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Diffusion-handout

Diffusion-handout
Diffusion-handout

7.0 TRANSPORT FUNDAMENTALS

?Up to this point we have only been concerned with distribution of contaminants under equilibrium conditions.

?In this section we will concentrate on the dynamic behavior of systems on their way to attain equilibrium.

?The second law of thermodynamics states that:

In the absence of an external energy source, entropy in a system is always increasing until equilibrium is reached

? A manifestation of this law in natural systems of interest to us is the attrition of spatial unevenness in the distribution of mass, heat, concentration, pressure and other system properties.

?For example if one of the properties mentioned above at a certain locality has a higher value than at some other locality then there must be a net flux of the property towards the direction of the property’s lower value so that equilibrium is attained.

?This net flux of the property will cease as soon as the property value becomes uniform throughout the entire system otherwise it would violate the 2nd law of thermodynamics.

?The magnitude of the net flux is directly proportional to the property value gradient (difference in value of property between two localities).

This is known as the gradient-flux law.

7.1 GRADIENT-FLUX LAW-FICK’S 1ST LAW OF DIFFUSION

?Experiment: Consider the following simple experiment:

?At time t = 0, in one end of a narrow transparent plastic pipe filled with standing water, add a drop of a iodine.

?At various times t0, t1, t2, …, t∞ we determine the iodine concentration along the length of the tube.

? Observation(s): (see Figure 1)

? As time goes by we observe the iodine spreading towards the other

end of the pipe.

? Eventually the iodine forms a uniform solution in water ? Comments:

? The iodine concentration profiles along the length of the tube at

times t 0, t 1, t 2, …, t ∞ are shown in Figure 1.

? The net flux of the iodine continues until the concentration is

uniform throughout the entire tube length as shown in the concentration profile at t = t ∞.

? The driving force for this diffusion process is the concentration

gradient ?C C C right left =?across the imaginary interface.

? The net flux is proportional of ?C :

[]

Flux Along F T x x =constant C M L -2≡?????1 [1] ? By convention if the concentration on the RHS of the imaginary

interface is higher, then ?C >0 (positive).

? Similarly if the flux is directed towards the positive x-direction a

positive sign is assigned.

? Thus Fick’s 1st Law of Molecular Diffusion is:

F D dC dx

x x

=? [2]

where F x =mass flux along the x-axis, mol/m 2?s

D x =molecular diffusion coefficient in the x-axis, m 2/s C =concentration, mol/m 3

x =distance, m

Figure 1: Diffusion of Iodine in Water Column

Add iodine

B

Iodine Concentration

C

Distance

x

x = 0

Imaginary Interface

x B

?The molecular diffusion coefficient (diffusivity) is specific to the compound being diffused as well as the carrier medium.

?For a given carrier medium the diffusivity depends on the average mean-square molecular displacement(x2), and the diffusion time (t) as described by the Einstein- Smoluchowski equation:

D

x

t

=

2

2

[3]

?The average mean-square molecular displacement is the distance between the original position of the particle and its position at time t.

?Another quantity of interest related to molecular diffusion is the mean free path (MFP) defined as the average distance traveled by

a molecule between two successive collisions.

?Note that:

MFP MFP MFP

gas liquid solid

>>

?Similarly for diffusivities:

D D D

gas liquid solid

>>

7.2 FICK’S 2nd LAW OF DIFFUSION

? Fick’s 2nd Law is summarized in the following statement: the time

change of a conservative property inside an elementary volume, ?V , is equal to the algebraic sum of the fluxes across all of its boundaries .

? A conservative property is a property that is not subjected to any in situ

sources or sinks.

? Consider a rectangular elementary volume, ?V , placed in the Cartesian

coordinate system with sides ???x,y,z . The three dimensional concentration field at a given locality and time is defined as

()C x y z t ,,,.

? For simplicity we will only consider concentration variation along the

x-axis. Thus the only fluxes for this system are those through the walls perpendicular to the x-axis, positioned at x = 0 and x = ?x. The area

a y z x =???.

? A mass balance for the elementary volume is:

x

()()()d dt C V dC dt

V a F Flux In a F x x x x x ?==????0from left Flux Out to right

1243412434 [4] ? Dividing by ??V a x x =, we get:

()()[]()()[]

dC dt a V

F F x x F F x x

x x x x =?=?????010 [5]

? As the volume of the element approaches zero:

()()[]

dC dt x F F x dF dx dC dt dF dx

x x x x x =?=?=?→lim ???01

0 [6]

? According to the last equation, the temporal concentration change is

equal to the negative spatial gradient of the flux .

? Using partial derivatives Eq. 6 becomes:

????C

t

F x

x const x t const ===.

.

[7]

? Substitution of Eq. 2 into 7 gives Fick’s 2nd Law:

??????????C t x D C x C t D C x

=??????

?=2

2 [8]

? Generalizing Eq. 7 for 3-dimensional diffusion, we get:

????????C x F t F t F t

F x y z =???=?? [9] ? Substitution of Fick’s 1st Law fir each dimension into Eq. 9 yields

Fick’s second law for 3-dimensional diffusion:

??????????????C t x D C x y D C y z D C z x y z =??????+??????+?????

[10]

?For uniform diffusivity along all three dimensions D D D D

x y z

===, Eq. 10 becomes:

???

?

?

?

?

?

C t

D

C

x

C

y

C

z

=++

?

?

?

?

?

2

2

2

2

2

2

[11]

?Eqs. 7-11 representing various forms of Fick’s 2nd Law are linear partial differential equations.

?The solution of these equations is generally exponential functions and depend greatly on:

?the shape of the volume within which diffusion occurs

?the initial and boundary conditions concerning C and its derivatives.

?Analytical solutions are frequently cumbersome. Frequently numerical approximations are used.

7.3 MOLECULAR DIFFUSIVITY ESTIMATION

?Primary parameters affecting molecular diffusivity include: ?Mean free path which in turn is associated with the mean molecular displacement

?Mean velocity which is inversely related to molecular volume and molecule x-sectional area.

?From the above it is apparent that molecular diffusivity must be inversely related to molecular mass and volume.

?Figures 2 and 3 show plots of molecular diffusivity in air and water respectively vs. molar volume and molecular mass for organic

compounds.

? Other parameters influencing molecular diffusivity include:

? Type of Carrier Medium : Media exhibiting more crowding

(higher viscosity) inhibit Brownian motion (e.g. diffusivity values in air are much higher than those in water)

? Temperature : As temperature increases so do diffusivities. ? Estimation of Air Diffusivities: A variety of semi-empirical equations

are available for estimation of air diffusivities:

? Fuller Correlation (Fuller et al .1:)

()()[]

[]

D T M M P V

V

a air =++?10

113

175

12

13

13

2

.// [12]

where D a = molecular diffusivity in air, cm 2/s

T = absolute temperature, K

M air = average molecular mass of air, 28.97 g/mol

M = molecular mass of diffusing species, g/mol

P = gas phase pressure, atm

V air = average molar volume of air, 20.1 cm 3/mol

V = molar volume of diffusing species, cm 3/mol ? Molar volumes can be obtained by summation of elemental

volume contributions given in Table 2.

? Alternatively molar volumes can be obtained from molecular mass

and liquid density data

1

Fuller, E. N., Schettler, P. D., Giddings, J. C., “A New Method for Prediction of Binary Gas-Phase Diffusion Coefficient”, Ind. Eng. Chem., 58, 19-27, 1966.

Table 2: Atomic Diffusion Volume Increments for Fuller’s Correlation

Atomic & Structure Diffusion Volume Increments, V (cm3/mol)Diffusion Volumes for Simple Molecules, V (cm3/mol)

Element V Molecule V

C

H

O

N

Cl

S Aromatic Ring Heterocyclic Ring 16.5

1.98

5.48

5.69

19.5

17.0

-20.2

-20.2

H2

D2

He

N2

O2

Air

Ar

Kr

CO

CO2

N2O

NH3

H2O

CClF2

SF6

Cl2

Br2

SO2

7.07

6.70

2.88

17.9

16.6

20.1

16.1

22.8

18.9

26.9

35.9

14.9

12.7

114.8

69.7

37.7

67.2

41.1

EXAMPLE 1

Using both methods of molar volume estimation determine the molecular diffusivity of benzene in air. Compute the deviation from the experimental value of D a = 0.096 cm 2/s

SOLUTION

A. From Density Data. 1.

The liquid density of benzene is ρB = 0.879 and the molecular mass is 78.1 g/mol.Thus:

V mol benzene =

=7810879889.../ g /mol g /cm

cm 3

3

2.

Compute the air diffusivity

()()[]

[]

()

()()()

[]

D T M M P V

V

D s

a air a =++=+?????

?+=??10

11102981

2917811201

88900903

17512

13

13

2

3

175

05

13

13

2

.//..//..../ cm 2

B. From Volume Contribution Data . 1.

The molar volume of benzene is:

()()()()V C H Aromatic R mol

benzene =+++?=666165620202908ing =. cm 3.../2. Compute the air diffusivity

()()[]

()

()()()

[]

D T M M P V

V

D s

a air a =++=+?????

+=??10

11102981

2917811201

90800893

175

12

13

13

2

3

175

05

13

13

2

.//..//..../ cm 2

C. Deviations from Experimental Value

(% Deviation)Method A = (0.096-0.090)*100/0.096=6.25% (% Deviation)Method B = (0.096-0.089)*100/0.096=7.30%

? Estimation of Water Diffusivities: Several semi-empirical equations

are available in the literature. The simplest one is the Haydouk and Laudie 2.

? Haydouk and Laudie Method D V

w =×?1326105

1140589

...μ [13]

where D w = water diffusivity, cm 2/s μ = solution dynamic viscosity, cP (10-2 g/cm ?s)

V = molar volume of diffusing compound, cm 3/mol

? Molar volumes for Eq. 13 are given in Table 3 EXAMPLE 2

Determine the diffusion coefficient of methanol in water at 25 o C. Water viscocity is 0.89 cP.

SOLUTION 1. Methanol : CH 3 OH, T = 25+273.2 = 2988.2 K, μ = 0.89 cP 2.

Estimate the mola volume of methanol:

C H O V cm mol =×==×==×==1148148

43714817474

3703....... 3.

Use the Haydouk and Laudie equation:

()

()

D V cm s

w =

×=

×=×???132610132610

0893709738105

1140589

5114

0589

42.........μ

2

Haydouk, W. and Laudie, H., “Prediction Diffusion Coefficients for Non-electrolytes in Dilute Aqueous Solutions”, J. AIChE , 20, 611-615, 1974.

Table 3: Atomic Diffusion Volume Increments for Haydouk and Laudie Method of Water Diffusivity Estimation

Atomic & Structure Diffusion Volume Increments, V (cm3/mol)Diffusion Volumes for Simple Molecules, V (cm3/mol)

Element V Molecule V

C

H

O (except as noted below) In Methyl Esters & Ethers In Ethyl Esters & Ethers In Higher Esters & Ethers

In Acids

Joined to S, P, N

N

Double Bonded

In Primary Amines

In Secondary Amines

Br

Cl

F

I

S

Ring, 3-membered

4-membered

5-membered

6- membered

Naphthalene

Anthracene 14.8

3.7

7.4

9.1

9.9

11.0

12.0

8.3

15.6

10.5

12.0

27.0

24.6

8.7

37.0

25.6

-6.0

-8.5

-11.5

-15.0

-30.0

-47.5

H2

O2

N2

Air

CO

CO2

SO2

NO

N2O

NH3

H2O

H2S

Cl2

Br2

I2

14.3

25.6

31.2

29.9

30.7

34.0

44.8

23.6

36.4

25.8

18.9

32.9

48.4

53.2

71.5

7.4 CORRECTION OF DIFFUSIVITY VALUES

? Similar Compound Principle:

? Air or Water :The air or water diffusivity of a compound can be

estimated from the known diffusivity of another compound with similar chemical structure by using the following equation:

D D

M M a w a q

,,.122105

??????

? [14]

where D a w ,1= unknown air or water diffusivity D a w ,2=known air or water diffusivity M 2, M 1 = molecular mass of compound with known or unknown diffusivity respectively

? Water: Unknown water diffusivities can be obtained from known

ones of chemically similar compounds by the following equation:

D

D V V w

w 12

210589

=?????

?. [15]

where D w 1

= unknown water diffusivity D w 2

= known water diffusivity V V 21,= molar volumes of compound with known or

unknown diffusivity respectively ? Temperature and Pressure Correction:

? Air or Gas: Air diffusivity values can be corrected for pressure

and temperature by the following equation:

D D T T P P AB condition AB condition ,,/12

=?????

??????

?

?1232

21 [16] ? Diffusivity of Organic Compounds in Gaseous Mixtures:

? The following equation is used to describe the diffusivity of a

gaseous component in a gas mixture:

D y D y D y D m n n

121231311

????=

′+′++′

(17)

where D 1-m = diffusivity of component 1 in the mixture

D 1-n = binary diffusivity between components 1 and n ′=y n mole fraction of component n in the gas mixture

evaluated on a 1-free basis ? For example ′y 2is defined as following:

′=

+++y y y y y n

22

23... EXAMPLE 3

Determine the diffusivity of carbon monoxide through a gas mixture with the following composition:

y O 2

020070010===... y y N CO 2

2

The temperature of the gas mixture is 298K and its pressure 2 atm. The following binary diffusivities are given:

D CO O ???=××2

01851001921044. m s at 273K, 1 atm

D =. m s at 288K, 1atm

2CO -N 2

2

SOLUTION

1.

Correct the binary diffusivities for 298k, 2 atm:

()()D m s CO O ?????=????

???????

?×=×?????

??????

?×=×2

298273120185

100105102982881201921001011032

424

32

44

/... m s D =. m s m s 2CO -N

/222

2.

Compute the oxygen and nitrogen molar fractions on a CO-free basis:

′=

?=′=

?=y y O N 22

020

1010022070

1010

078..

....

3. Substitute the corrected values of binary diffusivities and the molar

fractions into Eq. 17 to get

D D m s ft hr CO O N CO O

N ?????=

×+×=×=2

2

2

2

1

022*********

010110010210039544

422//....

.

./

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