1-43

Chapter 1

λ = 22 AB

T

P

κπσ

= ( ) ( )

( ) ( )23

210 5

1.38 10 373

2 3.635 10 10 1.013 10π

−

−

×

× × × = 8.65×10-9 m

λ = 86.5 o

A The Knudsen number is then:

Kn = λ/d = 86.5/4000 = 0.022 < 0.05. Therefore, diffusion inside the pores occurs only by ordinary diffusion. At a total pressure of 10 atm, DAB = 0.86/P = 0.86/10 = 0.086 cm2/s. The effective diffusivity is then

DAB,eff = ABDετ

= 0.4 0.086

2.5

× = 0.014 cm2/s.

1.5-b Knudsen diffusion in porous solid: When the molecular mean free path is much larger than the diameter of the pore in which the diffusing molecules reside, the molecules collide with the wall rather than colliding with other molecules. This mode of transport is known as Knudsen diffusion. In a straight cylindrical pore of diameter d and length L, the Knudsen diffusivity, DK,i, is given by

DK,i = 3

d

1/ 28

i

RT

Mπ

for Kn > 5 (1.5-4)

The Knudsen flux is given by

Ni = , 1 2( )K i i iD p p

RTL

− (1.5-5)

For binary gas mixtures, the species fluxes are related by

B

A

N

N = − A

B

M

M (1.5-6)

For Knudsen diffusion in porous solids of porosity ε and tortuosity τ,

DK,i,eff = K,iDετ

(1.5-7)

Ni = K,i,eff 1 2( )i iD p p

RTL

−

(1.5-8)

1-44

Example 1.5-21 ------------------------------------------------------------------------------ A mixture of O2 (A) and N2 (B) diffuses through the pores of a 2-mm-thick piece of unglazed porcelain at a total pressure of 0.1 atm and a temperature of 293 K. The average pore diameter is 0.1µm, the porosity is 30.5%, and the tortuosity is 4.39. Estimate the diffusion fluxes of both components when the mole fractions of O2 are 80% and 20% on either side of

the porcelain. The collision diameters for oxygen and nitrogen are 3.467 and 3.798 o

A , respectively. Solution ----------------------------------------------------------------------------------------------

σAB = 0.5(σA + σB) = 0.5(3.467 + 3.798) = 3.632 o

A

λ = 22 AB

T

P

κπσ

= ( ) ( )

( ) ( )23

210 5

1.38 10 293

2 3.632 10 0.10 1.013 10π

−

−

×

× × × = 6.81×10-7 m

λ = 6810 o

A The Knudsen number is then:

Kn = λ/d = 6810/1000 = 6.81 > 0.05. Therefore, diffusion inside the pores occurs mainly by Knudsen diffusion with diffusivity given by

DK,A = 3

d

1/ 28

i

RT

Mπ

= 710

3

−

1/ 2

8 8314 293

32π× ×

× = 1.47×10-5 m2/s = 0.147 cm2/s

The effective diffusivity is then

DK,A,eff = K,ADετ

= 0.305 0.147

4.39

× = 0.0102 cm2/s

The diffusion flux of oxygen is given by

NA = K,A,eff 1 2( )A AD p p

RTL

−=

4 4

3

0.0102 10 (0.8 0.2) 1.013 10

(8.314)(293)(2 10 )

−

−

× − × ××

= 1.274×10-3 mol/m2⋅s

The diffusion flux o nitrogen is given by

B

A

N

N = − A

B

M

M⇒ NA = − (1.274×10-3)

32

28 = − 1.362×10-3 mol/m2⋅s

1 Benitez, J. Principle and Modern Applications of Mass Transfer Operations, Wiley, 2009, p. 60

1-45

1.6 Solute Diffusion through Heterogeneous Medium We need to know the diffusivity of solute in biological systems that are heterogeneous medium. The discussion in the following section is general applicable to both the transport of solute from blood across the capillary wall and the transport of solute through a synthetic membrane.

Capillary pore

Solvent flows acrosssolute in small poresincreases drag

rr-a

Solute ofradius a

of radius r

Figure 1.6-1 Diffusion of solute through capillary pore

Let D be the diffusivity of solute in plasma. When the plasma flows through the capillary pore, the value of the diffusivity will change to Dm < D due to steric exclusion and hydrodynamic drag. Steric exclusion refers to the fact that only a fraction of the pore volume is available to the solute. This fraction is also known as the partition coefficient and is given

by the ratio K = 2

2)(

r

ar

ππ −

= (1 − a/r)2. As the solute diffuses through the pore it experiences

hydrodynamic drag since the solvent must flow over the solute surface to occupy the volume left behind by the solute. The Renkin equation provides a relation between the pore diffusivity Dm and the bulk diffusivity D

Dm/D = (1 − a/r)2

−

+

−53

95.009.21.21r

a

r

a

r

a (1.6-5)

The first term in equation (1.6-5) is the partition coefficient K(a/r) = (1 − a/r)2 The second term in equation (1.6-5) represents the increase in hydrodynamic drag as the solute diffuses through the pore.

ωr(a/r) =

−

+

−53

95.009.21.21r

a

r

a

r

a

1-46

The pore diffusivity can be written in terms of the bulk diffusivity, the steric exclusion, and the hydrodynamic drag as Dm = DK(a/r) ωr(a/r) A graphical depiction of Dm and D is given in Figure 1.6-2. Fick’s law for the diffusion of solute in pores is then given as

JS = − DmAPτ1

dx

dC = − DAP τ

ωK

dx

dC (1.6-6)

In this equation, τ is the tortuosity that accounts for the fact that the pore length is greater than the membrane thickness.

Solute

Capillary Tissue space

Capillary wall

PlasmaInterstitialfluid space

CellD

D m

D 0

Figure 1.6-2 Graphical depiction of Dm and D0

There is also a reduction of solute diffusivity as it diffuses through the interstitial fluid space due to the presence of a variety of macromolecules. The correlation by Brinkman can be used to estimate the diffusivity D0 of solute through the gel-like structure of interstitial fluid

D0/D = ωr(a/r) = 2)(

3

11

1

aa κκ ++

The above equation only accounts for the increase in drag as the solute diffuses through the interstitial fluid where the parameter κ depends on the composition of the interstitial fluid or the microstructure of the gel. κ must be determined from experiment. If the solute enters the gel from a bulk solution, the steric exclusion must be included in the correlation

D0/D = K(a/r)ωr(a/r) = exp

+−

2

1fa

aφ2)(

3

11

1

aa κκ ++

1-47

where the partition coefficient K(a/r) = exp

+−

2

1fa

aφ is developed by Ogston. In this

expression, af represents the radius of the macromolecules that are assumed to form very long cylindrical fibers and φ is the volume fraction of the macromolecules within the gel. Example 1.6-1. ---------------------------------------------------------------------------------- (Problem 2.8, Fournier) The following data was obtained by Iwata, Morikawa, and Ikada (1996) for the effective diffusivity of several solutes through a 5% agarose hydrogel. Show how well the Renkin equation

Dm/D = (1 − a/r)2

−

+

−53

95.009.21.21r

a

r

a

r

a

represent this data. What is the effective pore size of the gel based on the Renkin equation?

Solute Molecular weight Gel diffusivity, cm2/sec Glucose Vitamin B12 Myoglobin BSA IgG

180 1,200 17,000 69,000 150,000

1.6×10-6

1.7×10-6

4.0×10-7

1.0×10-7

1.3×10-7

The bulk diffusivity can be estimated from the correlation

D = 1.013×10-4 (MW)-0.46 (cm2/sec) The solute radius can be estimated from the formula

a = 3/1

4

3

AN

MW

πρ=

3/1

4

3

AN

MW

π

Solution ------------------------------------------------------------------------------------------ The solution to this problem requires knowledge of nonlinear curve fitting, a topic reviewed in appendix A. We first need to evaluate the experimental Dm/D or D0/D using the estimated solute bulk diffusivity.

1-48

MW =

150000

69000

17000

1200

180

⇒ D = 1.013×10-4 (MW)-0.46 =

×−

0421.0

0602.0

1147.0

3883.0

9294.0

10 5

Dm,exp =

×−

013.0

01.0

04.0

17.0

45.0

10 5 , therefore (Dm/D)exp =

3086.0

1661.0

3487.0

4378.0

4842.0

The solute radius is computed next

a = 3/1

4

3

AN

MW

π= 10-6×

3903.0

3013.0

1889.0

0781.0

0415.0

a) Find the effective pore radius r in the Renkin equation

Dm/D = (1 − a/r)2

−

+

−53

95.009.21.21r

a

r

a

r

a= f(r)

Since the expression depends on the ratio (a/r), we can multiply a by 107 and later divide the value of r obtained by 107. Therefore, let

a =

903.3

013.3

889.1

781.0

415.0

We need an initial guess for r. Since the largest solute radius is 3.903(×10-7), let r = 5.0. We can then calculate (Dm/D)cal

1-49

(Dm/D)cal =

0038.0

0184.0

1208.0

4843.0

6955.0

The goal of curve fitting is to minimize the error between the experimental data and the calculated values. We need to minimize the following expression Fobject = ∑ [(Dm/D)exp − (Dm/D)cal ]

2

The above equation is called the objective function. We need to choose a value of r so that the objective function is a minimum.

Figure 1.6-2 Effective diffusivity in 5% agarose hydrogel.

We can use the MATLAB function fminbnd or fminsearch to curve fit the data. We first must create a function that contains the expression to be minimized. In this case, the minimizing or objective function is ∑ [(Dm/D)exp − (Dm/D)cal ]

2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.80

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

a/r

Dm

/D

CorrelationExperiment

1-50

Table 1.6-2 lists the function poren used in fminbnd. We need to supply the name, poren, of the Matlab file that contains the function to be minimized and the range of the variable (4,20) to the function fminbnd.

______ Table 1.6-2 Matlab function for fminbnd ______ function y=poren(r) a=[.415 .781 1.889 3.013 3.903]; DmD=[.4842 .4378 .3487 .1661 .3086]; x=a./r; DmDc=((1-x).^2).*(1-2.1*x+2.09*x.^3-.95*x.^5); y=sum((DmD-DmDc).^2); >> rm=fminbnd('poren',4,20) rm = 6.3134 Example 1.6-2. ---------------------------------------------------------------------------------- 4A molecule with a radius of 2.5 nm enters a membrane having pores with a radius of 7.5 nm. For a temperature of 37oC, and assuming the fluid phase is water, what is the apparent diffusivity of the molecule in the pores of the membrane? Solution ------------------------------------------------------------------------------------------ The pore diffusivity can be calculated from

Dm/D = (1 − a/r)2

−

+

−53

95.009.21.21r

a

r

a

r

a

For a/r = 2.5/7.5 = 0.333333 ⇒ Dm/D = 0.166 The molecule molecular weight can be estimated from

a = 3/1

4

3

AN

MW

πρ⇒ MW =

4

3πρNAa3

MW = 4

3π(1)(6.023×1023)(2.5×10-7)3 = 39,420

The diffusivity of molecule in water is given by D (cm2/sec) = 1.013×10-4(MW)-0.46 = 1.013×10-4(39,420)-0.46 = 7.79×10-7 cm2/sec The pore diffusivity is then Dm = (0.166)(7.79×10-7) = 1.29××××10-7 cm2/sec 4 Fournier, R. L., “Basic Transport Phenomena in Biomedical Engineering”, Taylor & Francis, 2007, p. 219.