Modern Methods in Heterogeneous Catalysis Research
TRANSPORT PHENOMENA
10/31/08
Giulio Lolli
Outline
2

Part I - Theory
  Why
Transport?
  Fundamental equations
  Unified approach to transport phenomena
  Boundary Layer approach
  Dimensionless numbers

Part II - Practice
  Example:
Methanol Synthesis
Transport Phenomena
10/31/08
Further Reading
3

Part I
  Bird,
Steward, Lightfoot
  Transport
Phenomena
  Asano
Welty
Mass Transport

Fundamentals of M,H&M Transfer
Part II
  Handout
  Web:
  Perry’s
Document DB # 15375
Chemical Engineering Handbook
Transport Phenomena
10/31/08
4
Part I
Theory
Transport Phenomena
10/31/08
Why Transport? Catalytic Process
5

Thermodynamic
  Reaction
Equilibrium
  Phase equilibrium:
adsorption/desorption

Kinetic
  Reaction

Kinetic
Transport
  Reactants/Product
to/
from the active sites
  Heat to/from the
Transport Phenomena
active site
10/31/08
Transport Phenomena
6
nx
)
Type of transport
What is transferred
Specific direction x

Momentum Transport
!τxy
∂!uy
= −µ
∂x
Heat Transport

Kinetic Energy
∂T
∂CA
qx = −k
jA,x =Law
−DA
  Newton’s
∂x
∂x
Thermal Energy
1 Law (Fluxes)
∂T
∂CA
qx = −k
jA,x = −DA   Fourier’s Law
∂x
∂x
!
!Transport
!   Chemical
! A

T! Mass
= −µ∇!
u
!q = −k ∇T
j!A =Energy
−DA ∇C
∂CA
jA,x = −DA
  Fick’s Law
∂x

st

! = −σ ∇ϕ
! A
Φ
!
! A
!q = −k ∇T
j!A = −DA ∇C
Transport Phenomena
2nd Law (variation in space-time)
10/31/08
!τxy = −µ
First
Law
of
transport
Specific direction x
7
Specific direction x
∂!uy
!τxy = −µ
∂x
∂!
uy
!τxy = −µ
∂x
1st Law (Fluxes)
st
1 Law (Fluxes)
!
!u
T! = −µ∇!
!
!u
T! = −µ∇!
! = −σ ∇ϕ
! A
Φ
qx
∂x
st
1 q =Law
∂T (Fluxes)
−k
j = −D
x
∂x
∂T
qx = −k
∂x
A,x
∂CA
A
∂xA
∂C
jA,x = −DA
!!
!u
T!q ==−k∇T
−µ
∇!
!
j!
∂x
! A
= −DA ∇C
! A
j!A = −DA ∇C
!q =
A
!
!q = −k ∇T
! = −σ ∇ϕ
! A
Φ
! = −σ ∇ϕ
! A
Φ
2nd Law (variation in space-time)
2nd Law (variation in space-time)
D!u
!
ρ
= µ∇2!u+ρ!g − ∇p
nd Transport Phenomena
Dt
D!
u
2
!
ρ
= µ∇ !u+ρ!g − ∇p
Dt
2
Law (variation in sp
10/31/08
Math 101- Gradient
8
Math 101 Nabla, Gradient
 
 
∂
 ∂x 
∂S
 ∂x 
 ∂S 
∂!
!
∇ =  ∂y  ∇S =  ∂y 
 
 
∂
∂z
∂S
∂z
Math 102 Divergence, Laplacian
 
 
∂
 ∂x 
∂S
 ∂x 
∂Vx ∂Vy ∂Vz
∂!
 ∂S  ! !
!
∇ =  ∂y  ∇S =  ∂y ∇ · V =
+
+
∂x
∂y
∂z
 
 
∂
∂z
2
∂S
∂z
2
∂ S
2
Transport Phenomena
∂ S
2
∂ S
10/31/08
ρ
! = −σ ∇ϕ
! A
Φ
Dt
= µ∇ !u+ρ!g − ∇p
Second
Law
of
Transport
2 Law (variation in space-time)
nd
9
DT
2
D!u
Navier-Stokes
=
k∇ T +µφu
!
p
ρ
= µ∇ !u+ρ!g − ρC
∇p
Dt
Dt
DT
2 Fourier’s
ρC
= k∇ T +µφ − r∆H
DCA
Dt
2
=
D
∇
C
+ν
r
DC
A
A
A
= D ∇ C +ν r
2 Fick
Dt
Dt
Dϕ
= σ∇ ϕ
Dϕ
Dt
2
= σ∇ ϕ
Fundamental equations of the system
Dt

2
2
p
A
A
2
u
A
A
rxn
  nd
  nd
2
D!u
!
ρ
= µ∇2!u+ρ!g − ∇p
Dt
Fundamental equations
Transport Phenomena
10/31/08
 ∂S 
∂!
!
∇ =  ∂y  ∇S =  ∂y 
 
 
Math 102 – Divergence & Laplace
∂
∂z
10
∂S
∂z
Math 102 Divergence, Laplacian
 
 
∂
 ∂x 
∂S
 ∂x 
∂Vx ∂Vy ∂Vz
∂!
 ∂S  ! !
!
∇ =  ∂y  ∇S =  ∂y ∇ · V =
+
+
∂x
∂y
∂z
 
 
∂
∂z
∂S
∂z
2
2
2
∂
S
∂
S
∂
S
2
!
!
∇ S = ∇ · ∇S =
+ 2 + 2
2
∂x
∂y
∂z
Math 103 Substantial Derivative
 
 
∂
 ∂x 
∂S
 ∂x 
∂Vx ∂Vy ∂Vz
∂!
 ∂S  ! !
!
∇ =  ∂y  ∇S =  ∂y  ∇ · V =
+
+
∂y 10/31/08
∂z
 
 
Transport∂x
Phenomena
∂
∂S
∂z
∂z
2
2
2
∂
S
∂
S
∂
S
2
!
!
∇ S = ∇ · ∇S =
+ 2 + 2
2
∂x
∂y
∂z
Math 103 – Substantial Derivative
11
Math 103 Substantial Derivative
 
 
∂
 ∂x 
∂S
 ∂x 
∂Vx ∂Vy ∂Vz
∂!
 ∂S  ! !
!
∇ =  ∂y  ∇S =  ∂y  ∇ · V =
+
+
∂x
∂y
∂z
 
 
∂
∂z
∂S
∂z
2
2
2
∂
S
∂
S
∂
S
2
!
!
∇ S = ∇ · ∇S =
+ 2 + 2
2
∂x
∂y
∂z
DS
∂S
∂S
∂S
∂S
∂S
!
=
+ !u · ∇S =
+ ux
+ uy
+ uz
Dt
∂t
∂t
∂x
∂y
∂z
Transport Phenomena
10/31/08
Dt
= σ∇2 ϕ
Equations of Motion
12
Fundamental equations of the system
D!u
2
!
ρ
= µ∇ !u+ρ!g − ∇p
Dt
DT
2
ρCp
= k∇ T +µφu − r∆Hrxn
Dt
DCA
2
= DA ∇ CA +νA r
Dt
! · (ρ!u) = 0
∇
Transport Phenomena
10/31/08
Example
13

Start-up in circular tube
  Cylindrical
coordinates
  Only vz(r) ≠ 0
r
z
Example: startup in circular tube only uz (r) != 0
!
"
!
∂uz
∂u
1 ∂
∂uz
z
!
ρ
+ uz! = ∆p + µ
r
∂t
r ∂r
∂r
! ∂z
In S.S. conditions (∂/∂t = Transport
0)
Phenomena
θ
10/31/08
Example
Example: startup in circular tube only u (r) != 0
z
14
!
∂uz
∂u
z
!
ρ
+ uz! =
∂t
! ∂z
1 ∂
∆p + µ
r ∂r
!
∂uz
r
∂r
"
r
z
In S.S. conditions (∂/∂t = 0)
!
"
∂
∂uz
∆p
r
=−
r
∂r
∂r
µ
∂uz
∆p r"2 A"
r"
=−
+ "
∂r
µ 2 "r
∆p r2
uz = −
+B
µ 4
$
∆p # 2
2
uz =
R −r
4µ
B.C.
B.C.
θ
∂uz
= f inite
∂r
uz = 0
Transport Phenomena
r=0
r=R
10/31/08
θ
r
z
vz/vmax(r=0)
1.0
0.5
0.0
-1.0
0
0.0001
0.001
0.01
0.05
0.1
0.2
0.5
infinite
-0.8
-0.6
-0.4
-0.2
0.0
r/R
15
Transport Phenomena
10/31/08
0.2
0.4
0.6
0.8
1.0
Numerical Solution
16
Finite elements approach
  Software packages

  Fluent
  CFD
  Physics
Set up the correct problem and boundary conditions
  Many times is not necessary to know the exact T,C,v
of every single point in your system

Transport Phenomena
10/31/08
Boundary Layer
17
δ

All the phenomena happen in the boundary layer
  Everything
outside the boundary layer is in equilibrium
and constant

The boundary layer is very “thin”
  The
size of the boundary layer (δ) is smaller than the
characteristic size of the system
  Order of magnitude approach
  Taylor
series
Transport Phenomena
10/31/08
Reynold’s Number
18
Boundary layer approach
Reinolds Number
δ
D"u
ρ
= µ∇2"u
Dt
ρu
!
∆u
!
≈µ
!
∆u
!
x
δ2
! "2
δ
µ
≈
= 1/Re
x
ρux
Prandtl Number
ρux
Re =
µ
ρux
Re =
Transport Phenomena
10/31/08
µ
δ
x
µ
≈
= 1/Re
ρux
Prandtl’s Number
ρux
Re =
µ
Prandtl Number
19
uT = uδT /δ
DT
ρCp
= k∇2 T
Dt
1/2
!
!
∆T
∆T
!
!
ρCp uT
≈k 2
x
δT
δT
k
ρCp u
≈ 2
δx
δT
! "3
δT
k
≈
δ
Cp µ
δT
≈ P r−1/3
δ
δ
δT
Cp µ
Pr =
k
Cp µ
Pr =
Transport Phenomena 10/31/08
k
Dt
!
!
∆T
∆T
!
!
ρCp uT
≈k 2
x
δT
Schmidt’s Number…
20
Cp µ
Pr =
k
δT
k
ρCp u ≈ 2
x
δT
! "3
δT
k
≈
Re−1/2
δ
Cp µ
δT
≈ P r−1/3
δ
δ
… As usual everything is
the same
Schmidt Number
δC
≈ Sc−1/3
δ
Cp µ
Pr =
k
δC
ρDA
Sc = ρD
Sc =
µ µ
A
Transport Phenomena
10/31/08
Nusselt’s Number
21
Considering the
the fluxes:
fluxes:
Considering
δ
δT
∂T
∂T
= −k
−k
= h∆T
h∆T
qq =
=
∂x
∂x
!
!
∆T
!
∆T
! ≈ k!
!
∆T
!
hh!
∆T
≈k δ
δTT
hx
x δδ
hx
x
Nuu =
=
≈
N
≈
δ δδTT
kk
δ
!"#$
!"#$
!"#$ !"#$
1/2
1/3
Re1/2
Prr1/3
Re
P
In general
general
In
KA x
x δ
Sh =
≈
DA
δ δC
!"#$
!"#$
Re1/2 Sc1/3
Transport Phenomena
10/31/08
Sherwood’s Number
22
Considering the
the fluxes:
fluxes:
Considering
δ
δC
∂T
∂T
= −k
−k
= h∆T
h∆T
qq =
=
∂x
∂x
!
!
∆T
!
∆T
! ≈ k!
!
∆T
!
hh!
∆T
≈k δ
δTT
hx
x δδ
hx
x
Nuu =
=
≈
N
≈
δ δδTT
kk
δ
!"#$
!"#$
!"#$ !"#$
1/2
1/3
Re1/2
Prr1/3
Re
P
In general
general
In
KA x
x δ
Sh =
≈
DA
δ δC
!"#$
!"#$
Re1/2 Sc1/3
Transport Phenomena
10/31/08
hx
x
δ
Nu =
≈
k
δ δT
!"#$
!"#$
In General…
Re1/2 P r1/3
23
In general
hx
KA x
1/3
Nu =
= f (Re) P r
Sh =
= f (Re) Sc1/3
k
DA


1/2

Re
Plate in laminar flow






2 + Re1/2
Sphere in laminar flow
f (Re) =

0.8


0.005Re
Packed bed in turbulent flow





· · · · · · · · · · · ·
Extension to macroscopic balances
)
Transport Phenomena 10/31/08
*
*
Q̇ =
q dA = A U ∆T
1/U =
1/h +
s /k
Extension to Macroscopic Balances
24
Transport Phenomena
10/31/08
hx
KA x
1/3
Nu =
= f (Re) P r
Sh =
= f (Re) Sc1/3
k
DA

25

1/2

Re
Plate in laminar flow






2 + Re1/2
Sphere in laminar flow
f (Re) =

0.8


0.005Re
Packed bed in turbulent flow





· · · · · · · · · · · ·
Overall Transfer Coefficient
Extension to macroscopic balances
)
*
*
Q̇ =
q dA = A U ∆T
1/U =
1/hi +
si /ki
A
1/U = 1/hint + s/kmetal + 1/hext
Transport Phenomena
10/31/08
Nu =
hx
Nu =
= f (Re) P r1/3
k

26

1/2

Re






2 + Re1/2
f (Re) =

0.8


0.005Re





· · · · · · · · · · · ·
= f (Re) P r
k
KA x
Sh =  = f (Re) Sc1/3
D
A
1/2

Re
Plate






2 + Re1/2
Plate in laminar flow
Sphere
f (Re) =

0.8


Sphere in laminar flow
0.005Re
Packed





· · · ·flow
Packed bed in turbulent
········
Log-Mean Temperature Difference
Extension to macroscopic balanc
)
Extension to macroscopic balances
Q̇ =
q dA = A U ∆T
)
A
*
*
Q̇ =
q dA = A U ∆T
1/U =
1/hi +
si /ki
1/U = 1/hint + s/kmetal + 1/hext
A
1/U = 1/hint + s/kmetal + 1/hext
∆T1 − ∆T2
∆Tln =
ln (∆T1 /∆T2 )
Transport Phenomena
10/31/08
Conclusions
27

Momentum, Heat and Mass Transfer
  Highly
inter-correlated
  Similar
physical principles
  Similar driving force
  Similar mathematical formulas
  2nd
order PDE – numerical solutions
  Highly dependent from fluid-dynamic properties
  Simplifications
  Boundary
layer
  Dimensionless numbers
Transport Phenomena
10/31/08
28
PART II
Methanol Synthesis
Transport Phenomena
10/31/08
Methanol Production
29

Produced from SynGas
  Catalyst
Cu/ZnO-Al2O3
  Equilibrium limited
  P
= 50 – 100 bar
  T = 200 – 300 ˚C

Main Commodity
  20

Million ton/year
Typical plant size is 1 Million ton/year
  12
m3/s
Transport Phenomena
10/31/08
Catalyst specs
30
KATALCO 51-8
Methanol Synthesis Catalyst
Long life as a result of the optimized formulation incorporating a patented
use of a fourth component MgO
Product Benefits •
•
Low operating temperatures minimize the catalyst sintering rate
•
High catalyst selectivity gives very low impurities in the crude product
•
Close approach to equilibrium is achieved and maintained
•
Easy to reduce and start-up
Product Uses
•
Synthesis of methanol from H2, CO and CO2 mixtures arising from steam
reforming of hydrocarbons, coal gasification or POx
General
Description
•
KATALCO 51-8 is a copper catalyst on a ZnO-Al2O3 support with a MgO
promoter
Physical Properties (Typical)
KATALCO 51-8
Cylindrical pellet
5.4 mm
5.2 mm
1250 kg/m 3
80 kgf
Form
Diameter
Length
Bulk Density
Crush Strength
(axial)
Shipping &
Handling
Chemical Composition (Typical)
CuO:
Al2 O3 :
ZnO:
MgO:
64 wt%
10 wt%
24 wt%
2 wt%
•
Avoid contact with skin and clothing. Avoid breathing dust. Do not take
internally. Please refer to the relevant Material Safety Data Sheet for
further information.
•
KATALCO 51-8 catalyst is available in non-returnable polythene lined
mild steel drums or bulk bags for easy loading.
Transport Phenomena
10/31/08
Reactor types
31
Transport Phenomena
10/31/08
Multi-Bed Reactor
32
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)DEBFGHBG)@HGC)IAE>JKLDHGFM
6:27
6:25
6:24
N!C'$
6:29
6:26
6:67
6:65
6:64
6:69
6:66
456
476
866
896
846
#OPQ
856
Transport Phenomena
876
566
596
10/31/08
!
Reactor dimensioning
33
!"#$%&'()*+',"-#)
1 m/s
! wanted gas velocity
)
m2 section – approx 4 m diameter
#JKG<H
  Approx 7 m tall
  12
2244
2444
;44
L<E)3
:44
!<'$)=>?@ABC

944
L<E)6
844
744
344
L<E)5
644
544
L<E)2
244
4
4
2
5
6
3
7
8
9
D<E)@<FGHI)=>C
Transport Phenomena
10/31/08
!"#$%&'()'*+,-%'.,+/'+.'*&01-2+,'"2'.$230"+2'+.'4&5'5&601)'
!"#$%&'()*+',"-#)
!
)
Pressure Drop
34
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E786<)=>:)B5@@5G86<)B5;CE@DHI)
)
$# $
%
' %( "
%"
%"
)
) %&"$ ! %
*
&$ !
#" '
+ %'(&& # ,
,
& # ' " !% ( $ ! " . ' "
/
)
P
7QR) $F) 87)=>:
J>:;:)2)87)=>:)7E?:;B8A8D@)CD77)K:@5A8=L)M34)N<OC
  To have 1 bar pressure drop
D
=
5
mm
=>:)<D7)C8V=E;:)9:678=L)M34)N<OCUQR) %p)87)=>:)K87A578=L)M3SUU)34WH)
?:@@:=)98C:67856)M4S44H)CQS)
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+:7E@=7)=>D=)#4X)YP44)*DOC)X)4S4YP)FD;OC)
)
Transport Phenomena 10/31/08
%69)A56789:;86<)34)C)@:6<=>)
! ) !"#!
*
)
1<)E5)E7>8)7)B=>=BCB)5<<=6=5>6:)I78)845)J5D=>>=>D)A<)845)?5768A
B7K=BCBL)A<)MGN)845)B7K=BCB)#4=595)BA@C9C;);4AC9@)J5)OPQ3
)
!"#$%&'()*+',"-#)
!   At least 85% catalyst
*%&' +&'!
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Transport
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Transport Phenomena 10/31/08
)
Heat Exchanger
39
1st bed 580  490 K
  35
MW power to take out
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  16
  Shell
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596
Evaporation/condensation
5-22
HEAT AND MASS TRANSFER
40
For example, vapori
radiation and conve
result of heat absorb
the disengaging sur
when the superheat
also occurs from falli
flashing of liquids su
Pool boiling ref
heating surface is su
is not flowing at any
motion of the bubbl
of pool boiling are p
fluid temperature is
collapse of the bubb
pool boiling, with bu
resulting in net vapo
The general shape
to " tb, the tempera
temperature and the
metric relations tha
boiling curve was ori
[ J. Soc. Mech. Eng.
one of the great dile
faced with at least s
Pr water ≈ 7
Dukler plot showing average condensing-film coefficient as a function of physical properties of the condensate film and the terminal Reynolds
number. (Dotted line indicates Nusselt theory for Reynolds number < 2100.)
[Reproduced by permission from Chem. Eng. Prog., 55, 64 (1959).]
FIG. 5-10
1/2 Phenomena
1/2
Transport
hµl / kl !l1/2 = 0.065(N
Pr)l Fvc
2
10/31/08
(5-99a)
Heat Exchanger
41

1st bed 580  490 K
  35
MW power to take out
  Boiling water to produce low pressure steam
  16
  Shell
kg/s steam produced
tube heat exchanger: 2” (5cm) tubes
W/m2K
  U ≈ 200 W/m2K
  hout=220
  Low
pressure steam 1-2 atm 120˚C
  ΔTln=
140K
  Exchange Area 1300 m3 (about 2000 tubes)
  40 banks of 50 tubes (4-5 m in size)
Transport Phenomena
10/31/08
Further Reading
42

Part I
  Bird,
Steward, Lightfoot
  Transport
Phenomena
  Asano
Welty
Mass Transport

Fundamentals of M,H&M Transfer
Part II
  Handout
  Web:
  Perry’s
Document DB # 15375
Chemical Engineering Handbook
Transport Phenomena
10/31/08
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