selected topics in heat and mass transfer
TRANSCRIPT
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Selected Topics in heat and MassTransport
Jundika Candra KurniaAgus Pulung Sasmito
Sachin Vinayak Jangam
Hee Joo Poh
..A compilation of selected presentations
2011
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PREFACE
This e-book consists of selected topics to be covered as part of the postgraduate
course entitled ME6203 Mass Transport in the Mechanical Engineering Department ofthe National University of Singapore, given by Professor A. S. Mujumdar.
For the benefit of wider audience interested in the themes covered, this e-book is
being offered freely. It contains handouts of the PowerPoint presentations made by the
authors. We hope that this compilation will be useful to research students as well as
researchers in academia and industrial R&D. Professor Mujumdar recommended and
provided guidance to the authors of various chapters included in this e-book.
The authors would be happy to hear from readers about any related matter they
wish to discuss or seek clarification on.
Jundika Candra Kurnia
Agus Pulung Sasmito
Sachin Vinayak jangam
Hee Joo Poh
Singapore
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Index
Presentation
No
Title / Author names
01 Heat Transfer in Square duct
Jundika C. Kurnia
02 Computational Study of Energy-Efficient Thermal Drying Using
Intermittent Impinging Jets
Jundika C. Kurnia
03 Mass transport in a micro-channel T-Junction with coiled-base
channel design
Agus P. Sasmito
04 Mass Transport Considerations in PEM Fuel Cell ModelingHee Joo Poh
05 Heat Transfer in Fluidized Beds-An Overview
Sachin V. Jangam
06 Mass Transfer in Fluidized Beds - An Overview
Sachin V. Jangam
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Heat Transfer in
Square duct
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Heat Transfer in Square Duct
Prof. Arun S. Mujumdar
ME 6204: Convective Heat Transfer
Mathematical formulation
Governing equations:
Conservation of mass
Conservation of momentum
Conservation of energy
Constitutive equation (air)
Density Dynamic viscosity
Thermal conductivity
Specific heat
For water, properties are set
as constant
( ) 0 =u
( )( )( ) P = + + u u u u2
pc T k T = u
,absspecific
P
R T= ( )
6
2.67 10 ,MT
T
=
15 4 1,
4 15 3
pc MR
kM R
= +
.pR
cM
=
Pabs = Absolute pressure
Rspecific= Specific gas constant
cp = specific heat
= Collision diameter
= Collision integral
M = molecular weight
Nomenclature:
= fluid density
= fluid viscosity
u = fluid velocity
T = fluid temperature
kt = fluid thermal conductivity
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Mathematical Formulation
Turbulent model used in this simulation is k-model
3
( )
( )2
12
2 2 22 2 2
,
,
2 ,
tt
k
t t
k
t
kk k G
t
C GC
t k k
u v w u v u w w vG
x y z y x z x y z
+ = + +
+ = + +
= + + + + + + + +
u
u
2
,
k
C =
C1= 1.44
C2= 1.92
C = 0.09
k = 1.0
= 1.0
Nomenclature:
u, v, w = component velocity
t= turbulent viscosityk = turbulent kinetic energy
= turbulent dissipationG = turbulent generation rate
Mathematical Formulation
Nusselt number calculation
4
= Nusselt number
= Hydraulic diameter
= Conductive heat transfer
= mixed mean velocity
Nomenclature:
= mixed mean temperature
= surface temperature
= cross-section area
= convective heat transfer
= heat flux
1,
1,
,
c
c
mean c
c A
cc A
surf ace m ean
T T dAVA
V dAA
Qh
T T
hDNu
k
=
=
=
=
u
u
&
meanT
surfaceT
cA
hQ
NuD
kV
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Geometry
The flow configuration considered is full tubeflow inside square duct, as illustrated in figure
5
Schematic representation of flow in a) square duct and b) development
of a momentum boundary layer
Numerics
Finite-volume based solver: Fluent 6.3.
Mesh independence study ~10000 cells.
Pressure velocity coupling: SIMPLE (Semi-Implicit
Method for Pressure-Linked Equation).
Second-order upwind discretization.
Algebraic Multi-grid Method (AMG).
Relative residual ~10-6.
It took around one minute to converge in Quad-core 2.83
GHz with 8 GB RAM.
CFD analysis was carried out by Agus Pulung Sasmito
and Jundika Candra Kurnia (ME, NUS)
6
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7
LAMINAR FLOW
Boundary condition
Air Case 1: Constant heat flux at
wall
Inlet: air velocity = 1.6 m/s; T
air = 25 C.
Wall: no-slip; Q = 30 watt/m2.
Outlet: Pout = 1 atm; Q=0.
Re 1000.
Case 2: Constant wall
temperature
Inlet: air velocity = 1.6 m/s; T
air = 25 C.
Wall: no-slip; T wall = 50 C.
Outlet: Pout = 1 atm; Q=0.
Re 1000.
Water Case 3: Constant heat flux at
wall
Inlet: water velocity = 0.1 m/s;
T water = 25 C. Wall: no-slip; Q = 2870
watt/m2.
Outlet: Pout = 1 atm; Q=0.
Re 1000.
Case 4: Constant wall
temperature
Inlet: water velocity = 0.1 m/s;
T water = 25 C.
Wall: no-slip; T wall = 50 C.
Outlet: Pout = 1 atm; Q=0.
Re 1000.8
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Velocity in the middle channel
9
Water (Cases 3 and 4)
Air (Cases 1 and 2)
Temperature in the middle of the channel
10
Air (Case 1) Air (Case 2)
Water (Case 3) Water (Case 4)
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Temperature in the wall of the channel
11
Water (Case 3)
Air (Case 1)
Heat flux at the wall
12
Water (Case 4)
Air (Case 2)
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Nusselt Number
13
Water (Cases 3 and 4)Air (Cases 1 and 2)
Nusselt number asymptotic
Case 1: 2.8
Case 2: 2.7
Case 3: 2.7Case 4: 2.2
Wall and Mean temperature
14
Air (Case 1) Air (Case 2)
Water (Case 3) Water (Case 4)
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15
TURBULENT FLOW WITHRE=20000
Boundary condition
Air Case 1: Constant heat flux at
wall
Inlet: air velocity = 30 m/s; T
air = 25 C.
Wall: no-slip; Q = 497 watt/m2.
Outlet: Pout = 1 atm; Q=0.
Re 20000.
Case 2: Constant wall
temperature
Inlet: air velocity = 30 m/s; T
air = 25 C.
Wall: no-slip; T wall = 50 C.
Outlet: Pout = 1 atm; Q=0.
Re 20000.
Water Case 3: Constant heat flux at
wall
Inlet: water velocity = 2 m/s; T
water = 25 C. Wall: no-slip; Q = 170370
watt/m2.
Outlet: Pout = 1 atm; Q=0.
Re 20000.
Case 4: Constant wall
temperature
Inlet: water velocity = 2 m/s; T
water = 25 C.
Wall: no-slip; T wall = 50 C.
Outlet: Pout = 1 atm; Q=0.
Re 20000.16
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Velocity in the middle channel
17
Water (Cases 3 and 4)
Air (Cases 1 and 2)
Temperature in the middle of the channel
18
Air (Case 1) Air (Case 2)
Water (Case 3) Water (Case 4)
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Temperature in the wall of the channel
19
Water (Cases 3)
Air (Cases 1)
Heat flux at the wall
20
Water (Cases 4)
Air (Cases 2)
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Nusselt Number
21
Water (Cases 3 and 4)Air (Cases 1 and 2)
Nusselt number asymptotic
Case 1: 45
Case 2: 34
Case 3: 623
Case 4: 127
Wall and Mean temperature
22
Air (Case 1) Air (Case 2)
Water (Case 3) Water (Case 4)
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23
TURBULENT FLOW
WITH Re=60000
Boundary condition
Air Case 1: Constant heat flux at
wall
Inlet: air velocity = 80 m/s; T
air = 25 C.
Wall: no-slip; Q = 1272
watt/m2.
Outlet: Pout = 1 atm; Q=0.
Re 60000.
Case 2: Constant wall
temperature
Inlet: air velocity = 80 m/s; T
air = 25 C.
Wall: no-slip; T wall = 50 C.
Outlet: Pout = 1 atm; Q=0.
Re 60000.
Water Case 3: Constant heat flux at
wall
Inlet: water velocity = 6 m/s; T
water = 25 C. Wall: no-slip; Q = 433879
watt/m2.
Outlet: Pout = 1 atm; Q=0.
Re 60000.
Case 4: Constant wall
temperature
Inlet: water velocity = 6 m/s; T
water = 25 C.
Wall: no-slip; T wall = 50 C.
Outlet: Pout = 1 atm; Q=0.
Re 60000.24
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Velocity in the middle channel
25
Water (Cases 3 and 4)
Air (Cases 1 and 2)
Temperature in the middle of the channel
26
Air (Case 1) Air (Case 2)
Water (Case 3) Water (Case 4)
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Temperature in the wall of the channel
27
Water (Cases 3)
Air (Cases 1)
Heat flux at the wall
28
Water (Cases 4)
Air (Cases 2)
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Nusselt Number
29
Water (Cases 3 and 4)Air (Cases 1 and 2)
Nusselt number asymptotic
Case 1: 105
Case 2: 67
Case 3: 1700Case 4: 314
Wall and Mean temperature
30
Air (Case 1) Air (Case 2)
Water (Case 3) Water (Case 4)
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Summary of heat transfer calculation
31
Air (Laminar) Water(Laminar)
Air (Turbulent Re 20000) Water (Turbulent Re 20000)
32
LAMINAR FLOW WITH VARIABLE T
and Q BOUNDARY condition
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Boundary conditions-Variable T,Q
33
Air Constant heat flux at wall
Inlet: air velocity = 1.6 m/s; T
air = 25 C.
Wall: no-slip; Q = 30 watt/m2.
Outlet: Pout = 1 atm; Q=0.
Re 1000.
Constant wall temperature
Inlet: air velocity = 1.6 m/s; T
air = 25 C.
Wall: no-slip; T wall = 50 C.
Outlet: Pout = 1 atm; Q=0. Re 1000.
Temperature distribution
34
Distance from entrance: 5 cm Distance from entrance: 25 cm
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Temperature distribution
35
Distance from entrance: 50 cm Distance from entrance: 100 cm
36
LAMINAR FLOW WITH
HEATING/COOLING
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Boundary condition
Case 1: Heating
Inlet: air velocity = 1.6 m/s; T air = 25 C.
Wall: no-slip; T wall = 50 C, 100 C, 200 C.
Outlet: Pout = 1 atm; Q=0.
Re 1000.
Case 2: Cooling
Inlet: air velocity = 1.6 m/s; T air = 50 C, 100 C,200 C.
Wall: no-slip; T wall = 25 C. Outlet: Pout = 1 atm; Q=0.
Re 1000.
37
Nusselt number distribution in entry length
38
Heating Cooling
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39
LAMINAR FLOW WITH BOUYANCY
Boundary condition
Laminar flow (Air)
Boundary conditions
Inlet: air velocity = 0.15 m/s;
T air = 25 C.
Wall: no-slip; T wall = 200 C.
Outlet: Pout = 1 atm; Q=0.
Re 100.
Case 1: No gravity force
Case 2: Horizontal placement
Case 3: Tilted 45 Case 4: Vertical placement
Gravity force is applied for
case 2, 3 and 4.
40
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Velocity at entry region (z=2.5 cm)
41
Case 1 Case 2
Case 3 Case 4
Temperature at entry region (z=2.5 cm)
42
Case 1 Case 2
Case 3 Case 4
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Velocity at z= first 10 cm
43
Case 2
Case 1
Case 3
Case 4
Temperature at z= first 10 cm
44
Case 2
Case 1
Case 3
Case 4
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Nusselt number
45
46TAPERED DUCT
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Boundary condition
47
Laminar flow (Air)Case 1: Divergent duct
Inlet: air velocity = 16
m/s;
T air = 25 C. Wall: no-slip; T wall = 50
C. Outlet: Pout = 1 atm;
Q=0.
Re 1000.
Case 2: Convergent duct
Inlet: air velocity = 0.16
m/s; T air = 25 C. Wall: no-slip; T wall = 50
C. Outlet: Pout = 1 atm;
Q=0. Re 1000.
Velocity in the middle channel
48
Convergent duct (Case 2)
Divergent duct (Case1)
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Temperature in the middle channel
49
Convergent duct (Case 2)
Divergent duct (Case1)
Nusselt number
50
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51
POWER LAW FLUID
Boundary condition
52
Power law fluid
1
1
eff
eff
where:
= flow consistency index ( )
= shear rate or the velocity gradient ( )
= flow behaviour index
= apparent or effective viscosity ( )
n
n
n
uK
y
K Pa s
us
y
n
uK
y
Pa s
=
=
Laminar flow (Air)
Case 1 (Pseudoplastic n=0.5)
Inlet: air velocity = 1.6 m/s; T
air = 25 C.
Wall: no-slip; T wall = 50 C.
Outlet: Pout = 1 atm; Q=0.
Case 2 (Dilatant n=1.5)
Inlet: air velocity = 1.6 m/s; T
air = 25 C.
Wall: no-slip; T wall = 50 C.
Outlet: Pout = 1 atm; Q=0.
n Type of fluid
1 Dilatant
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Velocity in the middle channel
53
Dilatant (Case 2)
Pseudoplastic (Case1)
Temperature in the middle channel
54
Dilatant (Case 2)
Pseudoplastic (Case1)
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Nusselt number
55
56
PULSATING FLOW
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Boundary condition
Case 1 (Frequency = 5 Hz)
Inlet: air velocity = Vin; T air =
25 C.
Wall: no-slip; T wall = 200 C.
Outlet: Pout = 1 atm; Q=0.
Re 1000.
Case 2 (Frequency = 10 Hz)
Inlet: air velocity = Vin; T air =
25 C.
Wall: no-slip; T wall = 200 C.
Outlet: Pout = 1 atm; Q=0.
Re 1000.
57
Velocity in the middle channel
58
0 s 0.05 s
0.1 s 0.15 s
Case 1 (frequency 5 Hz)
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Temperature in the middle channel
59
0 s 0.05 s
0.1 s 0.15 s
Case 1 (frequency 5 Hz)
Velocity in the middle channel
60
0 s 0.01 s
0.02 s 0.03 s
Case 2 (frequency 20 Hz)
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Velocity in the middle channel
61
0.04 s 0.05 s
Case 2 (frequency 20 Hz)
Temperature in the middle channel
62
Case 2 (frequency 20 Hz)
0 s 0.01 s
0.02 s 0.03 s
Case 2 (frequency 20 Hz)
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Temperature in the middle channel
63
Case 2 (frequency 20 Hz)
0.04 s 0.05 s
Case 2 (frequency 20 Hz)
Average Nusselt number
64
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Summary
Flow inside square duct has been simulated for variety ofBCs and for buoyancy effect.
Several cases - laminar and turbulent flow are
considered
Cooling/heating, buoyancy effect, power law fluid,
tapered duct and pulsating inlet flow have also been
simulated; no analytical solution possible
Heat transfer distributions are calculated
65
References
W. Kays, M. Crawford, B. Weigand, Convective
heat and mass transfer 4th Edition, McGraw-Hill,
2005.
S. Kakac and Y. Yener, Convective Heat
Transfer, Hemisphere Pub, 1982.
A. Bejan, Convection heat transfer, Wiley, 2004.
F. P. Incropera and D. P. Dewitt,Fundamentals
of Heat and Mass Transfer, 5th Edition, Wiley,
2001.
J. H. Leinhard IV and J. H. Leinhard V,A Heat
Transfer Textbook, 3rd edition, 1980.66
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Computational Study ofEnergy-Efficient Thermal
Drying Using Intermittent
Impinging Jets
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1
Computational Study of Energy-Efficient
Thermal Drying Using Intermittent
ME 6203 Mass Transport
Impinging Jets
Prof. Arun S. Mujumdar
Email: [email protected]
Tel: +65-6516-4623, Fax: +65-6777-6235
2011
Guest lecturer
Jundika Candra Kurnia
Department of Mechanical Engineering
National University of Singapore
Outline
Overview of drying
Physical model
Problem description
Key assumptions
Numerical methodology Selected results
Case study
Velocit contours
Temperature contours
Drying kinetics
Summary
Q&A 2
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2
Drying Widely known as the most common way to preserve food
Essential operation in chemical, agricultural, biotechnology, food,
Overview
, , , ,
processing and wood processing industries
Involves simultaneous transport process
Transport processes in drying:
Mass transfer
Heat transfer
Flow
Occur simultaneously both internally and externally Induces deformation:
Shrinkage
Cracking
(Not modeled here) 3
Impinging jet drying
Various drying methods are available for
different material
One of this method is impinging jet drying
Offer high transport rate (mass and energy)
Effective for drying of continuous sheets (paperdrying); discrete flat/curved objects.
Impinging jets, however, have several
drawbacks
Non-uniform drying
High energy consumption compare to parallel flow
Further study in impinging jet drying is required
Pulsating jet is proposed to speed up drying kinetics
Not commonly used yet4
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3
Physical ModelAn orifice nozzle is used in this study- Axisymmetric case
55
Substrate dried is a potato chip. It is placed in a drying chamber under one
impinging jet. Pulsating and intermittent flow is applied on the inlet
Physical model
The aforementioned condition can be brought into computational domain as follows
6
Inlet
45
fixed pulsating and intermitent
17.5%
in
in
in
T C
V
RH
Drying chamber
0.4
0.02
L m
z m
Substrate dimension (chip)
30
0.5 and 5
s
s
L mm
H mm
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4
What phenomena occur during drying?
Transport mechanism of mass? Energy?7
Physical model
moisture diffusion from the inner drying
Basic mechanisms
,
evaporates
conductive heat transfer within the drying
substrate
eva oration and convection of the va or
from the surface of the drying substrate
into the drying air
convection heat transfer from drying air to
the surface of the drying substrate 8
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5
Why do we need assumptions?
What are the key assumptions?
Comment on validity of assumptions?
9
Assumptions
The drying substrate is compact and homogeneous with uniform
initial tem erature and moisture content.
In developing the mathematical model, several assumptions are
made- some for simplicity
Within the drying substrate, the diffusivity of water vapor is 100
times larger than the diffusivity of liquid water.
The thermophysiscal properties of the drying substrate aretemperature and moisture content-dependent and isotropic (equal in
all directions).
Variations in dependent variables in span wise direction are
,
height (reduction in dimensionality from three to two dimensions).
The shape of the drying substrate remains constant. No shrinkage
or deformation is accounted for.
Newtonian fluid
10
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6
How to translate these physical phenomena intomathematical model?
What conservation equations do we need to model
e p ys ca p enomena
Do we need more information?
11
Conservation equations
Conservation of
For drying substrate (chip)
lc
Transient term (time dependent)
Diffusion
Liquid water
Water vapor
,
,
lb l l
vvb v l
t
c
D c Kct
Evaporation
Transient term (time dependent)
Diffusion
energy
12
b pb bT
c k T qt
Transient term (time dependence; heat capacity)
Conduction
Cooling due toevaporation
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7
Conservation equation
Mass
For drying air (Impinging jet) unsteady case
0, u For incompressible fluidInertia/net rate Viscous
Momentum
Energy
' ,a apt
uu u u - u
' ' ,a pa a a paT
c T k T c T t
u u
Pressure gradient
ass o
water vapor
13
,v va v vc
D c ct
u
Convection
Transient term (storage) Diffusion
Advection/bulk motion Conduction ur u en
heating
What is a turbulence model?Why do we need one?
What is Reynolds averaging?
Basic /popular turbulence models
14
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8
Turbulence model
A turbulence model is a model which is used to approximates thephysical behavior of turbulent flows*
Turbulence model are necessary in numerical simulation due to
impracticality in computing all scales of turbulent motion. Therefore,
approx mate met o s tur u ence mo e s are ntro uce to s mp y
and reduce computational cost.
Reynolds averaging refers to the process of averaging a variable or
an equation in time. For example, if we have time dependent
variable , we can decompose this variable into an average partand fluctuating part in the following way:
'1
where T has to be long enough to phase out fluctuation part on (t).
Aside from time averaging, Reynolds averaging also deals with
space averaging and Ensemble averaging**
15*J.J. Bertin, J. Periaux, J. Ballmann, 1992, Advances in hypersonics v2: Modeling hypersonic flows, Birkhauser, Cambridge**J. Sodja, 2007, Turbulence model in CFD, Ljubljana, Slovenia (http://www-f1.ijs.si/~rudi/sola/Turbulence-models-in-CFD.pdf)
, .
T
T
Turbulence model
Various turbulence models have been developed. They
can be categorized as*:
Algebraic/Zero-equation models
One equation models
Two equation models
Second-order closure models
Among these models,
k-, k-, LES (Large Eddy Simulation), RSM (ReynoldsStress Model) are among the most popular turbulence
model used in computational fluid dynamics.
In this study, Reynolds Stress Model (RSM) is used as it
has been shown to be superior to k-and k-turbulence
models in steady impinging jets
16*D. C. Wilcox, 2006, Turbulence modeling for CFD, DCW Industries, La Canada, Clifornia** P. Xu, B. Yu, S. Qiu, H. J. Poh and A. S. Mujumdar. Turbulent impinging jet heat transfer enhancement due to intermittent pulsation. International Journal of
Thermal Science, 49 (7): 1247-1252, 2010.
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9
Turbulence model for impinging jet drying
ij
ij ij ij ij ij ij
RC P D
t
' 'u u
Reynolds stress model (5 equations)
ccumu a on
Convective
Production
Rotation
Diffusion
ij
t t
' 'ij a i jC u u Uj i
ij im jm
m m
U UP R R
x x
' ' ' '2ij k j m ikm i m jkmu u e u u e 2
with 0.09 and 1.0t k
D R C C
17
Dissipation
Pressure strain
interaction
tk
23
ij ij
1 2
1 2
2 2
3 3
with 1.8 and 0.6
ij ij ij ij ijC R k C P P k
C C
To solve this model, k- turbulence model is required
Governing equations
k- turbulence model
,t tk
k k G
u
2
12
2 2 22 2 2
,
2 ,
k
t t
k
C GC
t k k
u v w u v u w w vG
x y z y x z x y z
u
2k
18
t ,
C1= 1.44
C2= 1.92
C = 0.09
k = 1.0
= 1.0
Nomenclature:u, v, w = component velocity
t= turbulent viscosityk = turbulent kinetic energy
= turbulent dissipation
G = turbulent generation rate
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10
Constitutive relations
Density air
Dynamic viscosity of air
5 2 21.076 10 1.039 10 3.326air air air T T
15 3 11 2 8 75.21 10 4.077 10 7.039 10 9.19 10air air air air
T T T
Conductivity of air
Specific heat of air
Heat evaporation
Density of substrate
10 3 7 2 44.084 10 4.519 10 2.35 10 0.0147air air air air k T T T
1000 2.394( 273.15) 2502.1fgh T
, 1
1
b ref
b
X
SbX
6 3 3 2
, 4.647 10 4.837 10 1.599 1175
p air air air airc T T T
19
on uc v y o su s ra e
Specific heat of substrate
Diffusivity of water vapor and
liquid water inside substrate
3
. .exp
1 8.3143 10 273.15 335.15 1
b
s
k
T X
6 0.0725 20441.29 10 exp exp273.15
vb lb
s
D DX T
, 1750 2345
1p b
Xc
Constitutive relations
Heat of wetting (heat to evaporate
bound water)
Total heat of evaporation
6 4 6 3 5 2
4
8.207 10 4.000 10 6.161 10
2.368 10 1163for 0.01 0.2
wH X X X
X X
evap fg wh h H
Moisture content
Dry basis
Wet basis
Equilibrium moisture content
(GAB model)
mass of water
mass of dry product
mass of watermass of wet product 1
l
s
l l
s l b
X
XWX
,
1
0.0209, 0.976, 4.416
m we
w w w
m
X CKAX
KA KA CKA
X K C
20
Free moisture content
Cooling rate due to evaporation
Rate of water evaporation
Diffusivity of water vapor in air
evap l l q h M Kc
0
Ea
RTK K e
6 8 10 22.775 10 4.479 10 1.656 10va
D T T
free eX X X (free to be removed)
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11
Constitutive relations
Relation of moisture content to concentration of water inside substrate
,
,
1,
1 1
w b
b ref
w
W
XX
X SbX
,
2
,2 2
, ,2
,1
1 1 ,
1 1 0,
can be solved analytically for X and by neglecting wrong root th
b ref
w w
b ref
w w
b ref b ref
w w w w
X XM c
SbX X SbX
SbX Sb X X X M c
Sb X Sb X M c M c
2
e solution is
21
,
,
,2
,
1 ,
1.
b ref
w w
b ref
w w
Xa
where
a SbM c
b SbM c
c
Correlations
Local Nusselt number
Calculation of h, Nu, Nu distributions in impinging jets
( , ) x jet
luid
h DNu x t
k
Local heat transfer
coefficient
Local heat transfer flux
xx
jet wall
qh
T T
0
( )x fluid
y
T xq k
y
me average oca
Nusselt number
Time averaged Nusselt
number 22
0
1( ) ( , )avgNu x Nu x t dt
t
0 0
1 1( , )
x t
avgNu Nu x t dtdxx t
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12
Nomenclature
cl concentration of liquid water [mol m-3
] p Pressure [Pa]
cv concentration of water vapor [mol m-3
] Dva diffusivity of vapor on the drying air [m2s]
Dlb diffusivity of liquid inside the drying substrate [m2s] dynamic viscosity of the drying air [Pa s]
Dvb diffusivity of vapor inside the drying substrate [m2s] a density of the drying air [kg m
-3]
T temperature [K] cpa specific heat of the drying air [J kg-1
K-1
]
q cooling rate due to evaporation [W m-3
] ka thermal conductivity of the drying air [W m-2
K-1
]
K production of water vapor mass per unit volume Ea activation energy [kJ mol-1
]
b density of the drying substrate [kg m-3
] R universal gas constant [J K1
mol1
]
cpb specific heat of the drying substrate [J kg-1
K-1
] Ml molecular weight of water [kg kmol-1
]
23
kb thermal conductivity of the drying substrate [W m-
K-
] hevap total heat of evaporation [J g-
]
u mean velocity [m s-1] X moisture content (dry basis) [kg kg-1]
u fluctuate velocity [m s-1
] W moisture content (wet basis) [kg kg-1
]
Initial and boundary conditions
Substrate
Initial conditions:
,0
0.
where
,b
l b
l
Wc
M
0 0, 0,, , ,l l b v v bT T c c c c
Drying chamber inlet
Boundary conditions:
0,
0.
,0
0,
,
0,
1000 .1
v b
l a
a
v a
l
c
c
RHc
RH M
0,0, , , , .in in in v v au v v T T RH RH c c
0 0, 0,, , , 0,
l l a v v aT T c c c c u v
24
Drying chamber outlet
Drying chamber wall
, 0, 0.out vp p D c k T n n
0, 0, ( ) 0.v vu v D c c k T n u n
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13
Boundary conditions and parameters
Inlet velocity for various cases Steady laminar jet
vin = 2 m s-1
Pulsating laminar jet
Parameters needed to solvethe model are
3
, 1420 ,b ref kg m
vin = 1+1sin(2ft) m s-1
Intermitent laminar jet
vin= 2 m s-1(on),0 m s-1(off)
Steady turbulent jet
vin = 20 m s-1
Pulsating turbulent jet
1
0
3
,45
5 1 1
,45
1 1
,
0.018 ,
4.6,
1.110 ,
1.934 10
8.314 ,
in
l
a C
a C
M kg mol
X
kg m
kg m s
R J K mol
vin = + s n m s-
Intermitent laminar jet vin= 2 m s-1(on),0 m s-1(off)
25
1
48.7 ,1.4
1
120
Ea kJ molSb
f Hz
Numerics
Gambit: creating geometry, meshing,
labeling boundary condition
Fluent: solver based on finite volume
method
Domain is discretized onto a finite set of
control volumes (or cells).
General conservation (transport)equationsfor mass, momentum, energy, species, etc.
are solved on this set of control volumes.
V A A V
dV V dA dA S dV t
Partial differential equations are discretized
into a system of algebraic equations.
All algebraic equations are then solved
numerically to render the solution field.26
Convection Di ffusionUnsteady Generation
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14
Numerics
Flow chart of computational fluid dynamics (CFD)*
27
*http://progdata.umflint.edu/MAZUMDER/Fluent/Intro%20Training/L-1%20Introduction%20to%20CFD.pdf
Numerics
User Defined Scalars: solving for water liquid and vapor
User Defined Funct ions Macros
DEFINE_SOURCE, DEFINE_DIFUSIVITY, DEFINE_FLUX,
_ , _ ,
Three different mesh sizes of 2000, 4000, 8000 elements
were implemented and compared in terms of velocity,temperature and moisture content to ensure a mesh-
independent. We found that the result from the mesh
size 2000 deviates 5% and 4000 deviates 1% from 8000
28
. ,
chosen.
Relative residual 10-6 for all dependent variable.
It took around 30-50 min to converge in Quadcore 1.8
GHz with 8 GB RAM for 5 to 8 h drying time
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15
SELECTED RESULTS
29
Contours of velocity
Laminar steady jet
Turbulent steady jet
30
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16
Contours of temperature
Laminar steady jet
Turbulent steady jet
31
Temperature contours in substrate
32
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17
Moisture profile in substrate
33
Drying kinetics (0.5 mm substrate)
Impinging jet is not recomended
especially when drying cost is
considered (energy cost).
Effect of the pulsating and
intermittent flow can be seen on this
case Frequency and velocity have no
effect on this case, most likely due
to the thin substrate 34
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18
Drying kinetics (5 mm substrate)
Velocity slightly affect drying
kinetics. It is clearer compare to
that for thin substrate.
Effect of the pulsating andintermittent flow can be seen in this
case Frequency has no effect
35
Conclusion
Simple physical model to show how one
can use a math model consisting of
conservation equations and relevant
boundary conditions For gas-side, we use continuity,
momentum, energy and species equations
simple diffusion model for both water and
vapor. For low temperature only liquid
diffusion model is adequate.36
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19
References
[1] M. V. De Bonis, G. Ruocco, 2008, A Generalized Conjugate Model for
Forced Convection Drying Based on An Evaporative Kinetics, Journal of
Food Engineering, Vol. 89, pp: 232-240
2 M. R. Islam J. C. Ho A. S. Mu umdar 2003 Convective Dr in with Time-
Varying Heat Input, Drying Technology, Vol. 21(7), pp: 1333-1356
[3] J. Srikiatden a, J. S. Roberts, 2008, Predicting moisture profiles in potato
and carrot during convective hot air drying using isothermally measured
effective diffusivity, Journal of Food Engineering, Vol. 84, pp: 516-525
[4] W. Kays, M. Crawford, B. Weigand, 2005, Convective Heat and Mass
Transfer 4th ed., McGraw Hill, Singapore
[5] F. P. Incropera and D. P. Dewitt , 2001, Fundamentals of Heat and Mass
, ,
[6] P. Xu, B. Yu, S. Qiu, H. J. Poh, A. S. Mujumdar, 2010, Turbulent ImpingingJet Heat Transfer Enhancement Due to Intermittent Pulsation, International
Journal of Thermal Sciences. doi:10.1016/j.ijthermalsci.2010.01.020
[7] H. J. Poh, K. Kumar, A. S. Mujumdar, 2005, Heat transfer from a pulsed
laminar impinging jet, International Communications in Heat and Mass
Transfer, Vol. 32, pp:13171324 37
For Self-Study
How would you model
Case where jet temperature is 200 C?
The jet is superheated steam at atmospheric
pressure and 200 C?
The drying chamber is at very low (but finite)
pressure and dried by superheated steam
Will drying time be reduced if the slab is
flipped after some time? Why?
38
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Mass transport in a micro-
channel T-Junction with
coiled-base channel design
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Mass transport in a micro-channel
T-Junction with coiled-base channel design
ME 6203 Mass Transport Guest Lecture
Prof. Arun S. MUJUMDAR
Email: [email protected]
Tel: +65-6516-4623, Fax: +65-6777-6235
Guest lecturer
Agus Pulung SASMITO
Minerals Metals Materials Technology Center
National University of Singapore
2011
Outline
Overview of micro-channel T-Junction
Physical model
Problem description
Key assumptions
Numerical methodology
Selected results
Mass transport enhancement
Concluding remarks
2
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Micro-channel T-Junction Widely used in industry, especially pharmaceutical, for mixing
and reaction processes
Relatively easy to control the reactions, especially for highly
exothermic reaction
Involves simultaneous transport process
Transport process in T-Junction:
Momentum transfer
Mass transfer
Heat transfer
Occur simultaneously
Main phenomena:
Mixing
Surface reactions
Overview
3
Micro-Channel T-Junction
Passive mixing for various chemical reaction; it
does not require additional energy for mixing
processes
Micro-channel T-Junction, however, has several
drawbacks
Poor mixing, especially at short channel and high
Reynolds number
High pressure drop due to impingement effect
Various innovative designs are proposed to
improve mixing and reactions
Coiled channel
Channel with fins
Impinging jet channel4
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Typical geometry
5
Inlet
300
Re500
in
in
T K
V
=
Micro-channel T-junction
120 mm
1 mm
L
h
=
=
A typical micro-channel T-
junction is used to mix methane
and air; the channel surface iscoated with catalyst (platinum)
Example case of mixing and reaction of
methane oxidation in platinum surface
CH4H2T = 300K
O2N2T = 300K
CO
CO2H2O
Pt
surface;
T
1290K
Ptsurface;T1290K
L
h
4 2 2 2
2 2 2
2 3
2
Gas species: CH , O , H ,H O,
CO , HO , N
Surface species: Pt(s), H(s), O(s),OH(s), H O(s), H (s),
CH (s), CH(s), C(s),
CO( 2s), CO (s)
Solid species: Pt(b)
Innovative coiled-base design
6
Conical T-junctionTypical straight T-junction
In-plane spiral T-junctionHelical T-junction
Coiled-base channel design is
proposed to enhance heat and
mass transfer.
Coiled-base channel design
has been widely used in
industrial applications due to
compact structure, ease of
manufacture, higher heat and
mass transfer.
The presence of secondary
flow induced by coil curvatureand complex temperature and
concentration profiles caused
by curvature-induced torsionare among significant
phenomena observed in
coiled-base channel.
Length is kept constant for
comparison purpose.
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Physical model
Convective heat and mass transfer
Mixing in the opposing jet at T junction
Surface reactions at the channel wall which include
mass consumptions and generation,
heat release due surface reaction,
multi-step surface reactions including adsorption
reaction, surface reaction, and desorption reaction.
Surface species are calculated from site balance
equations
Surface reactions create sources of bulk phase, which
determines its deposition rate on a surface.
7
Basic mechanisms
Assumptions
The flow is steady-state, laminar, newtonian flow and
species mixture is follows ideal gas law.
There are three types of species: gas, surface (site) and
solid species. The model treats chemical speciesdeposited on surfaces as distinct from the same
chemical species in the gas
Thermo-physical properties of species mixture follows
mixing law of ideal gas with temperature dependent
effect.
Gas phase reaction are closely coupled with surface
reactions.
8
In developing the mathematical model, several assumptions are
taken:
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How to translate this physical phenomena into amathematical model?
What conservation equations do we need to interpret
the physical phenomena?
9
Conservation equations
Mass
Momentum
Species
Energy
0 =u
( )( ) ( )23
p = + +
Tu u u u u I
( ) ( )i i i iD R = +u
( ) ( )p eff tempc T k T S = +u
Inertia/net rate
Pressuregradient
viscous Effect of volumedillatation
convective diffusive reaction
4 2 2 2 2 2i: CH , O , H , H O, CO , HO
Compressible
flow
convective conductive Heat due to
reactions
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Detailed reactions
Theory: consider the rth wall surface reaction written in general forms' ' ' '' '' ''
, , , , , ,
1 1 1 1 1 1
g gb s b sr
N NN N N NK
i r i i r i i r i i r i i r i i r i
i i i i i i
g G b B s G g G b B s G= = = = = =
+ + + + where Gi, Bi, andSi represents the gas phase species, the solid species, and the
surface-adsorbed (or site) species, respectively. g, b, s are the stoichiometric
coefficients for each reactant species; g, b, and s are the stoichiometric coefficients
for each product species; and Kr is the overall reaction rate constant.
The rate of rth reaction is [ ] [ ]' ', ,
, wall wall1
g
i r i r
Ng s
r f r i i
i
k G S=
= The net molar rate of production or consumption of each species i is
given by ( )
( )
( )
rxn
rxn
rxn
'' '
,gas , ,
1
'' '
,bulk , ,1
'' '
,site , ,
1
1,2,3,...,
1, 2,3,...,
1, 2,3,...,
N
i i r i r r g
r
N
i i r i r r br
N
i i r i r r s
r
R g g i N
R b b i N
R s s i N
=
=
=
= =
= =
= =
Reaction rate constant is computed using Arrhenius expresion
/
,r rE RT
f r rk A T e =
Wall surface reaction boundary conditions
It is assumed that, on the reacting surface, the mass flux of each gasspecies is balanced with its rate of production/consumption
[ ]
,wall
wall dep i,wall , i,gas
walli,site
1,2,3,...,
1,2,3,...,
i
i w i g
i
s
D m M R i Nn
SR i N
t
= =
= =
The mass fraction at the wall is related to concentration by
[ ] wall i,wallwall
,
i
w i
GM
=
mdep is the net rate of mass deposition or etching as a result of surfacereaction
dep , ,
1
bN
w i i b ulk
i
m M R=
=
[Si]wall is the site species concentration at the wall, and defined as
[ ] sitewalli iS z=where is the site density and zi is the site coverage of species isite
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Multi-step reaction mechanism
1.74e701.88e18H2O(s) + Pt(s) => H(s) + OH(s)24
4.82e704.45e20H2O(s) + O(s) => OH(s) + OH(s)25
1.15e701.56e18OH(s) + Pt(s) => H(s) + O(s)23
1.84e801e17CO(s) + Pt(s) => C(s) + O(s)22
6.28e703.7e20C(s) + O(s) => CO(s) + Pt(s)21
2e703.7e20CH(s) + Pt(s) => C(s) + H(s)20
2e703.7e20CH2(s) + Pt(s) => CH(s) + H(s)19
2e703.7e20CH3(s) + Pt(s) => CH2(s) + H(s)18
00.52.3e16CH4 + 2Pt(s) => CH3(s) + H(s)17
1.05e803.7e20CO(s) + O(s) => CO2(s) + Pt(s)16
2.05e701e13CO2(s) => CO2 + Pt(s)15
1.25e801e13CO(s) => CO + Pt(s)14
00.57.85e15CO + Pt(s) => CO(s)13
4.82e703.7e20OH(s) + OH(s) => H2O(s) + O(s)12
1.74e703.7e20H(s) + OH(s) => H2O(s) + Pt(s)11
1.15e703.7e20H(s) + O(s) => OH(s) + Pt(s)10
1.93e801e13OH(s) => OH + Pt(s)9
00.53.25e8OH + Pt(s) => OH(s)8
4.03e701e13H2O(s) => H2O + Pt(s)7
00.52.37e8H2O + Pt(s) => H2O(s)6
2.13e803.7e202O(s) => O2 + 2Pt(s)5
00.52.01e14O2 + 2PT(s) => 2O(s)4
0-0.51.8e17O2 + 2Pt(s) => 2O(s)3
6.74e703.7e202H(s) => H2 + 2Pt(s)2
00.54.36e7H2 + 2Pt(s) => 2H(s)1
Er (J/kmol)rArReactionNo
Constitutive relations
Mixture density
Mixture viscosity
Effective thermal conductivity; heat capacity
/pM RT =
( )4 4 2 2 2 2 2 2 2 2 2 2
1
CH CH H H O O H O H O OH OH CO CO CO CO N N/ / / / / / / /M M M M M M M M M
= + + + + + + +
Mean molecular mass
4 2 2 2 2 2
,
with , = CH , H , O , H O, OH, CO, CO , Nx
x
=
2
1 11/2(g) 2 4
, (g)
11 1
8
MM
M M
= + +
eff i ik k = p i p,ii
c c=
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Constitutive relations (Contd)
in out FoMp
=
1,
c
c
c A
V dAA
= u
Figure of Merit is used to evaluate the effectiveness of themixing and reaction rate in the micro-channel T-Junction. It
defined as reactant conversion rate over pumping power
required. Since the mass flow rate is kept constant; hence
Mean velocity
Mixed-mean temperature
Mixed-mean mass fraction
1,
c
mean c
c A
T T dAVA= u
,
1
c
i mean i c
c A
dAVA
= u
Boundary conditions
At the air inletUin = 5 m/s (Re~500)
Tin = 300K
O2 = 0.21
N2 = 1-
O2
At the methane
inletUin = 5 m/s (Re~500)
Tin = 300K
CH4 = 0.9
H2 = 0.1
0i
T = =
At the walls No-slip condition
No species flux
Twall = 1290 K
At the outletPout = 101325
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Numerics
AutoCad for creating geometries Gambit for meshing and labeling boundary conditions:
fine structured mesh near wall to resolve boundary layer;increasingly coarser mesh to the middle of the channel to reducecomputational cost
Mesh independence test were carried out for three differentmesh sizecoarse, medium, finein terms of velocity, pressure,temperature and species.
Fluent for discretization and solving dependent variables Based on finite volume discretization method
Pressure-velocity coupling is solved by well-known SIMPLEmethod
Overall, it requires ~300 MB memory and 2 h solving time on
workstation with Quadcore 2.63 GHz processor for convergencecriteria 10-6 for all dependent variables.
ChemKIN for reaction kinetics To set up details multi-step reaction mechanism and thermo-
physical properties of gas species.
Flow chart for numerical solver
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Results and discussion
Velocity profiles at channel length 50mm
Fully developed
flow exists in
the straight
channel
Secondary flow
is developed in
the coiled base
channel.
Higher velocity
intensity existsin the outer wall
of the coiled
base channel
m/s
Oxygen mass fraction at channel length 50mm
Straight channel has
higher oxygen mass
fraction means that
lesser oxygen is
consumed for
reactions.
Among the coiledbase channel design,
helical coil gives
better conversion
rate.
Secondary flow
enhance mass
transport in the
surface reaction
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Methane mass fraction at channel length 50 mm
At straight channel, the
methane concentration
is higher at the right
side of the wall, it
means that the
methane is not mixed
well with air.
Helical coil yields the
best mixing and
reaction among others
The presence of
secondary flowimprove the mixing rate
of reactant species
Oxygen mass fraction along channel
Helical coil gives the best conversion rate among other designs.
Straight T-junction yields the lowest conversion rate due to poormixing.
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0 20 40 60 80 100 120
Length / mm
Oxygenmassfraction
helical
straight
in plane spiral
conical
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Mixed mean temperature along channel
Coiled base channel also gives higher heat transfer rate ascompared to straight channel
Coiled base channel is suitable for highly exothermic/endothermicreaction to control the desired environment/temperature.
300
400
500
600
700
800
900
1000
1100
1200
1300
0 20 40 60 80 100 120Length / mm
Temperature/K
conical
in plane spiral
helical
straight
Effect of Reynolds number
Lower mass flow rate performs better conversion rate
This is due to longer residence time of the species
0.03
0.05
0.07
0.09
0.11
0.13
0.15
0.17
0 20 40 60 80 100 120
Length / mm
Oxygen
massfraction
Re 100
Re 500
Re 1000Re increasing
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Effect of coil diameter
Smaller coil diameter produces slightly better conversion rate.
This is due to higher secondary flow produced in smaller coildiameter.
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0 20 40 60 80 100 120
Length / mm
Oxygenmassfraction
r 4r 3
r 5
diameter increasing
Pressure drop
Straight channel requires the lowest pressure drop;whereas, the helical coil has the highest pressure drop
Pressure drop increses as the mass flow increasing
0
2000
4000
6000
8000
10000
12000
14000
16000
18000
20000
0 200 400 600 800 1000
Reynolds
p/pa
straight
conical
in-plane
helical
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Figure of Merit
Straight channel has the highest figure of merit among otherdesigns; however, for industrial application where space is limitedand pumping power is not an issue, such as in pharmaceuticalindustry, coiled base channel design can be a desirable choice
0.00E+00
5.00E-05
1.00E-04
1.50E-04
2.00E-04
2.50E-04
3.00E-04
3.50E-04
100 500 1000Reynolds
FigureofMerit
straight
conical
in-plane
helical
Concluding remarks
Coiled base channel design can improve heatand mass transfer as compare to straight T-junction channel.
This improvement is due to the presence ofsecondary flow.
However, higher pressure drop is required forcoiled base channel design.
For industrial application where space is limited,conversion rate is the most important, andpumping power is not an issue, coiled basechannel design can be a desirable choice
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Nomenclature3
1
2
3
= density, kgm
= velocity, ms
= pressure, pa
= dynamic viscosity, Pas
= mass fraction of species i
= diffusivity of species i, ms
= reaction rate of species i, kgm
= specific heat
i
i
i
p
p
D
R
c
u
1 1
1
-3
temp
, Jkg K
= temperature, K
= effective thermal conductivity, WmK
= heat release due to reactions, Wm
= gas species, mol= bulk/solid species, mol
= surface-adsorbed/site species,
eff
i
i
i
T
k
S
GB
S
' ''
' ''
' ''
mol
, = stoichiometric coefficient for gas reactant, and product
, = stoichiometric coefficient for bulk reactant, and product
, = stoichiometric coefficient for site reactant, and p
i i
i i
g g
b b
s s roduct
,
= rate of rth reaction
= reaction rate constant using Arrhenius expression
= pre-exponential factor
= temperature exponent
= activation energy for the reaction, Jkgmol
= universal gas co
f r
r
r
r
k
A
E
R
-1 -1 -1
dep
nstant, Jkg mol K
= mean molecular mass
= net rate of mass deposition, kg
= mol fraction
M
m
x
References
[1] O. Deutschmann, L.i. Maier, U. Riedel, A.H. Stroemman, R.W. Dibble,Hydrogen Assisted Catalytic Combustion of Methane on Platinum,
Catalysis Today 59,141--150 (2000).
[2] V. Kumar, M. Paraschivoiu, K.D.P. Nigam, Single phase fluid flow and
mixing in microchannel, Chemical Engineering Science, 2011, in press.
[3] S. Vatisth, V. Kumar, K.D.P. Nigam,A review on the potential application
of curved geometries in process industry, Industrial Engineering Chemistry
Research 47, 3291-3337 (2008).[4] J.C. Kurnia, A.P. Sasmito, A.S. Mujumdar, Evaluation of heat transfer
performance of helical coils of non-circular tubes, J. Zhejiang University
Science: A, 2011, in press.
[5] J.C. Kurnia, A.P. Sasmito, A.S. Mujumdar, Laminar convective heat
transfer in coils of non-circular cross-section tube: a computational fluid
dynamics study, Thermal Science, 2011, accepted.
[6] J.C. Kurnia, A.P. Sasmito, A.S. Mujumdar, Numerical investigation of
laminar heat transfer performance of various cooling channel designs,
Applied Thermal Engineering, 2011, in press.
[7] Fluent user guide documentation, http://www.fluent.com30
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For Self Study
How can one reduce pressure drop in the coiledbase channel design?
What happens if the channel is not in squarecross-section, e.g. circular, triangle, star-shapeetc?
Will reaction and mixing rate improve if we addfins inside the channel?
What if the reactant species are in differentphase, e.g. gas and liquid? Is the modelpresented still valid?
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Mass Transport
Considerations in PEM
Fuel Cell Modeling
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1
1
ME6203 Mass TransportMass Transport Considerations
in PEM Fuel Cell Modeling
Prof. Arun S. Mujumdar, ME, NUS
Dr. Poh Hee Joo, IHPC
March 2010
2
Outline - Part 1 (Handout)
Fuel Cell Introduction
Fuel cell Mass Transport
Diffusive transport in electrode
Convective transport in flow structures
Analytical Modeling with MATLAB
Summary
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2
3
Why PEMFC Modeling in MassTransport course?
Fuel cells are becoming important in academicand industrial R&D. Some alreadycommercialized. Much more development isneeded to enhance performance cost-effectively
Excellent industrial illustration of a case wheremath modeling of transport phenomena-including mass transport is critically important
It is an excellent illustration of how very complextransport processes can be modeled and whatare the different levels of math models which arepossible
4
Why PEMFC model?
In ME6203 one objective is to look at advanced masstransport problems of real interest and examine how amodel can be developed based on fundamentals
It is also an example of complex interaction betweenvarious transport phenomena. Illustrates need forsignificant information needed for such a model
Due to time limitation, different levels of modeling e.g.1D, 2D,2.5D, 3D steady/unsteady, single phase/ twophase models etc are not discussed. Model is only asgood as assumptions made- they must be realistic.
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3
5
Mass Transport and PEMFC Whenever there is species movement causing
concentration changes there is mass transport Mechanisms are: diffusion, convection, electro-osmosis
etc PEMFC includes flow in channels, flow in porous media Involves proton and electron transfer, catalytic chemical
reactions, heat transfer etc An excellent-but complex-example for study of transport
phenomena
Here, please focus on the technique of math modelingrather than the complex details which are beyond thescope of this course.
Suitable for Term Paper Projects e.g. 1D analyticalmodeling of different types of fuel cells
6
Preamble
With this preamble , let us proceed to fuel cells..
Numerous resources are available on the web forself-study
Advanced models are being worked on at hundredsof labs around the world-useful for innovation!
Several excellent textbooks available as well No need to go beyond what is in this PPT-except for
those who choose to work on term papers on thissubject.
Poh Hee Joo will be happy to provide relevantresources and ideas to those interested
Caution: Some aspects are complex and areincluded only for completeness of coverage. You donot need to get into those details.
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5
9
Basic Fuel Cell Operation
1. Reactant transport
Efficient delivery of reactants by using flow field plates incombination with porous electrode structures.
2. Electrochemical reaction
Choosing right catalyst and carefully designing reactionzones
3. Ionic (and Electronic) Conduction
Thin electrolyte for ionic conduction, without fuel cross over
4. Product Removal
Flooding by product water can be major issue in PEMFC
10
Transportation
Stationary Power Generation
Residential
Portable Power Generation
Space and Defense
Applications of fuel cells
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6
11
Fuel Cells: Classification
PEM fuel cell Solid oxide fuel cell
Characterized by
Electrolyte materials
Operating temperature
Fuel used
PEM Fuel Cell 2H2 4H+ + 4e
Solid Oxide Fuel Cell H2 + O2-CO2 + H2O + 2e O2 + 2e O
2-
O2 + 4e + 4H+ 2H2O
Polymer membrane Ceramic membrane
800C 600 10000C
At Anode At Cathode
Direct Methanol Fuel Cell CH3OH + H2O CO2 + 6H+ + 6e 3/2O2 + 6H+ + 6e3H2O
12
Fuel Cells: Some Advantages
Replacement for IC Engines intransportation
Higher energy efficiency
Zero or ultra-low emission
Replacement for batteries in portableelectronics
Higher energy density
Nearly zero recharge time
Independent scaling between power(determined by fuel cell size) and capacity(determined by fuel cell reservoir)
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7
13
Fuel Cells: Some Limitations High cost of fuel cell
Low volumetric power density comparing to I.C.engines and batteries
Safety, Availability, Storage and Distribution ofpure hydrogen fuel
Alternative fuels (e.g. methanol, gasoline)
difficult to use directly and require reforming Susceptibility to environmental poisons
Operational temperature compatibility concerns
14
Challenges to Fuel Cell Commercialization
Simple question, but difficult answer
Prototype developed by SERC-PEMFC for2W portable battery charger fuel cell of
NOKIA mobile phone cost about $300
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8
15
Fuel cells: Interdisciplinary field ofscience and engineering:
Thermodynamics
Electrochemistry
Chemistry and Chemical Engineering
Fluid Mechanics
Heat and Mass Transfer
Material Science (metallurgy) and materials engineering
Polymer Science and specifically ionomer chemistry
Design, manufacturing and engineering optimization Solid mechanics and mechanical engineering
Electromagnetism and electrical engineering
Etc etc
16
PEMFC (Interdisciplinary!)
Membrane ScienceCatalysis
and
Electrochemistry
System Integration
High temperaturecation membrane
Reduce CO poisoningof catalyst)
Membrane for DMFC(prevent methanol crossover)
Anion membrane(use of low cost catalyst)
Alternative catalyst(reduce cost)
High catalyst utilization(reduce catalyst loading)
Improved performance(electro-oxidation/Reduction)
Measurement andCharacterization
(relate performance toelectrochemical processes)
Thermofluidsand
Component Design
Transport phenomena(molecular diffusion,
ion migration,convection)
Multi-phase physics(water management)
Heat transfer(performance stability)
Fluid dynamics(flow channel design)
System design & configuration(reduce cost, improved efficient)
Interconnection(increase power o/p)
Heat and water management(operational stability)
Courtesy of SERC Fuel Cell Project
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9
17
Schematic of a cross sectional view of PEM
Fuel Cell unit
Cathode
Bipolar Plate
Anode
Bipolar Plate
Cathode
Electrode
(GDL)
Anode
Electrode
(GDL)
Membrane
Cathode
Catalyst
Anode
Catalyst
O2 + 2H+ + 2e H2O H2 2H+ + 2e
H2channel
Air
channel
O2 H2H+
Loade-
18
Role of Each Component1. Cathode/Anode Bipolar plate
Electronic Conduction Heat Transport
2. Air/H2 channel Reactant Transport & Product Removal (Mass Transfer) Heat Transport
3. Cathode/Anode GDL
Ionic and Electronic Conduction Reactant Transport & Product Removal (Mass Transfer) Heat Transport
4. Cathode/Anode Catalyst Electrochemical reaction (Mass Transfer reactant consumption and
product generation) Ionic and Electronic Conduction
5. Membrane Ionic conduction Water transport
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10
19
Water Management in PEMFC
AnodeCathodeElectrolytemembrane
Water produced
within cathode
Water is dragged from
anode to cathode sides by
protons moving through
electrolyte (electro-osmoticdrag)
Water is back diffused from
cathode to anode, if cathode sideholds more water
Water is removed by
O2 depleted air
leaving the fuel cell
Water is removed by
circulating hydrogen
Water is supplied byexternally humidifying
air/O2 supply
Water is supplied byexternally humidifying
hydrogen supply
20
Why Mass Transfer is Important in PEMFCComponent Mass Transport Implication Where mass transport
limitation exists
Air/H2 channel To provide homogenous distributionof reactants across an electrodesurface while minimizing pressuredrop and maximizing water removalcapability
Reactant depletion fordownstream channel
Impurity contamination, e.g.N2
Cathode/Anode GDL Porous electrode support to
reinforce catalyst, allow easy gasaccess to catalyst layer, andenhances electrical conductivity
Liquid water flooding block the
pores for gas diffusion intocatalyst layer
Cathode/Anode Catalyst Electrochemical reaction takesplace at the catalyst layer,consume reactant (H2 and O2) andgenerate product (H2O)
Poor total reaction surfacearea (catalyst loading) foroptimal electrochemicalperformance
Membrane To separate the air and H2 whileallowing liquid water and ionictransport across membrane
Membrane dry-out at hightemperature, and loss of itsproton conducting capability
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11
21
Fuel Cell : Mass Transport To produce electricity, fuel cell must be continuallysupplied with fuel and ox idant. At the same time,products must be continuously removed so as toavoid strangling the cell. The process ofsupplying reactants and removing products istermed fuel cell mass transport.
Why is it important? Poor mass transport can leadto significant fuel cell performance loss, as thereactant depletion and/or product accumulation
within catalyst layer (not at the fuel cell in let) willadversely affect performance. This is calledconcentration or mass transport loss, and can beminimized by careful optimization of masstransport in the fuel cell electrodes and fuel cellflow structures
22
Transport in Electrode vs. FlowStructure
Difference between mass transport in fuel cell electrode and fuelcell flow structures in one of length scale, and this lead todifference in transport mechanism
For fuel cell flow structures, dimensions are generally on themillimeter or centimeters scale. Flow pattern typically consisted of
well-defined channel arrays. Gas transport in the channel isdominated by fluid flow and convection.
For fuel cell electrodes, it exhibit structure and porosity on themicrometer and nanometer length scale. The tortuous geometryof electrodes insulated gas molecules from convective forcespresent in flow channel. Gas transport within electrodes isdominated by diffusion.
*Velocity scale could also affect t ransport
mechanism
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12
23
Transport in Electrodes: Diffusive
Transport
An electrochemicalreaction on catalyst layerside of an electrode andconvective mixingon theother flow channel side ofthe electrode set upconcentration gradients,leading to diffusivetransport across theelectrode.
o
Rc
o
Pc
*
Pc
*
Rc
Flow
Structure
Flow
channel
Anode
Electrode
Catalyst
Layer
Electrolyte
Reactants (R) In
Products (P) Out
JR
JP
jrxn
Concen
tration
Reaction in catalyst
layer consumes R,
generates P
Schematic of mass transport situation withintypical fuel cell electrode
24
Faradays Law
From Faradays Law : current ievolved by anelectrochemical reaction is a direct measure ofthe rate of electrochemical reaction
n is the number of electrons transferred,
Fis Faradays constant, 96,485 C/mol,
is the rate of electrochemical reaction, mol/s.
*The current density
Jis the molar flux, mol/cm2s.
dt
dNnF
dt
dQi ==
dt
dN
nFJdt
dN
AnF
A
ij =
==1
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13
25
Transport in Electrode: Diffusive Transport
At steady state, the diffusion flux of reactants andproducts down the concentration gradient across theelectrode (diffusion layer) will exactly match theconsumption/production rate of reactants and productsat the catalyst layer.
Diffusion flux(kmol/m2s) of reactants to the catalystlayer may be described by
From Faradays Law
Using the flux balance equation, one can solve forreactant concentration in the catalyst layer
effRRnFD
jcc = 0*
dx
dcDJdiff =
o
RReff
diff
ccDJ
=*
o
RReff cc
nFDj
=
*
26
Transport in Electrode: DiffusiveTransport
Limiting Current Density,jL Limiting current density of fuel cell will be encountered
when reactant concentration in the catalyst layer drops
all the way to zero. Fuel cell mass transport design strategies focus onincreasing the limiting current density by:
1. Ensuring a high reactant concentration at flow channel bydesigning good flow structures that even distribute reactants
2. Ensuring that effective diffusivity is large and diffusion layerthickness is small by carefully optimizing fuel cell operatingconditions, electrode structure, and diffusion layer thickness.
Theoretical typicaljL are on the order of 1-10A/cm2
0
Reff
L
cnFDj =
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14
27
Question 11. Discuss the factors that determinejL,
limiting current density. List three ways toincreasejL.
28
Answer for Question 1
AnswerThe limiting current density is given byFactors determining thejL are reactant concentration at flowchannel, effective diffusivity and diffusion layer thickness. Wecould increasejL by
Ensuring a high reactant concentration at flowchannel by designing good flow structures thateven distr ibute reactants
Ensuring effective diffusivity is large;Ensuring diffusion layer thickness is small
by carefully optimizing fuel cell operatingconditions, electrode structure, and diffusionlayer thickness
0
Reff
L
cnFDj =
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15
29
Typical process of reactanttransport to reactant sites
If considering the convection mass transfer across electrode surface,how is the limiting current density being derived?
sRc
Flow channel
x
x = 0
x =H
HE
Gas Diffusion
Layer (porous)
Catalyst Layer
(porous)
Reactant molar
flux, JR
Convection, hm DiffusionDiffusion and
reactiono
Rc
*Rc
30
Process of reactant transport to reactant sites
Convection mass transfer at the electrode surface
Diffusion mass transport through the Gas Diffusion Layer
Combining Equation 1 & 2
( )sRoRm cchJ =
=
E
R
s
Reff
H
ccDJ
*
=m
R
o
R
R
ccJ
*
eff
E
m
mD
H
hR +=
1
(1)
(2)
From Faradays Law, current density is proportional to the rate ofelectrochemical reaction
(3)
nFJdt
dN
AnF
A
ij =
==1 ( )*1 RoReff
E
m
CCD
H
hnFj
+=
Limiting current density oReff
E
m
L CD
H
hnFj
1
1
+=
(4)
(5)
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16
31
Transport in Electrode: DiffusiveTransport
Concentration affects fuel cell performance through reactionkinetics.
This is because reaction kinetics also depend on the reactant andproduct concentration at the reaction sites.
Reactant depletion/product accumulation in the catalyst layerslead to fuel cell performance loss.
This is called fuel cell concentration (or mass transport) loss.
conc - Voltage loss due to reactant depletion in the catalyst layer IncreasingjL can greatly extend a fuel cells potential operating
range; therefore mass transport design is an active area ofcurrent fuel cell research.
jj
j
nF
RT
L
Lconc
+=
1
1
32
Question 22. Using the limiting current density equation, calculate
the limiting current density for a fuel cell cathoderunning on air at 1 Atm and 25C. Assume only O2 andN2 and ignore the presence of water vapor. Massfraction of O2 in air is 0.23. Assume the diffusion layeris 500m and has a porosity of 40%.
Hint* : Using Chapman-Enskog theory (Chp 5, Cussler)to find the binary diffusion coefficient, andBruggemann correction to account for the effectivediffusivity in porous structure. Molar concentration forO2 can be obtained by mole fraction of O2 multiply bythe total molar concentration for the air mixture. n isthe number of electrons consumed per mole of thereactant consumed. Molecular weight for N2 and O2are 28 and 32, respectively
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17
33
Answer for Question 2 In the H2-O2 fuel cell, the electrochemical reaction at
the cathode is given by O2 + 4H+ + 4e H2O. Hence
n = 4. F is the Faraday constant, 96,485 C/mol. Thebinary diffusion coefficient is given by Chapman-Enskog theory.O2= 3.467. N2= 3.798. O2-N2=3.6325. From the necessary calculation, = 0.9186.Therefore, Dij= 2x10-5 m
2/s. From Bruggemanncorrection Dij,eff= 5.06 x 10-6 m
2/s. From the ideal gasequation, total molar concentration for the mixture is =
40.9 mol/m3. Molecular weight for the mixture O2-N2 is= 28.92. Mole faction of O2 = = 0.20786. Therefore,molar concentration of O2 = 8.5015 mol/m3. Limitingcurrent density = 33,204 A/m2 = 3.32A/cm2
34
Transport in Flow Structure:Convective Transport
Fuel cell flow structures are designed to distributereactants across fuel cell
One could possibly use single-chamber structure, andencapsulate the entire fuel cell collector in a singlecompartment. Unfortunately, this would make reactantstend to stagnant inside the chamber, leading to poorreactant distribution and high mass transport losses;hence poor fuel cell performance
Conversely, employing intricate flow structure containingmany small flow channels keeps the reactants constantlyflowing across fuel cell, encouraging uniform convection,mixing and homogenous reactant distribution.
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18
35
Convective Transport Contd Analyzing convective gas transport in the
complex real world flow structures is only reallypossible with numerical methods. A commontechnique is to use CFD modeling
However, basic analysis of simple flow scenariosis still possible with the principle of fluid
mechanics, which can still yield great insight intofuel cell mass transport and flow structuredesign
36
Transport in Flow Structure:Convective Transport
Pressure difference between inlet and outlet drives the fluid flow. Although gas flowing in stream-wise direction along flow channel,
convective mass transport can also occur in transverse direction fromflow channel into (or out of) electrode. This happens whenconcentration of species iis different at the electrode surface versusthe flow channel bulk.
Inlet Outlet
u
JC
JD
Convection transfer at surface
Diffusion Electrode
Membrane
x
y
Dh
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19
37
Transport in Flow Structure:Convective Transport
Mass flux (kg/m2s) due to convective mass transfermay be estimated by
Mass transfer convection coefficient, hm, is dependenton the channel geometry, the physical properties ofspecies iandj, and the wall conditions. It can be foundfrom the nondimensional Sherwood number
( )isimiC hJ = ,,
h
ij
mD
DShh =
38
Transport in Flow Structure:Convective Transport
Gas is depleted along flow channel
As hydrogen or air is consumed continuously alonga flow channel, the reactants tend to becomedepleted, especially near the outlet. Depletion poses
adverse effect on fuel cell performance, sinceconcentration losses increase as reactantconcentrations decrease.
A simple 2D mass transport model for fuel cellcathode is developed. This is to determine how theoxygen concentration decreases along flow channelusing macro-scale mass flux balance.
*Refer to Note 1 for O2 mass concentration profilealong cathode catalyst layer.
* important
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20
39
Transport in Flow Structure: Convective
Transporty
x
Electrolyte
Cathode catalyst layer
Gas diffusion layer
Cathode flow channel
HC
HE
E
C
CONV
EyOJ =2inu
2O
2O
DIFF
EyOJ =2
RXN
CyOJ =2
Schematic of a 2D fuel cell transport model including diffusion and convection
Gas is depleted along flow channel
To find oxygen concentration profile along the catalystlayer
40
Convective Transport in Flow Structure :Assumptions
1. Steady state and isothermal operation
2. Flow channel has a square cross section.
3. The catalyst layer is infinitely thin.
4. Water exists only in vapor form.5. Diffusive mass transport dominates in
the diffusion layer. Furthermore, only y-direction diffusion is considered.
6. Convection mass transport dominates inthe flow channel
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21
41
Transport in Flow Structure:Convective Transport
From Faradays Law, if fuel cell is producing a current density at locationX,then the O2mass flux (kg/cm2s) that is consuming is given by
( )F
XjMJ O
rxn
CyXx
O422 ,
^
===
The O2 flux consumed by the electrochemical reaction must be provided bydiffusion in the gas diffusion layer, described by Ficks law
E
EyXxO
CyXxO
eff
O
diff
EyXx
OH
DJ ====
==
= ,,
,
^22
22
O2mass flux due to mass transport through the gas diffusion layer is providedby convective mass transport between the flow channel and gas diffusion layer
=
======
channelyXxO
EyXxOm
conv
EyXx
O hJ,,
,
^
222
(1)
(2)
(3)
42
Mass flux balance between convective transport in flowstructure and diffusion transport in GDL
1. O2mass flux consumed by theelectrochemical reaction at the catalystlayer
2. O2mass flux due to diffusion masstransport through the gas diffusion layer
3. O2mass flux provided by convectivemass transport between the flow channeland gas diffusion layer surface.
Mass f lux 1 Mass flux 2 Mass flux 3= =
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22
43
Transport in Flow Structure:Convective Transport
To maintain the flux balance, O2mass flux in equations 1, 2 & 3 must be same
conv
EyXx
O
diff
EyXx
O
rxn
CyXx
O JJJ======
==,
^
,
^
,
^
222
( )F
XjMJ O
diff
EyXx
O422 ,
^
===
( )eff
O
EOEyXxOCyXxO D
HF
XjM
2
222 4,,=
====
( )
m
OchannelyXx
OEyXx
OhF
XjM
1
4222 ,,=
====
The following relations can be derived
(4)
(5)
(6)
(7)
44
Transport in Flow Structure:Convective Transport
Couple ydirection O2mass transport in the diffusion layer to thexdirection O2mass transport in the flow channel by considering the overall flux balance inthe control volume (dotted box)
O2leaving out of the top of the control volume can be related to the currentdensity produced by fuel cell.
(8)
(9)
(10)
dxJHuHuX
conv
Ey
OchannelyXx
OCinchannelyx
OCin =
==== =
0
^
,,0 222
( ) =
=
X
O
Xconv
Ey
O dxF
xjMdxJ
00
^
422
Combining equations 6, 7, 8 & 9
( ) ( ) ( )
++= ====
X
Cin
eff
O
E
m
O
channelyXxO
CyXxO dx
Hu
xj
D
XjH
h
Xj
F
M
0,,
2
2
22 4
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23
45
Transport in Flow Structure:Convective Transport
Assume current density is constant along thexdirection
hm can be determined based on constant-flux Sherwood number
(11)
(12)
(13)
Final expression for oxygen concentration profile along the catalyst layer
++=
====Cin
eff
O
E
m
OchannelyXx
OCyXx
OHu
X
D
H
hF
jM
2
222
1
4,,
C
OF
mH
DShh 2=
++=
====Cin
eff
O
E
OF
C
OchannelyXx
OCyXx
OHu
X
D
H
DSh
H
F
jM
22
222 4,,
Linear profile
46
Transport in Flow Structure:Convective Transport
Three terms that affect O2concentration profile at thereaction site for fuel cell are1.Inlet flow velocity, uin
Supplying more O2 improves mass transport, thus
increasing O2concentration at the catalyst layer2.Diffusion layer thickness, HE
Decreasing diffusion layer thickness also increases theO2concentration at the catalyst layer.
3.Channel size, HC A little tricky as HCappears in both numerator of the
first term and denominator of third term in theparentheses. However, with constant volume flow rate,uinHC is constant. Therefore, decreasing channel sizewill increase the O2concentration.
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24
47
Transport in Flow Structure:Convective Transport
Flow Structure Pattern
Flow plate typically contain dozen or evenhundreds of fine channels (or groves) tohomogenously distribute gas flow over the fuelcell surface. The shape, size and pattern offlow channels can significantly affect fuel cellperformance.
In PEMFCs, flow field design effort often focuson the water removal capability of cathodeside.
48
Transport in Flow Structure:Convective Transport
Three basic flow structure patterns are1. Parallel flow
Low overall pressure drop between gas inlet andoutlet
However, when the width of the flow field is relativelylarge, flow distribution in each channel may not beuniform
2. Serpentine flow Excellent water removal capability, as only one flow
path exists in the pattern and liquid water is forced toexit the channel
However, in large area cell, serpentine design leadsto large pressure drop
3. Interdigitated flow Promotes forced convection of the reactant gases
through the gas diffusion layer. Far better water management, leading to improved
mass transport Significant pressure drop, but possible to be
overcome by employing extremely small rib spacing.
Inlet
Outlet
Inlet
Outlet
Inlet
Outlet
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Analytical Modelling with MATLAB Motivation
Shorter turnaround time
Accessibility
Working with basics: enhancing understanding
Issues:
Oversimplification
Disregard for physical factors
Incapable of complex reality simulations
Way out:
Compare results with numer