Transcript Diapositive 1

Chemical Reactor Analysis and Design
3th Edition
G.F. Froment, K.B. Bischoff†, J. De Wilde
Chapter 3
Transport Processes with Reactions
Catalyzed by Solids
Part one Interfacial Gradient Effects
Introduction
1. Transport of reactants A, B, ... from the
main stream to the catalyst pellet surface.
2. Transport of reactants in the catalyst pores.
3. Adsorption of reactants on the catalytic site.
4. Chemical reaction between adsorbed
atoms or molecules.
5. Desorption of products R, S, ....
6. Transport of the products in the catalyst
pores back to the particle surface.
7. Transport of products from the particle
surface back to the main fluid stream.
Steps 1, 3, 4, 5, and 7: strictly consecutive processes
Steps 2 and 6: cannot be entirely separated !
Chapter 2: considers steps 3, 4, and 5
Chapter 3: other steps
Reaction of a component of a fluid at the surface of a solid
Define reaction rate based on interfacial surface area:
For first-order rate: rAi  k r C Ai
with:
rAi = rate of reaction of A at surface, kmol/ m2p s
k r = rate coefficient for the reaction, m3f / m 2p s
C Ai = concentration of A at the interface, kmol/ m3f
Consumption of A at the interface: to be compensated for
by transport from the bulk fluid
N A  k g C A  C Ai 
with:
NA = mass flux with respect to the fixed solid surface, kmol/ m2p s
k g = mass transfer coefficient, m3f / m 2p s
CA = concentration of A in bulk stream, kmol/ m3f
Steady state
rA
rAi = N A = ~
Eliminate unmeasured
surface concentration CAi
Reaction of a component of a fluid at the surface of a solid
Steady state
C Ai 
kg
kr  k g
CA
1
1
 1

1
rA     C A
k

k
r 
 g
~r   1  1  C  k C
A
A
0 A
k

k
r 
 g
1
1
1
with k0 « overall » rate


k 0 Ccoefficient:
A
k0 k g kr
(for first order rate: add resistances
two consecutive linear processes)
Two limiting cases:
1) kg >> kr
« reaction controlling »
2) kr >> kg
« diffusion controlling »
k0  kr
CAi  CA
~
rA  k r C A
k0  kg
CAi  0
~r  k C
A
g A
Reaction of a component of a fluid at the surface of a solid
2
Same procedure: second order reaction: rAi  k r C Ai
2


k
k




~r  k  1   1  g  1   1  g


  1C A
A
g 
  2  kr C A 
  2  k r C A 

• Totally different form
• Concentration dependence neither first or second order
Two limiting cases:
1) kg >> kr
~
rA  k r C A2
(second order)
~
rA  k g C A
(first order)
« reaction controlling »
2) kr >> kg
« diffusion controlling »
~


r
A
~
For nth order reaction: rA  k r  C A  
kg 

n
~
Explicit solution rA
not possible
Iterative
Reaction of a component of a fluid at the surface of a solid
m
2
p

/ kg cat.
rA = am∙r~A
kmol /(kg cat. s)
kmol /(m s)
2
p
Specific experiments for determining:
• Intrinsic reaction kinetics
• Conditions such that global rate determined by reaction
 Lower T
 Increased turbulence
• Mass transfer coefficients
• Conditions such that global rate determined by mass transfer
 Higher T
+ Isothermal conditions (no heat transfer rate eq.)
Mass transfer resistance: Multicomponent diffusion in a fluid
General: flux given chemical species driven by:
• own concentration gradient
• concentration gradients all other species
N 1
N
N j   Ct D jk y k  y j  N k
k 1
j = 1, 2, …, N-1
k 1
depends on system studied
(e.g. ideal gas => kinetic
theory => Stefan-Maxwell)
bulk flow mixture
Remark: liquids & thermodynamic nonidealities: no complete theory yet
Form too complex for many engineering calculations:
Define a mean binary diffusivity for species j
diffusing through the mixture: Djm
N
N j  Ct D jmy j  y j  N k
k 1
(diagonalize matrix Djk)
Mass transfer resistance: Multicomponent diffusion in a fluid
Ideal gas
Stefan-Maxwell:
1
y k N j  y j N k 
 C t y j  
k 1 D jk
N
k j
Alternative (engineering) form:
N
N j  Ct D jmy j  y j  N k
k 1
N
1

D jm
1

k 1 D jk


N
k
 yk  y j



N
j


N
1 y j  Nk / N j
k 1
Equate driving
force, yj
Mass transfer resistance: Multicomponent diffusion in a fluid
N
1

D jm
1

k 1 D jk


N
k
 yk  y j



N
j


N
1 y j  Nk / N j
k 1
1) Classical use (unit operations): « Wilke equation »:
Diffusion of species 1 through stagnant 2, 3, …
N
yk
1
1


D1m 1  y1 k 2,3,... D1k
In reacting media: only appropriate for very dilute solutions !
(other species not necessarily stagnant)
2) Steady-state flux ratios of the various components:
determined by the reaction stoichiometry
Nj
 constant
j
Mass transfer resistance: Multicomponent diffusion in a fluid



k
 yk  y j




1
1
j




N
1  j y j
D jm
1  y j   k /  j 
2) cont.
N
1

k  j D jk
N
1

k  j D jk




k 
y

y
j
 k
 j 


k 1
Example: Use of mean binary diffusivity:
aA  bB  ...
rR  sS  ...
 1 
b 
1
 yB  y A  

1
1
a  D AR
 D AB 

D Am 1   A y A  1  y  s y  ...
 D AS  S a A 
with:  A 
r  s  ...  a  b  ...
a
r


 y R  y A 
a 



Mass transfer resistance: Multicomponent diffusion in a fluid
Example: Use of mean binary diffusivity: (cont.)
Then: flux expression (1D) becomes:
N A  Ct D Am
 Ct D Am
dyA
 y A N A  N B  N R  N S  ...
dz
dyA
 b r s

 y A N A 1     ...
dz
 a a a

 Ct DAm dyA
NA 
1   A y A dz
Consider steady state => NA = const.
Integrate across the film using an
average constant value for DAm
bulk fluid
(1D)
L
catalyst
particle
Ct DAm
1   A y A0
 Ct DAm  y A0  y A ( L)
NA 
ln


L A
1   A y A ( L)  L 
y fA
Mass transfer resistance: Multicomponent diffusion in a fluid
Example: Use of mean binary diffusivity: (cont.)
Ct DAm
1   A y A0
 Ct DAm  y A0  y A ( L)
NA 
ln


L A
1   A y A ( L)  L 
y fA
yfA : the “film factor”: aA  bB  ...
y fA 
with:
rR  sS  ...
1   A y A   1   A y Ass 
1  A yA
ln
s
1   A y As

r  s  ...  a  b  ...
A 
a
Written in terms of partial and total pressures: “film pressure factor pfA”
Equimolar counterdiffusion => δA = 0, yfA = 1
Mass transfer resistance: Mass transfer coefficients
Ct DAm
1   A y A0
 Ct DAm  y A0  y A ( L)
NA 
ln


L A
1   A y A ( L)  L 
y fA
Introduce mass transfer coefficient to be modeled:

s
N A  k g y A  y As
kg 
k
0
g
y fA
• DAm in Sc
• L (film thickness)
in terms of jD = f(Re)

In terms of non-dimensional numbers
k g0  j D
G
Sc 2 / 3
Mm
j D  f Re
Re  d p G / 
G  v
[kg/(m2s)]
(equimolar
counterdiffusion)
Sc  Schmidt number
 /f D
Mass transfer resistance: Mass transfer coefficients
Ct DAm
1   A y A0
 Ct DAm  y A0  y A ( L)
NA 
ln


L A
1   A y A ( L)  L 
y fA
Introduce mass transfer coefficient to be modeled:

s
N A  k g y A  y As
kg 
k
0
g
y fA

In terms of non-dimensional numbers
k g0  j D
G
Sc 2 / 3
Mm
j D  f Re
Re  d p G / 
G  v
[kg/(m2s)]
(equimolar
counterdiffusion)
Sc  Schmidt number
 /f D
Mass transfer resistance: Mass transfer coefficients:
j D  f Re
Example: packed beds:
Mass transfer between a fluid and a bed of particles (spheres; ε = 0.37). Curve 1:
Gamson et al. [1943]; Wilke and Hougen [1945]. Curve 2: Taecker and Hougen
[1949]. Curve 3: McCune and Wilhelm [1949]. Curve 4: Ishino and Otake [1951].
Curve 5: Bar Ilan and Resnick [1957]. Curve 6: DeAcetis and Thodos [1960]. Curve
7: Bradshaw and Bennett [1961]. Curve 8: Hougen [1961]; Yoshida, Ramaswami,
and Hougen [1962].
Mass transfer resistance: Mass transfer coefficients
j D  f Re
Example: packed beds of spheres with ε = 0.37:
for Re = dpG/μ < 190:
j D  1.66Re
0.51
for Re > 190:
j D  0.983Re
0.41
kg: Several driving force units are in common use:





s
s
s
N A  k g y A  y As
 k g C A  C As
 k g p A  p As

Heat transfer resistance: Heat transfer coefficients
Fluid-to-particle interfacial heat transfer resistances:
 H rA  hf am Tss  T 
Heat transfer coefficient: h f  jH cPG Pr2 / 3
with: jH = f(Re)
Heat transfer between a fluid and a
bed of particles (spheres; ε = 0.37).
Curve 1: Gamson et al. [1943];
Wilke and Hougen [1945]. Curve 2:
Baumeister and Bennett (a) for
dt/dp > 20, (b) mean correlation
[1958]. Curve 3: Glaser and
Thodos [1958]. Curve 4: deAcetis
and Thodos [1960]. Curve 5: Sen
Gupta and Thodos [1963]. Curve 6:
Hnadley and Heggs [1968].
Concentration or partial pressure and temperature differences
between bulk fluid and surface of a catalyst particle
One of the most important uses mass and heat transfer relationships:
determine external mass & heat transfer resistances for catalyst particles

k p
s
rA  am k g C A  C As
[kmol/(kg cat. s)]
 am
g
s

p
A
As


(kg in m3f /m2p s )
(kg in kmol / m 2p s bar)
[m2p/kg cat.]
If kg → ∞ with rA finite
C A  CsAs
p A  psAs
Concentration / partial pressure drop
over the external film may be neglected
Check whether allowed => e.g. experimental kinetic studies
Concentration or partial pressure and temperature differences
between bulk fluid and surface of a catalyst particle
Calculation of ΔpA is not straightforward
Calculation of the film pressure factor
s
pfA requires the knowledge of p As
s
Iterative: 1) Assume p As = pA or ΔpA = 0
pfA = pt + δApA
L’Hopital’s rule
kg(1)
(1)
ΔpA
2)
Better estimate pfA
kg(2)
etc..
External mass and heat transfer resistances in practice
• Usually, but not always, ΔpA is rather small
• More common: fairly large ΔT
• Significant ΔT, or ΔpA: especially likely
in laboratory reactors (low flow rates)
• Commercial reactors: commonly small external film
resistance (very high flow rates and Re numbers)
External mass and heat transfer resistances in practice
Simple estimate ΔTfilm in terms of ΔCfilm
T
s
s

T 
kg
hf
 H CA  CAss 
Gases in packed beds
[Smith, 1970]
 jD  Pr 2 / 3    H  C A
    

j
Sc

c
 H     f p  y fA
  H  p A
T  T  0.7 

 M m c p  p fA

s
s

Maximum possible actual temperature difference:
s
• complete conversion
p As
0
• very rapid reaction and heat release
T max
  H   ln1   A p A / pt 
 0.7 


M
c

 m p  
A

(Quick estimate)
Example:
INTERFACIAL GRADIENTS IN ETHANOL DEHYDROGENATION EXPERIMENTS
INTERFACIAL GRADIENTS IN ETHANOL
DEHYDROGENATION EXPERIMENTS:
(ethanol)
(acetaldehyde)
C2 H 5OH
A
CH 3CHO  H 2
RS
• Tubular reactor – fixed catalytic bed
• 275°C and 1 bar
• Molar feed rate ethanol, FA0 = 0.01 kmol/h
• Weight of catalyst, W = 0.01 kg
• Inside diameter reactor = 0.035m
• Catalyst particles: cylindrical:
diameter = height = d = 0.002 m
• Bulk density bed, ρB = 1500 kg/m3
• Void fraction, ε = 0.37, am = 1.26 m2/kg
• Conversion = 0.362 (measured)
• Reaction rate, rA = 0.193 kmol/(kg cat. h)
Example:
INTERFACIAL GRADIENTS IN ETHANOL DEHYDROGENATION EXPERIMENTS
Physicochemical data: estimates through general correlations
[Reid, Prausnitz, and Sherwood, 1977]
ESTIMATION OF THE PARTIAL PRESSURE DROP OVER THE
FILM
ESTIMATION OF VISCOSITIES
H2: Use the Lennard-Jones potential, with
 = 2.827 Å,
0
= 59.7 K,
k
 v = 0.8379
2  548 = 0.013195 cp = 0.0475 kg/m h
 H = 0.002669
2
2
2.827
 0.8379
= 1.32  10-5 kg/m s, or Pa s
C2H5OH: Use the Stockmayer potential, with
2  548 = 0.013195
GRADIENTS IN ETHANOL
DEHYDROGENATION
= 0.002669
cp = 0.0475EXPERIMENTS
kg/m h
Example: HINTERFACIAL
2
2.827  0.8379
2
ESTIMATION OF THE PARTIAL PRESSURE DROP OVER THE
FILM
= 1.32  10-5 kg/m s, or Pa s
ESTIMATION OF VISCOSITIES
C
Use
the Stockmayer
potential,
5OH:
H22H
: Use
the
Lennard-Jones
potential,
withwith
4.31Å,
Å,
2.827
 = =
0 
0
 = 0.3,
= 431
= K,
59.7 K,
k
k
1.422
v v==0.8379
0.002669 46
548
2 548
0.01604 cp
kg/mhh
Eth == 0.002669
==0.013195
cp==0.05775
0.0475 kg/m
2
H2
2
.31
2.4827
 10..422

8379
-5
-5 kg/m s, or Pa s
=
1.604
10

= 1.32  10 kg/m s, or Pa s
CH
Use the method of corresponding states, since the potential
C2H3CHO:
5OH: Use the Stockmayer potential, with
parameters are not available:
0
 = 0.3,
=
4.31
Å,
=p431
K, bar,
= 0.257
1.422
Zvc =
Tc= 461 K,
=
55.4
c
k
0.002669 46  548
=
= 0.01604
cp = 0.05775
kg/m h
GRADIENTS IN ETHANOL
DEHYDROGENATION
EXPERIMENTS
 EthINTERFACIAL
Example:
2
4.31  1.422
ESTIMATION OF THE PARTIAL PRESSURE DROP OVER THE
FILM
= 1.604  10-5 kg/m s, or Pa s
ESTIMATION OF VISCOSITIES
CH
CHO:
the method of
corresponding
H2: 3Use
theUse
Lennard-Jones
potential,
with states, since the potential
parameters are not available:
0
=
2.827
Å,
59.7bar,
K,
0.8379
 vZ=

Tc = 461 K,
pc ==55.4
c = 0.257
k
with
 Ac = 1.9Tr  0.29  104 Z c2 / 3
2  548 = 0.013195 cp = 0.0475 kg/m h
 H 2 = 0.002669
2
2.827

1/ 6
 0.8379 1 / 6
Tc
461
=
= 0.029078
 =
-5
1/ 2 2 / 3
2/3
= 1.32p c 10 kg/m
s,.7or Pa s
M
4454
1the Stockmayer
548 potential,
1

 with
4
C2H5OH:
Use
=
 0.29 10
1.9 
Ac
0.029078
461
0.2572 / 3

0
 = 0.3,
=
4.31
Å,
= 431 K,

 v = 1.422
k = 0.060293 kg/m h = 1.675 × 10-5 kg/m s, or
= 0.016748 cp
Pa s
Example:
INTERFACIAL GRADIENTS IN ETHANOL DEHYDROGENATION EXPERIMENTS
ESTIMATION OF THE PARTIAL PRESSURE DROP OVER THE
FILM
ESTIMATION OF VISCOSITIES
VISCOSITY OF THE GAS MIXTURE
H2: Use the Lennard-Jones potential, with
Composition of the reaction mixture:
0
= 59.7 K,
 v = 0.8379
 = 2.827 Å,
1  xA
k bar,
=
0.4684
yA = 0.4684
pA =
pt
1  xA
2  548 = 0.013195 cp = 0.0475 kg/m h
 H 2 = 0.002669
xA2  0.8379

2
.
827
p R = pS =
y R  y S = 0.2658
pt = 0.2658 bar,
1  xA
= 1.32  10-5 kg/m s, or Pa s
Since the hydrogen content cannot be neglected, Wilke’s method may yield
too highCa2Hvalue
of the
mixture.
Therefore, the viscosity is
Usethe
theviscosity
Stockmayer
potential,
with
5OH: for
computed from
0
=
4.31
Å,
=
= K,

 m431
 v = 1.422
 y j j = 0.3,
k
or
Example:
1  xA
INTERFACIAL GRADIENTS IN ETHANOL DEHYDROGENATION EXPERIMENTS
Since
the hydrogen
cannot bePRESSURE
neglected, Wilke’s
ESTIMATION
OF content
THE PARTIAL
DROPmethod
OVER may
THEyield
too
high a value for the viscosity of the mixture. Therefore, the viscosity is
FILM
computed
from OF VISCOSITIES
ESTIMATION
m =  y j  j
VISCOSITY OF THE GAS MIXTURE
H2: Use the Lennard-Jones potential, with
or
Composition of the reaction mixture:
 m = 0.4684 × 0.05775
 0 + 0.2658 × (0.0475 + 0.060293)
= 59.7 K,
 v = 0.8379
 = 2.827 Å,
1  xA
k bar,
=
0.4684
p A = = 0.0557
pt kg/m h = 1.547 × 10-5 kg/m s yA = 0.4684
1  xA
2  548 = 0.013195 cp = 0.0475 kg/m h
 H 2 = 0.002669
From Wilke’s method,
axA2value
of 0.06133 kg/m h is obtained.

2
.
827

0
.
8379
p R = pS =
y R  y S = 0.2658
pt = 0.2658 bar,
1  xA
-5= 0.4684 × 46 + 0.2658 × (44 + 2)
M m==1.32
 y j M10
j
kg/m s, or Pa s
Since the hydrogen content cannot be neglected, Wilke’s method may yield
= 33.77 kg/kmol
Usethe
theviscosity
Stockmayer
potential,
with
too highCa2Hvalue
of the
mixture.
Therefore, the viscosity is
5OH: for
computed from = M m 0 T0 = 33 .77 273 = 0.7510 kg/m3
m

0
=
4.31
Å,
=
431

 v = 1.422
 m.4= K,
548
y j j = 0.3,
22
V0 T
k
or
Example:
INTERFACIAL GRADIENTS IN ETHANOL DEHYDROGENATION EXPERIMENTS
ESTIMATION OF THE PARTIAL PRESSURE DROP OVER THE
FILM
ESTIMATION
OF VISCOSITIES
DIFFUSION
COEFFICIENTS
H2: Use
potential,
with
Since some
of the
theLennard-Jones
required potential
parameters
are not known, the
semiempirical relation of Fuller-Schettler-Giddings will be applied:
0
Å,
= 59.7 K,
 v = 0.8379
 = 2.827
3
k
vH 2 = 7.07 cm /mol
3
+ 6 H +2 0548
=2=
× 0.013195
16.5 + 6 ×cp
1.98
+
5.48
=
50.4
cm
/mol
 C H OH H= 2=0C.002669
= 0.0475 kg/m h
2.8272  0.8379
 CH CHO = 2 C + 4 H +  0 = 2 × 16.5 + 4 × 1.98 + 5.48 = 46.4 cm3/mol
2
5
3
2
= 1.32  10-5 kg/m s, or Pa s
1 1
1.75
0.00101 548

C2D
H5OH: =
Use
the=Stockmayer potential, with
46 2
D
AS
Eth H 2
1/ 3
1/ 3 2
1.013350.36  7.07
0
 = 0.3,
= 431 K,
 = 4.31 Å,
 v = 1.422
2
-4
2
k /s = 1.4235 × 10 m /s
= 1.4235 cm


Example:
INTERFACIAL GRADIENTS IN ETHANOL DEHYDROGENATION EXPERIMENTS
ESTIMATION OF THE PARTIAL PRESSURE DROP OVER THE
FILM
ESTIMATION
OF VISCOSITIES
DIFFUSION
COEFFICIENTS
Note: DAS has been experimentally measured at 340 K as 0.578 cm2/s. The
H2: Use
potential, with
Since
some
of the
theLennard-Jones
required
potential
are not known, the
Fuller-Schettler-Giddings
formula
yields forparameters
DAS at 340 K
semiempirical relation of Fuller-Schettler-Giddings will be applied:
1.75  0
Å,
= 59.7 K, 2
 v = 0.8379
 = 2.827
340
-4
2


3
D
=
=
0.6174
cm
/s
=
0.6174
×
10
m
/s
/mol
AS
1cm
.4235

vH 2 = 7.07
 k
 548
3
+ 6 H +2 0548
=2=
× 0.013195
16.5 + 6 ×cp
1.98
+
5.48
=
50.4
cm
/mol
Percentage
 C2 H5OH = 2=0C.002669
= 0.0475 kg/m h
H2
error:
2.8270.261742
 0.8379
 0.578
 100
+ 4=×6.82%
1.98 + 5.48 = 46.4 cm3/mol
 CH3CHO = 2 C + 4 H +  0 = 2 × 16.5
578 s, or Pa s
= 1.32  10-50.kg/m
1.75
11  11
1.75

0
.
00101

548
  with
0.00101 548
 2
C2H5OH: =
Use the=Stockmayer
potential,
46
44 46
DARAS=
DDEth
H2 = D
Eth-Ac
1/ 3
1 / 32 2
1
/
3
1
/

013346
50..436
507.36
.07  3


11..0133

0
 = 0.3,
= 431 K,
 = 4.31 Å,
 v = 1.422
-4
2
k 2/s = 1.4235
2
= 1.4235
cm
×
10
m
/s-2 m2/s
= 0.2466 cm /s = 0.2466 × 10


Example:
INTERFACIAL GRADIENTS IN ETHANOL DEHYDROGENATION EXPERIMENTS
ESTIMATION OF THE PARTIAL PRESSURE DROP OVER THE
FILM
ESTIMATION
OF VISCOSITIES
DIFFUSION
COEFFICIENTS
From (3.2.3-9),
H2: Use
potential,
with
Since
some
of the
theLennard-Jones
required potential
parameters
are not known, the
1
semiempirical relation of Fuller-Schettler-Giddings
will be applied:
1
y

y
y

y

 R

0
.
2658

0
.
4684
0
.
2658

0
.
4684
A
S
A

0 
Å,
=
59.7
K,
=
0.8379



 = 2.827



v
3
4
4
D
D
DvamH 2 ==7.07 ARcm /mol AS  k=  0.2466 10
1.4235 10





1  0.4684
1 yA
3

+ 6 H +2 0548
= 2=
× 0.013195
16.5 + 6 ×cp
1.98
+
5.48
=
50.4
cm
/mol
 C2 H5OH H= 2=0C.002669

= 0.0475 kg/m h

2
2
2.827  0.8379
3
2
= 2 C×+104-4 m
 CH3CHO
= 0.4203
/s 0 = 2 × 16.5 + 4 × 1.98 + 5.48 = 46.4 cm /mol
H+
= 1.32  10-5 kg/m s, or Pa s
1 1
1.75
0.00101 548

C2H5OH: =
Use the=Stockmayer potential, with
46 2
D
D
Eth H 2
AS

1.013350.36  7.07
0
 = 0.3,
= 431 K,
 = 4.31 Å,
k 2/s = 1.4235 × 10-4 m2/s
= 1.4235 cm
1/ 3

1/ 3 2
 v = 1.422
Example:
INTERFACIAL GRADIENTS IN ETHANOL DEHYDROGENATION EXPERIMENTS
ESTIMATION OF THE PARTIAL PRESSURE DROP OVER THE
FILM
ESTIMATION
OF VISCOSITIES
DIFFUSION
COEFFICIENTS
Now
and
Reynolds
may
be calculated:
HSchmidt
Lennard-Jones
potential,
with
Sincethe
some
of the
the
required numbers
potential
parameters
are not known, the
2: Use
semiempirical
Fuller-Schettler-Giddings
will be applied:
 105
 m relation 1of.547
2/3
0
Sc =  = 2.827
= Å,
=
0.490,
from
which
(Sc)
= 0.622
=
59.7
K,
=
0.8379


4
v
.751 0.4203k 10
D Am cm30/mol
vH 2 =m7.07
2
3
.01
=46
 0548
kg/m
s
+=6 H +02
2=
× 0.013195
16.5=+0.1328
6 ×cp
1.98
+
5.48
=
50.4
cm
/mol
 C H OH H= 2=0CG.002669
= 0.0475 kg/m h
2.82720.0035
.8379
2  3600
 CH CHO = 2 C + 44 H +  0 = 2 × 16.5 + 4 × 1.98 + 5.48 = 46.4 cm3/mol
2
5
2
= 1.32  10-5 kg/m s, or Pa
3 s
d pG
2.289.10  0.1328
Re =
=
1 = 119.65
1.75
5
548


1.547
 10
 m 0.00101
C2H5OH: =
Use the Stockmayer potential, with
46 2
DEth H 2 DAS =
1/ 3
1/ 3 2
50.36 should
 used:
1.0133
 7.07be
Since Re < 190, the following
jD correlation
0
 = 0.3,
= 431 K,
 = 4.31 Å,
 v = 1.422
-0.51
2
k 2/s = 1.4235
jD = 1.66(Re)
and× 10jD-4=m0.3635
= 1.4235
cm
/s
3


Example:
INTERFACIAL GRADIENTS IN ETHANOL DEHYDROGENATION EXPERIMENTS
ESTIMATION OF THE PARTIAL PRESSURE DROP OVER THE
FILM
ESTIMATION
VISCOSITIES
Now
the partialOF
pressure
drop can be calculated. Assuming that ΔpA = 0 and
R  S becomes
with δA = 1, the film pressure factor for a reaction A
H2: Use the Lennard-Jones potential, with
p fA = pt   A p A = 1  0.4684 = 1.4684 bar
0
= 59.7 K,
 v = 0.8379
 = 2.827 Å,
k
rA M m p fA
0.193  33.77  1.4684
2/3
=
p A =
 0.622
Sc
am GjD 2  548 3600  1.26  0.1328  0.3635
0
.
002669
= 0.013195 cp = 0.0475 kg/m h
 H2 =
2
2.827  0.8379
pA = 0.0272 bar
= 1.32  10-5 kg/m s, or Pa s
Substitution of this estimate for ΔpA in (3.2.1-3), written in terms of partial
pressures, leads to a better estimate for pfA:
C2H5OH: Use the Stockmayer potential, with
0.0272
= 1.4547
p fA =  0
  = 0.3,
431 /K,
ln1=.4684
1.4412
 = 4.31 Å,
 v = 1.422
k
No further iterations required
Example:
INTERFACIAL GRADIENTS IN ETHANOL DEHYDROGENATION EXPERIMENTS
ESTIMATION OF THE TEMPERATURE DROP OVER THE
FILM
Requires two further properties of the reaction mixture:
• The specific heat cp
• The thermal conductivity λ
cp values for the pure components:
• Literature
• Accurate estimation: correlation [Rihani & Doraiswamy, 1965]
Tb (K)
Tc (K)
H v (kJ/kmol)
S v (kJ/kmol K)
 b (kmol/m3)
Cp (kJ/kmol K)
 (kJ/m s K)
Ethanol
Acetaldehyde
Hydrogen
351.7
516.3
38604
109.8
294
461
27006
91.86
20.4
33.3
-
1000
63
106.5
4.61 × 10-5
783
44
81.2
3.977 × 10-5
29.3
27.2 × 10-5
Example:
INTERFACIAL GRADIENTS IN ETHANOL DEHYDROGENATION EXPERIMENTS
ESTIMATION OF THE TEMPERATURE DROP OVER THE
FILM
The heat capacity of the mixture may be computed accurately by means of
c pm =  y j c Pj
= 0.4684 × 106.5 + 0.2658 × (81.2 + 29.3) = 79.25 kJ/kmol K
Thermal conductivities pure components: estimated by Bromley’s method
The following details provide the basis for the numbers in the table:
C2H5OH (polar nonlinear molecule): From Perry and Chilton [1984],
H vb = 38604 kJ/kmol
Consequently,
S vb =
38604
H vb
=
= 109.78 kJ/kmol K
351 .7
Tb
Example:
INTERFACIAL GRADIENTS IN ETHANOL DEHYDROGENATION EXPERIMENTS
ESTIMATION OF THE TEMPERATURE DROP OVER THE
FILM
ρb, the density of liquid ethanol at the normal boiling point, is estimated using
Schroeders’ rule:
Vb = 9 × 7 = 63 cm3/mol
1000
kmol/m3
b =
63
 = 3b Svb  36.63  R ln Tb 
 = 3
1
109 .8  36.63  8.315  ln 351 .7  = 1.16
63
cint rot = 4.98

—CH2—OH
+
8.50 = 13.48 kJ/kmol K

CH3CH2—
Example:
INTERFACIAL GRADIENTS IN ETHANOL DEHYDROGENATION EXPERIMENTS
ESTIMATION OF THE TEMPERATURE DROP OVER THE
FILM
cv = cp – 8.37 = 106.47 – 8.37 = 98.1 kJ/kmol K
M

= 1.3cv + 15.07 – 0.3cint rot – 0.69
Tc
 3
T
516 .3
46  
= 1.3 × 98.1 + 15.07 - 0.3 × 13.48 – 2.89 ×
5
548
1.604  10
– 3 × 1.16 = 132.35 kJ/kmol K
 = 4.61 × 10-5 kJ/m s K
Example:
INTERFACIAL GRADIENTS IN ETHANOL DEHYDROGENATION EXPERIMENTS
ESTIMATION OF THE TEMPERATURE DROP OVER THE
FILM
CH3CHO (polar nonlinear molecule: ΔHvb has to be estimated,
Giacalone’s method is used:
H vb
8.315  294  461  ln 54.7
RTbTC ln pc
=
=
461  294
Tc  Tb
H vb = 27006 kJ/kmol
S vb =
27006
H vb
=
= 91.86 kJ/kmol K
294
Tb
 b is found in the literature:
 = 3
783
kmol/m3;
44
0.783
91.86  36.63  8.315  ln 294  = 0.426
44
Example:
INTERFACIAL GRADIENTS IN ETHANOL DEHYDROGENATION EXPERIMENTS
ESTIMATION OF THE TEMPERATURE DROP OVER THE
FILM
CH3CHO (cont.)
cint rot = 5.07 kJ/kmol K
cv = cp – 8.37 = 81.17 – 8.37 = 72.80 kJ/kmol K
44  
461
=
1.3  72.8  15.07  0.3  5.07  2.89 
5
548
1.675  10
– 3 × 0.426 = 104.48 kJ/kmol K
 = 3.977 × 10-5 kJ/m s K
H2 (nonpolar linear molecule):
M
Tc
= 1.3 cv  14.23  2.93
T

Example:
INTERFACIAL GRADIENTS
– IN
3 ETHANOL
× 0.426 DEHYDROGENATION
= 104.48 kJ/kmolEXPERIMENTS
K
-5
ESTIMATION
THE
TEMPERATURE
DROP OVER THE
 =OF
3.977
× 10
kJ/m s K
FILM
H2 (nonpolar linear molecule):
M
Tc
= 1.3 cv  14.23  2.93
T

33.3
2
=
1.3  20.91  14.23  2.93 
5
548
1.3195  10
= 41.235 kJ/kmol K
 = 27.20 × 10-5 kJ/m s K
Example:
INTERFACIAL GRADIENTS IN ETHANOL DEHYDROGENATION EXPERIMENTS
ESTIMATION OF THE TEMPERATURE DROP OVER THE
THERMAL CONDUCTIVITY OF THE GAS MIXTURE
FILM
THERMAL
CONDUCTIVITY
OF THE GASequation
MIXTURE
To
estimate the
factors Aij the Lindsay-Bromley
is appropriate and
Estimation
factors
will
be applied
here. Aij: Lindsay-Bromley equation
The required Sutherland constants are
SEth = 1.5 × 351.7 = 527.55 K
SEth-Ac = 482.34 K
SAc = 1.5 × 294 = 441 K
S EthhH2 = 204.15 K
S H 2 = 79 K
S AcH 2 = 186.65 K
The Lindsay-Bromley formula yields
A12 = 0.9615
A21 = 1.038
A13 = 0.3653
A31 = 3.1565
A23 = 0.3872
A32 = 3.0988
Example:
INTERFACIAL GRADIENTS IN ETHANOL DEHYDROGENATION EXPERIMENTS
ESTIMATION OF THE TEMPERATURE DROP OVER THE
FILM
THERMAL CONDUCTIVITY OF THE GAS MIXTURE
5
5
3
.
977

10
4
.
61

10
m =

0.4684
0.2658
0.2658
1  1.038
 0.387 1
1  0.9615
 0.3654
0.2658
0.4684
0.4684
27.20  105

0.4684
1  3.156
 3.0988 1
0.2658
m = 6.682 105 kJ/m s K
Note: If the thermal conductivity of the mixture had been considered as linear
in the composition λm would be given by
m =  y j  j = 10.446 105 kJ/m s K
or
50 % higher than the more correct estimate
Example:
INTERFACIAL GRADIENTS IN ETHANOL DEHYDROGENATION EXPERIMENTS
ESTIMATION OF THE TEMPERATURE DROP OVER THE
FILM
Then, the Prandtl number is:
c pm  m / M m

79.25 / 33.8  1.547 105
Pr =
=
= 0.5428
5
6.682 10
m
(Pr)2/3 = 0.665
From Fig. 3.2.2-1, a value of 0.60 may be chosen for jH at Re = 19.65. The
heat of reaction is calculated as follows:
 H = H Eth  H Ac  H H = 70338 kJ/kmol
2
so that
r  H Pr 
ΔT = A
am j H c p G
2/3
=
0.193  70338  0.665
= 10.6 K
79.25
3600  1.26  0.6 
 0.1328
33.77