7 7-1 nozzle thk inc.emvd (thk = 7/8 in.)

Paragraph L-7.8 + UG-45 : L-7 8.emvd

UHX-20.1 Examples of UHX-12 for U-Tube Tubesheets :

UHX-20.1.1 Example 1 : UHX-20-1-1.emvd



UHX-20. 2 Examples of UHX-13 for Fixed Tubesheets :



UHX-20. 3 Examples of UHX-14 for Floating Tubesheets :


Other calculations :

Taylor Forge calculation method for flange : Flange Welding Neck.emvd

Unreinforced bellows (Annex 26) : 26-14-1-1.emvd



Appendix L-2 [Thickness Calculation for shells under internal pressure with supplemental

loadings ]

L-2.3.1 – L-2.3.2

Conical shell (30.23) internal pressure.ASME VIII DIV.1 2010 - 2011a

t = minimum required thickness t n = nominal thickness E = Joint efficiency

P = internal pressure S = Allowable stress T = Temperature

D = inside diameter at large end Ca = corrosion + tolerance σ = circular stress

α = Half apex angle Tol % = tolerance for pipes P a = Max. allowable pressure

t min = t +Ca+Tol % shall be ≤ t n t a = (t n×Tol %)-Ca shall be ≥ t P h = Hydrostatic pressure

[UG-32(g)] t = P ( D+2Ca)/[2cosα (SE -0.6 P )] σ = P (( D+2Ca)+1.2t acosα )/(2 Et acosα ) P a =2SEt acosα /(( D+2Ca)+1.2t acosα )

SA387GR2CL2 Plate Schedule : / DN : /

t n = 0.4380 in D = 200 in Tol % = / PWHT : No Radiography : Spot

α = 30 ° Cor. = 0 in Tol. = 0 in UG-16(b) = 0.0625 in

P (psi) P h (psi) T (°F) S (psi) E t a (in) σ (psi) P a (psi) t (in) t min (in)

Operation N 50 0 650 17 500 0.85 0.4380 15 543 56.3 0.3889 0.3889

MAWP (650 °F, Corroded) = 56.3 psi MAWP (70 °F, new) = 64.4 psi

Appendix 1-5(d )Q L = PR L/2+ f 1 ArL = kQ L R L(1-∆/α ) tan α / (S s E 1) AeL = (t s-t ) ( R Lt s)

0.5 + (t c-t r ) ( R Lt c /cos )0.5

k = max(1, y/S r E r ) y = S s E s Reinforcing ring : Plate 15 × 0.5 Aring = 7.5 in P (psi) S s (psi) E 1 ∆ t (in) t c (in) t r (in) t s (in) R L (in)

Operation N 50 17 500 1 17.57 ° 0.2862 0.4380 0.3889 0.3130 100.000

k f 1(max) (lbf/in) f 1(min) (lbf/in) Q L(max) (lbf/in) Q L(min) (lbf/in) rL (in2) AeL (in2)

Operation N 1.207 250.0274 -283.0145 2 750.0260 2 216.9850 4.54 0.5the junction is adequately reinforced

Appendix 1-5(e)

QS = PRS /2+ f 2 ArS = kQS RS (1-∆/α ) / (S s E 1) AeS = 0.78 ( RS t s)0.5

[(t s-t ) + (t c-t r ) /cos )]

k = max(1, /S r E r ) = S s E s Reinforcing ring : Plate 15 × 0.5 Aring = 7.5 in2

P (psi) S s (psi) E 1 ∆ t (in) t c (in) t r (in) t s (in) RS (in)

Operation N 50 17 500 1 4.57 ° 0.1431 0.4380 0.1945 0.1880 50.000

k f 2(max) (lbf/in) f 2(min) (lbf/in) QS (max) (lbf/in) QS (min) (lbf/in) ArS (in2) AeS (in2)

Operation N 1.207 62.6374 -93.5602 1 312.6370 1 156.4390 2.22 0.78the junction is adequately reinforced

[UCS-79] : % extreme fiber elongation = 50 t n / R ( 1 - R f / Ro ) = 0.44 % ( R = 50.219 in ; Ro = +∞ )



Appendix L-3 [Vessels under External Pressure ]

Comment for reading of external pressure curves ASME curves for external pressure have been checked using stored vertex to redraw the curve on a transparency supportvisually compared to the Code original. Storage AutoPIPE Vessel® file is “courbvid.emsd”, located in“AutoPIPE Vessel\Calcul\Data” directory. AutoPIPE Vessel® uses logarithmic interpolation to determine the value of B.

L-3.1

External Pressure - Shell (Section No. 1) (in operation)

Ends of section :

Bottom : Support line level : 0.0000 in

top : Support line level : 39.0000 in

Elements considered :

TagDiameter Thickness

modulus of

elasticity

Vacuum

curveTemperature

(in) (in) (psi) (°F)

001 31.05 Virole 168.625 0.312524 500 060.

0CS-2 700 °F

ASME VIII DIV.1

External Pressure : P = 15 psi Design Temperature : 700 °F

Allowable stress : S = 14 257.2 psi modulus of elasticity : E = 24 500 060 psi

Unsupported shell length : 39 in Pipe, tolerance on the new thickness : c1 = /

Diameter of section : Do = 168.625 in Vacuum curve : CS-2

Checked thickness : t = 0.3125 in

UG-28 (c) Cylindrical shell with straight circular section

L = 39 in L/ Do = 0.231 Do/t = 539.6

Do/t ≥4 : A (Subpart 3 Section II Part D Fig.G) = 0.000491 Do/t ≥10 : P a =t D

B

o3

4

Do/t <4 : A =( )

2

o /D

1.10.10,min

t = / Do/t <10 : P a=

−

−

t Dt DS

Bt D ooo

11

2;0833.0

2.167min

B (Subpart 3 Section II Part D or AE /2) = 6 054.676 psi P a = 14.961 psi < P INSUFFICIENT THICKNESS Minimum required thickness = 0.313 in

CommentsSee “Comment for reading of external pressure curves”. Rounded value in ASME example of Do /t = 540 is used.



L-3.2

External Pressure - Hemispherical head (Section No.2)(in operation)



modulus of

elasticity

Vacuum

curveTemperature


002 03 72.000 0.500010 000 000.

0 NFA-1 100 °F

ASME VIII DIV.1 2010 – 2011a UG-33

External Pressure : P = 20 psi modulus of elasticity : E = 10 000 000 psi

Material : SB209-A93003-H112 Corrosion : 0 in

Vacuum curve : NFA-1 Tolerance : 0 in

Nominal thickness : t = 0.5 in External Diameter : Do = 73 in

Checked thickness : t = 0.5 in outside radius : Ro = 36.5 in

outside height : ho = / Axis ratio : Do/(2ho) = /

(c) Hemispherical heads

UG-28 (d ) : A =t Ro

125.0= 0.001712

B (Subpart 3 Section II Part D or 0.0625 E /( Ro/t )) =1 763.745 psi

P a =t R

B

o

= 24.161 psi ≥ P

CommentsSee “Comment for reading of external pressure curves”.



L-3.3 without stiffening ring

External Pressure – Cone (Section N°2)(in operation)

Ends of section :

Bottom : Cone-to-cylinder junction without stiffening ring Support line level : 250.0000 in

top : Cone-to-cylinder junction without stiffening ring Support line level : 380.0000 in



modulus of

elasticity

Vacuum

curveTemperature


005 30.23 Cône de calandre 0.000 1.250025 125 030.

0CS-2 650 °F

ASME VIII DIV.1 2010 – 2011a


Allowable stress : S = 15 000 psi modulus of elasticity : E = 25 125 030 psi

Unsupported shell length : 130 in half opening angle of conical shell : α = 29.98 °

Checked thickness : t = 1.25 in Vacuum curve : CS-2

Element diameter at large end : DL = 202.5 in Element diameter at small end : Ds = 50.75 in

UG-33( f ) Conical shell with straight circular section

L = 130 in Fig. UG-33.1⇒ Le = 81.29 in Le / D L = 0.401 t e = t cos(α ) = 1.083 in D L /t e = 187.027

DL/t e≥10 : A (Subpart 3 Section II Part D Fig.G) = 0.001447 DL /t e ≥ 10 : P a =e L t D

B

3

4

DL/t e<10 : A =( )

2

o /D

1.10.10,min

t = / DL /t e < 10 : P a =

−

−

e Le Le L t Dt DS

Bt D

11

2;0833.0

2.167MIN

B (Subpart 3 Section II Part D or AE /2) = 9 414.372 psi P a = 67.116 psi ≥ P



External Pressure – Large cylinder-to-cone junction (Section No. 2) (in operation)

ASME VIII DIV. 2010 - 2011a APPENDIX 1-8

Junction without knuckle, without stiffener

Internal corroded radius of knuckle : r = 0 in Angle : α = 29.98 °

Shell adjacent to the junction, element No. : 4 Calculation length of cylinders element : L L= 250 in

Cone adjacent to junction, element number : 5 Calculation length of cones element : L= 130 in

External Pressure : P = 50 psi

Cylindrical shell

Material : SA516GR70 Temperature : 650 °F

Nominal stress : S s = 17 500 psi modulus of elasticity : E s = 25 125 ksi

Vacuum curve : CS-2 Longitudinal Joint Efficiency : E 1 = 0.85

Nominal thickness : t s = 1.25 in Corrosion : 0 in

Checked thickness : t s = 1.25 in Tolerance : 0 in

External Diameter : D L = 202.5 in External radius : R L = 101.25 in

Conical shell


Nominal stress : S c = 15 000 psi modulus of elasticity : E c = 25 125 ksi

Vacuum curve : CS-2

Nominal thickness : t c = 1.25 in Corrosion : 0 in

Checked thickness : t c = 1.25 in Tolerance : 0 in

Stiffener

Type and dimensions : / Material : /

Nominal stress : S r = / Temperature : /

Vacuum curve : / modulus of elasticity : E r = /

Cross-sectional area : A s = / Diameter of shell at centroid elevation : De = /

Checking junction’s area

P /S s E 1 = 0.0034 ⇒ ∆ = 5.91 < α ⇒ Reinforcement of large end intersection

k =1 f 1 = 249.901 lbf/in Q L = PR L/2 + f 1 = 2 781.151 lbf/in

ArL =

α∆

−−

α

L

L L

s

L L

QQ PR

E S RkQ

411

tan

1

= 10.48 in2 AeL = ( )α+ cos55.0 c s s L t t t D = 23.566 in

2

AeL + A s = 23.566 in2 ≥ ArL

Checking the junction’s moment of inertia

M =α

−++

α−tan322

tan 22

L

s L L L

R

R R L R = 150.619 in f 1 = 249.901 lbf/in F L = α+ tan1

f PM = 7 675.119 lbf/in

Lc = ( )22 Rs RL L ++ = 150.522 in ATL = scc s L A

t Lt L++

22= 250.327 in

2 B =

TL

L L

A

D F

4

3 = 4 656.55 psi

A (Subpart 3 Section II Part D FIG.G) = 0.000368 I s=14

2

TL L A AD= / I s′=

9.10

2

TL L A AD= 346.49 in4

I ' = 64.39 in < I s '



Comments Angle of cone : α = arc tg[((200-50)/2) /130] = 29.98164 ° (30° in ASME example)tg a = [(200-50)/2) /130] = 0.576923(value of ASME based on angle of 30°) P/S s E 1 = 50/(17500x0.85) = 0.00336 According to Table 1-8.1, P/S s E 1 =0.002 ∆ = 5°

P/S s E 1 =0.005 ∆ = 7° For P/S s E 1 = 0.00336 , ∆ = 5 + (7-5) x [(0.00336-0.002)/(0.005-0.002)] = 5.9066° (ASME Example value is 5.93°). No additional area of reinforcement k =1 f1 obtains by load due to wind and external load. No detail in ASME example.Q L = 50x101.25/2 + 249.904 = 2 781.154 lbf/in

ArL =

−−

98.299066 .5

154.2781154.278125.101 x50

411

85.0 x17500576923.0 x25.101 x154.2781 x1

= 10.4798 in2

AeL = ( ) )30cos( 25.125.125.1 x5.20255.0 + = 23.5685 in2 ≥ ArL Junction area checked

M =576923.0 x25.101 x3

375.2525.1012

2502

576923.0 x25.101 22 −++

− = 150.61895 in

F L =50 x 150.61895 + 249.904 x 0.576923 = 7675.12286 lbf/in

Lc = ( ) ( )22 375.2525.101130 −+ = 150.522475 in

ATL =2

25.1 x522475.1502

25.1 x250+ = 250.32655 in2 (rounded value in ASME example)

B =32655.250

5.202 x12286 .7675 x

43

= 4656.55475

A from Fig CS-2 = 0.000368 (see “Comment for reading of external pressure curves”).

As there is no stiffener : I s′ =9.10

32655.250 x5.202 x000368.0 2=346.487 in4

Inertia of shell : Length to be considered : L = (1.10/2) x √(202.5x1.25) = 8.75 in I sh = (8.75 x1.253 ) /12 = 1.424 in4 Inertia of cone : Length to be considered on shell : L = (1.10/2) x √(202.5x1.25) = 8.75 in Length to be considered on cone: H = L/ cos( α ) = 8.75 /cos(29.98) = 10.1 in I co = (H tc3 cos( α )2 )/12+(H 3 tc sin( α )2 )/12 = 28.0317 in4 Position of centers of inertia from external line of shell :Shell = ts / 2 = 0.625 inCone = (tc/2) +(H sin(a)/2 + (tc/2)(1-cos( α )) = 3.0655 inSection of shell and cone = (1.25x8.75) + (1.25x10.1) = 23.5625 in2 Position of center of inertia of the combined section = [(0.625x1.25x8.75) + (3.0655x1.25x10.1)] / 23.5625 = 1.9326 in Inertia of combined section : I’ = 1.424 + 28.0317 + (1.9326-0.625)2 x(1.25x8.75) + (3.0655-1.9326)2 x(1.25x10.1)= 64.4 in4

Is’ > I’ large cylinder to cone junction not verified



External Pressure – Small cylinder-to-cone junction (Section No. 2) (in operation)


Junction without knuckle, without stiffener


Shell adjacent to the junction, element No. : 6 Calculation length of cylinders element : L s= 75 in



Cylindrical shell


Nominal stress : S s = 17 500 psi modulus of elasticity : E s = 25 125 030 psi

Vacuum curve : CS-2 Longitudinal Joint Efficiency : E 1 = 1



External Diameter : D s = 50.75 in External radius : R s = 25.375 in

Conical shell


Nominal stress : S c = 15 000 psi modulus of elasticity : E c = 25 125 030 psi

Vacuum curve : CS-2



Stiffener

Type and dimensions : / Material : /

Nominal stress : S r = / Temperature : /

Vacuum curve : / modulus of elasticity : E r = /

Cross-sectional area : A s = / Diameter of shell at centroid elevation : De = /


k =1 f 2 = 62.518 lbf/in Q s = PR s/2 + f 2 = 696.893 lbf/in

Ars =1

tan

E S

RkQ

s

s s α= 0.583 in

2 Aes = ( ) ( )( )α−+− cos55.0 r c s s s t t t t t D = 0.765 in

2

Required thicknesses of portions at junction :

Shell : t = 0.325 in Cone : t r = 1.017 in

Aes + A s = 0.765 in2 ≥ Ars


N =α

−++

αtan622

tan 22

s

s L s s

R

R R L R = 154.201 in f 2 = 62.518 lbf/in F s = α+ tan2

f PN = 7 746.115 lbf/in

Lc = ( )22 Rs RL L ++ = 150.522 in ATs = scc s s A

t Lt L++

22= 108.139 in

2 B =

Ts

s s

A

D F

4

3 = 2 726.46 psi

A (Subpart 3 Section II Part D FIG.G) = 0.000219 I s=

14

2

Ts s A AD= / I s′=

9.10

2

Ts s A AD= 5.6 in

4

I ' = 1.17 in4 < I s '



Comments No additional area of reinforcement k =1 f2 obtains by load due to wind and external load. No detail in ASME example.Q s = 50x25.375/2 + 62.088 = 696.463 lbf/in

Ars =1 x17500

576923.0 x375.25 x463.696 x1= 0.58262 in2

N =576923.0 x375.25 x6

375.2525.1012

752

576923.0 x375.25 22 −++ = 154.2 in

F S =50 x 154.2 + 62.088 x 0.576923 = 7745.82 lbf/in f1 obtains by load due to wind and external load. No detail in ASME example.

ATS =2

25.1 x522475.1502

375.0 x75+ = 108.139 in2 (rounded value in ASME example)

B =139.108

75.50 x82.7745 x

43

= 2726.3647

A from Fig CS-2 = 0.000219 (see “Comment for reading of external pressure curves”).

As there is no stiffener : I s′ =9.10

139.108 x375.50 x000219.0 2=5.514 in4

Inertia of combined section : I’ = 1.17 in4

Is’ > I’ large cylinder to cone junction not verified



L-3.3 with stiffening rings

External Pressure – Conical shell (Section No. 2) (in operation)

Ends of section :

Bottom :Cone-to-cylinder junction with stiffening ring

No. 1)Support line level : 250.0000 in

top :Cone-to-cylinder junction with stiffening ring

No. 2)Support line level : 380.0000 in



modulus of

elasticity

Vacuum

curveTemperature


005 30.23 Cône de calandre 0.000 1.250025 125 030.

0CS-2 650 °F

ASME VIII DIV.1 2010 - 2011a



Unsupported shell length : 130 in half opening angle of conical shell : α = 29.98 °


Element diameter at large end : DL = 202.5 in Element diameter at small end : Ds = 50.75 in


L = 130 in Fig. UG-33.1⇒ Le = 81.29 in Le / D L = 0.401 t e = t cos(α ) = 1.083 in D L /t e = 187.027


B

3

4

DL/t e<10 : A =( )

2

o /D

1.10.10,min

t = / DL /t e < 10 : P a =

−

−

e Le Le L t Dt DS

Bt D

11

2;0833.0

2.167MIN




External Pressure – Large cylinder-to-cone junction (Section No. 2) (in operation)


Junction without knuckle, reinforced by stiffener No. 1


Shell adjacent to the junction, element No. : 4 Calculation length of cylinders element : L L= 250 in



Cylindrical shell


Nominal stress : S s = 17 500 psi modulus of elasticity : E s = 25 125 ksi

Vacuum curve : CS-2 Longitudinal Joint Efficiency : E 1 = 0.85



External Diameter : D L = 202.5 in External radius : R L = 101.25 in

Conical shell


Nominal stress : S c = 15 000 psi modulus of elasticity : E c = 25 125 ksi

Vacuum curve : CS-2



Stiffener

Type and dimensions : WT8x18 7.93 × 6.99 × 0.295 ×

0.43Material : SA516GR60

Nominal stress : S r = 14 500 psi Temperature : 650 °F

Vacuum curve : CS-2 modulus of elasticity : E r = 25 125 030 psi

Cross-sectional area : A s = 5.28 in Diameter of shell at centroid elevation : De = 0 in


P /S s E 1 = 0.0034 ⇒ ∆ = 5.91 < α ⇒ Reinforcement of large end intersection

k =1.2069 f 1 = 250.002 lbf/in Q L = PR L/2 + f 1 = 2 781.251 lbf/in

ArL =

α∆

−−

α

L

L L

s

L L

QQ PR

E S RkQ

411

tan

1

= 12.649 in2 AeL = ( )α+ cos55.0 c s s L t t t D = 23.566 in2

AeL + A s = 28.846 in2 ≥ ArL


M =α

−++

α−tan322

tan 22

L

s L L L

R

R R L R = 150.619 in f 1 = 250.002 lbf/in F L = α+ tan1 f PM = 7 675.178 lbf/in

Lc = ( )22 Rs RL L ++ = 150.522 in ATL = scc s L A

t Lt L++

22= 255.607 in

2 B =

TL

L L

A

D F

4

3 = 4 560.4 psi

A (Subpart 3 Section II Part D FIG.G) = 0.00036 I s=14

2

TL L A AD= / I s′=

9.10

2

TL L A AD= 346.07 in

4

I ' = 368.75 in4 ≥ I s '

CommentsSame calculation as calculation without stiffener except for :k = y/Sr Er = 17500x25125000/14500x25125030 = 1.206895 As = area of stiffener f 1 = 250.003 lbf/in



External Pressure – Small cylinder-to-cone junction (Section No. 2) (in operation)


Junction without knuckle, reinforced by stiffener No. 2


Shell adjacent to the junction, element No. : 6 Calculation length of cylinders element : L s= 75 in



Cylindrical shell


Nominal stress : S s = 17 500 psi modulus of elasticity : E s = 25 125 030 psi

Vacuum curve : CS-2 Longitudinal Joint Efficiency : E 1 = 1



External Diameter : D s = 50.75 in External radius : R s = 25.375 in

Conical shell


Nominal stress : S c = 15 000 psi modulus of elasticity : E c = 25 125 030 psi

Vacuum curve : CS-2



Stiffener

Type and dimensions : Bar 3.5 × 0.375 Material : SA516GR60

Nominal stress : S r = 14 500 psi Temperature : 650 °F

Vacuum curve : CS-2 modulus of elasticity : E r = 25 125 030 psi

Cross-sectional area : A s = 1.313 in Diameter of shell at centroid elevation : De = 0 in


k =1.2069 f 2 = 62.5 lbf/in Q s = PR s/2 + f 2 = 696.875 lbf/in

Ars =1

tan

E S

RkQ

s

s s α= 0.704 in

2 Aes = ( ) ( )( )α−+− cos55.0 r c s s s t t t t t D = 0.765 in

2

Required thicknesses of portions at junction :

Shell : t = 0.325 in Cone : t r = 1.017 in

Aes + A s = 2.078 in2 ≥ Ars


N =α

−++

αtan622

tan 22

s

s L s s

R

R R L R = 154.201 in f 2 = 62.5 lbf/in F s = α+ tan2

f PN = 7 746.105 lbf/in

Lc = ( )22 Rs RL L ++ = 150.522 in ATs = scc s s A

t Lt L++

22= 109.452 in

2 B =

Ts

s s

A

D F

4

3 = 2 693.76 psi

A (Subpart 3 Section II Part D FIG.G) = 0.000216 I s=

14

2

Ts s A AD= / I s′=

9.10

2

Ts s A AD= 5.6 in

4

I ' = 4.64 in4 < I s '

CommentsSame calculation as calculation without stiffener except for :k = y/Sr Er = 17500x25125000/14500x25125030 = 1.206895 As = area of stiffener f 2 = 62.5 lbf/in



Appendix L-5 [Design of circumferential stiffening ring and attachment weld for a cylindrical

shell under External Pressure]

Comment for reading of external pressure curves ASME curves for external pressure have been checked using stored vertex to redraw the curve on a transparency supportvisually compared to the Code original. Storage AutoPIPE Vessel® file is “courbvid.emsd”, located in “AutoPIPEVessel \Calcul\Data” directory. AutoPIPE Vessel ® uses logarithmic interpolation to determine the value of B.



L-5.1 (using stiffening ring of 2 in. x 3.75 in.)

Ends of section :

Bottom : Support line level : 0.0000 in

top : Stiffening ring (ring No. 1) Support line level : 40.0000 in



modulus of

elasticity

Vacuum

curveTemperature


001 31.05 Virole 169.000 0.312524 500 060.

0CS-2 700 °F

ASME VIII DIV.1




Diameter of section : Do = 169 in Vacuum curve : CS-2



L = 40 in L/ Do = 0.237 Do/t = 540.8


B

o3

4

Do/t <4 : A =( )

2

o/D

1.10.10,min

t = / Do/t <10 : P a=

−

−

t Dt DS

Bt D ooo

11

2;0833.0

2.167min


Ends of section :

Bottom : Stiffening ring (ring No. 1) Support line level : 40.0000 in




modulus of

elasticity

Vacuum

curveTemperature


001 31.05 Virole 169.000 0.3125 24 500 060. CS-2 700 °F

ASME VIII DIV.1




Diameter of section : Do = 169 in Vacuum curve : CS-2



L = 40 in L/ Do = 0.237 Do/t = 540.8


B

o3

4

Do/t <4 : A =( )

2

o /D

1.10.10,min

t = / Do/t <10 : P a=

−

−

t Dt DS

Bt D ooo

11

2;0833.0

2.167min





Shell

External Pressure : P = 15 psi Temperature : 700 °F

Material : SA285GRC modulus of elasticity : E = 24 500 060 psi

Vacuum curve : CS-2 Corrosion : 0 in

Nominal thickness : t = 0.313 in Tolerance : 0 in


Calculation length of previous element : L1 = 40 in

Calculation length of next element : L2 = 40 in

Stiffener

Type and dimensions : Plate 3.75 × 2 Material : SA36

modulus of elasticity : E = 24 500 060 psi Vacuum curve : CS-2

Cross-sectional area : As = 7.5 in2 Diameter of shell at centroid elevation : Do = 169 in

UG-29 Stiffened elements of cylindrical shells

x1 = ( ( )2;55.0MIN 1o Lt D s = 3.997 in t 1 = 0.313 in x2 = ( ( )2;55.0MIN 2o Lt D s = 3.997 in t 2 = 0.313 in

B = s s

o L / At

D P

43

+= 3 802.5 psi Ls =

2

21 L L ++++= 40 in t eq = 0.313 in

A (Subpart 3 Section II Part D FIG.G or 2 B/ E ) = 0.00031 I s =

( )14

/ sss

2

o A L At L D +=

16.22 in4

I s ′ = ( )

9.10

/ sss

2

o A L At L D += /

I = 16.54 in4 ≥ I s

Comments ASME example checks only stiffening ring.The thickness of shell is not sufficient (no information about this in ASME example). Index 1 for x1 ,t 1 and L1 is used for shell before the stiffener and index 2 for x2 ,t 2 and L2 is used for shell after the stiffener.t 1 and t 2 should be read 0.3125 in, t eq is the average thickness calculated as follow : t eq = (L1e1+L2e2 )/(L1+L2 )



Appendix L-6 [Required thickness for formed heads with pressure on the convex side]

Comment for reading of external pressure curves ASME curves for external pressure have been checked using stored vertex to redraw the curve on a transparency supportvisually compared to the Code original. Storage AutoPIPE Vessel ® file is “courbvid.emsd”, located in “AutoPIPEVessel \Calcul\Data” directory. AutoPIPE Vessel ® uses logarithmic interpolation to determine the value of B.



L-6.1

External pressure – Elliptical Head (Section No.2)(in operation)



modulus of

elasticity

Vacuum

curveTemperature


002 05 169.000 0.562524 500 060.

0CS-2 700 °F



Material : SA285GRC Corrosion : 0 in

Vacuum curve : CS-2 Tolerance : 0 in


Checked thickness : t = 0.563 in outside radius : Ro = /

outside height : ho = 42.531 in Axis ratio : Do/(2ho) = 2

(d ) Elliptical heads

K o (Table UG-33.1) = 0.894 Equivalent outside radius Ro = K o. Do = 151.094 in

(a)(1)(a) : E = 1 P a(a) = 1.67 P = 57.173 psi

UG-28 (d ) : A =t Ro

125.0= 0.000465 P a(d) =

t R B

o

= 21.431 psi

B (Subpart 3 Section II Part D or 0.0625 E /( Ro/t )) =5 756.503 psi (a)(1)(b) P a = MIN ( P a(a) , P a(d)) = 21.431 psi ≥ P

Comments AutoPIPE Vessel ® checks UG-33(a) (1) (a) In ASME example value of K 1 from Table UG-37 is used and this is not correct. AutoPIPE Vessel ® uses Table UG-33.1 (UG-33 FORMED HEADS, PRESSURE ON CONVEX SIDE) to determine K 0.That gives a different value of RO

See “Comment for reading of external pressure curves”.



L-6.2 [t = 0.5 in.]

External pressure – Torispherical Head (Section No.2)(in operation)



modulus of

elasticity

Vacuum

curveTemperature


002 30.10 Fond 169.000 0.500024 500 060.

0CS-2 700 °F






Checked thickness : t = 0.5 in outside radius : Ro = 169 in


(e) Torispherical heads

(a)(1)(a) : E = 1 P a(a) = 1.67 P = 28.638 psi

UG-28 (d ) : A =t Ro

125.0= 0.00037 P a(e) =

t R B

o

= 13.592 psi


(a)(1)(b) : P a = MIN ( P a(a) , P a(e)) = 13.592 psi < P INSUFFICIENT THICKNESS

Comments AutoPIPE Vessel ® checks UG-33(a) (1) (a)See “Comment for reading of external pressure curves”.



L-6.2 [t = 0.5625 in.]

External Pressure – Torispherical Head (Section No.2)(in operation)



modulus of

elasticity

Vacuum

curveTemperature


002 08 169.125 0.562524 500 060.

0CS-2 700 °F





Nominal thickness : t = 0.563 in External Diameter : Do = 169.125 in



(e) Torispherical heads

(a)(1)(a) : E = 1 P a(a) = 1.67 P = 32.216 psi

UG-28 (d ) : A =t Ro

125.0= 0.000416 P a(e) =

t R B

o

= 17.22 psi

B (Subpart 3 Section II Part D or 0.0625 E /( Ro/t )) =5 175.545 psi (a)(1)(b) : P a = MIN ( P a(a) , P a(e)) = 17.22 psi ≥ P

Comments AutoPIPE Vessel® checks UG-33(a) (1) (a)t n and t should be read 0.5625See “Comment for reading of external pressure curves”.



L-6.3

External Pressure – Hemispherical head (Section No.2)(in operation)



modulus of

elasticity

Vacuum

curveTemperature


002 03 169.000 0.312524 500 060.

0CS-2 700 °F








(c) Hemispherical heads

UG-28 (d ) : A =t Ro

125.0= 0.000462


P a =t R

B

o

= 21.157 psi ≥ P

Commentst n and t should be read 0.3125See “Comment for reading of external pressure curves”.



L-6.4 [t = 0.75 in.]

External Pressure – Conical Head (Section No. 2)(in operation)

Ends of section :





modulus of

elasticity

Vacuum

curveTemperature


002 30.23 Cône 169.500 0.750024 500 060.

0CS-2 700 °F




Unsupported shell length : 204.605 in half opening angle of conical shell : α = 22.5 °


Element diameter at large end : DL = 169.5 in Element diameter at small end : Ds = 0 in


L = 204.604 in Fig. UG-33.1⇒ Le = 102.302 in Le / D L = 0.604 t e = t cos(α ) = 0.693 in D L /t e = 244.621


B

3

4

DL/t e<10 : A =( )

2

o /D

1.10.10,min

t = / DL /t e < 10 : P a =

−

−

e Le Le L t Dt D

S B

t D

11

2;0833.0

2.167MIN


CommentsSee “Comment for reading of external pressure curves”.The cone to cylinder junction at large end is ignored like in ASME example.



L-6.4 [t = 0.563 in.]

External Pressure – Conical Head (Section No. 2) (in operation)

Ends of section :





modulus of

elasticity

Vacuum

curveTemperature


002 30.23 Cône 169.126 0.563024 500 060.

0CS-2 700 °F




Unsupported shell length : 204.153 in half opening angle of conical shell : α = 22.5 °


Element diameter at large end : DL = 169.126 in Element diameter at small end : Ds = 0 in


L = 204.153 in Fig. UG-33.1⇒ Le = 102.077 in Le / D L = 0.604 t e = t cos(α ) = 0.52 in D L /t e = 325.152


B

3

4

DL/t e<10 : A =( )

2

o /D

1.10.10,min

t = / DL /t e < 10 : P a =

−

−

e Le Le L t Dt D

S B

t D

11

2;0833.0

2.167MIN


CommentsSee “Comment for reading of external pressure curves”.The cone to cylinder junction at large end is ignored like in ASME example. ASME example uses rounded values. Le =L/2 = 204.153/2 = 102.0765 in; Le / D L = 102.0765/169.126 = 0.603553 ; te = t cos( α ) = 0.563x 0.92388 = 0.52 in D L / te = 169.126/0.52 = 325.152 ≥ 10 A = 0.000378 ; B = 4692.673 psi Pa = 4/3 [B/(D L /te)] = 4/3 x [4692.673/325.152) = 19.243 psi ≥ P



Appendix L-7 [Openings and reinforcements]



L-7.1

Isolated Opening(s)

Figures for all configurations from FIG. UG-37.1 And FIG UG-40.

Shell ( α = 0 ) or Cone ( α > 0 ) : in longitudinal plane.

Nozzle with or without pad, set-in or set-on Selfreinforcing nozzle, set-in or set-on

Cylndrical or conical shell : circumferential plane

Head : in the plane that contains the axis of the nozzle and the longitudinal axis of vessel.

Nozzle with or without pad, set-in or set-on Selfreinforcing nozzle, set-in or set-on

β

h

d

R t

t e

L2

t p L1

δ >90° δ < 90°

P r o j e c t i o n

t n t rn

t r

L<2.5t x L≥2.5t xt e=0

t n

β

d

R t

P r o j e c t i o n

t n

t e

30°

t x

S e l f

h e i g h t

t x

γ

L1

β

t n

h R

t

t e

L2

t p

α

P r o j e c t i o n

d

t rn

t rδ >90°

δ < 90°

β

t n

h R

t

t e

α

P r o j e c t i o n

d

30°

L≥2.5t xt e=0

S e l f

h e i g h t

t n

t x

t x

L<2.5t x

γ



Opening 1 [ in operation Int.P. ] (Process) ASME VIII DIV.1

Nozzle without pad on Shell (No. 1) Set In

Pressure : P = 250 psi Temperature : 150 °F

Shell Material :SA516GR55 Allowable stress : S v = 13 700 psi

Joint efficiency : 1 E 1 = 1 Corrosion + tolerance : Cav = 0 in Allowable stress : S = 13 700 psi

Ext. Diameter : Do = 30.75 in Nominal thickness : 0.375 in Tolerance for seamless pipe : /

Nozzle Neck Material : SA516GR55 Allowable stress : S n = 15 000 psi

Joint efficiency : 1 Corrosion + tolerance : Can = 0 in Tolerance for seamless pipe : /

Ext. Diameter : Don = 5.5 in Nominal thickness : 0.75 in

External Projection : 7.875 in Internal Projection : 0 in Schedule :

Inclination : 0 ° Eccentricity : 0 in

Flange Material : / Type : /

Rating : / Height : / /

Pad Material : / Allowable stress : S p = /

Height : / Width : / Ext. Diameter : Dop = /

Weld Outside : leg 41= 0.375 in outer reinforcement : leg 42= / Inside : leg 43= /

fr 1 =min(1,S n/S v) = 1 fr 2 = min(1,S n/S v) = 1 fr 3 = min(1,min(S n, S p)/S v) = 1 fr 4 = min(1,S p/S v) = 1

Required thickness of the nozzle neck UG-27

t rn = P Rn / (S n E -0.6 P ) = 0.0337 in Rn = 2 in E = 1

The nozzle neck thickness is adequate per UG-27.


t a = t rn+Can= 0.034 in ; t b1 = max[t UG-27 , UG-16(b)]+Cav= 0.277 in ; t b3 = Table UG-45+Can= 0.226 in

t UG-45 = max [t a , min [t b3 , t b1 ] ] = 0.226 in


Dimensions FIG. UG-40angle of plane with longitudinal axis : Longitudinal plane : θ = 0° Circumferential plane : θ = 90°

angle of each side / vessel wall : δ = 90 ° δ = 90 ° δ = / δ = /

β = deflection angle / normal line 0 °

d = diameter of the finished opening 4 in

Rn = radius of the finished opening 2 in

t i = thickness of internal projection /

t p = width of reinforcing pad /

t x = thickness of selfreinforcing /

L = height of selfreinforcing / /

Configuration of the reinforcement : / /

t e = thickness or height of the reinforcement 0 in 0 in

t n = thickness of nozzle 0.75 in 0.75 in

h = height of the internal projection 0 in 0 in

Reinforcement checking UG-37 opening 1 [ in operation Int.P. ]

Required thicknesses UG-37(a)

t r = 0.2768 in [ UG-27(c) ] t = 0.375 in E = 1

t rn = P Rn / (S n E -0.6 P ) = 0.0337 in Rn = 2 in E = 1

Limits of reinforcement UG-40 :Longitudinal plane : θ = 0° Circumferential plane : θ = 90°

δ = 90 ° δ = 90 ° δ = / δ = /

UG-40 (b) : max [d , Rn+t n+t ] = 4 in 4 in

UG-40 (c) : min [ 2.5t , 2.5t n+t e ] = 0.938 in 0.938 in

Area required UG-37 (c) : Longitudinal plane : θ = 0° Circumferential plane : θ = 90°

F = Correction factor FIG.UG-37 1

A = d t r F + 2 t n/cos( β ) t r F (1- f r 1) 1.107 in



Lengths and heights of calculation of the areas :Longitudinal plane : θ = 0° Circumferential plane : θ = 90°

δ = 90 ° δ = 90 ° δ = / δ = /

L1 = min [ UG-40(b)-Radius , length available ] 2 in 2 in

L2 = min [ UG-40(c) , height available ] 0.938 in 0.938 in

L3 = min [ h , 2.5t , 2.5t i ] = 0 in 0 in

L5 = min [ UG-40 (b)- Ron , t p , length available] 0 in 0 in

Area available (in2) :

Longitudinal plane : θ = 0° Circumferential plane : θ = 90°

δ = 90 ° δ = 90 ° δ = / δ = /

A1 = L1 ( E 1 t -t r F ) - t n/cos( β ) ( E 1 t -t r F ) (1- f r 1) 0.196 0.196

A2 = L2 (t n-t rn) fr 2 0.672 0.672

A3 = L3 t i f r 2 0 0

A41 = leg 412 /2 f r 2 0.07 0.07

A42 = leg 42 /2 f r 4 0 0

A43 = leg 432 /2 f r 2 0 0

A5 = L5 t e f r 4 0 0

A1 + A2 + A3 + A41 + A42 + A43 + A5 =0.938 ≥ A/2 0.938 ≥ A/2

1.877 ≥ A The opening is adequately reinforced per UG-37.

Weld sizes check UW-16(c). opening 1 [ in operation Int.P. ]

Fig. UW-16.1(d) full penetration weld

Minimum throat required actual

t c,outward min[¼ in.(6 mm);0.7×t min]= 0.25 in

t in = min[ ¾ in.(19 mm) ; t , t n]0.7×leg 41 = 0.2625 in

t c,inward min[¼ in.(6 mm);0.7×t min]= /

t in = min[ ¾ in.(19 mm) ; t , t n]0.7×leg 43 − Can = /

Weld sizes are adequate

Weld loads check UG-41(b).

Strength calculations for attachments welds are not required for this opening in accordance with UW-15(b).

t n = 0.75

t = 0.375

t c,outward

t c,inward

leg 41

leg 43



L-7.2 (without reinforcing plate)

Opening 1 [ in operation Int.P. ]

ASME VIII DIV.1





Ext. Diameter : Do = 61,5 in Nominal thickness : 0,75 in Tolerance for seamless pipe : /

Nozzle Material : SA516GR55 Allowable stress : S n = 16 600 psi


Ext. Diameter : Don = 12,75 in Nominal thickness : 0,5 in

External Projection : 12 in Internal Projection : 0 in Schedule :


Flange Material : SA105 Type : WN

Rating : (ASME B16.5) 300 Height : 5,12 in NPS 12"



Weld Outside : leg 41= 0,375 in outer reinforcement : leg 42= / Inside : leg 43= /

fr 1 =min(1,S n/S v) = 1 fr 2 = min(1,S n/S v) = 1 fr 3 = min(1,min(S n, S p)/S v) = 0.923 fr 4 = min(1,S p/S v) = 0.923


t rn = P Rn / (S n-0.6 P ) = 0,0893 in Rn = 5,875 in E = 1




t UG-45 = max [t a , min [t b3 , t b1 ] ] = 0,328 in


Dimensions FIG. UG-40angle of plane with longitudinal axis : Longitudinal Plane : θ = 0° Circumferential Plane : θ = 90°



d = diameter of the finished opening 11,75 in

Rn = radius of the finished opening 5,875 in







t n = thickness of nozzle 0,5 in 0,5 in


Reinforcement checking UG-37 opening 1 [ in operation Int.P. ] Required thicknesses UG-37(a)

t r = 0.53 in [ UG-27(c) ] t = 0,75 in E = 1

t rn = P Rn / (S n E -0.6 P ) = 0,0893 in Rn = 5,875 in E = 1

Limits of reinforcement UG-40 :Longitudinal Plane : θ = 0° Circumferential Plane : θ = 90°

δ = 90 ° δ = 90 ° δ = / δ = /

UG-40 (b) : max [d , Rn+t n+t ] = 11,75 in 11,75 in

UG-40 (c) : min [ 2.5t , 2.5t n+t e ] = 1,25 in 1,25 in

Area required UG-37 (c) : Longitudinal Plane : θ = 0° Circumferential Plane : θ = 90°


A = d t r F + 2 t n/cos( β ) t r F (1- f r 1) 6,228 in²



Lengths and heights of calculation of the areas :Longitudinal Plane : θ = 0° Circumferential Plane : θ = 90°

δ = 90 ° δ = 90 ° δ = / δ = /

L1 = min [ UG-40 (b)-Radius , length available] 5,875 in 5,875 in

L2 = min [ UG-40 (c) , height available] 1,25 in 1,25 in

L3 = min [ h , 2.5t , 2.5t i ] = 0 in 0 in

L5 = min [ UG-40 (b) , t p , length available ] 0 in 0 in

Area available (in²) :Longitudinal Plane : θ = 0° Circumferential Plane : θ = 90°

δ = 90 ° δ = 90 ° δ = / δ = /

A1 = L1 ( E 1 t -t r F ) - t n/cos( β ) ( E 1 t -t r F ) (1- f r 1) 1,292 1,292

A2 = L2 (t n-t rn) fr 2 0,513 0,513

A3 = L3 t i f r 2 0 0

A41 = leg 412 /2 f r 2 0,07 0,07

A42 = leg 42 /2 f r 4 0 0

A43 = leg 432 /2 f r 2 0 0

A5 = L5 t e f r 4 0 0

A1 + A2 + A3 + A41 + A42 + A43 + A5 =1,876 < A/2 1,876 < A/2

3,752 < A The opening is not adequately reinforced per UG-37.

Weld sizes check UW-16(c). opening 1 [ in operation Int.P. ] FIG. UW-16.1(d) full penetration weld


t c,outward min(¼ in. (6 mm),0.7×t min) = 0,25 in

t in = min( ¾ in. (19 mm) , t , t n)0.7×leg 41 = 0,2625 in

t c,inward min(¼ in. (6 mm),0.7×t min) = /

t in = min( ¾ in. (19 mm) , t , t n)0.7×leg 43 = /


Weld loads check UG-41(b)


t n = 0,5

t = 0,75

t c,outward

t c,inward

leg 41

leg 43



L-7.2 (with reinforcing plate of 3 in. wide x 0.375 in. thick


ASME VIII DIV.1

Nozzle with pad (0,375×3) on Shell (No. 1) Set In












Pad Material : SA516GR60 Allowable stress : S p = 13 200 psi

Height : 0,375 in Width : 3 in Ext. Diameter : Dop = 18,75 in

Weld outward : leg 41= 0,375 in outer reinforcement : leg 42= 0,3125 in inward : leg 43= /















t p = width of reinforcing pad 3 in



Configuration of the reinforcement : sketch (b-1) sketch (b-1)

t e = thickness or height of the reinforcement 0,375 in 0,375 in




t r = 0.53 in [ UG-27(c) ] t = 0,75 in E = 1


Limits of reinforcement UG-40 :Longitudinal Plane : θ = 0° Circumferential Plane : θ = 90°

δ = 90 ° δ = 90 ° δ = / δ = /



Area required UG-37 (c) : Longitudinal Plane : θ = 0° Circumferential Plane : θ = 90°






δ = 90 ° δ = 90 ° δ = / δ = /

L1 = min [ UG-40 (b)-radius , length available] 5,875 in 5,875 in


L3 = min [ h , 2.5t , 2.5t i ] = 0 in 0 in



δ = 90 ° δ = 90 ° δ = / δ = /


A2 = L2 (t n-t rn) fr 2 0,667 0,667

A3 = L3 t i f r 2 0 0

A41 = leg 412 /2 f r 3 0,065 0,065

A42 = leg 42 /2 f r 4 0,045 0,045

A43 = leg 432 /2 f r 2 0 0

A5 = L5 t e f r 4 1,038 1,038

A1 + A2 + A3 + A41 + A42 + A43 + A5 =3,108 < A/2 3,108 < A/2


Weld sizes check UW-16(c). opening 1 [ in operation Int.P. ] FIG. UW-16.1 (d)+(a-1) full penetration weld


t c,inner min(¼ in. (6 mm),0.7×t min) = 0,25 in


t c,outer 0.5 × t min = 0,1875 in

t in = min(¾ in. (19 mm). , t e , t )0.7×leg 42 = 0,2188 in

t c min(¼ in. (6 mm),0.7×t min ) = /



Weld loads check UG-41(b). opening 1 [ in operation Int.P. ] FIG. UG-41.1 (a)full penetration weld

Allowable strengths UW-15(c) + UG-41(a) + UG-45(c) + L-7

Outer fillet weld in shear :

s ,outer = π/4 × Dop × leg 42 × 0.49 × min( S , S ) = 29 765 lbf

Inner fillet weld in shear :

s ,inner = π/4 × Don × leg 41 × 0.49 × min( S , S ) = 24 289 lbf

Inside fillet weld in shear :

sc = π/4 × Don × leg 43 × 0.49 × min( S , S ) = 0 lbf

Lower groove weld in tension :

sg,lower = π/4 × Don × t × 0.74 × min( S , S ) = 79 475 lbf

Upper groove weld in tension :

sg,upper = π/4 × Don × t e × 0.74 × min( S , S ) = 36 681 lbf

Nozzle wall in shear :

s = π/4 × ( Don − t ) × t × 0.70 × S = 55 899 lbf

Loads to be carried by welds Fig. UG-41.1(a)Longitudinal plane : θ = 0° Circumferential plane : θ = 180°

δ = 90 ° δ = 90 ° δ = δ =

Total : W = [ A − A1 + t n f r1 ( E 1 t − F t r )] S v 27 623 lbf 27 623 lbf

Path 1-1 : W 1-1 = ( A2+ A5+ A41+ A42) S v 25 967 lbf 25 967 lbf

Path 2-2 : W 2-2 = ( A2 + A3 + A41 + A43 +t n t f r1) S v 15 835 lbf 15 835 lbf

Path 3-3 : W 3-3 = ( A2+ A3+ A5+ A41+ A42+ A43+t n t f r1) S v 31 329 lbf 31 329 lbf

Check strength pathsLongitudinal plane : θ = 0° Circumferential plane : θ = 180°

δ = 90 ° δ = 90 ° δ = δ =

Path 1-1 : min(W ;W 1-1) ≤ sw,outer + s n yes yes

Path 2-2 : min(W ;W 2-2) ≤ sw,inner + sg,upper + sg,lower + sc yes yes

Path 3-3 : min( W , W 3-3 ) ≤ sw,outer + sg,lower + sc yes yes

Strength of welded joints is adequate

t = 0,75

t c,inner

t c

leg 41

leg 43

t e = 0,375

t c,outer

leg 42

t n = 0,5

1 3

2

32

1



L-7.2 (with reinforcing plate of 3.125 in. wide and 0.375 in. thick)


ASME VIII DIV.1

Nozzle with pad (0,375×3,125) on Shell (No. 1) Set In













Height : 0,375 in Width : 3,125 in Ext. Diameter : Dop = 19 in

Weld Outside : leg 41= 0,375 in outer reinforcement : leg 42= 0,3125 in Inside : leg 43= /















t p = width of reinforcing pad 3,125 in








t r = 0.53 in [ UG-27(c) ] t = 0,75 in E = 1



δ = 90 ° δ = 90 ° δ = / δ = /









δ = 90 ° δ = 90 ° δ = / δ = /



L3 = min [ h , 2.5t , 2.5t i ] = 0 in 0 in

L5 = min [ UG-40 (b) , t p , length available ] 3,125 in 3,125 in

Area available (in²) :Longitudinal plane : θ = 0° Circumferential plane : θ = 90°

δ = 90 ° δ = 90 ° δ = / δ = /


A2 = L2 (t n-t rn) fr 2 0,667 0,667

A3 = L3 t i f r 2 0 0

A41 = leg 412 /2 f r 3 0,065 0,065

A42 = leg 42 /2 f r 4 0,045 0,045

A43 = leg 432 /2 f r 2 0 0

A5 = L5 t e f r 4 1,082 1,082

A1 + A2 + A3 + A41 + A42 + A43 + A5 =3,151 ≥ A/2 3,151 ≥ A/2

6,303 ≥ A The opening is adequately reinforced per UG-37.



t c,inner min(¼ in. (6 mm),0.7×t min) = 0,25 in


t c,outer 0.5 × t min = 0,1875 in

t in = min(¾ in. (19 mm). , t e , t )0.7×leg 42 = 0,2188 in

t c min(¼ in. (6 mm),0.7×t min) = /
















s = π/4 × ( Don − t ) × t × 0.70 × S = 55 899 lbf


δ = 90 ° δ = 90 ° δ = δ =




Path 3-3 : W 3-3 = ( A2+ A3+ 5+ 41+ A42+ A43+t n t r1) S v 31 948 lbf 31 948 lbf


δ = 90 ° δ = 90 ° δ = δ =





t = 0,75

t c,inner

t c

leg 41

leg 43

t e = 0,375

t c,outer

leg 42

t n = 0,5

1 3

2

32

1



L-7.4


ASME VIII DIV.1

Nozzle Self Reinforcing (3,25×2 γ = 25 °) on Shell (No. 1) Set-on



Joint efficiency : 1 E 1 = 1 Corrosion + tolerance : Cav = 0,063 in Allowable stress : S = 11 400 psi

Ext. Diameter : Do = 100 in Nominal thickness : 2 in Tolerance for seamless pipe : /

Nozzle Material : SA105 Allowable stress : S n = 12 000 psi

Joint efficiency : 1 Corrosion + tolerance : Can = 0,063 in Tolerance for seamless pipe : /




Flange Material : SA105 Type : LN

Rating : (ASME B16.5) 600 height : 3.25 in NPS 16"


Height : 2 in Width : 3,25 in Ext. Diameter : Dop = 26 in







t a = t rn+Can= 0,354 in ; t b1 = max[t UG-27 , UG-16(b)]+Cav= 1,895 in ; t b3 = Table UG-45+Can= 0,391 in










t x = thickness of selfreinforcing 4,9375 in

L = height of selfreinforcing 2 in 2 in

Configuration of the reinforcement : sketch (e-1) sketch (e-1)

t e = thickness or height of the reinforcement 3.5 in 3.5 in




t r = 1,8328 in [ UG-27(c) ] t = 1,9375 in E = 1



δ = 90 ° δ = 90 ° δ = / δ = /









δ = 90 ° δ = 90 ° δ = / δ = /



L3 = min [ h , 2.5t , 2.5t i ] = 0 in 0 in

L5 = min [ UG-40 (c) , t e , height available ] 3.5 in 3.5 in


δ = 90 ° δ = 90 ° δ = / δ = /


A2 = L2 (t n-t rn) fr 2 6,761 6,761

A3 = L3 t i f r 2 0 0

A41 = leg 412 /2 f r 2 0,281 0,281

A42 = leg 42 /2 f r 4 0 0

A43 = leg 432 /2 f r 2 0 0

A5 = L5 (t x-t n) f r 2 8,937 8,937

A1 + A2 + A3 + A41 + A42 + A43 + A5 =16,824 ≥ /2 16,824 ≥ A/2


Weld sizes check UW-16(c). opening 1 [ in operation Int.P. ] FIG. UW-16.1(a) full penetration weld


t c min(¼ in. (6 mm),0.7×t min)= 0,25 in





t n = 1,688

t = 1,937 t c

leg 41



1.75

1 2

2

3.25

2 4

4 8 '

1 3

. 7 1 9 5 7

3 . 7

1 9 5 7

Ø16

Ø26

1

t e limited to self reinforcement height : 2 +1.5 = 3.5 inCalculation of A5 area :a) part of cylindrical reinforcement : 3.25 x 2 = 6.5 in2 b) part of cone : 3.25 x 1.5 /2 = 2.4375 in2

A5 = 6.5 +2.4375 = 8.9375 in2

t e

ccording to FIG UG-40 (e-1 )

Comments



L-7.5Opening 1 [ in operation Int.P. ]

ASME VIII DIV.1





Ext. Diameter : Do = 87 in Nominal thickness : 2 in Tolerance for seamless pipe : /


Joint efficiency : 1 Corrosion + tolerance : Can = 0,25 in Tolerance for seamless pipe : /

Ext. Diameter : Don = 16 in Nominal thickness : 0,75 in




Rating : (ASME B16.5) 300 height : 5,12 in NPS 12"


Height : 1,5 in Width : 6,125 in Ext. Diameter : Dop = 28,25 in




t rn = P Ron / (S n+0.4 P ) = 0,2878 in Ron = 8 in E=1 The nozzle neck thickness is adequate per UG-27.


t a = t rn+Can= 0,538 in ; t b1 = max[t UG-27 , UG-16(b)]+Cav= 1,808 in ; t b3 = Table UG-45+Can= 0,578 in






d = diameter of the finished opening 15 in

Rn = radius of the finished opening 7.5 in










t r = 1,5578 in [ UG-27(c) ] t = 1,75 in E = 1

t rn = P Ron / (S n E +0.4 P ) = 0,2878 in Ron = 8 in E = 1 ]


δ = 90 ° δ = 90 ° δ = / δ = /

UG-40 (b) : max [d , Rn+t n+t ] = 15 in 15 in








δ = 90 ° δ = 90 ° δ = / δ = /



L3 = min [ h , 2.5t , 2.5t i ] = 0 in 0 in



δ = 90 ° δ = 90 ° δ = / δ = /


A2 = L2 (t n-t rn) fr 2 0,584 0,584

A3 = L3 t i f r 2 0 0

A41 = leg 412 /2 f r 3 0,07 0,07

A42 = leg 42 /2 f r 4 0,383 0,383

A43 = leg 432 /2 f r 2 0 0

A5 = L5 t e f r 4 9,188 9,188

A1 + A2 + A3 + A41 + A42 + A43 + A5 =11,665 < A/2 11,665 < A/2




t c,inner min[¼ in.(6 mm);0.7×t min]= 0,25 in

t in = min[ ¾ in.(19 mm) ; t e , t n]0.7×leg 41 = 0,2625 in

t c,outer 0.5 × t min = 0,375 in

t in = min[ ¾ in.(19 mm) ; t e , t ]0.7×leg 42 = 0,6125 in

t c min[¼ in. (6 mm);0.7×t min]= /

t in = min[ ¾ in.(19 mm) ; t , t n]0.7×leg 43 − Can = /















s = π/4 × ( Don − t ) × t × 0.70 × S = 58 373 lbf


δ = 90 ° δ = 90 ° δ = δ =






δ = 90 ° δ = 90 ° δ = δ =





t = 1,75

t c,inner

t c

leg 41

leg 43

t e = 1,5

t c,outer

leg 42

t n = 0,49

1 3

2

32

1



Comments

Calculation of t r,n is normally done in AutoPIPE Vessel ® with formula of external diameter and consequently gives adifferent thickness than in ASME example.t rn = 500x8 / (13700+0.4x500) = 0,28777 in

t r = 500x41.75 / (13700-0.6x500) = 1,55784 in

Required area to be reinforced : A = d t r F + 2 t n /cos( β ) t r F (1-f r1 ) = 15x1.55784 = 23.3676 in2 (ASME example value is 23.4 in2 )

Available areas on each side : A1 = 7.5x(1.75-1.55784) = 1.4412 in2 (ASME example value is 2.85/2 =1.425 in2 ) A2 = 2.75x(0.5-0.28777) = 0.5836 in2 (ASME example value is 1.21/2 =0.605 in2 ) A3 = 0 in2 A41 = 0.3752 /2= 0.0703 in2 (ASME example value of the sum of A41 and A42 is 0.906/2 = 0.453 in2 ) A42 = 0.8752 /2= 0.3828 in2 ASME example value of the sum of A41 and A42 is 0.906/2 = 0.453 in2 ) A43 = 0 in2

A1 + A2 + A3 + A41 + A42 = A13 = 2.4779 in2 (ASME example value is 4.97/2 = 2.485 in2 )

A5 = 6.125x1.5 = 9.1875 in2 (ASME example value is 18.4/2 = 9.2 in2 )Total of available areas in AutoPIPE Vessel ® calculation : 2.4779+9.1875 = 11.6654 in2 < 23.3676/2 = 11.6835 in2 Total of available areas in ASME example is : 2.485+9.2 = 11.685 in2 < 23.4/2 =11.7 in2

ASME example conclusion is : “This is equal to the required area; therefore, opening is adequatly reinforced.” ASME calculation is not correct. There are too many rounded values resulting in an important error.



L-7.6


ASME VIII DIV.1

Nozzle without pad on Elliptical Head (No. 2) Set In




Ext. Diameter : Do = 24 in Nominal thickness : 0,188 in Tolerance for seamless pipe : /

Nozzle Material : SA106GRB Allowable stress : S n = 12 000 psi

Joint efficiency : 1 Corrosion + tolerance : Can = 0 in Tolerance for seamless pipe : 12,5 %

Ext. Diameter : Don = 8,625 in Nominal thickness : 0,25 in NPS 8"

External Projection : 9,906 in Internal Projection : 0,6 in Schedule : 20



Rating : / height : /



Weld Outside : leg 41= 0,25 in outer reinforcement : leg 42= / Inside : leg 43= 0,25 in

fr 1 =min(1,S n/S v) = 0.686 fr 2 = min(1,S n/S v) = 0.686 fr 3 = min(1,min(S n, S p)/S v) = 0.686 fr 4 = min(1,S p/S v) = 1

Required thickness of the nozzle neck Appendix 1-1

t rn = P Ron / (S n+0.4 P ) = 0,0536 in Ron = 4,313 in E=1 The nozzle neck thickness is adequate per Appendix 1-1.










t i = thickness of internal projection 0,25 in







h = height of the internal projection 0,6 in 0,6 in


t r = 0.0912 in [ UG-37(a)(c) + UG-27(d) ] t = 0,1875 in K 1 (Table UG-37) = 0,9 E = 1

t rn = P Ron / (S n E +0.4 P ) = 0,0536 in Ron = 4,313 in E = 1


δ = 90 ° δ = 90 ° δ = / δ = /









δ = 90 ° δ = 90 ° δ = / δ = /



L3 = min [ h , 2.5t , 2.5t i ] = 0,469 in 0,469 in



δ = 90 ° δ = 90 ° δ = / δ = /


A2 = L2 (t n-t rn) fr 2 0,063 0,063

A3 = L3 t i f r 2 0,08 0,08

A41 = leg 412 /2 f r 2 0,021 0,021

A42 = leg 42 /2 f r 4 0 0

A43 = leg 432 /2 f r 2 0,021 0,021

A5 = L5 t e f r 4 0 0

A1 + A2 + A3 + A41 + A42 + A43 + A5 =0,567 ≥ A/2 0,567 ≥ A/2


Weld sizes check UW-16(c). opening 1 [ in operation Int.P. ] FIG. UW-16.1(d) full penetration weld


t c,outward min[¼ in.(6 mm);0.7×t min]= 0,1313 in

t in = min[ ¾ in.(19 mm) ; t , t n]0.7×leg 41 = 0,175 in

t c,inward min[¼ in.(6 mm);0.7×t min]= 0,1313 in





Comments AutoPIPE Vessel ® does not take into account weld attachment following FIG.UW-16.1(i) and consequently does notcheck weld loading.

t n = 0,25

t = 0,188

t c,outward

t c,inward

leg 41

leg 43



L-7.7 (nozzle thickness = ½ in.)

Opening A [ in operation Int.P. ]

ASME VIII DIV.1

Nozzle without pad on Shell (No. 1) Set-in

Pressure : P = 1 000 psi Temperature : 150 °F






Ext. Diameter : Don = 5 in Nominal thickness : 0,5 in







Weld outward : leg 41= 0,5 in outer reinforcement : leg 42= / inward : leg 43= /



t rn = P Rn / (S n-0.6 P ) = 0,1389 in Rn = 2 in E=1 The nozzle neck thickness is adequate per UG-27.






angle of each side / vessel wall : δ = 90 ° δ = 90 ° δ = 43,34 ° δ = 136,66 °

β = deflection angle / normal line 0 ° 46,66 °

d = diameter of the finished opening 4 in 6,366 in

Rn = radius of the finished opening 2 in 3,183 in

t i = thickness of internal projection / /

t p = width of reinforcing pad / /

t x = thickness of selfreinforcing / /

L = height of selfreinforcing / / / /

Configuration of the reinforcement : / / / /

t e = thickness or height of the reinforcement 0 in 0 in 0 in 0 in

t n = thickness of nozzle 0,5 in 0,5 in 0,5 in 0,5 in

h = height of the internal projection 0 in 0 in 0 in 0 in

Reinforcement checking UG-37 opening A [ in operation Int.P. ] Required thicknesses UG-37(a)

t r = 1.1364 in [ UG-27(c) ] t = 1,5 in E = 1

t rn = P Rn / (S n E -0.6 P ) = 0,1389 in Rn = 2 in E = 1


δ = 90 ° δ = 90 ° δ = 43,34 ° δ = 136,66 °

UG-40 (b) : max [d , Rn+t n+t ] = 4 in 4 in 6,366 in 6,366 in

UG-40 (c) : min [ 2.5t , 2.5t n+t e ] = 1,25 in 1,25 in 1,25 in 1,25 in


F = Correction factor FIG.UG-37 1 0,5

A = d t r F + 2 t n/cos( β ) t r F (1- f r 1) 4,545 in² 3,617 in²




δ = 90 ° δ = 90 ° δ = 43,34 ° δ = 136,66 °

L1 = min [ UG-40 (b)-Radius , length available] 2 in 2 in 3,161 in 3,161 in

L2 = min [ UG-40 (c) , height available] 1,25 in 1,25 in 1,25 in 1,25 in

L3 = min [ h , 2.5t , 2.5t i ] = 0 in 0 in 0 in 0 in

L5 = min [ UG-40 (b) , t p , length available ] 0 in 0 in 0 in 0 in


δ = 90 ° δ = 90 ° δ = 43,34 ° δ = 136,66 °

A1 = L1 ( E 1 t -t r F ) - t n/cos( β ) ( E 1 t -t r F ) (1- f r 1) 0,727 0,727 2,946 2,946

A2 = L2 (t n-t rn) fr 2 0,451 0,451 0,451 0,451

A3 = L3 t i f r 2 0 0 0 0

A41 = leg 412 /2 f r 2 0,125 0,125 0,125 0,125

A42 = leg 42 /2 f r 4 0 0 0 0

A43 = leg 432 /2 f r 2 0 0 0 0

A5 = L5 t e f r 4 0 0 0 0

A1 + A2 + A3 + A41 + A42 + A43 + A5 =1,304 < A/2 1,304 < A/2 3,522 ≥ /2 3,522 ≥ /2

2,607 < A 7,044 ≥ A The opening is not adequately reinforced per UG-37.

Weld sizes check UW-16(c). opening A [ in operation Int.P. ] FIG. UW-16.1(d) full penetration weld





t in = min[ ¾ in.(19 mm) ; t , t n]0.7×leg 43 = /




Comments Differences are due to the fact that AutoPIPE Vessel ® calculates “d” using the nominal thickness of the shell (1.5 in.)instead of the calculated value “t r ”.

t r = (1000x15)/(13800-0.6*1000) = 1.1364 in (ASME value is 1.14in) Angles are different and consequently d. AutoPIPE Vessel ® calculation : Rm = 15 + (1.5/2) = 15.75 inα 1 = arc cos((12+2)/15.75) = 27.266°α 2 = arc cos((12-2)/15.75) = 50.586°α = 50.586-27.266 = 23.32°d = 2x15.75x√(1-cos( α /2)2 ) = 6.366 in ASME calculation is approximative taking account the rounding of t r value.

UG parts is based on the use of the internal diameter and figure UG-37.1 shows extra thickness for reinforcementoutside. If thickness is calculated according to Appendix 1-1 ,we can consider probably that extra thickness for reinforcement isinside. Consequently, d may be different for the same geometry following that the thickness is calculated with theinternal or the external diameter.

t n = 0,5

t = 1,5

t c,outward

t c,inward

leg 41

leg 43



L-7.7 (nozzle thickness = 7/8 in.)

Opening A [ in operation Int.P. ]

ASME VIII DIV.1


Pressure : P = 1 000 psi Temperature : 150 °F
















t rn = P Rn / (S n-0.6 P ) = 0,1212 in Rn = 2 in E=1 The nozzle neck thickness is adequate per UG-27.



t UG-45 = max [t a , min [t b3 , t b1 ] ] = 0.245 in



angle of each side / vessel wall : δ = 90 ° δ = 90 ° δ = 43,34 ° δ = 136,66 °

β = deflection angle / normal line 0 ° 46,66 °

d = diameter of the finished opening 4 in 6,366 in

Rn = radius of the finished opening 2 in 3,183 in

t i = thickness of internal projection / /

t p = width of reinforcing pad / /

t x = thickness of selfreinforcing / /

L = height of selfreinforcing / / / /

Configuration of the reinforcement : / / / /

t e = thickness or height of the reinforcement 0 in 0 in 0 in 0 in

t n = thickness of nozzle 0,875 in 0,875 in 0,875 in 0,875 in

h = height of the internal projection 0 in 0 in 0 in 0 in

Reinforcement checking UG-37 opening A [ in operation Int.P. ] Required thicknesses UG-37(a)

t r = 1.1364 in [ UG-27(c) ] t = 1,5 in E = 1

t rn = P Rn / (S n E -0.6 P ) = 0,1212 in Rn = 2 in E = 1


δ = 90 ° δ = 90 ° δ = 43,34 ° δ = 136,66 °

UG-40 (b) : max [d , Rn+t n+t ] = 4,375 in 4,375 in 6,366 in 6,366 in

UG-40 (c) : min [ 2.5t , 2.5t n+t e ] = 2,188 in 2,188 in 2,188 in 2,188 in


F = Correction factor FIG.UG-37 1 0,5

A = d t r F + 2 t n/cos( β ) t r F (1- f r 1) 4,545 in² 3,617 in²




δ = 90 ° δ = 90 ° δ = 43,34 ° δ = 136,66 °

L1 = min [ UG-40 (b)-Radius , length available] 2,375 in 2,375 in 3,161 in 3,161 in

L2 = min [ UG-40 (c) , height available] 2,188 in 2,188 in 2,188 in 2,188 in

L3 = min [ h , 2.5t , 2.5t i ] = 0 in 0 in 0 in 0 in

L5 = min [ UG-40 (b) , t p , length available ] 0 in 0 in 0 in 0 in


δ = 90 ° δ = 90 ° δ = 43,34 ° δ = 136,66 °

A1 = L1 ( E 1 t -t r F ) - t n/cos( β ) ( E 1 t -t r F ) (1- f r 1) 0,864 0,864 2,946 2,946

A2 = L2 (t n-t rn) fr 2 1,649 1,649 1,649 1,649

A3 = L3 t i f r 2 0 0 0 0

A41 = leg 412 /2 f r 2 0,125 0,125 0,125 0,125

A42 = leg 42 /2 f r 4 0 0 0 0

A43 = leg 432 /2 f r 2 0 0 0 0

A5 = L5 t e f r 4 0 0 0 0

A1 + A2 + A3 + A41 + A42 + A43 + A5 =2,638 ≥ A/2 2,638 ≥ A/2 4,719 ≥ /2 4,719 ≥ /2

5,275 ≥ A 9,439 ≥ A The opening is adequately reinforced per UG-37.

Weld sizes check UW-16(c). opening A [ in operation Int.P. ] FIG. UW-16.1(d) full penetration weld









CommentsSame as for L7-7 (nozzle thickness = ½ in)

t n = 0,875

t = 1,5

t c,outward

t c,inward

leg 41

leg 43



L-7.8 + UG-45


ASME VIII DIV.1






Nozzle Material : SA106GRB Allowable stress : S n = 12 000 psi

Joint efficiency : 1 Corrosion + tolerance : Can = 0,125 in Tolerance for seamless pipe : 12,5 %

Ext. Diameter : Don = 10,75 in Nominal thickness : 0,594 in NPS 10"

External Projection : 10,438 in Internal Projection : 0 in Schedule : 80




Reinforcement Material : SA516GR60 Allowable stress : S p = 15 000 psi

Height : 0,5 in Width : 2,75 in Ext. Diameter : Dop = 16,25 in


fr 1 =min(1,S n/S v) = 0.686 fr 2 = min(1,S n/S v) = 0.686 fr 3 = min(1,min(S n, S p)/S v) = 0.686 fr 4 = min(1,S p/S v)= 0.857

Required thickness of the nozzle neck Appendix 1-1

t rn = P Ron / (S n+0.4 P ) = 0,133 in Ron = 5,375 in E=1 The nozzle neck thickness is adequate per Appendix 1-1.



















t r = 0.3659 in [ UG-27(c) ] t = 0,4375 in E = 1

t rn = P Ron / (S n E +0.4 P ) = 0,133 in Ron = 5,375 in E = 1


δ = 90 ° δ = 90 ° δ = / δ = /









δ = 90 ° δ = 90 ° δ = / δ = /



L3 = min [ h , 2.5t , 2.5t i ] = 0 in 0 in



δ = 90 ° δ = 90 ° δ = / δ = /


A2 = L2 (t n-t rn) fr 2 0,252 0,252

A3 = L3 t i f r 2 0 0

A41 = leg 412 /2 f r 3 0,048 0,048

A42 = leg 42 /2 f r 4 0,06 0,06

A43 = leg 432 /2 f r 2 0 0

A5 = L5 t e f r 4 1,179 1,179

A1 + A2 + A3 + A41 + A42 + A43 + A5 =1,88 ≥ /2 1,88 ≥ /2




t c,inner min[¼ in.(6 mm);0.7×t min]= 0,25 in

t in = min[ ¾ in.(19 mm) ; t e , t n]0.7×leg 41 = 0,2625 in

t c,outer 0.5 × t min = 0,2187 in

t in = min[ ¾ in.(19 mm) ; t e , t ]0.7×leg 42 = 0,2625 in

t c min[¼ in. (6 mm);0.7×t min]= /
















s = π/4 × ( Don − t ) × t × 0.70 × S = 31 81 lbf


δ = 90 ° δ = 90 ° δ = δ =






δ = 90 ° δ = 90 ° δ = δ =





t = 0,4372

t c,inner

t c

leg 41

leg 43

t e = 0,5

t c,outer

leg 42

t n = 0,469

1 3

2

32

1



UG 45 NOZZLE NECK THICKNESS

case of pipe nozzleOuter Diameter of nozzle = 10.75 in

Nominal thickness of nozzle = 0.594 in

Pipe (yes =1) = 1

Tolerance (12.5 % nominal thickness for pipe) = 0.07425 in

Tolerance (other than pipe) 0 in

Corrosion allowance = 0.125 in

Design pressure = 300 psi

Allowable stress = 12000 psi

Minimum calculated shell thickness (internal pressure) = 0.3659 in

Minimum calculated shell thickness (external pressure) = 0 in

Standard nozzle thickness = 0.365 in

before release 2010

Thickness according UG45B1 = 0.4909 in (Mini shell thk +CA with E=1)

Thickness according UG16B = 0.1875 in (1/16 in + CA)

Thickness according UG45B2 = in (Ext pressure pres.thk +CA)

Thickness according UG45B4 = 0.444375 in Standard thk * 0.875 + CA

Thickness according UG45A = 0.258706 in (Mini nozzle thk +CA)

Minimum nozzle neck thickness requirement :

Max ( Min (Max ( UG45B1,UG16B,UG45B2 ),UG45B4),UG45A) mini = 0.44438 in

Minimum undertoleranced thickness of nozzle neck = 0.51975 in thickness OK per UG 45

release 2010

ta = 0.258706 in

tb1 = 0.4909 in

tb2 = in

TABLE UG-45 (undertoleranced standard

thickness) = 0.319 in

tb3 = 0.444 in

tb = min(tb3,max(tb1,tb2)) = 0.444 in

tUG45 = 0.444 in

Minimum undertoleranced thickness of nozzle neck 0.51975 in thickness OK per UG 45



Part UHX

General comment ASME examples have not been updated. AutoPIPE Vessel® examples are issued from ASME VIII div.1 2010-2011aConsequently, where W* is used for bolting connection and if W* is null, values of Q2 and resulting values are different.



UHX-20.1.1 Tubesheet Integral with Shell and Channel

Tubesheet, Loading conditions 2 [corroded normal condition].

ASME VIII DIV.1

2010 – 2011a

§[UHX-12]

TubesheetTubes Shell Tubeside

Tubeside Shellside

Pressure P t =140 psi P s=-10 psi

Corrosion ct =0 in c s=0 in 0 in 0 in

Material SA240GR316 SA213TP316 SA240GR316 SA240GR316

Temperature 500 °F (T ’ =/) 500 °F (T t,m=/) 500 °F (T s,m=/) 500 °F (T c’ =/)

Allowable Stress S = 18 000 psi S a = 20 000 psi S t = 18 000 psi S s = 18 000 psi S c = 18 000 psi

Yield Strength S y = 20 000 psi S y,t = 20 000 psi S ,s = 20 000 psi S y,c = 20 000 psi

modulus of elasticity E = 25 800.1 ksi E t = 25 800.1 ksi E s = 25 800.1 ksi E c = 25 800.1 ksi

Poisson’s ratio ν = 0.3 νt = 0.3 ν s = 0.3 νc = 0.3

Diameter A = 12.939 in d t =0.75 in 12.39 in 12.313 in

Nominal thicknesses 0.521 in t t =0.065 in 0.18 in 0.313 in

Tolerance 0 in

pattern Square N t =76 L = / OTL=11.6 in p=1 in

Configuration a

D s = 12.39 in t s = 0.18 in h = 0.521 in

Dc = 12.313 in t c = 0.313 in h p = 0.521 in

h g = 0 in

Extra thickness (periphery) : Tubeside : 0 in Shellside : 0 in

Tubesheet characteristics

Diameter of perforated region D0 = 2r 0+d t = 11.6 in r 0 = 5.425 in Tube expansion depth ratio ρ = l t,x /h = 0 l t,x = 0 in

Effective Tube Pitch

p* =( )[ ]2

4,MIN41

O

O L

D

p D A

p

⋅−

π

= 1.152 in

basic ligament efficiency : µ = ( p- d t )/ p = 0.25 effective ligament efficiency : µ

* = ( p*- d *)/ p* = 0.349

d * = max[(d t -2t t E t / E S t /S ρ ).(d t -2t t )]= 0.75 in

largest center-to-center distance between adjacent tube rows : U L = 2.25 in Unperforated Area : L = 26.1 in²

Effective elastic constants E * = 11 492 530 psi ν * = 0,254 (Fig. UHX-11.3 , Fig. UHX-11.4)

Conditions of applicability

Minimum thickness : TEMA 9th Ed. RCB 7-11 U L ≤ 4 p

Parameters

ρ s = D s/ D0 = 1.068 ρ c = Dc/ D0 = 1.061 M TS = D02/16 [( ρ s-1)( ρ s

2+1) P s-( ρ c-1)( ρ c

2+1) P t ]

β s = ( ) ( ) ssss t t D ⋅+−4 2112 ν = 1.209 in-1 β c = ( ) ( ) cccc t t D ⋅+−4 2

112 ν = 0.914 in-1

k s = β s E s t s3/ 6(1-ν s

2) = 33 304.05 lbf k c = β c E c t c3/ 6(1-ν c

2) = 132 498.6 lbf

λ s = 6 Ds k s [1+h β s+(h2 β s2/2)] / h3 = 31 999 900 psi λ c = 6 Dc k c [1+h β c+(h2 β c

2/2)] / h3 = 110 048 600 psi

δ s = [ D s2 / (4 E s t s)] (1-ν s /2) = 7.024388×10

-6 in

3/lbf δ c = [ Dc

2 / (4 E c t c)] (1-ν c /2) = 3.98953×10

-6 in

3/lbf

ω s = ρ s k s β s δ s (1+h β s) = 0.492109 in2 ω c = ρ c k c β c δ c (1+h β c) = 0.757528 in

2

W s = 0.0000 lbf W c = 0.0000 lbf

W m1 s = 0.0000 lbf W m1c = 0.0000 lbf W ax = max[W s , W c ] W m1max = max[W m1 s , W m1c ]

h’ g = max[(h g -ct ),(0)] = 0 in K = A/ D0 = 1.115 F =(1-ν

*)/ E * (λ s+λ c+ E ln K )= 9.404

bending moments

M *= M TS +ω c P t -ω s P s M p = ( M *- D02/32 F ( P s- P t ))/(1+ F )

M = max(| M p|;| M 0|) M 0 = M p + D02/64 (3+ν

*) ( P s- P t )

Case W * P s (psi) P t (psi) M TS (lb.in/in) M * (lb.in/in) M p (lb.in/in) M 0 (lb.in/in) M (lb.in/in) 1

2

3

0

0

0

0

-10

-10

140

0

140

-153.9086

-12.26169

-166.1703

-47.85456

-7.340597

-55.19516

527.5189

37.30293

564.8218

-430.2753

-31.11094

-461.3862

527.5189

37.30293

564.8218

h

Dc

Ds

t c

t s

h p



Bending stresses in the Tubesheet :

σ = 6 M / [µ * (h-h’ g )2] |σ | shall be ≤ S b = 2S

Shear stresses in the Tubesheet :

τ = 1/(4µ ) ( D0/h p) | P s- P t | |τ | shall be ≤ S τ = 0.8S τ = 1/(4µ ) ({4 A p/C p}/h p) | P s- P t | if | P s- P t | > 3.2S µ h p / D0

where : C p = 38.5 in and A p = 68.25 in²

Case σ (psi) S b (psi) τ (psi) S τ (psi)

1

2

3

33 595

2 361

35 756

36 000

36 000

36 000

3 117

223

3 340

14 400

14 400

14 400

Axial tube stress

a0 = D0 /2 x s = 1- N t (d t /2a0)2 xt = 1- N t ((d t -2t t)/2a0)

2 σ t,o = [ P t (1– xt ) – P s (1– x s )] / ( xt – x s )

[UW-20] Tube-To-Tubesheet Welds Fig. (c)

a f = 0.175 in a g = 0.175 in ac= a f + a g F f =min(0.55π a f (d t +0.67a f ) S w ; F t )=2 517.836 lbf

At = π (d t -t t )t t f w= S t / S w = 1 f f =1- F g /( f d F t ) F g =min(0.85π a g (d t +0.67a g ) S w ; F t )=2 517.836 lbf

La = |σ t,o| At F d = min ( Lactual ; F t ) f d = 1.0 S w = min (S t ; S ) F t = π (d t -t t ) S t = 2 517.836 lbf

[UW-20.4] Case (1-3) : Lmax= F t Case (4-7) : Lmax=2 F t La shall be ≤ Lmax

[UW-20.6] ar = 2 [ ( (0.75d t )2 + 1.07t t (d t -t t ) f w f d )

0.5 - 0.75d t ] ac shall not be < max(ar ,t t )Case σ t,o (psi) F d (lbf) f d f f ar (in) La (lbf) Lmax (lbf)

1

2

3

302

32

334

42.267

4.418

46.685

1.0000

1.0000

1.0000

0.0000

0.0000

0.0000

0.082

0.082

0.082

42.267

4.418

46.685

2 517.836

2 517.836

2 517.836

Shell stresses calculation

Membrane

stressσ s,m = ( ) sssss P t Dt D ⋅+4

2

Axial

bending

stress

σ s,b = ( )

−+

+⋅

−+ ts

2

0 p

s

3

0

*

*

ss2

s 3221

16

6 P P

D M

h

h

D

E P k

t s s

β ν δ β

Total axial

stressσ s = | σ s,m | + | σ s,b |

σ s shall be ≤ S mb,s =1,5S s otherwise perform the elastic-plastic calculation procedure if σ s ≤ S PS,s = 3 S ,s =54 000 psi

Case σ s,b (psi) σ s,m (psi) σ s (psi) S mb,s (psi)

1

2

3

-15 851

-1 753

-17 604

0

-170

-170

15 851

1 923

17 774

27 000

27 000

27 000

The shell shall have a uniform thickness adjacent to the

Tubesheet for a minimum length of l s. l s = ss. t D +81 = 2,688 in

Channel stresses calculation

Membrane

stressσ c,m= ( ) t cccc P t Dt D ⋅+4

2 bending

stressσ c,b= ( )

−+

+⋅

−+ ts

2

0

3

0

*

*

2 3221

16

6 P P

D M

h

h

D

E P k

t pc

ccccc

β ν δ β

Total axial

stressσ c = | σ c,m | + | σ c,b |

σ c shall be ≤ S mb,c =1,5S c otherwise perform the elastic-plastic calculation procedure if σ c ≤ S PS,c = 3 S ,c =54 000 psi

Case σ c,b (psi) σ c,m (psi) σ c (psi) S mb,c (psi)

1

2

3

23 785

1 524

25 309

1 343

0

1 343

25 127

1 524

26 651

27 000

27 000

27 000

The channel shall have a uniform thickness adjacent to the

Tubesheet for a minimum length of l c. l c= cc t D +80. = 3,534 in

Error(s) and/or warning(s)

The thickness is acceptable

The stresses in the shell and/or channel integral with tubesheet are acceptable



AutoPIPE Vessel checkingUHX-12.1 Configuration a

cylinder

UHX-12.3 ν factors 0.3 shell Ps = -10 psi

Ds = 12.39 ints = 0.18 in

Gs = 0 inSs = 18000 si

Sy,s = 20000 psi Es = 25800000 psi

Sps,s = psi

channel Pt = 140 psi Dc = 12.313 in

tc = 0.313 inGc = 0 inSc = 18000 psi

Sy,c = 20000 psi Ec = 25800000 si

Sps,c = psi

tubesheet h = 0.521 ind= in

Do = 11.6 in A = 12.939 in G1 = inC = 0 in W* = 0

hg = 0 in equal fillet and groove weldsct = 0 in expansion 0 %cs = 0 inS = 18000 si E = 25800000 psi

tube t = 76 tubes pitch : (90°) p = 1 indt = 0.75 in minimum U radius 1.125 intt = 0.065 in number of pitch for Cp 34

stT = 18000 psiSy,t = 20000 psi EtT = 25800000 psi

UHX-11.5 UL1 = 2.25 in µ = 0.25 d* = 0.75 in

UHX-12.5 LL1 = 11.6 in µ∗ = 0.349167893 p* = 1.15237093 in

AL = 26.1 in2 ρ = 0 h'g = 0 in

ltx = 0 in ro = 5.425 in

Fig. UHX-11.4 0.5 2 0.521

(pitch 90°) h/p = 0.521 α0 = 0.0394 0.0372 0.0393692

α1 = 1.3024 1.0314 1.298606

E*/E = 0.44544631 α2 = -1.1041 -0.6402 -1.0976054

α3 = 2.8714 2.6201 2.8678818

α4 = -2.3994 -2.1929 -2.396509



0.5 1ν* = 0.25392992 β0 = 0.3636 0.3527 0.3631422

β1 = -0.8057 -0.2842 -0.783797

β2 = 2.0463 0.4354 1.9786422

β3 = -2.2902 -0.0901 -2.1977958

β4 = 1.1862 -0.159 1.1297016

case 1 Ps = 0 psi Pt = 140 psicase 2 Ps = -10 si Pt = 0 sicase 3 Ps = -10 psi Pt = 140 psi

Step 1 Do = 11.6 in Step 7

µ = 0.25 case 1 Mp = 527.518947 lb.in / in

µ∗ = 0.34916789 Mo =-

430.275324 lb.in / in

h'g = 0 in M = 527.518947 lb.in / in

case 2 Mp = 37.3029323 lb.in / in

Step 2 ρ s= 1.06810345 o =-

31.1109442 lb.in / in

ρ c= 1.06146552 M = 37.3029323 lb.in / in

case 1 MTs = -153.90887 lb.in / in case 3 Mp = 564.82188 lb.in / in

case 2 MTs = -12.2616896 lb.in / in Mo =-

461.386269 lb.in / in

case 3 MTs = -166.17056 lb.in / in M = 564.82188 lb.in / in

Step 3 h/p = 0.521 Step 8

E*/E = 0.44544631 case 1 σ = 33394.8552 psi

ν * = 0.25392992 σ ≤ 2 S OK

case 2 σ = 2361.4811 psi

Step 4 β s = 1.20851423 β c = 0.914430279 σ ≤ 2 S OK

ks = 33303.9961 kc = 132498.4123 case 3 σ = 35756.3363 psi

λ s = 31999843.6 λ c = 110048441.2 σ ≤ 2 S OK

δ s = 7.0244E-06 δ c = 3.98954E-06

ω s = 0.49210905 ω c= 0.757528882

Step 5 K = 1.11543103 Step 9

F = 9.40444385 case 1 τ = 3117.08253 psi

τ ≤ 0.8 S OK

Step 6 case 2 τ = 222.648752 psi

case 1 M* = -47.8548267 lb.in / in τ ≤ 0.8 S OK

case 2 * = -7.34059906 lb.in / in case 3 τ = 3339.73129 psi

case 3 M* = -55.1954258 lb.in / in τ ≤ 0.8 S OK

Step 10

case 1 σ s,m = 0 psi σ s,b = -15850.60261 psi

σ s = |σ s,m| + |σ s,b| = 15850.60261 psi ≤ 1.5 Ss OK

σ c,m = 1342.72069 psi σ c,b = 23784.67283 psi

σ c = |σ c,m| + |σ c,b| = 25127.39352 psi ≤ 1.5 Sc OK

case 2 σ s,m = -169.619133 psi σ s,b = -1753.412641 psi


σ c,m = 0 psi σ c,b = 1523.888808 psi


case 3 σ s,m = -169.619133 psi σ s,b = -17604.01525 psi


σ c,m = 1342.72069 psi σ c,b = 25308.56164 psi




Comments for discrepancies between AutoPIPE Vessel and ASME example Elasticity modulus for Stainless Steel 316 (16Cr-12Ni-2Mo) has been input because read value from Table TM-1 ofSection II Part D (GroupG at 500°F) is 25,9 106 psi.Some of values in ASME example have been rounded like “parameters”, AutoPIPE Vessel ® calculation gives forexample : ρ s = 12.39/11.6 = 1.0681 ; ρ c = 12.313/11.6 = 1.0615 M TS = 11.6 2 /16 [( 1.0681 -1)(1.06812+1)x-10-(1.0615-1)(1.06152+1)x140] = -166.17026Consequently, results are slightly different.



UHX-20.1.2 Tubesheet gasketed with Shell and Channel


ASME VIII DIV.1

2010 – 2011a

§[UHX-12]


Tubeside Shellside

Pressure P t =135 psi P s=-15 psi

Corrosion ct =0,125 in c s=0 in 0 in 0 in

Material SA285GRCSB111-C70600-

H55SA516GR60 SA516GR60




modulus of elasticity E = 28 300,1 ksi E t = 15 400 ksi E s = 28300,1 ksi E c = 28300,1 ksi


Diameter A = 20 in d t =0.625 in 16.978 in 16.978 in


Tolerance 0 in

pattern Rotated Triangular N t =386 L = / OTL=16.8 in p=0.75 in

Configuration d D s = 16.978 in G s = 19 in h = 1.28 in

Dc = 16.978 in Gc = 19 in h p = 1.28 in

h g = 0 in t s = 0.375 in t c = 0.375 in

Extra thickness (periphery) : Tubeside = 0 in Shellside = 0 in


Diameter of perforated region D0 = 2r 0+d t = 16.8 in r 0 = 8.087 in Tube expansion depth ratio ρ = l t,x /h = 1 l t,x = 1.28 in


p* =( )[ ]2

4,MIN41

O

O L

D

p D A

p

⋅−

π

= 0.805 in

basic ligament efficiency : µ = ( p- d t )/ p = 0,167 effective ligament efficiency : µ

* = ( p*- d *)/ p*

= 0,28

d * = max[(d t -2t t E t / E S t /S ρ ),(d t -2t t )]= 0,58 in

largest center-to-center distance between adjacent tube rows : U L = 1.75 in Unperforated Area : A L = 29,4 in²

Effective elastic constants E = 7 506 417 psi ν * = 0.358 (Fig. UHX-11.3 , Fig. UHX-11.4)


Minimum thickness : TEMA 9th Ed. RCB 7-11 U L ≤ 4

Parameters

ρ s = G s/ D0 = 1.131 ρ c = Gc/ D0 = 1.131 M TS = D02/16 [( ρ s-1)( ρ s

2+1) P s-( ρ c-1)( ρ c

2+1) P t ]

β s = ( ) ( ) ssss t t D ⋅+−4 2112 ν = / β c = ( ) ( ) cccc t t D ⋅+−4 2112 ν = /

k s = β s E s t s3/ 6(1-ν s

2) = 0 lbf k c = β c E c t c3/ 6(1-ν c

2) = 0 lbf

λ s = 6 Ds k s [1+h β s+(h2 β s2/2)] / h3 = 0 psi λ c = 6 Dc k c [1+h β c+(h2 β c

2/2)] / h3 = 0 psi

δ s = [ D s2 / (4 E s t s)] (1-ν s /2) = 0 in3/lbf δ c = [ Dc

2 / (4 E c t c)] (1-ν c /2) = 0 in3/lbf

ω s = ρ s k s β s δ s (1+h β s) = 0 in2 ω c = ρ c k c β c δ c (1+h β c) = 0 in

2

W s = 0.3378×106 lbf W c = 0.3378×106 lbf

W m1 s = 0.0000 lbf W m1c = 0.6534×10 lbf W ax = max[W s , W c ] W m1max = max[W m1 s , W m1c ]


*)/ E * (λ s+λ c+ E ln K )= 0.422

h

Dc Gc

Gs

h p

Ds



bending moments

M *= M TS +ω c P t -ω s P s+((Gc-Gs)/2π D0)W * M p = ( M *- D02/32 F ( P s- P t ))/(1+ F )

M = max(| M p|;| M 0|) M 0 = M p + D02/64 (3+ν

*) ( P s- P t )

Case W P s (psi) P t (psi) M TS (lb.in/in) M (lb.in/in) M p (lb.in/in) M 0 (lb.in/in) M (lb.in/in) 1

2

3

W m1c

W m1 s

W m1max

0

-15

-15

135

0

135

-710.7117

-78.96794

-789.6796

-710.7117

-78.96794

-789.6796

-146.1635

-16.24037

-162.4039

-2 145.09

-238.3433

-2 383.433

2 145.09

238.3433

2 383.433

Bending stresses in the Tubesheet :

σ = 6 / [µ * (h-h’ g )

2] |σ | shall be ≤ S b = 2S

Shear stresses in the Tubesheet :

τ = 1/(4µ ) ( D0/h p) | P s- P t | |τ | shall be ≤ S τ = 0.8S τ = 1/(4µ ) ({4 A p/C p}/h p) | P s- P t | if | P s- P t | > 3.2S µ h p / D0

in which : C p = 54,5 in and A p = 189.103 in²

Case σ (psi) S b (psi) τ (psi) S τ (psi)

1

2

3

28 071

3 119

31 190

31 400

31 400

31 400

2 658

295

2 953

12 560

12 560

12 560

Axial tube stress

a0 = D0 /2 x s = 1- N t (d t /2a0)2 xt = 1- N t ((d t -2t )/2a0)

2 σ t,o = [ P t (1– xt ) – P s (1– x s )] / ( xt – x s )

Tube-to-Tubesheet joint

S min = min( S ; S t ) Welded tubes : S t,j = min[ ( S min g / t t ) ; S t ] (Weld : g = 0 in)

S t,j = 8 000 psi Expanded joint

(0 groove(s) )

no groove : S t,j = 0.5 S min min[ ( l t,x / d t ) ; 1.6 ]

|σ t,o| shall be ≤ S t,j one groove : S t,j = 0.6 S min

with grooves > 1 : S t,j = 0.8 S min



AutoPIPE Vessel checkingUHX-12.1 Configuration d

cylinderUHX-12.3 ν factors 0.3

shell Ps = -15 psi Ds = 16.978 in

ts = 0.375 inGs = 19 inSs = 17100 psi

Sy,s = 28300 psi Es = 28300000 si

Sps,s = psi

channel Pt = 135 psi Dc = 16.978 in

tc = 0.375 inGc = 19 inSc = 17100 psi

Sy,c = 28300 si Ec = 28300000 psi

Sps,c = psitubesheet hnew = 1.405 in

h = 1.28 in

d= in



Do = 16.8 in A = 20 in G1 = inC = 0 in W* = 0

hg = 0 in equal fillet and groove weldsct = 0.125 in expansion 0 %cs = 0 inS = 15700 psi E = 28300000 psi

tube Nt = 386 tubes pitch : (60°) p = 0.75 indt = 0.625 in minimum U radius 0.875 intt = 0.065 in number of pitch for Cp

stT = 10000 psiSy,t = 31600 si ltx = in EtT = 15400000 psi ltx/h = 1

UHX-11.5 UL1 = 1.75 µ = 0.166666667 d* = 0.57994137

UHX-12.5 LL1 = 16.8 µ∗ = 0.279846218 p* = 0.805302124

AL = 29.4 ρ = 1 h'g = 0

ro = 8.0875

Fig. UHX-11.3 0.5 2 1.706666667

(pitch

60°) h/p = 1.706666667 α0 = 0.0054 -0.0029 -0.00127689

α1 = 0.5279 0.2126 0.27425867

E*/E = 0.265244238 α2 = 3.0461 3.9906 3.80589778

α3 = -4.3657 -6.173 -5.81957244

α4 = 1.9435 3.4307 3.13986978

1 2ν* = 0.357565577 β0 = 0.9923 0.9966 0.99533867

β1 = -4.8759 -4.1978 -4.39670933

β2 = 12.3572 9.0478 10.0185573

β3 = -13.7214 -7.9955 -9.67509733

β4 = 5.7629 2.2398 3.27324267

case 1 Ps = 0 Pt = 135case 2 Ps = -15 Pt = 0case 3 Ps = -15 Pt = 135

Step 1 Do = 16.8 Step 7

µ = 0.166666667 case 1 Mp =-

146.1719455

µ∗ = 0.279846218 Mo =-

2145.098611

h'g = 0 = 2145.098611

case 2 Mp =-

16.24132727

Step 2 ρ s= 1.130952381 Mo =-

238.3442902

ρ c= 1.130952381 M = 238.3442902

case 1 MTs = -710.7227679 case 3 Mp =-

162.4132727

case 2 MTs = -78.96919643 Mo = -



2383.442902

case 3 MTs = -789.6919643 M = 2383.442902

Step 3 h/p = 1.706666667 Step 8

E*/E = 0.265244238 case 1 σ = 28071.08054

ν * = 0.357565577 σ ≤ 2 S OK

case 2 σ = 3119.008949

Step 4 β s = 0 β c = 0 σ ≤ 2 S OK

ks = 0 kc = 0 case 3 σ = 31190.08949

λ s = 0 λ c = 0 σ ≤ 2 S OK

δ s = 0 δ c = 0

ω s = 0 ω c = 0

Step 5 K = 1.19047619 Step 9

F = 0.42229237 case 1 τ = 2657.8125

τ ≤ 0.8 S OK

Step 6 case 2 τ = 295.3125

case 1 M* = -710.7227679 τ ≤ 0.8 S OK

case 2 M* = -78.96919643 case 3 τ = 2953.125

case 3 M* = -789.6919643 τ ≤ 0.8 S OK

CommentsSome of values in ASME example have been rounded like “parameters”.Consequently, results are slightly different.



UHX-20.1.4 Tubesheet gasketed with Shell and Integral with Channel extended as a flange


ASME VIII DIV.1

2010 -2011a

§[UHX-12]

TubesheetTubes Shell Channel

Tubeside Shellside

Pressure P t =650 psi P s=650 psi

Corrosion ct =0.125 in c s=0 in 0 in 0 in

Material SA516GR70 SA179 SA516GR70 SA516GR70




modulus of elasticity E = 27 700.1 ksi E t = 27 700.1 ksi E s = 27 700.1 ksi E c = 27 700.1 ksi


Diameter A = 37.25 in d t =0.75 in 31 in 31 in


Tolerance 0 in

pattern Square N t =496 L = / OTL=26,25 in p=1 in

Configuration e t s = 0.625 in t c = 0.625 in

D s = 31 in C = 35 in h = 3.5 in

Dc = 31 in G s = 32.375 in h p = 3.313 in

h g = 0 in Recessed face = 0.188 in t fl = 2.935 in

Extra thickness (periphery) : Tubeside = 0 in Shellside = 0.188 in


Diameter of perforated region D0 = 2r 0+d t = 26.25 in r 0 = 12.75 in Tube expansion depth ratio ρ = l t,x /h = 1 l t,x = 3.5 in


p* =( )[ ]2

4,MIN41

O

O L

D

p D A

p

⋅−

π

= 1,035 in


* = ( p*- d *)/ p* = 0.385

d * = max[(d t -2t t E t / E S t /S ρ ),(d t -2t t )]= 0.636 in

largest center-to-center distance between adjacent tube rows : U L = 1.375 in Unperforated Area : L = 36.094 in

2 Effective elastic constants E * = 12 224 340 psi ν

* = 0.318 (Fig. UHX-11.3 , Fig. UHX-11.4)



Parameters

ρ s = G s/ D0 = 1.233 ρ c = Dc/ D0 = 1.181 M TS = D02/16 [( ρ s-1)( ρ s

2+1) P s-( ρ c-1)( ρ c

2+1) P t ]

β s = ( ) ( ) ssss t t D ⋅+−4 2112 ν = / β c = ( ) ( ) cccc t t D ⋅+−4 2

112 ν = 0.409 in-1

k s = β s E s t s3/ 6(1-ν s

2) = 0 lbf k c = β c E c t c3/ 6(1-ν c

2) = 506 440.2 lbf

λ s = 6 Ds k s [1+h β s+(h2 β s2/2)] / h3 = 0 psi λ c = 6 Dc k c [1+h β c+(h2 β c

2/2)] / h3 = 7 591 004 psi

δ s = [ D s2 / (4 E s t s)] (1-ν s /2) = 0 in

3/lbf δ c = [ Dc

2 / (4 E c t c)] (1-ν c /2) = 11.79564×10

-6 in

3/lbf

ω s = ρ s k s β s δ s (1+h β s) = 0 in2 ω c = ρ c k c β c δ c (1+h β c) = 7.012692 in

2

W s = 0.6560×10 lbf W c = 0.0000 lbf

W m1 s = 0.6404×10 lbf W m1c = 0.0000 lbf W ax = max[W s , W c ] W m1max = max[W m1 s , W m1c ]


*)/ E * (λ s+λ c+ E ln K )= 0.964

bending moments

h

Dc

t fl

G s

C A

h p



M *= M TS +ω c P t -ω s P s+((C -Gs)/2π D0)W * M p = ( M *- D02/32 F ( P s- P t ))/(1+ F )

M = max(| M p|;| M 0|) M 0 = M p + D02/64 (3+ν

*) ( P s- P t )

Case W P s (psi) P t (psi) M TS (lb.in/in) M (lb.in/in) M p (lb.in/in) M 0 (lb.in/in) M (lb.in/in)

1

2

3

0

W m1 s

W m1 s

0

650

650

650

0

650

-12 129.92

16 467.23

4 337.309

-7 571.668

26 660.3

19 088.63

3 017.57

6 699.226

9 716.796

-20 202.12

29 918.92

9 716.796

20 202.12

29 918.92

9 716.796

Bending stress in the Tubesheet :

σ = 6 M / [µ *(h-h’ g )2] |σ | shall be ≤ S b = 2S

Shear stress in the Tubesheet :

τ = 1/(4µ ) ( D0/h p) | P s- P t | |τ | shall be ≤ S τ = 0.8Sτ = 1/(4µ ) ({4 A p/C p}/h p) | P s- P t | si | P s- P t | > 3.2S µ h p / D0

in which : C p = 94.75 in and A p = 457 in2

Case σ (psi) S b (psi) τ (psi) S τ (psi) 1

2

3

25 669

38 016

12 346

40 000

40 000

40 000

5 151

5 151

0

16 000

16 000

16 000

Axial tube stress

a0 = D0 /2 x s = 1- N t (d t /2a0) xt = 1- N t ((d t -2t )/2a0) σ t,o = [ P t (1– xt ) – P s (1– x s )] / ( xt – x s )

Tube-to-Tubesheet joint

S min = min( S ; S t ) Welded tubes : S t,j = min[ ( S min g / t t ) ; S t ] (weld height : g = 0 in)


(0 groove(s) )



grooves > 1 : S t,j = 0.8 S min

Channel stresses calculation

Membrane

Stressσ c,m= ( ) t cccc P t Dt D ⋅+42

bending

stressσ c,b= ( )

−+

+⋅

−+ ts

2

0

3

0

*

*

2 3221

16

6 P P

D M

h

h

D

E P k

t pc

ccccc

β ν δ β

Total axial

stressσ c = | σ c,m | + | σ c,b |

σ c shall be ≤ S mb,c = 1.5S c otherwise perform the elastic-plastic calculation procedure if σ c ≤ S PS,c = 2 S y,c =

65 000 psi

Case σ c,b (psi) σ c,m (psi) σ c (psi) S mb,c (psi) 1

2

3

54 418

-56 611

-2 192

7 901

0

7 901

62 319

56 611

10 093

30 000

30 000

30 000

The channel shall have a uniform thickness adjacent to the

Tubesheet for a minimum length of l c. l c= cct D8.1 = 7.923 in

perform a simplified elastic-plastic calculation :.

Configuration a-b-c : If S mb,s < σ s ≤ S PS,s S PS,s = / S mb s = 1.5S s

E s replace by E s*= s smb s S E σ, ; recalculation of k s , λ s

Configuration a-e-f : when S mb,c < σ c ≤ S PS,c S PS,c = 2 S y,c = 65 000 psi S mb,c = 1.5S c

E c replace by E c*= ccmbc S E σ, ; recalculation of k c , λ c

recalculation of σ , when σ ≤ S b = 2S the thickness is acceptable (h)

Case E s* (psi) E c

* (psi) k s (lbf) k c (lbf) λ s (psi) λ c (psi)

1

2

/

/

19 218 970

20 164 720

/

/

351 380.7

368 671.8

/

/

5 266 826

5 526 000

Case F M p (lb.in/in) M 0 (lb.in/in) M (lb.in/in) σ (psi) S b (psi)

1

2

0.835

0.849

2 241.579

7 988.715

-20 978.12

31 208.41

20 978.12

31 208.41

26 655

39 654

40 000

40 000



Tubesheet flanged extension :

t fl = 2.935 in S = 20 000 psi S a = 20 000 psi G = 32.375 in

W o = W m1 = 640 449 lbf W a = 655 968 lbf hG = 1.313 in

Minimum required thickness : hr =

GS hW

GS hW Go

a

Ga 9.1;

9.1max = 1.589 in

CommentsSome of values in ASME example have been rounded like “parameters”.Consequently, results are slightly different.



UHX-20.2.1 Tubesheet integral with Shell and gasketed with Channel extended as a flange

Tubesheet, Loading conditions 1 [corroded normal condition] (With expansion joint).ASME VIII DIV.1

2010 - 2011a§[UHX-

13]

Tubesheet

Tubes

Shell

TubesideTubeside Shellside far tubesheet near tubesheet

Pressure (psi) P t =400 P s =150

Corrosion ct =0 in c s=0 in 0 in 0 in 0 in

Material SA516GR70 SA214 SA516GR70 SA516GR70 SA516GR60

Temperature 700 °F 700 °F 700 °F 700 °F

Metal temperature T ’ =550 °F T t.m =510 °F T s.m =550 °F T c’ =550 °F

Allowable Stress S = 18 100 psi S t = 10 500 psi / 0.85 S s = 18 100 psi S s.1 = 18 100 psi S c = 15 300 psi

Yield Strength S y = 27 200 psi S y.t = 18 600 psi S y.s = 27 200 psi S y.s.1 = 27 200 psi S y,c = 22 900 psi

modulus of elasticity E = 25 500 ksi E t =25 500 ksi E s =25 500 ksi E s.1 =25 500 ksi E c =25 500 ksi

Nominal thicknesses 3.063 in t t =0.083 in 0.188 in 0.188 in 0,25 in

Diameter A = 40.5 in d t =1 in 34.75 in 34.75 in 38 in

Tolerance 0 in


Length : L = 161.875 in l 1 = 0 in l 1’ = 0 in

pattern Rotated Triangular t =649 OTL=34.25 in =1.25 in

Configuration b t s = 0,188 in t c = 0.25 in

D s = 34.75 in C = 38.875 in h = 3.063 in

Dc = 38 in Gc = 36,8125 in h p = 2.875 in

h g = 0 in Flange face = 0 in t fl = 2.56 in

Extra thickness (periphery) : Tubeside = 0.188 in Shellside = 0 in


Diameter of perforated region D0 = 2r 0+d t = 34.25 in r 0 = 16.625 in Tube expansion depth ratio ρ = l t,x /h = 0.95 l t,x = 2.909 in


p* =( )[ ]2

4,MIN41

O

O L

D

p D A

p

⋅−

π

= 1.25 in


* = ( p*

- d *)/ p* = 0.286


largest center-to-center distance between adjacent tube rows : U L = 0 in Unperforated Area : A L = 0 in


Minimum thickness : CODAP C7.1.4.2 U L ≤ 4 p

h’ g = max[(h g -ct ).(0)] a0 = D0 /2 = 17.125 in

a s = D s/2 = 17.375 in ac = Gc/2 = 18.406 in

ρ s = a s /a0 = 1.015 ρ c = ac /a0 = 1.075 x s = 1- N t (d t /2a0) = 0.447 xt = 1- N t ((d t -2t )/2a0) = 0.615

Shell axial stiffness, tube axial stiffness and axial rigidity of expansion joint :

K s =π ( D s +t s ) /[( L-l 1-l 1’ )/( E s t s )+ (l 1+l 1

’ )/ ( E s,1 t s,1 )]= 3 241 922 lbf/in K J = 11 388 lbf/in

K t = π t t (d t -t t ) E t / L = 37 666.64 lbf/in D J = 385 in Shell-tubes stiffness factor Ratio of expansion joint to shell axial rigidity

K s,t = K s /( N t . K t )= 0.133 J = 1/(1+ K s / K J ) = 0.004

Shell coefficients Channel coefficients

β s = [12(1-ν s2)]1/4 / [( D s+t s,1)t s,1]

1/2 = 0.7102 in-1 β c = [12(1-ν c2)] 1/4 / [( Dc+t c)t c]

1/2 = 0 in-1

k s = β s E s,1 t s,13

/ [6(1-ν s2)] = 21 865.51 lbf k c = β c E c t c

3/ [6(1-ν c

2)] = 0 lbf

λ s = 6 D s k s /h3 (1+h β s+h2

β s2/2) = 879 436 psi λ c = 6 Dc k c /h3

(1+h β c+h2 β c

2/2) = 0 psi

δ s = D s2

/4 E s,1 t ,1 (1-ν s /2) = 53.66955×10

-6 in³/lbf δ c = Dc

2/4 E c t c

(1-ν c /2) = 0 in³/lbf

X a = [24(1-ν 2

) N t E t t t (d t -t t )a02/( E * Lh3

)]1/4 = 3.963

Table UHX-13.1 Z d = 0.0246 Z v = 0.0643 Z m = 0.3715 Z w = 0.0643

K = A/ D0 = 1.182 F = (1-ν *)/ E * (λ s+λ c+ E ln K ) = 0.489

Φ = (1+ν *) F = 0.667 Q1 = ( ρ s-1- Φ Z v) / (1+ Φ Z m) = -0.023

h

D s

t fl

Gc

C A

h p



Q Z 1 = ( Z d + Q1 Z w ) X a4 / 2 = 2.856 Q Z 2 = ( Z v + Q1 Z m ) X a

4 / 2 = 6.888

U = [ Z w+( ρ s-1) Z m] X a4/ (1+Φ Z m) = 13.776

s,m(T s,m -T a) = 0.003504 in/in α s,m,1(T s,m -T a) = 0.003504 in/in t,m(T t,m -T a) = 0.003212 in/in

γ = αt,m(T t,m -T a) L-[ s,m(T s,m -T a)( L- l 1-l 1’ )+ α s,m,1(T s,m -T a)(l 1+l 1

’ )]

ω s = ρ s k s β s δ s (1+h β s) = 2.6851 in² ω c = ρ c k c β c δ c (1+h β c) = 0 in²

ω s*= a0

2 ( ρ s

2-1)( ρ s-1)/4 - ω s = -2.6536 in² ω c

*=a0

2[( ρ c

2+1)( ρ c-1)/4-( ρ s-1)/2]-ω c= 9.6816 in²

γ b = (Gc-C )/ D0 = -0.0602

P s’ = [ x s + 2(1- x s)ν t + 2/ K s,t ( D s / D0)

2ν s – ( ρ s

2-1)/( JK s,t ) – (1- J ) /(2 JK s,t ) [ D J

2-( D s)

2] / D0

2) ] P s

P t ’ = [ xt + 2(1- xt )ν t + 1/( JK s,t ) ] P t P γ = N t K t γ /(π a0

2)

P rim = – U /a02 (ω s

* P s-ω c* P t ) P = – U γ bW * / (2π a0

2)

W = 0.5129×10 lbf W m1 = 0.5125×10 lbfThe effect of radial differential thermal expansion adjacent to the Tubesheet is considered (loading case 4,5,6 et 7).

T s* = T s

‘ = 550 °F T c* = T c

‘ = 550 °F T r = T ‘ = 550 °F s

’(T s

*-T a) = 0.003504 in/in αc

’(T c

*-T a) = 0.003504 in/in α’

(T r -T a) = 0.003504 in/in P s

* = E s 1 t s 1 /a s[ s

’(T s

*-T a)-

’(T r -T a)] P c

* = E ct c/ac[

c’(T c

*-T a)-

’(T r -T a)]

P ω = U / a02(ω s P s

*-ω c P c*)

Equivalent Pressure : P e = JK s,t ( P s’ – P t ’ + P γ + P ω + P + P rim) / (1+ K s,t [Q Z 1+( ρ s-1)Q Z 2])

Case W * P s (psi) P t (psi) γ (in) P s’ (psi) P t

’ (psi) P W (psi) P γ (psi) P rim

(psi) P s* (psi) P c

* (psi) P ω (psi) P e (psi)

1

2

3

4

5

6

7

W m1

0

W m1

W W W W

0

150

150

0

0

150

150

400

0

400

0

400

0

400

0.000

0.000

0.000

-0.047

-0.047

-0.047

-0.047

0

-46 387

-46 387

0

0

-46 387

-46 387

862 001

0

862 001

0

862 001

0

862 001

231

0

231

231

231

231

231

0

0

0

-1 254

-1 254

-1 254

-1 254

182

19

201

0

182

19

201

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

-399

-21

-421

0

-400

-22

-421

Q2 = [(ω s* P s-ω c

* P t ) - (ω s P s*-ω c P t

*) + (γ bW */(2π )) ] / (1+ Φ Z m) Q3 = Q1 + 2Q2/( P e a0

2)


If P e≠ 0 : σ = (1.5 F m / µ *) (2a0 /(h-h g

’ ))2 P e

|σ | shall be ≤ σ allowable If P e = 0 : σ = (6 / µ *) (Q2 /(h-h g

’ ))2


τ = (1/(2µ )) (a0/h p) P e |τ | shall be ≤ τ allowable τ = (1/(4µ )) ({4 A p/C p}/h p) P e If | P e| > 3.2S µ h p / D0

with : C p = 112.5 in , A p = 815.958 in2

Case h’ g (in) Q2 (lbf) Q3 F m σ (psi) σ allowable (psi) τ (psi) τ allowable (psi)

1

2

3

4

5

6

7

0.000

0.000

0.000

0.000

0.000

0.000

0.000

-7 040.685

-319.030

-7 359.715

-3 940.284

-7 044.244

-4 259.314

-7 363.274

0.0976

0.0786

0.0966

56.6267

0.0975

1.2995

0.0965

0.0975

0.0901

0.0971

28.4182

0.0975

0.6705

0.0971

-25 544

-1 270

-26 812

-8 840

-25 570

-9 660

-26 837

1.5 S = 27 150

1.5 S = 27 150

1.5 S = 27 150

3 S = 54 300

3 S = 54 300

3 S = 54 300

3 S = 54 300

-5 948

-320

-6 268

-7

-5 957

-327

-6 277

0.8 S = 14 480

0.8 S = 14 480

0.8 S = 14 480

0.8 S = 14 480

0.8 S = 14 480

0.8 S = 14 480

0.8 S = 14 480



F q = [( Z d +Q3 Z w) X a4 ] / 2 r t = [d t

2+(d t -2t t )

2]

1/2/ 4 = 0.326 in l = 59 in

F s = min{2.0 ; max[(3.25-0.5 F q);1.25]} l t = k l = 59 in k = 1

Maximum permissible buckling stress limit for the tubes :

C t ≤ F t : S tb = min{ [π 2 E t / ( F s F t

2)] ; S t } C t = (2π

2 E t / S ,t )1/2 = 164.5

C t > F t : S tb = min{ [S y,t / F s (1- F t / (2C t ))] ; S t } F t = l t / r t = 181.2

Axial tube stress in the outermost tube row :

Si P e ≠ 0 : σ t,1 = [( P s x s – P t xt ) – P e F t,min ] / ( xt – x s) σ t,2 = [( P s x s – P t xt ) – P e F t,max ] / ( xt – x s)

F t,min et F t,max

suivantTable UHX-13.2

Si P e = 0 : σ t,1 = [( P s x s – P t xt ) – (2Q2 /a02) F t,min ] / ( xt – x s)

σ t,2 = [( P s x s – P t xt ) – (2Q2 /a02) F t,max ] / ( xt – x s)

σ t,max = max( |σ t,1| ; |σ t,2| ) σ t,max shall be ≤ σ t,o allowable Si σ t,1 ou σ t,2 < 0 : σ t,min = min( σ t,1 ; σ t,2 ) |σ t,min| shall be ≤ S t,b

Case F q F s F t,min F t,max σ t,1 (psi) σ t,2 (psi) σ t,o allowable (psi) S t b (psi) 1

2

3

4

5

6

7

3.81

3.66

3.80

451.80

3.81

13.33

3.80

1.35

1.42

1.35

1.25

1.35

1.25

1.35

-1.0817

-1.0116

-1.0781

-213.1982

-1.0813

-5.5205

-1.0777

3.8083

3.6577

3.8006

451.7987

3.8074

13.3336

3.7997

-4 026

269

-3 757

-600

-4 029

-322

-3 760

7 570

865

8 434

1 272

7 581

2 137

8 445

S t = 12 353

S t = 12 353

S t = 12 353

2 S t = 24 706

2 S t = 24 706

2 S t = 24 706

2 S t = 24 706

5 693

5 391

5 677

6 129

5 691

6 129

5 675

Tube-to-tubesheet joint



(0 groove(s) )




Axial membrane stress in shell.

σ s,m = ( a02 [ P e+( ρ s

2 –1)( P s – P t )] + a s2 P t ) / [t s ( D s+ t s)]

|σ s,m| shall be ≤ σ s allowable if σ s,m < 0 : |σ s m| shall be ≤ S ,b S s,b in accordance with [UG-23(b)]

Case t s (in) E s,w σ s,m (psi) σ s allowable (psi) S s,b (psi) 1

2

3

4

5

6

7

0.188

0.188

0.188

0.188

0.188

0.188

0.188

0.85

0.85

0.85

0.85

0.85

0.85

0.85

26

-765

-739

-21

0

-786

-765

S s E s.w = 15 385

S s E s.w = 15 385

S s E s.w = 15 385

3 S s.1 = 54 300

3 S s.1 = 54 300

3 S s.1 = 54 300

3 S s.1 = 54 300

8 770

8 770

8 770

8 770

8 770

8 770

8 770

σ s,b = 6k s /t s 12{ β s[δ s P s+a s

2 P s*/Ε s 1 t s 1]+6(1-ν

*2)/ E *(a0

3/h3

)(1+h β s/2)[ P e( Z v+ Z mQ1)+2 Z mQ2/a02]}

σ s,m = ( a02[ P e+( ρ s

2 –1)( P s – P t )] + a s

2 P t ) / [t s 1 ( D s+ t s 1)]

Stress in the shell integral with the Tubesheet : σ s = |σ s,m | + |σ s,b |

σ s shall be ≤ σ s allowable (otherwise, perform the elastic-plastic calculation procedure)

Case σ s,b (psi) σ s,m (psi) σ s (psi) σ s allowable (psi) 1

2

3

4

5

6

7

-42 440

19 214

-23 227

-10 581

-42 484

8 633

-23 271

26

-765

-739

-21

0

-786

-765

42 466

19 978

23 966

10 602

42 484

9 419

24 035

1,5 S s,1 = 27 150

1,5 S s,1 = 27 150

1,5 S s,1 = 27 150

3 S s,1 = 54 300

3 S s,1 = 54 300

3 S s,1 = 54 300

3 S s,1 = 54 300

The shell shall have a uniform thickness adjacent to the Tubesheet for a minimum

length of l s. l s = 1s,s8.1 t D = 4.595 in

Calculation Procedure for Effect of Plasticity at Tubesheet/Channel or Shell Joint (case 1,2, and 3)

S SP,s,1 = 3 S s,1 = 54 300 psi S SP,c = 3 S c = 45 900 psi

S s* = min[S y,s 1,(S SP,s,1/2)] = 27 150 psi S c

* = min[S y,c,(S SP,c/2)] = 22 900 psi



fact s = min[(1.4-0.4|σ s,b|/S s*),1] factc = min[(1.4-0.4|σ s,c|/S c

*),1]

E s* = E s 1 (fact s) E c

* = E c (factc)

σ s shall be ≤ S SP,s,1 σ c shall be ≤ S SP,c The design is acceptable if: fact s = 1 and factc = 1 (Configuration a) or fact s = 1 (Configurations b and c). Otherwise calculation with reduced values of E s 1 and E c : |σ | shall be ≤.σ allowable

Case fact s factc E s* (psi) E c


1 0.775 / 19 755 950 / 16939.82 / 681 323 /

Case F Φ Q1 Q Z 1 Q Z 2 U Q3 F m 1 0.470 0.641 -0.021 2.865 6.941 13.882 0.0985 0.0979

Case P (psi) P rim (psi) P e (psi) Q2 (lbf) σ (psi) σ allowable (psi)

1 233 183 -399 -7 094.819 -25 641 1,5 S = 27 150

Tubesheet flanged extension :

t fl = 2.56 in S = 18 100 psi S a = 20 000 psi G = 36.813 in

W o = W m1 = 512 473 lbf W a = 512 937 lbf hG = 1.031 in

Minimum required thickness : hr =

GS hW

GS hW Go

a

Ga 9.1;

9.1max = 1.228 in



The stresses in the shell and/or channel integral with tubesheet are acceptable

AutoPIPE Vessel checkingUHX-13.1 Configuration b

cylinderν factors 0.3

UHX-13.3

shell Ps = 150 psi channel Pt = 400 psi Ds = 34.75 in Dc = 38 in

ts = 0.1875 in tc = 0.0825 inGs = 0 in Gc = 36.8125 inSs = 18100 psi Sc = 15300 psi

Sy,s = 27200 psi Sy,c = 22900 psi Es = 25500000 si Ec = 25500000 psiSps = 54300 psi Spc = 45900 psi

tubesheet h = 3.0625 in G1 = ind= in Dj = 38.5 in

Do = 34.25 in expansion 95 % A = 40.5 in h'g = 0 inC = 38.875 in α t,m = 0.0000073

hg = 0 in Tt,m = 510 °Fct = 0 in α s,m = 0.0000073

cs = 0 in Ts,m = 550 °FS = 18100 psi Ta = 70 °F E = 25500000 psi Ap = 815.958 in2hp = 0.1875 in Cp = 112.5 in

tube Nt = 649 tubes pitch : (60°) p = 1.25 indt = 1 in welded ott = 0.083 in Kj = 11388

stT = 12352.94118 psi UL1 = inSy,t = 18600 si Lt = 168 in EtT = 25500000 psi L = 161.875 in

lt = 59 inUHX-13.5 µ = 0.2 d* = 0.892372441 in



UHX-11.5 LL1 = 34.25 in µ∗ = 0.286102047

AL = 0 in2 ρ = 0.95

ltx = 290.9375 in ro = 16.625 in

p* = 1.25 in

Fig. UHX-11.3 2 2 2.45

(pitch 60°) h/p = 2.45 α0 = -0.0029 -0.0029 -0.0029

α1 = 0.2126 0.2126 0.2126

E*/E = 0.262995887 α2 = 3.9906 3.9906 3.9906

α3 = -6.173 -6.173 -6.173

α4 = 3.4307 3.4307 3.4307

2 2ν* = 0.363965607 β0 = 0.9966 0.9966 0.9966

β1 = -4.1978 -4.1978 -4.1978

β2 = 9.0478 9.0478 9.0478

β3 = -7.9955 -7.9955 -7.9955

β4 = 2.2398 2.2398 2.2398

case 1 Ps = 0 psi case 2 Ps = 150 psi Pt = 400 si Pt = 0 psi



case 7 Ps = 150 psi Pt = 400 psi

Step 1 Step 4 K = 1.182481752

ao = 17.125 F = 0.488770192

as = 17.375 φ = 0.666665731

ac = 18.40625 Q1 = -0.022635292

ρ s = 1.01459854 Qz1 = 2.855594459

ρ c = 1.074817518 Qz2 = 6.887968085

xs = 0.446747296 U = 13.77593617

xt = 0.615181762

Step 2 Step 5 γ Ks = 3241928.027 case 1 0 in

Kt = 37666.71098 case 2 0 in

Ks,t = 0.132617541 case 3 0 in

= 0.003500428 case 4 -0.0472675 in

β s = 0.710246514 case 5 -0.0472675 in

ks = 21865.55297 case 6 -0.0472675 in

λ s = 879437.2064 case 7 -0.0472675 in

δ s = 5.36694E-05

β c = 0

kc = 0

λ c = 0

δ c = 0

Step 3 Xa = 3.962989113

(AutoPIPEVessel)

FIG. UHX-13.2 TABLE UHX-13.1 ω s = 2.685050221

Zd = 0.025 0.024609 ω∗ s= -2.653572119

Zv = 0.065 0.064259 ω c= 0

Zw = 0.065 0.064259 ω∗ c= 9.681584045



Zm = 0.37 0.371463 γ b = -0.060218978

Steps 6 - 7 - 8

P's P't P W* Pw Primcase 1 0 862002.1601 0 5.1247E+05 230.719586 181.9141044

case 2 -46387.08793 0 0 0.0000E+00 0 18.69743861

case 3 -46387.08793 862002.1601 0 5.1247E+05 230.719586 200.6115431

case 4 0 0 -1254.163035 5.1290E+05 230.911825 0

case 5 0 862002.1601 -1254.163035 5.1290E+05 230.911825 181.9141044

case 6 -46387.08793 0 -1254.163035 5.1290E+05 230.911825 18.69743861

case 7 -46387.08793 862002.1601 -1254.163035 5.1290E+05 230.911825 200.6115431

Pe Q2 Q3 Fm(AutoPIPE

Vessel)(FIG. UHX-13.3.1

/ 2) TABLE UHX-13.1

case 1 -399.4174331 -7046.192369 0.097673209 0.1 0.09752845

case 2 -21.49555321 -319.2801565 0.078660671 0.09 0.0907486

case 3 -420.9129863 -7365.472525 0.09670226 0.1 0.09714

case 4 -0.474360887 -3943.083603 56.66599298 / 28.41821

case 5 -399.9987514 -7049.475068 0.097554333 0.1 0.0974828

case 6 -21.96991409 -4262.363759 1.300459583 / 0.6705008

case 7 -421.4943046 -7368.755225 0.096590785 0.1 0.09709733

σ τσ ≤ σ

allowable τ ≤ τ

allowable case 1 -25544.51072 -5947.846558 OK : (1.5 Ss) OK : (0.8 Ss)

case 2 -1279.16866 -320.0968249 OK : (1.5 Ss) OK : (0.8 Ss)

case 3 -26812.02863 -6267.943383 OK : (1.5 Ss) OK : (0.8 Ss)

case 4 -8839.848901 -7.063852336 OK : (Sps) OK : (0.8 Ss)

case 5 -25569.71461 -5956.503146 OK : (Sps) OK : (0.8 Ss)

case 6 -9659.772744 -327.1606773 OK : (Sps) OK : (0.8 Ss)

case 7 -26837.26463 -6276.599971 OK : (Sps) OK : (0.8 Ss)

Step 9 Ftmin Ftmax σ t,1 σ t,2 σ t,max σ t,max

(AutoPIPEVessel)

(AutoPIPEVessel) ≤ σ allowable

case 1 -1.081718 3.808283 -4026.074634 7569.839738 7569.839738 OK

case 2 -1.011577 3.657727 268.7554882 864.6505852 864.6505852 OK

case 3 -1.078136 3.800594 -3757.319201 8434.489638 8434.489638 OK

case 4 -213.1982 451.7987 -600.4286979 1272.397727 1272.397727 OK

case 5 -1.081297 3.807379 -4028.808178 7580.836453 7580.836453 OK

case 6 -5.520454 13.33355 -322.2131845 2137.02724 2137.02724 OK

case 7 -1.077741 3.799747 -3760.051717 8445.487086 8445.487086 OK

rt = 0.325533792 in Ft = 181.2407848 Ct = 164.5047675

Wt Fs Stb abs(σ t,min) case 1 1.345484493 5694.418233 4026.074634 OK

case 2 1.420821018 5392.481764 268.7554882 OK

case 3 1.349331844 5678.181731 3757.319201 OK

case 4 1.25 6129.401142 600.4286979 OK

case 5 1.345955535 5692.425363 4028.808178 OK

case 6 1.25 6129.401142 322.2131845 OK

case 7 1.349773558 5676.323542 3760.051717 OK

Steps 10 to 12



σ s,m σ s,b σ s σ s ≤ σ allowable elastic-plastic

case 1 26.08037779 -42455.00525 42481.08563 FALSE Yes

case 2 -764.8182186 19212.84469 19977.66291 OK No

case 3 -738.7378408 -23242.16056 23980.8984 OK No

case 4 -21.23620628 -10588.37904 10609.61525 OK /

case 5 0.055895981 -42498.12134 42498.17723 OK /

case 6 -786.0544249 8624.465651 9410.520076 OK /

case 7 -764.7623226 -23285.27665 24050.03897 OK /

ls = 4.594630018 in

σ c,m σ c,b σ c σ c ≤ σ

allowable

case 1case 2case 3case 4case 5case 6case 7

Elastic- lastic calculation

Ss* = 27150 Sc* = 22900 facts factc Es* Ec* ks kc

case 1 0.774511893 19750053.27 16935.13082

case 2 0

case 3 0

λ s λ c F φ Q1 Qz1case 1 681134.5756 0 0.469963174 0.641013605 -0.021314095 2.866064851

case 2 0

case 3 0

Qz2 U Fm Pw ω s ω∗ s (AutoPIPE

Vessel)case 1 6.948494443 13.8819591 0.0979 232.4952597 2.07960333 -2.048125228

case 2case 3

ω c ω∗ c Prim Pe Q2 σ

case 1 0 9.681584045 183.3141593 -399.413858 -7100.248932 25641.59708

case 2case 3



UHX-20.2.3 Fixed Tubesheet Exchanger , Configuration a

Tubesheet, Loading conditions 1 [corroded normal condition] (Without expansion joint).ASME VIII DIV.1

2010 – 2011a §[UHX-

13]

Tubesheet

Tubes

Shell

TubesideTubeside Shellside far tubesheet near tubesheet

Pressure (psi) P t =200 P s =325

Corrosion ct =0 in c s=0 in 0 in 0 in 0 in

Material SA240GR304L SA249TP304L SA240GR304L SA240GR304L SA516GR70

Temperature 400 °F 400 °F 400 °F 300 °F

Metal temperature T ’ =151 °F T t,m =113 °F T s,m =151 °F T c’ =151 °F

Allowable Stress S = 15 800 psi S t = 13 400 psi / 0,85 S s = 15 800 psi S s,1 = 15 800 psi S c = 20 000 psi

Yield Strength S y = 17 500 psi S y,t = 17 500 psi S y,s = 17 500 psi S y,s,1 = 17 500 psi S y,c = 33 600 psi

modulus of elasticity E = 26 400,1 ksi E t =26 400,1 ksi E s =26 400,1 ksi E s,1 =26 400,1 ksi E c =28 300,1 ksi

Nominal thicknesses 1.375 in t t =0.049 in 0.563 in 0.563 in 0.375 in

Diameter A = 43.125 in d t =1 in 42 in 42 in 42.125 in

Tolerance 0 in


Length : L = 237.25 in l 1 = 0 in l 1’ = 0 in

pattern Rotated Triangular t =955 OTL=41.25 in =1.25 in

Configuration a

D s = 42 in t s = 0.563 in h = 1.375 in

Dc = 42.125 in t c = 0.375 in h p = 1.375 in

h g = 0 in

Extra thickness (periphery) : Tubeside : 0 in Shellside : 0 in




p* =( )[ ]2

4,MIN41

O

O L

D

p D A

p

⋅−

π

= 1.25 in


* = ( p*

- d *)/ p* = 0.271


largest center-to-center distance between adjacent tube rows : U L = 0 in Unperforated Area : A L = 0 in²


Minimum thickness : CODAP C7.1.4.2 U L ≤ 4 p

h g = max[(h g -ct ),(0)] a0 = D0 /2 = 20.625 in

a s = D s/2 = 21 in ac = Dc/2 = 21.063 in

ρ s = a s /a0 = 1.018 ρ c = ac /a0 = 1.021

x s = 1- N t (d t /2a0)2 = 0.439 xt = 1- N t ((d t -2t )/2a0)

2 = 0.543

Shell axial stiffness, tube axial stiffness and axial rigidity of expansion bellows :

K s =π ( D s +t s ) /[( L-l 1-l 1’ )/( E s t s )+ (l 1+l 1

’ )/ ( E s,1 t s,1 )]= 8 369 473 lbf/in K J = 0 lbf/in

K t = π t t (d t -t t ) E t / L = 16 290.14 lbf/in D J = / Shell-tubes stiffness factor Ratio of expansion bellows to shell axial rigidity

K s,t = K s /( N t . K t )= 0.538 J = 1/(1+ K s / K J ) = 1

Shell coefficients Channel coefficients

β s = [12(1-ν s2)]1/4 / [( D s+t s,1)t s,1]

1/2 = 0.3715 in-1 β c = [12(1-ν c2)] 1/4 / [( Dc+t c)t c]

1/2 = 0.4554 in-1

k s = β s E s,1 t s,13 / [6(1-ν s

2)] = 319 713.1 lbf k c = β c E c t c3 / [6(1-ν c

2)] = 124 461.1 lbf

λ s = 6 D s k s /h3 (1+h β s+h2

β s2/2) = 50 868 080 psi λ c = 6 Dc k c /h3

(1+h β c+h2 β c

2/2) = 22 049 150 psi

δ s = D s2

/4 E s,1 t s,1 (1-ν s /2) = 25,24237×10

-6 in³/lbf δ c = Dc

2/4 E c t c

(1-ν c /2) = 35,53205×10

-6 in³/lbf

X a = [24(1-ν *2) N t E t t t (d t -t t )a0

2/( E * Lh3)]1/4 = 6.976


h

Dc

Ds

t c

t s

h p



K = A/ D0 = 1.045 F = (1-ν * )/ E * (λ s+λ c + E ln K ) = 6.732

Φ = (1+ν * ) F = 9.024 Q1 = ( ρ s-1-Φ Z v) / (1+Φ Z m) = -0.059

Q Z 1 = ( Z d + Q1 Z w ) X a4 / 2 = 3.754 Q Z 2 = ( Z v + Q1 Z m ) X a

4 / 2 = 10.15

U = [ Z w+( ρ s-1) Z m] X a4/ (1+Φ Z m) = 20.301

s,m(T s,m -T a) = 0.000713 in/in α s,m,1(T s,m -T a) = 0.000713 in/in t,m(T t,m -T a) = 0.000372 in/in

γ = αt,m(T t,m -T a) L-[ s,m(T s,m -T a)( L- l 1-l 1’ )+ s,m,1(T s,m -T a)(l 1+l 1

’ )]

ω s = ρ s k s β s δ s (1+h β s) = 4.6123 in² ω c = ρ c k c β c δ c (1+h β c ) = 3.344 in²

ω s*= a0

2 ( ρ s -1)( ρ s-1)/4 - ω s = -4.5413 in² ω c*=a0

2[( ρ c

2+1)( ρ c-1)/4-( ρ s-1)/2]-ω c= -2.6027 in²

γ b = 0

P s’ = [ x s + 2(1- x s)ν t + 2/ K s,t ( D s / D0)

2ν s – ( ρ s

2-1)/( JK s,t ) – (1- J ) /(2 JK s,t ) [ D J

2-( D s)

2] / D0

2) ] P s

P t ’ = [ xt + 2(1- xt )ν t + 1/( JK s,t ) ] P t P γ = N t K t γ /(π a0

2)

P rim = – U /a02 (ω s

* P s-ω c* P t ) P = – U γ bW * / (2π a0

2)

W m1 = 0.0000 lbf W = 0.0000 lbfThe effect of radial differential thermal expansion adjacent to the Tubesheet is considered (loading case 4,5,6 et 7).

T s* = T s

‘ = 151 °F T c

* = T c

‘ = 151 °F T r = T ‘ = 151 °F

s’ (T s

*-T a) = 0.000713 in/in αc’ (T c

*-T a) = 0.000535 in/in α’ (T r -T a) = 0.000713 in/in P s

* = E s 1 t s 1 /a s[ s’ (T s

*-T a)- ’ (T r -T a)] P c

* = E ct c/ac[

c’ (T c

*-T a)- ’ (T r -T a)]

P ω = U / a02(ω s P s

*-ω c P c

*)

Equivalent Pressure : P e = JK s,t ( P s’ – P t ’ + P γ + P ω + P + P rim) / (1+ JK s,t [Q Z 1+( ρ s-1)Q Z 2])

Case W * P s (psi) P t (psi) γ (in) P s’ (psi) P t

’ (psi) P W (psi) P γ (psi) P rim

(psi) P s* (psi) P c

* (psi) P ω (psi) P e (psi)

1

2

3

4

5

6

7

0

0

0

0

0

0

0

0

325

325

0

0

325

325

200

0

200

0

200

0

200

0.000

0.000

0.000

-0.081

-0.081

-0.081

-0.081

0

606

606

0

0

606

606

535

0

535

0

535

0

535

0

0

0

0

0

0

0

0

0

0

-942

-942

-942

-942

-25

70

46

0

-25

70

46

0

0

0

0

0

0

0

0

0

0

-90

-90

-90

-90

0

0

0

14

14

14

14

-97

117

20

-160

-257

-43

-140

Q2 = [(ω s* P s-ω c

* P t ) - (ω s P s*-ω c P t

*) + (γ bW */(2π )) ] / (1+ Φ Z m) Q3 = Q1 + 2Q2/( P e a02)


If P e≠ 0 : σ = (1.5 F m / µ *) (2a0 /(h-h g

’))

2 P e

|σ | shall be ≤ σ allowable If P e = 0 : σ = (6 / µ *) (Q2 /(h-h g

’ ))2


τ = (1/(2µ )) (a0/h p) P e |τ | shall be ≤ τ allowable τ = (1/(4µ )) ({4 A p/C p}/h p) P e If | P e| > 3.2S µ h p / D0

with : C p = 135 in , A p = 1 217.848 in2 Case h’

g (in) Q2 (lbf) Q3 F m σ (psi) σ allowable (psi) τ (psi) τ allowable (psi) 1

2

3

4

5

6

7

0.000

0.000

0.000

0.000

0.000

0.000

0.000

180.999

-513.192

-332.193

-104.399

76.600

-617.591

-436.592

-0.0680

-0.0798

-0.1372

-0.0561

-0.0606

0.0079

-0.0445

0.0340

0.0399

0.0686

0.0280

0.0303

0.0374

0.0222

-16 349

23 184

6 836

-22 337

-38 685

-8 069

-15 501

1.5 S = 23 700

1.5 S = 23 700

1.5 S = 23 700

3 S = 47 400

3 S = 47 400

3 S = 47 400

3 S = 47 400

-3 623

4 373

750

-5 998

-9 621

-1 625

-5 248

0.8 S = 12 640

0.8 S = 12 640

0.8 S = 12 640

0.8 S = 12 640

0.8 S = 12 640

0.8 S = 12 640

0.8 S = 12 640



F q = [( Z d +Q3 Z w) X a4 ] / 2 r t = [d t

2+(d t -2t t )

2]

1/2/ 4 = 0.337 in l = 48 in

F s = min{2.0 ; max[(3.25-0.5 F q);1.25]} l t = k l = 48 in k = 1



2)] ; S t } C t = (2π

2 E t / S y,t )1/2 = 172.6



Si P e ≠ 0 : σ t,1 = [( P s x s – P t xt ) – P e F t,min ] / ( xt – x s) σ t,2 = [( P s x s – P t xt ) – P e F t,max ] / ( xt – x s)

F t,min et F t,max

suivantTable UHX-13.2

Si P e = 0 : σ t,1 = [( P s x s – P t xt ) – (2Q2 /a02) F t,min ] / ( xt – x s)


σ t,max = max( |σ t,1| ; |σ t,2| ) σ t,max shall be ≤ σ t,o allowable Si σ t,1 ou σ t,2 < 0 : σ t,min = min( σ t,1 ; σ t,2 ) |σ t,min| shall be ≤ S t,b

Case F q F s F t,min F t,max σ t,1 (psi) σ t,2 (psi) σ t,o allowable (psi) S t b (psi) 1

2

3

4

5

6

7

3.54

3.24

1.82

3.83

3.72

5.41

4.12

1.48

1.63

2.00

1.34

1.39

1.25

1.25

-0.2703

-0.2432

-0.1926

-0.3031

-0.2900

-0.5616

-0.3404

3.5361

3.2425

2.1252

3.8297

3.7191

5.4102

4.1164

-1 288

1 634

361

-463

-1 750

1 130

-131

2 227

-2 252

-82

5 855

8 082

3 604

5 831

S t = 15 765

S t = 15 765

S t = 15 765

2 S t = 31 529

2 S t = 31 529

2 S t = 31 529

2 S t = 31 529

6 931

6 306

5 135

7 693

7 387

8 217

8 217

[A-2] Allowable Loads for Tube-To-Tubesheet Joints (Joint Type = i )

a = 0 in S y,m = 22 700 psi P 0 = 0 psi P T = 0 psi f r = 0.7 S y,t,m = 24 400 psi f e = 1 S a = k S t At = π (d t -t t )t t La = |σ t,o| At f y = min ( 1 ; S ym / S y,t,m ) = 0.93 f T = ( P 0 + P T ) / P 0 = 1 La shall be ≤ Lmax = At S a f e f r f y f T Case σ t,o (psi) k S a (psi) La (lbf) Lmax (lbf)

1

2

3

4

5

6

7

2 227

2 252

361

5 855

8 082

3 604

5 831

1

1

1

2

2

2

2

15 765

15 765

15 765

31 529

31 529

31 529

31 529

325.991

329.616

52.862

857.190

1 183.181

527.574

853.564

1 502.956

1 502.956

1 502.956

3 005.911

3 005.911

3 005.911

3 005.911

Axial membrane stress in shell.

σ s,m = ( a02[ P e+( ρ s

2 –1)( P s – P t )] + a s

2 P t ) / [t s (D s+ t s)]|σ s,m| shall be ≤ σ s allowable if σ s,m < 0 : |σ t,min| shall be ≤ S s,b S s,b in accordance with [UG-23(b)]

case t s (in) E s,w σ s,m (psi) σ s allowable (psi) σ s,b (psi)

1

2

3

4

5

6

7

0,562

0,562

0,562

0,562

0,562

0,562

0,562

0,85

0,85

0,85

0,85

0,85

0,85

0,85

1 837

2 284

4 121

-2 842

-1 005

-558

1 279

S s E s,w = 13 430

S s E s,w = 13 430

S s E s,w = 13 430

2 S s = 31 600

2 S s = 31 600

2 S s = 31 600

2 S s = 31 600

6 773

6 773

6 773

6 773

6 773

6 773

6 773



σ s,b = 6k s /t s 12{ β s[δ s P s+a s

2 P s*/Ε s 1 t s 1]+6(1-ν *2)/ E *(a0

3/h3)(1+h β s/2)[ P e( Z v+ Z mQ1)+2 Z mQ2/a02]}

σ s,m = ( a02[ P e+( ρ s

2 –1)( P s – P t )] + a s

2 P t ) / [t s,1 ( D s+ t s,1)]

Stress in the shell integral with the Tubesheet : σ s = |σ s,m | + |σ s,b |

σ s shall be ≤ σ s allowable (otherwise, perform the elastic-plastic calculation procedure)

Case σ s,b (psi) σ s,m (psi) σ s (psi) σ s allowable (psi) 1

2

3

4

5

6

7

-12 223

27 825

15 603

-27 649

-39 872

176

-12 047

1 837

2 284

4 121

-2 842

-1 005

-558

1 279

14 060

30 109

19 724

30 491

40 877

734

13 326

1.5 S s,1 = 23 700

1.5 S s,1 = 23 700

1.5 S s,1 = 23 700

3 S s,1 = 47 400

3 S s,1 = 47 400

3 S s,1 = 47 400

3 S s,1 = 47 400


length of l s.l s = 1s,s8.1 t D = 8.749 in

σ c b = 6k c /t c2{ β c[δ c P t +ac

2 P c*/Ε c t c]–6(1-ν

*2)/ E *(a0

3/h3

)(1+h β c /2)[ P e( Z v+ Z mQ1)+2 Z mQ2/a02]}

σ c,m = ac2 P t ) / [t c ( Dc+ t c)]

Stress in the channel integral with the tubesheet : σ c = |σ c m | + |σ c,b |

σ c shall be ≤ σ c allowable (otherwise, perform the elastic-plastic calculation procedure)

case σ c,b (psi) σ c,m (psi) σ c (psi) σ c allowable (psi) 1

2

3

4

5

6

7

28 381

-8 563

19 818

16 254

44 635

7 691

36 072

5 567

0

5 567

0

5 567

0

5 567

33 948

8 563

25 385

16 254

50 202

7 691

41 640

1,5 S c = 30 000

1,5 S c = 30 000

1,5 S c = 30 000

3 S ,c = 60 000

3 S ,c = 60 000

3 S ,c = 60 000

3 S c = 60 000


length of l c. l c = cct D8.1 = 7.154 in

Calculation Procedure for Effect of Plasticity at Tubesheet/Channel or Shell Joint (case 1,2, and 3)

S SP,s,1 = 3 S s,1 = 47 400 psi S SP,c = 3 S c = 60 000 psi

S s = min[S y,s 1,(S SP,s,1/2)] = 17 500 psi S c = min[S y,c,(S SP,c/2)] = 30 000 psi

fact s = min[(1.4-0.4|σ s,b|/S s*),1] factc = min[(1.4-0.4|σ s,c|/S c

*),1]

E s* = E s 1 (fact s) E c

* = E c (factc)

σ s shall be ≤ S SP,s,1 σ c shall be ≤ S SP,c The design is acceptable if: fact s = 1 And factc = 1 (Configuration a) or fact s = 1 (Configurations b and c). Otherwise calculation with reduced values of E s 1 and E c : |σ | shall be ≤.σ allowable

Case fact s factc E s* (psi) E c


1

2

1.000

0.764

1.000

1.000

26 400 060

20 169 460

28 300 050

28 300 050

319 713.2

244 258.7

124 461.1

124 461.1

50 868 080

38 862 860

22 049 150

22 049 150

Case F Φ Q1 Q Z 1 Q Z 2 U Q3 F m 1

2

6.732

5.641

9.024

7.561

-0.059

-0.054

3.754

3.874

10.150

11.350

20.301

22.700

-0.0680

-0.0777

0.0340

0.0389

Case P (psi) P rim (psi) P e (psi) Q2 (lbf) σ (psi) σ allowable (psi)

1

2

0

0

-25

79

-97

115

180.999

-573.843

-16 349

22 292

1.5 S = 23 700

1.5 S = 23 700



AutoPIPE Vessel checkingUHX-13.1 Configuration a

cylinderν factors 0.3

UHX-13.3

shell Ps = 325 psi channel Pt = 200 psi Ds = 42 in Dc = 42.125 in

ts = 0.5625 in tc = 0.375 inGs = 0 in Gc = 0 inSs = 15800 psi Sc = 20000 psi

Sy,s = 17500 psi Sy,c = 33600 psi Es = 26400000 psi Ec = 28300000 psiSps = 47400 psi Spc = 60000 psi

tubesheet hnew = 1.375 inh = 1.375 in G1 = ind= in Dj = 38.5 in

Do = 41.25 in expansion 90 % A = 43.125 in h'g = 0 inC = 0 in at,m = 0.000008652

hg = 0 in Tt,m = 113 °Fct = 0 in as,m = 0.000008802

cs = 0 in Ts,m = 151 °FS = 15800 psi Ta = 70 °F E = 26400000 psi Ap = 1217.848 in2hg = 0 in Cp = 135 in

tube Nt = 955 tubes pitch : 60 °dt = 1 in p = 1.25 intt = 0.049 in welded No

stT =15764.7058

8 psi Kj = 0Sy,t = 17500 psi UL1 = in EtT = 26400000 psi Lt = 240 in

lt = 48 in L = 237.25 in

ltx = 1.25 in Effect of Radial Differential Thermal Expansion Adjacent to the tubesheet : Yes

α 's = 0.000008802α 'c = 0.000006602α ' = 0.000008802

UHX-13.5 µ = 0.2 d* = 0.911108103

UHX-11.5 LL1 = 41.25 µ∗ = 0.271113518

AL = 0 ρ = 0.909090909

ltx = 1.25 ro = 20.125

p* = 1.25

Fig. UHX-11.3 0.5 2 1.1

h/p = 1.1 α0 = 0.0054 -0.0029 0.00208

α1 = 0.5279 0.2126 0.40178

E*/E = 0.274983607 α2 = 3.0461 3.9906 3.4239

α3 = -4.3657 -6.173 -5.08862

α4 = 1.9435 3.4307 2.53838

1 2ν* = 0.340360504 β0 = 0.9923 0.9966 0.99273

β1 = -4.8759 -4.1978 -4.80809



β2 = 12.3572 9.0478 12.02626

β3 = -13.7214 -7.9955 -13.14881

β4 = 5.7629 2.2398 5.41059

case 1 Ps = 0 case 2 Ps = 325 Pt = 200 Pt = 0



case 7 Ps = 325 Pt = 200

Step 1 Step 4 K = 1.045454545

ao = 20.625 F = 6.732232476

as = 21 φ = 9.023618515

ac = 21.0625 Q1 = -0.059159042

ρ s= 1.018181818 Qz1 = 3.753824734

ρ c= 1.021212121 Qz2 = 10.15043261

xs = 0.438751148 U = 20.30086522

xt = 0.543365689

Step 2 Step 5 γ Ks = 8369456.077 case 1 0

Kt = 16290.11595 case 2 0

Ks,t = 0.537984406 case 3 0

= 1 case 4 -0.08088469

β s = 0.371518831 case 5 -0.08088469

ks = 319712.5586 case 6 -0.08088469

λ s = 50867972.91 case 7 -0.08088469

δ s = 2.52424E-05

β c = 0.45535023

kc = 124460.9629

λ c = 22049113.04

δ c = 3.55321E-05

Step 3 Xa = 6.976242513

(AutoPIPEVessel)

FIG. UHX-13.2 TABLE UHX-13.1 ω s = 4.612272297

Zd = 0.01 0.00440434 ω∗ s= -4.54132059

Zv = 0.02 0.0208699 ω c= 3.343988236

Zw = 0.02 0.0208699 ω∗ c= -2.60273882

Zm = 0.205 0.2078966 γ b = 0

Steps 6 - 7 - 8

P's P't P γ W* Pw Primcase 1 0 535.2272424 0 0.0000E+00 0 -24.8420379

case 2 605.634798 0 0 0.0000E+00 0 70.4355899

case 3 605.634798 535.2272424 0 0.0000E+00 0 45.593552

case 4 0 0 -941.577569 0.0000E+00 0 0

case 5 0 535.2272424 -941.577569 0.0000E+00 0 -24.8420379

case 6 605.634798 0 -941.577569 0.0000E+00 0 70.4355899

case 7 605.634798 535.2272424 -941.577569 0.0000E+00 0 45.593552

P*s P*c P ω Pe Q2 Q3

case 1 0 0 0 -96.6108419 180.9984202 -0.067967319



case 2 0 0 0 116.6208033 -513.191814 -0.079848341

case 3 0 0 0 20.00996137 -332.193393 -0.137211543

case 4 0 -89.7874184 14.32870228 -159.948593 -104.398539 -0.05609033

case 5 0 -89.7874184 14.32870228 -256.559435 76.59988063 -0.060562767

case 6 0 -89.7874184 14.32870228 -43.3277903 -617.590353 0.007856579

case 7 0 -89.7874184 14.32870228 -139.938632 -436.591933 -0.044490729

Fm AutoPIPE

Vessel

(FIG. UHX-13.3.1 / 2) TABLE UHX-13.1 σ τ σ≤σallowable τ≤τ allowable

case 1 0.03 0.0339837

-

16348.54564 -3622.90657 OK : (1.5 Ss) OK : (0.8 Ss)

case 2 0.04 0.03992419 23184.34007 4373.280123 OK : (1.5 Ss) OK : (0.8 Ss)

case 3 0.07 0.0686057 6835.795255 750.3735515 OK : (1.5 Ss) OK : (0.8 Ss)

case 4 / 0.0280452 -22336.8312 -5998.07226 OK : (Sps) OK : (0.8 Ss)

case 5 0.1 0.03028139 -38685.3378 -9620.97883 OK : (Sps) OK : (0.8 Ss)

case 6 / 0.037398 -8068.58752 -1624.79213 OK : (Sps) OK : (0.8 Ss)

case 7 0.1 0.02224539 -15501.0188 -5247.69870 OK : (Sps) OK : (0.8 Ss)

Step 9

Ftmin Ftmax σ t,1 σ t,2 σ t,max σ t,max

(AutoPIPE Vessel)(AutoPIPE

Vessel) =<σallowable

case 1 -0.2703342 3.536119 -1288.44758 2226.786961 2226.786961 OK

case 2 -0.2431864 3.242469 1634.139142 -2251.55331 2251.553312 OK

case 3 -0.1926081 2.125221 361.0881005 -82.2505625 361.0881005 OK

case 4 -0.3030955 3.82967 -463.412623 5855.307726 5855.307726 OK

case 5 -0.2900353 3.71913 -1750.08587 8082.095923 8082.095923 OK

case 6 -0.5616265 5.410164 1130.436425 3603.749256 3603.749256 OK

case 7 -0.3403612 4.116364 -131.040058 5830.540641 5830.540641 OK

rt = 0.336675289 in Ft= 142.5706062 Ct = 172.5630506

Wt Fs Stb abs(σ t,min) case 1 1.481939944 6930.645233 1288.447582 OK

case 2 1.628765084 6305.881744 2251.553312 OK

case 3 2 5135.400003 82.25056251 OK

case 4 1.335164629 7692.534527 463.4126238 OK

case 5 1.390434811 7386.754075 1750.085875 OK

case 6 1.25 8216.640004 1130.436425 OK

case 7 1.25 8216.640004 131.040058 OK

Steps 10 to 12

σ s,m σ s,b σ s σ s ≤ σallowable elastic-plastic

case 1 1837.017346 -12222.6477 14059.66505 OK No

case 2 2284.011336 27825.23134 30109.24268 FALSE Yes

case 3 4121.028682 15602.58365 19723.61233 OK No

case 4 -2841.964732 -27649.41404 30491.37877 OK /

case 5 -1004.947386 -39872.06173 40877.00912 OK /

case 6 -557.9533965 175.8173092 733.7707057 OK /

case 7 1279.06395 -12046.83039 13325.89434 OK /

ls = 8.748999943

σ c,m σ c,b σ c σ c



≤σ allowable elastic-plasticcase 1 5567.107843 28381.11121 33948.21906 FALSE Yes

case 2 0 -8562.889306 8562.889306 OK No

case 3 5567.107843 19818.22191 25385.32975 OK No

case 4 0 16254.21505 16254.21505 OK /

case 5 5567.107843 44635.32626 50202.43411 OK /

case 6 0 7691.325742 7691.325742 OK /

case 7 5567.107843 36072.43696 41639.5448 OK /

Elastic-plastic calculation

Ss* = 17500 Sc* = 30000 facts factc Es* Ec* ks kc

case 1 1 1 OK 26400000 28300000 319712.5586 124460.9629

case 2 0.763994712 1 20169460.4 28300000 244258.7042 124460.9629

case 3 0 0

λ s λ c F φ Q1 Qz1case 1 50867972.91 22049113.04 6.732232476 9.023618515 -0.05915904 3.753824734

case 2 38862862.32 22049113.04 5.641389889 7.561496195 -0.05428668 3.874249576

case 3 0

Qz2 U Pw ω s ω∗ s Fm

case 1 10.15043261 20.30086522 0 4.612272297 -4.54132059 0.03399

case 2 11.35005096 22.70010193 0 4.612272297 -4.54132059 0.03886

case 3

δ s ω c ω∗ c Q3case 1 2.52424E-05 3.343988236 -2.60273882 -0.06796731

case 2 3.30401E-05 3.343988236 -2.60273882 -0.07770044

case 3

Prim Pe Q2 σ

case 1 -24.8420379 -96.61084191 180.9984202 -16351.5763

case 2 78.75994708 115.2295308 -573.842856 22297.1413

case 3

Comments In ASME example, the tubes design temperature is equal to channel design temperature. In AutoPIPE Vessel ®, the tubes design temperature is the maximum of the channel design temperature and the shelldesign temperature( values for allowable stress and elasticity modulus may be input).



UHX-20.3.1 Stationary Tubesheet Gasketed With Shell and Channel; Floating Tubesheet Gasketed, Not

extended as a FlangeTubesheet, Loading conditions [corroded normal condition].

ASME VIII DIV.1

2010 – 2011a

§[UHX-14]


Tubeside Shellside

Pressure P t =150 psi P s=250 psi

Corrosion ct =0 in c s=0 in 0 in 0 in

Material SA105 SA179 SA516GR60 SA516GR60


Allowable stress S = 19 000 psi S t = 13 350 psi S s = 17 100 psi S c = 17 100 psi

Yield strength S y = 36 000 psi S y,t = 20 550 psi S ,s = 32 000 psi S y,c = 32 000 psi

Modulus of elasticity E = 27 000 ksi E t = 27 000 ksi E s = 29 400 ksi E c = 29 400 ksi


Diameter A = 33.071 in d t =0.75 in 27.265 in 27.265 in

Nominal thickness 1.75 in t t =0.083 in 0.375 in 0.375 in

Tolerance 0 in

pattern Triangular N t =466 L = 256 in OTL=25.8 in p=1 in

Configuration d D s = 27.265 in G s = 29.375 in h = 1.75 in

Dc = 27.265 in Gc = 29.375 in h p = 1.356 in

h g = 0.197 in t s = 0.375 in t c = 0.375 in

Extra thickness (periphery) : Tubeside = 0.197 in Shellside = 0.197 in

Configuration C

A = 26.89 in Dc = 26.036 in h = 1.75 in

Gc = 26.496 in h = 1.75 in hg = 0 in

G1 = 26.463 in

Extra thickness (periphery) Tubeside : 0 in Shellside : 0 in




p* =( )[ ]2

4,MIN41

O

O L

D

p D A

p

⋅−

π

= 1.068 in

basic ligament efficiency : µ = ( p- d t )/ p = 0.25

effective ligament efficiency : µ * = ( p*- d *)/ p* = 0.385


largest center-to-center distance between adjacent tube rows : U L = 2.496 in Unperforated Area : L = 64.397 in²



a0 = D0 /2 = 12.9 in h g = max[(h g -ct ).(0)]

x s = 1- N t (d t /2a0) = 0.606 xt = 1- N t ((d t -2t )/2a0) = 0.761Stationary Tubesheet Floating Tubesheet

a s = G s/2 = 14.688 in a s = ac = 13.248 in ac = Gc/2 = 14.688 in ac = Gc/2 = 13.248 in

ρ s = a s /a0 = 1.139 ρ s = a s /a0 = 1.027

ρ c = ac /a0 = 1.139 ρ c = ac /a0 = 1.027

h

Dc Gc

Gs

h p

Ds

Gc

A

Dc

h hp



Shell and channel coefficients Stationary Tubesheet Floating Tubesheet

β s = [12(1-ν s2)]

1/4/ [( D s+t s)t s]

1/2 = 0 in- 0 in-

k s = β s E s t s3

/ [6(1-ν s2)] = 0 lbf 0 lbf

λ s = 6 D s k s /h3 (1+h β s+h2

β s2/2) = 0 psi 0 psi

δ s = D s2 /4 E s t s

(1-ν s /2) = 0 in³/lbf 0 in³/lbf

β c = [12(1-ν c2)] 1/4 / [( Dc+t c)t c]

1/2 = 0 in-1 0 in

-1

k c = β c E c t c3/ [6(1-ν c

2)] = 0 lbf 0 lbf

λ c = 6 Dc k c /h3 (1+h β c+h2

β c2/2) = 0 psi 0 psi

δ c = Dc2/4 E c t c

(1-ν c /2) = 0 in³/lbf 0 in³/lbf

X a = [24(1-ν *2

) N t E t t t (d t -t t )a02/( E * Lh3

)]1/4 = 3.602


Stationary Tubesheet Floating Tubesheet

K = A/ D0 = 1.282 1.042

F = (1-ν * )/ E * (λ s+λ c + E ln K ) = 0.425 0.071

Φ = (1+ν * ) F = 0.557 0.093

Q1 = ( ρ s-1- Φ Z v) / (1+ Φ Z m) = 0.077 0.019

Stationary Tubesheet Floating Tubesheet

ω s = ρ s k s β s δ s (1+h β s) = 0 in² 0 in²

ω s*= a0

2 ( ρ s2-1)( ρ s-1)/4 - ω s = 1.7083 in² 0.0614 in²

ω c = ρ c k c β c δ c (1+h β c ) = 0 in² 0 in²

ω c *=a0

2[( ρ c

2+1)( ρ c -1)/4-( ρ s-1)/2]-ω c = 1.7083 in² 0.0614 in²

γ b = (Gc-G s)/ D0 = 0 (Gc-G1)/ D0 = 0.0013

W s = 0.2147×106 lbf /

W c = 0.2147×10 lbf 0.1026×10 lbf

W m1 s = 0.2017×106 lbf /

W m1c = 0.1268×106 lbf 0.9644×10

5 lbf

W = / /

W ax = max[W s , W c ] W m1max = max[W m1 s , W m1c ]

Equivalent Pressure : P e = P s - P t T s

* = T s

‘ = 70 °F T c

* = T c

‘ = 70 °F T r = T ‘ = 70 °F

s’(T s

*-T a) = 0 in/in αc

’(T c

*-T a) = 0 in/in ’

(T r -T a) = 0 in/in The effect of radial differential thermal expansion adjacent to the Tubesheet is not considered.

P s* = E st s/a s[ s

’(T s

*-T a)-

’(T r -T a)] P c

* = E ct c/ac[αc

’(T c

*-T a)- α

’(T r -T a)]

Case P s (psi) P t (psi) Stationary Tubesheet Floating Tubesheet

P e (psi) P s* (psi) P c

* (psi) P e (psi) P s

* (psi) P c

* (psi)

1

2

3

0

250

250

150

0

150

-150

250

100

0

0

0

0

0

0

-150

250

100

0

0

0

0

0

0

bending stress :

Q2 = [(ω s* P s-ω c

* P t ) - (ω s P s*-ω c P c

*) + (γ bW /(2π )) ] / (1+ Φ Z m) Q3 = Q1 + 2Q2/( P e a0

2)

σ = 6 Q2 / [ µ * (h-h g

’) 2 ] ( P e = 0 )

|σ | shall be ≤ σ allowable σ = (1.5 F m / µ

*) (2a0 /(h-h g ’ ))2 P e ( P e ≠ 0 )

Shear stress in the stationary Tubesheet :

τ = (1/(2µ )) (a0/h p) P e τ = (1/(4µ )) ({4 A p/C p}/h p) P e if | P e| > 3.2S µ h p / D0

in which : C p = 84.992 in , A p = 415.595 in2

|τ | shall be ≤ τ allowable



Stationary Tubesheet

Case W h g (in) Q2 (lbf) Q3 F m σ (psi) σ allowable (psi) τ (psi) τ allowable (psi) 1

2

3

W m1c

W m1 s

W m1max

0.197

0.197

0.197

-207.470

345.783

138.313

0.0932

0.0932

0.0932

0.1016

0.1016

0.1016

-16 382

27 303

10 921

1.5 S = 28 500

1.5 S = 28 500

1.5 S = 28 500

-2 854

4 757

1 903

0.8 S = 15 200

0.8 S = 15 200

0.8 S = 15 200Floating Tubesheet

Case W * h’ g (in) Q2 (lbf) Q3 F m σ (psi) σ allowable (psi)

1

2

3

W m1c

0

W m1c

0.000

0.000

0.000

9.972

14.764

24.736

0.0181

0.0196

0.0219

0.0742

0.0747

0.0755

-9 423

15 813

6 390

1.5 S = 28 500

1.5 S = 28 500

1.5 S = 28 500

r t = [d t 2+(d t -2t t )

2]

1/2/ 4 = 0.238 in S ,t = 20 550 psi k = 1

F s = min{ 2.0 ; max[ (3.25-0.25 [ Z d +Q3 Z w] X a4 ) ; 1.25 ] } ( P e ≠ 0 ) l t = k l = 15.375 in

F s = 1.25 ( P e = 0 ) l = 15.375 in



2)] ; S t } C t = (2π

2 E t / S ,t )1/2 = 161



If P e ≠ 0 : σ t,1 = [( P s x s – P t xt ) – P e F t,min ] / ( xt – x s) σ t,2 = [( P s x s – P t xt ) – P e F t,max ] / ( xt – x s)

F t,min and F t,max

FromTable UHX-13.2

If P e = 0 : σ t,1 = [( P s x s – P t xt ) – (2Q2 /a02) F t,min ] / ( xt – x s)


σ t,max = max( |σ t,1| ; |σ t,2| ) σ t,max shall be ≤ σ t,o allowable If σ t,1 or σ t,2 < 0 : σ t,min = min( σ t,1 ; σ t,2 ) |σ t,min| shall be ≤ S t,b

Case F s F t,min F t,max σ t,1 (psi) σ t,2 (psi) σ t,o allowable (psi) S t b (psi) 1

2

3

1.55

1.55

1.55

-1.0091

-1.0091

-1.0091

3.3962

3.3962

3.3962

-1 713

2 605

892

2 549

-4 499

-1 950

S t = 13 350

S t = 13 350

S t = 13 350

10 582

10 582

10 582

Tube-to-tubesheet joint



(2 groove(s) )






Comments In ASME example, the values of Do and r o are not coherent :dt = 0.75 in. ; Do = 25.8 in. and ro = 12.5 in. AutoPIPE Vessel® example keeps values of dt and Do ro = 12.525 in.



Other calculations



Welding Neck Flange ( Taylor Forge Bulletin)

Body flange(s) and cover(s)

Flange 1801 in operation.ASME VIII DIV.1 2010 – 2011a

Design pressure P = 400 psi Corrosion : 0 in

Design Temperature : 500 °F Tolerance : 0 in

Material

Flange Bolt

SA105 SA193GRB7

Uncorroded dimensions Flange thickness (in the bolt circle) : tC = 2 in

Inside diameter of flange : Bf = 32 in

Calcutated as integral type

Raised face (male) height = 0.188 in

Bolt nb = 36 db = ∅ 1 in A b = 19.84 in²

Gasket N = 0.75 in t N = 0.118 in y = 3 700 psi yc = 0 psi m = 2.75 b = 0.306 in b0 = 0.375 in

Hub dimensions (uncorroded) : length = 2.75 in ; thickness at the end = 0.5 in ; thickness at back of flange = 1.125 in

Corroded dimensions B = 32 in t = 2 in (1) g0 = 0.5 in g1 = 1.125 in

D = 32 in A = 39.125 in C = 37 in G = 33.888 in h = 2.75 in

(1) : Flange face over thickness, tolerance is taken into account.

Gasket and bolt calculation :

BSmin = 2.25 in BS = 3.14 C / nb = 3.229 in

BSmax = 2 db + 6 t / (m + 0.5) = 5.692 in C0 = max [1, (BS / (2.db+t) ) 0.5] = 1

H = 0.785 G2 P = 360 771 lbf H p = 2 P b ×3.14 G m = 71 715 lbf S b = 25 ksi

Wm1 = H + H p = 432 486 lbf Wm1x = max(Wm1) = 432 486 lbf Wm1x/A b = 21.8 ksi ≤ S b Am = max (Wm2/Sa, Wm1/S b) = 17.299 in² Amx = max (Am) = 17.299 in²

Wm2 = 3.14 b G y = 120 612 lbf W = Sa (A b + Am) / 2 = 0.4642×106 lbf Sa = 25 ksi

Nmin = W / (2 × 3.14 G y) = 0.589 in W/A b = 23.4 ksi ≤ Sa Load to apply on bolting : W/nb = 12 894 lbf

Loads, lever arms and moments

HD = 0.785 B2 P = 321 699 lbf hD = (C - B - g 1) / 2 = 1.,938 in MD=HD.hD= 623 291.9 lbf.in

HG = Wm1x - H = 71 715 lbf hG = (C - G) / 2 = 1.556 in MG=HG.hG= 111 603.3 lbf.in

HT = H - HD = 39 072 lbf hT = (2 C - B - G) / 4 = 2.028 in MT=HT.hT= 79 241.77 lbf.in

MO=MD+MG+MT= 814 136.9 lbf.in MA = W.hG = 722 375.7 lbf.in R = ( C - B ) / 2 - g1

Stress Factors

h0 = √ (B g0) = 4 in K = A/B = 1.223 g1/g0 = 2.25 h/h0 = 0.688

F = 0.777 T = 1.83 U = 10.74 V = 0.162

Y = 9.773 Z = 5.041 f = 1 B1 = B

e = F/h0 = 0.1942 in-1 L = ((t e + 1)/T) + (t3/((U h0 g02)/V)) = 0.,879

Stresses : ( in operation : M=MO ; Sf = 19.6 ksi ;Sn = 17.1 ksi )

SH = (f M C0)/(L g12 B1) = 22.87 ksi ≤ min (1.5 Sf , 2.5 Sn) = 29.4 ksi (78%)

SR =(1.33 t e + 1) M C0/ L t2 B = 10.97 ksi ≤ Sf = 19.6 ksi (56%)

ST = ((Y M C0)/(t2 B)) - Z SR = 6.85 ksi ≤ Sf = 19.6 ksi (35%)

SHR = (SH + SR )/2 = 16.92 ksi ≤ Sf = 19.6 ksi (86%)

SHT = (SH + ST)/2 = 14.86 ksi ≤ Sf = 19.6 ksi (76%)

Stresses : ( seating : M=MA ; Sf = 20 ksi ;Sn = 17.1 ksi )

SH = (f M C0)/(L g12 B1) = 20.29 ksi ≤ min (1.5 Sf , 2.5 Sn) = 30 ksi (68%)

SR =(1.33 t e + 1) M C0/ L t2 B = 9.74 ksi ≤ Sf = 20 ksi (49%)

ST = ((Y M C0)/(t2 B)) - Z SR = 6.08 ksi ≤ Sf = 20 ksi (30%)

SHR = (SH + SR )/2 = 15.01 ksi ≤ Sf = 20 ksi (75%)

SHT = (SH + ST)/2 = 13.18 ksi ≤ Sf = 20 ksi (66%)

Flange Rigidity (Moments without correction for bolt spacing) Rigidity Factor : K I = 0.3

JP = (52,14 |MO| V)/(L EO g02 h0 K I) = 0.959 shall be ≤ 1,0 with : EO = 27100 ksi

JA = (52,14 |MA| V)/(L EA g02 h0 K I) = 0.79 shall be ≤ 1,0 with : EA = 29200 ksi

A C

G B

HG HD

HT

W

g0 g1

h

t

D



t A = 39.125 in

h B = 32 in

W C = 37 in

A C D = 32 in

t (at bolt circle) = 2 in

HG g1 h = 2.75 in

HT HD g0 = 0.5 in

G g0 g1 = 1.125 in

Ext. Gasket dia. 34.5 in

B D Gasket width (w) 0.75 in

raised face male 0.188 in

Design Pressure 400 psi

Design Temperature 500 °F

Corrosion allowance 0 in

Flange Material SA105

Bolt material SA193GRB7

Shell material

Type of gasket Grooved metal (Soft Iron or Steel)

Gasket factor m 2.75

Min. Design Seating stress 3700 psi

Bolt diameter 1 in UNC 8

number of bolts 36

Bolt section 0.55 in2 from BoltUN.doc

total section Ab 19.84 in2

Mechanical characteristics

Allowable Stress at temp. Allowable Stress at seating

Shell Sf 19600.00 psi 20000.00 psi

Flange Sn 17100.00 psi 17100.00 psi

Bolt 25000.00 psi 25000.00 psi

Elasticity Modulus 27100000.00 psi 29200000.00 psi

B (corroded) 32 in Gasket dimensions

g0 (corroded) 0.5 in b0 (w/2) = 0.375 in

g1 (corroded) 1.125 in b = 0.306186218 in

G = 33.88762756 in

Bolt loads and section

H = 360771.485 lbf

Hp = 71713.2421 lbf

Wm2 = 120608.634 lbf

Wm1 = 432484.727 lbf

Am = 17.2993891 in2 OK Ab >= Am

Flange moments

HD = 321699.088 lbf

HT = 39072.3968 lbf

hD = 1.9375 in

hG = 1.55618622 in MA = 722369.7583 lbf in



hT = 2.02809311 in

W = 464192.363 lbf Mo = 814133.6003 lbf in

Stresses calculation

deltab = 3.22885912 in

C0 = 1

K = 1.22265625

h0 = 4

T = 1.82975561 M (seating) = 722369.7583 lbf in

U = 10.7398932

Y = 9.77332415 M (operation) = 814133.6003 lbf in

F = 0.77693343 E6 = -1.246983257

V = 0.16155154 E4 = 1.338307677

e = 0.19423336 for which A = 1.25

L = 0.87916386 C = 9.758283691

Z = 5.04131594 C36 = 2.074097577

operation (stresses in psi)

SH = 22864.98009 ≤ 29400.00 min(1.5 Sf ; 2.5 Sn) OK

SR = 10972.46802 ≤ 19600.00 Sf OK

ST = 6846.756232 ≤ 19600.00 Sf OK

0.5 (SH + SR ) 16918.72405 ≤ 19600.00 Sf OK

0.5 (SH + ST ) 14855.86816 ≤ 19600.00 Sf OK

seating (stresses in psi)

SH = 20287.78831 ≤ 30000.00 min(1.5 Sf ; 2.5 Sn) OK

SR = 9735.722817 ≤ 20000.00 Sf OK

ST = 6075.03442 ≤ 20000.00 Sf OK

0.5 (SH + SR ) 15011.75556 ≤ 20000.00 Sf OK

0.5 (SH + ST ) 13181.41137 ≤ 20000.00 Sf OK

Flange Rigidity KI = 0.3

Jo = 0.95943917 ≤ 1 OK

JA = 0.79007398 ≤ 1 OK



ASME VIII Div.1 2010 App. 26 Unreinforced Bellows

Pressure P = 150 psi

Bellows allowable stress S = 20 000 psi

Bellows yield strength S y = 25 000 psi

Bellows modulus of elasticity E b = 27 600 ksi

Bellows modulus of elasticity

Ambient E o = 28 300 ksi

Collar allowable stress S c = /

Collar modulus of elasticity E c = /

No. of convolutions : N = 8Convolution Height : w =1.125 in

Convolution Pitch : q =1.125 in

No. of plies : n = 1 Ply thickness : t = 0.05 in Collar Thickness : t c = /

Int. Diameter : Db = 24 in Lt = 0.75 in Lc = / Lb = N q = 9 in

Mean diameter of convolution : Dm = Db+w+nt = 25.175 in C 1 = q / 2w = 0.5

Mean diameter of collar : Dc = Db+2nt +t c = / C 2 = pm t D2.2q = 0.461

Ply thickness (corrected for thinning) : t p = mb D Dt = 0.049 in Figure 26-4 C p = 0.659

Cross section of 1 convolution : A = [ (( Π - 2)/2)q + 2w] n t p = 0.141 in2 Figure 26-5 C f = 1.686

Figure 26-6 C d = 1.718

k = ( ){ }t D5.1 L;1min bt = 0.456 K f = 3.0 Joint efficiency : C wc = 1

Sum of combining movements : etot = 1 in [∆q = etot /N ] No. of cycles : N spe = 1 000

Spring rate (1 conv.) :( ) f

3 p

mb2b

b C 1

w

t D E

N n

12 K

−=

ν

Π = 7 267.063 lbf/in ( ν b = 0.3)

Stresses due to pressure (n = 1)

Bellows circumferential membrane( )

( )( ) P

k E L Dt E Lnt Dnt

k E Lnt D

21

S ccccbt b

bt 2

b1 ++

+= = 16 466 psi ≤ S = 20 000 psi

Collar circumferential membrane( )

P k E L Dt E Lnt Dnt

k E L D

21

S ccccbt b

ct 2c

1 ++=′ = / C wc.S c = /

End convolutions circumferential

membrane P

Lt Lnt A )nt D( LqD

21

S cct p

bt m E ,2 ++

++= = 19 555 psi ≤ S = 20 000 psi

Intermediate convolutions

circumferential membrane P

A

qD

21

S m I ,2 = = 15 044 psi ≤ S = 20 000 psi

Bellows meridional

Membrane S 3, Bending S 4 P

nt 2w

S p

3 = = 1 728 psi P C t w

n21

S p

2

p4

= = 26 239 psi

S 3+S 4 = 27 968 psi ≤ K .S = 60 000 psi

Meridional stresses due to deflection

membrane : q f

3

2 pb

5C w

t E

2

1S ∆= = 1 713 psi bending : q

d 2

pb6

C w

t E

3

5S ∆= = 129 129 psi

Fatigue Life

S t = 0.7(S 3+S 4)+(S 5+S 6 ) = 150 420 psi N spe ≤

2

ot b

o g

oalw

S S E E K

K N

−= = 2 012

Kg = 1 Ko = 5 200 000 psi So = 38 300 psi

w

nt

Db

Lt Lb Lc

t c

q



Limiting Internal

Design Pressure for

buckling

Nq K 34.0 P b sc π = = 862 psi P ≤ P sc

α Π q D AS )2( P m* y si −= = 205 psi P ≤ P si

S y* = 2.3 S δ = 1/3 S 4 /S 2,I α = 1+2δ 2+(1-2δ 2+4δ 4 )0.5 = 2.5597

asme viii div1 2010-2011a

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