Several problems have been covered in various lectures. A few of these with problem statements with solutions are below. 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) Pellet Radius
Thiele modulus and effectiveness factor for each
(mol/g cat s) x 105
(m)
pellet and (b) how 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)
−rA' ( obs ) R 2 ρc
De C As
Suppose
= ηφ12 = 3 (φ1 coth φ1 − 1) [1]
φ11 and φ12 are the Thiele Moduli at Run 1 and Run 2 with − rAʹ′1 and −rAʹ′ 2 being the
corresponding observed reaction rates, R1 and R2 being the corresponding radii. Using Eq. (1), we
obtain
−rA' 2 R22 φ12 coth φ12 − 1
=
−rA' 1 R12 φ11 coth φ11 − 1
[2]
Taking the ratio of the Thiele module for runs 1 and 2, we obtain
φ11
=
φ12
R1
R2
−rAs' ρc
DeC As
'
As
−r ρc
DeC As
=
R1
R
0.01m
⇒ φ11 = 1 φ12 =
φ12 = 10φ12
R2
R2
0.001m
[3]
Using Eqs. (2) & (3) and introducing the information in Table 1, we obtain
0.05 =
φ12 coth φ12 − 1
10φ12 coth (10φ12 ) − 1
[4]
φ12 = 1.65 and φ11 = 10φ12 = 16.5 . The corresponding effectiveness factors
obtained using Eq. (1) are η2 = 0.856;η1 = 0.182 Solving which gives
Part (b)
Suppose that operating at an effectiveness factor of 0.95 is sufficient to eliminate most of internal
diffusion resistance.
Using Eq. (1), that is, ηφ12 = 3 (φ1 coth φ1 − 1) ,
which η = 0.95 . Using Eq. (2), R3 = R1
φ13 = 0.9 , where subscript 3 refers to the radius R3 at
φ13
⎛ 0.9 ⎞
−4
= ( 0.01) ⎜
⎟ = 5.5*10 m = 0.55mm .
φ11
⎝ 16.5 ⎠
Lect. No.: 16
Problem : 16A
Time : 05:40
For the reaction C + CO2 → 2CO conducted in a catalytic reactor containing particles of radius
R = 0.7cm with bulk concentration being C As = 1.22 *10−5 mol / cm3 , the observed reaction rate is
−r ʹ′(obs ) ρc = 4.67 *10−9 mol / cm3 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.
The rate law, in concentration units is −rA =
kC A
where, CA is the concentration of CO2
1 + K 2CD + K 3C A
(species A) and CD is the concentration of CO at the surface. The constants
K 2 = 4.15 *109 cm3 / mol and K3 = 3.38 *105 cm3 / mol . k is the rate constant. Diffusivity of the
species in the catalyst is given by DeA = 0.1cm 2 / sec .
Solution
Weisz-Prater parameter (CWP) under the given conditions is
CWP
− rA' ( obs ) ρc R 2 4.67 *10−9 *0.7 2
=
=
= 1.88*10−3 << 1
−5
DeAC As
0.1*1.22*10
[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 CWP 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 = DeD and that concentration of CO at surface
CDs ≈ 0 , the rate expression can be rewritten as,
−rA' =
kCA
(1 + 2K2CAs ) + ( K3 − 2K2 ) CA
[2]
Assuming the pellet was infinitely long with C A,eq = 0 , the modified parameter
2
Φ = ηφ =
−rA' ( obs ) R 2 ρc ( −rAs' )
C As
2 ∫ DeA ( −rA' )dC A
0
−r ' ( obs ) R 2 ρc
= A
2 DeA
⎧⎪1 + K3C As
⎨
⎩⎪ K 3 − 2 K 2
⎡
⎡ 1 + K 3C As ⎤ ⎤ ⎫⎪
1 + 2 K 2C As
ln ⎢
⎢1 −
⎥ ⎥ ⎬
⎣ C As ( K 3 − 2 K 2 ) ⎣1 + 2 K 2C As ⎦ ⎦ ⎭⎪
= 2.5 > 1
So, as observed experimentally, there is a strong internal diffusion limitation.
−1
[3] 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 = 4.94atm . Assume DeA = 2.68 ×10−8 m 2 / sec ,
k ʹ′ʹ′ = 51m6 / m2 .mol.sec , ρb = 2.1×106 g / m3 , S a = 410m 2 / g , d p = 0.38cm . Assume rate law
2
− rAʹ′ʹ′ = k ʹ′ʹ′C Ab
Solution
The inlet concentration C Ab 0 =
P
4.94
=
= 0.113gmol / l
RT 0.082 × 533
Mole balance for the reactor is given by
DeA
d 2 CAb
dC
2
− U Ab − Ωk " Sa ρb CAb
=0
2
dz
dz
[1]
where Ω is the overall effectiveness factor. It should be noted that in general, for a second order
reaction explicit expression for Ω 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,
d 2 C Ab
dC Ab
, Eq (1) can be simplified to
DeA
<< U
2
dz
dz
dCAb
S ρ C2
− Ωk " a b Ab = 0
dz
U
[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 Ω 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 η and assumed constant. Under this approximation,
Eq. (2) can integrated to obtain the length required to achieve the desired conversion as
L=
U
⎛ 1
⎞
− 1⎟
⎜
"
Ωρb k Sa CAb 0 ⎝ 1 − X
⎠
[3]
Using the expression for φ2 for a second order reaction, the effectiveness factor
12
12
12
⎛ 2 ⎞ 3 ⎛ 2 ⎞ 3 ⎛ 2 ⎞
η = ⎜
= ⎜
= ⎜
⎟
⎟
⎟
⎝ n + 1 ⎠ φn ⎝ 2 + 1 ⎠ φ2 ⎝ 2 + 1 ⎠
3
= 9.47 ×10−8
7
2.59 ×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. η << 1 implies strongly internal diffusion limited, therefore
approximating Ω ≈ η = 9.47 ×10−8
L=
U
X
4
0.81
=
= 3.62 ×10−2 m
"
−8
6
Ωρb k Sa C Ab 0 1 − X 9.47 ×10 × 2.1×10 × 51× 410 × 0.113 (1 − 0.81)
Lect. No.: 22
Problem : 22A
Time : 13:25
Ref.: It is required to determine the value of kL and â for a batch absorber using the reaction
A( g ) + 2 B(l ) → C (l ) which is first order in A. kL and â are expected to be about 10-4 m/s and 200 m2/m3 respectively. Da =
2.5x10-9 m2/s. A choice of liquid phase reactants is available with different rate constants. Determine
what value of k will suit the purpose.
Solution:
Given:
kL = 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:
δ=
DA 2.5*10−9
≅
≅ 2.5*10−5 m −4
kL
1*10
δ . â = 200 * 2.5 * 10-5 = 5 * 10-3
by assuming slow reaction regime,
M≈
δ 2 k1
DA
2.52 *10−10 * k1
=
≈ 0.25k1 2.5*10−9
We know that,
(
)
P = M / (δ. â) = 0.25k1 / 5*10−3 = 50k1 Chosen a value of k1 ≈ 0.2sec −1 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
⎛ P ⎞
RA = k L aC A* ⎜
⎟ ⎝ P + 1 ⎠
⎛ 10 ⎞
= k L aC A* ⎜ ⎟ ≈ k L aC A* ⎝ 11 ⎠
Hence, the value of k1 ≈ 0.2sec −1 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 + ν B → P , which is first order in oxygen(A) is carried out in a stirred cell
with a flat gas-liquid interface of 132 cm2 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. (DAB = 2.1x10-5
cm2/s, concentration of B = 0.01 mol/cm3).
Solution:
Given:
âVL = 132 cm2
RAVL = 1.23*10−5 mol / sec CA* = H * pO2 = 5.8*10−7 mol / cm3
DA = 2.1*10−5 cm 2 / sec CBb = 0.01mol / cm3
To find the rate constant of the reaction
We consider fast reaction regime, for an given information which suggest that, kL various with RPM
leads to RAVL independent of RPM, kL and VL
So, rate reaction expressed as,
RAVL = DA k1 * C A* * âVL
1.23*10−7 = 2.1*10−5 * k1 *5.8*10−7 * 132
k1 = 1.229.13sec −1
Lect. No.: 23
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 cm3. 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: pO2 = 0.21atm
CA* = H * pO2 = 5.8*10−7 *0.21 = 1.218*10−7 mol / cm3
Total dispersion volume = 1700 cm3
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 =
DA k1
kL
=
(
M >3
)
2.1*10−5 *1229.13
4 *10−2
M = 4.02 > 3 The rate of absorption in fast reaction regime is,
RAVL = DA k1 * C A* * âVL
3.95*10−5 = 2.1*10−5 *1229.13 *1.218*10−7 * âVL
Total interfacial area (âVL) = 2018.56 cm2
Interfacial area per unit volume of dispersion (âVL) =
Lect. No.: 26
2018.56
= 1.19cm2 / cm3 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 CO2 is 1 bar, and the concentration of NaOH 0.5 kmol/m3.
Other data are as follows: kL = 10-4 m/s; interfacial area per unit volume of packed space is 100 m-1;
C A* = 0.04 kmol/m3; second order rate constant of the reaction k = 104 m3/kmol s, DA = 1.8 x 10-9 m2/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 → Na2 CO3 + H 2 O Solution:
Given:
CBb = 0.5kmol / m3 ; C A* = 0.04kmol / m3 ; k L = 10−4 m / sec k1 = 104 m3 / kmol sec ; DA = 1.8 *10−9 m 2 / sec ; DB = 3.06 *10−9 m 2 / sec DB DA = 1.7 (a) To find the maximum enhancement possible
We know that, q =
DB CBb 1.7 * 0.5
=
= 10.625
ν DACA* 2 * 0.04
Maximum enhancement factor,
E∞ ≅
DA
(1 + q ) = 1.7 *11.625 DB
E∞ ≅ 8.91 (b) To find actual enhancement
We know that,
M =
DA k1CBb
kL
1.8*10−9 *104 * 0.5
=
1*10−4
M = 30 > 10.625 ( = q ) Actual enhancement factor,
E= M
E∞ − E
E∞ − 1
⎛
E − E ⎞
tanh ⎜ M ∞
⎟⎟ ⎜
E
−
1
∞
⎝
⎠
First approximation:
⎛
⎜
⎝
For a larger value of M and E , tanh ⎜ M
∞
E≅ M
E∞ − E ⎞
⎟ ≅ 1 which lead to
E∞ − 1 ⎟⎠
E∞ − E ⎛ 30 ⎞
= ⎜
⎟ E∞ − E E∞ − 1 ⎝ 7.91 ⎠
E = 8.30 (by trial and error)
Second approximation:
⎛
⎛
E∞ − E ⎞
8.91 − 8.30 ⎞
tanh ⎜ M
= tanh ⎜⎜ 30
⎟
⎟⎟ ≅ 1 ⎜
⎟
E
−
1
8.91
−
1
∞
⎝
⎠
⎝
⎠
E= M
E∞ − E
E∞ − 1
⎛
E − E ⎞
tanh ⎜ M ∞
⎟ ≅ 8.30
⎜
E∞ − 1 ⎟⎠
⎝
(c) To find actual absorption rate, in units of kmol per sec per unit volume of packed space
The Rate of absorption is
RA = k L C A* E = 1*10−4 * 0.04 *8.3 = 3.32 *10−5 kmol / m 2 sec RA a = 3.32*10−3 kmol / m3 sec
Lect. No.: 12
Ref.: Scott Fogler, pg.: 858
Problem : 12A
Time : 38:35
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/dm3, the diameter
(2R) is 2 x 10-3 cm, and the diffusion coefficient is 0.1cm2/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/dm3; dp = 2 x 10-3 cm; De = 0.1 cm2/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,
ψ=
CA
1 ⎛ sinh φ1λ ⎞
= ⎜
⎟ -­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐-­‐> (1) C AS λ ⎝ sinh φ1 ⎠
0.1 =
1 ⎛ sinh φ1 0.5 ⎞
⎜
⎟ ⇒ φ1 = 6 (by trial & error method)
0.5 ⎝ sinh φ1 ⎠
Dimensionless radius of the catalyst expressed in the form of λ=
r R − 3*10−4 1*10−3 − 3*10−4
=
=
= 0.7 R
R
1*10−3
Substituting value of λ and φ1 in eq.(1), we get
ψ=
CA
CA
1 ⎛ sinh φ1λ ⎞
1 ⎛ sinh(6 * 0.7) ⎞
= ⎜
=
⎟ =
⎜
⎟ CAS λ ⎝ sinh φ1 ⎠ 0.001 0.7 ⎝ sinh 6 ⎠
C A = 2.36 *10−4 mol / dm3 Lect. No.: 13
Ref.: Scott Fogler, pg.: 858
Problem : 12A (Cont.)
Time : 00:00
(b) To what diameter should the pellet be reduced if the effectiveness factor is to be 0.8
The Thiele modulus is,
φ=R
k1 ρc sa
k
= R 1r De
De
k1r
⇒ k1r = 3600000sec−1 0.1
6 = 1*10−3
Calculating Thiele modulus for an effectiveness factor 0.8 is
η = 0.8 =
3
φ12
[φ1 coth φ1 − 1] ⇒ φ1 = 2 The corresponding Thiele modulus expression to calculate diameter of the catalyst particle is,
φ =2=R
k1r
3600000
=R
⇒ R = 3.4 *10−4 cm De
0.1
d p = 6.8 *10−4 cm Lect. No.: 38
Ref.: Scott Fogler, pg.: 971
Problem : 38A
Time : 28:40
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
9
C(t)
0
1
5
8
10
8
6
4
3
2.2
(a) To calculate conversion using the closed vessel dispersion model
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 tm and σ2
time
C(t)
E(t)
tE(t)
t2E(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 tm, we first find the area under the C curve, which is
∞
∫ C (t ) dt = 50 g min 0
∞
τ = tm = ∫ tE ( t ) dt = 5.15 min Then
0
Using Simpson rule, we find,
∞
⎛ 1 ⎞
∫ t E (t ) dt = ⎜⎝ 3 ⎟⎠ ⎡⎣( 0.0 + 3.0 ) + 2 ( 0.4 + 3.2 + 4.32 + 3.84 ) + 4 (0.02 + 1.44 + 4.0 + 3.92 + 3.6 )⎤⎦
2
0
⎛ 2 ⎞
⎝ 3 ⎠
+ ⎜ ⎟ ⎡⎣( 3.0 + 0.0 ) + 4 (1.68 )⎤⎦ = 32.63min 2 To obtain the variance, we substituting these values
∞
∞
2
2
σ = ∫ ( t − τ ) E ( t ) dt = ∫ t 2 E ( t ) dt − τ 2 0
0
2
σ 2 = 32.63 − (5.15) = 6.10min 2 Dispersion in a closed vessel is represented by
σ2
2
= 2 ( Pe − 1 + exp ( − Pe ) ) 2
τ
Pe
=
6.1
(5.15)
2
= 0.23 =
2
( Pe − 1 + exp ( − Pe )) Pe 2
Solving for Pe by trial and error, we obtained Pe = 7.5 (
)
Next we need to calculate Da, Da = τ k = (5.15min ) 0.25min −1 = 1.29 Using the equation for q and X gives
q = 1+
4 (1.29 )
4 Da
= 1+
= 1.30 Pe
7.5
Then, substitute q and Pe value in conversion expressed for a dispersion model
X = 1−
X = 1−
4q exp ( Pe 2 )
(1 + q )
2
2
exp ( qPe 2 ) − (1 − q ) exp ( − qPe 2 )
4 (1.30 ) exp ( 7.5 2 )
(1 + 1.30 )
2
2
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 ( −τ k ) = 1 − exp ( − Da ) = 1 − exp ( −1.29 ) X = 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
τ 2 ( 5.15)
n= 2 =
= 4.35 6.1
σ
To calculate the conversion for first-order for n tanks in series is
X = 1−
1
(1 + τ i k )
n
= 1−
1
(1 + (τ n ) k )
n
= 1−
1
(1 + (5.15 / 4.35) 0.25)
4.35
X = 0.677
67.7% conversion achieved for the tanks-in-series model
(d) Conversion for CSTR:
For a single CSTR,
X =
τk
1.29
=
1 + τ k 2.29
X = 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:
k e = 1.6kJ / m / hr / K De = 5x10 −5 m 2 / hr hT = 160kJ / hr / m2 / K k m = 300m / hr ΔH = −160kJ / molA C Ab = 20mol / m3 robs = −105 mol / m3 / hr L = 4x10 −4 m Answer the following questions: Is external mass transfer important to consider? Are there significant limitations due to pore diffusion? Do we expect significant temperature gradients within the pellet & outside? 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
As a first approximation one may neglect pressure drop. The entering concentration of A is 1.0M. Calculate the catalyst weight necessary to achieve 60% conversion? Kinematic viscosity: 0.02 cm2/sec; Particle diameter: 0.1 cm Superficial velocity 10 cm/s; Catalyst surface area /mass of the catalyst bed: 60 cm2/g. cat Diffusivity of A 10-­‐2 cm2/sec. Specific rate constant (k) is 0.01 cm3/sec g cat with E= 4000 cal/mol SOLUTION 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.
(b) The above reaction is performed in a fluidized bed reactor which received the feed at 100 kmol/hr and a conversion of 10% is realized. Predict the conversion if the original particle radius of 1.8cm of the same catalyst is reduced by half under otherwise similar conditions. Fluidized bed reactor can be considered to be a perfectly back-­‐mixed reactor for all practical purposes. SOLUTION: 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
2m/s, a temperature of 2000C 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.
Data given: The intrinsic reaction rate constant calculated by performing experiments with very small particle size of the same catalyst is 0.0003 m3/g cat. sec. Effective diffusivity: 1.5 x 10-­‐8 m2/s Catalyst density: 2 gm/cm3 SOLUTION: