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//....
.
./