!"#"$%& ($)*&"+, -%#" *"". /)#"$"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 7?@512>-2 25 A5-.7B01C ,10 23010 .78-747A>-2 67?72>275-. B:0 25 @510 B744:.75-C D5 /0 0=@0A2 .78-747A>-2 20?@01>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 B15@G 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 %
R@0A747A 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 0=@017?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:
%