Conversion CSTR and PFR Problems

 

TU UT TOR ORI AL2SOLUTI ONS Lect ur er :Mi s sAni sAt i kahAhmad Tel :+6049768190 Emai l :ani s at i kah@uni map. edu. my

 

QUESTI ONS Thee x o t he r mi mi cr e ac t i o n  

A>B+C

  wasc ar r i e do utadi abat i c a l l yandt hef o l l o wi ngdat ar e c o r de d:

X -r A 

0

0.2

0.4

0.45

0.5

0.6

0.8

0.9

(mol/dm3.

1.0

1.67

5.0

5.0

5.0

5.0

1.25

0. 91

min Thee nt er i ngmo mol arflo wr at eofAwas300mol / mi n.

( a)Whatar et hePFRandCSTRvol umesnec es s aryt oac hi ev e40% c onve r s i on? ( b)Ove rwhatr angeo fc onver s i onswoul dt heCSTRandPFRr eac t orvo l umesbe i de nt i c al ? ( c )Wha ti st hema ma x i mum c o n v e r s i o nt ha tc a nb ea c hi e v e di na10 5dm3  CSTR? ( d)Wha tc o nv e r s i o nc anbeac hi e v e di fa72dm3  PFRi sf o l l o we di ns e r i e sb ya24 dm3CSTR? ( e )Whatc o nv e r s i o nc a nbeac hi e v e di fa2 4dm3  CSTRi sf o l l o we di nas e r i e sb ya 72dm3PFR? ( f )Pl o tt hec o nv e r s i o nandr at eo fr e ac t i o nasaf unc t i o no fPFRr e ac t o rv o l umeup t oav o l umeo f1 00dm3.

 

mes PART( A)WhatarethePFR andCSTR volume

neces s aryt oachi eve40% convers i on?



PFRVo l ume me :  X 

V  PF  PFR R

dX 

=  F  A0 ∫   − r  A 0

Rec al lSi mps onOneThi r dRul eFor mul a:

Us i ngSi mps onOneThi r dRul e;  X 2

0.4 − 0    2

h=  X 0

= 0,

V  PFR

∫ 

 f  ( X ) dX  =

= 0.2 = ∆ X 

 X 0

where h =

 X 1  =  0.2, X 2

h 3

[  f  ( X 0 ) + 4 f  ( X 1 ) +  f  ( X 2 ) ]

 X 2 −  X 0 2

= 0.4,

  F  0  F  0 ∆ X    F  0 = +4 +   3  − r  ( X  = 0) − r  ( X  = 0.2) − r  ( X  = 0.4)   A

 A

 A

 A

 A

 A

 X 1

=  X 0 + h

 

PART( A)  X  A r  ( mo l / dm . mi n) 3

F  / r  A 0 A ( dm3)

0

0. 2

0. 4

0. 45

0. 5

0. 6

0. 8

0. 9

1. 0

1. 67

5. 0

5. 0

5. 0

5. 0

1. 25

0. 91

300

180

60

60

60

60

240

330

 X  

V  PFR

 = ∆3  − r  ( X 0 = 0) + 4 − r  ( X  0= 0.2) + − r  ( X  0= 0.4)     F 

 F 

 A

 A

 F 

 A

 A

 A

 A

0 .2

=

400

3 [ 300 + 4(180 ) + 60 ]

300

FA0/ r A

= 72 dm

Le ve l s pi e l Pl o t

3

200 100 V  PFR 0 0

0. 2 0. 4 0. 6 0. 8

 X

1

 

PART( A)

400 300



FA0 / r A

CSTRVo l ume me :

V CSTR

=

 F   X     A 0

200 100

V0 CSTR 0

0. 2 0. 4 0. 6 0. 8

− r  A

 X

Subs t i t ut i ngt hev al ueo  F f A  /  – r  a  X t =0. 4; 0  A   3 ( 0 .4 ) V CSTR   = 60dm

= 24dm

1

3

Le v e l s pi e l P l o t

 

PART( B)Overwhatrangeofconversionswouldthe

CSTR andPFR r eac t orvol umesbei dent i cal ?

Fr o m Le v e l s p i e lPl o t ; 350

V  PF    PFR R

300

0.6

 

0.4

= V CSTR  

0.6 0.4

250 200

0.6

0.4

100 50 0 0 0. 1 0. 2 0. 3 0. 4 0. 5 0. 6 0. 7 0. 8 0. 9 1

 X

Thevol umeof bot hreact ors ar ei dent i c al at  X=0. 40. 6  ,  

 F  A0

∫ − r  dX  =

FA0 / r A 150

 A

   A0 ( 0.6 − 0.4)  F 

− r  A

 

PART( B)Overwhatrangeofconversionswouldthe

CSTR andPFR r eac t orvol umesbei dent i cal ?

Pr o vi ngbyc al c ul at i o n; Us i ngSi mps o n OneThi r dRul e

Vo l umeo fPFR

h=

0.6 − 0.4    2

= 0.1 = ∆ X 

 X 0

= 0,

 X 1  =  0.1, X 2

= 0.2

 ∆ X    F  A0  F  A 0  F  A0 +4 + V  PFR =  3  − r  A ( X  = 0.4) − r  A ( X  = 0.5) − r  A ( X  = 0.6)  0 .1  

 [ 60 + 4( 60 ) + 60 ]  

=     3  

  3 12 dm =

Vol ume meo fCSTR  F  V CSTR

=

 X 

 A0

2

−  X 

1

( −( r  A ) exit  )

 F  =

 A 0  

( − r  A ) exit  

(  X 2

−  X 1

)

=

60dm 3 0. 2

(

)

  = 12 dm 3

 

ma axi mum conver s i ont hatcanbe PART( C)Whatisthem achi evedi na105dm3  CSTR?

Byt r i alande r r o r ,c a l c ul at ewhati  Xt  X s t hatgi v e sa CSTRv o l ume meo f105dm3  Tr i al1:X=0. 6,  F  A0 X  V CSTR = −

dm = 60  

3

( 0. 6) = 36  dm 3

r  A

Tr i al2:X=0. 8, V CSTR

=

 F  A0 X  −

400 =

240dm ( 0.8  )  

3

=

 

192dm

3

300

FA0 / r A 200

r  A

100

Tr i al3:X=0. 7 ,  

V CSTR

=

 F  A0 X  −

dm = 150  

3

  3 ( 0.7  ) = 105 dm

0 0 0. 20. 4 . 6 0. 8 1   X0 = 0 . 7

 X

r  A

Them ma axconver s i oncanbeac hi eve di n105dm3  ofCSTR i s0. 7

 

3 W h a t c o n v e r s i o n c a n b e a c h i e v e d i f a 7 2 d m   PART( D) PFR i sf ol l owe wedi ns eri esbya24dm3CSTR?

F 300mo mol / dm3  A0=  A

 VPFR=72dm3

 X1=0. 4

Fr om part( a)

 VCSTR=24dm3

 X Byt r i alande r r o r ,c a l c ul at ewh whati s 2 

 X2=?

t hatgi ve saCSTRvo l umeo f24dm3  Tr i al1:X=0. 6,

V CSTR

=

 F  A0 (  X 2 −  X 1 ) ( − r  A ) exit 

Tr i al2:X=0. 7 ,  F  A  X  0 2 V CSTR = − r  A

(

(

−  X  1

)

exit 

No w,wekno w t hatt he c o n v e r s i o nmus t b eb e t we e n0. 6 0 . 7 .

  60dm 3 ( 0.6 −  0.4) = 12  dm 3

= 

)

3 =  150dm

 

 

( 0.7 −  0.4) = 45dm

3

 

3 W h a t c o n v e r s i o n c a n b e a c h i e v e d i f a 7 2 d m   D) PART( PFR i sf ol l owe wedi ns eri esbya24dm3CSTR?

F 300mo mol / dm3  A0=  A

 VPFR=72dm3

 X1=0. 4

 VCSTR=24dm3

Tr i al3:X=0. 64, V CSTR

=

 F  A0 (  X    2

(

=

−

−

 X2=?  X 1 )

400 300

r   A

exit 

) 3 100dm ( 0.64   0.4) 24  dm 3  −

FA0/ r A

=

200 100 0 0 0. 20  . 4= 0 . 6 0. 8 1 X 0 . 64

3

 X=0. 64canbeachi evedi fa72dm   PFR i s f ol l owe wedi nseri esbya24dm3CSTR.

 X

 

3 W h a t c o n v e r s i o n c a n b e a c h i e v e d i f a 2 4 d m   PART( E) CSTR i sf ol l owe wedi nas eri esbya72dm3PFR?

3  A0 l / dm F =300mo

 VCSTR=24dm3

Fr om part( a)

 X1=0. 4

 VPFR=72dm3

 X2=? 3

Byt r i alande r r o r ,c a l c ul at ewhati s 2  t hatgi ve saPFRvo l umeo f72dm    X Tr i al1:X=0. 8,

h=

0 .8 − 0. 4   

= 0.2 = ∆ X 

 X 0

= 0.4,

 X 1  = 0.6, X 2

2

V  PF  PFR R

  F  0  F  0  F  0 ∆ X   = +4 +  − r   X  = − r  ( X  = 0.6) − r  ( X  = 0.8)  3  ( 0.4)  A

 A

 

 A

 A

= 0.2 [ 60 + 4( 60 ) + 240 ] = 36  dm 3 3

 A

 A

= 0.8,

 

3 W h a t c o n v e r s i o n c a n b e a c h i e v e d i f a 2 4 d m   E) PART( CSTR i sf ol l owe wedi nas eri esbya72dm3PFR?

3  A0 F =300mo l / dm

 VCSTR=24dm3

 X1=0. 4

 VPFR=72dm3

 X2=? 3

Byt r i alande r r o r ,c a l c ul at ewhati s 2  t hatgi ve saPFRvo l umeo f72dm    X Tr i al2:X=0. 9,

h=

0 .9 − 0.4    

= 0.25 = ∆ X 

 X 0

= 0.4,

  0.65, X 2  X 1  =

2

V  PFR

  F  0  F  0  F  0 ∆ X   = +4 +  − r   X  =  − r  ( X  = 0.65) − r  ( X  = 0.9)  3  ( 0.4)  A

 A

 

=

 A

 A

 A

0.25 [ 60 + 4(107 ) + 330 ] 3

 A

= 68.17 dm  

3

= 0.9,

 

3 W h a t c o n v e r s i o n c a n b e a c h i e v e d i f a 2 4 d m   PART( E) CSTR i sf ol l owe wedi nas eri esbya72dm3PFR?

Tr i al3:X=0. 91 ,

h= V  PF  PFR R

0.91 − 0.4     2

= 0.255 = ∆ X 

 F  A0  − r  A ( X  = 0.4)

∆ X  

= 3

+4

 X 0

= 0.4,

 F  A 0 − r  A ( X  = 0.655)

+

 X 1 =     0.655, X 2

= 0.91,

 F  A 0 − r  A ( X  = 0.91)  400

0.25   = 3

[

60 + 4 11 110 0

(

)

+ 346 =

 

300

3

71.91dm

]

FA0 / r A

200 100

 X=0. 91canbeachi evedi fa24dm3  CSTR i sf ol l owe wedi naser i esbya72

dm3PFR

0 00 . 10 . 20 . 30 . 40 . 50 . 60 . 70 . 80 . 91

 X

 

PART( F)Plottheconversionandrateofreactionasa

f unct i onofPFR r eact orvol umeupt oavol ume of100dm3?

 X r   A

0

0. 2

0. 4

0. 45

0. 5

0. 6

0. 8

0. 9

1. 0

1. 67

5. 0

5. 0

5. 0

5. 0

1. 25

0. 91

300

180

60

60

60

60

240

330

0

48

72

76. 5

80

84

104 130. 5

( mo l / dm3. mi n)

 / r F  A 0 A ( dm ) 3

 VPFR  ( dm3)

 

PART( F)Plottheconversionandrateofreactionasa

f unct i onofPFR r eact orvol umeupt oavol ume of100dm3?

0 . 7 6 . 0 0 0 . 6 5 . 0 0 0 . 5 4 . 0 0

0 . 4  X

0 . 3

rA( mol / dm3. mi n)

3 . 0 0

0 . 2

2 . 0 0

0 . 1

1 . 0 0

0

0 . 0 0

0

10

2 0

3 0

4 0

5 0

6 0

 VPFR ( dm3 )

7 0

8 0

9 0 1 0 0

0 1 02 0 30 4 05 06 07 08 09 01 0 0  VPFR ( dm3 )

 

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