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Mass transport MechanismsMass transport equation

Mass transfer at interfacesExercises

Mass Transport and Transfer

Yannick Hallez

LGC-UPS

18 octobre 2011




Introduction

Quantity of interest

As far as mass of a solute (molecule, ion, small particle...) is concer-ned the quantity of interest is the mass of this solute per unit vo-lume : its concentration C in kg/m3 in SI units. More generally ithas the dimension of Mass/Lenght3, formally denoted ML−3.

Remark

This chapter could be recast in terms of number of moles or numberof molecules (number density) instead of mass, and concentrationwould be in moles/m3 or number of particles/m3.




Introduction

Flux density

The mass flux density will be denoted j. It has a dimensionMT−1L−2 where T stands for Time. In SI units it becomeskgs−1m−2. It is a vector, as denoted by the bold font.




AdvectionDiffusionMigrationReaction

Plan

1 Mass transport Mechanisms

2 Mass transport equation

3 Mass transfer at interfaces

4 Exercises





Mechanisms : an overview

Mass source or sink

Mainly chemical reactions

Mass transport mechanisms

Advection

Diffusion

Electrophoresis (movement of a charged species in an electricfield)

Sedimentation (movement of a massive species in agravitational field)

etc





1 Mass transport MechanismsAdvectionDiffusionMigrationReaction



4 Exercises





Transport by advection

Idea

Consider a blob of ink inside a clear stream with a velocity fieldu. At any point inside the fluid, the advective mass flux density isj = Cu. It means that the ink spot is just transported from a pointto another with a velocity u. Since u has a dimension LT−1 andC has a dimension ML−3, the flux density clearly has a dimensionML−2T−1, as expected.






Remark

Note that on the scheme below, u is constant in the whole spacefor clarity and therefore the ink spot is just translated. In reality, thevelocity and concentration fields both depend on the position x sothat the ink spot can also be distorted by advection.






Characteristic advective time scale

If the movement of mass takes place on a length scale L and witha velocity scale U , it is possible to define a characteristic advectivetime scale by

Ta =L

U(1)








4 Exercises





Transport by diffusion

Idea (1/2)

Imagine a molecule of solute or a small particle immersed in a fluid.Every molecules/particles move because of thermal agitation. Theiraverage position does not change, but they can move around be-cause of hits from the other molecules/particles. This process canbe modelled simply by the random walk (or drunkard’s walk) : adrunkard moves with a constant step at some constant frequency,but the direction of the step is random and any direction is equipro-bable. Therefore the drunkard’s average position 〈xp〉 will always beat the bar where he started his walk.






Idea (2/2)

However he’s able to explore places farther and farther from thebar as the observation time grows : the standard deviation from hisaverage position (the bar, say x = x0) grows as the square rootof time

√〈(xp − x0)2〉 ∼

√t (see next slides for images). In other

words, his mean squared displacement varies linearly with time :〈(xp − x0)2〉 ∼ t.






Back to a macrosopic description

The macroscopic description is recovered by considering that thebar closes and a large number of drunkards is assembled in frontof the door of the bar. It is a very highly concentrated drunkardzone. As time passes, every drunkard walks through the city. Thedrunkard concentration will always be maximum near the bar, butthe drunkards will visit zones farther and farther away from the bar.The typical distance between visited locations and the bar is thestandard deviation and varies like t1/2. This 1/2 exponent is typicalof diffusive processes.






The red curve is√

number of steps.


One particle trajectory





The red curve is√

number of steps.


Five particle trajectories





The red curve is√

number of steps.


1000 particle trajectories





Fick’s law

The simplest macroscopic model for the diffusive flux density ofsome species is based on the idea that particles go from highlypopulated zones to less populated ones, so that the flux is alongthe concentration gradient, and in the opposite direction. So let Dbe the proportionality factor, and we get

j = −D∇C, (2)

where ∇ is the gradient (vector !) operator (the symbol is called”nabla“). This is Fick’s law, and D is called the moleculardiffusivity coefficient of the species of concentration C inside somemedium. The molecular diffusivity coefficient has a dimensionL2T−1 (m2/s in SI units).






Try to draw the diffusive flux vector on these simulation images !Red is for high concentration, blue is for low concentration.






Characteristic diffusive time scale

Since a diffusive time scale depends on the length scale used forobservation (dimension L) and of the diffusivity coefficient D(dimension L2T−1), an obvious time scale can be built as

Td =L2

D(3)

Note that we recover the idea that a diffusive length scale growswith time as

Ld = (DT )1/2 (4)





Comparison of advection and diffusion

The Peclet number (1/2)

It is possible to compare the diffusive and advective time scales usingtheir ratio, called the Peclet number

Pe =TdTa

=UL

D(5)

When Pe � 1, the advective time scale is much shorter than thediffusive time scale, so that advection is the prevailing phenomenon.When Pe � 1, diffusion is much faster than advection and is thedominant phenomenon.






The Peclet number (2/2)

Note that the Peclet number can also be seen as the ratio of anadvective to a diffusive flux :

Pe =jajd

=CU

DC/L=UL

D(6)

When Pe � 1 the advective flux is much larger than the diffusiveflux so that advection dominates, and vice-versa.






Because of the linear and quadratic Ldependencies of Ta and Td, whateverthe values of the velocity scale and thediffusivity coefficient, for very shortlength scales diffusion is always fas-ter than advection, and at large lengthscales advection is always faster thanadvection. The critical length scale Lcwhere the transport regime is swit-ched from diffusive to advective cor-responds to equal advective and dif-fusive times (or fluxes), i.e. Pe = 1.Then we get Lc = D/U .

0.0 0.5 1.0 1.5 2.0L

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

T

Ta

Td





Transport by diffusion and advection

Let’s keep in mind that ...

In advective processes, characteristic distances vary like t.

In diffusive processes, characteristic distances vary like t1/2.

The Peclet number Pe = UL/D compares the relativeimportance of advective and diffusive mass transportphenomena. If Pe� 1 advection dominates, and if Pe� 1diffusion dominates.

An advective flux is written j = Cu, where u is the fluidvelocity (vector).

A diffusive flux is given by j = −D∇C (Fick’s law), where Dis the molecular diffusivity coefficient (dimension L2T−1).








4 Exercises





Migration

Introduction

Migration could be recast into some special advective or diffusiveflux, but let’s say a few words of transport by migration (or drift) dueto external forces such as those due to an electrical or gravitationalfield.





Electrophoresis

If a particle (ion, colloid...) with a charge q is placed in an electricfield E, it experiences a force qE and starts to move. As it moves,it experiences a friction force, or drag, from the surrounding fluid.In the limit of small Reynolds numbers (small particles or slow mo-vement or viscous fluid, see lesson on hydrodynamics) this force onone isolated spherical is object is −6πµaV, where µ is the dynamicviscosity of the fluid, a is the particle radius and V is the particlevelocity. At steady state, the sum of these forces is zero, so that

Velectrophoresis =qE

6πµa. (7)

If particles are not too concentrated, the particle flux is thusj = CVelectrophoresis (An advective flux, with velocity being the elec-trophoresis velocity).





Sedimentation

If a particle with a density ρp and volume V is placed in a gravita-tional field g, and in a fluid of density ρ, it experiences a net force∆ρgV (weight+buoyancy), where ∆ρ = ρp− ρ and starts to move.As it moves, it experiences a friction force, or drag, from the sur-rounding fluid. In the limit of small Reynolds numbers this force onone isolated spherical is object is −6πµaV, where µ is the dynamicviscosity of the fluid, a is the particle radius and V is the particlevelocity. At steady state, the sum of these forces is zero, so that

Vsedimentation =2

9

(ρp − ρ)a2g

µ. (8)

If particles are not too concentrated, the particle flux is thusj = CVsedimentation (An advective flux, with velocity being the sedi-mentation velocity).








4 Exercises





Reaction

Reaction rate

Reaction is not really a transport mechanism, but rather asource/sink of mass. Therefore is will appear into a mass balanceand the transport equation for mass. Here we just provide a simpleform for a reaction rate (dimension ML−3T−1) in an homogeneousreaction

r = kCn, (9)

where k is the reaction rate coefficient with dimension(ML−3)1−nT−1 (Wow !) and n is the order of the reaction. Forfirst order reactions, k has dimension T−1 (s−1 in SI units).





Reaction

Reaction time scale

A reaction time scale will depend on k and on a characteristicconcentration C0 :

Tr =1

kCn−10

, (10)

as suggested by the reaction rate being r ∼ ∂C/∂t = kCn





Reaction

Comparison of reaction and ”fluid“ time scales

Damkohler numbers are based on a comparison of the reaction and”fluid“ (advective or diffusive) time scales denoted τ .

Da =τ

1/(kCn−10 )= τkCn−10 . (11)

For example, if advection is compared to reaction (can a flame useoxygen faster than it is brought to it by a pipe...) Da = LkCn−10 /U .If reaction is compared to diffusion (in a diffusive boundary layernear a catalytic surface for instance, or near a droplet evaporating),Da = L2kCn−10 /D.




From mass balance to transport equationAbout the mass transport equationSteady state solutionsUnsteady diffusion equation solutionsNon-dimensional form of the transport equation

Plan




4 Exercises






2 Mass transport equationFrom mass balance to transport equationAbout the mass transport equationSteady state solutionsUnsteady diffusion equation solutionsNon-dimensional form of the transport equation


4 Exercises





Sketch

Let us consider a control volume Vof fluid or solid, delimited by a closedsurface S. Anywhere on the surface itis possible to define a unit normal vec-tor n directed arbitrarily outside thevolume V . In the whole fluid or soliddomain, a non-constant flux of massj exists.





From mass balance to transport equation

A mass balance on the volume V saysthat ”the accumulation of mass insidethe volume is equal to the mass fluxentering the volume minus the flux ofmass leaving the volume (net massflux) plus the production of mass in-side the volume“.






If a concentration field C is defi-ned throughout the domain, the totalmass inside the volume V is∫

VCdV, (12)

and its variation in time is

∂

∂t

∫VCdV (13)






The mass flux density at any point isj. Hence the mass flux through a smallsurface element with unit normal nand surface area dS is dJ = j · ndS.So the net flux through the full closedsurface S is

J = −∮Sj · ndS. (14)

The minus sign is due to our conven-tion of n being directed outward thesurface : a positive value of J meansmass is entering the volume.






Finally, if a reaction takes place withreaction rate r, the total mass sourcein the volume V is∫

VrdV (15)






Then, the classical mass balance on the volume V becomes

∂

∂t

∫VCdV = −

∮Sj · ndS +

∫VrdV (16)

or ∫V

∂C

∂tdV = −

∮Sj · ndS +

∫VrdV (17)

(note that the volume boundaries are held at fixed positions).






With the help of Green’s formulae (divergence theorem), the firstsurface integral on the right-hand-side can be rewritten as a volumeintegral : ∫

V

∂C

∂tdV = −

∫Vdiv(j)dV +

∫VrdV, (18)

where div(j) ≡ ∇ · j is the divergence of j.






Let’s continue... ∫V

[∂C

∂t+ div(j)− r

]dV = 0. (19)

Since the mass balance presented above is valid for any volume V ,the integrand is necessarily zero in the whole space, and we get themass transport equation valid everywhere

∂C

∂t+ div(j) = r (20)





Other forms

Let’s make appear our favorite mass fluxes, namely the advectiveand diffusive ones. In this case

j = Cu−D∇C (21)

and we often encounter the equation under the form of an ”advection-diffusion equation“

∂C

∂t+ div(Cu) = div(D∇C) + r (22)

Moreover if the fluid is incompressible (∇·u = 0, see hydrodynamicslessons) and if the diffusivity coefficient is constant we get

∂C

∂t+ u · ∇C = D∆C + r (23)

where ∆ ≡ ∇2 is the laplacian.








4 Exercises





The mass transport equation

In this course we will most often use this last form∂C

∂t+ u · ∇C = D∆C + r (24)

The second term on the left hand side is the ”advective term“, thefirst on the right hand side is the ”diffusive term“ and the last oneon the right hand side is the ”source term“.This equation can be seen as a constraint between the velocity andconcentration fields that ensures that the first principle of massconservation is respected.It is a partial differential equation (PDE) of second order and mostof the time non-linear since the velocity field depends itself on C.In some flows it is possible to consider that the solute does notinfluence the flow, it behaves as a so-called ”passive scalar“. In thiscase the equation is linear (if r is also linear in C or absent !).






Boundary conditions

This PDE must be complemented by initial and boundary conditions.Some of them are

Constant concentration on a surface : C = Ci

Constant flux through a surface : ∂C/∂n = constant (where nis a coordinate normal to the surface)

Symmetry (or zero flux) : ∂C/∂n = 0






Some special cases of the transport equation are :

The advection equation

∂C

∂t+ u · ∇C = 0 (25)

The diffusion equation (used for example in solids)

∂C

∂t= D∆C (26)

stationary equations ...

Now, let’s try to solve the mass transport equation in some simplecases !








4 Exercises





Exercise : Steady diffusion equation in planar geometry

Solve the mass transport equation througha plane wall of thickness e, with boundaryvalues C = C1 and C = C2 on the planesat x = 0 and x = e respectively. Once theconcentration profile is obtained, computethe mass flux density j(x) and the mass fluxthrough a wall with surface area S.






Since it is a steady problem, ∂tC = 0. Since it is a solid wall, u = 0.There is no reaction (r = 0). Therefore the transport equation (24)simplifies to ∆C = 0. Variations of C exist only in the x directionso that in the laplacian ∂2yy ≡ 0 and ∂2zz ≡ 0. So we have to solve∂xxC = 0. The solution has the form C(x) = ax + b. By applyingboundary conditions C(x = 0) = C1 and C(x = e) = C2, we findb = C1 and a = (C2 − C1)/e. Hence the solution is

C(x) =C2 − C1

ex+ C1

or in non-dimensional form

C∗ = x∗ , where C∗ =C(x)− C1

C2 − C1and x∗ =

x

e






The flux density is a diffusive one : j = −D∇C.There are variations only along x so that

j = −D∂xCex = D(C1 − C2)/eex

or in non dimensional form

j∗ = j/(D(C1 − C2)/e) = 1.

The flux density is than integrated on the trans-verse surface area to get the total mass fluxthrough the wall :

J =C1 − C2

eDS

0.0 0.2 0.4 0.6 0.8 1.0x ∗

0.0

0.2

0.4

0.6

0.8

1.0

C∗

C

0.0 0.2 0.4 0.6 0.8 1.0x ∗

0.94

0.96

0.98

1.00

1.02

1.04

1.06

j∗

j





Exercise : Steady diffusion equation in cylindrical geometry

Solve the mass transport equation through acylindrical tube wall of thickness e, with boun-dary values C = Ci and C = Co on the sur-faces at r = ri and r = re respectively. Oncethe concentration profile is obtained, computethe mass flux density j(r) and the mass fluxthrough a piece of tube of length L.A gaz with helium molar fraction xi = 0.02flows inside the tube. Only the helium is ableto diffuse through the tube wall. Outside thetube, the helium molar fraction is xo = 0.Data : D = 1.08 10−8m2s−1, ri = 0.03m,re = 0.031m and L = 1m.






The transport equation (24) simplifies to

∆C =1

r∂r(r∂rC) +

1

r2∂2θθC + ∂2zzC = 0

with ∂θ ≡ ∂z ≡ 0 since C varies with r only. So ∂r(r∂rC) = 0.Hence r∂rC = a, with a a constant. Finally C has the form :

C(r) = a ln(r) + b

Using the boundary conditions C(ri) = Ci and C(ro) = Co we findthe values of the constants and the solution is

C(r) = (Co − Ci) ln(r/ri)/ ln(ro/ri) + Ci


C∗ =ln(r∗)

ln(r∗o)where C∗ =

C(r)− CiCo − Ci

, r∗o =rori

and r∗ =r

ri






The flux density is j = −D∇C = −D∂rCer.Finally

j =D(Co − Ci)r ln(ro/ri)

or normalizing distances by ri and flux densitiesbyD(Co−Ci)/ri, we get j∗ = 1/(r∗ ln r∗o). Thetotal flux through the wall of a tube of length Lis found by multiplying the flux density by thesurface area crossed by the mass flux 2πrL and

J =Co − Ci(ln(ro/ri)2πLD

)Plots on the right were obtained for ro = 2ri

1.0 1.2 1.4 1.6 1.8 2.0r ∗

0.0

0.2

0.4

0.6

0.8

1.0

C∗

C

1.0 1.2 1.4 1.6 1.8 2.0r ∗

0.7

0.8

0.9

1.0

1.1

1.2

1.3

1.4

1.5

j∗

j






If the helium molar fraction inside the tube xi is known, we need toswitch to a (molar) concentration Ci. Let’s use the perfect gas law :PHeV = nHeRT with PHe = xiP and Ci = nHe/V . Thus

Ci = xiP

RT

With the numerical values provided and R = 8.31,Ci = 0.808mol/m3 and then J = 1.67 10−6mol/s.





Exercise : Steady diffusion equation in spherical geometry

Solve the mass transport equation through aspherical shell wall of thickness e, with boun-dary values C = Ci and C = Co on the sur-faces at r = ri and r = re respectively. Oncethe concentration profile is obtained, computethe mass flux density j(r) and the mass fluxthrough the spherical shell.






The transport equation (24) simplifies to

∆C =1

r2∂r(r

2∂rC) + 0 + 0 = 0

since ∂θ ≡ ∂z ≡ 0. So ∂r(r2∂rC) = 0. Hence r2∂rC = a, with a a

constant. Finally C has the form :

C(r) = −ar

+ b

Using the boundary conditions C(ri) = Ci and C(ro) = Co we findthe values of the constants and the solution is

C(r) = (Co − Ci)ri(ro/r − 1)/(ri − ro) + Co


C∗ =r∗i (1− r∗)r∗(1− ri∗)

where C∗ =C(r)− C0

Co − Ci, r∗i =

riro

and r∗ =r

roMass Transport and Transfer





The flux density is j = −D∇C = −D∂rCer.Finally j = D(Ci − Co)riro/(r

2(ro − ri)) ornormalizing distances by ro and flux densitiesby D(Ci − Co)/ro, we get j∗ = r∗i /(r

∗(1 −r∗i )). The total flux through the spherical shellis found by multiplying the flux density by thesurface area crossed by the mass flux 4πr2 and

J =Ci − Co(ro−ri

4πDriro

)Plots on the right were obtained for ro = 2ri

0.5 0.6 0.7 0.8 0.9 1.0r ∗

0.0

0.2

0.4

0.6

0.8

1.0

C∗

C

0.5 0.6 0.7 0.8 0.9 1.0r ∗

1.0

1.2

1.4

1.6

1.8

2.0

j∗

j





The concept of resistance to mass transfer

Using an analogy with electricity and Ohm’s law, it is possible todefine a resistance to mass transfer for steady state problems. Thedifference of concentration through a medium is equivalent to adifference of electrical potential and the mass flux is equivalent tothe current. So

J =C1 − C2

Rand R =

e

DA(27)

where e is the medium thickness, and A is homogeneous to a surfacearea.

Geometry thickness e definition of A

plane xo − xi A = S

cylindrical ro − ri A = Ao−Ailn(Ao/Ai)

, Ao = 2πroL, Ai = 2πriL

spherical ro − ri A =√AoAi, Ao = 4πr2o , Ai = 4πr2i








4 Exercises





Exercise : solve the unsteady diffusion equation for ageneral initial condition

Problem :

Find the solution of the time-dependent 1D diffusion equation

∂C

∂t= D∂

2C

∂x2

with a general initial condition

C(x, t = 0) = C0(x)

We assume that C0(x) is provided. Later the solutions for C0(x) =δ(x) (the dirac distribution) and C0(x) = H(x) (the heaviside dis-tribution, i.e. a ”step“) will be provided as examples.





Solution with the Fourier transform method

Let’s recall that the Fourier transorm f(k) of a function f(x) isdefined by

f(k) =

∫ +∞

−∞f(x)e−ikxdx

From this definition we see that ∂f∂t (x, t) = ∂f

∂t (k, t) and that ∂f∂x (x, t) =

(−ik)f(k, t) (use integration by parts + the property limx→±∞ f(x, t) =

0), and thus∂2f

∂x2(x, t) = −k2C(k, t). Then, if we take the Fourier

transform of the diffusion equation we get

∂C

∂t(k, t) = −k2DC(k, t)

Not that with this process, we have switched from a PDE on C toan ODE on C !






It is straightforward to integrate the last equation against t to obtain

C(k, t) = A(k)e−k2Dt,

where A(k) is some function not depending on t. To fix it we mustuse the initial condition C(x, t = 0) = C0(x). But since we areworking in Fourier space, let’s transform it in :

C(k, t = 0) = C0(k)

Using this in the general solution we see that A(k) = C0(k). Thesolution to our diffusion problem is thus (in Fourier space)

C(k, t) = C0(k)e−k2Dt






We need to switch back to real space. For this we will use theso-called convolution theorem : if f(k) = g(k)h(k), then f(x) =(g ∗ h)(x) =

∫ +∞−∞ g(x − y)h(y)dy where ∗ represents the convolu-

tion operation. Since we obtained C(k, t) = C0(k)e−k2Dt, we could

use the theorem if we find S(x, t) such that S(k, t) = e−k2Dt. By

definition of the inverse Fourier transform,

S(x, t) =1

2π

∫ +∞

−∞S(k, t)eikxdk =

1

2π

∫ +∞

−∞e−k

2Dt+ikxdk

Let’s complete the square in the exponent :

S(x, t) =1

2π

∫ +∞

−∞e−Dt[k−

ix2Dt ]

2− x2

4Dtdk =e−

x2

4Dt

2π

∫ +∞

−∞e−Dt[k−

ix2Dt ]

2

dk






To continue, we perform the change of variable K =√Dt[k − ix

2Dt],

then dK =√Dtdk and

S(x, t) =e−

x2

4Dt

2π√Dt

∫ +∞

−∞e−K

2dK

As you all know ;-), the value of the integral is√π and we finally

get

S(x, t) =1√

4πDte−

x2

4Dt

(this is a normal law with zero-mean and standard deviation√

2Dt)






Now that we know the real form of S(x, t), we can apply the convolu-tion theorem to get the real form of C(x, t) : C(x, t) = (S∗C0)(x) =∫∞−∞ S(x − y)C0(y)dy, so that the general solution to the time-

dependent diffusion equation (for a ’not-to-exotic’ initial condition)is

C(x, t) =1√

4πDt

∫ ∞−∞

e−(x−y)2

4Dt C0(y)dy






Since it is a little disapointing obtain a solution with an integral tocompute, let’s apply this result to the case C0(x) = δ(x). Physi-cally it represents the diffusion of a very concentrated spot initiallylocated at x = 0. Using the property

∫∞−∞ f(x)δ(x)dx = f(0), we

immediately get the result to our spot diffusion problem :

C(x, t) =1√

4πDte−

x2

4Dt






Conclusion : the solution to thetime diffusion problem of a spotis a Gaussian function (or normaldistribution) with zero-mean anda standard deviation σ =

√2Dt.

Keep in mind that the standarddeviation of this law is a mea-sure of the width of the function.This solution tells us that the spotwill be spread by diffusion isotropi-cally and on a characteristic lengthscale evolving as

√Dt. Do this t1/2

behaviour reminds you on something ?

1.0 0.5 0.0 0.5 1.0x

0.0

0.5

1.0

1.5

2.0

2.5

3.0

C

−σ +σ

2σ

C





Solution with an educated-guess-type of change ofvariables method

Another way of solving the diffusion equation consists in finding achange of variables that would transform it in an ODE. We have saidthat diffusion occurs on length scales of the order of Ld =

√Dt.

Therefore, imagine we rescale the space coordinate x in X = x/Ld.A solution varying on a length scale Ld when using coordinate xwould not vary anymore using coordinate X.This kind of solution iscalled a self-similar solution. Using the chain-rule, let us express thederivatives of C against t and x as ones against X :

∂C

∂t=∂C

∂X

∂X

∂t= −1

2

∂C

∂X

x√Dt3/2

∂C

∂x=∂C

∂X

∂X

∂x=∂C

∂X

1√Dt






Then

∂2C

∂x2=

∂

∂x

[∂C

∂x

]=

∂

∂x

[∂C

∂X

1√Dt

]=

∂

∂X

[∂C

∂X

1√Dt

]∂X

∂x=∂2C

∂X2

1

Dt

Now we reinject everything into the diffusion equation :

∂tC = D∂2xxC ⇔ −1

2∂XC

x√Dt3/2

= D∂2XXC1

Dt

after some simplifications, and since now X is the only variable theequation reads :

2C ′′(X) +XC ′(X) = 0






Now let Y (X) = C ′(X), and the equation becomes

2Y ′(X) +XY (X) = 0

which can be solved easily :

Y (X) = Ae−X2/4

where A is a constant to determine with the initial condition. Thenback to C :

C ′(X) = Ae−X2/4

and so C(X) has the form

C(X) =

∫ X

0Ae−X

′2/4dX ′ +B






Recall the definition of the error function :

erf(x) =2√π

∫ x

0e−x

′2dx′

Then the solution we have found is

C(X) = A

√π

2erf(X/2) +B






The solution will be given for an initial condition being a step withvalue CL for x < 0 and CR for x > 0. When x < 0 and t → 0,X → −∞ and C = CL whatever X is, so

CL = A

√π

2erf(−∞) +B = −A

√π

2+B

When x > 0 and t→ 0, X → +∞ and C = CR whatever X is, so

CR = A

√π

2erf(∞) +B = A

√π

2+B

Let’s substract these two results to get A, and add them to obtainB, so that we find :

A =CR − CL√

πand B =

CR + CL2






Inserting these values in the general solution we finally have :

C(X) =CR + CL

2+CR − CL

2erf(X/2)

or

C(x, t) =CR + CL

2+CR − CL

2erf

(x

2√Dt

)or again

C∗ =C(x, t)− CLCR − CL

=1

2

[1 + erf

(x

2√Dt

)]






Solution for three successive times plotted against x (left) andagainst x/(2

√Dt) (right).As expected, all the curves collapse on

the unique self-similar solution when the X variable is used.

1.0 0.5 0.0 0.5 1.0x

0.0

0.2

0.4

0.6

0.8

1.0

C

C(x,t=0)

C(x,t=t1 )

C(x,t=t2 )

C(x,t=t3 )

3 2 1 0 1 2 3x/(2

√Dt )

0.0

0.2

0.4

0.6

0.8

1.0

C

C(x,t=0)

C(x,t=t1 )

C(x,t=t2 )

C(x,t=t3 )








4 Exercises





Introduction of non-dimensional variables

Let us denote the characteristic time scale of a mass transfer phe-nomena T , its length scale L, its concentration scale C0, its velocityscale U and its reaction rate scale kCn0 . We now introduce the fol-lowing non-dimensional variables :

t∗ = t/T , x∗ = x/L, u∗ = u/U , C∗ = C/C0 and r∗ = r/(kCn0 )

By construction, all the variables with a star are of order one in ourproblem. Do not forget that the ∇ operator is based on one spacederivative and is thus of order to 1/L. Similarly, the ∆ operator isbased on two successive derivatives in space, and is thus of order1/L2. We then introduce the scaled operators :

∇∗ = ∇/(1/L) and ∆∗ = ∆/(1/L2)





Introduction of non-dimensional variables

The mass transport equation is recalled here

∂C

∂t+ u∇C = D∆C + r

We can express all the variables and operators in this equation asfunctions of the non-dimensional ones :

t = t∗T , x = x∗L, u = u∗U , C = C∗C0, and r = r∗(kCn0 )

and for the operators

∇ = ∇∗/L and ∆ = ∆∗/L2

Then inject everything in the transport equation :

∂C∗

∂t∗C0

T+ u∗∇∗C∗UC0

L= D∆∗C∗

C0

L2+ r∗kCn0





The non-dimensional mass transport equation

Multiply both sides by L/(UC0) to obtain[L

UT

]∂C∗

∂t∗+ u∗∇∗C∗ =

[DLU

]∆∗C∗ +

[kCn−10 L

U

]r∗

or1

St

∂C∗

∂t∗+ u∗∇∗C∗ =

1

Pe∆∗C∗ +Dar∗

We recognize the aforementioned Peclet and Damkohler numbers.The first one is new and is called the Strouhal number

St =TU

L

It compares some time scale T to the advective time scale. Usually,if the Strouhal number is needed, the time scale T is imposed by anexternal forcing.





The non-dimensional form : comparing different effects

If T not linked to something external, it is either an advective timescale and St = 1, or a diffusive time scale and St = Pe or a reactiontime scale and 1/St = Da (check all this !).Let’s recall the non-dimensional mass transport equation :

1

St

∂C∗

∂t∗+ u∗∇∗C∗ =

1

Pe∆∗C∗ +Dar∗

By construction, every term involving only variables or derivativeswith star superscripts is of order one. On the other hand, non-dimensional numbers have values fixed by the characteristic scalesof the problem (and are generally not of order one). Therefore, theabove non-dimensional form provides an effective and easy way toevaluate the relative weight of the different terms in the solution.






Example

Particles of 10−6m are transported with a flow velocity of 10m/sin water (D = 10−8m2/s). There is no reaction and no externalforcing. The general form of the mass transport equation in thiscase is

∂C

∂t+ u∇C = D∆C

With this problem, both diffusion and advection exist. But shouldbe consider both or may we neglect one of the two phenomena ?






The mass transport is governed by the following non-dimensionalequation :

∂C∗

∂t∗+ u∗∇∗C∗ =

1

Pe∆∗C∗

The Peclet number is here Pe = UL/D = 10.10−6/10−8 = 1000Thus we see that the temporal and advection terms are of order one(only “star” variables) and the diffusion term ifs of order 1/Pe =10−3. Then we conclude that the diffusion term is negligibly smallcompared to the two other ones and the problem is governed by theequation

∂C∗

∂t∗+ u∗∇∗C∗ = 0

Yes, it is possible to neglect diffusion.





The non-dimensional form : a proof of the allmightiness ofnon-dimensional numbers

Consider a mass transport problem with velocity scale U = 0.1m/s,length scale L = 10 cm and diffusivity D = 10−5m2/s (somethinghappening at large scale in a gas). The Peclet number is Pe = 10.Consider a second mass transport problem with scale U = 1 cm/s,length scale L = 1µm and diffusivity D = 10−9m2/s (somethinghappening at very small scale in a water). The Peclet number isPe = 10. Do you believe the solution is exactly the same be causethese two problems have the same Peclet number ? They both obeyto exactly the same equation

∂C∗

∂t∗+ u∗∇∗C∗ =

1

100∆∗C∗,

so yes, the (non-dimensional) solution is the same !




Mass boundary layerIntroduction of the mass transfer coefficientThe Sherwood numberExamples of mass transfer coefficient correlations

Plan




4 Exercises







3 Mass transfer at interfacesMass boundary layerIntroduction of the mass transfer coefficientThe Sherwood numberExamples of mass transfer coefficient correlations

4 Exercises





Mass boundary layer

Consider an interface between two phases(liquid/gaz, solid/liquid, solid/gaz) onwhich the concentration Ci is constant.The concentration far from the interfacein the fluid is C∞. If there is a flow alongthe interface and a diffusion process in thefluid between the interface and the bulkof the fluid, a so-called boundary layerappears near the interface. It correspond toa thin fluid zone in which the variation ofconcentration is localized. The extent ofthis zone δm is called the mass boundarylayer thickness.





Mass boundary layer

The film model

Sometimes, if the transport through theboundary layer is mainly diffusive (low Pe),the assumption that the concentration pro-file is linear in the boundary layer is done.It is called the film model (or film theory).In this case, the diffusive flux through theboundary layer is exactly

jd = −DC∞ − Ciδm








4 Exercises





Mass transfer coefficient

Mass transfer across the boundary layer

Since at the interface the fluid velocity va-nishes (adherence), sufficiently close to itadvection becomes dominated by diffusionand the mass flux density is j = −D∇C.In particular, it is true right at the interface,where the mass flux density through the wallis

jw = −D ∂C

∂x

∣∣∣∣w

,

where x is the direction normal to the wall.







The derivative in this expression is the slopeof the concentration profile at the wall.Hence, knowledge of the concentration pro-file through the boundary layer is required tocompute the mass transfer between the twophases. To obtain this profile, it is necessaryto solve the full mass transport equation andhydrodynamics equations. This is rarely pos-sible analytically, and often very complicatednumerically.







Instead, we an use an ersatz : since the massflux is driven by a concentration variation,we write it as

j = k(Ci − C∞),

where k is the mass transfer coefficient.Note that it could also be written

j = (Ci − C∞)/R,

where R is a resistance to mass transfer.






The dimension of the mass transfer coefficient k is LT−1, expressedin m/s in SI units. Note that if the boundary layer is dominated bydiffusion (Pe is low), the film model can be used and the mass fluxdensity is

j = k(Ci − C∞) = jd = −DC∞ − Ciδm − 0

= DCi − C∞δm

which shows that

k =Dδm

Hence the smaller the boundary layer thickness, the stronger thetransfers (and vice-versa).





Form of correlations for the mass transfer coefficient

For practical problems, semi-empirical correlations can be found inthe litterature for k, or rather its non-dimensional form kL/D = Shcalled the Sherwood number.To guess what these correlations look like, let us use the BuckinghamΠ theorem. The transfer through the boundary layer depends on theflow and on the fluid/solute properties. The flow depends on thevelocity scale U , a length scale L and the kinematic viscosity ν. Theproperties of the fluid/solute are the fluid viscosity ν and the solutemolecular diffusivity in the fluid D. So our problem is then to find acorrelation of the form

k = f(U,L, ν,D)






We have a relation between 5 variables (k, U , L, ν and D) and allthe dimensions of these variables can be expressed with units oflength L and time T (2 dimensions). Hence the Buckingham Πtheorem says that the correlation can be written with only5− 2 = 3 non-dimensional numbers. To find them, let’s choose ourtwo independent variables. For example L and D ([L] = L and[D] = L2T −1 so it is really two independent variables 1). Forinstance [D] = [ν] = L2T −1 so these two are not independent andthey would not be a correct choice. If L and D are our independentvariables, all the dimensions of the other variables can be recoveredby combinations of powers of L and D of the form LαDβ.

1. [X] means “dimension of X“Mass Transport and Transfer





Hence the three remaining variables (k, U and ν) can be non-dimensionalized by something of the form LαDβ. This process willlead to three non-dimensional variables (or numbers).If we try to non-dimensionalize k, we must verify [k] = [LαDβ] ormore precisely

LT −1 = Lα(L2T −1)β

For this to be possible, the power of L on the right hand side shouldbe 1, so that 1 = α + 2β. Similarly, the power of T on the righthand side should be −1, so that −1 = −β. After solving this system,we get β = 1 and α = −1. So it means [k] = [L−1D1], and thatwe can define a non-dimensional number Π1 = k

L−1D1 . Actually wehave found Π1 to be the so-called Sherwood number Sh,

Π1 = Sh =kL

DMass Transport and Transfer





Now we apply the procedure to U : we must ensure [U ] = [LαDβ](with new α and β exponents of course !). But in this precise case,since the dimension of U is the same as the dimension of k, we aregoing to find the same result as for k, namely α = −1 and β = 1and the second non-dimensional number is Π2 = U

L−1D1 . Actually,we have found Π2 to be the Peclet number

Π2 = Pe =UL

D






Finally, we apply the procedure to ν. We must have [ν] = [LαDβ],or

L2T −1 = Lα(L2T −1)β

so we have the system : 2 = α + 2β and −1 = −β. It is solvedin β = 1 and α = 0. Then the third non-dimensional number isΠ3 = ν

L0D1 = νD . This is the Schmidt number Sc :

Π3 = Sc =ν

D






The Buckingham Π theorem states that the dimensional relation

k = f(U,L, ν,D)

can be recast in the non-dimensional form Π3 = f(Π1,Π2) or here

Sh = f(Pe, Sc)

Since Pe = ULD = UL

ννD = Re.Sc, the above relation an also be

writtenSh = f(Re, Sc) or Sh = f(Re, Pe)

The existence of these different possibilities is linked to thedifferent possibilities in chosing the independent variables. But inthe end any choice is correct.








4 Exercises





The Sherwood number

From the previous slides, it is obvious that the Sherwood numberis the non-dimensional version of the mass transfer coefficient. Butit is also more than that. It is actually a measure of the ratio of aconvective mass transfer (mass transfer through a boundary layercaused by a non-zero fluid velocity, i.e. both advective and diffusive)to a purely diffusive mass transfer (through the boundary layerif advection was ignored). We have seen that the flux density of theconvective mass transfer is jc = k(Ci − C∞) and we know that adiffusive flux density is jd = −D∇C ∼ D(Ci − C∞)/L where L isthe length scale on which concentration gradients spread. Then

jcjd

=k(Ci − C∞)

D(Ci − C∞)/L=kL

D= Sh








4 Exercises





Examples of mass transfer coefficient correlations

Important note

Whenever you need a correlation for the mass transfer coefficient k,look for a correlation for the Sherwood number Sh = kL/D. It isexactly the same !

Warning

Before using any of these correlations, check in a textbook the exacthypotheses and the adequation with your flow/geometry !






Liquid flowing inside a tube

Consider a tube of length L and diameter d.

For L/d > 60 :

if Re < 2100 : Leveque formula Sh = 1.86(Re.Sc. dL )1/3

if 0.7 < Sc < 100, 104 < Re < 1.0 105 : Colburn formulaSh = 0.023Re0.8Sc1/3

if 0.5 < Sc < 1.5 and 104 < Re < 106 :Sh = 0.0214(Re0.8 − 100)Sc0.4

For 20 < L/d < 60 :Sh = ShL/d>60

[1 + 6 dL

]For 2 < L/d < 20 :Sh = ShL/d>60

[1 +

(dL

)0.7]






Fluid flowing over a flat plate (forced convection)

Consider a flat plate, and that we are intersted in the flux densityat distance x from the upstream point of the plate (Shx innon-dimensional form) or in the total flux between the upstreampoint of the plate and another point at distance L (ShL).

For a laminar flow :

if Sc < 0.1 : Shx = 0.565Re1/2x Sc1/2, where Rex = Ux/ν and

ShL = 1.13Re1/2L Sc1/2, where ReL = UL/ν

if Sc > 0.1 : Shx = 0.332Re1/2x Sc1/3 and

ShL = 0.664Re1/2L Sc1/3

For a turbulent flow (Rex > 3 105) and for 0.5 < S < 50 :

Shx = 0.029Re4/5x Sc1/3 or ShL = 0.036Re

4/5L Sc1/3






Fluid flowing around a cylinder (forced convection)

Consider a cylinder of diameter d. The Reynolds number is definedas Re = Ud/ν, the Sherwood number corresponds to an averageflux density (the real flux density depends on the position on thecylinder)

If 1 < Re < 4000 : Sh = 0.43 + 0.53Re1/2Sc0.31

If 4000 < Re < 40000 : Sh = 0.43 + 0.193Re0.618Sc0.31

If 40000 < Re < 400000 : Sh = 0.43 + 0.0265Re0.805Sc0.31






Fluid flowing around a sphere (forced convection)

Consider a sphere of diameter d. The Reynolds number is definedas Re = Ud/ν,if 1 < Re < 70000 and 0.6 < Sc < 400 : Sh = 2 + 0.6Re1/2Sc1/3






Free convection

The previous examples were based on flows generated by forcedconvection, in other words generated by an external source. In thecase of free convection, i.e. the flow is generated by buoyancy, thecorrelations are different and involve the Grashof number Gr orthe Rayleigh number Ra. They are the equivalents of the Reynoldsnumber and Re.Sc for free convection flows.




Gas diffusion through a tube wallDiffusion of O2 in waterEstimate of the human lung surface with a study of oxygen transfer in bloodDiffusion boundary layer

Plan




4 Exercises








4 ExercisesGas diffusion through a tube wallDiffusion of O2 in waterEstimate of the human lung surface with a study of oxygentransfer in bloodDiffusion boundary layer





Already done.





Diffusion of O2 in water

Consider an air/water interface. The pressure is P = 1 bar. TheO2 molar fraction in the air is x = 0.2. Henry’s law is recalled :the equilibrium concentration of O2 in the water Ce is given byHeCe = pO2 , where He is Henry’s constant and pO2 = xP is thepartial pressure of dioxygen in the air.

Determine Ce.

Data : for dioxygen in water, He = 28 bar.m3/kmole.





Diffusion of O2 in water

The liquid is supposed to occupy the semi-infinite half space z >= 0.At the interface with air, the concentration in the liquid is C(z =0, t) = Ce. Far away from the interface, C(z → ∞, t) = C∞ =10−4mol/L.

Determine C(z, t).

Determine the penetration depth zp such that(C(zp, t)− C∞)/C∞ = 0.01.

Determine the quantity of O2 transfered inside the water perunit surface area in 10 minutes.

Data : For diffusion of dioxygen in water, D = 1.95 10−9m2/s.





Estimate of the human lung surface with a study of oxygentransfer in blood

A 100 kg person climbs upstairs (3m) in 20 seconds.

Compute the O2 consumption (in mol/s) considering thatenergy is provided by glucose oxydation with a 80% efficiceny.

Data : glucose oxydation is C6H12O6 + 6O2 → 6CO2 + 6H2O.Enthalpy of oxydation is ∆H = 2808 kJ/mol of C6H12O6






We consider the oxygen transfer from the air inside the lungs to theblood can be decomposed in three successive steps.

1 Absorption of O2 from the air on the external alveolocapillarymembrane

2 Diffusion of O2 dissolved in blood from the external to theinternal surface of the alveolocapillary membrane

3 Diffusion of O2 from the internal surface of thealveolocapillary membrane to the bulk of the circulating blood






Determine Ce using Henry’s law. For oxygen in bloodHe = 28 bar.m3/kmol.

Determine the relation between the O2 diffusion flux throughthe alveolocapillary membrane and Ce − Ci.Determine the relation between the O2 flux at the innerinterface and Ci − C0.

Estimate the mass transfer coefficient for a blood velocity of1mm/s

Determine the total O2 flux from the lungs to the bloodconsidering C0 = 0 (due to oxygen complexation withhemoglobin).






Determine the corresponding lung surface necessary forgetting 100 kg upstairs in 20 seconds. Actually, the lungs havebetween 300 and 400 million alveoli (diameter between 0.1and 0.3 mm) and a total surface of 100m2. Comment.

Reproduce the computations with blood velocity of 5mm/s.Comment.

Data : blood density : ρ = 1000 kg/m3, dynamic viscosityµ = 0.01 kg/m/s. Diffusion coefficient of oxygen in bloodD = 1.8 10−9m2/s. Vein dimensions : length is 10 cm, internalradius is ri = 1mm, alveolocapillary membrane thickness ise = 1µm.





Diffusion boundary layer

To do when fluid mechanics is finished...


mass transport and transfer - yannick.hallez.free.fryannick.hallez.free.fr/data/masstransfer.pdf ·...

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