proiect pc

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II. BILANT DE MATERIALE 1. Debitul de stiren produs de instalatie: D ST = D STanual τ functionare = 84000 × 10 3 8000 = 10500 kg / h 2. Debitul de stiren produs de instalatie: D ST.pur =D ST × p st 100 =10 500 × 99,4 100 =10437 kg / h 3. Debitul de etilbenzen care reactioneaza: x 10437 C 6 H 5 -C 2 H 5 → C 6 H 5 -CH=CH 2 + H 2 106 104 η= C u C t × 100= 41 48 × 100=85,41 % D EB = D STtotal M ST × M EB η = 9443 × 106 104 × 0 , 8541 =12453,91 kg / h 4. Debitul de etilbenzen care tine cont de conversie: D EB. alim = D EB C t = 12453,91 0 , 48 =25945,64 kg / h 5. Debitul de etilbenzen recirculat: D EB.rec =D EB .alim D EB =13491,73 kg / h 6. Debitul de materie prima proaspata: D mp. proaspata =D EB +D B =D EB + 0 , 1 100 ×D EB =124 78,81 kg / h 7. Debitul de materie prima recirculata: D mp rec =D EB. rec +D B . rec +D T .rec +D S T .rec =D EBrec + ( 0,1 100 + 0,2 100 + 0,2 100 ) ×D mp . rec

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Calculul Tehnologic al Unui Reactor Izoterm pentru Fabricarea Stirenului prin Dehidrogenarea Etilbenzenului

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Page 1: Proiect PC

II. BILANT DE MATERIALE

1. Debitul de stiren produs de instalatie:

DST=D STanual

τ functionare=84 000×103

8000=10 500kg/h

2. Debitul de stiren produs de instalatie:

DST . pur=D ST×pst

100=10500×

99,4100

=10437kg /h

3. Debitul de etilbenzen care reactioneaza:

x 10437

C6H5-C2H5 → C6H5-CH=CH2 + H2

106 104

η=Cu

C t

×100=4148×100=85,41%

DEB=DSTtotal

M ST

×M EB

η= 9443×106

104×0 ,8541=12453,91kg/h

4. Debitul de etilbenzen care tine cont de conversie:

DEB .alim=DEB

Ct=12453,91

0 ,48=25945,64 kg /h

5. Debitul de etilbenzen recirculat:

DEB .rec=DEB . alim−DEB=13491,73kg /h

6. Debitul de materie prima proaspata:

Dmp . proaspata=DEB+DB=DEB+0 ,1100

×DEB=124 78,81kg /h

7. Debitul de materie prima recirculata:

Dmprec=DEB .rec+DB.rec+DT .rec+D ST . rec=DEBrec+( 0,1

100+ 0,2

100+ 0,2

100 )×Dmp .rec

Dmp . rec=DEB .rec+0,5100

×Dmp .rec❑⇒ Dmp . rec=DEB . rec

1−0,005=13 559 ,53kg /h

DBrec=0,1100

×13559,53=13 ,56 kg/h

DTrec=0,2100

×13559,53=2 7,11kg /h

DSTrec=0,2100

×13559,53=27,11 kg/h

Page 2: Proiect PC

Tabelul nr.1

COMPONENTMP proaspata MP recirculata

%gr. kg/h kmol/h %gr. kg/h kmol/h

ETILBENZEN 99,8 12453,91 117,49 99,5 13491,73 127,28

BENZEN 0,2 24,91 0,32 0,1 13,56 0,17

TOLUEN - - - 0,2 27,12 0,33

STIREN - - - 0,2 27,12 0,26

TOTAL 100 12478,81 117,81 100 13559,53 128,05

Dabur=aburmp

× DEB . alim=3 ,51×25945,64=90809,73kg /h

8. Calculul cantitatii de etilbenzen in reactii secundare:

CRS=C t−Cu=0,4 8−0,4 1=0,0 7 %

1.DEBrs 1=DEBalim×CRS×CS1=544,86 kg

2.DEBrs 2=DEBalim×CRS×CS2=817,29kg /h3.DEBrs 3=DEBalim×CRS×CS3=454,05kg /h

9. Calculul produsilor de reactie:

9.1. Reactia principala

10437 x

C6H5-C2H5 → C6H5-CH=CH2 + H2

104 2

x = 200,7 kg/h

9.2. Reactia secundara 1

544,86 y z

C6H5-C2H5 → C6H6 + C2H4

106 78 28

y = 400,93 kg/h benzen

z = 143,92 kg/h etena

Page 3: Proiect PC

9.3. Reactia secundara 2

817,29 x y z

C6H5-C2H5 + H2 → C6H5-CH3 + CH4

106 2 92 16

x = 15,42 kg/h H2

y = 709,34 kg/h C5H6-CH3

z = 123,36 kg/h CH4

9.4. Reactia secundara 3

454,05 x y z

C6H5-C2H5 + 16H2O → 8CO2 + 21H2

106 16.18 8.44 2.2

x = 1233,64 kg/h H2O

y = 1507,78 kg/h CO2

z = 179,91 kg/h H2

DST=DST total+D ST rec

=10437+27,12=10464,12kg /h

DH 2=DH 2 rs1

+DH 2 rs2−DH 2rs3

=200,7+123,36−179,91=365,20kg /h

Dabur=DABalim−DH 2O

=90809,73— 1507,78=89576,09kg /h

Tabelul nr.2 BILANT DE MATERIALE PE REACTOR

COMP.Mi,

kg/kmol

INTRARI IESIRI

kg/h kmol/h %mol kg/h kmol/h %mol

EB 106 25945,64 244,77 4,62634 12453,91 117,49 2,17

B 78 38,47 0,49 0,00932 13,56 0,17 0,00

T 92 27,12 0,29 0,00557 27,12 0,29 0,01

St 104 27,12 0,26 0,00493 10464,12 100,62 1,85

C2H4 28 - - - 143,92 5,14 0,09

CH4 16 - - - 123,36 7,71 0,14

CO2 44 - - - 1507,78 34,27 0,63

H2 2 - - - 365,20 182,60 3,37

ABUR 18 90809,73 5044,99 95,35392 89576,09 4976,45 91,74

TOTAL - 116848,08 5290,80 1 91562,2 5424,74 100,00

Page 4: Proiect PC

III. PROPRIETATI TERMICE

1.Calculul caldurilor specifice: (kcal/kmol.K)

Cp=a+b⋅T+c⋅T 2+d⋅T3

T = 660 + 273 = 918 K

COMP. a b c d Cp

CH4 4,75 0,012 3,03E-06 -2,63E-09 16,45

C2H4 0,944 0,03735 -1,99E-05 4,22E-09 21,87

CO2 6,85 0,008533 -2,48E-05 0 -6,73

H2 6,88 0,000066 2,79E-07 0 7,18

H2O 6,89 0,003283 -3,43E-07 0 9,65

B -8,65 0,1158 -7,54E-05 1,85E-08 48,81

T -8,213 0,1336 -8,23E-05 1,92E-08 60,39

EB -8,4 0,1593 -0,0001 2,40E-08 72,63

St -5,968 0,1435 -9,15E-05 2,20E-08 66,14

c)

a = 6,85

b = 0,008533

c = -2,475

d=0

Cp=6 ,85+0 ,0853⋅T−2 ,475⋅10−6⋅T 2

T(K)=918

Cp=12,58kcal/kmol*K

Page 5: Proiect PC

2.Calculul entalpiilor de formare:

Component EB ST B T CH4 C2H4 CO2 H2O H2H kcal/kmol 7,12 35,2

219,82 11,95 -17,889 12,496 -94,052 -57,798 0

1.C6H5-C2H5 → C6H5-CH=CH2 + H2

a -8,4 -5,968 6,88 a=9,312

b 0,1593 0,1435 0,000066 b=-0,0158

c −10−4 −9 ,15⋅10−5

2 ,7910−7 c= 0 ,878⋅10−5

d 2 ,395⋅10−8 2,2 ¿10−8

- d= −0 ,195⋅10−8

a= ∑ ν pr⋅apr−∑ νr⋅ar

2.C6H5-C2H5 → C6H6 + C2H4

a -8,4 -8,65 0,944 ∆a = 0,694

b 0,1593 0,1158 0,03735 ∆b = -0,00615

c −10−4 −7 ,54⋅10−5

−1 ,993⋅10−7 ∆c = 0 ,0467⋅10−5

d 2 ,395⋅10−8 1,854 ¿10−8

4 ,22⋅10−9 ∆d = −1 ,19⋅10−9

Page 6: Proiect PC

3. C6H5-C2H5 + 16H2O → 8CO2 + 21H2

a -8,4 6,89 6,85 6,88 ∆a = 97,44

b 0,1593 0,003283 0,08533 0,000066 ∆b = -0,142

c −10−4 −3 ,43⋅10−7

−2 ,475⋅10−7 −2 ,475⋅10−7

∆c = 9 ,15⋅10−5

d 2 ,395⋅10−8 0 0 0 ∆d =

−2 ,395⋅10−8

4. C6H5-C2H5 + H2 → C6H5-CH3 + CH4

a -8,4 6,88 -8,213 4,75 ∆a = -1,943

b 0,1593 0,000066 0,1336 0,012 ∆b = -0,0138

c −10−4 −2 ,475⋅10−7

−8 ,23⋅10−5 0 ,303⋅10−5

∆c =2 ,051⋅10−5

d 2 ,395⋅10−8 0 1 ,92⋅10−8

−2 ,63⋅10−9 ∆d = −7 ,38⋅10−9