Processing Equipment Design

Transcription

1 Processing Equipmen Design Exercise No. 5: Reinforcemen of fla cover of shell and ube hea exchanger Lecurer: Pavel Hoffman hp://fsine.fsid.cvu.cz/cz/u18/peoples/hoffman/index.hm

2 Design of reinforcemen of fla cover of shell and ube hea exchanger (HE) b Solved as a fixed beam Given: Shell inner diameer Calculaed diameer of cover sealing Pich diameer of cover bols Cover exernal diameer Overpressure in HE Maximal allowable cover deflecion PED - exercise 5 D o = 1000 mm D = 1050 mm D r = 1085 mm D C = 110 mm p = 300 kpa y = 1 mm Type and size of reinforcing profiles I 140 Secion modulus of he profile for bending W o = 81,9 cm 3 Momen of ineria of he profile J = 573 cm 4 fd T fd o fd C fd baffle wih sealing s (ighness of baffles sealing)

3 Examples of ubular HEs wih several passes, fla cover and baffles VT s pevnými rubkovnicemi, přepážkami v mezirubkovém prosoru a dvěma ahy v rubkách HE wih fixed ube-plaes and baffles in iner mediae ube space and passes in ubes cross flow + couner flow gap for drainage Přepážka Baffle plae 6 Trubky Tubes Paka Suppor, fooing 7 Pevná rubkovnice Fixed ube plae (Welded like flange) 3 Výsupní hrdlo Oule neck 8 Příruba Cover wih flange 4 Segmenová přepážka Segmen baffle 9 Vsupní hrdlo Inle neck 5 Plášť Shell neck reinforcing wih he ring cross flow + in upper par is couner and down is parallel flow baffle Four passes HE baffles keep flow in proper ubes baffles ube plaes fron chamber back PED - exercise 5 3

4 Beams of I profile and heir parameers momen of ineria needed from he deflecion poin of view secion modulus needed from he srengh poin of view Type Mass Type Mass PED - exercise 5 4

5 Beams of U profile and heir parameers momen of ineria o bending axis x - x secion modulus o bending axis x - x momen of ineria o bending axis y - y secion modulus o bending axis x - x Type Mass Type Mass PED - exercise 5 5

6 Profile widh b = 66 mm b Profile specific mass s m P1 = 14,4 kg/m baffle wih sealing fd Modulus of elasiciy o E = 06 GPa fd T fd Maximal allowed sress σ D = 156 MPa fd C Maerial densiy ρ = 7850 kg/m 3 Minimal fla cover hickness specified s = 10 mm from echnological reasons (E.g. shell welding on he cover, flange srengh...) Task is o specify: Pich of reinforcing profiles heir number Maximal bending sress in cenral profiles Maximal deflecion of cenral reinforcing profiles Mass of cover ogeher wih reinforcing profiles Comparison wih unreinforced fla cover Appreciaion of possibiliy of he reinforcemen opimizaion =? mm σ Bmax =? MPa y max =? mm M o =? kg M =? kg PED - exercise 5 6

7 During our calculaion we neglec a hin fla cover rigidiy and will calculae only wih rigidiy of reinforcing profiles (i is ha all forces acs only on hese profiles). Our resuls will be on a side of beer safey as he cover is able o wihsand some loading. We consider he case of a fixed beam. A skech of he cover wih profiles is on fig.1. PED - exercise 5 7

8 Fig.1 Skech of fla circular cover reinforcemen U profiles View of he longiudinal and ransversal reinforcing beams b s n longiudinal profiles fd o fd T fd fd C L i I profiles effecive area for 1 profile = L i * (he profile mus wihsand a force equal o he pressure acing on his surface) n ransversal profiles PED - exercise 5 8

9 Examples of fla covers of shell and ube HEs reinforcemen PED - exercise 5 9

10 1. Loading course and equaions used for fixed beam wih lengh L and loaded wih coninuous load q. L q fixed beam wih coninuous load (in our example = inernal overpressure) For simpliciy of our calculaion we will consider he reinforcemen as he fixed beam. The real siuaion of he load will be beween he fixed and suppored beam closer o he fixed one. M BL/ M Bmax course of bending momen in fixed beam wih coninuous load A B PED - exercise 5 10

11 Equaions needed for he reinforcing beams calculaion Remember lecures of elasiciy and srengh M B ( L / ) q* 4 L bending momen in he beam cener M Bmax M A M B q L 1 * bending momen in he place of he beam fixaion y( L / ) 4 q * L 384* E * J deflecion in he beam cener Coninuous load of he beam q = N/m * m = N/m PED - exercise 5 11

12 . Specificaion of beams (I profiles) pich for he mos loaded profile (near he axis) p Noe: Profiles are crosswise herefore is he loading divided by. Specificaion of he coninuous loading referenced o 1 m of profile lengh q = 1 / * p * 1 * (N/m; Pa, m, m) Maximal bending momen in he longes beam 1 m M Bmax = q * L max / 1 (N*m; N/m, m ) =? For he 1 s approximaion we suppose ha L max D (for odd number of profiles i is valid, for even number is L max a few less han D we are on he safey side) PED - exercise 5 1

13 Real sress in beam mus be equal or lower han allowed sress Sress = bending momen / secion modulus σ Breal = M Bmax / W O = (q * L max ) / (1 * W O ) σ Breal = 1 / *(p * * L max ) / (1 * W O ) σ D Theoreical maximal pich of profiles is max = * (1 * σ D * W O ) /(p * L max ) max = * (1* 156*10 6 * 81,9*10-6 ) / (300*10 3 * (1050*10-3 ) = max = 0.97 m = 97 mm Toal maximal heoreical force acing on he longes profile F TPmax = p * * L 1 / 300 * 0.97 * / = kn PED - exercise 5 13

14 Toal force acing on he cover is force = area * pressure F TC = * D / 4 * p = * 1.05 / 4 * 300 = 59.8 kn Maximal specific loading of he profile neares he axis is (1 s ieraion = in he axis) q max = p * 1 * max / = 300 * 1 * 0,97 / = kn/m Maximal bending momen in he profile locaed neares he axis is M Bmax = q max * L max / 1 = * / 1 = 1.78 kn.m Maximal heoreical bending sress in he profile locaed neares he axis is Bmax = M Bmax / W o = 1.78 / (81.9*10-6 ) = kpa = D (checking if we couned correcly) Maximal heoreical deflecion of he profile locaed neares he axis is y max = q * L 4 max / (384 * E * J) = 139.1*10 3 * / (384 * 06*10 9 * 573*10-8 ) y max = m 0.4 mm 1 mm OK (J = momen of ineria) PED - exercise 5 14

15 3. Specificaion of real number of profiles and loading Number of profiles in one direcion n P = D / max = 1050 / 97 = 1.13 pcs Chosen number of profiles (minimally a nex higher whole number) n Pch = pcs Real pich of profiles real = D / n Pch = 1050 / = 55 mm Lengh of he longes profile (near he axis) For odd number of profiles L 1 = D L unreinforced pars of he plae For even number of profiles: Equaion for deermining he lengh of he chord of he arc Pro sudý D poče výzuh D D * *( 0,5* )* ( 0,5* ) 1 PED - exercise 5 15 v L real

16 For our example is n =... even number. Then is a lengh of he longes profile L1 * *( 0,5*55)* ( 0,5*55) 909mm Real specific loading of he longes profile is (for he chosen No. of beams) q real = 1 / * p * 1 * real = 1 / * 300 * 1 * 0.55 = kn/m Real maximal bending momen in he longes profile is M Bmaxreal = q real * L 1 / 1 = * / 1 = 5.4 kn.m Real maximal bending sress in he longes profile is (for specified secion modulus) Bmaxreal = M Bmaxreal / W o = 5.4 / 81.9*10-6 = kpa = 66.0 MPa Maximal real deflecion of he longes profile is 66 MPa < 156 MPa reserve.36 x y maxreal = q real * L 14 / (384* E* J) = 78.75*10 3 * / (384 * 06*10 9 * 573*10-8 ) y maxreal = m = 0.1 mm < 1 mm O.K. reserve 8.3 x PED - exercise 5 16

17 4. Specificaion of he reinforced cover mass Mass of he fla cover is M FC = * D C / 4 * s * = * 1.1 / 4 * * 7850 = 77.3 kg Mass of longiudinal profiles Lengh of i-h profile for odd number of profiles L i D D D * *( 0,5* ( i 1)* )* ( 0,5* ( i 1)* ) v i Lengh of i-h profile for even number of profiles L i D D D * *( ( i 1)* )* ( ( i 1)* ) PED - exercise 5 17

18 Real pich of profiles L L 1 L 1 / / / / D D n = 3 (odd) number of profiles real = D / n n = (even) number of profiles real = D / n L max = L 1 = D L max = L 1 D PED - exercise 5 18

19 In our example is n = profiles, i = 1 (only symmeric profiles) so ha we can use he above calculaed resul L 1 = 909 mm. Mass of he 1 s profile M P1 = L 1 * m P1 = * 14.4 = 13.1 kg (m * kg/m = kg) Mass of all longiudinal profiles M Plongi = L i = * L 1 = * 13.1 = 6. kg Analogical we can specify mass of ransversal profiles (we mus subrac places where is he maerial of longiudinal profiles ransversal profiles are welded among longiudinal ones) M Prans 4.5 kg PED - exercise 5 19

20 Toal mass of reinforcing profiles M Po = M Plongi + M Prans = = 50.7 kg Mass of he fla cover is (circular plae) M FC = * D C / 4 * s * M FC = * 1.1 / 4 * 0.01 * 7850 = 77.3 kg Toal mass of he reinforced cover M Co = M FC + M Po = = 18.0 kg PED - exercise 5 0

21 5. Specificaion he of non-reinforced cover mass In he case we can specify he minimal calculaion hickness of he fla circular cover according Czech sandard ČSN , par 4.9. A skech of he cover wih marking according he ČSN is on he figure. s 1 s f D 0 p f D PD f D C Equaion for cover hickness specificaion s K 1 R * K 0 * D R * [mm; -, -, mm, MPa, MPa, -] p * D Calculaed diameer (pich diameer of flange bols) D PD = 1085 mm Coefficien of cover weakening wih holes for necks K 0 = 1 (in he cover are no holes for necks) Coefficien of a ype of cover periphery fixaion K = 0.40 Coefficien of cover weakening wih weld = 1 (no weld in he cover) for circular cover forming flange PED - exercise 5 1

22 Minimal calculaed cover hickness 0,3 s R 0,4*1*1085* 19, 1mm 1 156*1 Real cover hickness s 1 = s 1R + c s 1 1 mm c = allowances for corrosion ec. Mass of non-reinforced cover M CN = * D C / 4 * s 1 * M CN = * 1.1 / 4 * 0.01 * 7850 = 16.3 kg PED - exercise 5

23 Mass reducion if reinforcing is used M = M CN M Co = = 34.3 kg or M = (M CN M Co ) / M CN * 100 M = 100 * ( ) / 16.3 = 1.1 % As you can see from previous resuls he heoreical number of profiles was 1.1 and he chosen number was n =. Therefore are profiles oo overdesigned. I is affirmed by he calculaed value of real sress omaxreal = 66.0 MPa ha is much lower han allowed value 156 MPa. Therefore we can do an opimizaion of a choice of profiles size. Calculaions were performed in Excel and he resul was ha he bes soluion is using of profiles I 100. PED - exercise 5 3

24 6. Calculaion for opimized size of reinforcing profiles I 100 insead original I 140. Parameers of profile I 100: secion modulus of bending W o = 34. cm 3 momen of ineria J = 171 cm 4 profile specific mass m P1 = 8.3 kg/m profile widh b = 50 mm for I Theoreical maximal pich of profiles max = * (1 * σ D * W O ) /(p * L max ) M Bmax = q*l max /1 = σ B *W o q = p*/ σ B σ D max = * (1 * 156*10 6 * 34.*10-6 ) / (300*10 3 * (1050*10-3 ) max = m = 387 mm 97 PED - exercise 5 4

25 Beams of I profile momen of ineria needed from he deflecion poin of view secion modulus needed from he srengh poin of view Type Mass Type Mass PED - exercise 5 5

26 Toal max. heoreical force acing on he profile neares he axis F TPmax = p * * L 1 / 300 * * / = 61.0 kn Toal force acing on he cover 146 F CT = * D / 4 * p = * 1,05 / 4 * 300 = 59.8 kn Maximal specific loading of he profile neares he axis 59.8 q max = p * 1 * max / = 300 * 1 * / = 58.1 kn/m Maximal bending momen in he profile locaed neares he axis is M omax = q max * L max / 1 = 58.1 * / 1 = 5.34 kn.m 1.78 PED - exercise 5 6

27 Maximal heoreical bending sress in he profile locaed neares he axis is (for given secion modulus) I 100 I 140 Bmax = M Bmax / W o Bmax = 5.34 / (34.*10-6 ) = kpa = 156 MPa = D Maximal heoreical deflecion of he profile locaed neares he axis is y max = q * L max 4 / (384 * E * J) y max = 58.1*10 3 * / (384 * 06*10 9 * 573*10-8 ) 156 y max = m 0.5 mm 1 mm OK 0.4 PED - exercise 5 7

28 6.. Specificaion of real number of profiles and loading Number of profiles in one direcion n P = D / max = 1050 / 387 =.7 pcs I 100 I Chosen number of profiles (minimally a nex higher number) n Pch = 3 pcs Real pich of profiles real = D / n Pch = 1050 / 3 = 350 mm 55 Lengh of he longes profile (near he axis) For odd number of profiles For even number of profiles L 1 = D L * *( 0,5* )* ( 0,5* D D D 1 ) PED - exercise 5 8

29 For our example is n = 3... odd number. Then is a lengh of he longes profile L 1 = D = 1050 mm Real specific loading of he longes profile q real = 1 / * p * 1 * real = 1 / * 300 * 1 * = 5.5 kn/m Real maximal bending momen in he longes profile M Bmaxreal = q real * L 1 / 1 = 5.5 * 1.05 / 1 = 4.8 kn.m Real maximal bending sress in he longes profile I 100 I Bmaxreal = M Bmaxreal / W O = 4.8 / 34.*10-6 = kpa Bmaxreal = MPa < 156 MPa O.K. Maximal real deflecion of he longes profile y maxreal = q real * L 1 4 / (384 * E * J) 1 mm reserve is 1.1 x reserve is x y maxreal = 5.5*10 3 * / (384 * 06*10 9 * 171*10-8 ) = m 0.5 mm beer maerial uilizaion reserve was.56 x reserve was 8.3 x 0.1 PED - exercise 5 9

30 6.3. Specificaion of he reinforced cover mass Mass of he fla cover M FC = * D C / 4 * s * = * 1.1 / 4 * * 7850 = 77.3 kg Mass of longiudinal profiles Lengh of i-h profile for even number of profiles L i D D D * *( 0,5* ( i 1)* )* ( 0,5* ( i 1)* ) Lengh of i-h profile for odd number of profiles L i D D D * *( ( i 1)* )* ( ( i 1)* ) In our example is n = 3 profiles, i = (only 1 profile in axis and symmerical profiles). For he axial profile we can use above calculaed resul L 1 = 1050 mm. PED - exercise 5 30 I 100 I

31 Mass of he1 s profile M P1 = L 1 * m P1 = * 8.3 = 8.7 kg Lengh of he nd profile I 100 I L * *( ( 1)*350)* ( ( 1)*350) L = 783 mm 1050 Mass of he nd profile M P = L * m P1 = * 8.3 = 6.5 kg 0 Mass of all longiudinal profiles M Plongi = L i = 1 * L 1 + * L = 1 * * 6.5 = 1.7 kg 6. PED - exercise 5 31

32 Analogical we can specify mass of ransversal profiles (we mus subrac places where is he maerial of longiudinal profiles ransversal profiles are welded among longi. ones) M Prans 0. kg Toal mass of reinforcing profiles I 100 I M Po = M Plongi + M Prans = = 41.9 kg 50.7 Toal mass of he reinforced cover M Coop = M FC + M Po = = 119. kg Toal mass of non-reinforced cover Is he same as in he previous case, i is M CN = 16.3 kg. PED - exercise 5 3

33 6.5. Mass reducion for his opimized varian Mass reducion if he varian is used compared wih non- reinforced cover I 100 I 140 M = M CN M Coop = = 43.1 kg M = (M CN M Coop ) / M CN * 100 M = 100 * ( ) / 16.3 = 6.5 % or or compared wih reinforcing profiles I 140 i is M = M Co M Coop = = 8.8 kg M = (M Co M Coop ) / M Co * 100 M = 100 * ( ) / 18.0 = 6.9 % or PED - exercise 5 33

34 Recapiulaion of resuls Similar calculaions can be performed for oher reinforcing profiles. Way of cover reinforcing Toal cover mass (kg) Non-reinforced cover 16.3 I 140 x pcs of profiles 18.0 I x 3 pcs of profiles 119. leas maerial I x 1 pc of profile 14.4 smalles laboriousness U 00 3 x 3 pcs of profiles PED - exercise kg hemispherical cover s real = = 1.5 mm 18.5 U profile has lower bending secion modulus! heory, in fac he hickness would have o be greaer from echnological aspecs (welding of flange or baffles...) Noe: During a decision making i is necessary o ake ino accoun no only he maerial saving bu for example a labor consumpion of a varian (e.g. more welds). Therefore we mus judge every varian (no only like i is in he very simple example) from more poins of view. 34

35 op cover Drawing of he acual ube heaer wih basic dimensions (producer ZVU Hradec Králové) Handling wih heavy covers: axis of cover roaion connecing bar Mass of he lower cover lifs he op cover (hrough he lever mechanism) A he op is he axis of he cover roaion before he connecing bar, down is he axis of he cover roaion behind he connecing bar boom cover PED - exercise 5 35