!"#"$%& ($)*&"+, -%#" *"". /)#"$"0 1. #%$1)2, &"/32$",4 5 6"7 )6 3-"," 713- ($)*&"+ ,3%3"+".3, 713- ,)&231)., %$" *"&)74

Lect. No.: 15

Problem : 15A

Time : 12:26

The first-order reaction A B was carried out and the following experimental data were obtained (Table 1). All other conditions for these experiments were same. Assuming negligible Table 1: Experimental data external mass transfer resistance, (a) estimate the Measured Rate (obs) ( obs) Pellet Radius Thiele modulus and effectiveness factor for each (mol/g cat s) x 10 5 (m) pellet and (b) how ho w small should the pellets be made Run 1 3.0 0.01 to eliminate nearly all internal diffusion resistance? Run 2 15.0 0.001 !

Solution: Part (a)

$r A' ( obs ) R2 ! c

2

=

De C As

"#1

=

3

(#

1

coth #1

$ 1)

[1]

Suppose ! 11 and ! 12 are the Thiele Moduli at Run 1 and Run 2 with

" r !

A1

and

"r !

A 2

being the

corresponding observed reaction rates, R rates, R1 and R and R2 being the corresponding radii. Using Eq. (1), we obtain

"r ' 2 R22 "r ' 1 R12 A

=

A

!12 coth ! 12 " 1 !11 coth ! 11 " 1

[2]

Taking the ratio of the Thiele module for runs 1 and 2, we obtain '

# r

R1

" 11

! c

As

DeC As

=

" 12

R1

$

=

'

# r

As

R2

R2

! c

"11

R1 =

R2

0.01m

"12

=

0.001m

"12

=

10"1 2

[3]

DeC As

Using Eqs. (2) & (3) and introducing the information in Table 1, we obtain 0.05

=

!12 coth ! 12 " 1

) "1

(

10!12 coth 10! 12

[4]

Solving which gives ! 12 1.65 and !11 =

obtained using Eq. (1) are !2

=

=

0.856;! 1

10! 12 =

=

16.5 . The corresponding effectiveness factors

0.182

Part (b)

Suppose that operating at an effectiveness factor of 0.95 is sufficient to eliminate most of internal diffusion resistance. 2

Using Eq. (1), that is, !"1 which !

=

=

3

(" coth " # 1) , ! 1

0.95 . Using Eq. (2), R3

13

1

=

R1

! 13 =

! 11

=

radius R3 at 0.9 , where subscript 3 refers to the radius R

# 0.9 $ & ' 16.5 (

(0.01) %

=

5.5 *10

"4

m

=

0.55mm .

Lect. No.: 16

Problem : 16A

For the reaction C + CO2 R

=

!

2CO conducted in a catalytic reactor containing particles of radius

concentration being C As 0.7cm with bulk concentration

"r #

(obs ) ! c

=

4.67 *10

"

9

Time : 05:40

mol / cm

3

=

1.22 * 10

!5

3

mol / cm , the observed reaction rate is

sec . After the reaction was conducted, the particles were cut

open and the reacted carbon profiles were measured. These profiles suggested strong diffusional effects to be present. Verify this observation.

!r A =

The rate law, in concentration units is

kC A 1 + K 2C D

+

K 3C A

where, C A is the concentration of CO 2

(species A) and C D is the concentration of CO at the surface. The constants K 2

9

=

3

5

4.15 * *1 10 cm / mol and K 3

=

species in the catalyst is given by DeA

3

is the rate constant. Diffusivity of the 3.38 * 10 cm / mol . k is 2

=

0.1cm / sec .

Solution

Weisz-Prater parameter (C ( C WP WP ) under the given conditions is C WP

=

2 obs ) ! c R "r A' ( ob

DeAC As

4.67*1 .67*10 0 =

"9

2

*0.7 *0.7 "5

0.1*1.22*10

=

1.88 *10

"3

<<

1

[1]

indicating no internal diffusion limitations present. However the experimental observations suggest otherwise. Poor prediction by the Weisz-Prater method is due to the fact that C WP WP in Eq. (1) uses Thiele modulus expression for a first order reaction when the actual reaction is not first-order. Therefore, this problem warrants the use of Generalized Thiele Modulus. Assuming equimolar counter diffusion i.e.; DeA C Ds

!

=

DeD and that concentration of CO at surface

0 , the rate expression can be rewritten as,

!r A' =

kC A

(1

+

2 K2C As

+

3

2

A

Assuming the pellet was infinitely long with C A,eq 2

% = "#

=

"#$

) ( K ! 2 K ) C

=

0 , the modified parameter

$r A' ( obs ) R2 ! c ( $rAs' ) C As

3 D ( $r )dC '

2

eA

A

A

0

$r A' ( obs ) R 2 ! c &*1 + K3CAs ( ( 1 + K3CAs ) ) '* 1 + 2 K 2 CAs ln = + -1 $ .., $ $ 2 DeA 2 2 1 + 2 K C K K C K K ( ) 3 2 A s 3 2 2 A s / 0 0 2* / 1* =

2.5 > 1

So, as observed experimentally, there is a strong internal diffusion limitation.

$1

"%$

Lect. No.: 18

Problem : 18A

Time : 00:00

Design a packed bed reactor in which the reaction A B + 2C is being conducted under internal diffusional limiting conditions and the exit conversion is 0.81. The fluid is being pumped into the reactor at a superficial velocity of U 4m / sec . The reaction is being conducted at temperature !

=

T = 260°C = 533K and at inlet pressure of P

k !!

6

=

" r A!!

2

, ! b 51m / m .mol.sec

=

=

2.1"10

6

=

4.94atm . Assume DeA

g / m3 , S a

2

=

410m /

g , d p

=

=

2.68 "10

8

!

2

m / sec ,

0.38cm . Assume rate law

2

k !!C Ab

Solution

P

The inlet concentration C Ab 0

=

RT

4.94 =

0.082 ! 533

=

0.113 gmol / l

Mole balance for the reactor is given by 2

DeA

d C Ab 2

dz

dCAb

"U

dz

"

2

" #k Sa ! b C Ab

=

0

[1]

where ! is the overall effectiveness factor. It should be noted that in general, for a second order reaction explicit expression for ! is usually not available and will be a function of the local concentration of species A and as a result will be a function of position as well. Assuming the flow rate through the bed is very large and the axial diffusion can be neglected, that is, 2

DeA

d C Ab 2

dz

dC Ab dz

U

<<

dCAb dz

, Eq (1) can be simplified to

2

"

" #k

Sa ! b CAb =

U

0

[2]

along with the condition at the entrance of the reactor C Ab

=

C Ab 0 @ z

=

0. Analytical solution for

Eq. (2) is usually unavailable due to the dependence of the overall effectiveness factor ! whose explicit dependence on the concentration is a priori unknown. However, the reaction under the specified conditions is internal diffusion controlling. In this regime, the overall effectiveness factor may be approximated to the effectiveness factor ! and assumed constant. Under this approximation, Eq. (2) can integrated to obtain the length required to achieve the desired conversion as L

U =

' ! b k

"

Sa C Ab 0

" 1 # % $ $ 1& ( 1 X )

[3]

Using the expression for ! 2 for a second order reaction, the effectiveness factor 12

$ 2 % 3 ! = ' ( ) n +1* "

n

12

$ 2 % 3 = ' + ( ) 2 1 * " 2

12

$ 2 % = ' + ( ) 2 1*

3 7

2.59 & 10

=

#8

9.47 & 10

Note that the Thiele Modulus will be a function of position. For the chosen parameters, as the variation with respect to position is negligible, the Thiele Modulus is evaluated at the inlet concentration and is assumed constant. ! << 1 implies strongly internal diffusion limited, therefore approximating # $ !

=

"8

9.47 %10

L

U =

"

X

$ ! b k Sa C Ab 0

1" X

4 =

"8

0.81

6

9.47 # 10

# 2.1# 10 # 51#

Lect. No.: 22

410# 0.113 (1" 0.81)

Problem : 22A

=

3.62# 10"2 m

Time : 13:25

Ref.: -

It is required to determine the value of k L and â for a batch absorber using the reaction

A( g ) + 2B (l )

!

C (l )

which is first order in A. k L and â are expected to be about 10 -4 m/s and 200 m 2/m3 respectively. Da = 2.5x10-9 m 2/s. A choice of liquid phase reactants is available with different rate constants. Determine what value of k will suit the purpose. Solution: Given:

k L = 1 x 10 -4 m/s; â = 200 m2/m3; Da = 2.5x10 -9 m2/s To find k value at which given condition will satisfies Thickness of the film: !

D A =

k L

2.5*10 #

1*10

9

"

2.5*10

#

4

"

5

"

m

! . â = 200 * 2.5 * 10 -5 = 5 * 10 -3 by assuming slow reaction regime, 2

2

! k1

M

#

D A

2.5 *10 =

2.5*10

P = M / ( !. â)

We know that,

10

=

(0.25k

1

* k 1

"

#

9

"

!3

/ 5 *10

0.25k 1

)

=

50k 1

1

Chosen a value of k 1 0.2sec gives M = 0.05 and P=10, which satisfies the conditions such !

"

as M << 1 and P >> 1 , also In general, rate of mass transfer for slow reaction regime is R A

=

=

! P " # $ % P + 1 & * ! 10 " * k L aC A $ % # k L aC A & 11 ' *

k L aC A

1

Hence, the value of k 1 0.2sec satisfies the condition for slow reaction regime. "

Lect. No.: 23

!

Problem : 23A

Time : 06:20

Ref.: Part:1 Rate constant of an unknown reaction An oxidation reaction A + ! B " P , which is first order in oxygen(A) is carried out in a stirred cell with a flat gas-liquid interface of 132 cm 2 at atmospheric pressure with pure oxygen. Over a stirrer speed range of 60-200 RPM, the rate of absorption was measured to be nearly constant at 1.23x10 -5 mol/s, as measured by the difference in the flow rates of gas at inlet and outlet; it was also

independent of the volume of liquid in the vessel. The solubility of A in the liquid phase follows Henry’s law with H = 5.8x10 -7 mol/cm3/atm. Find the rate constant of the reaction. (D AB = 2.1x10-5 cm2/s, concentration of B = 0.01 mol/cm 3). Solution: Given: âV L = 132 cm2 R AVL

C A*

=

D A

C Bb

=

1.23 *10

H * pO

=

!5

mol / sec !7

5.8 *10

2

=

2.1*10

=

!5

mol / cm3

2

cm / sec

0.01mol / cm

3

To find the rate constant of the reaction We consider fast reaction regime, for an given information which suggest that, k L various with RPM leads to R AVL independent of RPM, k L and V L So, rate reaction expressed as, R AVL

!7

1.23 *10

=

*

DA k1 * C A * âV L

!5

!7

* k 1 * 5.8 *10

2.1*10

=

k 1 1.229.13sec =

Lect. No.: 23

* 132

!1

Problem : 23B

Time : 19:30

Ref.: Part:2 Interfacial area by the chemical method. The same reaction is now conducted in an agitated, bubbling stirred tank, with air instead of oxygen. From a measurement of the oxygen content in the gas leaving, a rate of absorption of 3.95x10 -5 mol/s was determined with a total dispersion volume of 1700 cm 3. Determine the specific interfacial area per unit volume of the dispersion. Mass transfer co-efficient in such equipment usually varies in the range of 2-4x10-2 cm/s.

Solution: Given:

Assume: pO

=

2

C A*

=

H * pO

0.21atm

=

!7

5.8 *10

* 0.21

2

=

1.218 *10

!7

3

mol / cm

Total dispersion volume = 1700 cm 3 k L

=

2 ! 4 *10

!2

cm / sec

To find the specific interfacial area per unit volume of the dispersion Assumed k L

=

4 *10

!2

cm / sec for an fast reaction regime

We know that,

M

!5

D A k 1 =

k L

(

2.1*10

M

>

3

)

*1229.13

=

4*10

!2

M

=

4.02 > 3

The rate of absorption in fast reaction regime is, R AVL

3.95*10

!5

*

DA k1 * C A * âV L

=

2.1*10

=

!5

*1229.13 *1.218*10

!7

* âV L

Total interfacial area (âV L ) = 2018.56 cm 2

Interfacial area per unit volume of dispersion ( âV L )

Lect. No.: 26

2018.56 =

2

=

3

1.19cm / cm

1700

Problem : 26A

Time : 24:10

Ref.: Maximum and actual enhancement factors CO2 is being absorbed from a gas into a solution of NaOH at 20ºC, in a packed tower. At a certain point in the tower, the partial pressure of CO 2 is 1 bar, and the concentration of NaOH 0.5 kmol/m 3. Other data are as follows: k L = 10-4 m/s; interfacial area per unit volume of packed space is 100 m -1; *

3

4

3

-9

2

C A = 0.04 kmol/m ; second order rate constant of the reaction k = 10 m /kmol s, D A = 1.8 x 10 m /s

and DB = 3.06x10-9m2/s. Find the maximum enhancement possible and the actual enhancement. Find also the actual absorption rate, in units of kmol per sec per unit volume of packed space. The reaction is: CO2

+

2 NaOH

!

Na2CO3

H2O

+

Solution: Given: 3

*

0.5kmol / m & C A

3

C Bb

=

k1

10 m / kmol sec & D A

4

=

D B DA

=

0.04kmol / m & k L

3

=

1.8 *10

=

!9

=

10

!4

m / sec

2

m / sec & D B

=

3.06 *10

!9

1.7

(a) To find the maximum enhancement possible

We know that, q

=

D B C Bb D AC A*

1.7*0.5 =

!

=

10.625

2*0.04

Maximum enhancement factor,

E

!

"

D A D B

(1 E

+

!

q ) = 1.7 *11.625

"

8.91

(b) To find actual enhancement

We know that,

M

D A k1C Bb =

1.8*10

!9

=

k L

4

*10 *0.5

1*10

!4

2

m / sec

M

=

(

30 > 10.625

=

q)

Actual enhancement factor, E

M

=

" % '

$E E! $ 1

E!

tanh %

M

$ E # & E ! $ 1 & (

E!

First approxim ation:

" & (

For a larger value of M and E , tanh & M "

E

%

" E E ! " 1

E!

M

E

=

=

$ E # ' % 1 which lead to E ! $ 1 ' )

E!

# & (

$ ' 7.91 ) 30

" E

E!

8.30 (by trial and error)

Second approximation:

" & (

tanh & M

E

=

M

$ E # ' E ! $ 1 ' )

E!

$E E! $ 1

E!

=

" & (

tanh & 30

" & (

tanh &

M

8.91 $ 8.30 8.91 $ 1

# '' % 1 )

$ E # ' % 8.30 E ! $ 1 ' )

E!

(c) To find actual absorption rate, in units of kmol per sec per unit volume of packed space The Rate of absorption is R A

*

=

kL CA E

=

1*10

R A a

Lect. No.: 12

!4

=

* 0.04 *8.3

3.32 *10

!3

=

3.32 *10

!5

2

kmol / m sec

3

kmol / m sec

Problem : 12A

Time : 38:35

Ref.: Scott Fogler, pg.: 858

A first-order heterogeneous irreversible reaction is taking place within a spherical catalyst pellet which is plated with platinum throughout the pellet. The reactant concentration halfway between the external surface and the centre of the pellet (i.e., r = R/2) is equal to one-tenth the concentration of pellet’s external surface. The concentration at the external surface is 0.001 g mol/dm 3, the diameter (2R) is 2 x 10-3 cm, and the diffusion coefficient is 0.1cm 2/s.

A

!

B

(a) What is the concentration of reactant at a distance of 3 x 10 -4 cm in from the external pellet surface? (b) To what diameter should the pellet be reduced if the effectiveness factor is to be 0.8?

Solution:

Given: CA / CAS = 0.1; CAS = 0.001 g mol/dm 3; d p = 2 x 10 -3 cm; D e = 0.1 cm 2/s; (a) To find the concentration of reactant at a distance of 3 x 10 -4 cm in from the external pellet surface We know that,

#

C A =

C AS

0.1

$ sinh !1" % & ' " ( sinh ! 1 ) 1

=

'''''''''''( )*+

" sinh ! 1 0.5 # % & $ ! 1 0.5 ' sinh ! 1 ( 1

=

=

6 (by trial & error method)

Dimensionless radius of the catalyst expressed in the form of !

R " 3 *10

r =

=

R

"4

"3

1*10 =

" 3 *10"4

1*10

R

"3

=

0.7

Substituting value of # and $1 in eq.(1), we get

#

C A =

=

C AS

C A

=

1 $ sinh !1" %

& ' " ( sinh ! 1 )

2.36 *10

Lect. No.: 13

!4

C A =

mol / dm

1 $ sinh(6 * 0.7) % & ' 0.7 ( sinh 6 )

=

0.001

3

Problem : 12A (Cont.)

Time : 00:00

Ref.: Scott Fogler, pg.: 858

(b) To what diameter should the pellet be reduced if the effectiveness factor is to be 0.8 The Thiele modulus is, k1 ! c sa

" R =

6

=

De

1*10

k 1r

3

!

R

=

0.1

k 1r De

"

k 1r

=

3600000sec

1

!

Calculating Thiele modulus for an effectiveness factor 0.8 is

!

=

0.8

3 =

2

" 1

["

1

coth "1

#

]

1

$

"1

=

2

The corresponding Thiele modulus expression to calculate diameter of the catalyst particle is,

!

=

d p

2

=

=

R

k 1r De

6.8*10

!4

=

R

cm

3600000 0.1

#

R

=

3.4 *10

4

"

cm

Lect. No.: 38

Problem : 38A

Time : 28:40

Ref.: Scott Fogler, pg.: 971 Conversion using Dispersion and Tank-in-Series Models:

The first-order reaction

A

B

!

is carried out in a 10 cm diameter tubular reactor 6.36 m in length. The specific reaction rate is 0.25 min-1. The results of a tracer test carried out on this reactor are shown in Table T38A-1. Table T38A-1. Effluent tracer concentration as a function of time time(min) C (mg/L)

0 0

1 1

2 5

3 8

4 10

5 8

6 6

7 4

8 3

9 2.2

10 1.5

12 0.6

14 0

Calculate conversion using (a) the closed vessel dispersion model, (b) PFR, (C) the tank-in-series model, and (d) a single CSTR. Solution:

Given: d = 10 cm, k = 0.25 min -1 time 0 1 2 3 4 5 6 7 8 C(t) 0 1 5 8 10 8 6 4 3 (a) To calculate conversion using the closed vessel dispersion model

9 2.2

10 1.5

12 0.6

10 1.5 0.03 0.3 3.0

12 0.6 0.012 0.14 1.68

14 0

Table T38A-2. Calculation to determine t m and ! 2 time C(t) E(t) tE(t) t E(t)

0 0 0 0 0

1 1 0.02 0.02 0.02

2 5 0.1 0.2 0.4

3 8 0.16 0.48 1.44

4 10 0.2 0.8 3.2

5 8 0.16 0.80 4.0

6 6 0.12 0.72 4.32

7 4 0.08 0.56 3.92

8 3 0.06 0.48 3.84

9 2.2 0.044 0.40 3.60

14 0 0 0 0

To find E(t) and then t m, we first find the area under the C curve, which is !

" C (t ) dt

=

50 g min

0

"

Then

!

=

tm

# tE (t ) dt

=

=

5.15min

0

Using Simpson rule, we find, !

, t

2

()

E t dt =

0

"1# & ' $(( 0.0 + 3.0 ) + 2 (0.4 + 3.2 + 4.32 + 3.84 ) + 4 (0.02 + 1.44 + 4.0 + 3.92 + 3.6 )%) * 3+

+

=

!2" % 3 & #'( 3.0 + 0.0 ) + 4 (1.68 )$( ) *

32.63min

2

To obtain the variance, we substituting these values

# !

# 2

% (t $ ) E (t ) dt % t E (t ) dt $

2

2

"

=

=

0

0

2

!

2

"

=

(

2

)

32.63 " 5.15

2

=

6.10 min

Dispersion in a closed vessel is represented by 2 ! =

2 "

2

( Pe # 1

2

Pe

=

6.1

( 5.15 )

2

Solving for Pe by trial and error, we obtained Pe Next we need to calculate Da, Da

=

! k

+

exp ( # Pe) )

=

0.23 =

=

7.5

2 Pe

2

( Pe ! 1

(5.15min ) ( 0.25min"

1

=

)

=

+

exp ( ! Pe ) )

1.29

Using the equation for q and X gives

q

=

1+

4 Da

Pe

(

4 1.29 1+

=

)

7.5

1.30

=

Then, substitute q and Pe value in conversion expressed for a dispersion model

X = 1 !

X

=

1!

4q exp ( Pe 2 )

(1

+

2

2

q ) exp ( qPe 2) ! (1 ! q ) exp ( ! qPe 2)

4 (1.30) exp (7.5 2 )

(1

+

2

2

1.30 ) exp ((1.30 * 7.2 ) 2 ) ! (1 !1.30 ) exp ( ! (1.30 * 7.2 ) 2 ) X

=

0.68

When dispersion effects are present in this tubular reactor, 68% conversion is achieved. (b) Conversion for Plug flow reactor: If the reactor were operating ideally as a plug-flow reactor, the conversion would be X

=

1 " exp ( "! k ) 1 " exp ( " Da) =

X

=

=

1 " exp ( "1.29)

0.725

72.5% conversion would be achieved in an ideal plug-flow reactor. (c) Conversion for tank-in-series: First calculate the number of tanks in series,

2

(5.15 )

2

! n

=

=

2

=

4.35

6.1

"

To calculate the conversion for first-order for n tanks in series is X

=1

1

"

(1

)

+ ! k i

n

=

1"

1

(1 (

+ !

n

) )

n k

X

=

1"

1 4.35

(1 (5.15 / 4.35) 0.25) +

0.677

=

67.7% conversion achieved for the tanks-in-series model (d) Conversion for CSTR: For a single CSTR, X

X

! k =

=

1 + ! k

1.29 =

2.29

0.563

56.3% conversion achieved for the single CSTR.

ADDITIONAL PROBLEMS WITH SOLUTIONS

1. Consider the first order decomposition of A. The following data is given: L

=

hT

4 x10

!4

k e

m 2

=

C Ab

160kJ / hr / m / K

=

20mol / m

3

=

k m r obs

1.6kJ / m / hr / K

=

300m / hr

=

De " H

=

=

!5 2 5 x10 m / hr !160kJ / molA

!105 mol / m3 / hr

,-./01 230 45665/7-8 9:0.275-.; <. 0=201->6 ?>.. 21>-.401 [email protected]>-2 25 A5-.7B01C ,10 23010 .78-747A>-2 67?72>275-. B:0 25 @510 B744:.75-C D5 /0 [email protected] .78-747A>-2 [email protected]>2:10 81>B70-2. /7237- 230 @06602 E 5:2.7B0C SOLUTION:

2. The irreversible gas-phase reaction A B is carried out isothermally over a packed bed of solid catalyst particles. The reaction is first order in the concentration of A on the catalyst surface. The feed consists of 50% (mole) A and 50% inerts and enters the bed at a temperature 300 K. The entering volumetric flow rate is 10 lit/sec The relation between Sh and Re is Sh= 100 (Re) 0.5 ,. > 471.2 >@@15=7?>275- 5-0 ?>F -0860A2 @10..:10 [email protected] H30 0-2017-8 A5-A0-21>275- 54 *GIJG K>6A:6>20 230 A>2>6F.2 /07832 -0A0..>1F 25 >A370L0 MIN A5-L01.75-C

! 7.

#

O7-0?>27A L7.A5.72F; IGI# A? P.0A& Q>127A60 B7>?0201; IG* A? #

R:@0147A7>6 L065A72F *I A?P.& K>2>6F.2 .:14>A0 >10> P?>.. 54 230 A>2>6F.2 S0B; MI A? P8G A>2 '#

#

D744:.7L72F 54 , *I A? P.0AG %

[email protected] 1>20 A5-.2>-2 )T+ 7. IGI* A? P.0A 8 A>2 /723 UV WIII A>6P?56 !89:;<8=

3. (a) Following is the observed reaction rate in an isothermal reactor as a function of particle size for an elementary first order liquid phase reaction. The bulk concentration (1 mol/lit) is same in each case. Find the approximate value of effective intra-particle diffusivity. Catalyst density is 1 gm/cc.

)S+ H30 >S5L0 10>A275- 7. @01451?0B 7- > 46:7B7X0B S0B 10>A251 /37A3 10A07L0B 230 400B >2 *II T?56P31 >-B > A5-L01.75- 54 *IN 7. 10>67X0BG Q10B7A2 230 A5-L01.75- 74 230 51787->6 @>127A60 1>B7:. 54 *GYA? 54 230 .>?0 A>2>6F.2 7. 10B:A0B SF 3>64 :-B01 52301/7.0 .7?76>1 A5-B7275-. G Z6:7B7X0B S0B 10>A251 A>- S0 A5-.7B010B 25 S0 > @0140A26F S>AT'?7=0B 10>A251 451 >66 @1>A27A>6 @:[email protected] !89:;<8=>

4. A first order irreversible cracking reaction A = B is performed in a fixed bed reactor on a catalyst particle size of 0.15 cm. Pure A enters the reactor at a superficial velocity of 0 2m/s, a temperature of 200 C and pressure of 1 atm. Under these conditions, the reaction is severely affected by internal diffusion effects. Calculate the length of bed necessary to achieve 60% conversion. D>2> 87L0-;

H30 7-217-.7A 10>A275- 1>20 A5-.2>-2 A>6A:6>20B SF @01451?7-8 [email protected]?0-2. /723 L01F .?>66 % @>127A60 .7X0 54 230 .>?0 A>2>6F.2 7. IGIII% ? P8 A>2G .0AG 'Y

#

U440A27L0 B744:.7L72F; *G[ = *I ? P. K>2>6F.2 B0-.72F; # 8?PA? SOLUTION:

%

Our partners will collect data and use cookies for ad personalization and measurement. Learn how we and our ad partner Google, collect and use data. Agree & close