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FEMLAB Chemical Engineering COPYRIGHT 1994 - 2000 by COMSOL AB. All right reserved The software described in this document is furnished under a license agreement. The software may be used or copied only under the terms of the license agreement. No part of this manual may be photocopied or repro-duced in any form without prior written consent from COMSOL AB.

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Printing History: October 2000 First printing FEMLAB 2.0

Contents

1Model Library

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-2Model Library Guide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-3

Chemical Engineering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-4Diffusion of Gas Through a Membrane . . . . . . . . . . . . . . . . . . . 1-4Diffusion in Isothermal Laminar Flow Along a Flat Plate . . . 1-10Monolithic Reactors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-19Concentration Distribution in a Porous Catalyst Pellet . . . . . 1-28Boat Reactor for Low Pressure Chemical Vapor Deposition . . 1-37Tubular Reactor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-52Cylindrical Tubular Reactor with Cooling . . . . . . . . . . . . . . . . 1-59Absorption in a Falling Film . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-64The Chlor-alkali Membrane Cell . . . . . . . . . . . . . . . . . . . . . . . 1-78Model of a Fuel Cell Cathode . . . . . . . . . . . . . . . . . . . . . . . . . . 1-90

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I-1

i

ii Contents

1
Model Library

1 Model Library

1-2

IntroductionThe model library presented here is the first package of FEMLAB models devoted solely to chemical engineering. The models are obtained from the classical chemical engineering literature and, in a few cases, from publications in scientific papers. As a Chemical Engineer, you will feel familiar with the notations and the vocabulary used in this library.

The examples are intended to introduce you to modeling of transport-reaction processes in FEMLAB. You will see that FEMLAB is very well-suited for modeling these type of processes because of its capability of coupling nonlinear equations. In addition, this model library can provide some analogies to your own modeling work and in this way show you how to implement it in FEMLAB.

The examples show you how to model transport by diffusion, convection, and migration. Heterogeneous and homogeneous reactions are introduced to exemplify the relative simplicity of working with these kind of processes in FEMLAB.

Although most of the examples are relatively simple, you will find them fairly realistic in the sense that they couple the true physical processes encountered in chemical systems.

This model library does not give you exact instructions of how to enter a model in FEMLAB. You can find detailed step-by-step instructions, of how to introduce a model in FEMLAB, in the FEMLAB Users’s Guide and Introduction Volume. You can read some further comments on the style used here in the introduction to the general FEMLAB Model Library. For more information on how to use the Graphical User Interface, see the FEMLAB Reference Manual. You can also solve the examples, presented in this library, by using command-line functions, which are described in the FEMLAB User’s Guide and Introduction.

Enjoy your modeling!

Eduardo Fontes

Introduction

Model Library Guide

Model Page

chlor_alkali 1-78 √ √ √ √ √

cvd_reactor 1-37 √ √ √ √ √ √ √ √

falling_film 1-64 √ √ √ √ √

mcfc 1-90 √ √ √ √ √

membrane 1-4 √ √ √

monolithic_reactor 1-19 √ √ √ √ √

pellet 1-28 √ √ √ √ √ √

plate 1-10 √ √ √ √ √

tubular_reactor 1-52 √ √ √

tubular_reactor_cooling 1-59 √ √ √ √

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1 Model Library

1-4

Chemical Engineering

Diffusion of Gas Through a MembraneDiffusion is one of the main transport mechanisms in chemical systems. Molecular diffusion is in many cases the only transport mechanism in microporous catalysts and in some types of membranes. Moreover, the diffusion equation is one of the first partial differential equations that a chemical engineer encounters during his or her education.

In this example, we will model the diffusion of a chemical species through a membrane. The system that we will treat is similar to the one used to measure the permeability of gases in membranes. This type of equipment consists of a two-compartment cell divided by a membrane. The permeability of the membrane is measured by introducing the species of interest in a carrier gas on one side of the membrane and the pure carrier gas on the other side of the membrane. The composition in the two compartments, on each side of the membrane, is measured as a function of time in order to estimate the permeability of the species diluted in the carrier gas.

The problem that we will treat will account for the transport through diffusion in a unit cell of the type of equipment sketched above. The unit cell is a small part of the cell that is representative for the whole system.

membrane

compartment 1 compartment 2

In 1 In 2

membrane support


We will model the concentration profile as a function of time in the membrane, supported by a mesh structure, and in the free gas confined in the mesh support. The supporting structure could be a mesh made of a dense and rigid polymeric material.

Definition of the problemThe domain that we model in this example includes the membrane, the two compartments that are divided by this membrane and the membrane support. We assume that the membrane support makes perfect contact to the membrane and constitutes an obstacle for the diffusion process. We further assume that the compartment between the support and the wall of the cell is closed. This implies that the compartments on each side of the membrane are confined by the membrane support at the top and bottom and by the walls of the measuring cell on the sides. The boundaries that confine our domain are drawn with thick dashed lines in the below figure.

membrane support

compartment 1 compartment 2

Ωmem

Ωcom2Ωcom1

∂Ω

cellwall

cellwall

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One example of a system of the type described above could be used to measure the hydrogen permeability through a membrane. In compartment 1 we would introduce nitrogen with a low concentration of hydrogen. Only nitrogen would be present in compartment 2 at the beginning of the experiment.

The model of our system is obtained by making a mass balance, as a function of time, for hydrogen diluted in nitrogen. We can use the diffusion equation if we neglect the transport of nitrogen through the membrane and the change in pressure due to the transport of hydrogen from one compartment to the other. Furthermore, we neglect convection in the compartments and assume that the membrane contains the same amount of hydrogen per unit volume as the compartment.

The assumptions mentioned above make it possible to describe our system with the following equations:

In these equations c denotes concentration (mole m-3), t time (s), D and Dm the diffusion coefficients in the compartments and in the membrane, respectively (m2 s-1). We can deduct the initial conditions from the above description of the experiment, which gives

We can see from these equations, that the diffusing species, i.e. hydrogen, is introduced in compartment 1 at the beginning of the experiment. We assume that the gases are entrapped in our domain and, in this respect, our system is closed, which results in

where Di in this case represents D or Dm.

c∂t∂

----- D– ∇2c 0= in Ωcom1 and Ωcom2

c∂t∂

----- Dm– ∇2c 0= in Ωmem

c 0 x,( ) co= in Ωcom1

c 0 x,( ) 0= in Ωmem

c 0 x,( ) 0= in Ωcom2

Di∇c n⋅ 0= at Ω∂


Model Library Chemical_Engineering/membrane

Solving the problem using the Graphical User InterfaceSelect the Diffusion Time dependent physics mode.

Options and Settings

• Set axis and grid settings according to following table:

• Enter the following variable names, for later use, in the Add/Edit Variables dialog box

Draw Mode

• Draw a rectangle, R1, with the lower corner at (-0.0005,0) and its upper right corner at (0,0.001).

• Draw a second rectangle, R2, with the lower corner at (0,-0.0005) and its upper right corner at (0.0005,0.0015).

Axis Grid

X min -0.001 X spacing 0.0005

X max 0.002

Y min -0.001 Y spacing 0.0005

Y max 0.002

Name Expression

D 1e-5

Dm 1e-9

c1o 0.3

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• Draw a third rectangle, R3, with the lower corner at (0.0005,0) and its upper right corner at (0.001,0.001).

• Press the Zoom Extents button to adjust the coordinate system with respect to the composite object.

Boundary Mode

• Select all boundaries and set the quantities q and g equal to 0.

This condition implies that there is no flux out of our system.

PDE Mode

• Enter the PDE coefficients, in the subdomains, according to the following table.

We are now ready to mesh our domain.

Mesh Mode

• Initialize the mesh.

• Refine the mesh once.

Solve Mode

• Specify the initial conditions according to following table:

• Specify the Timestepping Output Times in the Solver Parameters dialog box to 0:30:300.

• Solve the problem by pressing the Solve button.

Subdomain 1 2 3

D D Dm D

Q 0 0 0

Subdomain 1 2 3

c(t0) c1o 0 0


Plot Mode

The default plot shows the concentration distribution in our system after 300 seconds:

This figure shows that the variations in concentration are fairly small when the diffusion process has been allowed to take place during 5 minutes.

We can visualize the transport of the diffusing species after 90 seconds by making an arrow plot on top of the surface plot.

• On the Arrow page, select flux (flux_x) and flux (flux_y) as Arrow data for the x and y-expressions, respectively.

• Press the Color button and select white as the arrow color.

• Press OK.

• Set both the Arrow parameters x spacing and y spacing to 30.

• In the General page select Solution at time 90.

• Check the Arrow plot checkbox and press OK.

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Diffusion in Isothermal Laminar Flow Along a Flat Plate

IntroductionIn most industrial systems, the species transport in laminar flow differs by several orders of magnitude between diffusion and convection. In every day life, we can see this phenomenon when we make tea with a tea bag. Without stirring, it takes quite a long time for the leached tea to spread throughout the cup.

This can alternatively be visualized by modeling these transport mechanisms in the simplest possible geometry. Diffusion of a leached species from a flat plate into laminar flow is neatly shown in [1]. The model is similar to an idealized model of the transport of tea, from the surface of the tea bag to the laminar flow that arises in parallel to this surface when we gently lift the tea bag in the cup. Below we can see the

0


idealized velocity profile relative to the surface of the tea bag. The problem can be treated analytically, which however requires a fairly large degree of simplification and does not even provide a straight-forward solution.

In this example, we will treat a similar type of problem with a minimum of simplification. This simple model serves as an introduction to the modeling of systems where a mass balance is coupled to a momentum balance, and where the flux of dissolved species in a fluid is given by diffusion and convection [2].

We will study the concentration and flow distributions along a flat plate in a parallel channel. We will assume that a constant flux of the dissolved species is generated along the flat plate. The solution of the problem will generate the developing structures of the viscous and diffusion layers in the parallel channel.

Definition of the problemWe treat the problem with one diffusing species, dissolved in water at room temperature. The geometry of the domain in this model is the simplest possible; a rectangle of 6 mm times 20 mm, see the below figure. The fluid inlet to our system is situated at the left boundary and the outlet at the right. The plate is represented by the lower boundary and symmetry is assumed at the top. The diffusing species is produced at the surface of the plate.

We use the Navier-Stokes equations, in combination with the continuity equation and a mass balance equation, for one dissolved species. The transport of this species takes place by diffusion and convection. We further assume that the production of the diffusing species, at the

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surface of the plate, does not influence the viscosity and density of the fluid.

The assumptions listed above result in the following equations, for the momentum balance in the domain:

where µ denotes the dynamic viscosity (kg m-1 s-1), v the velocity vector (m s-1), ρ the density of the fluid (kg m-3) and p the pressure (Pa).

We solve these equations of motion coupled to the mass balance equations:

where D denotes the diffusion coefficient (m2 s-1) and c the concentration (mol m-3). We can see that the flux vector, in the brackets on the left hand side of the equation, has a diffusive and a convective contribution.

We obtain the boundary conditions for the equations of motion by assuming a uniform velocity profile at the inlet and constant pressure at

Ω

∂Ωplate

∂Ωfluid

∂Ωoutlet∂Ωinlet

µ∇2v ρ v ∇⋅( )v ∇p+ + 0= in Ω

∇ v⋅ 0= in Ω

∇ D c cv+∇–( )⋅ 0= in Ω

2


the outlet. Furthermore, we assume symmetry along the boundary towards the free fluid and impose that laminar flow is fully developed at the outlet. These assumptions give the following boundary conditions for the equations of motion:

We additionally obtain the corresponding boundary conditions, for the mass balance equations, by assuming that the concentration is known at the inlet and at the symmetry boundary. We also know the production rate of the diffusing species at the surface of the plate, and assume that the dominating transport process in the direction of the flow, at the outlet, is transport by convection. This can be formulated by the following equations:

The condition that determines the concentration at the symmetry boundary is only valid in the case when the diffusion layer does not reach this boundary. This assumption will be validated or falsified in the solution.

Model Library Chemical_Engineering/plate

Solving the problem using the Graphical User InterfaceSelect Incompressible Navier-Stokes, from the Multiphysics page in the Model Navigator, and add it as an application mode by moving it to the right field with the arrow button.

v n⋅ v0= at ∂Ωinlet

v n⋅ 0= at ∂Ωfluid

v 0 0,( )= at ∂Ωplate

p 0= at ∂Ωoutlet

vy 0= at ∂Ωoutlet

c c0= at ∂Ωinlet

c c0= at ∂Ωfluid

D c cv+∇–( ) n⋅ k= at ∂Ωplate

D c cv+∇–( ) n⋅ cv( ) n⋅= at ∂Ωoutlet

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1 Model Library

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• Set the axis and grid settings according to the following table:

• Enter the following variable names, for later use, in the Add/Edit Variables window

Draw Mode

• Draw a rectangle, R1, with lower left corner at (0,0) and upper right corner at (0.02,0.006).

Boundary Mode

• Enter boundary coefficients according to the following table:

Axis Grid


X max 0.021


Y max 0.008

Name Expression

rho 1e3

miu 1e-3

D 5e-9

flux 5.2e-2

vo 0.01

c1o 0

4


• Choose View as Boundary Coefficients in the Boundary menu.

• For boundary 4, enter the value 1 in the edit field for the element component (2,2) of the h coefficient.

• For boundary 4, enter the expression -v in the edit field for the second element component of the r coefficient.

PDE Mode

We can start by defining the coefficients for the Navier-Stokes equations, which are the density and viscosity of the fluid, in this case water.

• Enter the PDE coefficients, in subdomain 1, according to the following table.

We are now ready to define the mass transport equations for the diffusing species that is being produced at the surface of the plate.

Multiphysics Mode

• Choose PDE, general form, label your application mode with the name, massbal, and your dependent variable, c1.

• Add the new application by moving it to the right field with the arrow button.

Boundary 1 2 3 4

Type Inflow No-slip Slip Outflow

u vo

v 0

p 0

Subdomain 1

ρ rho

η miu

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1 Model Library

1-1

Boundary Mode

• Deselect View as Boundary Coefficients in the Boundary menu.


PDE Mode

We define the flux vector, in the PDE Mode, as an expression that consists of a diffusion and a convection term.

• Enter the PDE coefficients according to the following table:

Mesh Mode

This example requires a fairly dense mesh since the Reynolds number is relatively high. For that purpose, we define a maximum element size for the edge that represents the flat plate. This requires that we identify the edge number of our geometry, see Option/Labels, Show Edge Labels in the Boundary Mode.

• From the Mesh menu, choose Parameters and press the button labeled More.

• In the edit field Max element size for edges, set the maximum element size by entering 2 1e-4.

• Press Remesh.


Boundary 1,3 2 4

G 0 flux -c1.*u

R -c1+c1o 0 0

Subdomain 1

Γ -D.*c1x+c1.*u -D.*c1y+c1.*v

F 0

da 0

6


Solve Mode

• Choose Stationary nonlinear from the Solver Parameters dialog box.

• Set the Tolerance to 1e-8.

• Turn streamline diffusion on.

• Solve the problem

Plot Mode

The default plot shows the concentration of the reacting species in the solution (in mole m-3). The most interesting result from this simulation is a comparison between the thickness of the viscous layer and of the diffusion boundary layer, which is often given in the Schmidt number (Sc). We can obtain a notion of this relation by plotting the results in a 3-D surface plot. To do this, choose to plot the velocity field as Surface expression, and the concentration as Height expression in the Plot Parameters dialog box. After rotating we obtain the resulting plot:

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The difference between the viscous and diffusion layers can be clearly seen, in the figure above, by the amount that they extend into the fluid. This difference can be seen even more clearly be if we reverse the plotting instructions. In order to do this, use the concentration as Surface expression and the velocity field as Height expression in the Plot Parameters dialog window. Rotate freely to generate a suitable view.

In addition, we can see in the above figure that we have a parabolic velocity profile at the outlet of the domain, which supports our assumption of fully developed laminar flow along this boundary.

References [1] R. Bird, W. Stewart and E. Lightfoot, “Transport Phenomena”, John Wiley & Sons, New York, 1960.

[2] G. F. Froment and K. B. Bischoff, “Chemical Reactor Analysis and Design“, Second Edition, John Wiley & Sons, 1990.

8


Monolithic Reactors1

The importance of monolithic reactors has grown rapidly the last two decades. They serve mainly as tools for environmental protection, but the development of new heat generation techniques offers a new role for them.

The most well-known example of a monolithic reactor is the three-way catalytic converter for automobiles. This reactor simultaneously transforms unburned hydrocarbons, carbon monoxide, and nitrogen oxides from the exhaust gases into carbon dioxide, nitrogen, and water.

Catalytic purification of emissions from other sources has also been developed and commercialized in the last decade. One important area of applications is the abatement of volatile organic compounds (VOC) from processes such as spray painting, offset printing and coating operations. Selective catalytic reduction (SCR) of nitrogen oxides by ammonia is another area where monolithic reactors are used. This technique has been widely applied to purify effluent gases from coal burning boilers and electric power plants.

Another challenging application for monolithic reactors is the current development of high-temperature catalytic combustion for heat and power generation. This technology has proven to be a promising alternative to ordinary flame combustion for converting chemical energy into heat or mechanical power with a minimum level of emissions from combustion by-products.

A monolithic reactor consists of thin parallel straight channels of arbitrary shape, through which the gas, containing the reactants, flows. The walls of the channels are coated with a porous ceramic containing the catalyst layer. The transition from reactants to products involves transport of the reactants by convective flow in the channels and molecular diffusion towards the channel walls. Simultaneous diffusion and reaction occur inside the porous washcoat whereby the products diffuse back into the gas and are transported out from the reactor.

There is obviously a large number of variables in optimizing the reactor performance. Examples of parameters that are important for the general

1. Model provided by Dennis Papadias, The Royal Institute of Technology, Stockholm.

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behavior of the reactor are temperature, channel geometry, flow rate, properties of the washcoat and catalytic activity.

3-D Reactor ModelThe following model simulates the catalytic abatement of a volatile organic compound (VOC), in this case a contaminant in a waste gas. For simple geometries the reactor can be described by a 2-D model using time as the z-axis. 3-D modeling is necessary if the channels have irregular shape or if the level of active coating is non-uniform along the monolith channels.

Model Assumptions

Oxidation for one type of VOC is considered in this model. Since the concentrations of contaminants from the waste gases usually are very low, the heat release from the oxidation is negligible. The flow in the monolith channel is laminar and the axial diffusion of the reactants is small compared to the convective flow and can thus be neglected.

We assume identical conditions within each channel so the model of the whole monolith is reduced to one channel. The transversal cross-section of the channel is described by a circle surrounded by a square. The area between the circle and the square represents the washcoat of the reactor. We will choose a radius of the monolith channel of 0.625 mm and the square will have sides of length 0.645 mm. Due to symmetry effects, only one octant of the channel will be studied in the simulation.

A) Monolithic reactor, B) Channel wall, C) Porous ceramic (washcoat) with catalyst. Figure from [1].

0


Boundary and Inlet Conditions

The conditions at the channel inlet, z=0, are uniform concentration and a fully developed velocity profile. The velocity can be analytically derived using the Hagen-Poiseuille law. For a circular tube we get the parabolic profile given by the expression:

In this equation Vz describes the velocity (m s-1) profile, and Um is the average velocity given in dimensionless coordinates x and y. For all outer boundaries, Neumann boundary conditions for symmetry are used.

Gas Phase Equation

The equation describing the process for the gas in the monolith channel is

where U(x,y) is the velocity profile in the monolith channel, D the diffusion coefficient in the free gas (m2 s-1), and c the concentration (mol m-3).

Solid Phase Equation

The equation for the washcoat,

describes the simultaneous reaction and diffusion in the porous network of the washcoat. The washcoat is treated as a homogeneous medium where Deff, which denotes the effective diffusion coefficient, reflects the pore properties of the washcoat. Since the concentration of the contaminant species is very low, the reaction term can be assumed to be of first order (linear) with respect to the concentration, c. This is expressed in the rate equation where Q denotes the reaction rate, k the pre-exponential factor, E the activation energy, T temperature, and R

Vz 2Um 1 x2– y2

–( )=

U x y,( ) c∂z’∂------ x’∂

∂ D c∂x’∂-------

y’∂∂ D c∂

y’∂------- += Ω1

x’∂∂ Deff c∂

x’∂-------

y’∂∂ Deff c∂

y’∂-------

Q+ + 0= Ω2

Q ke

ERT---------–

c–= Ω2

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the gas constant. The temperature dependence of the rate of the reaction is accounted through the Arrhenius law.

Scaling

We choose the following scaled variables: V(x,y)=U(x,y)/Um, x=x’/Ra, y=y’/Ra, z=z’/L and C=c/C0 where Um is the mean inlet velocity, L the reactor length (m), Ra the radius (or maximum length) of the monolith channel, and C0 the inlet concentration. By rearranging the equations, we obtain the following set of equations:

In this example we will use the following parameter values: K1=0.39, K2=11.2, D=0.88 (in this case cm2 s-1), and Deff=0.02 (cm2 s-1).

Due to the absence of a second derivative in the z-axis, the DAE-system (differential algebraic equation system) can be integrated by using a time dependent model in FEMLAB with time representing the z-axis.

Note that in case of other geometric configurations for the monolith channel, the velocity profile has to be computed separately and the solution used in the equation for V(x,y).

Model Library Chemical_Engineering/monolithic_reactor

V x y,( )K1C∂z∂

-------x∂

∂ D C∂x∂

-------

y∂∂ D C∂

y∂-------

+= Ω1

x∂∂ Deff C∂

x∂-------

y∂

∂ Deff C∂y∂

------- K2C–+ 0= Ω2

K1UmRa2

L--------------------= Ω1

K2 ke

ERT---------–

Ra2= Ω2

2


Using the Graphical User InterfaceSelect the 2-D, Coefficient, Time dependent PDE mode in the Model Navigator.


• Set axis and grid settings according to the following table:

• Enter the following variable names and expressions:

Draw Mode

• Draw a circle, C1, centered at (0,0) with a radius of 1.

• Draw a triangle, CO1, with the corner points (0,0), (1.032,1.032), (0,1.032).

• Form the composite object CO2 using the set formula CO1+CO1*C1.

Axis Grid

X min -1.5 X spacing 1

X max 1.5 Extra X 1.032

Y min -1.1 Y spacing 1

Y max 1.1 Extra Y 1.032

Name Expression

D 0.88

Deff 0.02

K1 0.39

K2 11.2

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• Press the Zoom Extents button.

Boundary Mode

• Enter boundary coefficients according to the following table.

PDE Mode

• Enter PDE coefficients according to the following table.

Boundary All

Type Neumann

q 0

g 0

Subdomain 1 2

c D Deff

a 0 K2

4


Mesh Mode

We need a finer mesh in subdomain 2 where the reaction takes place.

• In the Mesh Parameters dialog box, enter 1 0.1 2 0.015 as Max element size for subdomains.


Solve Mode

• Set the initial condition to 1 for both subdomains.

• Verify that Output times is 0:0.1:1 and that Timestepping algorithm is ode15s, and set Relative Tolerance to 0.001.

f 0 0

da K1*2*(1-x.^2-y.^2) 0

Subdomain 1 2

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• Solve the problem.

Plot Mode

The default plot shows the outlet concentration at time 1, i.e., at the outflow of the reactor. Observe that the concentration in the solid-phase drops rapidly towards zero close to the monolith wall. This is because the reaction in this model is quite rapid and the reacting gas is consumed near the outer surface. It would therefore be enough to coat the channels with a smaller amount of washcoat.

• To see the concentration along the whole channel (from z=0 to z=1), press the Animation button.

Post Processing

To produce a slice plot of the full 3-D geometry, export the FEM structure as fem and type

clf; ax=gca;for i=0:10h=postplot(fem,’tridata’,’u’,’solnum’,i+1,...’tridlim’,[0 1],’parent’,ax);set(h,’zdata’,ones(size(get(h,’xdata’)))*i/10)

end

6


view(3), axis off

It is important to investigate the amount of the contaminant gas flowing out from the reactor. The purpose is to convert almost all contaminant into carbon dioxide and water. The quantity we really can measure out from the reactor is the bulk (mean) concentration of the contaminant. It is given by the following equation,

This integration for the bulk concentration is done in FEMLAB by typing

postint(fem,’u.*(1-x.^2-y.^2)’,’sdl’,1)/postint(fem,’(1-x.^2-y.^2)’,’sdl’,1)

<C>VC sd

s∫V sd

s∫---------------------=

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giving a dimensionless outlet mean concentration of 0.1588. To improve the catalytic purification of the reactor several parameters can be altered, e.g., the length of the reactor, the temperature, or the shape of the channels.

References [1] Berg, M., Catalytic combustion over high temperature stable metal oxides, 1996, Licentiate of engineering thesis, KTH, TRITA-KET R50.

Concentration Distribution in a Porous Catalyst Pellet

IntroductionPorous catalyst particles are widely used in the chemical industry and are extensively treated in the chemical engineering literature, see [1-5]. The catalyst pellets are, in most cases, fluidized by the action of a gas or liquid flowing through a reactor. The fluid enters at the bottom of a bed of catalyst particles and the particles are fluidized by the shear force that the fluid exerts on their surface. At the same time, species dissolved in the fluid react on the surface and inside the porous catalyst particles.

In this study, we will model a cylindrical catalyst particle, subjected to the flow of a reacting gas. We will resolve the flow pattern and concentration field in the free fluid, surrounding the particle, and the concentration field both inside and outside of the particle.

Definition of the problemThe below figure shows the cross-section of a cylindrical catalyst pellet with internal and external mass transfer resistances; a typical problem from the chemical reaction engineering literature. This problem is usually solved by assuming a uniformly thick diffusion layer around the porous particle [1-5]. This requires a mass transfer coefficient, kg (m s-1), to describe transport through the diffusion layer, which is estimated from the transport properties of the fluid along with the operational conditions.

8


We will resolve the flow pattern around the porous particle using the Navier-Stokes and continuity equations (equations of motion). We will couple the equations of motion and the mass balance equations through the multiphysics feature. This will result in us obtaining the structure of the diffusion layer from the simultaneous solution of these equations.

Below is a schematic representation of the domain that we will model. The gas is introduced at the bottom of the picture and exits at the top. Symmetry is assumed at the lateral boundaries, implying symmetry in the y-axis and that the modeled unit cell is typical for the system. Furthermore, we assume that the density of the fluid does not change due to the chemical reaction in the pellet.

δ

c

cs

cb

∂Ωpl

∂Ωinlet

∂Ωoutlet

∂Ωff,2∂Ωff,1

Ωff

Ωpl

∂Ωpl,1

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The equations that we will use in the free fluid sub-domain, denoted by Ωff, are

where µ denotes the dynamic viscosity (kg m-1 s-1), v the velocity vector (m s-1), ρ the density of the fluid (kg m-3), and p denotes pressure (Pa).

We will couple the equations of motion to the mass transport equation, for the reactant in the fluid at steady-state, through:

where D denotes the diffusion coefficient (m2 s-1) and c the concentration (mol m-3). The expression within the brackets on the left hand side is the flux vector for the reacting species. We define the mass balance in the pellet by using an effective diffusion coefficient in the porous media:

where Deff denotes the effective diffusion coefficient in the pellet (m2 s-1) and k the rate constant for the second order reaction (m3 s-1 mol-1). Deff is related to the porosity and tortuosity of the pellet.

We will define the boundary conditions from the assumptions of symmetry and from the direction of the gas flow, from the bottom to the top of the domain. For the equations of motion we obtain the following boundary conditions:

where ∂Ωpl is a subdomain boundary, since the Navier-Stokes and continuity equations are not defined in the pellet domain.

µ∇2v ρ v ∇⋅( )v ∇p+ + 0= in Ωff

∇ v⋅ 0= in Ωff

∇ D c cv+∇–( )⋅ 0= in Ωff

∇ Deff c∇–( ) kc2–⋅ 0= in Ωpl

v n⋅ v0= at ∂Ωinlet

v n⋅ 0= at ∂Ωff 1, and ∂Ωff 2,

v 0 0,( )= at ∂Ωpl 1,

p 0= at ∂Ωoutlet

0


We can find the boundary conditions for the mass transport equation by assuming that we know the inlet concentration. In addition, we can assume that the reactant transport, in the gas at the outlet, is mainly driven by convection, i.e., we neglect diffusion in the main direction of the convective flow:

We will not scale the above equations, which we could have done to get dimensionless variables and a better conditioned problem. However, in this case, we will see that this is not needed.

Model Library Chemical_Engineering/pellet

Solving the problem using the Graphical User InterfaceSelect Incompressible Navier-Stokes from the Multiphysics page in the Model Navigator, and add it as an application mode by moving it to the right field with the arrow button.



Axis Grid


X max 0.003 Extra X 0.0009


Y max 0.007 Extra Y 0.0021 0.0039

c c0= at ∂Ωinlet

D c cv+∇–( ) n⋅ 0= at ∂Ωff 1, and ∂Ωff 2,

Deff c∇–( ) n⋅ 0= at ∂Ωpl

D c cv+∇–( ) n⋅ cv( ) n⋅= at ∂Ωoutlet

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Draw Mode

• Deselect solid by double-clicking on SOLID on the Status bar.

• Draw an arc by clicking at the corners (0,0.0039), (0.0009,0.0039), (0.0009,0.003), (0.0009,0.0021) and (0,0.0021).

• Select solid by double-clicking on SOLID on the Status bar.

• Draw a rectangle, R1, with the lower left corner at (0,0) and upper right corner at (0.002,0.006).

PDE Mode

In order to deactivate the Navier-Stokes and Continuity equations in the pellet subdomain, we will enter the PDE mode before specifying the boundary conditions.


Name Expression

rho 0.66

miu 2.6e-5

D 1e-5

Deff 1e-6

k 100

vo 0.1

c1o 1.3

Subdomain 1

ρ rho

η miu

2


• Deactivate the Navier-Stokes and continuity equations in subdomain 2 by uncheking the Active in this subdomain option.

By this procedure, we automatically get the surface of the pellet as a boundary for the equations of motion and continuity.

Boundary Mode


We are now ready to define the mass transport equations for the reacting species in the gas phase. We will do this by adding a new model equation in the Multiphysics Mode.

Multiphysics Mode

• Choose PDE, general form, label your application mode with the name massbal, and your dependent variable c1.


PDE Mode

In order to activate the transport equations in the pellet subdomain, we will again enter the PDE mode before specifying the boundary conditions.

Boundary 2 1,4,6 5 7-8

Type Inflow Slip Outflow No-slip

u 0

v vo

p 0

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By this procedure, we automatically get the proper boundaries for the mass balance equations.

Boundary Mode


Mesh Mode

This example requires a fairly dense mesh since the Reynolds number is relatively high. For that purpose, we will define the maximum size for some of the geometry edges. This requires that we identify the edge number of our geometry, see Option/Labels, Show Edge Labels from the Boundary Mode. In Mesh Mode:

Subdomain 1

Γ -D.*c1x+c1.*u -D.*c1y+c1.*v

F 0

da 0

Subdomain 2

Γ -Deff.*c1x -Deff.*c1y

F -k.*c1.^2

da 0

Boundary 1,3,4,6 2 5

G 0 0 -c1.*v

R 0 -c1+c1o 0

4


• Choose Parameters and press the More button.

• In the Max element size for edges edit field, set the maximum edge size for edge by entering 6 1e-4 7 5e-5 8 5e-5.

• Press Remesh.


Solve Mode

• Choose Stationary nonlinear from the Solver Parameters window.

• Set the Tolerance to 1e-6.


Plot Mode

The default plot shows the concentration of the reacting species in the free fluid and in the pellet (in mole m-3). We can see from this figure that the diffusion layer, in the free fluid surrounding the particle, is of non-uniform thickness. This results in a non-symmetrical concentration

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distribution along the outer surface of the particle. The diffusion layer at the particle surface, facing the inlet of the gas, is far thinner than that obtained behind the particle. Moreover, depletion of the reacting gas, along the main direction of flow, influences the concentration around the surface of the particle.

The flow pattern around the particle is shown in the plot below. The color bar represents the modulus of the velocity vector, while the arrows symbolize the flow velocity vector. We can see from this figure that the expected maximum velocity occurs in the contraction between the particle and the symmetry line on the right hand side. There are two stagnation zones in front and behind the particle, where the zone behind the particle is somewhat more extended.

We can conclude from this example that it is easy to model diffusion-convection-reaction processes by using the Multiphysics feature. We have also shown that it is simple to exclude selected equations in certain subdomains.

6


References[1] H. Scott Fogler, “Elements of Chemical Reaction Engineering”, Third Edition, Prentice Hall, 1999.

[2] R. B. Bird, W. E. Stewart and E. N. Lightfoot, “Transport Phenomena”, John Wiley & Sons, 1960.


[4] O. Levenspiel, “Chemical Reaction Engineering”, Third Edition, John Wiley & Sons, 1990.

[5] R. Aris, “Introduction to the Analysis of Chemical Reactors“, Prentice Hall, 1965.

Boat Reactor for Low Pressure Chemical Vapor DepositionChemical vapor deposition (CVD) is an important step in the process of manufacturing microchips. One of the common applications is the deposition of silicon on wafers at low pressure. Low pressure reactors are used to get a high diffusivity of the gaseous species, which results in a uniform deposition thickness, since the process becomes limited by the deposition kinetics [1,2].

In this example we will model the momentum and mass transport coupled to the reaction kinetics for the deposition process. We will treat

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a low pressure boat reactor, and the goal of the simulation is to describe the rate of deposition as a function of the fluid mechanics and kinetics in the system. The figure above shows a schematic picture of a boat reactor and the principle of the deposition process.

The gas, in our case silane, enters the reactor from the left and reacts on the wafers to form hydrogen and silicon. The remaining mixture leaves the reactor at the outlet on the right. The deposition of silicon, on the surface of the wafers, depends on the local concentration of silane, which is determined by the operational conditions of the reactor.

More details about this example can be obtained in “Elements of Chemical Reaction Engineering” by H. Scott Fogler [1].

Definition of the problemIn order to obtain a fairly transparent model we have to first make a few simplifying assumptions in the definition of our system. We assume that the density of the gas is constant throughout the reactor, which implies that the reacting gas is either diluted in a carrier gas or that the conversion in the reactor is small. Moreover, we only account for the mass balance of the reactant gas, in this case silane, but it is simple to extend the model and include the mass balance of hydrogen. We also assume constant temperature in the reactor.

It is very time consuming to geometrically model every single wafer in the wafer bundle, and to simplify the geometrical description we treat the wafer bundle as an anisotropic porous medium. We assume in this description that there is no axial diffusivity in the wafer bundle, since silane gas can not diffuse through the wafers. Furthermore, we correct the diffusivity in the radial direction according to the degree of packing in the bundle. We neglect the transport by convection inside the wafer bundle. The wafers are supported mechanically by a support boat that holds the wafers in the center of the reactor. If we neglect the influence of the support boat on the transport process we obtain a rotational symmetrical problem that reduces our 3-D model to a 2-D problem. The rotational symmetric domain that we will treat is shown in the below figure.

8


In the figure, our domain is bounded by the walls of the reactor, ∂Ωwall, the inlet, ∂Ωin, the outlet, ∂Ωout, the symmetry line, ∂Ωsym, and the inner walls that represent the wafer support on both ends of the wafer bundle, ∂Ωiw. In addition, our domain consists of two subdomains; the free fluid subdomain, Ωff, and the wafer subdomain, Ωwf.

The chemical reaction that we account for in this example is given below:

The rate of this reaction is dependent on the partial pressure of silane and the temperature in the reactor.

The assumptions mentioned above in combination with the chemical reaction for the deposition process makes it possible to define our equation system. The momentum balance and the continuity equations for laminar flow in cylindrical coordinates gives:

SiH2 g( ) Si s( ) H2 g( )+=

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where µ denotes the viscosity (kg m-1 s-1), ρ the density of the gas (kg m-3), u the component of the velocity vector in the x-direction, v the component of the velocity vector in the r-direction (m s-1), and p pressure (Pa).

We can define the mass balance equations in cylindrical coordinates by the following equation in the free fluid subdomain

where D denotes diffusivity (m2 s-1) and c the concentration of silane (mole m-3). We obtain the corresponding mass balance equation for the wafer bundle subdomain by neglecting transport by convection and adding the rate of the reaction of silane. This gives us:

where the effective diffusivity, Deff, has components in the x and r directions. In the equation above, k denotes the rate constant for the reaction (m s-1), and Sa the specific surface area (m2 m-3).

We can see in the equations above that the momentum balance is not defined in the wafer subdomain and that the mass balance equations are different in the two subdomains.

We will solve the system of equations defined above by using the proper boundary conditions. In laminar flow we obtain no-slip conditions for the surface of the reactor wall:

x∂∂ µr u∂

x∂------–

r∂∂ µr u∂

r∂------– ρru u∂

x∂------ ρrv u∂r∂------ r p∂

x∂------+ + + + 0= in Ωff

x∂∂ µr v∂

x∂------–

r∂

∂ µr v∂r∂

------– µ

r---v ρru v∂

x∂------ ρrv v∂

r∂------ r p∂

r∂------+ + + + + 0= in Ωff

r u∂x∂

------v∂r∂

------+ v+ 0= in Ωff

x∂∂ Dr c∂

x∂-----–

r∂

∂ Dr c∂r∂

-----– ru c∂

x∂----- rv c∂

r∂-----+ + + 0= in Ωff

x∂∂ Dx

effr c∂x∂

-----–

r∂∂ Dr

effr c∂r∂

-----– rkSac+ + 0= in Ωwf

0


In this equation the first boundary denotes the boundary between the reactor wall and the free fluid, the second denotes the boundary between the inner wall, and the free fluid, and the third the boundary between the wafer bundle and the free fluid. In addition we have:

The last three conditions for the momentum balance and continuity equations are

For the mass balance equations we have:

where Di represent the diffusivity in the free fluid and inside the wafer bundle. This equation implies that we have defined no-flux conditions since there is no transport by convection perpendicular to these boundaries. At the inlet we know the composition of the gas, which yields

At the outlet we assume that the transport of species takes place mainly by convection and neglect the concentration gradients perpendicular to this boundary:

At this stage we have defined our domain equations and all the boundary conditions.

u v( , ) 0 0( , )= at Ω∂ wall, Ω∂ iw ff, and Ω∂ wf ff,

v 0= at Ω∂ sym

u uo= at Ω∂ in

v 0= at Ω∂ in

p po= at Ω∂ out

Dxi c∂

x∂----- Dr

i c∂r∂

-----–,– n⋅ 0= at Ω∂ wall, Ω∂ sym and Ωiw∂

c co= at Ω∂ in

D c∂x∂

----- D c∂r∂

-----–,– n⋅ 0= at Ω∂ out

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Before we solve our problem in FEMLAB, we have to treat some of the parameters that we are going to use as input to our model. These parameters are the specific surface for the wafer bundle, the effective diffusivity in the wafer bundle, and the rate of reaction for the deposition process.

We can calculate the specific surface area, which is the area per unit volume, of the wafer bundle by assuming a certain pitch between the wafers. This gives us the following expression:

where Sunit cell denotes the surface area of the unit cell, Vunit cell the volume and Sa the specific surface area.

We further make an estimation of the effective diffusivity, in the radial direction of the wafer bundle, by calculating the cross-sectional area of free gas in relation to the total cross sectional area. In our case this relation is fairly simple, since there is no tortuosity between the wafer bundles. We can therefore use the following approximation for the effective diffusivity in the radial direction:

We can obtain the deposition rate of silicon, as a function of the partial pressure of silane at 600 oC and a total pressure of 25 Pa, from [2]. If we assume a first order reaction, which is a fairly good assumption for our

Dreff

dwwdcc

-----------D= in Ωwf

2


case according to the experimental data [2], we get a rate constant, k, of 8.06⋅10-3 m s-1.

Model Library Chemical_Engineering/gallery/cvd_reactor

Solving the problem using the Graphical User InterfaceSelect PDE, general form from the Multiphysics page in the Model Navigator, and enter the Application mode name ns and the Dependent variables u v p in the respective edit fields. Add these equations as an application mode by moving it to the right field with the arrow button.

We have to add a second application mode of the PDE, general form for the mass balance equation of silane. The reason for adding this equation in a separate application mode, and not as a fourth variable in the above system, is that we will define the rotational symmetric Navier-Stokes equations and the mass balance equation in different subdomains. To add the mass balance equation; select PDE, general form from the Multiphysics page in the Model Navigator, and enter the Application mode name massbal and the Dependent variables c1 in the respective edit fields. Add this equation as an application mode by moving it to the right field with the arrow button and press OK. Select 3 Variables general form PDE (ns) from the Multiphysics menu.



Axis Grid


X max 0.2 Extra X 0.035 0.145


Y max 0.1 Extra Y

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Name Expression

dcc 2.5e-3

Dn 5e-5

dw 5e-4

k1 8.06e-3

miu 3.1e-5

Mn2 28e-3

Msil 30e-3

Pr 25

Ra 0.04

Rg 8.31

Te 600

vo 1

c1o 0.2*Pr/Rg/(Te+273)

D Dn*1.013e5/Pr

Deff D*(dcc-dw)/dcc

rho Pr*(0.2*Msil+0.8*Mn2)/Rg/(Te+273)

Re Ra*vo*rho/miu

Sa 2*(Ra^2+Ra*dw)/Ra^2/dcc

4


Draw Mode

• Press the Draw Line button on the draw toolbar and draw a line by clicking at the positions (0,0.06) and (0.14,0.06) with the left mouse button.

• Press the Draw Arc button and draw an arc by clicking at the positions (0.14,0.06), (0.18,0.06) and (0.18,0.02) with the left mouse button.

• Press the Draw Line button and click at the positions (0.18,0.02), (0.18,0) and (0,0).

• Click the right mouse button to create a solid object, CO1.

• Draw a rectangle, R1, with the lower left corner at (0.035,0) and upper right corner at (0.04,0.04).

• Draw second rectangle, R2, with the lower left corner at (0.04,0) and upper right corner at (0.14,0.04).

• Draw a third rectangle, R3, with the lower left corner at (0.14,0) and upper right corner at (0.145,0.04).

• Press the Draw Point button and click at the position (0,0.02).

• In the Draw menu select Create Composite Object..., for the object type Solids, and enter CO1+R2-R1-R3 in the Set formula: edit field.

PDE Mode

In order to deactivate the Navier-Stokes and continuity equations in the wafer subdomain, we will enter the PDE mode before specifying the boundary conditions.

• Enter the Γ vector, in subdomain 1, according to the following table.

Γ Expression

ga(1) -miu.*y.*ux -miu.*y.*uy

ga(2) -miu.*y.*vx -miu.*y.*vy

ga(3) 0 0

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• Enter the F vector, in subdomain 1, according to the following table.

• Deactivate the Navier-Stokes and continuity equations in subdomain 2 by uncheking the Active in this subdomain option.

Boundary Mode

• Enter boundary coefficients according to the following table in the corresponding edit fields.

PDE Mode

• Select 1 variable general form PDE (massbal) in the Multiphysics menu.

F Expression

F(1) y.*(-rho.*u.*ux-rho.*v.*uy-px)

F(2) -miu.*v./y+y.*(-rho.*u.*vx-rho.*v.*vy-py)

F(3) y.*(ux+vy)+v

Boundary 1 2,13 3-6,9,11,12,15 14

G(1) 0 0 0 0

G(2) 0 0 0 0

G(3) 0 0 0 0

R(1) -u+vo 0 -u 0

R(2) -v -v -v 0

R(3) 0 0 0 -p

6




Boundary Mode


Mesh Mode

This example requires a fairly dense mesh and we will therefore define the maximum mesh edge size for some of the geometry edges. This requires that we identify the edge number of our geometry, use Option/Labels, Show Edge Labels from the Boundary Mode. In Mesh Mode:

• Choose Parameters and press the More button.

Subdomain 1

Γ -D.*y.*c1x -D.*y.*c1y

F -y.*(u.*c1x+v.*c1y)

da 0

Subdomain 2

Γ 0 -Deff.*y.*c1y

F -(k1*Sa).*y.*c1

da 0

Boundary 1 2-8,10-15

G 0 0

R -c1+c1o 0

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• In the Max element size for edges edit field, set the maximum edge size for edges 1, 9 and 14, by entering 1 3e-3 9 3e-3 14 3e-3.

• Press Remesh.

• Refine and jiggle the mesh and repeat these procedures once.

Solve Mode


• Set the Nonlinear Tolerance to 1e-13 in the Nonlinear page.

In order to get a faster convergence we will define some initial values for the non-linear solver.

• Select Specify Initial Conditions from the Solve menu.

• Enter the initial values according to the following table:


Plot Mode

The default plot shows the concentration profile of silane in the reactor. We can see from this graph that the conversion of silane is fairly small in the reactor, which strengthens our assumption that the consumption of silane and the production of hydrogen do not affect the volume of the gas mixture to a larger extent. We can see from the graph that the outlet concentration of silane is 75% of the inlet concentration.

Subdomain 1 2

u(t0) 0

v(t0) 0

p(t0) Pr.*(0.18-x)./0.18

c1(t0) c1o c1o

8


We can view the flow distribution in the reactor by plotting the modulus of the velocity vector and the velocity vector.

• Select the Surface page in the Plot Parameters dialog box.

• Enter sqrt(u.^2+v.^2) in the Surface expression edit field.

• Select the Arrow page.

• Enter u and v in the x and y expression edit fields, respectively.


• Press OK.



We can see from this graph that the absolute value of the velocity vector decreases at the periphery of the reactor, compared to the inlet, since the cross sectional area increases with the square of the radius.

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Post Processing

We are interested in the deposition rate on the surface of the wafers, which are only defined in subdomain 2. We can obtain the deposition rate, in the domain of interest, by interpolating the concentration profile on the position of the wafer bundle and calculating the rate of deposition from the reaction kinetics. We do this by exporting our solution to the MATLAB work space as fem. This MATLAB script generates the proper plot:

r1=rect2(0.04,0.14,0,0.04);femR.geom=r1;femR.mesh=meshinit(r1);femR.mesh=meshrefine(femR);femR.mesh=meshrefine(femR);femR.mesh=meshrefine(femR);re=posteval(fem,’c1’);cR=postinterp(fem,re,femR.mesh.p);femR.sol.u=((8.06e-3*60*28e-3/2e3*1e9).*cR)’;postsurf(femR);xlabel(’x [m]’);ylabel(’r [m]’);

0


grid on;view(30,40);axis([0.035 0.14 0 0.045 min(femR.sol.u) max(femR.sol.u)]);title(’Deposition rate, [nm/min]’)

The figure above gives us the deposition rate in nm min-1 in the wafer bundle. We can see that the highest deposition rate is obtained at the position of the entrance of the silane gas in the annular region of the reactor. There is also a tendency of higher deposition rate along the annular part throughout the whole reactor. The difference in deposition rate is about 0.1 nm min-1 between the outer part and the center of the wafers. The difference along the reactor, in the main direction of the convective flow, is about 0.5 nm min-1.

The rate of transport by diffusion is of the same order of magnitude as the transport by convection, at the operational pressure of 25 Pa. This is shown in the below figure, where we have changed the inlet flow rate of

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gas from 1 m s-1 to 2 m s-1. The rate of deposition is not dramatically changed by the change in flow rate in the reactor.

References[1] H. Scott Fogler, “Elements of Chemical Reaction Engineering”, Third Edition, Prentice Hall, 1999.

[2] A.T. Voutsas and M. K. Hatalis, “Structure of As-Deposited LPCVD Silicon Films at Low Deposition Temperatures and Pressures“, J. Electrochem. Soc., Vol. 139, No. 9, 1992.

Tubular ReactorIn a tubular gas reactor, chemical reactions take place while the reactants are carried in a gas stream from the inlet to the outlet. Mass and energy transport occur through a convection-diffusion process. In this example we will treat a simplified model of a tubular reactor.

2


In the model, we will assume that radial variations in compositions and temperature are small, due to plug flow and heat-insulated walls.

The gas velocity is v, and the density and specific heat are ρ and C. The concentration c of the reactant is influenced by convection, diffusion and reaction as described by

where D denotes the diffusion coefficient and Q the reaction term. We further assume that the reaction can be modeled by a first order rate law where the temperature dependence is given by the Arrhenius equation. In this equation, E denotes the activation energy and k the pre-exponential factor. The assumption of first order kinetics is plausible in combustion of low concentration fuel in an oxygen-rich environment.

Heat transfer by convection and conduction, where thermal conductivity is denoted λ, and heat production, given by the reaction enthalpy ∆H, is modeled by

The influx gas has temperature T0 and concentration c0. If we assume that the main transport of reactants into the reactor takes place by convection, we obtain following boundary condition for the flux at the inlet:

v

(c,T)

x

t∂∂c v

x∂∂c

+x∂

∂ Dx∂

∂c Q,–= Q kce

ERT---------–

=

ρcp t∂∂T v

x∂∂T

+

x∂∂ λ

x∂∂T

Q H∆+=

vc 0 t,( ) Dx∂

∂c– vc0=

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Similarly, for the heat flux, we will have:

The gradients at the exit are small, i.e.:

We choose intrinsic scales for the variables, length scale = L, time scale = D/L2, temperature scale T0, and concentration scale c0.

We can asses the importance of the various processes in the model from the values of the coefficients in the scaled, dimensionless problem:

Thermal conduction/diffusion

“Thermal” Peclet number, convection/conductive heat transfer

Peclet number, convection rate/diffusion rate

Reaction heat production/diffusion

Reaction production rate/diffusion rate

Model-characteristic temperature constant

ρCT 0 t,( ) λx∂

∂T– ρCvT0=

x∂∂c 0

x∂∂T 0=,=

r0λ

ρCD-------------=

r1vLρC

λ---------------

r2r0-----= =

r2vLD-------=

B1∆Hc0L2k

DρCT0-------------------------=

B2kL2

D----------=

q ERT0-----------=

4


Thus, the scaled equations of the PDE system become

with boundary conditions:

with T=T(x,t), c=c(x,t), x0=0, x1=1, and T0=1.

We can reduce the model to a pure diffusion problem by the introduction of coordinates that move with the fluid, e.g.:

For a pure initial value problem with

we can conclude that two different scenarios can arise:

• The reactant will be consumed completely and the final state will be

• The process will not go to completion in finite time.

The relevance of this model, in real-life, lies in a process involving a tube reactor of finite length operating during a finite period of time.

t∂∂T r2 x∂

∂T+ r0

x2

2

∂

∂ T B1ce

qT----–

+=

t∂∂c r2 x∂

∂c+

x2

2

∂

∂ c B2– ce

qT----–

=

r0 x∂∂T x0 t,( )– r2 T0 T x0 t,( )–( );=

x∂∂T x1 t,( ) 0=

x∂∂c x0 t,( )– r2 c0 c x0 t,( )–( );=

x∂∂c x1 t,( ) 0,=

ξ x r2t τ,– t= =

T x 0,( ) T0 0 and c ξ 0,( ) 0 given for ∞– ξ ∞,< <>>=

c ξ ∞,( ) 0 T ξ ∞,( ), T0

B1B2-------c0+= =

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The design parameters are L, v, T0, and c0, and, in order to minimize the investment costs, we require a large flow of the reaction product from the reactor.

Furthermore, we have to remember that if T0 is low, v is large or the reactor is short, the time during which the gas resides in the reactor will not suffice for a high degree of conversion. However, a small velocity v means a small flow of reaction products through the reactor, and a large L implies that the reactor becomes excessively large and expensive. For this reason, we are forced to compromise during the process of designing our reactor.

In some cases, several steady-state concentration and temperature profiles are obtained for a given set of L, T0, v, and c0. If this occurs, effects of stability and sensitivity to perturbations, during operation, become important.

We can obtain guidelines for design and operation of the tubular reactor, from the point of view of the topics discussed above, by simulating its dynamical operation.

In this example, we will investigate the following set of parameters:

Model Library Chemical_Engineering/tubular_reactor

Using Command-Line FunctionsLet us model the tubular reactor using command-line functions. Start by clearing the FEM structure to eliminate any previously created variables:

clear fem

r0 1= B1 7.8 107⋅=

r1 30= B2 1.2 108⋅=

r2 30= q 17.6=

6


Specify the dimension of the problem (number of dependent variables) and the form in which the equations are given:

fem.dim=2fem.form=’general’

In this case, a PDE system is specified using the general form.

The interval on the real axis studied in this model is [0,1]. The geometry is specified by:

fem.geom=solid1([0 1]);

The function solid1 creates a one-dimensional solid object on the real axis.

The following creates an initial 1-D mesh:

fem.mesh=meshinit(fem,’hmax’,0.01);

Next, specify the parameters of the problem:

fem.variables=’r0’ 1 ’r1’ 30 ’r2’ 30 ...’B1’ 7.8e7 ’B2’ 1.2e8 ’q’ 17.6;

Now specify the PDE system (recall that r1=r2/r0):

fem.dim=’T’,’c’;fem.equ.ga=’-r0*Tx’ ’-cx’;fem.equ.f=’-r2*Tx+B1.*c.*exp(-q./T)’ ...

’-r2*cx-B2.*c.*exp(-q./T)’;fem.equ.da=1 1;fem.bnd.g=’r1*(1-T)’ ’r2*(1-c)’ 0;

In order to solve the nonlinear PDEs, the Jacobian must be formed by symbolic differentiation using the function femdiff:

fem=femdiff(fem);

The field fem.init contains the initial values for each of the two solution components. Try two different initial conditions. The first one corresponds to a start-up of a cold reactor:

fem.init=1 1;

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This is a time-dependent problem for which the solver function femtime is used. Specify that a progress report is desired and also specify the time intervals for the simulation:

fem.sol=femtime(fem,’report’,’on’,’tlist’,0:0.002:0.1);

Depending on the memory size of your system, you might need to reduce the number of output times, e.g., try 0:0.005:0.1.

A progress report is displayed.

Animate T and c:

postmovie(fem,’lindata’,’T’,’liny’,’c’);

Also try an initial value slightly above the stationary value of the previous run:

fem.init=’1+0.15*x’ 1;fem.sol=femtime(fem,’report’,’on’,’tlist’,0:0.002:0.1);

Animate T and c again:

postmovie(fem,’lindata’,’T’,’liny’,’c’);

Notice that small upsets are created in the reactor, and these have to pass through before the system settles.

ReferencesW. Fred Ramirez, Computational Methods for Process Simulation, Butterworths, 1989, ISBN 0-498-90184-9.

8


Cylindrical Tubular Reactor with Cooling2

In this example we will study a modification of the previous model. This implies that we will incorporate the cooling effect of the walls in a cylindrical reactor.

In the model, we still assume plug flow along the reactor, whereas heat conduction and molecular diffusion takes place in radial and axial directions. In this case, the scaled equations become

with the boundary conditions

where T=T(x,y,t), c=c(x,y,t), LR=10, κ=100 and TR=C0=T0=1.

2. This model was provided by Bernt Nilsson, Chemical Engineering, Lund University.

v

(c,T)

x

y

outside

symmetryaxis

t∂∂T r2 x∂

∂T+ r0

x2

2

∂

∂ T r0

LRy

-------y∂

∂ yy∂

∂T B1ce

qT----–

+ +=

t∂∂c r2 x∂

∂c+

x2

2

∂

∂ c LRy

-------y∂

∂ yy∂

∂c B2e

qT----–

–+=

r0 x∂∂T 0 y t, ,( )– r2 T0 T 0 y t, ,( )–( )=

x∂∂T 1 y t, ,( ) 0=

x∂∂c 0 y t, ,( )– r2 c0 c 0 y t, ,( )–( )=

x∂∂c 1 y t, ,( ) 0=

y∂∂T x 0 t, ,( ) 0=

y∂∂T x 1 t, ,( ) κ TR T x 1 t, ,( )–( )=

y∂∂c x 0 t, ,( ) 0=

y∂∂c x 1 t, ,( ) 0=

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TR is the scaled temperature of the cooling medium and κ is a scaled heat transfer coefficient.

LR = (Lz/Lr)2

The initial conditions are:

T(x,y,0) = 1, c(x,y,0) = 1

Model Library Chemical_Engineering/tubular_reactor_cooling

Using the Graphical User Interface

Model Navigator

• In the Model Navigator, select General Time-dependent from the PDE modes, using 2 dependent variables.

• Specify the number of dependent variables by entering it in the edit field for No. of dependent variables at the bottom of the New page.

• To alter the default names of the dependent variables, press the More button.

• Enter T c instead of u1 u2 in the Dependent variables edit field.

• Change the Application mode name to, e.g., convection_diffusion.

• Press OK.


• Open the Axis and Grid Settings dialog box from the Options menu. Set the axis limits to [-0.25 1.25] for the x-axis range and [-0.25 1.25] for the y-axis range.

0


• Open the Add/Edit Variables dialog box from the Options menu. Enter the following variable names and expressions:

Draw Mode

• Press the Draw Rectangle button on the draw toolbar and draw a square from the origin to (1,1).

Boundary Mode

• Choose Specify Boundary Conditions from the Boundary menu. Enter the boundary coefficients according to the following table:

Name Expression

r0 1

r1 30

r2 30

B1 7.8e7

B2 1.2e8

q 17.6

LR 10

Kappa 20

TR 1

Boundary 1 2,4 3

G(1) r1*(1-T) 0 Kappa*(TR-T)

G(2) r2*(1-c) 0 0

R(1) 0 0 0

R(2) 0 0 0

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You can try both Kappa=0 and Kappa=20. Kappa=0 makes the problem equivalent to the previous 1-D command-line model. Kappa=20 introduces some cooling on the outside of the cylinder.

PDE Mode

• Choose PDE Specification from the PDE menu. Enter the PDE coefficients according to the table below.

Mesh Mode

• Select Initialize Mesh from the Mesh menu.

Solve Mode

• Choose Specify Initial Conditions from the Solve menu. Set the following initial conditions:

Subdomain 1

Γ1 -r0*Tx -r0*LR*Ty

Γ2 -cx -LR*cy

F1 -r2*Tx+r0*LR./y.*Ty+B1.*c.*exp(-q./T)

F2 -r2*cx+LR./y.*cy-B2.*c.*exp(-q./T)

da11 1

da21 0

da12 0

da22 1

Subdomain 1

T 1+0.15*x

c 1

2


• Choose Parameters from the Solve menu. Select the Timestepping page and set the time steps to 0:0.005:0.1.


Plot Mode

• On the Surface page in the Plot Parameters dialog box, select T for Surface data and c for Height data (3-D). Press OK to visualize.

• Press the Animation toolbar button to animate the result.

• On the General page, you can visualize different times by selecting from the Solution at time menu.

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You can rotate the 3-D graph, by clicking and dragging in the GUI.

Absorption in a Falling Film3

The absorption process of gases in liquids is thoroughly treated in the chemical engineering literature [1,2,3]. In modeling of reactors and equipment for unit operations, the chemical engineer is dependent on experimental data for the relation between gas and liquid composition. In the process of measuring absorption, the research engineer has to design his or her experiments in such a way that relevant data is obtained.

In this example, we will model an experimental set-up for studies of gas absorption, more specific the absorption of chlorine, diluted in hydrogen, in a caustic solution containing low amounts of carbonate. Hydrogen is generated in many electrolysis processes, at the cathode, and is often contaminated by chlorine, which is produced at the anode in chloride solutions. The chlorine contaminated hydrogen is cleaned in caustic soda scrubbers.

There are a number of different experimental methods for determining the absorption of gas in a liquid. The principle of these methods is

3. This model was developed in cooperation with Johan Sundquist, Eka Chemicals R&D.

4


however the same in most cases; to obtain a well-defined velocity profile in the liquid phase in order to obtain an analytical estimation of the diffusion boundary layer [1,2,3]. This well-defined velocity profile can be obtained by using a laminar jets, rotating drums with thin liquid films, falling films, etc., see [4]. The mass transport properties in the gas phase are estimated from the relative velocity between the liquid flow and the gas flow. In most cases the magnitude of the forced convection in the gas is negligible and the transport properties can be obtained from the free convection that is induced in the gas phase by the flow of the liquid.

We will model a measuring equipment for gas absorption based on the principle of a falling laminar film. The principle for this apparatus is sketched in the figure above. A liquid film is formed by a guiding funnel that creates an annular flow with a free liquid surface facing the center of a tube. The gas is introduced in the middle of the tube and is absorbed on the free surface of the liquid. The composition of the liquid and the gas is measured before and after entering the apparatus. The contact surface between the gas and the liquid is confined to the free surface inside of the tube. This implies that the contact area and the convection in this phase boundary is well-defined.

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The model will show that it is relatively simple, in FEMLAB, to treat fairly large system of nonlinear partial differential equations.

Definition of the problemWe will model a falling film apparatus for the absorption of low amounts of chlorine in hydrogen. The gas is further contaminated with low amounts of carbon dioxide, which in some cases can limit the absorption rate of chlorine. In order to obtain a more transparent model, we have to make some simplifying assumptions about our system:

• We assume that chlorine in hydrogen is diluted to such an extent that the volume of the gas is not influenced by the absorption process.

• We do not account for condensation and evaporation of water in the system.

• We neglect any homogeneous gas phase reactions.

• We assume laminar flow in the liquid phase.

• The radius of the tube is large enough, in comparison to the thickness of the falling film, for us to neglect effects of curvature in the tube.

• We assume that the contribution of diffusion, to the flux of species, is negligible in the direction of the convective flow, i.e., in the vertical direction.

• We assume that our system is isothermal.

The thermodynamic and kinetic data for our system, as well as a detailed description of the chemistry, is found in [5]. The chemical species that we will model in this example are tabulated below.

Species Index number

Cl2 1

OH - 2

HClO 3

Cl - 4

H + 5

6


The chemical reactions that we will incorporate in the mass balances, for the species above, are the following:

We will account for the kinetics in every reaction above, i.e., we will not assume equilibrium in any of these reactions. We denote the kinetic parameters in the equations above ki and k_i, for the forward and backwards reaction, respectively, where i denotes the number of the reaction. We denote the equilibrium constant for the reactions Ki.

ClO - 6

HCO3- 7

CO2 8

CO32- 9

H2CO3 10

Species Index number

Cl2 OH-+ HClO Cl-

+= 1( )

H+ O+ H- H2O= 2( )

HClO OH-+ ClO- H+ 2O= 3( )

Cl2 H2O+ HClO HCl+= 4( )

Cl2 HCO3-

+ CO2 HClO Cl-+ += 5( )

Cl2 CO32-

+ CO2 ClO- Cl-+ += 6( )

CO2 H+ 2O H2CO3= 7( )

CO2 OH-+ HCO3

-= 8( )

HCO3- O+ H- CO3

2- H+ 2O= 9( )

HCO3- H++ H2CO3= 10( )

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The assumptions listed above, together with the chemical reactions, make it possible for us to define the mass balance equations for our system. In addition, we can neglect the component of the flow velocity vector in the horizontal direction, which allows for us to obtain an analytical expression for this vector [2]. The result of this is that we can reduce our 3-D problem to a 1-D time dependent model, where we will use time to represent the space coordinate in the vertical direction. The domain that we will treat in the model is shown in the sketch below.The domain is confined by the gas phase on the left hand side and by the wall of the tube on the right hand side. It is also bounded by the inlet at the top and the outlet at the bottom.

In our domain we obtain the following general expression for the flux vector of every species:

where Ni represents the flux of species i (mole m-2 s-1). Di denotes the diffusion coefficient of i in the liquid phase, ui the concentration (mole m-3), x the space coordinate in the horizontal direction (m), and vy the component of the velocity vector in the vertical direction (m s-1). We can obtain the velocity profile, vy, from [1]

where vav represents the average velocity and δ the thickness of the film. The space coordinate x is 0 at the gas phase boundary and δ at the wall of the tube. The coordinate y is 0 at the inlet and equal to the length of the tube at the outlet.

Ni Di

ui∂x∂--------– uivy,

= in Ω where i = 1,..,10

vy 1,5vav 1 xδ---–

2⋅=

8


We are now ready to define our mass balance at steady-state for our species:

In this equation Rj represents the reaction rate for each of the reactions j that the species i takes part in. We can obtain the kinetic expressions for these reactions from [4] and they are also listed in the modeling instructions below. The equation above can be expressed in the following form:

We can rewrite this equation by using the transformation y = vavt. This gives us the final system of equations in our domain:

This transformation implies that the boundary conditions at y = 0 become initial conditions. The concentration for all the involved species, at the inlet, is 0 except for the hydrogen ion and the hydroxide ion. This gives us the following initial conditions:

The corresponding boundary conditions for our system are

∇ N⋅ i ΣRj– 0= in Ω where i = 1,..,10

Di

∂2ui

x2∂-----------– vy

ui∂y∂

--------+ Rj∑– 0= in Ω where i = 1,..,10

1,5 1 xδ---–

2⋅

ui∂t∂

-------- Di

∂2ui

x2∂-----------– Rj∑–⋅ 0= in Ω where i = 1,..,10

ui x 0,( ) 0= for i 1 3 4 6 10–, , ,=

u2 x 0,( ) COH=

u5 x 0,( ) CH=

Di

∂ui∂x--------– 0 t,( ) 0= for i 2 7 9 10, ,–=

D1

∂u1∂x

---------– 0 t,( ) kga Cl2, pCl2u1HCl2

–( )=

D8

∂u8∂x

---------– 0 t,( ) kga CO2, pCO2u1HCO2

–( )=

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where kga denotes the mass transport coefficient in the gas phase, p the partial pressure, and H the Henry constant for the respective species chlorine and carbon dioxide.

The boundary conditions at the tube wall are

which implies that there is no flux of species out of our domain at this boundary.

We are now ready to enter our system of equations, with corresponding initial and boundary conditions.

Model Library Chemical_Engineering/falling_film

Using the Graphical User InterfaceSelect 1-D, General, Time dependent PDE mode. Specify the number of variables to 10 in the No. of dependent variables edit field.


• Set the axis limits to [-1e-5 11e-5] for the x-axis range.

• Define the following variable expressions:

Name Expression

D1 1.47e-9

D2 3.43e-9

D3 1.54e-9

D4 1.565e-9

D5 3.43e-9

D6 3.43e-9

D7 1.5e-9

Di

∂ui∂x--------– δ t,( ) 0= for i 1 ... 10, ,=

0


D8 1.92e-9

D9 1.5e-9

D10 1.5e-9

k1 1.565e6

K1 4.491e10

k_1 k1/K1

k2 1e9

K2 1.002e-8

k_2 k2/K2

k3 2.79e12

K3 2.79e3

k_3 k3/K3

T 298

k4 1.4527e10*exp(-6138.6/T)

K4 4.5e10

k_4 k4/K4

Vav 0.0362

delta 1e-4

Coh 10

Ch K2/Coh

pCl2 3e-3

kgaCl2 0.244

pCO2 1e-3

k5 5.63e7*exp(-4925/T)

Name Expression

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Draw Mode

• Choose Specify Geometry in the Draw menu. Specify 0 and 0.0001 as Start and Stop, and press the Set button.

K5 1.092e6

k_5 k5/K5

k6 k5

K6 6.569e9

k_6 k6/K6

k7 10^(329.85-110.541*log(T)/log(10)-17265.4/T)

K7 2.06e-3

k_7 k7/K7

k8 10^(13.635-2895/T)/1000

K8 4.112e4

k_8 k8/K8

k9 1e10

K9 4.640

k_9 k9/K9

k10 k9/K9

K10 5

k_10 k10/K10

HCl2 0.01636

HCO2 0.0266

kgaCO2 0.244

Name Expression

2


Boundary Mode

• Enter boundary conditions for Boundary selection 1 according to following expression for the first and eighth component of G:

• All other components of G and R set to 0 for Boundary selection 1 and 2.

PDE Mode

• Enter the Γ vector according to the following table.

G Expression

G(1) kgaCl2*(pCl2-u1*HCl2)

G(8) kgaCO2*(pCO2-u8*HCO2)

Γ Expression

ga(1) -D1*u1x

ga(2) -D2*u2x

ga(3) -D3*u3x

ga(4) -D4*u4x

ga(5) -D5*u5x

ga(6) -D6*u6x

ga(7) -D7*u7x

ga(8) -D8*u8x

ga(9) -D9*u9x

ga(10) -D10*u10x

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• Enter the F vector according to the following table:

The time dependence in this model is given by the same expression for all variables, which results in a diagonal da matrix.

• Enter the diagonal components of the da matrix according to the following expression:

F Expression

F(1) (-k1*u1.*u2+k_1*u3.*u4-k4*u1+k_4*u3.*u4.*u5)-(k5*u1.*u7-k_5*u3.*u4.*u8)-(k6*u1.*u9-k_6*u4.*u6.*u8)

F(2) (-k1*u1.*u2+k_1*u3.*u4+k2-k_2*u2.*u5-k3*u2.*u3+k_3*u6)-(k8*u2.*u8-k_8*u7)-(k9*u2.*u7-k_9*u9)

F(3) (k1*u1.*u2-k_1*u3.*u4-k3*u2.*u3+k_3*u6+k4*u1-k_4*u3.*u4.*u5)+(k5*u1.*u7-k_5*u3.*u4.*u8)

F(4) (k1*u1.*u2-k_1*u3.*u4+k4*u1-k_4*u3.*u4.*u5)+(k5*u1.*u7-k_5*u3.*u4.*u8)+(k6*u1.*u9-k_6*u4.*u6.*u8)

F(5) (k2-k_2*u2.*u5+k4*u1-k_4*u3.*u4.*u5)-(k10*u5.*u7-k_10*u10)

F(6) (k3*u2.*u3-k_3*u6)+(k6*u1.*u9-k_6*u4.*u6.*u8)

F(7) -(k5*u1.*u7-k_5*u3.*u4.*u8)+(k8*u2.*u8-k_8*u7)-(k9*u2.*u7-k_9*u9)-(k10*u5.*u7-k_10*u10)

F(8) (k5*u1.*u7-k_5*u3.*u4.*u8)+(k6*u1.*u9-k_6*u4.*u6.*u8)-(k7*u8-k_7*u10)-(k8*u2.*u8-k_8*u7)

F(9) -(k6*u1.*u9-k_6*u4.*u6.*u8)+(k9*u2.*u7-k_9*u9)

F(10) (k7*u8-k_7*u10)+(k10*u5.*u7-k_10*u10)

da Expression

da(i,i), i= 1,...,10 1.5*(1-(x/delta).^2)

4


Mesh Mode


• In the Mesh Parameter dialog box, press the More button.

• Set the Mesh growth rate by entering 1.02.

• Set the Max element size for points by entering 1 3e-10 in the corresponding edit field.

• Press the Remesh button.

We have obtained a proper mesh, which is dense where it will be required.

Solve Mode

We have to remember that this example is fairly extensive and may take some time to solve on a conventional PC. This implies that you will have to wait between 10-30 minutes before a solution is computed by FEMLAB.

• Enter initial conditions for the second and fifth components according to the expressions below:

The remaining components will be zero, by default, at the beginning of the process.

• On the Timestepping page, specify 0:0.02:1 under Output times and set the Relative tolerance to 1e-6.


Plot Mode

• Press the Zoom Extents button.

Variable Value

u2(t0) Coh

u5(t0) Ch

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The default plot shows the concentration distribution of chlorine along the width of the film at time 1, which corresponds to the outlet of the tube.

We can see from this graph that chlorine does not penetrate the film to a larger extent. This is explained by chlorine’s reaction with hydroxide giving hypochlorous acid and hypochlorite. This reaction is very fast and limits the spreading of chlorine in the falling film. The low pH at the surface of the film, and thus the low content of hydroxide ions, allows for a high chlorine concentration.

Post Processing

The reaction products from chlorine’s reaction with hydroxide and water penetrate the falling film during the absorption process. We can look at the concentration profile of hypochlorite, from the inlet to the outlet of the tube, by exporting the FEM structure as fem and typing

figure(1), clf, hold onfor isol=1:5:prod(size(fem.sol.tlist)) postplot(fem,’liny’,’u6’,’solnum’,isol);

6


endhold offxlabel(’film width’)ylabel(’concentration’)

We can see in this figure that the hypochlorite ion increases in concentration and extension throughout the absorption process. At a certain stage, we can detect a maximum in the profile. This maximum is caused by the consumption of hypochlorite, through the formation of hypochlorous acid, at the low pH at the surface of the film (x = 0). The low pH is caused by the reaction of chlorine with hydroxide and water.

References[1] R. Perry, D. Green, “Perry´s Chemical Engineering Handbook“, Seventh Edition, McGraw-Hill, 1997.

[2] R. B. Bird, W. E. Stewart and E. N. Lightfoot, “Transport Phenomena”, John Wiley & Sons, 1960.


0 0.2 0.4 0.6 0.8 1

x 10−4

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

film width

conc

entr

atio

n

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[4] C. W. Spalding, “Reaction Kinetics in the Absorption of Chlorine into Aqueous Media”, AIChE J., 27, 856 (1981).

[5] S. S. Ashour, E. B. Rinker and O. C. Sandall, “Absorption of Chlorine into Aqueous Bicarbonate Solutions and Aqueous Hydroxide Solutions“, AIChE J., 42, 671 (1996).

The Chlor-alkali Membrane Cell4The chlor-alkali membrane process is one of the largest processes in industrial electrolysis with productions of around 40 million metric tons of both chlorine and caustic soda [1]. Chlorine’s largest use is in the production of vinyl chloride monomer, which in turn is used for the production of poly vinyl chloride (PVC). Among the applications of PVC are as electrical insulator in cables and as construction material for pipes, floor carpets etc. The production of chlorine implies a simultaneous production of caustic soda (alkali), which is widely used in the chemical industry for alkalisation and neutralization of acidic streams. Caustic soda is also used in alkaline batteries.

The traditional process for manufacturing chlorine and caustic soda is the mercury cell process. This technology has been partly replaced by the diaphragm process and in the later years the membrane process has been the dominating process in retrofit and for new plants. The purpose of the mercury, diaphragm and membrane is to separate the products chlorine and caustic soda, which otherwise would react to produce hypochlorite and hydrochloric acid. Chlorine and caustic soda are produced at the anode and cathode, respectively. We can see a schematic picture of the process in the below figure.

The current density in the membrane cell technology has increased dramatically during the last decade. The reason for this increase is that it is possible to produce a larger amount of chlorine and caustic, at constant surface area, which gives a lower investment cost. However, the increase in current density implies an increase in power consumption, if nothing is done to damp the voltage increase. Advances in cell design, by increased internal convection in the cells, decreased ohmic losses, and better membranes have allowed for large increases in current density with small increases in cell voltage. One of the important parameters in the design of modern membrane cells is the current density distribution

4. This model was developed in cooperation with Olof Parhammar, Eka Chemicals R&D.

8


on the surface of the electrodes. It is important, from the point of view of life-time of the catalyst and minimization of losses, that the current density on the frontal surface of the electrodes is as uniform as possible.

In this example we will study the current density distribution in a realistic structure for the anodes and cathodes in a membrane cell. The membrane cell is a fairly large cell and we will limit our model to one unit cell of the whole cell. This unit cell is drawn on the right hand side in the figure below.

The anode and cathode ribs are separated by the membrane, a cat ion selective membrane. The membrane is forced to adapt its shape to fit within the inter electrode distance. The membrane prevents mixing between brine and chlorine in the anolyte and caustic and hydrogen on the catholyte.

We can find a detailed description of the process of chlor-alkali electrolysis in reference [1].

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Definition of the problemIn this example we model the current and potential distribution in a unit cell in the membrane cell sketched in the figure above. Our model is a secondary current distribution model [2], which implies that we take into account the dependence of the electron transfer on the local potential and that we assume constant composition in the subdomains. The electron transfer reactions at the anode and cathode surfaces are the following:

The domain that we will treat is half of the unit cell shown above, which for a symmetric system can be used to build up the current density distribution along the whole width of the cell.

Note that we have rotated our unit cell 90o compared to the preceding figure. The figure above shows that our domain consists of three subdomains, the anolyte, Ωa, the membrane, Ωm, and the catholyte Ωc. Moreover, our domain is bounded by the anode surface, ∂Ωa, the cathode surface, ∂Ωc, and the outer surface of the unit cell, ∂Ωouter.

2Cl- Cl2 g( ) 2e-+= at the anode

2H2O 2e-+ 2OH- H2 g( )+= at the cathode

∂Ωc

∂Ωa

Ωa

Ωc

Ωm

3.5 mm

∂Ωouter

0


The chemical reactions above show that we have gas evolution in both the anodic and cathodic compartments. The chlorine gas in the anolyte and the hydrogen gas in the catholyte create a vigorous internal convection in the respective compartments. This makes it possible for us to simplify our model by neglecting the concentration gradients in the anolyte and catholyte. The simplification implies that the transport of ionic current inside the cell takes place through migration, i.e. a flux of ions is induced by the electrical field. For this reason, we do not need to model the complex problem of internal free convection of the two phase flow in order to get an estimation of the current density distribution in the cell.

The equation that gives the transport of charged species in the electrolyte is the following:

where Ni denotes the transport vector (mole m-2 s-1), ci the concentration in the electrolyte (mole m-3), zi the charge number for the ionic species, ui the mobility of the charged species (mole s kg-1), F Faraday’s constant (As mole-1), φ the potential in the electrolyte (V) and v the velocity vector (m s-1). If we neglect the transport by diffusion in the electrolyte, which is natural due to the vigorous internal convection, we get

We can obtain the current density vector from the transport of charged species through Faraday’s law:

In an electrochemical system, the potential gradients are fairly moderate, which implies that no separation of charges takes place in the electrolyte. This means that the electroneutrality condition is valid in the bulk of the electrolyte:

Ni Di∇ci– ziuiFci∇φ– civ+=

Ni ziuiFci∇φ– civ+=

i F ziNi∑–=

zici∑ 0=

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If we combine the above equation with the two preceding equations we get

We can rewrite this equation to get Ohm’s law for electrolytes

where κ is the conductivity of the electrolyte (S m-1). We can now set up a balance of current in our three subdomains at steady-state:

This gives for the three subdomains

The indexes in the conductivity denote the anode, cathode and the membrane. We solve these equations with the proper boundary conditions. The anode reaction is a very fast reaction and small changes in potential give large changes in current density. This implies that we can assume a constant potential at the surface of the anode, which gives an error in potential of about 20 mV at the anode surface [3]. A small change in potential gives also a large change in current density at the cathode but this relation is not at all as steep as it is at the anode. For the cathodic reaction we have to account for the potential distribution around the electrode surface to get a proper current density distribution. From the above discussion we can formulate following boundary conditions:

i F zi2uiFci∇φ–∑–=

i κ∇φ–=

∇ i⋅ 0=

∇ κa∇φ–( )⋅ 0= in Ωa

∇ κm∇φ–( )⋅ 0= in Ωm

∇ κc∇φ–( )⋅ 0= in Ωc

φ φa= at ∂Ωa

i n⋅ i0–αF

RgT-----------φ

exp= at ∂Ωc

i n⋅ 0= at ∂Ωouter

2


In the above equations i0 denotes the exchange current density (A m-2) for the cathodic reaction, α the transfer coefficient, Rg the gas constant (J mole-1 K-1), and T temperature (K). The last equation implies that we have no flow of current out of our domain at the outer walls of the unit cell.

We are now ready to solve the problem, which will give us the potential in the electrolyte and the current distribution in the electrolyte and at the surface of the electrodes.

Model Library Chemical_Engineering/gallery/chlor_alkali

Solving the problem using the Graphical User InterfaceSelect the application mode Conductive media DC, Nonlinear stationary in the Model Navigator.



Axis Grid

X min -4e-3 X spacing 1e-3

X max 1e-3 Extra X -3.5e-3 -1.5e-3-1.25e-3 -0.5e-3

Y min -4.5e-3 Y spacing 1e-3

Y max 6.5e-3 Extra Y 0.8e-3 1.2e-3 1.5e-3 2.2e-3 2.7e-3 3.3e-3

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Draw Mode

• Press the Draw Cubic Bezier Curve button.

• Draw a Bezier curve by clicking at the points (-0.0035,0.0022), (-0.002,0.0022), (-0.0015,0.001) and (0,0.001).

• Continue by pressing the Draw Line button and drawing a line between the points (0,0.001) and (0,0.0008).

• Draw a second Bezier curve by clicking at the points (0,0.0008), (-0.0015,0.0008), (-0.002,0.002) and (-0.0035,0.002).

• Right click to create a solid object, CO1.

• Double click on the solid object to open the Object Properties window.

• Change the position of the third point in the Point Selection to (-0.0021,0.002).

• Press OK.

At this stage we have the geometry of the membrane. We will continue by drawing the cathode:

• Press the Draw Arc button.

• Draw an arc by clicking at the points (0,0.0012), (-0.0005,0.0012) and (-0.001,0.0015).

Name Expression

alfap 0.5*96487/8.31/(273+90)

fio 1.19

io 1e-3

Ka 50

Kc 100

Km 3

4


• Continue by pressing the Draw Line button and drawing lines between the points (-0.001,0.0015), (-0.002,0.0022), (-0.00125,0.0033) and (0,0.0027).

• Right click to create a solid object, CO2.

We have now created the cathode and we will use this object to create the anode:

• Select the solid object CO2 and select Copy from the Edit menu.

• Select Paste from the Edit menu and enter 0 in the X-axis displacement edit field and 0.0006 in the Y-axis displacement edit field.

• Rotate the solid object CO3 by selecting Linear Transformation, Rotate in the Draw menu.

• Enter 180 in the Rotation edit field and [0,1.8e-3] in the Center coordinates edit field.

• Move the solid object CO3 by selecting Linear Transformation, Move in the Draw menu.

• Enter -3.5e-3 in the X-axis displacement edit field and 0 in the Y-axis displacement edit field.

We are now ready to create the electrolyte:

• Draw a rectangle, R1, with lower left corner at (-0.0035,-0.004) and upper right corner at (0,0.006).

• In the Draw menu, select Create Composite Object... and enter R1+CO1-CO2-CO3 in the Set formula edit field.

Boundary Mode


Boundary 3,8,9,17 10-12,18 1,2,4-7,13-16

V fio 0

g -io.*exp(alfap.*V)

g=q=0 select insulation

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PDE Mode

• Enter the PDE coefficients according to the following table:

Mesh Mode


• Refine and jiggle the mesh.

Solve Mode

Since we have a highly nonlinear expression at the cathode boundary, our problem will require a special solver for optimal convergence.

• Select a solution for highly nonlinear problems by checking the Highly nonlinear problem option in the Nonlinear page in the Solver Parameters window.

• Enter 40 in the Maximum number of iterations edit field.

• Enter 1e-4 in the Nonlinear tolerance edit field.

• Specify the initial conditions in the respective subdomains according to the following table:

This will give a proper initial guess for the nonlinear solver.


Subdomain 1 2 3

σ Ka Km Kc

Q 0 0 0

Subdomain Value

1 fio

2 0.85*fio

3 0.77*fio

6


Plot Mode

The default plot gives the potential distribution in our unit cell. We can see from this plot that the largest Ohmic losses are obtained in the membrane, as expected from its low conductivity.

We can visualize the current density distribution by adding arrows for the current density vector in our domain.

• On the Arrow page, select current density (Jx) and current density (Jy) as Arrow data for the x and y-expressions, respectively.


• Press OK.



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We can clearly see in this graph that the current density distribution is more uniform on the cathode surface compared to the anode surface.

The modulus of the current density vector gives us, the “hot spots” in the electrolyte, where the current density is extremely large. These “hot spots” are obtained at the edges of the electrodes. At these edges, we will loose the catalyst due to accelerated wear.

8


Post Processing

The large Ohmic losses in the membrane can be seen even more clearly in the figure below.

• Select electric potential (V) as Surface expression and Height expression in the Surface page in the Plot Parameters window.

• Select winter in the Colormap edit field.

• In the General page, uncheck the Arrow, Geometry boundaries and Plot in main GUI axes boxes.

• Press OK.

• Enter the following commands in the MATLAB Command Window:

view(220,40)axis([-5e-3 2e-3 -5e-3 7e-3 0.85 1.25])cameramenu on

• Select Orbit Scenelight in Mouse Mode in the Camera menu in the figure window.

• Orbit the scene light to get a proper view.

The plot shows the potential profile in our domain, and we can see that the steepest potential profile is located at the position of the membrane.

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If we look closer, we can also see that the variations in potential around the cathode surface are of the order of magnitude of 80 mV.

References[1] H. S. Burney, “Past Present and Future of the Chlor-Alkali Industry”, Chlor-Alkali and Chlorate Technology: R.B. Macmullin Memorial Symposium, Proceedings Volume 99-21, The Electrochemical Society, NJ, 1999.

[2] J. S. Newman, “Electrochemical Systems” 2nd edn., Prentice Hall, Englewood Cliffs, NJ, 1991.

[3] P. Bosander, P. Byrne, E. Fontes and O. Parhammar, “Current Distribution on a Membrane Cell Anode”, Chlor-Alkali and Chlorate Technology: R.B. Macmullin Memorial Symposium, Proceedings Volume 99-21, The Electrochemical Society, NJ, 1999.

Model of a Fuel Cell CathodeThe development of fuel cells has accelerated during the last decade. The field of fuel cells for traction applications has experienced the largest boost in development, due to the efforts put in by car manufacturers. Despite funding being more limited for stationary applications, these fuel cells have also gained from this general development and have proven to give excellent performance in small scale applications.

The fuel cell is a continuous electrochemical reactor that converts chemical energy to electrical energy. This takes place by physically separating the oxidation of hydrogen and reduction of oxygen at separate electrodes, the anode and cathode, respectively. Separation occurs through an ion conducting electrolyte.

The fuel cell presented in this example is a high temperature fuel cell for stationary applications. Fuel cells are named after their electrolyte medium, and in this case we will study the Molten Carbonate Fuel Cell (MCFC). This cell operates at 650oC where the fuel is oxidized at the anode, and can be either hydrogen or natural gas. Oxygen is used as the oxidant, being reduced at the cathode together with carbon dioxide.

The cathode is the electrode that gives the largest contribution to energy losses in the cell, i.e., it has the largest overvoltage. The overvoltage consists of ohmic overvoltages, in the electrolyte and catalyst, activation

0


overvoltage, for the electron transfer reaction, and concentration overvoltage, due to mass transfer resistance in the porous electrode.

This example will again demonstrate the applicability of the Multiphysics feature in limiting the definition of variables to certain subdomains. We will solve a model for the porous oxygen reducing cathode, in the fuel cell, and the electrolyte between the anode and cathode. The electrolyte is defined in the porous electrode and in the free electrolyte, while the solid phase is present only in the porous electrode subdomain.

The difference between the potential in the liquid and solid phases is proportional to the distribution of the electron transfer reaction in the electrode, the property that we will study.

Definition of the problemThe porous fuel cell cathode consists of three different phases; the solid catalyst, the electrolyte that transfers into the porous phase through the action of capillary forces, and the gas phase in the large pores of the electrode. This is detailed in the figure below. We will individually treat these three phases as homogeneous, defined over the whole electrode.

Anode

Cathode

O2

H2

Electrolyte Magnification

Unit cell

∂Ωs,1 ∂Ωs,2

∂Ωl,1 ∂Ωl,2

∂Ωl

∂Ωo ∂Ωcc

Ωsl

Ωl

Detailed structure

∂Ωls

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Furthermore, the liquid phase will also be defined in the free electrolyte between the anode and cathode.

The electron transfer reaction, shown in the sketch above, transfers the current from ionic conduction, in the electrolyte, to electronic conduction in the solid catalyst. The solid catalyst makes electronic contact with the current collector, which is the perforated plate on the top of the cathode. The perforations are required so as to allow gas to be supplied to the electrode.

The catalyst agglomerates are sketched as dotted particles, in the detailed figure, the electrolyte is shaded grey while the gas filled macro-pores are white. Oxygen has to firstly dissolve into the electrolyte in the porous cathode, to further react on the catalyst particles that make up the agglomerates.

We can see that there is a mass transfer resistance in the electrolyte surrounding as well as inside the catalyst agglomerates, which consists of tightly packed smaller catalyst particles with electrolyte in the remaining space between them. The mass transport resistance is considered in the homogenization of the system, by a function that is linearly dependent on the oxygen concentration in the gas phase. This function has previously been found to give good approximations of mass transfer resistance in agglomerates contained in the cathode [1].

currentcollector

ioniccurrent

electronic current

Electron transfer through:

CO2+1/2O2+2e- CO32-

2


We will assume that gas in the porous electrode is transported by diffusion and that current is conducted through the migration of ions in the molten carbonate electrolyte and electronic conduction in the catalyst. The model is based on a mass balance for oxygen in the gas phase and charge balances in the solid catalyst and liquid electrolyte phases.

Furthermore, by assuming that the concentration and potential gradients, through the depth of the cell, are small in comparison with the local gradients, we can state that symmetry is obtained perpendicular to the page in the figure above. This assumption simplifies our problem from a 3-D to a 2-D model.

The model equations in the subdomains are the following:

In the equations above, Deff denotes the effective diffusion coefficient in the gas phase inside the porous electrode (m2 s-1), c denotes concentration in the gas inside the electrode (mol m-3), and k denotes the effective exchange current density (S mol-1). The potential in the catalyst is denoted by φs (V), while in the liquid phase it is denoted by φl. The effective conductivities are denoted by κs,eff and κl,eff in the catalyst and liquid phases (S m-1), respectively. The conductivity in the free electrolyte is denoted by κl,free. All parameters are given in SI-units. Overall, we now have a system of four coupled equations to be solved in the domain.

The boundary conditions required to solve our problem are obtained by assuming symmetry for the lateral boundaries of the unit cell, i.e., we assume that the solution is identical for any unit cell of the cathode.

∇ Deff c∇–( ) koc φs φl–( )+⋅ 0= in Ωsl

∇ κs eff, φs∇–( ) kc φs φl–( )+⋅ 0= in Ωsl

∇ κl eff, φl∇–( ) k– c φs φl–( )⋅ 0= in Ωsl

∇ κl free, φl∇–( )⋅ 0= in Ωl

Deff c∇–( ) n⋅ 0= at ∂Ωs 1, and ∂Ωs 2,

κs eff, φs∇–( ) n⋅ 0= at ∂Ωs 1, and ∂Ωs 2,

κl eff, φl∇–( ) n⋅ 0= at ∂Ωs 1, and ∂Ωs 2,

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The only defined variable in the free electrolyte is the potential in the liquid phase:

At the position of the perforation in the collector plate, current is assumed to be unable to leave the catalyst or liquid phases. Furthermore, we assume that the concentration of oxygen in the gas stream is known. This gives us the following boundary conditions:

We further know that gas is unable to be transported to or from the electrode at the position of the current collector:

At the position of the current collector we may arbitrarily set the potential in the solid catalyst to zero (since potential is relative). Furthermore, we know that no current is transferred from the liquid phase to the current collector, since a catalyst is needed for this to happen. Therefore:

We will assume that the potential is known for the boundary facing the anode, which means that the total overvoltage is given at the position of this boundary. Therefore, total overpotential is obtained by the difference between the potentials of the current collector and the liquid phase at this boundary.

κl free, φl∇–( ) n⋅ 0= at ∂Ωl 1, and ∂Ωl 2,

c c0= at ∂Ωo

κs eff, φs∇–( ) n⋅ 0= at ∂Ωo

κl eff, φl∇–( ) n⋅ 0= at ∂Ωo

Deff c∇–( ) n⋅ 0= at ∂Ωcc

φs 0= at ∂Ωcc

κl eff, φl∇–( ) n⋅ 0= at ∂Ωcc

φl φl o,= at ∂Ωl

4


In addition to the boundary conditions, we need the subdomain boundary conditions for the variables not defined in the free electrolyte. These variables are the concentration of oxygen in the gas phase and potential in the solid catalyst. The transport of oxygen from the gas phase to the electrolyte is neglected in the model. Furthermore, we will assume that transport of current from the catalyst to the free electrolyte is negligible, since this direct contact surface is minimal:

At this stage, we have defined our problem and are ready to implement it using the Graphical User Interface (GUI).

Model Library Chemical_Engineering/mcfc

Solving the problem using the Graphical User InterfaceSelect PDE, general form, from the Multiphysics menu in the Model Navigator.

• Label your application mode with the name potl, and your Dependent variable fil.




Axis Grid


X max 0.002 Extra X 0.0008 0.0012

Deff c∇–( ) n⋅ 0= at ∂Ωls

κs eff, φs∇–( ) n⋅ 0= at ∂Ωls

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Draw Mode

• Draw a rectangle, R1, with lower left corner at (0,0) and upper right corner at (0.0012,0.0005).

• Define a point, PT1, at the position (0.0008,0.0005).

• Draw a rectangle, R2, with lower corner at (0,-0.0005) and upper right corner at (0.0012,0).

• Choose Zoom Extents from the Option menu.


Y max 0.002 Extra Y

Name Expression

kl 5

ks 20

klf 142

Far 96487

k 6.7e7

ko k/2/Far

c1o 1.98

D 2.8e-5

Axis Grid

6


Boundary Mode


PDE Mode

• Enter the PDE coefficients, in subdomain 1, according to the following table:

• Enter the PDE coefficients, in subdomain 2, according to the following table:

We are now ready to define the charge balance in the solid catalyst phase in the electrode. We will do this by adding a new model equation in the Multiphysics mode.

Boundary 1,3,5-8 2

G 0 0

R 0 -fil+0.15

Subdomain 1

Γ -klf.*filx -klf.*fily

F 0

da 0

Subdomain 2

Γ -kl.*filx -kl.*fily

F k/c1o.*c1.*(fis-fil)

da 0

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Multiphysics Mode

• Choose PDE, general form, label your application mode with the name pots, and your Dependent variable fis.


PDE Mode

In order to deactivate the subdomain of the free electrolyte, we will enter the PDE Mode before specifying our boundary conditions.

• Deactivate the charge balance equations for the solid catalyst phase, in subdomain 1, by uncheking the Active in this subdomain option.


By this procedure, we automatically obtain the proper boundaries for the subdomain.

Boundary Mode

• Enter the boundary coefficients according to the following table.

We are now ready to introduce the mass balance equation for oxygen in the electrode.

Subdomain 2

Γ -ks.*fisx -ks.*fisy

F -k/c1o.*c1.*(fis-fil)

da 0

Boundary 3-5,8 6

G 0 0

R 0 -fis

8


Multiphysics Mode

• Choose PDE, general form, label your application mode with the name massbal, and your Dependent variable c1.


PDE Mode

• Deactivate the mass balance equations for the gas phase, in subdomain 1, by uncheking Active in this subdomain option.


Boundary Mode


Mesh Mode

• The model requires a fairly dense mesh. Refine the mesh twice.

Solve Mode



Subdomain 2

Γ -D.*c1x -D.*c1y

F ko/c1o.*c1.*(fis-fil)

da 0

Boundary 3,4,6,8 5

G 0 0

R 0 -c1+c1o

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Plot Mode

The distribution of the electron transfer reaction is the most interesting property to study.

• Plot the reaction rate per unit volume, in the porous electrode, by entering the following expression; -ko/c1o.*c1.*(fis-fil).

We can see from this figure, that the reaction rate is greater close to the free electrolyte and on the right hand side. The low conductivity of the electrolyte, inside the porous electrode, forces the current to be transferred from the electrolyte to the catalyst phase, at the lower boundary of the electrode. Furthermore, the relatively low conductivity of the solid phase forces the reaction to take place on the right hand side of the electrode, where the current collector is positioned. The oxygen concentration is fairly uniform and does not determine the distribution of the reaction. The average current density, which can be obtained by plotting nga1 in Line plot, is around 1600 A m-2.

00


The below figure, shows the potential distribution in volts, within the liquid phase. We can see the large difference in conductivity between the electrolyte, in the porous electrode, and the free electrolyte, by the large difference in the potential gradient in the two subdomains.

This example shows that FEMLAB is able to easily model coupled systems in diffusion-reaction problems that often arise in porous media in electrochemistry and chemical engineering.

References

[1] E. Fontes, C. Lagergren, G. Lindbergh and D. Simonsson, “Influence of Gas Phase Mass Transfer Limitations on Molten Carbonate Fuel Cell Cathodes“, Journal of Applied Electrochemistry, 27, 1149-1156, 1997.

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Index
Index
Aabsorption 1-64Arrhenius law 1-22, 1-53

Ccarbonate 1-64catalyst agglomerate 1-92catalyst layer 1-19catalyst pellet 1-28catalytic combustion 1-19catalytic converter 1-19cathode 1-90caustic solution 1-64charge balance 1-93chlor-alkali membrane cell 1-78combustion 1-53command-line functions 1-56concentration 1-6, 1-53conductivity 1-93continuity equation 1-11, 1-29convection 1-10convection-diffusion process 1-52CVD 1-37cylindrical coordinates 1-39

Ddensity 1-12diffusion 1-55diffusion boundary layer 1-17, 1-28diffusion coefficient 1-6diffusion equation 1-4diffusion-convection-reaction process 1-36dynamic viscosity 1-12

Eeffective diffusion coefficient 1-21, 1-30electrolysis 1-78electron transfer reaction 1-80, 1-91, 1-92equations of motion 1-29exchange current density 1-83

Ffalling film 1-64flux 1-11flux of species 1-68flux vector 1-12fuel cell cathode 1-90

Hheat transfer, equation-based model 1-53Henry constant 1-70hydrogen 1-64hydrogen ion 1-69hydrogen permeability 1-6hydroxide ion 1-69hypochlorite 1-76hypochlorous acid 1-76

Iion conducting electrolyte 1-90

Llaminar film 1-65laminar flow 1-10low pressure chemical vapor deposition 1-37

I-1

Index

I-2

Mmass balance 1-11mass transfer coefficient 1-28membrane 1-4membrane cell 1-78migration 1-81molten carbonate fuel cell 1-90momentum balance 1-11monolithic reactors 1-10, 1-19

NNavier-Stokes equations 1-11, 1-29

OOhm’s law for electrolytes 1-82overvoltage 1-90

Ppermeability of gases 1-4plug flow 1-59porous electrode 1-91pressure 1-12

SSchmidt number 1-17secondary current distribution 1-80silane 1-39silicon 1-37slice plot 1-26

Ttemperature 1-53tortuosity 1-30

transfer coefficient 1-83transport mechanisms 1-4tubular reactor 1-52

Vvelocity profile 1-11viscous layer 1-17volatile organic compounds 1-19

Wwafer bundle 1-42washcoat 1-19

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