6 (2)External Pressure Vessel 外压容器—— vessels where the media pressureinside the vessel is lower than that outside(gauge pressure). When the internalpressure < 0.1 MPa (absolute pressure),such vessels are called Vacuum Vessel.
7 ii. According to temperature Normal temperature vessel-20< T ≤200℃Medium temperature vessel—— between normal T & high T vesselsLow temperature vessel< -20 ℃High temperature vessel—— where the wall temperature isabove the creep temperature.
9 iii. According to management Grade (I)Grade (II)Grade (III)P, P*VmediaimportanceFactorDegree of danger:I < II < III
10 Design of Pressure Vessels 2.2Basic Law andCommon Standard forDesign of Pressure Vessels1.Conditions:i. The maximum working pressurePw≥ 0.1 MPa(neglecting the net pressure of liquid )
11 ii. Internal diameter Di (equal the maximum dimension or size in the non-circularsections) ≥ 0.15m, and V ≥ m3iii. With the medium as gas or liquefied gas,or liquid whose maximum workingT ≥ standard b.p.
13 2.3 Standardization of Pressure Vessel Parts 1.Significance of Standardization:标准化的重要性i. It consolidates and harmonizes the variousactivities in the manufacture and social life.ii. It’s the importance means to organize themodernization production.
14 iii. It’s the importance component in scientific management.iv. It makes for the development of newproducts, assures the interchangeabilityand common-usability and is convenient touse and maintain.v. It’s helpful in the interchange of internationalscience & technology, culture and economy.
15 2.Active Chemical Vessels and Standards of Equipments Components: Cylinder HeadsVessel FlangePipe Connecting (Nozzle) Flange
16 Support Stiffening Ring Manhole HandholeSight (level) Glass Liquid Leveler (LG)Corrugated expansion jointHeat Exchanger Tube TrayFloat (floating) Valve (FV)Bubble (bubbling) cap Packing etc.
17 3.Basic parameters of standardization: i. Nominal Diameter —— DN (Dg) 公称直径 is a typical dimension.(1)Rolled cylinder and head from steelsDN = Di (inside diameter)DN Standard of pressure vessels:…… 600048 grades in total.
18 (2)Cylinder made by seamless pipes —— DN = Do (outside diameter)Six grades:(3)Seamless pipes—— DN≠Di & DN ≠ Do,but DN is a certain value that is smaller thanDo. When the DN is a certainty, Do is to be acertainty, while Di depends on the thickness.Denotation of seamless pipes:such as 252.5 (outside × thickness)
19 Check the standards according to DN. Comparison of DN and Do of seamless pipes/mmDN1015202532405065Do141838455776S33.54
20 (4)DN of flanges—— consistent to their suitable cylinders,heads and tubes.i.e. DN of flange = DN of cylinderDN of headDN of tube
21 ii. Nominal Pressure —— PN (Pg) 公称压力Prescriptive standard pressure gradesFor example ——PN (MPa) of pressure vessel flange:0.25 、 0.6 、 1.0 、 1.6 、 2.5 、 4.0 、 6.4
22 iii. ApplicationIn standard designing:(1)Diameters of cylinders, heads and tubesmust be close to the standard grades.e.p. Diameter of cylindershould be 、600 、700 …shouldn’t be 、645 、750 …
23 (2)When selecting the standard vessel parts, the design pressure at the workingtemperature should be regulated to acertain grade of PN.Then choose the parts according to PNand DN.
24 4.Classification of standards: i. Chinese StandardSymbol: GB (Guo biao)ii. Standard issued by MinistryJB —— Ministry of Mechanical IndustryYB —— Ministry of Metallurgical IndustryHB —— Ministry of Chemical IndustrySY —— Ministry of Petroleum Industry
25 iii. Specialty Standard iv. Trade Standardv. Plant Standard
26 & Contents of Vessel Design 2.4 Basic Requirements& Contents of Vessel Design1.Basic requirements:基本要求i. Enough strength —— no breakageii. Enough rigidity —— limit deformation
27 iii. Enough stability —— no failure, ?, drape iv. Durability —— assuring certain usage lifev. Tightness —— no leakagevi. Saving materials and easy to manufacturevii. Convenient to be installed, transported,operated and maintainedviii. Rational technical economy index in total
28 2.Basic Contents:i. Selection of materialsSelecting the materials ofequipment according to the technicalindexes t, p, media and the principles ofmaterial selection.ii. Structure design
29 iii. Calculation of strength and thickness (including the cylinders and heads)iv. Strength verification in hydraulicpressure testv. Seal design; selecting or designing flanges
30 vi. Selection of support & the verification of strength and stabilityvii. Design and calculation of reinforcementfor openingviii. Selection of other parts and accessoriesix. Other special designx. Plotting the equipment drawingsxi. Compiling the equipment specifications
31 Chapter 3 Stress Analysis of Thin-walled Internal-P Vessel of thin-walled Cylinders Subjected to Internal Pressure
32 2.stress characteristics: 1.Thin-walled vessels薄壁容器(1)Thin-walled vessels:S / Di < 0.1 (Do / Di = K < 1.2)(2) Thick-walled vessels:S / Di ≥ 0.12.stress characteristics:There are always two kinds ofstress in pressure vessels.
33 moments and conditions of deformational compatibility i. membrane stress—— membrane (shell)theoryii. boundary stress—— shell theory withmoments and conditionsof deformationalcompatibilityP
34 Rotary Shells’ Stress Analysis 3.2 Membrane Theory ——Rotary Shells’ Stress Analysis1.Basic conceptions and hypothesis:i. Basic conceptions(1)rotary curved-surface & shell
35 Geometry figure, endured load (2)Axial SymmetryGeometry figure, endured loadand restrictions of shell are all symmetryto the revolving shaft (OA).Several basic conceptions:generatrix, meridian, normal, latitude,longitudinal radius, tangential radius.
36 The plane curve which forms the curved surface. BB’(3)Generatrix (AB)The plane curve which forms the curved surface.(4)Longitude (AB’)Section passing OA and intersecting with shell, the cross-line is AB’.
37 is vertical with midwall surface. The extension of (5)Normal (n)The line passingpoint M in meridian andis vertical with midwallsurface.The extension ofnormal must intersect withOA.(6)Latitude (CND)The cross-line formed by the conic surface passing point K2’ intersects with the rotary curved surface.AOMnK2K1K2’CND
38 (7)Longitudinal radius (R1) *The radius of curvature of meridianwhich passes point M in midwall surfaceis called the longitudinal radius of pointM in meridian.*The center K1 of curvature of the roundwith diameter R1 must be in the extensionof normal passing point M.For example:Longitudinal radius of point M:R1 = M K1
39 (8)Tangential radius (R2) *The plane which is vertical to the normalpassing the point M in meridianintersects with the mid-wall surface, theresulted cross line (EMF) is a curve, theradius of curvature of this curve in pointM is called tangential radius.
40 *The center K2 of curvature of the round with diameter R2 must be in theextension of normal passing point Mand in the revolving shaft.For example:Longitudinal radius of point M:R2 = M K2
41 ii. Basic hypothesis:Small displacement hypothesisStraight linear law hypothesisNon-extrusion hypothesis
42 2.Free body balance equation —— calculation formula of radial (meridional) stress:i. Intercepting shell—— uncovering the radial stress mii. Choosing separation bodyiii. Analysis of stress
44 3.Infinitesimal balance equation —— calculation formula of hoop stress:i. Intercepting shell—— uncovering the radial stress mii. Choosing separation bodyiii. Analysis of stress
45 iv. Constitute balance equation k2k1d2d1pmSdl2Sdl1
46 Basic calculation equation of membrane stress: Illustration of symbols:m —— radial stress of a random point inrotary thin shell, MPa —— hoop stress of a random point in
47 P —— internal pressure, MPa S —— thickness of wall, mmR1 —— longitudinal radius of required stresspoint in the mid-wall surface of therotary shell, mmR2 —— tangential radius of required stress
48 v. Application range of membrane theory ☆Applicable to axial symmetric thin-walledshell without bending stress☆No bending stress —— only normal stress(tensile stress & compression stress)☆Thin-walled shell—— S / Di < 0.1 ( Do / Di = K < 1.2 )☆Axial symmetry and continuous—— Geometry, loads, physical properties☆Free supporting boundary
49 3.3 Application of Membrane Theory Calculation equations: S pR = s S p 2sSpRm=+21qs
50 1.Cylindrate shell subjected to uniform gas internal pressure: ∵ R1 = ∞ R2 = D / 2Putting them into theprevious equations:sDp
51 2.Spherical shell subjected to uniform gas internal pressure: ∵ R1 = R2 = D / 2Putting them intoequations (3-3) and (3-4):DS
52 3.Elliptical shell subjected to uniform gas internal pressure: Example:Known:Major semiaxis - a Short semiaxis - bThickness - S Internal Pressure - PFind the m and of a random point on the elliptical shell.
53 radius of curvature of A yxabA(x,y).k2k1Solution:(1)Find R1 and R2 of point A:R1:radius of curvature of A(a)By the elliptical equation:getting y’ and y’’, then put them into (a).result is:
54 R2: R2 = K2 A = x / sin (b)here:putting them into (b), getting:
55 (2)Find m and of point A: Putting R1 and R2 into (3-3) and (3-4), getting:
56 Stress of special points on elliptical shell: (1)x=0 (Top of elliptical shell)(2)x=a (Boundary or equator of elliptical shell)
57 Standard elliptical heads: The elliptical heads whose ratio of major and short semiaxis a / b = 2 are called standard elliptical heads.a / b = 2 ——x=0 (Top):x=a (Boundary):ma / b = 2
58 4.Conic shell subjected to uniform gas internal pressure: Example:Known:Diameter of tapered bottom - DHalf tapered angle - Thickness - SInternal Pressure - PFind the m and of a random point on the conic shell.
59 . Solution: (1)Find R1 and R2 of point A: R1 = ∞ R2 = A K2 = r / cos (2)Find m and of point A:Putting R1 and R2 into (3-3) and (3-4)respectively, getting:Dr.Ak2
60 Characteristics of stress distribution of conic shell: m
61 . 5.Cylindrate shell subjected to liquid static pressure: i. Supporting along the boundary of bottomExample:Known:Gauge pressure – Po (Pa)Liquid level – H (m)Density of liquid - (N/m3)Find the m and of a randompoint on the wall of cylindrical shell.DH.AxPo
62 . Solution: (1)Meridional stress: Cutting along section B-B, taking thelower part as the separation body.DH.AxPoBB-B(H-x)(Po+x)mN
63 Establishing the balance equation of axial stress:
64 (2)Hoop stress:Infinitesimal balance equation (3-4):For point A:R1 = ∞ R2 = D/ P = Po + xPutting them into (3-4),getting:when x=H:
65 . ii. Supporting along the boundary of top Example: Known: Gauge pressure – Po (Pa)Liquid level – H (m)Density of liquid - (N/m3)Find the m and of a randompoint on the wall of cylindrical shell.DH.AxPo
66 . Solution: (1)Meridional stress: Cutting along section B-B, taking thelower part as the separation body.PoDH.AxBB-BH-x(H-x)m(Po+ x)
67 Establishing the balance equation of axial stress:
68 (2)Hoop stress:Infinitesimal balance equation (3-4):For point A:R1 = ∞ R2 = D/ P = Po + xPutting them into (3-4),getting:when x=H:
69 6.Examples: i. A certain cylindrical vessel with a spherical upper head and a semi-elliptical lower head,its ratio of major semiaxis to short semiaxisis a / b = 2. The average diameter D=420mm,thickness of all cylindrical shell and headsare 8mm. The working pressure P=4MPa.Calculating:(1)Find m and of the shell body.(2)Find the maximum stress on the both theheads and their position respectively.
70 (2)Upper head —— spherical Solution:(1)m and :DSP(2)Upper head —— spherical
71 (3)Lower head —— elliptical When a / b = 2:a = D/2 = 210 mm b = a/2 = 105 mmx=0 (Top):x=a (Bottom):
72 3.4 Conception of Boundary Stress 1.Forming of boundary stress: 边界—— The joint and its vicinity oftwo parts with different geometryshape, load, material and physicalconditions, i.e. discontinuous point.
74 Boundary stress forming not for balancing the loads but for receivingrestrictions from self or exterior. It’s agroup of internal force with same valuebut contrary direction occurringbetween two parts which are forced torealize transfiguration harmonization.
75 2.Characteristics of boundary stress: 边缘应力特性 i. Distributing along the wall non-evenlyii. Different joint boundary forming differentboundary stressiii. It’s local stress, i.e. only forming large stresslocally and decaying apparentlyiv. Value of boundary stress can be 3~5 timesof that of membrane stressv. Self-constrained
76 3.Treatments to boundary stress: i. Treatments locally in structure(1)Improving the structure of joint boundary
78 (3)Assuring the quality of welding line at boundary(4)Decreasing the remnant stress at local andprocessing the heat treatment to eliminatethe stress(5)Avoiding the local stress added to theboundary region overlap with connaturalstressii. Materials are of high plasticity
79 Chapter 4 Strength Design of Cylinders and Heads subjected to Internal-Pressure4.1 Basic Knowledgeof Strength Design
80 < [ ] 1.Criterions of elasticity failure: eqSafety Allowance kept for therequirements of safety:s=noeqeq —— equivalent stresso —— limiting (ultimate) stress, can bes、 b、 n、 D, etc. —— allowance stressn —— safety coefficient
81 2.Strength Theory: i. The first strength theory —— the maximum tensile stress theoryapplying to brittle materialsii. The second strength theory—— the maximum major strain theoryiii. The third strength theory—— he maximum shear stress theory231maxst-=
82 Shear limit:Failure condition:Strength condition:Applying to the plastic materials
83 iv. The fourth strength theory —— the maximum deformation energy theoryStrength condition:Applying to the plastic materials
84 4.2 Strength Calculation of Thin-walled Cylinder Subject to Internal Pressure1.Strength calculating equation:i. Determining the major stress
85 ii. Determining the equivalent stress According to the third strength theory:iii. Strength condition
86 iv. Strength calculation equation Solving equation (4-4) as following:(1)Replacing medium diameter with internaldiameter: D + Di = SCalculating pc from p.Putting them into equation (4-4), getting:
87 (3)Eliciting the corrosion allowable thickness C2: (2)Introducing the welded joint efficiency :This is the calculated thickness.(3)Eliciting the corrosion allowable thickness C2:This is the design thickness.
88 (4)Adding negative deviation C1: Getting the nominal thickness which indicated on the drawing.(5)Calculating the effective thickness:
89 v. Equation of strength verification vi. Calculating equation of pw
90 2.Strength calculating equation of thin-walled spherical vessels:
91 *Equation of strength verification: *Equation of [pw] —— the maximum allowable working pressure:*Scope of application of previous equation:cylinder: P≤0.4 t (Do / Di ≤1.5)spherical shell: P≤0.6 t (Do / Di ≤1.35)
92 Illumination of symbols: Pc —— Calculated pressure MPaDi 、 Do —— Internal & external diameters ofcylinder mmS —— Calculated thickness mmSd —— Design thickness mmSn —— Nominal thickness mmSe —— Efficient thickness mm
93 C1 —— Negative deviation mm C2 —— Corrosion allowable thickness mmC —— Additional value of wall thickness mm —— Welded joint efficiencyt —— Allowable stress at design temperatureMPat —— Calculated stress at design temperature[Pw] —— The maximum allowable pressure atdesign T MPa
94 3.Determination of design parameters: i. Pressure P(1)Working pressure Pw工作压力—— the maximum pressure at the top ofvessel and under normal operatingcondition
95 (2)Design pressure P计算压力—— the maximum pressure at thespecified top of vesselThe design pressure P and thecorresponding design temperature T areconditions of designing load, and its valueis not less than working pressure.
96 (3)Calculated pressure Pc 计算压力—— the pressure which is used to determinethe thickness at correspondingdesign temperatureIncluding the liquid (column) staticpressure, when the liquid (column) staticpressure ＜ 5% design pressure, it can beneglected.
97 Choosing the value of design pressure Illustrating in the following chart
98 Determined by the charging proportion and Tmax ConditionsEvaluation of Design PWith safety devicesP≤(1.05~1.1)PwSingle vessel(no safety devices)P≥PwWith explosive mediaand rupture diskP≤(1.15~1.3)PwWith liquefied gasDetermined by the charging proportion and TmaxExternal Pressure VesselUnder normal working condition, P≥△P=P2-P1Vacuum Pressure VesselWith safety valve: P=1.25△PWithout SV: P=0.1MPaJacketed VesselAs external P vessel
99 Design P and design T both are the design load condition. ii. Design Temperature T设计温度—— the enacted temperature of metalliccomponents under normal operatingconditionDesign P and design T bothare the design load condition.
100 iii. Allowable stressNormal T VesselMedium T VesselHigh T Vessel
101 σb σts σtD σtn Material iv. Safe (Safety) coefficient n Strength PerformanceσbσtsσtDσtnSafetyCoefficientnbnsnDnnCarbon SteelLow Alloy Steel≥3.0≥1.6≥1.5≥1.0High Alloy Steel
102 v. Welded joint efficiency ( ) (1)Double welded butt or completely weldedbutt which is the same as double one.NDE 100% = 1.0NDE Local = 0.85Double welded butt(2)Single welded buttNDE 100% = 0.9NDE Local = 0.8Single welded butt
103 vi. Additional value of wall thickness C (1)Negative deviation of plate and tube C1Referring to the teaching material page95, figure 4-7 & 4-8, selecting according tothe nominal thickness Sn.(2)Corrosion allowable thickness C2C2 = Ka BKa —— corrosion rate, mm/yearB —— design life of utility, year
104 Generally speaking:Ka < 0.05 mm/yearSingle corrosion C2 = 1 mmDouble corrosion C2 = 2 mmKa = 0.05~0.1 mm/yearSingle corrosion C2 = 1 ~ 2 mmDouble corrosion C2 = 2 ~ 4 mmFor stainless steel,when the media is little corrosive C2 = 0
105 4.Pressure Test and Strength Verification of vessels: i. Purpose(1)Verifying the macro-strength anddeformation of vessels(2)Verifying the tightness of vessels
106 ii. Time(1)For new vessels, the Pressure Test andStrength Verification should be proceededafter completely welded and heat treatment.(2)For inservice vessels, the Pressure Test andafter examination and repair, and beforeputting into production.
107 iii. Media in Test(1)Water —— the most commonly usedStipulation to T of water:Carbon steels、16MnR、normalizing15MnVR —— T≮ 5 ℃Other low alloy steels —— T≮ 15 ℃Stainless steels ——content of [Cl-] in water ≤ 25ppm
108 (2)For the vessels which cannot be filled with liquid, something like dry andclean air, nitrogen gas or other inertgases can be used to fill these vessels.
109 iv. Determination of Pressure for Testing (1)Internal Pressure VesselHydrostatic TestPneumatic Test(2)External Pressure VesselHydrostatic TestPneumatic Test
110 Interpretation of symbols: P —— Design pressure, MPaPT —— Test pressure, MPa —— Allowable stress at testtemperature, MPat —— Allowable stress at design
111 v. Pressure Testing Methods (1)Hydrostatic Test水压测试*Filling the vessel in test with liquid.*Slowly increasing P to the test pressure PT .*Keeping this pressure more than 30minutes.*Decreasing P to 80% of PT .
112 *Checking the welded seam and connection, reducing P to repair them if existingleakage.*Repeating the previous test until upping tograde.*After testing, discharging the liquid anddrying the vessel with compressed air.
113 (2)Pneumatic Test气体测试*Slowly increasing P to 10% of PT as well as≤0.05MPa.*Keeping this P for 5 minutes and have anprimary inspection.*If up to grade, continue to slowly increase Pto 50% PT, then by the △P=10% PTdegree difference increasing slowly P to PT.
114 *Keeping this P for 10 minutes. *Decreasing P to 87% PT, then keeping itand examining and repairing.*Repeating the previous test until upping tograde.
115 (3)Air (gas) Tight Test泄漏性测试*Slowly increasing P to PT.*Keeping this P for 10 minutes.*Decreasing P to the design pressure.*Examining the sealing condition.
116 vi. Stress verification before pressure test (1)Hydrostatic test(2)Pneumatic testT —— Calculating stress at testing pressure,MPas —— Yielding point at testing temperature,
117 5.Examples: i. There is a boiler barrel whose Di=1300mm, working pressure Pw=15.6MPa and it hasa safety valve. Also know that the designT=350ºC, the material is 18MnMoNbR, itis double welded butt with 100% NDE.Try to design the thickness of this boilerbarrel.
118 Solution: (1)Determining the parameters Pc = 1.1PW = 1.1×15.6 = MPa(with the safety valve)Di = 1300mmt = 190MPa (Design T = 350ºC) = 190 Mpa (At normal T, S > ) = 1.0 (Double welded butt, 100% NDE)C2 = 1 mm (Single corrosion, Low alloy steel)
119 (2)Calculating the thickness Design thicknessSd = S + C2 = = 62.5 mm
120 Choosing C1 = 1.8 mm (P95 Figure 4-7) Additional value of wall thicknessC = C1 + C2 = = 2.8 mmNominal thicknessSn = S + C + round-of value= round-of value= 65 mm
121 (3)Hydrostatic test for strength verification *Parameters:*Efficient thickness:Se = Sn - C = = 62.2 mms = 410 MPa
122 *Stress:*Stress verification:That is to say the strength in hydrostatic test is enough.
123 ii. There is a oxygen cylinder which has been kept in storage for a long time, withDo=219mm using 40Mn2A and rolled byseamless steel. The actual Sn= 6.5mm andb = 784.8MPa, s = MPa, 5 = 18%,the design T is normal T.If the working pressure Pw=15MPa, try tofind whether the thickness is enough or not. Ifnot, what is the maximum allowable working pressure in this cylinder?
124 Solution:it is the problem about strength verification—— Whethert ≤ t or not(1)Determining the parametersPc = 15MPa Do = 219 mm Sn = 6.5 mmChoosing the little one: i.e. t = 261.6MPa
125 = 1.0 (for seamless steel) C2 = 1 mmC1 = 0 (for the minimum thickness,negative deviation is neglected.)Se = Sn - C = = 5.5 mm
126 (2)Strength verification Obviously, t > t = MPaSo, 15MPa is too large, should be reduced.
127 (3)Determining the maximum allowable working P So, the maximum safety P for thiscylinder is MPa
128 4.3 Designing Heads subject to Internal Pressure
129 Classification according to the shape: i. Convex headsSemi-spherical headElliptical headDished head (spherical head with hem)Spherical head without hem
130 ii. Conical headsConical head without hemConical head with hemiii. Flat heads
131 1.Semi-spherical head i. Molding of heads Small diameter and thin wall (整体热压成型?) ——Integrally heat-pressingmoldingLarge diameter—— Spherical petal welding球瓣式组焊DiS
133 2.Thickness calculating equation of elliptical head hi (b)hoDiSRi (a)i. Calculating equationfor thickness:For the elliptical heatwhose m = a / b ≤ 2
134 Under the condition about strength: 2. The maximum stress should be at the top point:Putting m = a / b, a = D / 2 into the equation,getting:Under the condition about strength:Then:
135 (1)Replacing P with Pc(2)Multiplying t with welded joint efficiency (3)Substituting D with Di, D = Di + S(4) m = a / b = Di / 2 hiPutting these conditions into the equation:getting:m = a / b = Di / 2 hi
136 For the standard elliptical head whose m=2: For the elliptical head whose m>2: at boundary » and m at the top pointThen introducing the stress strengtheningcoefficient K to replace (Di / 4hi)
137 In this equation:For standard elliptical head: K=1This is the common equation for calculatingthe wall thickness of elliptical heads.
138 Beside these conditions: for standard elliptical heads Se ≮ 0.15% Difor common elliptical heads Se ≮ 0.30% DiThe straight side length of standardelliptical heads should be determinedaccording to P103, Figure 4-11
139 iii. Working stress and the maximum allowable working pressure
140 M —— Shape factor of dished head i. StructureContaining three parts:Sphere: RiTransition arc (hem): rStraightedge: ho (height)DirshoRiii. Calculating equation for thicknessandM —— Shape factor of dished head
141 iii. Working stress and the maximum allowable working pressure
142 iv. Dished headWhen Ri = 0.9 Di & r = 0.17 Dithe dished head is standard dished headand M = 1.325So the equation is:
143 4.Conical head i. Structure *without hem (suitable for ≤ 30 o ) without local strengthwith local strength*with hem (suitable for > 30 o )—— Adding a transition arc and astraightedge between the jointof head and cylinder
144 ii. Calculating equation for thickness The maximum stress is in the main aspectof conical head.mAccording to the strength condition:
145 ThenReplacing P with Pc, considering , andchanging D into Dc ，D=Dc+S
146 This equation only contains the membrane stress but neglects the boundary stress at thejoint of cylinder and head. Therefore thecomplementary design equation should beestablished:(1)Discriminating whether the joint ofcylinder and head should be reinforcedor not.(2)Calculation for the local reinforcement.
147 Conical head without hem ( ≤ 30 o ) (1)Not require reinforcing(consistent thickness for the whole head)main aspect:small aspect:
148 (2)Require reinforcing (for the thickness of joint,the reinforcement region)Main aspect:Small aspect:
149 Interpretation: Dc —— inside diameter of main aspect Di.s —— inside diameter of small aspectDi —— inside diameter of cylinderQ —— coefficient (Consulting the Figure4-16 or 4-18 in book)
150 Conical head with hem ( > 30 o ) (1)Thickness of hem at the transition section(2)Thickness of conical shell at the joint withtransition sectionK —— coefficient (Consulting Figure4-13)f —— coefficient (Consulting Figure4-14)
151 4.Flat head i. Structure The geometric form of flat heads: rotundity, ellipse, long roundness,rectangle, square, etc.ii. Characteristics of loadRound flat with shaft symmetry which issubjected to uniform gas pressure
152 (1)There are two kinds of bending stress states, distributing linearly along the wall.(2)Radial bending stress r and hoop bendingstress t distributing along the radius.
153 max = r.max The maximum stress is at the edge of disk. ▲Fastening the peripherymax = r.maxThe maximum stress isat the edge of disk.SRptr.maxr
154 max = r.max = t.max The maximum stress is in the center of disk. ▲ Periphery with simply supported endsmax = r.max = t.maxThe maximum stress isin the center of disk.SRPtr.maxr
155 iii. Calculation equation for thickness From the condition of strength max ≤ t ,getting:Fastening the peripheryPeriphery with simply supported ends
156 In fact, the supporting condition at boundary of flat head is between the previous two.After introducing the coefficient K which iscalled structure characteristics coefficient andconsidering the welded joint efficient , gettingthe calculating equation for thickness of round disk:
157 5.Examples Design the thicknesses of cylinder and heads of a storage tank. Calculating respectively thethickness of each heads if it’s semi-spherical,elliptical, dished and flat head as well ascomparing and discussing the results.Known: Di = 1200 mm Pc = 1.6Mpamaterial: 20R t = 133Mpa C2 = 1 mmThe heads can be punch formed by a completesteel plate.
158 Solution: (1)Determining the thickness of cylinder = 1.0 (Double welded butt, 100% NDE)C1 = 0.8 mm (Checking Figure 4-7)Sd + C1 = = mmRound it of, getting: Sn = 10 mm
159 (2)Semi-spherical head = 1.0 (wholly punch forming)C1 = 0.5 mm (Checking Figure 4-7)Sd + C1 = = mmRound it of, getting: Sn = 6 mm
160 (3)Standard elliptical head = 1.0 (wholly punch forming)C1 = 0.8 mm (Checking Figure 4-7)Sd + C1 = = mmRound it of, getting: Sn = 10 mm
161 (4)Standard dished head = 1.0 (wholly punch forming)C1 = 0.8 mm (Checking Figure 4-7)Sd + C1 = = mmRound it of, getting: Sn = 12 mm
162 (5)Flat headK = 0.25; Dc = Di = mm; t = 110 Mpa = 1.0 (wholly punch forming)C1 = 1.8 mm (Checking Figure 4-7)Sd + C1 = = mmRound it of, getting: Sn = 80 mm
163 Comparison: Selection: Head-formSn mmkgSemi-sphe.EllipticalDishedFlat6106101280137163662Selection:It’s better to use the standard elliptical head whose thickness is the same to that of cylinder.
164 Chapter 5 Design of Cylinders and Formed Heads subjected to External-Pressure5.1 Summarization
165 1.Failure of External Pressure Vessel 外压容器失效 Under the effect of external pressure, thevessels may deform when the pressure is largerthan a certain value. This kind of damage iscalled the failure of external pressure vessels.2.Classification of FailureSide bucking —— the main form of failureAxial buckingLocal bucking
166 5.2 Critical Pressure 1.Critical pressure and critical compressive stress临界压力The pressure that makes the externalpressure vessels fail is called the criticalpressure, indicating by Pcr.At the moment that exists Pcr, the stressinside the vessels is called the criticalcompressive stress, indicating by cr .
167 2.Factors affect the critical pressure i. Geometric dimension of cylinder 903500.3(3) 903500.3(4)Degree of vacuum in failure 901750.51(1) 901750.3(2)Pcrmm HO2500300120~150
168 Comparison and analysis for the experimental resultsFigure (1) and Figure (2):*When the value of L / D is equal, the larger the value of S / D, the higher the Pcr.Figure (2) and Figure (3):*When the value of S / D is equal, the smaller the value of L / D, the higher the Pcr.Figure (3) and Figure (4):*When the value of S / D and L / D are equal, having the stiffening ring as well, high Pcr.
169 ii. Materials’ Performance of the cylinders (1)The critical pressure (Pcr) hasn’t directrelation with the strength ( s) of thematerials.(2)The critical pressure (Pcr) depends of theflexural rigidity of the cylinders in someaspects.The stronger the flexural rigidity, the moredifficult for the failure.
170 iii. The differential in the dimension at the process of vessels’ manufacturingMainly reflecting on the “ellipticity” (椭圆度),which is the processing differential in thedimension of the cylindrical section.
171 *Large ellipticity e can make the critical DmaxDmin*Large ellipticity e can make the criticalpressure Pcr decrease and failure happen inahead.*Regulated as in the engineering, ellipticitye ≤ 0.5% when vessels subjected to theexternal pressure are made.
172 3.Long cylinder, short cylinder and rigid cylinder, the calculating equations of their critical pressure
173 Calculating equation of the critical stress: i. Long cylinder—— cylinders with large L / DoCalculating equation of the critical P:For steel cylinders: = 0.3Calculating equation of the critical stress:
174 ii. Short cylinder—— cylinders with small L / DoCalculating equation of the critical P:Calculating equation of the critical stress:
175 iii. Rigid cylinder—— cylinders with small L / Do, large Se / DoDesigning criterion:Only need to satisfy the strength condition:compression ≤ [ ]tcompressioni.e.
176 4.Critical Length临界长度Critical length —— which is used to classify the long cylinder and short cylinder; and it is the critical dimension of the short cylinder and rigid cylinder.L > Lcr Long cylinderL’cr < L < Lcr Short cylinderL < L’cr Rigid cylinder
177 i. Critical length Lcr of long and short cylinder: ii. Critical length L’cr of short and rigid cylinder:
179 Pc —— Calculating Pressure, MPa Pcr —— Critical Pressure, MPa[p] —— Allowable External Pressure, MPam —— Stable safety coefficientFor cylinders, m = 3at the same time, 椭圆度 e ≤ 0.5%
180 2.Nomograph for the thickness designing of the external-P cylinders i. Calculating StepsStep 1: L、 Do、 Se → Drawing the curve
181 Step 2: Find the relationship between and [P] For cylinder m=3 andThen getting the relationship curve B = f ()
182 ii. Steps of nomograph for the thickness designing of the external-P cylinders (Tubes) For the cylinders and tubes whose Do/Se ≥20: (1)Supposing Sn, Se = Sn - C, calculating the values of L / Do and Do / Se.(2)Calculating the value of (value of A), checking the Figure (5-5).If L / Do > 50, checking the figure usingL / Do = 50.If L / Do < 0.05, checking the figure usingL / Do = 0.05.
183 (3)Calculating the value of B According to the used material, choosingthe relevant graphs from Figure (5-7) andFigure (5-14) and then finding the point Afrom abscissa.
184 Two situations maybe encountered: *Point A with that certain value lies at the rightof the curve and intersects with the curve,then the value of B can be found directly in thefigure.*Point A with that certain value lies at the leftof the curve and has no joint with the curve,then the value of B is calculated by thefollowing equation:
185 (4)Calculating [P]Putting the value of B into Equation (9) →[P]
186 i.e. the supposed Sn is usable, safe (5)Comparingi.e. the supposed Sn is usable, safei.e. the supposed Sn is too large andshould be decreasedappropriately, repeating the previous calculatingsteps until satisfying the first condition.i.e. the supposed Sn is too small andshould be increased appropriately,repeating the previous calculating steps untilsatisfying the first condition.
187 3.Pressure test of external-P vessels 外压容器的压力测试 i. Pressure test of external-P vessels and vacuum vessels is processing as the hydrostatic pressure test.Testing pressure:PT = 1.25 PP —— design pressure
188 ii. Vessels with jackets (Jacketed Vessels) 夹套容器(1)Welding the jacket before the hydrostatic test to the cylindrical parts of jacketed vessels is assured to be completely qualified.(2)Taking another pressure test to the jacket after welding the jacket.Testing pressure: PT = 1.25 P(3)At the cause of pressure test to the jacket, the stability of the cylindrical part should be guaranteed. If necessary, charging pressure into the cylinder to make the internal-external pressure difference less than the design pressure.
189 4.Example and discussion Design the thickness of an external-P cylinder.Known:Calculating pressure: Pc = 0.2 MPaDesign temperature: t = 250℃Inside diameter: Di = 1800 mmCalculating length: L = mmAdditional value of wall thickness: C = 2 mmMaterial: 16MnR; Et = 103 Mpa
190 Solution: (1)Assuming Sn = 14 mm Then Do = Di + 2 Sn = 1828 mm Se = Sn - C = 12 mmFinding out:L / Do = 1828 = 5.7Do / Se = 12 = 152(2)Calculating the value of (A)Checking the Figure 5-5, getting:A =
191 (3)Calculating the value of B From Figure 5-9, we can see that point A is at the left of the curve, then the calculating equation is like following:
192 (4)Calculating [P](5)Comparing [P] and Pc[P] < Pc = 0.2 MPa unsatisfiedReassuming Sn, or setting the stiffening ring.
193 Calculation under the condition that supposes there have two stiffening rings: (1)Thickness is the same: Sn = 14 mmAfter setting two stiffening rings,the calculating length is like following:
194 (2)Calculating the value of (A) Checking the Figure 5-5, getting:A =(3)Calculating the value of BFrom Figure 5-9, we can see that point A is at the right of the curve, getting B = MPa
195 (4)Calculating [P](5)Comparing [P] and Pc[P] > Pc = 0.2 MPa satisfied
196 Calculation under the condition that supposes to increase the thickness: (1)Assuming: Sn = 20 mmThen Do = Di + 2 Sn = 1840 mmSe = Sn - C = 18 mmFinding out:L / Do = 1840 = 5.6Do / Se = 18 = 102
197 (2)Calculating the value of (A) Checking the Figure 5-5, getting:A =(3)Calculating the value of BFrom Figure 5-9, we can see that point A is at the right of the curve, getting B = MPa
198 (4)Calculating [P](5)Comparing [P] and Pc[P] > Pc = 0.2 MPa and closingSo, we can use the steel plate withSn = 20 mm, whose material is 16MnR.
199 Spherical Shell and Convex Head 5.4 Design of External-PSpherical Shell and Convex Head1.Design of external-P sphericalshell and semi-spherical headi. Assuming Sn, and Se = Sn － C.Calculating the value of Ro / Se.
200 ii. Calculating the value of (A) iii. Calculating the value of B and [P]According the used material, choosingthe relevant graph from Figure 5-7 andFigure 5-14 and finding out the point A at theabscissa.
201 Two situations maybe encountered: (1)If point A is at the right of the curve, the value of B can be found from the figure directly.(2)If point A is at the left of the curve, directly calculating:
202 iv. Comparisoni.e. the original assuming Sn isusable, and safety.i.e. the original assuming Sn istoo small, Sn should beincreased appropriately,repeating the previouscalculating steps until satisfyingthe first condition.
203 2.Design of external-P convex head The method of designing the external-Pconvex head is the same to that of designingexternal-P spherical head. But the Ro in thedesigning of spherical head should be adjustedlike following:
204 i. For elliptical headRo —— the equivalent spherical diameter ofelliptical head; Ro = K1DoK1 —— coefficient; depending on a / b,checking P141, Figure 5-3
205 ii. For dished headRo —— the equivalent spherical diameterof the dished head; it’s the outsidediameter of the spherical part atthe dished head.
206 5.5 Design of the Stiffening Ring in External-P Vessels 1.Function of stiffening ring
207 From the previous equations, we can know the methods to increase [P]: i. Increasing Sii. Decreasing the calculating length L∴ Function of stiffening ring加强圈的功能—— decreasing calculating length toincrease [P]
208 2.Space length and number of stiffening ring Assuming the space length of stiffening ring is LsFrom the design criterions of external-P:Pc ≤ [P] and [P] = Pcr / mMaking Pc = [P] then Pcr = m Pc (a)From the equation for the critical pressure of short cylinder:
209 Putting equation (a) in, getting: Then putting m=3 in, getting:
210 (Ls)max —— Under the condition that Do and Se of the cylinder is determined,the maximum space lengthbetween the needed stiffeningrings working safely under thecalculating external pressure Pc,mm.
211 The actual space length between stiffening rings Ls ≤ (Ls)max is indicating safety. The number of stiffening rings:In the above equation:L —— the calculating length of cylinder beforesetting the stiffening rings, mmLs—— the space length between stiffeningrings, mm
212 3.Connection of stiffening rings and cylinders Connection Demands连接要求Must assure all the cylinder and stiffeningring are under the load together.ii. Connection Methods连接方法Welding —— Continuous Weld (连续焊接)Tack Weld (间断焊接)
213 iii. The stiffening rings should not be randomly crippled or cutoff. If those must be done, thelength of the arc that are crippledor cut off should not be largerthan the values shown in Figure 5-19.
214 For example:There is a horizontal external pressure vessel.When the stiffening ring is set inside thecylinder, in order not to affect the fluid flowingor fluid discharging, we must leave a hole豁口atthe lowest position of the stiffening ring or set athoroughfare of fluid.
215 As illustrating like the following two figures
216 Chapter 5 Components and Parts of Vessels 6.1 Flanges Connection1.The Sealing Theory (密封原理) and Connection Structure of Flanges
217 连接结构 Connection Structure Three parts: (1)Connected parts —— a couple of flanges(2)Connecting parts—— several couples ofbolts and nuts(3)Sealing parts—— gasket
218 ii. Sealing Theory密封原理Taking the bolts’ forced sealing as an example to illustrate the Sealing Theory:(1)Before butting(2)After butting(3)After chargingmedium
219 Classification of Flanges 2.The Structure andClassification of FlangesAccording to the connection ways of flanges and equipment (pipelines)(1)Integrated flange—— S.O.flange (slip on flange)W.N.flange (welding neck flange)
223 3.Factors effect the sealing of flanges 影响密封因素 i. Bolt load under pretension condition(bolt load for gasket sealing)The bolt load is too small to seal specific pressure (顶紧密封比压); the bolt load is too large to avoid the gasket being pressed or extruded.
224 Increasing the bolt load appropriately can strengthen the sealing ability of gasket. So under the condition of certain bolt load, decreasing the diameter of bolts or increasing the number of them are both beneficial for sealing.
225 ii. The types of sealing face (1)plain (face) flange(2)M&F (male and female)(3)T&G (tongue and groove face)(4)Conical face(5)Trapezoidal groove face
227 iii. Properties of gasket (1)The common-used materials of gasket*Non-metal Material—— Rubber, Asbestos, Synthetic resins.Advantages: soft and corrosion resistantDisadvantages: the properties of high-Tresistance and pressureresistance is inferior to themetallic materials.Used in: Common and Medium T; Flangesealing of Medium and Low Pdevices and pipes.
228 *Metal (Metallic) Material 金属垫片—— soft aluminum, copper, iron (softsteel), 18-8 stainless steel.Advantages: high-T resistant, with high strengthDemands: Excellent soft toughnessUsed in: Medium and high T; Flange sealingsubjected to medium and high P
229 (2)Gasket Types (Classifying according to the properties of materials)*Non-metal Gasket—— such as rubber gasket, asbestos-rubbergasket.*Compound Gasket (Metal and non-metalcompound gasket)—— such as metal jacketed gasket (金书包垫片) and Metal spirotallic [spiral-wound] gasket
230 Metal jacketed gasket (金书包垫片), i.e. wrapping the metal slice around the asbestosgasket or asbestos-rubber gasketMetal spirotallic [spiral-wound] gasket (金属缠绕垫片), i.e. making by alternately rollingthin steel belt and asbestos
231 *Metal gasket—— such as octagon ring gasket, ellipticalgasket, lens ring (washer) [groovedmetallic gasket]
232 (3)Selection of gasket垫片选择*Factors of working pressure and temperatureMedium and low P; common and medium T—— Non-metal gasketMedium P; Medium T—— Metal and non-metal compound gasketHigh P; high T —— Metal gasketHigh vacuum; cryogenic —— Metal gasket
233 *Degree of demands for sealing *Demands for the types of sealing face*Properties of gasketConcrete selection should be referred toJB , JB , JBAt the same time, the practical experienceshould be taken into account.
234 iv. Rigidity (刚度) of flange (1)If the rigidity of flange is not enough, therewill occur the serious buckling [翘曲]deformation, as well the specific pressure willdecrease and the sealing face will be loose, asa result, the sealing will fail.(2)Measures to increase the rigidity of flange(3)Strengthening the rigidity of flange toincrease the weight of flange as well the valueof whole-flange’s sealing.
235 Corrosive Characteristics Penetrant Characteristics v. Effect of working conditionsof mediumTemperaturePressureCorrosive CharacteristicsPenetrant CharacteristicsCombined effectGreatly affecting the sealing
236 4.Standard and Selection of Flanges i. Standard number of pressure vessel flangesJB / T ∼ JB / T
237 Standard types and marks of pressure vessel flanges A-S.O.FlangeJB/TB-S.O.FlangeJB/TW.N.FlangeJB/TPlainM&FT&GMFTGPASCTypeSealingCodeWithout lined ringWith lined ring (C)faceStandard types and marks of pressure vessel flanges
239 StandardCodeNominalPressureMPaC-C 800 — 1.6 JBCode of Flange TypeNominalDiametermmCode of Sealingface Type
240 ii. Dimension of pressure vessel flanges Dimension of flanges is only confirmed bytwo standardized parameters PN and DN offlanges.Confirmation of Nominal Pressure PN offlanges: JB (Book, P160)
241 iii. Selection steps for pressure vessel flanges 法兰选用步骤(1)According to the design task, confirmingthe types of flanges (S.O. or W.N.).(Referring to P157 Table 6-2)(2)According to the nominal diameter DN offlanges , working temperature, designpressure, material of flanges, confirmingthe nominal diameter DN and nominalpressure PN of flanges.(Referring to P160 Table 6-4; P332 Appendix 12)
242 (3)Confirming the sealing face types of flanges and the types of gaskets.(Referring to P155 Table 6-1)(4)According to the types of flanges, DN andPN of flanges, checking and finding out thedimension of flanges; number of bolts andtheir specification.(Referring to P336 Appendix 14)
243 (5)Confirming the material of bolts and nuts. (Referring to P163 Table 6-6; P333Appendix 13)(6)Portraying the unit drawing of flanges.
244 Example: There are flanges to connect the body of a fractionating (rectifying) tower and the heads.Knowns:Inside diameter of tower: Di = 1000mmWorking temperature: t = 280℃Design Pressure: P = 0.2MPaMaterial of tower: Q235-AR
245 Solution: (1)From P157 Table 6-2, A-S.O.Flange is selected. (2)Confirming the nominal diameter DN andnominal pressure PNDN = 1000 mm(Equal to the inside diameter of tower)From P160 Table 6-4, choose the material oftower as that of flanges, i.e. Q235-ARt = 280℃
247 (3)Confirming the sealing face types of flanges From P155 Table 6-1, choosing plain sealingface, spirotallic [spiral-wound] gasket(4)According to the DN and PN of flanges, fromAppendix 14, Table 32, finding out thedimension of every part of flanges.Specification of bolts: M20; Number: 36
248 (5)From P163 Table 6-6, finding out: Material of bolts: 35 steelMaterial of nuts: Q235-A(6)Portraying the unit drawing of flanges(Omitting)
249 Standard of tube flanges (New Standard issued by Chemical Ministry)European: HG — 97 ~ HG — 97American: HG — 97 ~ HG — 97
250 6.2 Support for vessels Support for horizontal vessels Saddle support, ring support, leg, etc.Support for vertical vesselsSkirt support, hanging support, etc.
251 1.Double-saddle support i. The structure of double-saddle support120°GasketWeb-plateAnchor boltSub-plate
252 Model Type Nominal Diameter Type ii. Position of support (A)安座位置A≤Do/4 & < 0.2L. The maximum value < 0.25Liii. Standard and selection of double-saddlesupportType —— Stationary type: F Movable type: SModel Type —— Light-duty: A Heavy-duty: BMark —— JB / T SupportModel TypeNominal DiameterType
253 2.Checking calculation of stress in double-saddle horizontal vessels i. Load analysis for horizontal vessels
255 ii. Reserved force to support In this equation:q —— Mass load/unit length of vessels,N / mmL —— Distance between the T.L.(tangent lines) of two heads, mmhi —— Height of curved surface of heads,mm
256 iii. The maximum radical bending moment The section across the middle point of momentThe section at support
257 iv. Calculation for radical stress of cylinder —— to the vessels subjected to positive pressure(1)Stress across the middle sectionThe most highest point in section (Point 1):The most lowest point in section (Point 2):
258 (2)Stress in the section of support The most highest point in section (Point 3):The most lowest point in section (Point 4):
259 v. Checking calculation for radical stress of cylinderRadical tensile stressRadical compressive stress
260 In these two equations: t —— The allowable stress of material at thedesign T, MPac r —— The allowable compressive stress ofmaterial, MPaB —— Calculation method is the same with thatin design of external pressure, see P172
261 6.3 Reinforcement for opening of vessels 1.The phenomena and reason for opening stress concentration
262 Stress concentration factor: max —— The maximum stressat the boundary of opening* —— The maximum basicstress of shellSmall openingin plate
263 Reasons for stress concentration: (1)Material of vessel wall is deteriorate(2)The continuity of structure is damaged
264 2.Opening reinforcement’s Designing i. Designing Criterions(1)Equi-area criterion of reinforcement(2)Plastic failure criterion of reinforcement
265 ii. Reinforcement Structure (1)Structure of Stiffening Ring Nozzle (Connecting Tube)ShellStiffening Ring
266 (2)Structure of 加强元件Method —— Taking theparts of nozzles or vicinity ofshells’ openings which needto be reinforced as the 加强元件, then welding theseparts with nozzles or shells.
267 (3)Structure of Integral Reinforcement Method —— Taking the connecting parts of nozzles and shells as the integral forgings, at the same time thickening them, then welding them with nozzles and shells.
268 iii. Diameter Range of the openings that need not to be reinforcedWhen the following requirements are all met, the reinforcement is out of need.(1)Design Pressure P ≤ 2.5 MPa(2)The distance between two mid-points of two nearby openings (taking length of are as the length of curved surface) should be larger than 2× (D1+D2), D1, D2 are the diameters of the two openings respectively.
269 (3)Nominal Outside Diameter of connecting tubes ≤ 89 mm (4)The minimum wall thickness δminof connecting tubes should meetthe following requirements:(mm)δmin2532384548576576893.54.05.06.0
270 Metallic areas in local reinforcement 3.Designing methods ofequi-area reinforcementMetallic areas in local reinforcement≥the area of sections which arethe position of openings
271 i. Confirmation of the effective range of opening and reinforcement areas Bh1h2d
273 In these equations: Sn —— Nominal thickness of cylinders Sn.t —— Nominal thickness of connectingtubes (nozzles)d —— Diameter of openings d = di +2Cdi —— Inside diameter of openingsC —— Additional value of wall thickness
274 Computation of metallic areas for effective reinforcement (1)The area of the sections on shell whichare the positions of openings A: A = S×d(2)The unnecessary metallic area A1 on shell orheads which is larger than calculating thickness S:A1 = (B – d) (Se – S) – 2 (Sn.t – C) (Se – S) (1 – fr)
275 (3)The unnecessary metallic area A2 on nozzles which is larger than the calculating thickness S:A2 = 2 h1 ( Sn.t – St –C ) fr + 2 h2 ( Sn.t – C – C2 ) fr(4)The metallic area of welding seam in thereinforcement region A3:A3 = according to the actual dimension
276 ii. Designing Steps in Reinforcement for openings (1)Getting the following data from the strengthcalculation:Calculating wall thickness of cylinders or heads SNominal wall thickness of cylinders or heads SnCalculating wall thickness of nozzles StNominal wall thickness of nozzles Sn.tAdditional value of wall thickness C = C1+ C2
277 (2)Calculating the effective reinforcement range B, h1, h2(3)Calculating the necessary reinforcement areaA according to P183 Table 6-17(4)Calculating the available reinforcement areaA1, A2, A3
278 (5)Judging whether it is necessary to add some reinforcement areaIf A1 + A2 + A3 ≥ Areinforcement not requiredIf A1 + A2 + A3 < Areinforcement required
279 (6)If reinforcement is required, calculating the added reinforcement area A4A4 = A －（ A1 + A2 + A3 ）(7)ComparisonFinally getting A1 + A2 + A3 + A4 ≥ A
280 6.4 Attachment of vessels 1.Man Hole and Hand Hole i. Nominal Diameter of standard man-holeDN ：ii. Nominal Diameter of standard hand-holeDN ：
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