AD2000-薄管板换热器强度计算示例

1、Component: Front TubesheetAD-Merkblatt B 5 - Unstayed and Stayed Flat Ends and Plates - January 19956.7.1 Circular Flat Plates Mutually Stayed by the Tubes and the Shell Tubesheet Material: 1.4404 X 2 CrNiMo 17 13 2 BlechTube Material:1.4404 X 2 CrNiMo 17 13 2Shell Outside Diam. Tube Outside Diam. T

2、ube Inside Diam. Design Diameter Number of TubesDa = 1404 mm da = 25 mm di = 21 mmD1 = 1408 mmn = 1526Shell Wall ThicknessTube Wall Thickness Tube PitchDesign Diameterse = 2 mmse = 2 mm t = 32 mm d2 = 281.6 mmCondition:OperatingTestingTemperatureT =3020CStrenth Value, TubesheetK = 220.2221.8N/mm2Str

3、ength Value, TubesK = 222.3225N/mm2Safety FactorS = 1.51.1Modulus of Elasticity, ShellE = 198725-N/mm2Modulus of Elasticity, TubesE = 198200-N/mm2Geometric DataActual Thickness of Tubesheetse = 23.8mmShell Material:1.4301 X 5 CrNi 8 10 BlechComponent: Front TubesheetAD-Merkblatt B 5 - Unstayed and S

4、tayed Flat Ends and Plates - January 1995 Boundry Tubes (approx. the two outermost tube rows)nt = 2*PI(Da-2*se-2*t)/t = 263Cross-sectional Area, Shell AM =PI*(Da*2-(Da-2*se)*2)/4 = 8809 mm2Cross-sectional Area, ntAR = nt*PI*(da*2-(da-2*st)*2)/4 = 38007.1 mm2AD S 3/7 - Allowing For Thermal Stresses i

5、n Heat Exchangers - October 1991Shell TubesThermal Expansion Coef. (*10*6 1/deg. C) alpha =16 16.5Temperature, C theta = 30 20Thermal Expansion Difference, xdxd = |(alphaM*(thetaM-20)-alphaR*(thetaR-20)/10*6| = 0.00016 Thermal Stress in Shell sM = -xd /(1/EM)+AM/(AR*ER)*F = -258 bar Thermal Stress i

6、n Tubes sR = xd /(1/ER)+AR/(AM*EM)*F = 59.8 bar F, % = 100Component: Front TubesheetAD-Merkblatt B 5 - Unstayed and Stayed Flat Ends and Plates - January 19956.7.1.4 - Tube ForcesDes. Pressure, Shell pu = 1 barDes. Pressure, Tubes pi = 2 barPressurep1 =pu+sR= 60.8 barPressurep2 =pi-sR =2 barMaximum

7、Pressurep = 60.8 bar6.7.1.1 - Required Wall Thickness, mmTensile Stress, Tubes sR = 59.8 bar Comp. Stress, Tubes sR = -Test Pressure, Shell p1= 1 barTest Pressure, Tubes p2= 1 barExpanded Lengthlw = 0 mmOperating Testing 2.5s = 0,4*d2*SQRT(p*S/(10*K) = 22.9(12)6.7.1.5 - Required Tubesheet Thickness(

8、FRcmax Fk)Loaded AreaAR = PI/4*(Da-2*se)*2-n*da*2)/n =517.9 mm2Bild 12Effective AreaAW = (da-di)*lw= 0 mm2(13)AW = 0.1*da*lw= 0 mm2(14)Tube Forces, NOperatingTestingTensile ForceFR =AR*p1/10 =314952(21)Compressive ForceFR= AR*p2/10= 10452(21)Maximum ForceMFR =314952Allowable Loading of the Expanded

9、JointFR/AW =00N/mm2Component: Front TubesheetAD-Merkblatt B 5 - Unstayed and Stayed Flat Ends and Plates - January 1995B5 - 6.5.3 - Calculation of the Buckling Loads and Wall ThicknessBuckling Length lk = 1400 mm Design Factor C = 0.4 J = 0,049*(da*4-di*4)= 9611Weakening Factorv = 0.219v = (t-da)/tS

10、lenderness RatioLa = 4*lk/SQRT(da*2-di*2) = 171.5(9a)Slenderness Ratio RangeLa0 = PI*SQRT(E/K)= 93.8 (9b)Operating Testing(8a)3197 3207 (8b)Allowable Buckling Load, NFk = PI*2*E*J/(lk*2*3.0) orFk = (K/S)*PI*(da*2-di*2)/4)*(1-(La/La0)*(1-S/3) =Maximum Buckling LoadFRcmax = 104 52Smallest Weld Thickne

11、ss g = 0,4*FR*S/(da*K) =0.3 0 mm (15)s = C*SQRT(D1*2-n*di*2)/v)*(pi*S/(10*K) =36.121.8 mm (16)Component: Front TubesheetAD-Merkblatt B 5 - Unstayed and Stayed Flat Ends and Plates - January 19956.7.5 - Circular, Flat Tubesheets on Heat Exchangers with Expansion JointsExp. Joint DiameterDk = 1964 mmO

12、uter Tube Limitl = 656.5 mmDesign DiameterD3 = (Dk+Da)/2= 1684 mmDesign Pressurep = pi+pu*(D3*2-4*l*2)/D1*2 = 2.6 bar(26)Fig. 16 RatioDesign Factorl/D1 = 0.47C5 = 0.19Ratio dt/dDTable 1 Type= 1.03Required Tubesheet Thickness(27)s = C5*D1*SQRT(p*S/(10*K*v) = 23.8 mmComponent: Rear TubesheetAD-Merkbla

13、tt B 5 - Unstayed and Stayed Flat Ends and Plates - January 19956.7.1 Circular Flat Plates Mutually Stayed by the Tubes and the ShellTubesheet Material:1.4404 X 2 CrNiMo 17 13 2 BlechTube Material:1.4404 X 2 CrNiMo 17 13 2Shell Material:1.4301 X 5 CrNi 8 10 BlechCondition:OperatingTestingTemperature

14、T =3020CStrenth Value, TubesheetK = 220.2221.8N/mm2Strength Value, TubesK = 222.3225N/mm2Safety FactorS = 1.51.1Modulus of Elasticity, ShellE = 198725-N/mm2Modulus of Elasticity, TubesE = 198200-N/mm2Geometric DataActual Thickness of Tubesheetse = 23.9mmDa = 1404 mm da = 25 mm di = 21 mmD1 = 1400 mm

15、n = 1526se = 2 mmse = 2 mm t = 32 mm d2 = 280 mmShell Outside Diam. Tube Outside Diam. Tube Inside Diam.Design Diameter Number of TubesShell Wall ThicknessTube Wall ThicknessTube PitchDesign DiameterComponent: Rear TubesheetAD-Merkblatt B 5 - Unstayed and Stayed Flat Ends and Plates - January 1995 B

16、oundry Tubes (approx. the two outermost tube rows)nt = 2*PI(Da-2*se-2*t)/t = 263Cross-sectional Area, Shell AM =PI*(Da*2-(Da-2*se)*2)/4 = 8809 mm2Cross-sectional Area, ntAR = nt*PI*(da*2-(da-2*st)*2)/4 = 38007.1 mm2AD S 3/7 - Allowing For Thermal Stresses in Heat Exchangers - October 1991Shell Tubes

17、Thermal Expansion Coef. (*10*6 1/deg. C) alpha =16 16.5Temperature, C theta = 30 20Thermal Expansion Difference, xdxd = |(alphaM*(thetaM-20)-alphaR*(thetaR-20)/10*6| = 0.00016 Thermal Stress in Shell sM = -xd /(1/EM)+AM/(AR*ER)*F = -258 bar Thermal Stress in Tubes sR = xd /(1/ER)+AR/(AM*EM)*F = 59.8

18、 bar F, % = 100Component: Rear TubesheetBild 12AD-Merkblatt B 5 - Unstayed and Stayed Flat Ends and Plates - January 19956.7.1.4 - Tube ForcesDes. Pressure, Shell pu =1 barTensile Stress, Tubes sR = 59.8 barDes. Pressure, Tubes pi =2 barComp. Stress, TubessR= -Pressure p1 = pu+sR= 60.8 barTest Pressure, Shellp1=1 barPressure p2 = pi-sR =2 barTest Pressure, Tubesp2=1 barMaximum Pressurep = 60.8 barExpanded Lengthlw = 0 mm6.7.1.1 - Required Wall Thickness, mmOperatingTestings=