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Transcript of ChE 333 : Mass transfer Textbook:Fundamentals of Momentum, Heat and Mass transfer. J.R. Welty, R.E....
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ChE 333 : Mass transfer
Textbook: Fundamentals of Momentum, Heat and Mass transfer. J.R.Welty, R.E.Wilson and C.E.Wicks. 5 th Edition , John Wiley (2007).
Reference: Fundamentals of Heat and Mass transfer. Theodore L. Bergman, Adrienne S. Lavine, Frank P. Incropera and David P. DeWitt.
7 th Edition , John Wiley (2011)
Dr. Sharif Fakhruz ZamanDepartment of Chemical and Materials Engineering, Faculty of Engineering, King Abdulaziz University,
Jeddah, KSA
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About myself
Dr. Sharif Fakhruz ZamanAssistant Professor at KAU.
PhD in Chemical and Biochemical Engineering, University of British Columbia, Vancouver Canada (2010).MS in Chemical Engineering, KFUPM, KSA .
BSc in Chemical Engineering, BUET, Bangladesh.
Field of interest : Heterogeneous catalysis and molecular modeling of catalyst and catalytic reactions, reaction kinetics, reactor design, diffusion in zeolites etc.
Office address :Room 216, Building 45, Department of Chemical and Materials Engineering
Faculty of Engineering, King Abdulaziz university, Jeddah, KSA.
E-mail : [email protected] or [email protected] : cell : 0563063594, office : 6402000-ext-68044
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Topic 1 : Fundamentals principles of mass transfer
Introduction to mass transfer and its industrial applications. Week 1
Topic 2 : Diffusion coefficients; mass transfer coefficients
Molecular mass transfer, Fick's rate equation, Diffusion coefficient, Gas mass
diffusivity, Liquid mass diffusivity.Week 2
Pore diffusivity, Knudsen diffusion, Solid mass diffusivity, Convective mass
transfer.Week 3
Topic 3: Differential equations of mass transfer
Modeling mass transfer phenomena, Special form of mass transfer equation, Fick's
second law. Week 4
Commonly encountered boundary conditions, steps for modeling process
involving molecular diffusion. Steps to solve mass transfer problems.Week 5
Major exam 1
Course syllabus
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Topic 4 : Steady state molecular diffusion
One dimensional mass transfer independent of chemical reaction, Pseudo
steady state diffusion.Week 6
One dimensional system associated with reaction (heterogeneous system) Week 7
One dimensional system associated with reaction (homogeneous system),
film theory and penetration theory.Week 8
Major exam 2
Course syllabus
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Topic 5 : Unsteady state molecular diffusion
Unsteady state diffusion and Fick's law, transient diffusion in a semi-
infinite medium.Week 9
Transient diffusion in a finite medium,
Concentration time chart for simple geometric shapes Week 10
Topic 6 : Convective mass transfer
Fundamental consideration of convective mass transfer, Significant
parameters in convective mass transfer, dimensionless analysis for
convective mass transfer, exact analysis of the laminar concentration
boundary layer.
Week 11
Approximate analysis of the laminar concentration boundary layer.
Mass transfer and momentum transfer analogy, Chilton Colburn
analogy.
Week 12
Major exam 3
Course syllabus
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Topic 7 : Convective mass transfer correlations
Mass transfer for plates sphere and cylinders Week 13
Mass transfer through pipes, wetted wall columns,
mass transfer in packed and fluidized bedWeek 14
Final exam Week 15
Course syllabus
Class Schedule
Lectures : Sunday – Tuesday 8:00 – 9:30 am
Tutorial : Sunday : 2:30 -5:20 pm
Class Room : 220 , Building 45
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Grading
My goal is that you to learn the material and make a high grade in the course!
Homeworks ----------------------------------------------------------- 5%
Midterm I and II and III ------------------------------------------ 40%
Weekly in-lecture quizzes + projects--------------------------- 15%
{ Based on class content or core homework problems +
Diffusion lab experiment report}
Written final exam --------------------------------------------------- 40%
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Mass Transfer
When a system consists of two or more components whose concentration vary from point to point , there is a natural tendency of mass transfer minimizing the concentrationdifference within the system.
The transport of one constituent from one region of higher concentration to that of a lower concentration is called mass transfer.
What happens if a lump of sugar added to a cup of black coffee.
Sugar will eventually dissolvediffuse uniformly throughout the coffee.
How long it will take to have uniform concentration of sugar in the coffee cup??
-It depends up on the process- - 1) Quiescent (being at rest/ quite / still)- -2) Mechanically agitated by a spoon.
Mechanism of mass transferDepends on the dynamics of the system in which it occurs
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Mechanism of mass transferDepends on the dynamics of the system in which it
occurs.
Random molecular motion in quiescent fluid.
Transferred from a surface into a moving fluid aided by the dynamic characteristics of the flow.
Mass transfer
Molecular mass transfer. Convective mass transfer.
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Molecular mass transfer examples
(1)Biological process(a)oxygenation of blood(b)Transportation of ions across membranes
within kidney
(2) Chemical processes(c) Chemical vapor deposition(d)Aeration of waste water(e)Purification of ores and isotopes
(3) Chemical separation processes(f) Adsorption(g)Crystallization(h)Absorption(i) Liquid liquid extraction
Component remains at the interface
Component penetrates to the interface and the transfer to the bulk of the 2nd phase.
We will talk little about interface mass transfer : Chapter 29
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Molecular mass transfer
First observed by Parrot 1815.
A gas mixture contains two or more molecular species whose relative concentration varies from point to point, an apparently natural process which tends to diminish any inequalities of composition. This microscopic transport of mass, independent of any convection within the system is called molecular mass transfer.
Kinetic theory of gases can explain mass transfer in gaseous mixture in specific case.
At temperature above absolute zero, individual molecules are in a state of continual yet random motion.
Within dilute gas mixture each solute molecule behaves independently of the other solute molecule, since it seldom encounters them. Collision between solute and solvent molecules are continually occurring. As a result of collision the solute molecules move along a zigzag path sometime towards a region of higher concentration sometime towards a lower concentration.
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Mass transfer refers to mass in transit due to a species concentration gradient in a mixture.
Must have a mixture of two or more species for mass transfer to occur.
The species concentration gradient is the driving potential for transfer.
Mass transfer by diffusion is analogous to heat transfer by conduction.
• Physical Origins of Diffusion:
Transfer is due to random molecular motion.
Consider two species A and B at the same T and p, but initially separated by a partition.
– Diffusion in the direction of decreasing concentration dictates net transport of A molecules to the right and B molecules to the left.
– In time, uniform concentrations of A and B are achieved.
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HomeworkChapter 24 (WWWR) : 1,8,12,13,15,22
TutorialChapter 24 (WWWR) : 3,4,11,13,17,21,22
Example Chapter 24 (WWWR) : 1,2,3,4,5,6,7,8
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Definitions:iCMolar concentration of species i. 3kmol/m
:iMass density (kg/m3) of species i.
:iMMolecular weight (kg/kmol) of species i.
i i iC M* :iJMolar flux of species i due to diffusion. 2kmol/s m
Transport of i relative to molar average velocity (v*) of mixture.
:iN Absolute molar flux of species i. 2kmol/s m Transport of i relative to a fixed reference frame.
:ijMass flux of species i due to diffusion. 2kg/s m Transport of i relative to mass-average velocity (v) of mixture.
Transport of i relative to a fixed reference frame.
:ix Mole fraction of species i / .i ix C C
:im Mass fraction of species i / .i im
Absolute mass flux of species i. 2kg/s m:in
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Concentration :
In multi component mixture , the concentration of a molecular species can be expressed in many ways.
Mass concentration also known as density (gm/cm3)
Mass fraction , ωA:
Mass transfer occurs in mixtures, so it is important to evaluate the effect of each component in the transfer process. To explain the role of a component in the mixture we will use the following definitions.
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Molecular concentration:
Mole fraction: (liquids,solids) ,
(gases)
RT
p
V
n
m
mol
molkg
mkg
Mc AA
A
AA
3
3
/
/
c
cy
c
cx A
AA
A For gases,
Velocity: mass average velocity,
molar average velocity,
Velocity of a particular species relative to mass/molar average is the diffusion velocity.
P
p
RTP
RTpy AAA
n
iii
n
ii
n
iii
1
1
1
vvv
c
cn
iii
1
vV
For GAS only
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mol
Example # 1
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Flux: A vector quantity denoting amount of a particular species that passes per
given time through a unit area normal to the vector, given by Fick’s First Law, for basic molecular diffusion.
Flux can be expressed in different ways , three different expression(1) Flux reference to a coordinate that are fixed in space (Total/absolute flux)
nA = Mass flux and NA = molar flux.
(2) Flux reference to a coordinate that are moving with the mass average velocity (jA).
(3) Flux reference to a coordinate that are moving with the molar average velocity (JA).
Fick’s first law defines the molar flux relative to molar average velocity.
or, in the z-direction,
JA,z = Molar flux of component A in z direction relative to molar average velocity.
DAB = Proportionality factor, Mass diffusivity, diffusion coefficient for component A diffusing through component B .
AABA cD J
dz
dcDJ AABzA ,
Isothermal, Isobaric system
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For a general relation in a non-isothermal, isobaric system,
dz
dycDJ A
ABzA ,
Mass flux relative to Mass average velocity, jA,z.
dz
dDJ AABzA
,Concentration gradient in terms of mass fraction
Initial experimental investigation were unable to verify Fick’s 1st law
Why??
Since mass is transferred by two means:(1) concentration differences (concentration gradient)
and (2) convection differences from density differences (bulk motion)
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For binary system with constant average velocity in z direction Vz,
Thus,
Rearranging to
)( ,, zzAAzA VvcJ
dz
dycDVvcJ A
ABzzAAzA )( ,,
zAA
ABzAA Vcdz
dycDvc ,
)(1
,, zBBzAAz vcvcc
V
)( ,, zBBzAAAzA vcvcyVc
Multiply by cA and rearrange
Molar average velocityc
cn
iii
1
vV
Which substituted, becomes
)( ,,, zBBzAAAA
ABzAA vcvcydz
dycDvc
Defining molar flux, N as flux relative to a fixed coordinate,
AAA c vN
)( ,,, zBzAAA
ABzA NNydz
dycDN
)( BAAAABA yycD NNN Generalized Eqn.
Finally
Bulk motion contributionConcentration gradient contribution
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Average velocity
Stairs : Fixed coordinate
Escalator : Moving coordinate
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Related molecular mass transfer
Defined in terms of chemical potential, molar diffusion velocity:
Nernst-Einstein relation
dz
d
RT
D
dz
duVv cABcAzzA
,
dz
d
RT
DcVvcJ cABAzzAAzA
)( ,,
Mobility of component A
dz
dcDJ AABzA ,
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DIFFUSION COEFFICIENT
Fick’s law proportionality/constant
Similar to kinematic viscosity, n (momentum transfer) m2/s
and thermal diffusivity, (a heat transfer) m2/s
t
L
LLMtL
M
dzdc
JD
A
zAAB
2
32, )
1
1)((
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Gas mass diffusivity {Sutherland – Jeans – Chapman - Cowling}
Theoretical expression for gas mass diffusivity for low density gas mixture Based on Kinetic Gas Theory Assumptions (i) Rigid sphere(ii) No intermolecular forces(iii) Elastic collision l = mean free path length, u = mean speed
uDAA l3
1* 2/1
3
22/3
2/3
* )(3
2
AAAA M
N
P
TD
MA = molecular weight ( gm/mol)N = Avogadro’s number = 6.022 x 10 23 molecules/moleP = system pressure (atm)T = absolute temperature (K)k = Boltzmann constant ( 1.38 x 10-16 erg/mol)σ AB = Lennard Jones diameter of the spherical molecule
Species ‘A’ diffusing through its isotopes ‘A*’
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Hirschfelder’s equation: For Non polar and Non reacting molecule.
Molecular weight (periodic Table)
DAB
BAAB P
MMT
D
2
2/1
2/3 11001858.0
P in atm
Diffusivity in cm2/s.
Collision diameterA Lennard Jones parameter (Å)
Collision integral
Temperature in Kelvin
For binary gas mixture
Diffusion coefficient for gases: DAB = DBA
Gas phase diffusion coefficient , DAB = f(P,T)
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Lennard-Jones parameters and e from tables, or from empirical relations
for binary systems, (non-polar, non-reacting)
ΩD = f(T, intermolecular potential field for one molecule of A and one molecule of B). From Table : K-1;
Extrapolation of diffusivity up to 25 atmospheres
2BA
AB
BAAB eee
2
1
1,12,2
2/3
1
2
2
1
TD
TD
ABAB T
T
P
PDD
PTPT
2/3
1
2
2
1
1,12,2
T
T
P
PDD
PTPT ABAB
Quick estimation
Temperature dependency of collision integral is very nominal/small.
Temperature andPressure correction
of diffusivity
kkBA
AB
eee
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BINARY GAS-PHASE LENNARD-JONES “COLLISION INTEGRAL”
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If Lennard Jones parameter values are not reported :
Empirical relation to estimate Lennard Jones parameter for PURE COMPONENT.
Vb = Molecular volume at normal boiling point (cm3/g mol); Table 24.4
Tc = Critical temperature(K)
Pc = Critical pressure (atm)
Vc = Critical volume (cm3/g mol)
Tb = Normal boiling point temperature (K)
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Gas constant R or Rg :
0.08205746(14) L atm K−1 mol−1
1.9858775(34) cal K−1 mol−1
1.9858775(34)×10−3 kcal K−1 mol−1
8.3144621(75)×107 erg K−1 mol−1
8.3144621(75) L kPa K−1 mol−1
8.3144621(75) m3 Pa K−1 mol−1
8.3144621(75) cm3 MPa K−1 mol−1
8.3144621(75)×10−5 m3 bar K−1 mol−1
8.205746×10−5 m3 atm K−1 mol−1
82.05746 cm3 atm K−1 mol−1
Unit conversion of pressure
Convert from these to pascals (Pa) multiply by
standard atmosphere (atm) 101 325
bar (bar) 100 000kilopascal (kPa) 1 000
megapascal (MPa) 1 000 000millibar (mbar) 100
std centimeter of mercury (cmHg) 1 333.224
millimetre of mercury (mmHg) 133.322 4
Unit conversion of viscosity (μ)
1 poise = dyne·s/cm² = g/cm·s = 1/10 Pa·s1 Pa·s = 1 N·s/m² = 1 kg/m·s
1 cP = 1 mPa·s = Pa·s/1000 = poise/100
Unit of work or energy1 erg = 10-7 J = 10-7 N-m
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Interpolation
If you know the atomic wt. you should be able to calculate the molecular wt if you know the correct chemical formula. Periodic Table – keep one with you.
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With no reliable s or e, we can use the Fuller-Schettler and Giddings correlation,
Careful about addition of structural correction; i.e. aromatic ring, for calculation of diffusion volume from atomic volume.
Calculation of volume of benzene (C6H6) [vapor/gas phase] from table 24.3.
Atomic diffusion volume : C = 16.5; H = 1.98 Diffusion volume = 16.5*6 +1.98*6 +aromatic ring correction
= 99 + 11.88- 20.2 = 90.68
23/13/1
2/1
75.13 1110
BA
BAAB
vvP
MMT
D
Gas Diffusivityin cm2/s
Pressure in atm
Atomic diffusion volume (Table 24.3) cm3/g mol
For binary gas mixture
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Calculation of volume of benzene (C6H6) [vapor/gas phase] from table 24.3.Atomic diffusion volume : C = 16.5; H = 1.98 Diffusion volume = 16.5*6 +1.98*6 +aromatic ring correction
= 99 + 11.88- 20.2 = 90.68
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where
bb
PBAAB TV
232/1 1094.1,
ABTT e /* 2/1
e
e
e BAAB
bT23.1118.1/ e
)exp()exp()exp( ****0 HT
G
FT
E
DT
C
T
ABD
For binary gas with polar compounds, we calculate by
*
2196.00 T
ABD
2/1BAAB
3/1
23.11
585.1
bV
Example of polar gases:NH3 , SO2, H2S, PH3
For binary gas mixture containing POLAR component
Suggested by Brokaw (1969)Hirschfelder equation is valid but need to estimate the collision integral (ΩD) in a
different way.
DAB
BAAB P
MMT
D
2
2/1
2/3 11001858.0
μb = dipole moment, Debye.Vb = Liquid molar volume of the specific compound at its boiling point, cm3/g mol, Table 24-4 and 24-5.Tb = Normal boiling point in K.
Problem Exercise 24-11
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For gas mixtures with several components (multiple components),
nnmixture DyDyDy
D
1'
31'321
'2
1 /...//
1
1
2
32
2'2 1... y
y
yyy
yy
n
1
3
32
3'3 1... y
y
yyy
yy
n
Calculate D1-2, D1-3 --- using any empirical equation or you can get it from literature i.e values reported in Appendix J-1. You may need to perform temperature correction of the diffusivity value if you use the value from the App. J.1.
In appendix J.1 : diffusivity values are reported in the form of => DAB.P [cm2 .atm/s]
you need to divide the value with the system pressure to get the actual value at the reported temperature.
2/3
1
2
1,12,2
T
TDD
PTPT ABAB
132
'
1... y
y
yyy
yy n
n
nn
yn’= Mole fraction of component ‘n’ in the gas mixture evaluated on a component -1-free basis.
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2
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Liquid mass diffusivity No rigorous theories Diffusion as molecules or ions Eyring theory Hydrodynamic theory
Stokes-Einstein equation
Equating both theories, we get Wilke-Chang equation
BAB r
TD
6
6.0
2/18104.7
A
BBBAB
V
M
T
D
Viscosity in centipoises (cP) unit 1 cP = 10−2 P = 10−3 Pa·s = 1 mPa·s
Non electrolyte solute in low concentration solution
Association parameter for solvent, B
Molar volume atNormal boiling point
A = soluteB = solvent
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If data for computing the molar volume of solute at its normal boiling point, VA is not available, Tyne and Calus (1975) recommended the following correlation :
Vc = critical volume of species A in cm3/g. mol. Values are tabulated in literature {Reid, Prausnitz and Sherwood, 1977}.
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Molar volume of ethanol
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6.0
2/18104.7
A
BBBAB
V
M
T
D
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For infinite dilution of non-electrolytes in water, W-C is simplified to Hayduk-Laudie eq.
Scheibel’s equation eliminates FB, simplified form of W-C
Exceptions:1) for benzene as a solvent, if VA < 2VB , use K = 18.9x10-8.
2) For other organic solvents, if VA < 2.5 VB, use K = 17.5x 10-8.
Liquid diffusion coefficient in concentrated solution:
Combine the infinite dilution coefficient DAB and DBA =>
Where, DAB = infinitely dilute diffusion coefficient of A in solvent B.
For associating compound , i.e alcohols,
589.014.151026.13 ABAB VD
3/1A
BAB
V
K
T
D
3/2
8 31)102.8(
A
B
V
VK
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As diffusivity changes with temperature, extrapolation of DAB is by
Tc = Critical temperature of solvent B in Kelvin. Temperatures are in Kelvin unit.
n = Exponent related to the heat of vaporization of solvent(B), ΔHv, at its normal boiling point temperature {Text book page 419 - table for the value of n}.
For diffusion of univalent salt in dilute solution, we use the Nernst equation:
DAB = Diffusion coefficient based on the molecular concentration of A in cm/s2.
R = Gas constant 8.316 Joules/K/g mol. F = faraday’s constant 96500 coulombs/g equivalent
= Limiting ionic conductance in (amp/cm2 )(volt/cm)(g equivalent/cm3)
n
c
c
ABT
ABT
TT
TT
D
D
1
2
)(
)(
2
1
F
RTDAB )/1/1(
200
ll
00 /1/1 ll and
polyvalent salt solution
Replace the constant 2 in the univalent salt diffusion equation with where n+ and n- are the valances of the cation and anion of the polyvalent salt.
nn
11
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Pore diffusivityDiffusion of molecules within pores of porous solids
example Heterogeneous catalysts are porous solids containing active
material inside the pore wall. So reactants need to diffuse through the pore and reach the active metal surface to convert into products.
Separation of solute from dilute solution by the process of adsorption.
Knudsen diffusion for gases in cylindrical pores(1)Pore diameter smaller than mean free path, and (2) density of gas is low Knudsen number
If Kn>>>1 then Knudsen diffusion is important.
From Kinetic Theory of Gases,
diameterpore
speciesgdiffuofpathfreemean
dKn
pore
sin
l
AAA M
NTuD
ll 8
33*
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But if Kn >1, then
If both Knudsen and molecular diffusion exist, then
with
For non-cylindrical pores, we estimate
Apore
A
poreporeKA M
Td
M
NTdu
dD 4850
8
33
KAAB
A
Ae DD
y
D
111
A
B
N
N1
AeAe DD 2' e
dpore in cm
Diffusivity value in cm2/s
Void fraction in the solid, ε = void volume /(total volume of solid + void)
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Types of porous diffusion. Shaded areas represent nonporous solids
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EXAMPLE 6
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Updating the diffusivity value to a different temperature and pressure
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Hindered diffusion for solute in solvent-filled pores
Diffusion of a solute molecule through tiny capillary pore filled with liquid solvent. A general model is
DAB0 = Liquid liquid diffusivity.
F1 and F2 are correction factors, function of pore diameter, values between 0 and 1.
If φ >1, solute molecule is greater than pore diameter, This phenomena is known as solute exclusion. It is used to separate large biomolecules such as proteins from dilute aqueous mixture.
)()( 21 FFDD oABAe
pore
s
d
d
ds = diameter of solute molecule
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F2 is the hydrodynamic hindrance factor, one equation is by Renkin, range in between 0<=φ=>0.6
532 95.009.2104.21)( F
Assumptions for the model:
-Spherical rigid solute in straight cylindrical pore.
-Ignore electrostatic or other energetic solute, solvent and pore wall interaction; polydispersity of solute diameter and noncircular pore diameter.
o F1 is the stearic partition coefficient: geometric hindrance
22
1 2
( )( ) (1 )pore s
pore
d dF
d
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EXAMPLE 7
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CONVECTIVE MASS TRANSFER
Mass transfer between moving fluid with surface or another fluid
Forced convection Free/natural convectionRate equation analogy to Newton’s cooling equation
AAScAcA CCkckN
Surface concentration (mol/m3)
Bulk fluid concentration
Mass transfer coefficient (m/s)
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Film mass transfer and film mass transfer co-efficient:
Fluid flowing past a surface, there is a thin layer close to the surface where the fluid is laminar and the fluid particles next to the solid boundary are at rest.
Laminar boundary layer
No slip condition
Mechanism of mass transfer between the surface and stagnant and laminar layer is by molecular mass transfer.
Controlling resistance to the convective mass transfer sometime is from this laminar “film” due to molecular diffusion.
Momentum transfer/ Fluid Mechanics
h
wL
W
L
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EXAMPLE 8
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Diffusion in solids
Diffusion of atoms within solids. Mainly covered in the Materials Engineering course.
Examples:Semiconductor manufacturing process :Impurity atoms : DOPANTS are introduced to the solid silicon to control the conductivity in a semiconductor device.
Hardening of still (i.e. Carburization) : Diffusing carbon(C) atom through iron (Fe).
Solid diffusion mechanism :
(1)Vacancy Diffusion(2)Interstitial diffusion
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Vacancy diffusion:
Transported atoms JUMPS from a lattice position of the solid into the neighboring unoccupied solid site or vacancy.
Atoms continues to diffuse by a series of jumps into the neighboring vacancy.
This normally requires a distortion of lattice or lattice defect sites.
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Interstitial Diffusion
the diffusing atom is not on a lattice site but on an interstice. The diffusing atom is free to move to any adjacent interstice, unless it is already occupied.
The rate of diffusion is therefore controlled only by the ease with which a diffusing atom can move into an interstice.
Appendix J-3 : Values of binary diffusion coefficient in solids
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Do = proportionality constantQ = activation energy , kcal/mol, J/mol etc.R = Gas constant = 8.314 J/mol/K
Effect of temperature :
Diffusion coefficient value increases with temperature according to Arrhenius equation
There is an energy barrier to change the places of atom.Eyring “unimolecular rate theory” concept explains the mechanism of the
diffusion.
Problem : 24.19
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Linear Interpolation formula
X1 Y1
X ( @ known x value) Y (unknown)
X2 Y2
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