single reactions in continuous isothermal reactions

7/29/2019 Single reactions in Continuous Isothermal Reactions

1/41

Isothermal Reactors

1

CSTR constant densit & SS single reaction

Reversible reaction

Plug Flow 1/r plot concept

Semi-batch reactors

Variable density

Reactors in series CSTR in series

Autocatalytic reactions Reversible Reactions

2


2/41

CSTR Continuous stirred tank reactor (CSTR)

ompos on s e same everyw ere n e reac or an

the exit pipe.o ume =

[L]

o v

CA

-*

3

Exit and tank composition same

Steady-state

Constant Density Reaction

liquid or

ases, no mole chan e, e. . A+2B3C

4


3/41

CSTR Mass Balance:

dNr

dtjjjo += time

Molar flow rates of species j

time

moles

joojo

CvF =*

L mole

5 s L V

CSTR at Steady-State & Const Vol. Flow

= o

=0*

rV

vjjo =ovv =

dCj

Batch

rCC jjjo +=0. .

Flow dj

=

rCC jjjo = rCCjjo=

6j


4/41

.1=AA roducts

rCC AAo =

jjjo

nth order reactionn

AkCr=

rs or er ne cs n=

AAo CC *

kCC += 1AkC

CC Ao=

*

7k+1

r=kCA

Single Reaction, AB AAoB CCC =

CCC AAoB =

CkCCCCC AoAoAoAoAoB

+==

*

8


5/41

.

nA

rjjjo =

=r

o pro uc s

For 2nd order (n=2)

0

411

AA

Ck

C+

=*

02 =+ AoAA CCkC

9

Fractional Conversion X AB Use A as the limiting reactant with stoichiometric

System is at constant densityr=kCA

n

X varies from 0 to 1

= XCCC= XCC AoA )1()(

=XCAo +=

kCC AoA )

11()(

*

=>= )1( XkX +=

k

Ao

1 kkXX =+

Xk = 1 == XCC AoB )(kX +

=1

*

10(X) or X() Cj()


6/41

CkCkr =

AB

=r*

CB=0 =

=CC AAo

*

rrCC jjjo =

AAoAAobAf

CCCCkCk = )(1+

bA kC

AAoAobAbAf

CkCCCkCk

CCCkCkCk

+=++

=+ 1)(0 ++ bfA kkC

11

oo So, CA()

CCC =

bAoAo

kCC =

1)(0 ++=

bfA

A

kkC

)( bAofbAo

fb

kCkkC +As

)( fb kk +*

)( bf

fAo

kk +=

fB kC = (the conc ratio = Keq

12bA


7/41

Same residence time for all molecules Assumptions

Constant Diameter

Steady state

Single reaction

13

ud =

VL ==

3

u (m/s)

v (m3/s)

14z


8/41

Constant Densit Plu Flow

joojo CvF = jj CvF=

2

4

D

v

A

v

u==

z Lz+dz

Mass Balance on j

0)()( =++ rdzdzzFzF jtjj jtjj

Tube Area=

dz

r

dz

dC

dzzCzCjj

=

+lim

tt

udzz

dz

0

15r

dzu j

j= r

dj

j

=uz=

0=++ rdzAdzzFzF

- =

0)()( =++ rdVdVVFVF

rVFdVVF jj

=+ )()(

dF

dv0

16

rdV

j=V V+dV


9/41

For reactant species A, A=-1

AC

AdCVrdCj

=

==AoCo

rvd

r

d

A =

ddCA =

17

First Order in Constant Density PF

=dCj =

C

d

r

dA= dk

CC A

A=

0

kCdC

AA =

kCA =ln

keCC=kd

dCA =

Ao

o

CA

18

than t)


10/41

Exs. 3-1 & 3-3 PF versus CSTR (lst Order)

The reaction AB, r=kCA, occurs with 90% conversion.If k=0.5 min-1 C =2 moles/liter and v=4 liters/min

what residence time and reactor volume will be

re uired? How lon a 2-cm diameter tube would be

required for this conversion and what would be the

fluid velocit ?

Assuming CSTR =CC AAo rCC jjjo =r

*

19

Exs. 3-1 & 3-3 PF versus CSTR (first order)

rdC

j

j=

AssumingPFTR

k

AoA eCC=

*

litersxvVPF 4.1861.44 === ma er an

CSTR

( )mcm

x

DL 6.585860

4/2

.

4/ 22====

min/12705860

cmL

u ===

May be

too LONG

2061.4


11/41

Exs. 3-2 & 3-4 (2nd Order)

The reactionAB r=kCA2 occurs with 90%. . , A0

and v=4 liters/min, what residence time and reactor

*

literslitersxvVCSTR 360904 ===

21

Exs. 3-2 & 3-4

PFTR rdCj

= r=kCA2

=AC

AdC

AoC r

min9111111

2= = ==

AdCAo

A

PF ==

22


12/41

PF versus Batch Reactor In PFTR, time t that a molecule spends in the reactor

is z/u. Time for molecule to leave reactor is L/u.

Residence time in a constant-density, constant cross-

u

L

v

VPFTR ==

By using udzdt / d We can convert Batch Reactor to PFTR Design

e uation:r

ddzu

dt

AAA ==

a) Aproductsdzdt,PFTR

23

Completely mixed inside 1 slice

z

L

(Constant Density)

nth order kinetics in a batch reactor

111)1(1 nnAoAoA tkCnCC +=

nth order kinetics in a PFTR is same

[ ] )1/(11)1(1 nnAoAoA kCnCC+=

batchPFTR t=

24


13/41

(FIRST order irreversible reactions with constant density)

AA

C

A

batchPFTR C

C

kC

C

kC

dC

kt

A

0

ln

1

ln

11

==== rd

dC

PFA

=:Ao

AA CC = 0AkC

1= XCA

1

1

1

0

0

=

C

C

A

A

1/ XCCCCV AAoAAoCSTRCSTR =

=

==

A

25

00 AAAAAPFTRPFTR

Com arison between 3 reactors

[ ])1/(1ln)1()/ln(1/

)/ln( 0

0

0

0

XX

X

CC

CC

CCC

CC

V

V

AA

AA

AAA

AA

PFTR

CSTR

PFTR

CSTR

=

=

==

X=(CA0-CA)/CA0PFTR

CSTR

0.0 1.0

. .

0.9 3.91

0.95 6.34

0.99 21.5

0.999 145

26

high conversion for these kinetics.


14/41


AB

- a c or w g ve e same convers on

- CSTR gives a lower conversion for the same reaction

27

me a c or res ence me con nuous

Difference between PF and CSTR becomes larger as

the order of reaction becomes larger.

CC r=kCA2

AA

AAPFTR

CkC 0

0= 2A

CSTRkC=

1C

)1( XCAPFTR ==

28


15/41

10.00

12.00

Fn=2

6.00

8.00

R/Tau_

P

2.00

4.00

Tau_

CS

0.00

0 0.2 0.4 0.6 0.8 1

X

29

Com arison between 3 reactors

th

are the same.

Negative order the CSTR requires a smaller volume.

Not true for non-isothermal.

Positive order the CSTR requires a larger volume

30


16/41


Batch CSTR PFTRReactor Size for a given

conversion

+ - +

Simplicity and cost + + -

Continuous operation - + +

Large throughput - + +

Cleanout + + -On-line analysis - + +

Product certification + - -

31

1/r Plot Aproducts

1AAo CCCC

=

=

A = -1*

1C

AA rr

AdC

C

CA

APFTR Cr

A

Ao

=CSTRrd

j

=

)( 0 AC

A

A

PFTR CCdCr

Ao

=

PF

= ydx*

(area under curve)

XA=0 XA=1

CSTR very inefficient

for high X (For first

32d(CA0-CA)=-dCA

orders)


17/41

1/r Plot for r=kC n

r is monotonically decreasing for n (+)

r increases if n


18/41

A 3B r=kC 2

At complete conversion, v=3voRT

nv =

, o

In Batch, we use X instead of C for variable

35

Mass Balance dNj

dtjjjo

=

jjo

X X

- jjjo

Molar flow rate varies with conversion

AoA

XV

=*

XV

AA0

36XrFA=

0

1=A


19/41

XV A products

XrFA0No inert

)1( Xvv o +=Since X is constant throughout,

the volumetric flow rate is

A(g)3B(g)+C(g)e.g. 3113 =+== j

*

rrvv

A

o

A

o

00 ===

37

PFTR Variable Densit For variable density and cross-sectional area.

Similarity of PF and CSTR (one is integral form)z z+dzj

zFj(z+dz)

=X

AA

r

dXFV 0A=-10)()( =++ rdzAdzzFzF

jtjj

==X

AA

dXC

V0AA

dVrdzzFzF +=+ )()(

jtjj0

1

)1(

AA

AAA

CX

XCC

=

=

o rv 0

ACA r

dF =

0

AA

A

C

dCdX =

==AC

A

o rv0

ConstdensityXFF )1( =AdX =

38AAA dXFdF 0=AA

dV0


20/41

Ex. 3-6 CBA + 2 r=kCA2

Find the expression for the reactor volume V for specified and

feed flow F for the reaction amon ideal ases with no

diluent in a CSTR and PFTR.

2)21( XXFX Ao +22 )1( XkCXr Ao

AoCSTR)1(0 XFF AA =

= = Xr

dXFV

X

AoPF

00

0

XFF AC

AB

=

+=

dX

XFV

X

A

PF

)21(

2

2

2

0

0A +=

)21( Xvv +=

++=

=

XXFA

XAo

9)1ln(12940

0

*

39

Ao

Ex. 3-6 CBA + 2

+ XX )21( 2 2+ +2- +4

( )= XX

X 102 4x2- 8x+4

12x-3

( )

+= =dX

XX 14

0

214

+= BAX

+= dX

X

XX

1

1434

2)1(14

22

+=

BXAX

+

+=

=

dX

X3

34

34)1(14 +=

BXAX

= XXX

90 3,1

==BX

40 =

X1,


21/41

Ex. 3-7 Variable Vol An B , r=kC Find the reactor volume V required to obtain 90%

CSTR and in a PFTR with no diluent fornB=2, 1 and-1 A0 , . , o

liters/min. (high pressure reactor, p=49 atm).

2=Bn*

=CA

)1(0 XFF AA =

)1(0

0

XFFF ABA

AB

+=+

=

41

=r)1( Xvv o +=

Ex. 3-7 Variable Vol A2BCSTR

AkCr=

XXvXXFXFV oAAoCSTR

)1()1(0

+=

+==

Ao

)9.01(9.04 +CSTR .

)9.01(5.0

42


22/41

Ex. 3-7 Variable Vol A2BAkCr=PFTR

dXXFdXFV

X

A

X

AoPFTR )1(0+

== -

Xv

r

X

Ao

1

00

+

-x+1x

-

XkPFTR

)1(0

=+1

dX

XXk

vV oPFTR

)1(

11

1

1

+

=

[ ] litersXXv

VPFTR 6.29)1ln(20 ==

43

Ex. 3.7 (Continued) Variable Vol A1B

1=Bn 00F

C AA =AA XFF = 0 )1(

)1( XkCkCr AoA

o

==

ABA

AB

FFF =+

=

0

0

o =

litersX

X

k

v

X

X

kC

F

Xr

XFV o

Ao

AACSTR 72

1.0

9.0

5.0

4

11

00 ==

=

==

LXk

v

X

X

kC

F

Xr

XFV o

Ao

AAPFTR 4.1810ln81

1ln1

0

0

0

0 ===== 44


23/41

Ex. 3.7 Continued2/1=n [AB]

)1()1(0 XCXF

C A

=

=)1(0 XFF AA =

)1(

)1()1( 2121

X

XXo

o

)1( 2102

1

XFFF

XFF AB

=+

=

)1( 21 XAoA

==)1( 21 Xo =

Xk

v

XkCXrFV o

Ao

AACSTR

11

2200

=

==

liters6.399.01

)45.01(9.0

5.0

4=

=

[ ]XXv

dXXFdX

FV oX

A

X

APF )1ln()1( 210

0 +=

==

45L

Ao

8.12)9.010(ln4

00

=+=

- AB. V is larger than with no density change if the

react on pro uces more mo es nB= ecause t s

dilutes the reactant, while V is smaller if the

react on re uces t e num er o mo es nB=An B

AkCr=nB CSTR PFTR

2 136.8 29.6*

*

39.6 12.8

46


24/41

, B Fig 3-6,

47

, B

valid when density varies with conversion.

Constant-density used normally as an

approximation when:

E ual moles of reactants and roducts

Liquids PFBatch* u e gases

EstimationsConstant

48Variable


25/41

Reactor residence time defined based on the

inlet conditions V *

vo=

STV

vSV o ==

In variable density reactions, the velocity

c anges w convers on. ang ng an

along reactor can also change v.

49

PFTRs in series A Products

1 2 N

X

ith

1 PFTR: XdXV

1 2 N-1

==A

o rv 00

Consider N plug flow reactors connected in series. Let

X X X be the fractional conversion of A leavin

reactor 1,2,..N. Based on the material balance for ith

reactor

==iX

Ai

i

dXCV 0

50iX 10


26/41

For N reactors in series

nXXX dXdXdXV 21

=nXXXA

rrrF110

...0

If no change of composition of intermediate streams,

*

then PFTRs in series is same as 1 large PFTR

nX

=rF

51

0

CSTR in Series lst Order Kinetics*

1 3n

2

01

CC AA =111110 AAAA kCCrCC ==

1

222221 AAAA kCCrCC ==

012

CCC AAA ==

52

212


27/41

s r er ne cs

333332 AAAA r ==

CC

)1)(1)(1()1( 32133

kkkkCA

+++=

+=

==

AnA CC 0)1(

nnnnnn

=

+=+= n

Ankk )1()1(

53

,

have the same residence time ( )

o a res ence men

1

oncen ra on rom n reac or n erms o A0is given by C

nn

An kC

)1( +=

54


28/41

-

AAo CC 1AAo

AA

CSTRCrCr

qua res ence me = same rec ang e area

55A series of CSTR has a performance close to PFTR

PFTR+CSTRFirst order kinetics

A1

then CA2.

o owe yAoC 1AC 2AC

1

12

1 kC AA+=

1keCC=

1

2

0

2

12

11

kAAA ekkC

+=+=

56


29/41

PFTR+CSTR first order kinetics CSTR followed by PFTR

0AC 21AC 2AC

1

0CA=

22 0 kAkC

1

11 k

A+

=

1

121

AAk+

kinetics, the expressions are identical for bothconfigurations. So which reactor is first is

57

nconsequen a .

high) then PFTR to minimize total reactor

.

en

CSTR

en

PFTR

58


30/41

. The reaction AB, r=kCA occurs in n equal volume

s n ser es, eac w t res ence t me w t

90% overall conversion. Ifk=0.5/min, CA0=2mo es ter, an v= ters m n, w at res ence t mes

and reactor volumes will be required forn=1,2, and 4?

/1 n

AoCnnAo

An

CC =

CSTR

Total

residence

time

An

n

Ckn

nAn k

C += 1 n

AoCPF

A kC

C=1ln

Cn 1

59n

AoC+=

-.min18

2.00.2)110(

11

1 01 =

=

==

==

CCC AAAo

...

22 2/12/1

==

== AoC

.

5.0 ACk3/1

min92.6)110(5.0

133/1

3 ==

==A

Ao

Ck

min22.6)110(4

14

4 4/14/1

4 ==

== Ao

C

.A

11

60

.10

n5.0=

=PF


31/41

.For n=1, we have single CSTR.

as n to approach PFTR for which =4.61min

Total Reactor volumes:

n=1, V1=72 L, (=4 x 18)vV =

n=2, V2=34.6 L, (=4 x 8.65)

n=3, V3=27.7 L, (=4 x 6.92)

n=4, V4=24.9 L, (=4 x 6.22)

The total reactor volume decreases as n.

61

BAr= ,

a e o orwar reac on s en ance yconcentration of a product. Can also be written by

CkCrBBA =+ ,2

B . , B , , B .

CA0CA0CB0=0

62

A0

CB0=0CAfCBf0 (TANK reaction happens)

B0


32/41

Autocatal tic Reactions Rate )( 00 ABAABAA CCCkCCkCCr +== B0= ,

)(0 AAA

CCkCr =

r versus CA0-CA is a parabola. PF requires timesince r=0 at feed (pure A). CSTR is feasible.

63

Autocatal tic Reactions PFTR

==AC

AdCV

+=

C

A

CCCkC

dCA

AoCo

rv

+=C

dCdCA

A

111

0

+

++ ABAC ABA

CCCC

CCCCCCkA

0000

1

0

( ) +

+=

ABAABA CCCCCCk 000000nn

( ) ++= ABAA

A

B

BA CCC

C

C

C

CCk 0000

00ln

1

64


33/41

Partial fractions 00

1)( AABA BCCCCA =++

C

AdC

A

=

00 ,1)(,0 BAA CCACLet =+=

AA

C

C ABAA

dCB

dCAA

A

+=

000

1

00

1

BA CCA+

=

C

AABA

ABAC A

dCBCCCCAA

A

++=

+0

00

00

)(1

00 ,1, ABAA BCCCCLet =+=

C ABAACCCCk

A+

000

00

1

BA CC

B+

=

AA

C

dCdCA

+= 111

65

ABAC ABA A0000

0

00 AAAACCCC

=

=)( 00 ABAAA CCCkCCr +

oss e

Tubulareac ors

66


34/41

Reversible Reactions Rate goes to 0 near. For PF and CSTR, as

.

CSTR needs a longer (like irreversible reaction)

r=kCACSTR

r=kfCA-kbCB PF

67

Ex. 3-9 Reversible Reactions

BbAf CkCkr =AB

For the reaction, find the residence times for 50%

conversion in a CSTR and in a PFTR ifkf=0.5 min-1,

kb=0.1 min-1, CAO=2 moles/liter, v=4 liters/min, andCBo=0.

For the CSTR

*

68


35/41

Ex. 3-9 Reversible Reactions

For the PFTR

C

AdCA

=C AAobAf

CCkCkAo

)(

AobAbf

Aof

bf

PFTRCkCkkkk ++

=)(

ln

25.01 .21.016.06.0

69

Ex. 3-9 continuedFor this reaction calculate the residence time for90%

AAo CC =AAobAf CCC 2.00.2=

=

Ck1

)2.00.2(1.02.05.0 18.010.0 =

AobAbfbf

PFTRCkCkkkk ++

=)(

n

)ln(21.02.06.0

.

ln6.0 =

=

7008.0=


36/41

Ex. 3-9 (continued)

For this reaction calculate the residence time for90%

= in either reactor. We go beyond the equilibriumconvers on, w c s

CCCk

,

,

,

, ===eqAeqAb CCk

so a A,eq= . mo es er, B,eq= . mo es er, an

the equilibrium conversion is 83%.

This is the maximum conversion obtained for these

71

ne cs n any s ng e reac or

-

Assume spatially uniformFjo

dNj =

Fj

dto

Constant Density (AProducts) Constant Volume

VrCCdCV AAoA = 0

72


37/41

Semi Batch Reactors A may be added continuously but nothing

removed to iverVFF

dt

jjjo

j+=

o ume o reac or ncreases near y w me

tvVV += We obtain

rVFdC

VdV

CdVC AA

+=+=dtdtdt

o

dCA =

dCA

dtjoAooA

73dt

ooAAoo 0

-

Bioreactors

Solids processing

74


38/41

Durin startu and shutdown

Mass-balance equation for species A ssume reac or vo ume an ow ra e are

constant and density is unchanged 1=A

A CVrFFdN

= AdC

AdC

t

AAAodt

=

AAAodt

75CSTR MASS BALANCE

Transients in Continuous Reactors

Solvent replacement (No Reaction)=

Initially t=0

CA0 is inlet concentration

C is concentration within tank

Ai

AdC =

CAi is initial concentration within tank

Let u=C -C

AAAo

dt

= AAoA CC

dC)( AAi

AA

i

t

CC

CC

u

u=

=

0

0lnln

=tC

A dtdCA 1

t

AA eCC

CC =

0

C AA

CCAi 00

tu

t

=

76

=u

dtu

i 0

i


39/41

Transient Reaction in CSTR Tank initially filled with pure solvent.At t=0, feed

A0 A

dCkdu

CkCu AA

+=

+=

1

)1( 0 kCCCdt

AAAoA = )(

dudCA =1

t

kCCCAAAo

A =

dtdu

=1

dt

kCC

dCA =

+ 1 uk+1

77

Transient Reaction in CSTRdtdu

tu

=1 t

k

u

u)

1exp(

+=

tu

uu i

=1

0

tkCkC AA )

1exp(

)1( 0 +=+

uk i+1

k1 +

AAi 0

tAiAAA exp11 00

+=+

1

)exp()1(00

k

tkCCC

CAiAA

A+

+=

At long times0CA SS lst order in CSTR

78)1( k

A+


40/41

te.g. A is salt

AiAAA eCCCC = 00CAo

CA

CAi=0 at t=0

)1(

0

kC AA+=

CA With reaction

CAi=0 at t=0= , A= Ai=

t=, CA Pure solvent

79

dV

productsA

ccumu a on = n u + enera on

dNA =

dVrdVVFVFdC

dVdV

CVCd

dt

AA +=+=)(

( dVrdVVFVFdC

dV

dtdtdt

A += )(

rFC

t

A =

rC

uC AA =

80


41/41

-

dN1 dFVariable Density

rdtV

A= rAA = 0 rdVA=

VXFA = 0

dVdXF AA = 0

rA rA

Constant Densit

rdC

j

j= rCC jjjo = r

dCj

j=

81

THE END

single reactions in continuous isothermal reactions

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