Report

Section 2 Materials of Chemical Vessels Chapter 2 Panorama 2.1 Structure and Classification of Vessels 1.Conception of Vessels: 容器概念 Chemical Vessels are the external shells of various equipments in the chemical process. 2.Structure of vessels: 3.Classification of vessels: i. According to pressure and its type (1)内压容器 Internal Pressure Vessel —— vessels where the media pressure inside the vessel is larger than that outside (gauge pressure). Low pressure vessel (L): 0.1≤P < 1.6 MPa Medium pressure vessel (M): 1.6 ≤P < 10 MPa High pressure vessel (H): 10 ≤ P < 100 MPa Ultra-high pressure vessel (U): P ≥100 MPa (2)External Pressure Vessel 外压容器 —— vessels where the media pressure inside the vessel is lower than that outside (gauge pressure). When the internal pressure < 0.1 MPa (absolute pressure), such vessels are called Vacuum Vessel. ii. According to temperature Normal temperature vessel -20< T ≤200℃ Medium temperature vessel —— between normal T & high T vessels Low temperature vessel < -20 ℃ High temperature vessel —— where the wall temperature is above the creep temperature. High temperature vessel Carbon steel & Low-alloy steel T> 420 ℃ Alloy steel (Cr-Mo steel) T> 450 ℃ Austenite stainless steel (Cr-Ni) T> 550 ℃ iii. According to management P, P*V Factor media importance Degree of danger: I < II < III Grade (I) Grade (II) Grade (III) 2.2 Basic Law and Common Standard for Design of Pressure Vessels 1.Conditions: i. The maximum working pressure Pw≥ 0.1 MPa (neglecting the net pressure of liquid ) ii. Internal diameter Di (equal the maximum dimension or size in the non-circular sections) ≥ 0.15m, and V ≥ 0.025 m3 iii. With the medium as gas or liquefied gas, or liquid whose maximum working T ≥ standard b.p. 2.Basic Law and Common Standard: i.《 Security Technique Supervising Rules for Pressure Vessel》1999 ii. GB 150 — 1998 《Steely Pressure Vessel》 iii. GB 151 — 1999 《Pipe-shell Heat Exchanger》 iv. JB 4710 — 2000 《Steely Tower Vessel》 2.3 Standardization of Pressure Vessel Parts 1.Significance of Standardization: 标准化的重要性 i. It consolidates and harmonizes the various activities in the manufacture and social life. ii. It’s the importance means to organize the modernization production. iii. It’s the importance component in scientific management. iv. It makes for the development of new products, assures the interchangeability and common-usability and is convenient to use and maintain. v. It’s helpful in the interchange of international science & technology, culture and economy. 2.Active Chemical Vessels and Standards of Equipments Components: Cylinder Heads Vessel Flange Pipe Connecting (Nozzle) Flange Support Stiffening Ring Manhole Handhole Sight (level) Glass Liquid Leveler (LG) Corrugated expansion joint Heat Exchanger Tube Tray Float (floating) Valve (FV) Bubble (bubbling) cap Packing etc. 3.Basic parameters of standardization: i. Nominal Diameter —— DN (Dg) 公称直径 is a typical dimension. (1)Rolled cylinder and head from steels DN = Di (inside diameter) DN Standard of pressure vessels: 300 350 400 450 500 …… 6000 48 grades in total. (2)Cylinder made by seamless pipes —— DN = Do (outside diameter) Six grades: 159 219 273 325 377 426 (3)Seamless pipes —— DN≠Di & DN ≠ Do, but DN is a certain value that is smaller than Do. When the DN is a certainty, Do is to be a certainty, while Di depends on the thickness. Denotation of seamless pipes: such as 252.5 (outside × thickness) Check the standards according to DN. Comparison of DN and Do of seamless pipes/mm DN 10 15 20 25 32 40 50 65 Do 14 18 25 32 38 45 57 76 S 3 3 3 3.5 3.5 3.5 3.5 4 (4)DN of flanges —— consistent to their suitable cylinders, heads and tubes. i.e. DN of flange = DN of cylinder DN of head DN of tube ii. Nominal Pressure —— PN (Pg) 公称压力 Prescriptive standard pressure grades For example —— PN (MPa) of pressure vessel flange: 0.25 、 0.6 、 1.0 、 1.6 、 2.5 、 4.0 、 6.4 iii. Application In standard designing: (1)Diameters of cylinders, heads and tubes must be close to the standard grades. e.p. Diameter of cylinder should be 500 、600 、700 … shouldn’t be 520 、645 、750 … (2)When selecting the standard vessel parts, the design pressure at the working temperature should be regulated to a certain grade of PN. Then choose the parts according to PN and DN. 4.Classification of standards: i. Chinese Standard Symbol: GB (Guo biao) ii. Standard issued by Ministry JB —— Ministry of Mechanical Industry YB —— Ministry of Metallurgical Industry HB —— Ministry of Chemical Industry SY —— Ministry of Petroleum Industry iii. Specialty Standard iv. Trade Standard v. Plant Standard 2.4 Basic Requirements & Contents of Vessel Design 1.Basic requirements: 基本要求 i. Enough strength —— no breakage ii. Enough rigidity —— limit deformation iii. Enough stability —— no failure, ?, drape iv. Durability —— assuring certain usage life v. Tightness —— no leakage vi. Saving materials and easy to manufacture vii. Convenient to be installed, transported, operated and maintained viii. Rational technical economy index in total 2.Basic Contents: i. Selection of materials Selecting the materials of equipment according to the technical indexes t, p, media and the principles of material selection. ii. Structure design iii. Calculation of strength and thickness (including the cylinders and heads) iv. Strength verification in hydraulic pressure test v. Seal design; selecting or designing flanges vi. Selection of support & the verification of strength and stability vii. Design and calculation of reinforcement for opening viii. Selection of other parts and accessories ix. Other special design x. Plotting the equipment drawings xi. Compiling the equipment specifications Chapter 3 Stress Analysis of Thin-walled Internal-P Vessel 3.1 Stress Analysis of thin-walled Cylinders Subjected to Internal Pressure 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 ≥ 0.1 2.stress characteristics: There are always two kinds of stress in pressure vessels. i. membrane stress —— membrane (shell) theory ii. boundary stress —— shell theory with moments and conditions of deformational compatibility P 3.2 Membrane Theory —— Rotary Shells’ Stress Analysis 1.Basic conceptions and hypothesis: i. Basic conceptions (1)rotary curvedsurface & shell (2)Axial Symmetry Geometry figure, endured load and restrictions of shell are all symmetry to the revolving shaft (OA). Several basic conceptions: generatrix, meridian, normal, latitude, longitudinal radius, tangential radius. O (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’. B B’ A (5)Normal (n) The line passing point M in meridian and is vertical with midwall surface. The extension of · K1 normal must intersect with OA. C (6)Latitude (CND) The cross-line formed by the conic surface passing point K2’ intersects with the rotary curved surface. O K2 · K2’ ·M · D N A n (7)Longitudinal radius (R1) *The radius of curvature of meridian which passes point M in midwall surface is called the longitudinal radius of point M in meridian. *The center K1 of curvature of the round with diameter R1 must be in the extension of normal passing point M. For example: Longitudinal radius of point M: R1 = M K 1 (8)Tangential radius (R2) *The plane which is vertical to the normal passing the point M in meridian intersects with the mid-wall surface, the resulted cross line (EMF) is a curve, the radius of curvature of this curve in point M is called tangential radius. *The center K2 of curvature of the round with diameter R2 must be in the extension of normal passing point M and in the revolving shaft. For example: Longitudinal radius of point M: R2 = M K 2 ii. Basic hypothesis: Small displacement hypothesis Straight linear law hypothesis Non-extrusion hypothesis 2.Free body balance equation —— calculation formula of radial (meridional) stress: i. Intercepting shell —— uncovering the radial stress m ii. Choosing separation body iii. Analysis of stress iv. Constitute balance equation Fz 0 4 Pz N z 0 m .k 2 D p m DS sin 0 2 D/2 sin R2 puttingit C p into theaboveequation,getting pR2 m 2S (3 - 3) z D R2 C’ 3.Infinitesimal balance equation —— calculation formula of hoop stress: i. Intercepting shell —— uncovering the radial stress m ii. Choosing separation body iii. Analysis of stress iv. Constitute balance equation k1 d1 k2 Sdl1 d2 Fn 0 Pn Nm.n N .n 0 p d1 pdl1dl 2 2 m Sdl 2 sin 2 d 2 2 Sdl1 sin 0 2 整理得： m p R1 R2 S (3 - 4) mSdl2 Basic calculation equation of membrane stress: pR2 m 2S m p R1 R2 S Illustration of symbols: m —— radial stress of a random point in rotary thin shell, MPa —— hoop stress of a random point in rotary thin shell, MPa P —— internal pressure, MPa S —— thickness of wall, mm R1 —— longitudinal radius of required stress point in the mid-wall surface of the rotary shell, mm R2 —— tangential radius of required stress point in the mid-wall surface of the rotary shell, mm v. Application range of membrane theory ☆Applicable to axial symmetric thin-walled shell 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 3.3 Application of Membrane Theory Calculation equations: pR2 m 2S m p R1 R2 S 1.Cylindrate shell subjected to uniform gas internal pressure: D pD m 4S pD 2S s ∵ R1 = ∞ R2 = D / 2 Putting them into the previous equations: p 2.Spherical shell subjected to uniform gas internal pressure: ∵ R1 = R2 = D / 2 Putting them into equations (3-3) and (3-4): pD m 4S pD 4S 3.Elliptical shell subjected to uniform gas internal pressure: Example: Known: Major semiaxis - a Short semiaxis - b Thickness - S Internal Pressure - P Find the m and of a random point on the elliptical shell. Solution: (1)Find R1 and R2 of point A: R1: radius of curvature of A 1 y ' y 2 '' 3 2 y . A(x,y) .k 1 (a) By the elliptical equation: a .k 2 x2 y2 2 1 2 a b getting y’ and y’’, then put them into (a). result is: 3 1 4 2 2 2 2 R1 4 a x a b a b x R2: here: R2 = K2 A = x / sin tg sin 1 tg 2 (b) 2 b x tg y 2 a y ' 2 b 2 y b 2 x a putting them into (b), getting: 1 4 2 2 2 R2 a x a b b 2 2 (2)Find m and of point A: Putting R1 and R2 into (3-3) and (3-4), getting: p 4 2 2 2 m a x a b 2Sb p 4 2 2 2 a x a b 2Sb 4 a 2 4 2 2 2 a x a b Stress of special points on elliptical shell: (1)x=0 (Top of elliptical shell) m pa a 2S b (2)x=a (Boundary or equator of elliptical shell) pa m 2S pa a 2 2 2S b 2 Standard elliptical heads: The elliptical heads whose ratio of major and short semiaxis a / b = 2 are called standard elliptical heads. a / b = 2 —— pa pD m x=0 (Top): S 2S x=a (Boundary): m a/b=2 pa pD m 2S 4S pa pD S 2S 4.Conic shell subjected to uniform gas internal pressure: Example: Known: Diameter of tapered bottom - D Half tapered angle - Thickness - S Internal Pressure - P Find the m and of a random point on the conic shell. D Solution: (1)Find R1 and R2 of point A: k2 . R1 = ∞ r A R2 = A K2 = r / cos (2)Find m and of point A: Putting R1 and R2 into (3-3) and (3-4) respectively, getting: pr 1 m 2 S cos pr 1 S cos Characteristics of stress distribution of conic shell: m 5.Cylindrate shell subjected to liquid static pressure: x i. Supporting along the boundary of bottom Example: Po Known: .A Gauge pressure – Po (Pa) Liquid level – H (m) D Density of liquid - (N/m3) Find the m and of a random point on the wall of cylindrical shell. H Solution: (1)Meridional stress: Cutting along section B-B, taking the lower part as the separation body. Po m .A x B (Po+x) B H (H-x) D N B-B m Establishing the balance equation of axial stress: Fx 0 2 2 2 ( po x) D ( H x) D m D S H D 4 4 4 m po D 4S (2)Hoop stress: Infinitesimal balance equation (3-4): m R1 R2 P S For point A: R1 = ∞ R2 = D/2 P = Po + x Putting them into (3-4), when x=H: getting: po x D po H D max 2S 2S ii. Supporting along the boundary of top Example: Known: P Gauge pressure – Po (Pa) Liquid level – H (m) D 3 Density of liquid - (N/m ) Find the m and of a random point on the wall of cylindrical shell. o x H .A Solution: (1)Meridional stress: Cutting along section B-B, taking the lower part as the separation body. Po B x .A H D B-B m (Po+ x) (H-x) m H-x B Establishing the balance equation of axial stress: Fx 0 2 2 H x D po x D m D S 0 4 4 m po H D 4S (2)Hoop stress: Infinitesimal balance equation (3-4): m R1 R2 P S For point A: R1 = ∞ R2 = D/2 P = Po + x Putting them into (3-4), getting: when x=H: po x D po H D max 2S 2S 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 semiaxis is a / b = 2. The average diameter D=420mm, thickness of all cylindrical shell and heads are 8mm. The working pressure P=4MPa. Calculating: (1)Find m and of the shell body. (2)Find the maximum stress on the both the heads and their position respectively. Solution: (1)m and : pD 4 420 m 52.5 (MPa ) 4S 48 pD 2 m 105 2S P (MPa) (2)Upper head —— spherical pD m 52.5 4S (MPa) D S (3)Lower head —— elliptical When a / b = 2: a = D/2 = 210 mm b = a/2 = 105 mm x=0 (Top): pa a pa pD m 105 2S b S 2S (MPa) pa pD x=a (Bottom): m 52.5 (MPa) 2S 4S 2 pa a pa pD 2 2 105 (MPa) 2S b S 2S 3.4 Conception of Boundary Stress 1.Forming of boundary stress: Boundary 边界 —— The joint and its vicinity of two parts with different geometry shape, load, material and physical conditions, i.e. discontinuous point. Boundary stress forming not for balancing the loads but for receiving restrictions from self or exterior. It’s a group of internal force with same value but contrary direction occurring between two parts which are forced to realize transfiguration harmonization. 2.Characteristics of boundary stress: 边缘应力特性 i. Distributing along the wall non-evenly ii. Different joint boundary forming different boundary stress iii. It’s local stress, i.e. only forming large stress locally and decaying apparently iv. Value of boundary stress can be 3~5 times of that of membrane stress v. Self-constrained 3.Treatments to boundary stress: i. Treatments locally in structure (1)Improving the structure of joint boundary (2)Strengthening the boundary locally (3)Assuring the quality of welding line at boundary (4)Decreasing the remnant stress at local and processing the heat treatment to eliminate the stress (5)Avoiding the local stress added to the boundary region overlap with connatural stress ii. Materials are of high plasticity Chapter 4 Strength Design of Cylinders and Heads subjected to Internal-Pressure 4.1 Basic Knowledge of Strength Design 1.Criterions of elasticity failure: eq < t s Safety Allowance kept for the requirements of safety: o eq n eq —— equivalent stress o —— limiting (ultimate) stress, can be s、 b、 n、 D, etc. [] —— allowance stress n —— safety coefficient 2.Strength Theory: i. The first strength theory —— the maximum tensile stress theory I eq 1 [ ] applying to brittle materials ii. The second strength theory —— the maximum major strain theory iii. The third strength theory —— he maximum shear stress theory t 1 3 max 2 Shear limit: s Shear limit : t 2 Failure condition: 0 t max t 1 s i.e. ( 1 - 3 ) 2 2 0 or 1 3 s Strength condition: 1 3 [ ] III eq Applying to the plastic materials iv. The fourth strength theory —— the maximum deformation energy theory IV eq 1 2 2 2 [( 1 2 ) ( 2 3 ) ( 3 1 ) ] 2 Strength condition: IV eq [ ] Applying to the plastic materials 4.2 Strength Calculation of Thin-walled Cylinder Subject to Internal Pressure 1.Strength calculating equation: i. Determining the major stress 1 pD 2S 2 m 3 r 0 pD 4S ii. Determining the equivalent stress eq f (1, 2 , 3 ) According to the third strength theory: III eq pD pD 1 3 0 2S 2S iii. Strength condition III eq pD [ ]t 2S (4 - 3) iv. Strength calculation equation pD S (4 - 4) t 2 [ ] Solving equation (4-4) as following: (1)Replacing medium diameter with internal diameter: D + Di = S Calculating pc from p. Putting them into equation (4-4), getting: pc Di S t 2[ ] pc (2)Introducing the welded joint efficiency : pc Di S (4 - 5) t 2[ ] pc This is the calculated thickness. (3)Eliciting the corrosion allowable thickness C2: pc Di Sd C2 t 2 [ ] p c This is the design thickness. (4 - 6) (4)Adding negative deviation C1: pc Di Sn C2 C1 round of value (4 - 7) t 2[ ] pc Getting the nominal thickness which indicated on the drawing. (5)Calculating the effective thickness: Se Sn (C1 C2 ) pc Di round of value t 2[ ] pc (4 - 8) v. Equation of strength verification pc ( Di Se ) t [ ] 2 Se t (4 - 9) vi. Calculating equation of pw 2[ ]t Se [ pw ] Di Se (4 - 10) 2.Strength calculating equation of thin-walled spherical vessels: pc Di S t 4[ ] pc pc Di Sd C2 t 4[ ] pc pc Di Sn C2 C1 round of value t 4[ ] pc *Equation of strength verification: pc ( Di Se ) t t [ ] 4 Se *Equation of [pw] —— the maximum allowable working pressure: 4[ ]t Se [ pw ] Di Se *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) Illumination of symbols: Pc —— Calculated pressure MPa Di 、 Do —— Internal & external diameters of cylinder mm S —— Calculated thickness mm Sd —— Design thickness mm Sn —— Nominal thickness mm Se —— Efficient thickness mm C1 —— Negative deviation mm C2 —— Corrosion allowable thickness mm C —— Additional value of wall thickness mm —— Welded joint efficiency []t —— Allowable stress at design temperature MPa t —— Calculated stress at design temperature MPa [Pw] —— The maximum allowable pressure at design T MPa 3.Determination of design parameters: i. Pressure P (1)Working pressure Pw 工作压力 —— the maximum pressure at the top of vessel and under normal operating condition (2)Design pressure P 计算压力 —— the maximum pressure at the specified top of vessel The design pressure P and the corresponding design temperature T are conditions of designing load, and its value is not less than working pressure. (3)Calculated pressure Pc 计算压力 —— the pressure which is used to determine the thickness at corresponding design temperature Including the liquid (column) static pressure, when the liquid (column) static pressure ＜ 5% design pressure, it can be neglected. Choosing the value of design pressure Illustrating in the following chart Conditions Evaluation of Design P With safety devices Single vessel (no safety devices) With explosive media and rupture disk With liquefied gas P≤(1.05~1.1)Pw P≥Pw External Pressure Vessel Vacuum Pressure Vessel Jacketed Vessel P≤(1.15~1.3)Pw Determined by the charging proportion and Tmax Under normal working condition, P≥△P=P2-P1 With safety valve: P=1.25△P Without SV: P=0.1MPa As external P vessel ii. Design Temperature T 设计温度 —— the enacted temperature of metallic components under normal operating condition Design P and design T both are the design load condition. iii. Allowable stress 0 limit Stress t [ ] Safe Coefficient n Normal T Vessel b [ ] nb Medium T Vessel t t t b s [ ] , ns min nb High T Vessel , s ns min t t t s n D t [ ] , , nD nn min ns iv. Safe (Safety) coefficient n Material Strength Performance Safety Coefficient Carbon Steel Low Alloy Steel High Alloy Steel σb σts σtD σtn nb ns nD nn ≥3.0 ≥1.6 ≥1.5 ≥1.0 ≥3.0 ≥1.5 ≥1.5 ≥1.0 v. Welded joint efficiency ( ) (1)Double welded butt or completely welded butt which is the same as double one. NDE 100% = 1.0 NDE Local = 0.85 Double welded butt (2)Single welded butt NDE 100% = 0.9 NDE Local = 0.8 Single welded butt vi. Additional value of wall thickness C (1)Negative deviation of plate and tube C1 Referring to the teaching material page 95, figure 4-7 & 4-8, selecting according to the nominal thickness Sn. (2)Corrosion allowable thickness C2 C2 = Ka B Ka —— corrosion rate, mm/year B —— design life of utility, year Generally speaking: Ka < 0.05 mm/year Single corrosion C2 = 1 mm Double corrosion C2 = 2 mm Ka = 0.05~0.1 mm/year Single corrosion C2 = 1 ~ 2 mm Double corrosion C2 = 2 ~ 4 mm For stainless steel, when the media is little corrosive C2 = 0 4.Pressure Test and Strength Verification of vessels: i. Purpose (1)Verifying the macro-strength and deformation of vessels (2)Verifying the tightness of vessels ii. Time (1)For new vessels, the Pressure Test and Strength Verification should be proceeded after completely welded and heat treatment. (2)For inservice vessels, the Pressure Test and Strength Verification should be proceeded after examination and repair, and before putting into production. iii. Media in Test (1)Water —— the most commonly used Stipulation to T of water: Carbon steels、16MnR、 normalizing15MnVR —— T≮ 5 ℃ Other low alloy steels —— T≮ 15 ℃ Stainless steels —— content of [Cl-] in water ≤ 25ppm (2)For the vessels which cannot be filled with liquid, something like dry and clean air, nitrogen gas or other inert gases can be used to fill these vessels. iv. Determination of Pressure for Testing (1)Internal Pressure Vessel Hydrostatic Test [ ] PT 1.25P Pneumatic Test [ ]t [ ] PT 1.15P [ ]t (2)External Pressure Vessel Hydrostatic Test PT 1.25P Pneumatic Test PT 1.15P Interpretation of symbols: P —— Design pressure, MPa PT —— Test pressure, MPa [] —— Allowable stress at test temperature, MPa []t —— Allowable stress at design temperature, MPa 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 30 minutes. *Decreasing P to 80% of PT . *Checking the welded seam and connection, reducing P to repair them if existing leakage. *Repeating the previous test until upping to grade. *After testing, discharging the liquid and drying the vessel with compressed air. (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 an primary inspection. *If up to grade, continue to slowly increase P to 50% PT, then by the △P=10% PT degree difference increasing slowly P to PT. *Keeping this P for 10 minutes. *Decreasing P to 87% PT, then keeping it and examining and repairing. *Repeating the previous test until upping to grade. (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. vi. Stress verification before pressure test (1)Hydrostatic test PT ( Di Se ) T 0.9 s 2S e (2)Pneumatic test PT ( Di Se ) T 0.8 s 2S e T —— Calculating stress at testing pressure, MPa s —— Yielding point at testing temperature, MPa 5.Examples: i. There is a boiler barrel whose Di=1300mm, working pressure Pw=15.6MPa and it has a safety valve. Also know that the design T=350ºC, the material is 18MnMoNbR, it is double welded butt with 100% NDE. Try to design the thickness of this boiler barrel. Solution: (1)Determining the parameters Pc = 1.1PW = 1.1×15.6 = 17.16 MPa (with the safety valve) Di = 1300mm []t = 190MPa (Design T = 350ºC) [] = 190 Mpa (At normal T, S > 60-100) = 1.0 (Double welded butt, 100% NDE) C2 = 1 mm (Single corrosion, Low alloy steel) (2)Calculating the thickness pc Di S 2[ ]t pc 17.16 1300 61.5 mm 2 190 1 17.16 Design thickness Sd = S + C2 = 61.5 + 1 = 62.5 mm Choosing C1 = 1.8 mm (P95 Figure 4-7) Additional value of wall thickness C = C1 + C2 = 1.8 + 1 = 2.8 mm Nominal thickness Sn = S + C + round-of value = 61.5 + 2.8 + round-of value = 65 mm (3)Hydrostatic test for strength verification *Parameters: [ ] PT 1.2 5P [ ]t 190 1.25 17.6 21.45 MPa 190 *Efficient thickness: Se = Sn - C = 65 - 2.8 = 62.2 mm s = 410 MPa *Stress: T PT ( Di S e ) 2Se 21.45 1300 62.2 2 62.2 234.9 MPa *Stress verification: 0.9 s 0.9 4101 369 MPa T < 0.9 S That is to say the strength in hydrostatic test is enough. ii. There is a oxygen cylinder which has been kept in storage for a long time, with Do=219mm using 40Mn2A and rolled by seamless steel. The actual Sn= 6.5mm and b = 784.8MPa, s = 510.12MPa, 5 = 18%, the design T is normal T. If the working pressure Pw=15MPa, try to find whether the thickness is enough or not. If not, what is the maximum allowable working pressure in this cylinder? Solution: it is the problem about strength verification —— Whethert ≤ []t or not (1)Determining the parameters Pc = 15MPa Do = 219 mm Sn = 6.5 mm b 784.8 [ ] 261.6 MPa nb 3 s 510.12 [ ] 318.8 MPa ns 1.6 Choosing the little one: i.e. []t = 261.6MPa = 1.0 (for seamless steel) C2 = 1 mm C1 = 0 (for the minimum thickness, negative deviation is neglected.) Se = Sn - C = 6.5 - 1 = 5.5 mm (2)Strength verification pc ( Do Se ) t 2S e 15 219 5.5 291 .1 MPa 2 5.5 Obviously, t > []t = 261.6 MPa So, 15MPa is too large, should be reduced. (3)Determining the maximum allowable working P t 2 [ ] Se [ p] Do Se 2 261 .6 1 5.5 13.48 MPa 219 5.5 So, the maximum safety P for this cylinder is 13.48 MPa 4.3 Designing Heads subject to Internal Pressure Classification according to the shape: i. Convex heads Semi-spherical head Elliptical head Dished head (spherical head with hem) Spherical head without hem ii. Conical heads Conical head without hem Conical head with hem iii. Flat heads 1.Semi-spherical head i. Molding of heads Small diameter and thin wall (整体热压成 型?) —— Integrally heat-pressing molding Large diameter —— Spherical petal welding 球瓣式组焊 Di ii. Calculating equation for thickness pc Di S t 4 [ ] pc pc Di Sd C2 t 4 [ ] pc pc Di Sn C2 C1 round of value t 4[ ] pc i. Calculating equation for thickness: For the elliptical heat whose m = a / b ≤ 2 max ho hi (b) 2.Thickness calculating equation of elliptical head Ri (a) Di pa a m 2S b S 2. The maximum stress should be at the top point: Putting m = a / b, a = D / 2 into the equation, getting: mpD max 4S Under the condition about strength: max Then: mpD [ ]t 4S pD m S t 4 [ ] (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 hi Putting these conditions into the equation: getting: Di pc Di m S t m p 2 2 [ ] 0 . 5 p 4 h c i 2[ ]t c 2 m = a / b = Di / 2 hi pc Di For the standard elliptical head whose m=2: pc Di S 2[ ]t 0.5 pc For the elliptical head whose m>2: at boundary » and m at the top point Then introducing the stress strengthening coefficient K to replace (Di / 4hi) K pc Di S t 2 [ ] 0.5 pc In this equation: Di 1 2 K 2h 6 i 2 For standard elliptical head: K=1 This is the common equation for calculating the wall thickness of elliptical heads. Beside these conditions: for standard elliptical heads Se ≮ 0.15% Di for common elliptical heads Se ≮ 0.30% Di The straight side length of standard elliptical heads should be determined according to P103, Figure 4-11 iii. Working stress and the maximum allowable working pressure pc KDi 0.5Se 2Se t 2[ ] Se [ p] KDi 0.5Se t i. Structure r Containing three parts: Di Sphere: Ri Transition arc (hem): r Straightedge: ho (height) ii. Calculating equation for thickness M pc Ri and S t 2[ ] 0.5 pc 1 M 3 4 M —— Shape factor of dished head ho 3.Dished head Ri r iii. Working stress and the maximum allowable working pressure pc MRi 0.5Se 2Se t 2[ ]t Se [ p] MRi 0.5Se iv. Dished head When Ri = 0.9 Di & r = 0.17 Di the dished head is standard dished head and M = 1.325 So the equation is: 1.2 pc Di S 2[ ]t 0.5 pc 4.Conical head i. Structure *without hem (suitable for ≤ 30 o ) without local strength with local strength *with hem (suitable for > 30 o ) —— Adding a transition arc and a straightedge between the joint of head and cylinder ii. Calculating equation for thickness The maximum stress is in the main aspect of conical head. m max max pD 1 2S cos According to the strength condition: pD 1 max max [ ]t 2S cos Then pD 1 S t 2[ ] cos Replacing P with Pc, considering , and changing D into Dc ，D=Dc+S pc Dc 1 S t 2[ ] pc cos This equation only contains the membrane stress but neglects the boundary stress at the joint of cylinder and head. Therefore the complementary design equation should be established: (1)Discriminating whether the joint of cylinder and head should be reinforced or not. (2)Calculation for the local reinforcement. Conical head without hem ( ≤ 30 ) (1)Not require reinforcing (consistent thickness for the whole head) main aspect: pc Dc 1 S t 2[ ] pc cos o small aspect: pc Dis 1 S t 2[ ] pc cos (2)Require reinforcing (for the thickness of joint, the reinforcement region) Main aspect: Q pc Di Sr 2[ ]t pc Small aspect: Q pc Dis Sr t 2[ ] pc Interpretation: Dc —— inside diameter of main aspect Di.s —— inside diameter of small aspect Di —— inside diameter of cylinder Q —— coefficient (Consulting the Figure 4-16 or 4-18 in book) Conical head with hem ( > 30 ) (1)Thickness of hem at the transition section K pc Dis S t 2[ ] 0.5 pc (2)Thickness of conical shell at the joint with transition section f pc Dis S t [ ] 0.5 pc o K —— coefficient (Consulting Figure4-13) f —— coefficient (Consulting Figure4-14) 4.Flat head i. Structure The geometric form of flat heads: rotundity, ellipse, long roundness, rectangle, square, etc. ii. Characteristics of load Round flat with shaft symmetry which is subjected to uniform gas pressure (1)There are two kinds of bending stress states, distributing linearly along the wall. (2)Radial bending stress r and hoop bending stress t distributing along the radius. ▲Fastening the periphery D r .max 0.188P S 2 R PD 0.75 S 2S S R t 0 r r.max max = r.max The maximum stress is at the edge of disk. p ▲ Periphery with simply supported ends max = r.max = t.max The maximum stress is in the center of disk. S P R D r .max 0.31P S R PD 1.24 S 2S t r 0 r.max 2 iii. Calculation equation for thickness From the condition of strength max ≤ []t , getting: Fastening the periphery 0.188P SD t [ ] Periphery with simply supported ends 0.31P SD [ ]t In fact, the supporting condition at boundary of flat head is between the previous two. After introducing the coefficient K which is called structure characteristics coefficient and considering the welded joint efficient , getting the calculating equation for thickness of round disk: S p Dc K Pc [ ]t Sn S p C2 C1 round of value 5.Examples Design the thicknesses of cylinder and heads of a storage tank. Calculating respectively the thickness of each heads if it’s semi-spherical, elliptical, dished and flat head as well as comparing and discussing the results. Known: Di = 1200 mm Pc = 1.6Mpa t material: 20R [] = 133Mpa C2 = 1 mm The heads can be punch formed by a complete steel plate. Solution: (1)Determining the thickness of cylinder pc Di 1.6 1200 S 7.26 mm t 2 [ ] pc 2 133 1.0 1.6 = 1.0 (Double welded butt, 100% NDE) Sd S C2 7.26 1.0 8.26 mm C1 = 0.8 mm (Checking Figure 4-7) Sd + C1 = 8.26 + 0.8 = 9.06 mm Round it of, getting: Sn = 10 mm (2)Semi-spherical head pc Di 1.6 1200 S 3.62 t 4 [ ] pc 4 133 1.0 1.6 = 1.0 (wholly punch forming) Sd S C2 3.62 1.0 4.62 mm C1 = 0.5 mm (Checking Figure 4-7) Sd + C1 = 4.62 + 0.5 = 5.12 mm Round it of, getting: Sn = 6 mm mm (3)Standard elliptical head pc Di 1.6 1200 S 7.24 mm t 2[ ] 0.5 pc 2 133 1.0 0.5 16 = 1.0 (wholly punch forming) Sd S C2 7.24 1.0 8.24 mm C1 = 0.8 mm (Checking Figure 4-7) Sd + C1 = 8.24 + 0.8 = 9.04 mm Round it of, getting: Sn = 10 mm (4)Standard dished head 1.2 pc Di 1.2 1.6 1200 S 8.69 mm t 2[ ] 0.5 pc 2 133 1.0 0.5 1.6 = 1.0 (wholly punch forming) Sd S C2 8.69 1.0 9.69 mm C1 = 0.8 mm (Checking Figure 4-7) Sd + C1 = 9.69 + 0.8 = 10.49 mm Round it of, getting: Sn = 12 mm (5)Flat head t K = 0.25; Dc = Di = 1200 mm; [] = 110 Mpa K Pc 0.251.6 S p Dc 1200 72.36 mm t [ ] 1101.0 = 1.0 (wholly punch forming) Sd S C2 72.36 1.0 73.36 mm C1 = 1.8 mm (Checking Figure 4-7) Sd + C1 =73.36 + 1.8 = 75.16 mm Round it of, getting: Sn = 80 mm Comparison: Head-form Semi-sphe. Sn mm kg Elliptical Dished Flat 6 10 12 80 106 137 163 662 Selection: It’s better to use the standard elliptical head whose thickness is the same to that of cylinder. Chapter 5 Design of Cylinders and Formed Heads subjected to External-Pressure 5.1 Summarization 1.Failure of External Pressure Vessel 外压容器失效 Under the effect of external pressure, the vessels may deform when the pressure is larger than a certain value. This kind of damage is called the failure of external pressure vessels. 2.Classification of Failure Side bucking —— the main form of failure Axial bucking Local bucking 5.2 Critical Pressure 1.Critical pressure and critical compressive stress 临界压力 The pressure that makes the external pressure vessels fail is called the critical pressure, indicating by Pcr. At the moment that exists Pcr, the stress inside the vessels is called the critical compressive stress, indicating by cr . 2.Factors affect the critical pressure i. Geometric dimension of cylinder 90 (1) Pcr mm HO2 500 90 90 0.3 350 0.3 175 175 Degree of vacuum in failure 0.51 350 0.3 90 (2) (3) (4) 300 120~150 300 Comparison and analysis for the experimental results Figure (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. ii. Materials’ Performance of the cylinders (1)The critical pressure (Pcr) hasn’t direct relation with the strength ( s) of the materials. (2)The critical pressure (Pcr) depends of the flexural rigidity of the cylinders in some aspects. The stronger the flexural rigidity, the more difficult for the failure. iii. The differential in the dimension at the process of vessels’ manufacturing Mainly reflecting on the “ellipticity” (椭圆度), which is the processing differential in the dimension of the cylindrical section. Dmin Ellipticity: Dmax Dmax Dmin e 100 % DN *Large ellipticity e can make the critical pressure Pcr decrease and failure happen in ahead. *Regulated as in the engineering, ellipticity e ≤ 0.5% when vessels subjected to the external pressure are made. 3.Long cylinder, short cylinder and rigid cylinder, the calculating equations of their critical pressure i. Long cylinder —— cylinders with large L / Do Calculating equation of the critical P: 3 3 2 E Se t S e For steel cylinders: 2.2 E Pcr 2 = 0.3 1 Do D o t Calculating equation of the critical stress: Pcr D Pcr Do t Se cr 1.1E 2Se 2Se Do 2 ii. Short cylinder —— cylinders with small L / Do Calculating equation of the critical P: Se / Do L / Do 2.5 P 2.59 E ' cr t Calculating equation of the critical stress: ' cr P Do t 1.3 E 2S e ' cr Se / Do L / Do 1.5 iii. Rigid cylinder —— cylinders with small L / Do, large Se / Do Designing criterion: Only need to satisfy the strength condition: t compression ≤ [ ] compression i.e. Pc Di Se t 压 [ ]t压 2Se 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 cylinder ’ L cr < L < Lcr Short cylinder ’ L < L cr Rigid cylinder i. Critical length Lcr of long and short cylinder: Lcr 1.17 Do ii. Critical length ’ L cr of Do Se short and rigid cylinder: t ' cr L 1.3E S e Do t comp . Se 5.3 Engineering Design of External-P Vessels 1.Designing criterions Pcr Pc [ P ] m Pc —— Calculating Pressure, MPa Pcr —— Critical Pressure, MPa [p] —— Allowable External Pressure, MPa m —— Stable safety coefficient For cylinders, m = 3 at the same time, 椭圆度 e ≤ 0.5% 2.Nomograph for the thickness designing of the external-P cylinders i. Calculating Steps Step 1: L、 Do、 Se → Drawing the curve f Do Se , L Do Step 2: Find the relationship between and [P] 2 t Making : B E m For cylinder m=3 and 2 t B E 3 Then getting the relationship curve B = f () So : Se [ P] B Do 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 using L / Do = 50. If L / Do < 0.05, checking the figure using L / Do = 0.05. (3)Calculating the value of B According to the used material, choosing the relevant graphs from Figure (5-7) and Figure (5-14) and then finding the point A from abscissa. Two situations maybe encountered: *Point A with that certain value lies at the right of the curve and intersects with the curve, then the value of B can be found directly in the figure. *Point A with that certain value lies at the left of the curve and has no joint with the curve, then the value of B is calculated by the following equation: 2 t B 3 EA (4)Calculating [P] Putting the value of B into Equation (9) →[P] Se B [ P] B Do Do Se （9） (5)Comparing ~ If [ P] ~ Pc i.e. the supposed Sn is usable, safe i.e. the supposed Sn is too large and If [P] Pc should be decreased appropriately, repeating the previous calculating steps until satisfying the first condition. i.e. the supposed Sn is too small and If [ P] < Pc should be increased appropriately, repeating the previous calculating steps until satisfying the first condition. 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 P P —— design pressure 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. 4.Example and discussion Design the thickness of an external-P cylinder. Known: Calculating pressure: Pc = 0.2 MPa Design temperature: t = 250℃ Inside diameter: Di = 1800 mm Calculating length: L = 10350 mm Additional value of wall thickness: C = 2 mm Material: 16MnR; Et = 186.4 103 Mpa Solution: (1)Assuming Sn = 14 mm Then Do = Di + 2 Sn = 1828 mm Se = Sn - C = 12 mm Finding out: L / Do = 10350 1828 = 5.7 Do / Se = 1828 12 = 152 (2)Calculating the value of (A) Checking the Figure 5-5, getting: A = 0.000102 (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: 2 2 t B E A 186 .4 10 3 0.000102 12.78 MPa 3 3 (4)Calculating [P] [ P] B Do Se 12.78 0.0834 MPa 152 (5)Comparing [P] and Pc [P] < Pc = 0.2 MPa unsatisfied Reassuming Sn, or setting the stiffening ring. Calculation under the condition that supposes there have two stiffening rings: (1)Thickness is the same: Sn = 14 mm After setting two stiffening rings, the calculating length is like following: L Lorigin 3 10350 3450 m m 3 3450 Then L D o 1.9 1828 (D o Se 152 ) (2)Calculating the value of (A) Checking the Figure 5-5, getting: A = 0.00035 (3)Calculating the value of B From Figure 5-9, we can see that point A is at the right of the curve, getting B = 42.5 MPa (4)Calculating [P] [ P] B Do Se 42 .5 0.28 152 (5)Comparing [P] and Pc [P] > Pc = 0.2 MPa satisfied MPa Calculation under the condition that supposes to increase the thickness: (1)Assuming: Sn = 20 mm Then Do = Di + 2 Sn = 1840 mm Se = Sn - C = 18 mm Finding out: L / Do = 10350 1840 = 5.6 Do / Se = 1828 18 = 102 (2)Calculating the value of (A) Checking the Figure 5-5, getting: A = 0.00022 (3)Calculating the value of B From Figure 5-9, we can see that point A is at the right of the curve, getting B = 27.5 MPa (4)Calculating [P] B 27 .5 [ P] 0.27 Do Se 102 MPa (5)Comparing [P] and Pc [P] > Pc = 0.2 MPa and closing So, we can use the steel plate with Sn = 20 mm, whose material is 16MnR. 5.4 Design of External-P Spherical Shell and Convex Head 1.Design of external-P spherical shell and semi-spherical head i. Assuming Sn, and Se = Sn － C. Calculating the value of Ro / Se. ii. Calculating the value of (A) 0.125 A Ro S e iii. Calculating the value of B and [P] According the used material, choosing the relevant graph from Figure 5-7 and Figure 5-14 and finding out the point A at the abscissa. 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. Then B [P] Ro Se (2)If point A is at the left of the curve, directly calculating: t 0.0833E [ P] ( Ro Se ) 2 iv. Comparison ~~ i.e. the original assuming Sn is If [ P] P c usable, and safety. If [ P] < i.e. the original assuming Sn is Pc too small, S should be n increased appropriately, repeating the previous calculating steps until satisfying the first condition. 2.Design of external-P convex head The method of designing the external-P convex head is the same to that of designing external-P spherical head. But the Ro in the designing of spherical head should be adjusted like following: i. For elliptical head Ro —— the equivalent spherical diameter of elliptical head; Ro = K1Do K1 —— coefficient; depending on a / b, checking P141, Figure 5-3 ii. For dished head Ro —— the equivalent spherical diameter of the dished head; it’s the outside diameter of the spherical part at the dished head. 5.5 Design of the Stiffening Ring in External-P Vessels 1.Function of stiffening ring And Se Pcr 2.2 E Do t Pcr [ P] m 3 Pcr' 2.59 E t Se Do L Do 2.5 From the previous equations, we can know the methods to increase [P]: i. Increasing S ii. Decreasing the calculating length L ∴ Function of stiffening ring 加强圈的功能 —— decreasing calculating length to increase [P] 2.Space length and number of stiffening ring Assuming the space length of stiffening ring is Ls From the design criterions of external-P: Pc ≤ [P] and [P] = Pcr / m Making Pc = [P] then Pcr = m Pc (a) From the equation for the critical pressure of short cylinder: 2.5 Se Do ' t Pcr 2.59 E L Do Putting equation (a) in, getting: P 2.59 E ' cr t Se Ls Do Do 2.5 m pc Then putting m=3 in, getting: ( Ls ) max Do 0.86E Pc t Se Do 2.5 (Ls)max —— Under the condition that Do and Se of the cylinder is determined, the maximum space length between the needed stiffening rings working safely under the calculating external pressure Pc, mm. The actual space length between stiffening rings Ls ≤ (Ls)max is indicating safety. The number of stiffening rings: L n 1 Ls In the above equation: L —— the calculating length of cylinder before setting the stiffening rings, mm Ls—— the space length between stiffening rings, mm 3.Connection of stiffening rings and cylinders Connection Demands 连接要求 Must assure all the cylinder and stiffening ring 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) i. iii. The stiffening rings should not be randomly crippled or cut off. If those must be done, the length of the arc that are crippled or cut off should not be larger than the values shown in Figure 5-19. For example: There is a horizontal external pressure vessel. When the stiffening ring is set inside the cylinder, in order not to affect the fluid flowing or fluid discharging, we must leave a hole豁口at the lowest position of the stiffening ring or set a thoroughfare of fluid. As illustrating like the following two figures Chapter 5 Components and Parts of Vessels 6.1 Flanges Connection 1.The Sealing Theory (密封原理) and Connection Structure of Flanges i. Connection Structure 连接结构 Three parts: (1)Connected parts —— a couple of flanges (2)Connecting parts —— several couples of bolts and nuts (3)Sealing parts —— gasket 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 charging medium 2.The Structure and Classification of Flanges According to the connection ways of flanges and equipment (pipelines) (1)Integrated flange —— S.O.flange (slip on flange) W.N.flange (welding neck flange) Pipeline Flange Vessel Flange S.O.flange W.N.flange (2)Simple [loose (type), lap joint, lapped] flange On the welding ring Interlink on the turn-down rims (3)Screwed flange Square flange Elliptical flange 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. 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. 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 iii. Properties of gasket (1)The common-used materials of gasket *Non-metal Material —— Rubber, Asbestos, Synthetic resins. Advantages: soft and corrosion resistant Disadvantages: the properties of high-T resistance and pressure resistance is inferior to the metallic materials. Used in: Common and Medium T; Flange sealing of Medium and Low P devices and pipes. *Metal (Metallic) Material 金属垫片 —— soft aluminum, copper, iron (soft steel), 18-8 stainless steel. Advantages: high-T resistant, with high strength Demands: Excellent soft toughness Used in: Medium and high T; Flange sealing subjected to medium and high P (2)Gasket Types (Classifying according to the properties of materials) *Non-metal Gasket —— such as rubber gasket, asbestos-rubber gasket. *Compound Gasket (Metal and non-metal compound gasket) —— such as metal jacketed gasket (金书包 垫片) and Metal spirotallic [spiralwound] gasket Metal jacketed gasket (金书包垫片), i.e. wrapping the metal slice around the asbestos gasket or asbestos-rubber gasket Metal spirotallic [spiral-wound] gasket (金属 缠绕垫片), i.e. making by alternately rolling thin steel belt and asbestos *Metal gasket —— such as octagon ring gasket, elliptical gasket, lens ring (washer) [grooved metallic gasket] (3)Selection of gasket 垫片选择 *Factors of working pressure and temperature Medium and low P; common and medium T —— Non-metal gasket Medium P; Medium T —— Metal and non-metal compound gasket High P; high T —— Metal gasket High vacuum; cryogenic —— Metal gasket *Degree of demands for sealing *Demands for the types of sealing face *Properties of gasket Concrete selection should be referred to JB4704-92, JB4705-92, JB4706-92. At the same time, the practical experience should be taken into account. iv. Rigidity (刚度) of flange (1)If the rigidity of flange is not enough, there will occur the serious buckling [翘曲] deformation, as well the specific pressure will decrease and the sealing face will be loose, as a result, the sealing will fail. (2)Measures to increase the rigidity of flange (3)Strengthening the rigidity of flange to increase the weight of flange as well the value of whole-flange’s sealing. v. Effect of working conditions Temperature Pressure Corrosive Characteristics of medium Penetrant Characteristics Greatly affecting the sealing Combined effect 4.Standard and Selection of Flanges i. Standard number of pressure vessel flanges JB / T 4701-2000 ∼ JB / T 4703-2000 Standard types and marks of pressure vessel flanges Sealing face Type Code Without lined ring Plain With lined ring (C) M&F M F T&G T G Plain M&F M F T&G T G A-S.O.Flange JB/T 4701-2000 P A T S C P A T S C B-S.O.Flange JB/T 4702-2000 P A T S C P A T S C W.N.Flange JB/T 4703-2000 P A T S C P A T S C For example: PN=1.6MPa, DN = 800mm, T&G B-S.O.Flange with lined ring T Flange: C-S 800 — 1.6 JB4702-92 G Flange: C-C 800 — 1.6 JB4702-92 Standard Code Nominal Pressure MPa C-C 800 — 1.6 JB4702-92 Code of Flange Type Code of Sealing face Type Nominal Diameter mm ii. Dimension of pressure vessel flanges Dimension of flanges is only confirmed by two standardized parameters PN and DN of flanges. Confirmation of Nominal Pressure PN of flanges: JB4700-92 (Book, P160) iii. Selection steps for pressure vessel flanges 法兰选用步骤 (1)According to the design task, confirming the types of flanges (S.O. or W.N.). (Referring to P157 Table 6-2) (2)According to the nominal diameter DN of flanges , working temperature, design pressure, material of flanges, confirming the nominal diameter DN and nominal pressure PN of flanges. (Referring to P160 Table 6-4; P332 Appendix 12) (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 and PN of flanges, checking and finding out the dimension of flanges; number of bolts and their specification. (Referring to P336 Appendix 14) (5)Confirming the material of bolts and nuts. (Referring to P163 Table 6-6; P333 Appendix 13) (6)Portraying the unit drawing of flanges. Example: There are flanges to connect the body of a fractionating (rectifying) tower and the heads. Knowns: Inside diameter of tower: Di = 1000mm Working temperature: t = 280℃ Design Pressure: P = 0.2MPa Material of tower: Q235-AR Solution: (1)From P157 Table 6-2, A-S.O.Flange is selected. (2)Confirming the nominal diameter DN and nominal pressure PN DN = 1000 mm (Equal to the inside diameter of tower) From P160 Table 6-4, choose the material of tower as that of flanges, i.e. Q235-AR t = 280℃ When PN = 0.25 Mpa, Pallowable = 0.14 MPa < Pdesign = 0.2 Mpa When PN = 0.6 Mpa, Pallowable = 0.33 MPa > Pdesign = 0.2 Mpa So, the nominal pressure of flanges is: PN = 0.6 MPa (3)Confirming the sealing face types of flanges From P155 Table 6-1, choosing plain sealing face, spirotallic [spiral-wound] gasket (4)According to the DN and PN of flanges, from Appendix 14, Table 32, finding out the dimension of every part of flanges. Specification of bolts: M20; Number: 36 (5)From P163 Table 6-6, finding out: Material of bolts: 35 steel Material of nuts: Q235-A (6)Portraying the unit drawing of flanges (Omitting) Standard of tube flanges (New Standard issued by Chemical Ministry) European: HG 20592 — 97 ~ HG 20600 — 97 American: HG 20615 — 97 ~ HG 20621 — 97 6.2 Support for vessels Support for horizontal vessels Saddle support, ring support, leg, etc. Support for vertical vessels Skirt support, hanging support, etc. 1.Double-saddle support i. The structure of double-saddle support Gasket 120° Web-plate Sub-plate Anchor bolt ii. Position of support (A) 安座位置 A≤Do/4 & < 0.2L. The maximum value < 0.25L iii. Standard and selection of double-saddle support Type —— Stationary type: F Movable type: S Model Type —— Light-duty: A Heavy-duty: B Mark —— JB / T 4712-92 Support Model Type Nominal Diameter Type 2.Checking calculation of stress in double-saddle horizontal vessels i. Load analysis for horizontal vessels q A F A F Shearing Force Diagram M1 Bending Moment Diagram M3 M2 ii. Reserved force to support mg F 2 or mg q 4 F L hi 2 2 3 In this equation: q —— Mass load/unit length of vessels, N / mm L —— Distance between the T.L. (tangent lines) of two heads, mm hi —— Height of curved surface of heads, mm iii. The maximum radical bending moment The section across the middle point of moment 2 Rm2 hi2 1 2 FL 4A L M1 4 1 4hi L 3L (N mm) The section at support 2 2 A Rm hi 1 L 2 AL M 2 FA1 4hi 1 3L (N mm) iv. Calculation for radical stress of cylinder —— to the vessels subjected to positive pressure (1)Stress across the middle section The most highest point in section (Point 1): Pc Rm M1 1 2Se Rm2 Se The most lowest point in section (Point 2): P c Rm M1 2 2 2Se Rm Se (2)Stress in the section of support The most highest point in section (Point 3): Pc Rm M2 3 2 2Se K1 Rm Se The most lowest point in section (Point 4): Pc Rm M2 4 2Se K 2 Rm2 Se v. Checking calculation for radical stress of cylinder Radical tensile stress comb.tensile max t Radical compressive stress comb . comp . max t thesmaller value cr B In these two equations: []t —— The allowable stress of material at the design T, MPa []c r —— The allowable compressive stress of material, MPa B —— Calculation method is the same with that in design of external pressure, see P172 6.3 Reinforcement for opening of vessels 1.The phenomena and reason for opening stress concentration Stress concentration factor: max K max —— The maximum stress at the boundary of opening * —— The maximum basic stress of shell Small opening in plate Reasons for stress concentration: (1)Material of vessel wall is deteriorate (2)The continuity of structure is damaged 2.Opening reinforcement’s Designing i. Designing Criterions (1)Equi-area criterion of reinforcement (2)Plastic failure criterion of reinforcement ii. Reinforcement Structure (1)Structure of Stiffening Ring Nozzle (Connecting Tube) Stiffening Ring Shell (2)Structure of 加强元件 Method —— Taking the parts of nozzles or vicinity of shells’ openings which need to be reinforced as the 加强 元件, then welding these parts with nozzles or shells. (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. iii. Diameter Range of the openings that need not to be reinforced When 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. (3)Nominal Outside Diameter of connecting tubes ≤ 89 mm (4)The minimum wall thickness δmin of connecting tubes should meet the following requirements:(mm) δmin 25 32 3.5 38 45 4.0 48 57 65 5.0 76 6.0 89 3.Designing methods of equi-area reinforcement Metallic areas in local reinforcement ≥ the area of sections which are the position of openings i. Confirmation of the effective range of opening and reinforcement areas h2 h1 d B A A1 A2 A3 A4 Effective width: B 2d B d 2 S n 2 S n. t max Effective [working] height: Outside height h1 d S n.t h1 act ualoverhangheight of nozzle min Inside height h2 d S n.t h2 act ualembedded height of nozzle min In these equations: Sn —— Nominal thickness of cylinders Sn.t —— Nominal thickness of connecting tubes (nozzles) d —— Diameter of openings d = di +2C di —— Inside diameter of openings C —— Additional value of wall thickness Computation of metallic areas for effective reinforcement (1)The area of the sections on shell which are the positions of openings A: A = S×d (2)The unnecessary metallic area A1 on shell or heads which is larger than calculating thickness S: A1 = (B – d) (Se – S) – 2 (Sn.t – C) (Se – S) (1 – fr) (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 the reinforcement region A3: A3 = according to the actual dimension ii. Designing Steps in Reinforcement for openings (1)Getting the following data from the strength calculation: Calculating wall thickness of cylinders or heads S Nominal wall thickness of cylinders or heads Sn Calculating wall thickness of nozzles St Nominal wall thickness of nozzles Sn.t Additional value of wall thickness C = C1+ C2 (2)Calculating the effective reinforcement range B, h1, h2 (3)Calculating the necessary reinforcement area A according to P183 Table 6-17 (4)Calculating the available reinforcement area A1, A2, A3 (5)Judging whether it is necessary to add some reinforcement area If A1 + A2 + A3 ≥ A reinforcement not required If A1 + A2 + A3 < A reinforcement required (6)If reinforcement is required, calculating the added reinforcement area A4 A4 = A －（ A1 + A2 + A3 ） (7)Comparison Finally getting A 1 + A 2 + A3 + A 4 ≥ A 6.4 Attachment of vessels 1.Man Hole and Hand Hole i. Nominal Diameter of standard man-hole DN ： 400 450 500 600 ii. Nominal Diameter of standard hand-hole DN ： 150 250 2.Connecting Tubes (Nozzles) [接管] 3.Flg (Flange, Flanch) [凸缘] 4.Sight (Level) Glass [视镜]