ChE 351 Mass Trans Transfer fer Processes Processes
E .G. A n ku de dey y, PhD Ph D Dept of Chemical Engineering, KNUST September 2011
1
W h a t t h e c o u r s e i s ab ab o u t
Mass Transfer Operations deal with “unit operations” involving “mass transfer” (a microscopic process on a macroscopic scale)
Mass transfer transfer is defined defined as the transportation transportation of one (or more) component from one phase to another
Motivation: in many industrial processes we use mass transfer to achieve separation (enrichment or removal ) of a substance from
a mixture
Emphasis is placed on separation processes that involve equilibrium between the phases 2
W h a t t h e c o u r s e i s ab ab o u t
Mass Transfer Operations deal with “unit operations” involving “mass transfer” (a microscopic process on a macroscopic scale)
Mass transfer transfer is defined defined as the transportation transportation of one (or more) component from one phase to another
Motivation: in many industrial processes we use mass transfer to achieve separation (enrichment or removal ) of a substance from
a mixture
Emphasis is placed on separation processes that involve equilibrium between the phases 2
C o u r s e C o n t en en t
Importance of mass transfer Molecular diffusion Diffusion in binary mixtures Mass transfer models Interphase Inter phase mass trans transfer fer Mass transfer with laminar and turbulent flow
Distillation Batch distillation Rayleigh equation Fractionation Binary distillation McCabe-Thiele method Ponchon-Sava Poncho n-Savarit rit metho method d
Liquid solid extraction.
Absorption in packed towers Effect of temperature on absorption
3
W h y i s t h i s c o u r s e i m p o r t an t ?
Mass transfer operations are largely the responsibility of chemical engineers
Chemical plants usually have from 50 to 90% of their capital invested in separation equipment
There is virtually no industrial process that does not involve purification of raw materials or final separation of products
4
C o u r s e O b j ec t i v es The student is expected to:
Recognize the various modes of mass transfer
Estimate diffusion coefficients under different physical conditions
Determine mass transfer rates using Fick’s Law for one-dimensional molecular diffusion in gases and liquids
Use the rate equation with mass transfer coefficients to determine mass transfer rates
Calculate film mass transfer coefficients using literature correlations or material balance findings
Calculate overall mass transfer coefficients from film mass transfer 5
C o u r s e O b j ec t i v es –cont’d
Use analogies between heat transfer and mass transfer to solve problems Determine liquid and gas flow rates required for absorption and stripping columns
Calculate the packing height for gas absorption towers.
Obtain practical information including physical properties from the literature relevant to contactor design
Use McCabe and Thiele method to analyze binary distillation problems
Determine the number of stages and feed plate location in simple binary distillations
Understand the concept of tray and overall efficiency and be able to determine the height of distillation column. 6
References Principles and Modern Applications of Mass Transfer Operations 2nd Edition, © 2009
Jaime Benitez Mass Transfer: Fundamentals and Applications © 1985
Anthony L. Hines and Robert N. Maddox
Unit Operations Of Chemical Engineering
©2004 Julian C. Smith, Peter Harriott, Warren McCabe
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References Transport Phenomena © 2006 Edwin N. Lightfoot, R. Byron Bird, Warren E. Stewart
Separation Process Principles
© 2005
E.J. Henley, Ernest J. Henley, J. D. Seader
Mass Transfer Operations 3 rd Edition © 1980
Robert Ewald Treybal
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C o u r s e O r g a n i s a t io n
4 hours regular lectures per week 2 hours tutorial session per week or as needed 8-12 Homework Assignments (10-20%) (10-30%) Two mid-semester exams Final Examination (50-70%) O ff i c e h o u r s : Open door (feel free to walk in)
TA :
M r. D er r i c k A m o a b e n g
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Prerequisites What do we need?
Differential Equations (MATH 251/252) Material/Energy Balances (ChE 251/252) Thermodynamics (ChE 253/254) Fluid Transport (ChE 255) Heat Transfer (ChE 256) Kinetics and Rate of reactions (??) 10
In t r o d u c t i o n t o M as s Tr an s f er
Transfer of material from one homogeneous phase to another
Driving force is concentration difference or difference in thermodynamic activity
Results from differences in solubility, vapour pressure or diffusivity. Independent of density and particle size
Examples are distillation, gas absorption, extraction, adsorption, dehumidification
A limit
to mass transfer occurs when the concentration in the two phases is the same. i.e. the two phases come to equilibrium 11
M as s t r an s f e r o p er at i o n s
The transfer of mass within a fluid mixture or across a phase boundary is a process that plays a major role in many industrial processes. Examples of such processes are:
Dispersion of gases from stacks
Removal of pollutants from plant discharge streams by absorption
Stripping of gases from waste water
Neutron diffusion within nuclear reactors
Air conditioning 12
M as s t r an s f e r o p er at i o n s
Many of our day-by-day experiences also involve mass transfer, for example: A lump of sugar added to a cup of coffee eventually dissolves and then eventually diffuses to make the concentration uniform.
Water evaporates from ponds to increase the humidity of passing-air-stream
Perfumes presents a pleasant fragrance which is imparted throughout the surrounding atmosphere.
The mechanism of mass transfer involves both molecular diffusion and convection. 13
P r o p er t i es o f M i x t u r es
Mass transfer always involves mixtures. Consequently, we must account for the variation of physical properties which normally exist in a given system.
When a system contains three or more components, as many industrial fluid streams do, the problem becomes complicated very quickly.
The conventional engineering approach to problems of multi-component system is to attempt to reduce them to representative binary (i.e., two component) systems.
Explanation of definitions and relations which are often used to explain the role of components within
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C o n c e n t r a ti o n o f Sp e c i es
Concentration of species in multi-component mixture can be expressed in many ways. For species A, mass concentration denoted by A is defined as the mass of A, m A per unit volume of the mixture. ρ A
mA V
….……. (1)
The total mass concentration density is the sum of the total mass of the mixture in unit volume: ρ
ρ i i
where i is the concentration of species i in the mixture.
15
C o n c e n t r at a t i o n o f s p e c i es es Molar concentration of, A, C A is defined as the number of moles of A present per unit unit volume of the mixture. By definition, mass of A Number of moles molecular weight of A n A
m A
………….. (2)
M A
Therefore from (1) & (2) C A
n A V
A
……... (3)
M A 16
C o n c e n t r a ti t i o n o f Sp S p e c i es es For ideal gas mixtures, n
A
p A V
…….…. (4)
R T
[ from Ideal gas law PV = nRT] C A
n A pA V R T
where p A is the partial pressure of species A in the mixture. V is the volume of gas, T is the absolute temperature, and R is the universal gas constant. The total molar concentration or molar density of the mixture is given by
C
C i
………. (5) 17
Velocities
In a multi-component system the various species move at different velocities
Evaluation of velocity of mixture requires the averaging of the velocities of each species s pecies present.
If I is the velocity of species i with respect to stationary fixed coordinates, then mass-average velocity for a multi-component mixture defined in terms of mass concentration is:
ρ ν ρ ν ν ρ ρ i
i
i
i
i
i
…………. (6)
i
i
18
Velocities Similarly, molar-average velocity of the mixture * is
C i V i
* i
…………. (7)
C For most engineering problems, there will be little difference in * and and so the mass average velocity, , will be used in most discussions.
The velocity of a particular species relative to the massaverage or molar average velocity is termed as diffusion velocity (i.e.) Diffusion velocity = i - 19
M o l e an d m as s f r ac t i o n s The mole fraction for liquid and solid mixture, x A ,and for gaseous mixtures, y A, are the molar concentration of species A divided by the molar density of the mixtures.
x A
y A
C A C C A
For solids and liquids
………. (8)
For gases
C
The mole fraction for liquid and solid mixture, x A ,and for gaseous mixtures, y A, are the molar concentration of species A divided by the molar density of the mixtures.
x i 1
y i 1
…….…. (9)
i i Similarly, the mass fraction of component A in the mixture is;
w A
A
……….. (10) 20
Example The molar composition of a gas mixture at 273 K and 1.5 x 10 5 Pa is: O2 7% CO 10% CO 2 15% N2 68% Determine •the composition in weight percent •average molecular weight of the gas mixture •density of gas mixture •partial pressure of O 2. 21
Solution Consider 1 mole of gas mixture (basis). Then O2 = 0.07 mol CO = 0.10 mol CO 2 = 0.15 mol N2 = 0.68 mol Molecular weight of the constituents are: O2 = 2 * 16 = 32 g/mol CO = 12 + 16 = 28 g/mol CO 2 =12+2*16=44g/mol N2 = 2 * 14 = 28 g/mol Weight of the constituents are: (1 mol of gas mixture) O2 = 0.07 * 32 = 2.24 g CO = 0.10 * 28 = 2.80 g CO 2 = 0.15 * 44 = 6.60 g N2 = 0.68 * 28 = 19.04 g Total weight of gas mixture = 2.24 + 2.80 + 6.60 + 19.04 22
Composition in weight percent: O2
2.24 * 100 7.30% 30.68
2.80 * 100 9.13% CO 30.68 CO 2
N 2
6.60 * 100 21.51% 30.68
19.04 * 100 62.06% 30.68
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Average molecular weight of the gas mixture M
Weight of gas mixture Number of moles
30.68 30.68 g mol M 1
Assuming that the gas obeys ideal gas law, PV = nRT n P V RT
n molar density m V Therefore, density (or mass density) = mM Where M is the molecular weight of the gas . Density m
PM 1.5 * 10 5 * 30.68 M kg m 3 8314 * 273 RT 24
Partial pressure of O 2 = [mole fraction of O 2] * total pressure
7 * 1.5 * 10 5 100
= 0.07 * 1.5 * 10 5 = 0.105 * 10 5 Pa
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M o d es o f m as s t r an s f er
Two mechanisms of mass transfer Molecular diffusion and convective transfer
The total mass (or molar) flux of a given species in a direction perpendicular to a stationary plane is a vector quantity and is given by the product of the volumetric average velocity and the mass (or molar) concentration
N A C A A
…………. (11)
The total flux consists of the diffusive flux and the convective flux 26
Diffusivities Diffusivities
are defined with respect to a plane moving at the volume average velocity v*. By definition,
ν
ν *
C ν i
i
C
i
…..…. (12)
27
D iff u s i o n f lu x The total flux of A could be written in terms of diffusion velocity of A, (i.e., A - ) and average velocity of mixture, , as
N A C A ( A ) C A
By definition
*
………. (13)
C
i i
i
C
Therefore, equation (13) becomes
C A N A C A ( A ) C
C
i i
i
C A ( A ) y A C i i
…….. (14)
i
28
D iff u s i o n f lu x For systems containing two components A and B,
N A
C A ( A ) y A (C A A C B B) C A ( A ) y A (N A N B )
N A
C A ( A ) y A N
………… (15)
..………. (16)
The first term on the right hand side of this equation is diffusional molar flux of A, and the second term is flux due to bulk motion. 29
Fl u x n o t a ti o n s Mass flux of species i with respect to fixed coordinates Total mass flux
ni i v i
n v Molar flux of species i with respect to fixed coordinates
Total molar flux
N i c i v i N cv * 30
Fl u x n o t a ti o n s Mass diffusion flux of species i with respect to the mass-average velocity
j i i (v i v ) Molar diffusion flux of species i with respect to molaraverage velocity
J i c i (v i v *) The mass flux ni is related to the molar diffusion flux by
ni j i i v j i i n The molar flux is related to the molar diffusions flux by
N i J i c i v * J i y i N
Fick’s law An empirical relation for the diffusional molar flux, first postulated by Fick and, accordingly, often referred to as Fick’s first law, defines the
diffusion of component A in an isothermal, isobaric system. For diffusion in only the Z direction, the Fick’s rate equation is
J A D A B
d C A d Z
………. (17)
where D AB is diffusivity or diffusion coefficient for component A diffusing through component B, and dC A / dZ is the concentration gradient in the Z-direction. A more general flux relation which is not restricted to isothermal, isobaric system could be written as
J A C D A B
d y A d Z
………. (18)
using this expression, Equation (16) could be written as
N A C D A B
d y A d Z
y A N
………. (19) 32
R el at i o n am o n g m o l ar f l u x e s For a binary system containing A and B, N A
J A y A N
………. (20)
J A N A y A N
.……... (21)
J B N B y B N
………. (22)
Addition of Equation (21) & (22) gives,
J A
J B N A N B ( y A y B ) N
…… (23) 33
R el at i o n a m o n g m o l ar f l u x e s By definition N = N A + N B and y A + y B = 1. Therefore equation (23) becomes, J A + J B = 0 J A = -J B C D AB
From
d y A d z
C D BA
………..…. (24) d y B d Z
…............... (25)
y A + y B = 1
dy A = - dy B Therefore Equation (25) becomes, D AB = D BA
………………….. (26) ……………... (27)
This leads to the conclusion that diffusivity of A in B is equal to diffusivity of B in A.
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Diffusivity
Fick’s law proportionality, D AB, is known as mass diffusivity
(simply as diffusivity) or as the diffusion coefficient.
D AB has the dimension of L 2 / t, identical to the fundamental dimensions of the other transport properties:
Kinematic viscosity, = ( / ) in momentum transfer Thermal diffusivity, (= k / C ) in heat transfer.
Diffusivity is normally reported in cm 2 / sec; the SI unit being m2 / sec.
Diffusivity depends on pressure, temperature, and composition of the system.
In literature, some values of D AB are given for a few gas, liquid, and solid systems. 35
Table 1a: Mass Diffusivities in Gases System Air Ammonia Aniline Benzene Bromine Carbon dioxide Carbon disulfide Chlorine Diphenyl Ethyl acetate Ethanol Ethyl ether Iodine Methanol Mercury Naphthalene Nitrobenzene n-Octane Oxygen Propyl acetate Sulfur dioxide Toluene Water Ammonia Ethylene Carbon dioxide Benzene Carbon disulfide Ethyl acetate
T, K
2
D AB P,m -Pa/s
273 298 298 293 273 273 273 491 273 298 293 298 298 614 303 298 298 273 315 273 298 298
2.006 0.735 0.974 0.923 1.378 0.894 1.256 1.621 0.718 1.337 0.908 0.845 1.641 4.791 0.870 0.879 0.610 1.773 0.932 1.236 0.855 2.634
293
1.793
318 318 319
0.724 0.724 0.675
System Carbon dioxide Ethanol Ethyl ether Hydrogen Methane Methanol Nitrogen Nitrous oxide Propane Water Hydrogen Ammonia Argon Benzene Ethane Methane Oxygen Pyridine Nitrogen Ammonia Ethylene Hydrogen Iodine Oxygen Oxygen Ammonia Benzene Ethylene Water
T, K
2
D AB P, m -Pa/s
273 273 273 273 298.6 298 298 298 298
0.702 0.548 5.572 1.550 1.064 1.672 1.185 0.874 1.661
293 293 273 273 273 273 318
8.600 7.800 3.211 4.447 6.331 7.061 4.427
293 298 288 273 273
2.441 1.651 7.527 0.709 1.834
293 296 293 308.1
2.563 0.951 1.844 2.857
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Table 1b: Mass Diffusivities in Liquids at Infinite Dilution 5
System
T, K
Chloroform (solvent B)
Acetone Benzene Ethanol Ethyl ether Ethyl acetate MEK
298 288 288 298 298 298
2.35 2.51 2.20 2.13 2.02 2.13
Acetic acid Acetone Ethyl benzoate MEK Nitrobenzene Water
293 293 293 303 293 298
2.18 3.18 1.85 2.93 2.25 3.20
1.49 2.00 1.92 1.25 2.00 1.02 1.00 1.20 2.10 0.58 0.92 1.64
298 298 298 281 298 288 298 353 303 281 298 281
2.09 1.96 1.38 1.45 2.09 2.25 2.28 4.25 2.09 1.19 1.85 1.77
Methane Air Carbon dioxide Chlorine Argon Benzene Ethanol Ethane Oxygen Pyridine Aniline Ammonia
298 298 298 298 298 298 288 298 298 288 293 298
1.87
2.92 4.04 2.62 3.77 4.56
Ethylene Allyl alcohol Acetic acid Benzoic acid Propionic acid Vinyl chloride
298
288 313 298 298 298
288 293 298 298 298
0.90 1.19 1.00 1.06 1.34
298 298
1.81 1.24
Ethylbenzene Sulfuric acid Nitric acid
293 298 298
0.81 1.73 2.60
Ethanol (solvent B)
Benzene Water
2
Water (solvent B)
Acetone (solvent B)
Acetic acid Acetic acid Benzoic acid Formic acid Water
5
System T,K DAB x 10 , cm/s
Ethyl acetate (solvent B)
Benzene (solvent B)
Acetic acid Aniline Benzole acid Bromobenzene Ciclohexane Ethanol Formic acid n-Heptane MEK Naphthalene Toluene Vinyl chloride
2
DAB x10 , cm /s
Sources: Reid, R. C., J. M. Prausnitz, and B. E. Poling, The Properties of Gases and Liquids,4th ed., McGraw -Hill, New York (1987); Cussler, E. Diffusion, L., Mass Transfer in Fluid Systems, 2nd ed., Cambridge University Press, Cambridge, UK (1997).
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Table 1c: Mass Diffusivities in the Solid State System
Hydrogen in iron
Hydrogen in nickel Carbon monoxide in nickel Aluminum in copper Uranium in tungsten Cerium in tungsten Yttrium in tungsten Tin in lead Gold in lead Gold in silver Antimony in silver Zinc in aluminum Silver in aluminum Bismuth in lead Cadmium in copper Carbon in iron Helium in silica Hydrogen in silica Helium in Pyrex
T, K -9
283
1.66 x 10
323 373 358 438 1223 1323 1123 293 2000 2000 2000 558 558 1033 293 773 323 293 293 1073 1373 293 773 473 773 293 773
11.4 x 10 -9 124 x 10 -9 11.6 x 10 -8 10.5 x 10 -8 4.00 x 10 -8 14.0 x 10 -9 2.20 x 10 -30 1.30 x l0 -11 1.30 x 10 -11 95.0 x 10 -11 1820 x 10 -10 1.60 x 10 -10 4.60 x 10 -10 3.60 x 10 3.60 x 10 -10 -9 2.00 x 10 -9 1.20 x 10 l.l0 x 10 -16 -15 2.70 x 10 1.50 x 10 -8 -8 45.0 x 10 -10 4.00 x 10 -8 7.80 x 10 6.50 x 10 -10 1.30 x 10 -8 -11 4.50 x 10 -8 2.00 x 10
-9
Sources: Barrer, R. M., Diffusion in and through Solids, Macmillan, New York (1941); American Society for Metals, Diffusion, ASM (1973); Cussler, E. L., Diffusion, Mass Transfer in Fluid Systems, 2nd ed., Cambridge University Press, Cambridge, UK (1997).
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Diffusivities
Diffusivities of gases at low density are almost composition independent, increase with the temperature and vary inversely with pressure.
Liquid and solid diffusivities are strongly concentration dependent and increase with temperature.
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R an an g e o f d i f f u s i v i t y v a l u e s
Gases : Liquids : Solids :
5 X 10 –6 10 –6 5 X 10 –14
----------------
1 X 10-5 10-9 1 X 10-10
m2 / s. m2 / s. m2 / s.
In the absence of experimental data, semi theoretical expressions have been developed which give approximation, sometimes as valid as experimental values, due to the difficulties encountered in experimental measurements
40
G en e n e r al a l ex e x p r es es s i o n s Pressure dependence of diffusivity is given by D
AB
1 p
(for moderate ranges of pressures, up to 25 atm).
…. (28)
And temperature temperature dependency is according to 3
2 D AB T AB
……………………. (29)
In the absence of experimental data, semi-empirical expressions have been developed to estimate the diffusion coefficient.
41
B i n ar a r y i d e a l g as a s s y s t em em s Based on the kinetic theory of gases at moderate pressures Lennard-Jones potential to evaluate the influence of intermolecular forces The diffusion coefficient of gas pairs of nonpolar, nonreacting molecules is given by the Wilke-Lee equation
3.03 0.98 / M 10 T 1/ 2
D AB
3
3 2
AB
1 2 2 PM AB AB D
42
1 1 M AB 2 M M B A
D AB M A, MB T P σ AB ΩD
1
= diffusion coefficient, cm 2/s = molecular weights of A and B = temperature, K = pressure, bar = collision diameter, diameter, a Lennard-Jones parameter, parameter, Å = diffusion collision integral, dimensionless
The collision integral is a function of the temperature and the intermolecular potential field for one molecule of A and one molecule of B. It is tabulated as a function of T* = κT/ε AB, where κ is the Boltzmann constant and ε AB is the energy of molecular interaction of the binary system A and B 43
An approximation approximation of ΩD is obtained from the following equation D
a
T
* b
c e g * * exp exp dT exp exp fT exp exp hT *
where T* = κT/ε AB a b c d
= = = =
1.06036 0.15610 0.19300 0.47635
e f g h
= = = =
1.03587 1.52996 1.76474 3.89411
44
For
a binary system of nonpolar molecular pairs, the Lennard-Jones parameters may be obtained as follows AB
A B 2
AB A B
The
interaction parameters for the pure components are usually obtained from viscosity data
45
In the absence of experimental data, the values of the parameters for pure components may be estimated from the following empirical correlations
1.18V b
13
A
1.15T b
where Vb is the molar volume of the substance as a liquid at its normal boiling point in cm3/gmol and Tb is the normal boiling temperature 1.048
V b 0.285V c
where Vc is the critical volume Otherwise, the atomic volume of each element present are added together 46
47
48
49
50
Fu l ler m et h o d D AB
1.00 x 10
7
T 1.75(1 / M A 1 / M B )0.5
v v
P
1
2
1
3
A
3
B
Σv A = sum of structural volume increments of
component A
P = absolute pressure in atm
51
52
M u l t i -c o m p o n e n t d i ff u s i v i t ies Diffusivity of a component in a mixture of components can be calculated using the diffusivities for the various binary pairs involved in the mixture. The relation given by Wilke is D 1 mixture
1 y 2 D12
y 3 D13
...........
y n
…. (30)
D1n
Where D 1-mixture is the diffusivity for component 1 in the gas mixture; D 1-n is the diffusivity for the binary pair, component 1 diffusing through component n; and y’n is the mole fraction of component n in the gas mixture evaluated on a component –1 – free basis, that is
y 2
y 2 y 2 y 3 ....... y n
…..…. (31) 53
Example
Determine the diffusivity of CO 2 (1), O 2 (2) and N 2 (3) in a gas mixture having the composition: CO2 : 28.5 %, O2 : 15%, N 2 : 56.5%, The gas mixture is at 273 K and 1.2 * 10 5 Pa. The binary diffusivity values are given as: (at 273 K) D 12 P = 1.874 m 2 Pa/s D 13 P = 1.945 m 2 Pa/s D 23 P = 1.834 m 2 Pa/s
54
Diffusivity of CO 2 in mixture D 1m
y 2
where
D 12
y 2
y 2
y 2 y 3
y 3 Therefore
1
y 3 y 2 y 3
D 1m P
y 3
D 13
0.15 0.21 0.15 0.565
0.565 0.79 0.15 0.565 1 0.21 1.874
0.79 1.945
= 1.93 m 2.Pa/s Since P = 1.2 * 10 5 Pa, D1m
1.93 5
1.61* 10 5 m 2 s
55
Diffusivity of O 2 in the mixture, D
2m
1
y 1 D
Where
y 1
21
y 1 y 1 y 3
y 3 D
23
0.285 0.335 0.285 0.565
(mole fraction on-2 free basis).
y 3
y 3 y 1 y 3
0.565 0.665 0.285 0.565 56
D 21 P = D 12 P = 1.874 m 2.Pa/sec Therefore 1 D 2m P 0.335 0.665 1.874 1.834 = 1.847 m 2.Pa/sec D 2m
1.847 1.2 * 10 5
1.539 * 10 5 m 2 sec
By similar calculations Diffusivity of N 2 in the mixture can be calculated, and is found to be, D 3m = 1.588 * 10 –5 m 2/s. 57
D iff u s i v i t y i n l i q u i d s Diffusivities
in liquids are nearer to 10-5 cm2 / s, and about ten thousand times slower than those in dilute gases.
This
characteristic of liquid diffusion often limits the overall rate of processes occurring in liquids
Diffusivity limits the rate of acid-base reactions;
Diffusion is responsible for the rates of liquid-liquid extraction.
58
D iff u s i v i t y i n l i q u i d s
Certain compounds diffuse as molecules,
Others which are designated as electrolytes ionize in solutions and diffuse as ions.
For example, sodium chloride (NaCl), diffuses in water as ions Na + and Cl-.
Though each ions has a different mobility, the electrical neutrality of the solution indicates the ions must diffuse at the same rate;
It is not possible to speak of a diffusion coefficient for molecular electrolytes such as NaCl. 59
D iff u s i v i t y i n l i q u i d s
If several ions are present, the diffusion rates of the individual cations and anions must be considered, and molecular diffusion coefficients have no meaning.
Diffusivity varies inversely with viscosity when the ratio of solute to solvent ratio exceeds five.
In extremely high viscosity materials, diffusion becomes independent of viscosity.
60
D iff u s i v i ty i n s o l i d s : An outstanding characteristic of these values is their small size, usually thousands of times less than those in a liquid, which are in turn 10,000 times less than those in a gas. Diffusion plays a major role in catalysis and is important to the chemical engineer. The diffusion of atoms within a solid is of prime importance to the metallurgist.
61
Steady State Diffus ion Steady-state molecular mass transfer through simple systems in which the concentration and molar flux are functions of a single space coordinate. In a binary system, containing A and B, the molar flux in the direction of z, as given by Eqn (16) is
d y A N A C D AB d z
y A (N A N B )
……. (32)
62
D iff u s i o n t h r o u g h a s t ag n an t g as f i lm The diffusivity or diffusion coefficient for a gas can be measured experimentally using a diffusion cell. This cell is illustrated schematically in figure. Gas B z=z2
Δz
z=z1=0
Liquid A
Liquid A maintained at constant temperature and pressure
Gas B has negligible solubility in A
No reaction between A and B 63
D if fu s i o n t h r o u g h a s t ag n an t g as f i lm
Component A vaporizes and diffuses into the gas phase
The rate of vaporization may be physically measured and may also be mathematically expressed in terms of the molar flux.
Consider the control volume S z, where S is the cross sectional area of the tube.
Mass balance on A over this control volume for a steady-state operation yields
64
M as s b al an c es [Moles of A leaving at z + z] – [Moles of A entering at z] = 0. (i.e.)
S N A
z z
S N A
z
0.
……. (32)
Dividing through by the volume, S Z, and evaluating in the limit as Z approaches zero, we obtain the differential equation
d N A 0 d z
……….…………. (33)
This relation stipulates a constant molar flux of A throughout the gas phase from Z1 to Z2. A similar differential equation could also be written for component B as, d N B ………….…. (34) 0, d Z and accordingly, the molar flux of B is also constant over the entire diffusion path 65
M as s b al an c e s Since the gas B is insoluble in liquid A, we realize that N B, the net flux of B, is zero throughout the diffusion path; (alternatively, B is a stagnant gas along the diffusion path). From equation 19
N A
d y A C D AB d z
y A (N A N B )
……. (35)
Since N B = 0,
N A
d y A C D AB d z
y A N A
…….…. (36)
Rearranging,
C D AB d y A N A 1 y A d z
…………………. (37) 66
M as s b al an c e s This equation may be integrated between the two boundary conditions: at z = z1 Y A = y A1 And at z = z2 Y A = y A2 Assuming the diffusivity is to be independent of concentration, and realizing that N A is constant along the diffusion path, by integrating equation (37) we obtain Z 2
N A
d z C D AB
Z 1
N A
C D AB Z 2 Z 1
y A 2
d y A 1 y A y A1
……. (38)
1 y A2 ln 1 y A1
……. (39) 67
,
The log mean average concentration of component B is defined as y B 2 y B1 ……. (40) y B, lm y B 2 ln y B1 y B 1 y A
Since
y B, lm
(1 y A 2 ) (1 y A1 ) y B 2 ln
y B 1
y A1 y A 2 y B 2 ln
y B
……. (41)
1
Substituting Eqn (41) in Eqn (39), N A
( y A1 y A2 ) y B, lm z 1
C D AB Z 2
……. (42) 68
D iff u s i o n f l u x o f g a s es For an ideal gas C
n V
p R T
……. (43)
and for mixture of ideal gases y
A
p A P
……. (44)
Therefore, for an ideal gas mixture equation. (6) becomes N A
D AB RT (z 2 z 1)
( p A1 p A2 ) p B, lm
……. (45)
This is the equation of molar flux for steady state diffusion of one gas through a second stagnant gas 69
Applications Many mass-transfer operations involve the diffusion of one gas component through another non-diffusing component; absorption and humidification are typical operations defined by these equation.
The concentration profile (y A vs. z) for this type of diffusion is shown in figure:
70
Example Oxygen is diffusing in a mixture of oxygen-nitrogen at 1 std atm, 25C. Concentration of oxygen at planes 2 mm apart are 10 and 20 volume % respectively. Nitrogen is nondiffusing. •Derive the appropriate expression to calculate the flux
oxygen. Define units of each term clearly. •Calculate the flux of oxygen. Diffusivity of oxygen in nitrogen = 1.89 * 10 –5 m 2/s.
71
Solution
Let us denote oxygen as A and nitrogen as B. Flux of A (i.e.) N A is made up of two components, namely that resulting from the bulk motion of A (i.e.), Nx A and that resulting from molecular diffusion J A:
N A Nx A J A
……. (a)
72
Solution From Fick’s law of diffusion, J A D AB
d C A
……. (b)
d z
Substituting this equation (a)
N A Nx A D AB
d C A
……. (c)
d z
Since N = N A + N B and x A = C A / C equation (c) becomes N A N A N B
C A C
D AB
d C A d z
…….. (d) 73
Rearranging the terms and integrating between the planes between 1 and 2,
d z cD AB
C A2
dC A
A1
N AC C A N A N B
C
……. (e)
Since B is non diffusing N B = 0. Also, the total concentration C remains constant. Therefore, equation (4) becomes z CD AB
dC A
C A 2
C A1
N AC N AC A
C C A2 1 ln N A C C A1
Therefore, N A
CD AB z
ln
……. (f)
……. (g)
C C A2 C C A1
……. (h) 74
Replacing concentration in terms of pressures using Ideal gas law, equation (h) becomes
N A
D AB P t RTz
ln
P t P A2 P t P A1
……. (i)
where D AB = molecular diffusivity of A in B P T = total pressure of system R = universal gas constant T = temperature of system in absolute scale z = distance between two planes across the direction of diffusion P A1 = partial pressure of A at plane 1, and P A2 = partial pressure of A at plane 2 Given:
D AB = 1.89 * 10 –5 m2/sec P t = 1 atm = 1.01325 * 10 5 N/m 2 T = 2 5C = 2 7 3 + 2 5 = 2 9 8 K z=2mm=0.002m P = 0 2 * 1 = 0 2 atm (From Ideal gas law and additive pressure rule) 75
Substituting these in equation (i)
N A
1.89 * 10 5 1.01325 * 10 5 1 0.1 ln 83142980.002 1 0.2
= 4.55 * 10 –5 kmol/m 2.s
76
Eq u i m o l ar c o u n t er d i ff u s i o n
A physical situation which is encountered in the distillation of two constituents whose molar latent heats of vaporization are essentially equal
The flux of one gaseous component is equal to but acting in the opposite direction from the other gaseous component; that is, N A = - NB.
The molar flux N A, for a binary system at constant temperature and pressure is described by N A
or
N A
C D AB
d y A d z
y A (N A N B )
……. (48)
D AB
d C A d z
y A (N A N B )
……. (49)
with the substitution of N B = - N A, Equation (49) becomes,
D AB
N A
d C A d z
………………. (50)
For steady state diffusion Equation. (50) may be integrated, using the boundary conditions: at z = z1 C A = C A1 and z = z2 C A = C A2 Giving, Z 2
N A
d z D AB
Z 1
C A 2
d C A
C A1
from which
N A
D AB z 2 z 1
(C A1 C A2 )
……. (51)
For ideal gases,
C A
n A pA V R T
. Therefore Equation. (51) becomes
N A .
D AB (P A1 R T ( z 2 z 1)
P A2 )
……. (52)
This is the equation of molar flux for steady-state equimolar counter diffusion. Concentration profile in these equimolar counter diffusion may be obtained from, d (N A ) 0 ……. (53) d z (Since N A is constant over the diffusion path). And from equation. (50)
N A D AB
d C A d z
……. (54)
Therefore
d d z
.
or
d
2
d C A D AB 0 d z C A
d z
2
0.
……. (55)
This equation may be solved using the boundary conditions to give
C A C A1 C A1
z z 1 z 1 z 2 C A2
……….. (56)
Equation, (56) indicates a linear concentration profile for equimolar counter diffusion.
Example Methane diffuses at steady state through a tube containing helium. At point 1 the partial pressure of methane is p A1 = 55 kPa and at point 2, 0.03 m apart, P A2 = 15 KPa. The total pressure is 101.32 kPa, and the temperature is 298 K. At this pressure and temperature, the value of the diffusivity is 6.75 * 10 –5 m 2/s. Calculate the flux of CH 4 at steady state for equimolar counter diffusion. Calculate the partial pressure at a point 0.02 m apart from point 1. Solution
For steady state equimolar counter diffusion, molar flux is given by
N A
D AB R T z
p A 1 p A 2
Therefore; N A
6.75 * 10 5 55 15 kmol 8.314 * 298 * 0.03 m 2 . sec
3.633 * 10
kmol
5
m
2
sec
And from (1), partial pressure at 0.02 m from point 1 is: 3.633 * 10 5
6.75 * 10 5 55 p A 8.314 * 298 * 0.02
p A = 28.33 kPa
Example In a gas mixture of hydrogen and oxygen, steady state equimolar counter diffusion is occurring at a total pressure of 100 kPa and temperature of 20C. If the partial pressures of oxygen at two planes 0.01 m apart, and perpendicular to the direction of diffusion are 15 kPa and 5 kPa, respectively and the mass diffusion flux of oxygen in the mixture is 1.6 * 10 –5 kmol/m 2.s, calculate the molecular diffusivity for the system.
Solution For equimolar counter current diffusion:
N A
D AB RTz
p A1 p A2
where N A = molar flux of A (1.6 * 10 –5 kmol/m 2.s): D AB = molecular diffusivity of A in B R = Universal gas constant (8.314 kJ/kmol.k) T = Temperature in absolute scale (273 + 20 = 293 K) z = distance between two measurement planes 1 and 2 (0.01 m) P A1 = partial pressure of A at plane 1 (15 kPa); and P A2 = partial pressure of A at plane 2 (5 kPa)
Substituting these in the equation
1.6 * 10
5
D AB
8.3142930.01
Therefore, D AB = 3.898 * 10 –5 m 2/sec
15 5
Example A tube 1 cm in inside diameter that is 20 cm long is filled with CO2 and H2 at a total pressure of 2 atm at 0C. The diffusion coefficient of the CO2 – H2 system under these conditions is 0.275 cm2/sec. If the partial pressure of CO2 is 1.5 atm at one end of the tube and 0.5 atm at the other end, find the rate of diffusion for: steady state equimolar counter diffusion (N A = - N B) steady state counter diffusion where N B = -0.75 N A, and steady state diffusion of CO2 through stagnant H2 (NB = 0)
N A C D AB
d y A d z
y A N A N B
Given N B = - N A
C D AB
N A (For ideal gas mixture
C A
d y A d z
D AB
d C A d z
p A R T
where p A is the partial pressure of A; such that p A + p B = P) Therefore
N A
D A B
d pA RT d z
For isothermal system, T is constant Therefore D AB d p A N A RT
d z
(i.e.)
Z 2
N A
d z
Z 1
N A
D AB RT z
D AB RT
P A2
d p A P A1
p A1 p A2
where Z = Z 2 – Z 1 Given: D AB = 0.275 cm 2/sec = 0.275 * 10 –4 m 2 / s e c ; T = 0C = 2 7 3 K N A
0.275 * 10 4 1.5 * 1.01325 * 10 5 0.5 * 1.01325 * 10 5 8314 * 273 * 0.2
6.138 * 10
6
k mol m
2
sec
Rate of diffusion = N A S Where S is surface area Therefore rate of diffusion = 6.138 * 10-6 * r 2 = 6.138 * 10 –6 * (0.5 * 10 –2) 2 = 4 821 * 10 –10 k mol/sec
ii) N A C D AB
d y A d z
y A N A N B
given: N B = - 0.75 N A Therefore
N A C D AB
C D A B
d y A d z
d y A d z
y A N A 0.75 N A
0.25 y A N A
N A 0.25 y A N A C D AB
N A d z C D AB
f
N
dC
d y A d z
d y A 1 0.25 y A
Z 2
N A
d z CD AB
Z 1
y A 2
y A1
d y A 1 0.25 y A
d x 1 a b x b ln a b x
1 N A z C D AB ln 1 0.25 y A y y A2 A1 0.25
N A
4 CD AB z
1 0.25 y A 2 ln 1 0.25 y A 1
Given:
2 * 1.01325 * 10 5 C 0.0893 K mol m 3 R T 8314 * 273 p
y A 1
y A 2
p A 1 P
1 .5 0.75 2
p A 2
0.5 0.25 2
P
Substituting these in equation (2),
N A
4 * 0.0893 * 0.275 * 10 4 0 .2
7.028 * 10
6
1 0.25 * 0.25 ln 1 0.25 * 0.75
kmol m 2 sec
Rate of diffusion = N A S = 7.028 * 10 –6 * * (0.5 * 10 –2) 2 = 5.52 * 10 –10 kmol/sec = 1.987 * 10 3 mol/hr.
iii)
N A
Given: N B = 0
N A
CD AB
d y A
CD AB
d z
d y A d z
Z 2
N A
y A N A N B
y A N A y A2
d y A
d z CD AB 1 y A Z 1 y A1
CD AB Z
1 y A2 ln 1 y A1
0.0893 * 0.275 * 10 4 0. 2
1.349 * 10 Rate of diffusion
5
1 0.25 ln 1 0.75
kmol m 2 . sec
= 1.349 8 10 –5 * * (0.5 * 10 –2) 2 = 1.059 Kmol / sec = 3.814 mol/hr
Prac tic e Pro b lem
Water in the bottom of a narrow metal tube 0.5 cm in diameter is held at a constant temperature of 293K. The total pressure of air is 1 atm and the temperature is 293K. Water evaporates and diffuses through the air in the tube and the diffusion path (z 2-z1) is 0.1524 m long. Calculate the rate of evaporation at steady state in kmol/s. The diffusivity of water at 293K and 1 atm is 0.250x10-4 m2/s. Assume that the system is isothermal.
93
D iff u s i o n i n t o an i n f i n i te s t an d ar d medium
Problems involving diffusion from a spherical particle into an infinite body of stagnant gas.
How to set up differential equations that describe the diffusion in these processes.
The solutions developed here for these problems actually represent a special case of the more common situation involving both molecular diffusion and convective mass transfer.
E v ap o r at i o n o f a s p h er i c a l D r o p l et Consider the evaporation of spherical droplet such as a raindrop or sublimation of naphthalene ball. The vapor formed at the surface of the droplet is assumed to diffuse by molecular motions into the large body of stagnant gas that surrounds the droplet. At any moment, when the radius of the drop is r 0, the flux of water vapor at any distance r from the center is given by
N A C D AB
d y A d r
y A N A N B ……….. (57)
Here N B = 0 (since air is assumed to be stagnant)
N A C D AB
d y A d r
y A N A
Rearranging, N A
C D AB d y A 1
y A
……….. (58)
d r
The flux N A is not constant, because of the spherical geometry; it decreases as the distance from the center of sphere increases. But the molar flow rate at r and r + r are the same. This could be written as,
A N A 4 r 2 N A
r r
r
A N A
r r
……….. (59)
4 r 2 N A 0 r
where A = surface area of sphere at r or r + r. Substituting for A = 4 r 2 in equation (59), r 2 N A lim
r
r 0
d
r dr Integrating,
r r
r 2 N A
2
N A
r
0
0
r 2 N A constant
……….. (60) ………... (61)
From equation (61),
r 2 N A
r 0 2 N A 0
Substituting for N A from equation (58),
r 2 C D AB d y A r 02 N A 0 1 y A d r 2 r 0
N A 0
d r r
2
C D AB
d y A 1 y A
………... (62)
Boundary condition : At r = r 0 y A = y AS and At r = y A = y A Therefore equation (62) becomes,
r 02 N A 0
y A 1 C D y ln 1 AB A y AS r r 0
Simplifying, N A
0
C D AB r 0
1 y A ln 1 y A S
………... (63)
Time required for complete evaporation of the droplet may be evaluated from making mass balance.
Moles of water diffusing moles of water leaving the droplet unit time unit time 4 r 02
d 4 3 L N A 0 r 0 dt 3 M A
………... (64)
L d r 0 4 r 0 M A d t 2
Substituting for N A0 from equation (63) in equation (64), C D AB r 0
1 y A L d r 0 ln 1 y AS M A d t
………... (65)
Initial condition : When t = 0
r 0 = r 1
Integrating equation (65) with these initial condition, t
d t 0
L M A
t
1 C D AB
L
M A
0
1
ln
r 0 1 y A r 1
d r 0
1 y A S
1 2 C D AB
r 12
1 y A ln 1 y A S
………... (66)
Equation (66) gives the total time t required for complete evaporation of spherical droplet of initial radius r 1.
C o m b u s t i o n o f a c o al p ar t ic l e The problem of combustion of spherical coal particle is similar to evaporation of a drop with the exception that chemical reaction (combustions) occurs at the surface of the particle. During combustion of coal, the reaction C+ O2 CO 2 occurs. According to this reaction for every mole of oxygen that diffuses to the surface of coal (maximum of carbon), react with 1 mole of carbon, releases 1 mole of carbon dioxide, which must diffuse away from this surface. This is a case of equimolar counter diffusion of CO 2 and O 2. Normally air (a mixture of N 2 and O 2) is used for combustion, and in this case N 2 does not take part in the reaction, and its flux is zero. N N 0 2 The molar flux of O 2 could be written as N O 2 Where
D
C D O 2 gas
d y O 2 d r
y O 2 N O 2 N CO 2 N N 2
is the diffusivity of O 2 in the gas mixture.
…... (67)
Since
N N 2 0
and from stoichiometry equation (67) becomes
N O 2 N CO 2 N O 2
For steady state conditions,
C D O 2 gas
d y O 2 d r
………... (68)
d r 2 N O 2 0 d r
Integrating, 2 r 2 N O 2 constant r 0 N O 2 s
………... (69)
Where r 0 is the radius of coal particle at any instant, and N O 2 s is the flux of O 2 at the surface of the particle. Substituting for N O , 2 2
r C D O 2 gas
d y O 2 d r
2 r 0
N O 2 s ………... (70)
Boundary condition s: At
r r 0 r
y O y O s y O 2 y O 2 2
2
With these boundary condition, equation (70) becomes 2
r 0 N A 0 .
which yields
N O 2 s
d r
r 0
r
2
y O 2
C D O 2 gas
C D O 2 gas r 0
y O
2s
y O s 2
d y O 2
y O 2
………... (71)
For fast reaction of O 2 with coal, the mole fraction of O 2 at the surface of particle is zero. (i.e.,) y 0 O2s
And also at some distance away from the surface of the particle
y O 2 y O 2 0.21 (because air is a mixture of 21 mole % O 2 and 79 mole % N 2) With these conditions, equation (6) becomes,
N O 2 s
0.21 C D O 2 gas r 0
………... (72)
Example
A sphere of naphthalene having a radius of 2mm is suspended in a large volume of shell air at 318 K and 1 atm. The surface pressure of the naphthalene can be assumed to be 0.555 mm Hg. The D AB of naphthalene in air at 318 K is 6.92 * 10 –6 m 2/sec. Calculate the rate of evaporation of naphthalene from the surface.
Solution Steady state mass balance over a element of radius r and r + r leads to
S N A
r
S N A
r r
0
where S is the surface are (= 4 r 2) dividing (1) by Sr, and taking the limit as r approaches zero, gives: d r 2N A d r
0
Integrating r 2 N A = constant (or) 4 r 2 N A = constant We can assume that there is a film of naphthalene – vapor / air film around naphthalene through which molecular diffusion occurs. Diffusion of naphthalene vapor across this film could be written as, d y A N A CD AB y A N A N B d r
N B = 0 (since air is assumed to be stagnant in the film)
N A
CD AB
d y A
N A CD AB
N A CD AB
d r d d r
y A N A y A 1 y A
d ln 1 y A d r
W A = Rate of evaporation = 4 r 2 N A R = constant. 4 r 2 CD AB d ln 1 y A
W A W A
d r
d r r
2
4 D AB C d ln 1 y A
Boundary condition: At r = R 0.555 y A 760
7.303 * 10 4 ln (1 – y A) = - 7.3 * 10 –4 y A = 0 ln (1-y A) = 0
At r = Therefore
W A
d r
R
r
2
4 D AB C
0
d ln 1 y A
7 3 *10 4
1 0 W A 4 D AB C ln 1 y A 4 7 . 3 * 10 r R
1 W A 0 4 D AB C 0 7.3 * 10 4 R W A = 4 R D AB C * 7 . 3 * 1 0 –4 5 P 1.01325 * 10 C R * T 8314 * 318
= 0.0383 kmol/m 3 Therefore W A = 4 * 3.142 * 2 * 10 –3 * 6.92 * 10 –6 * 0.0383 * 7.3 * 10 –4 = 4.863 * 10 –12 kmol/s = 1.751 * 10 –5 mol/h.
D iff u s i o n i n L i q u i d s The equations derived for diffusion in gases equally apply to diffusion in liquids with some modifications. Mole fraction in liquid phases is normally written as ‘x’ (in gases as y). The concentration term ‘C’ is replaced by average molar density,
M av
For steady – state diffusion of A through non diffusivity B: N A = constant , N B = 0 N A
D AB z x BM
x A1 x A2 …….. (73) M av
where Z = Z 2 – Z 1, the length of diffusion path; and X BM
X B 2 X B 1
X B 2 ln X B 1
…………. (74)
For steady – state equimolar counter diffusion : N A = - N B = const
N A
D AB
D AB C A1 C A2 x A 1 x A 2 Z Z M av
… (75)
Example Calculate the rate of diffusion of butanol at 20C under unidirectional steady state conditions through a 0.1 cm thick film of water when the concentrations of butanol at the opposite sides of the film are, respectively 10% and 4% butanol by weight. The diffusivity of butanol in water solution is 5.9 * 10 –6 cm 2/sec. The densities of 10% and 4% butanol solutions at 20C may be taken as 0.971 and 0.992 g/cm3 respectively. Molecular weight of butanol (C 4 H 9 OH) is 74, and that of water 18.
Solution For steady state unidirectional diffusion, N A
D AB z
C
x A1 x A2 x B, lm
where C is the average molar density.
M avg Conversion from weight fraction the Mole fraction:
x A1 x A2
0.1 74
0.1 74 0.9 18 0.04 74
0.04 74 0.96 18
0.026 0.010
Average molecular weight at 1 & 2:
M 1
M 2
1
0.1 74 0.9 18 1
kg Kmol 19.47
0.04 74 0.96 18
18. 56 kg Kmol
1 M 1 2 M 2 2 M avg
0.971 19.47 0.992 18.56 2 = 0.0517 gmol / cm 3 = 51.7 kmol/m 3 x B,lm
x B 2 x B1 ln x B 2 x B1
1 x A2 1 x A1
1 x A2 ln 1
(i.e.) x B,lm
1 0.01 1 0.026 1 0.01 1 0.026
ln
0.016 0.982 0.0163
Therefore
N A
D AB x A1 x A2 2 M avg x B, lm
5.9 * 10 6 * 10 4 * 51.7 0.1 * 10
2
7
4.97 * 10
*
0.026 0.010
kmol m 2 s
0.982
1.789
gmol m 2 .hr .
1.789 * 74
132.4
g m 2 . hr .
g 2
m . hr .
M as s d i ff u s i o n w it h h o m o g en eo u s c h em i c a l r eac t i o n
Absorption operations involve contact of a gas mixture with a liquid and preferential dissolution of a component in the contacting liquid.
Depending on the chemical nature of the involved molecules, the absorption may or may not involve chemical reaction.
The following analysis illustrates the diffusion of a component from the gas phase into the liquid phase accompanied by a chemical reaction in the liquid phase.
M as s d i ff u s i o n w it h h o m o g en eo u s c h em i c a l r eac t i o n Consider a layer of absorbing medium (liquid) as shown in diagram.
At the surface of the liquid, the composition of A is C A 0. The thickness of the film, is so defined, that beyond this film the concentration of A is always zero ; that is C A = 0. If there is very little fluid motion within the film, 118
N A D AB
d C A d z
C A C
N A N B
….. (76)
If the concentration of A in the film, C A is assumed small, equation (76) becomes
N A D AB
d C A d z
…………... (77)
The molar flux N A changes along the diffusion path. This change is due to the reaction that takes place in the liquid film. This changes could be written as d N A r A 0 …………. (78) d z
where –r A is the rate of disappearance of A. For a first order reaction, k
A B
r A k C A
……….. (79)
with the substitution from equation (79) and (77) in equation (78), d C A d D AB k C A 0 d z d z
For constant Diffusivity,
D AB
d 2 C A d z 2
k C A 0
….. (80)
….. (81)
which is a second order ordinary differential equation. The general solution to this equation is
k k C A C 1 cos h z C 2 sin h z ….. (82) D AB D AB The constants of this equation can be evaluated from the boundary conditions: at Z=0 C A = C A0 And at Z = C A = 0.
The constant C 1 is equal to C A0 , and C 2 is equal to C A0 k tan h D AB
….. (83)
with this substitution equation (82) becomes, k C A C A cos h D AB 0
k C A s in h z D A B z k tan h D A B 0
….. (84)
This equation gives the variation of concentration of A with z (i.e concentration profile of A in the liquid). The molar flux at the liquid surface can be determined by differentiating equation (84), and evaluating the derivative d C A d z
at z 0
Differentiating C A with respect to z,
d C A d z
C A 0 k D
A B
sin h k D AB
z
C A 0
k k z cos h D A B D AB
….. (85)
k D AB k tan h D AB
….. (86)
tan h
k D AB
Substituting z = 0 in equation (85) and from equation (77),
N A
Z 0
D AB C A 0
For absorption with no chemical reaction, the flux of A is obtained from equation (77) as N
A
D AB C A 0
…………………. (87)
which is constant throughout the film of liquid. On comparison of equation (86) and (87), it is apparent that the term
k D AB
k tan h D AB
shows the influence of the chemical reactions. This term is a dimensionless quantity, is often called as Hatta Number.
D if fu s i o n i n s o l id s In certain unit operations of chemical engineering such as in drying or in absorption, mass transfer takes place between a solid and a fluid phase. If the transferred species is distributed uniformly in the solid phase and forms a homogeneous medium, the diffusion of the species in the solid phase is said to be structure independent. In this case diffusivity or diffusion coefficient is direction – independent. At steady state, and for mass diffusion which is independent of the solid matrix structure, the molar flux in the z direction is : N A D AB
d C A d z
constant
….. (88)
as given by Fick’s law. Integrating the above equation,
N A
D AB C A1 C A2 z
….. (89)
which is similar to the expression obtained for diffusion in a stagnant fluid with no bulk motion (i.e. N = 0).
D iff u s i o n i n p r o c es s s o l id s
In some chemical operations, such as heterogeneous catalysis, an important factor, affecting the rate of reaction is the diffusions of the gaseous component through a porous solid.
The effective diffusivity in the solid is reduced below what it could be in a free fluid, for two reasons.
tortuous nature of the path increases the distance, which a molecule must travel to advance a given distance in the solid.
free cross – sectional area is restricted. For many catalyst pellets, the effective diffusivity of a gaseous component is of the order of one tenth of its value in a free gas.
Di ff u s i o n i n p r o c es s s o l id s
If the pressure is low enough and the pores are small enough, the gas molecules will collide with the walls more frequently than with each other.
This is known as Knudsen flow or Knudsen diffusion.
Upon hitting the wall, the molecules are momentarily absorbed and then given off in random directions.
The gas flux is reduced by the wall collisions.
By use of the kinetic flux, the concentration gradient is independent of pressure ; whereas the proportionality constant for molecular diffusion in gases (i.e. Diffusivity) is inversely proportional to pressure.
Knudsen diffusion occurs when the size of the pore is of the
Tran s ien t Diffu s ion Transient processes, in which the concentration at a given point varies with time, are referred to as unsteady state processes or time – dependent processes. This variation in concentration is associated with a variation in the mass flux. These generally fall into two categories:
the process which is in an unsteady state only during its initial startup the process which is in a batch operation throughout its operation.
Tran s ien t Diffu s ion In
unsteady state processes there are three variablesconcentration, time, and position
Therefore
the diffusion process must be described by partial rather than ordinary differential equations
Many
solutions are for one-directional mass transfer as defined by Fick’s second law of diffusion
C A t
D AB
2 C A z 2
……... (90)
Tran s ien t Diffu s ion
This partial differential equation describes a physical situation in which there is no bulk –motion contribution, and there is no chemical reaction.
This situation is encountered when the diffusion takes place in solids, in stationary liquids, or in system having equimolar counter diffusion.
Due to the extremely slow rate of diffusion within liquids, the bulk motion contribution of flux equation (i.e., y A N i) approaches the value of zero for dilute solutions ; accordingly this system also satisfies Fick’s second law of diffusion.
Tran s ien t Diffu s ion
The solution to Fick’s second law usually has one of the two standard forms.
It may appear in the form of a trigonometric series which converges for large values of time, or it may involve series of error functions or related integrals which are most suitable for numerical evaluation at small values of time.
These solutions are commonly obtained by using the mathematical techniques of separation of variables or Laplace transforms.
Con v ec tive Mas s Trans fer Coefficien t
Diffusive mass transfer results from a concentration gradient within a system.
In systems involving liquids or gases, however, it is very difficult to eliminate convection from the overall mass-transfer process.
Mass transfer by convection involves the transport of material between a boundary surface (such as solid or liquid surface) and a moving fluid or between two relatively immiscible, moving fluids.
131
Con v ec tive Mas s Trans fer Coefficien t There are two different cases of convective mass transfer:
Mass transfer takes place only in a single phase either to or from a phase boundary, as in sublimation of naphthalene (solid form) into the moving air.
Mass transfer takes place in the two contacting phases as in extraction and absorption.
132
C o n v e c t i v e M as s Tr an s f e r C o ef fi c i en t In the study of convective heat transfer, the heat flux is connected to heat transfer coefficient as
Q A q h t s t m
….. (91)
The analogous situation in mass transfer is handled by an equation of the form
N A k c C As C A
….. (92)
The
driving force is the difference between the concentration at the phase boundary, C AS and the concentration at some arbitrarily defined point in the fluid medium, C A .
The
convective mass transfer coefficient kC is a function of geometry of the system and the velocity and properties of the fluid similar to the heat transfer coefficient, h. 133
S ig n i f i c a n t P ar a m et er s i n C o n v e c t i v e Mas s Trans fer
Dimensionless parameters are often used to correlate convective transfer data. Reynolds number and friction factor (momentum transfer) Prandtl and Nusselt numbers (convective heat transfer)
The molecular diffusivities of the three transport process (momentum, heat and mass) have been defined as: Momentum diffusivity
Thermal diffusivity
k C p
Mass diffusivity D AB
….. (93)
134
Correlations It
can be shown that each of the diffusivities has the dimensions of L2 / t, hence, a ratio of any of the two of these must be dimensionless.
The
ratio of the molecular diffusivity of momentum to the molecular diffusivity of heat (thermal diffusivity) is designated as the Prandtl Number Momentum diffusivity Thermal diffusivity
Pr
C p
….. (94)
K
The analogous number in mass transfer is Schmidt number given as
Momentum diffusivity Mass diffusivity
Sc
D AB
….. (95)
D AB 135
Correlations The ratio of the molecular diffusivity of heat to the molecular diffusivity of mass is designated the Lewis Number , and is given by
k Thermal diffusivity Le C p D AB D AB Mass diffusivity
….. (96)
Lewis number is encountered in processes involving simultaneous convective transfer of mass and energy.
136
Sh er w o o d n u m b er
Consider the mass transfer of solute A from a solid to a fluid flowing past the surface of the solid.
For such a case, the mass transfer between the solid surface and the fluid may be written as
N A
k c C As C A
….. (97)
137
Sh er w o o d n u m b er Since the mass transfer at the surface is by molecular diffusion, the mass transfer may also be described by
N A D AB
d C A d y
….. (98) y 0
When the boundary concentration, C As is constant, equation (98) may be written as d C A C A s ….. (99) N A D AB d y
y 0
Equation (97) and (99) may be equated, since they define the same flux of component A leaving the surface and entering the fluid k c C A s C A D AB
d C A C A s d y
….. (100) y 0
138
Sh er w o o d n u m b er This relation may be rearranged into the following form: k c D AB
d C A C A s d y
C A C A
….. (101) y 0
Multiplying both sides of equation(100) by a characteristic length, L we obtain the following dimensionless expression: k c L D AB
d C A C A s d y
C A S C A L
y 0
….. (102)
The right hand side of equation (102) is the ratio of the concentration gradient at the surface to an overall or reference concentration gradient; accordingly, it may be considered as the ratio of molecular mass-transport resistance to the convective mass-transport resistance of the fluid. This ratio is generally known as the Sherwood number , Sh and analogous to the Nusselt number Nu in heat transfer 139
A p p l ic at io n o f D i m e n s i o n l es s A n al y s i s t o M as s Tr an s f e r
One of the methods of obtaining equations for predicting mass-transfer coefficients is the use of dimensionless analysis.
Dimensional analysis predicts the various dimensionless parameters which are helpful in correlating experimental data.
There are two important mass transfer processes, which we shall consider, the transfer of mass into a stream flowing under forced convection and the transfer of mass into a phase which is moving as the result of natural convection associated with density gradients. 140
Tr an s f er i n t o a s t r eam f lo w i n g u n d e r f o r c ed c o n v ec t i o n Consider the transfer of mass from the walls of a circular conduit to a fluid flowing through the conduit. The mass transfer is due to the concentration driving force C As – C A . The important variables, their symbols and their dimensions are listed in the table Variable
Symbol
Dimensions
tube diameter
D
L
fluid density
ρ
M/L3
fluid viscosity
μ
M/Lt
fluid velocity
υ
L/t
mass diffusivity
D AB
L2 /t
mass-transfer coefficient k
L/t
141
D im en s i o n l es s an a ly s i s
These variables include terms descriptive of the system geometry, the flow and fluid properties and the quantity of importance, k c.
By the Buckingham method of grouping the variables, the number of dimensionless groups is equal to the number of variables minus the number of fundamental dimensions.
Hence the number of dimensionless group for this problem will be three.
With D AB, and D as the core variables, the three groups to be formed are
a 1 D AB
b D c k c
2 D
e
d AB
f
D v
….. (102)
….. (103)
and g
3 D AB h D i
….. (104)
Substituting the dimensions for , a
1 D AB b D c k c 2
L 1 t
a
M 3 L
b
L
c
L t
….. (105) 143
Equating the exponents of the fundamental dimensions on both sides of the equation, we have L : 0 = 2a – 3b + c + 1 t : 0 = – a – 1 M:0=b Solving these equations, a = –1, b = 0 and c=1 Thus k c D ……….. (106) 1 D AB which is the Sherwood number. The other two groups could be determined in the same manner, yielding D …………... (107) 2 and D AB
3
D AB
which is termed as Schmidt Number
S c
….. (108) 144
Dividing 2 by 3, we get 2 3
D D AB
D Re D AB
….. (109)
which is the Reynolds Number The result of the dimensional analysis of mass transfer by forced convection in a circular conduit indicates that a correlating relation could be of the form,
Sh Re, Sc
….. (110)
Which is analogous to the heat transfer correlation
Nu Re, Pr
….. (111)
145
M as s Tr an s f e r w i t h n at u r al convection
Natural convection currents develop if there exists any variation in density within the fluid phase.
The density variation may be due to temperature differences or to relatively large concentration differences.
In the case of natural convection involving mass transfer from a vertical plane wall to an adjacent fluid, the variables of importance are listed in the table.
146
M as s Tr an s f er w i th n a t u r al c o n v ec t i o n According to Buckingham theorem, there will be three dimensionless groups. Choosing D AB, L and as the core variables, the groups to be formed are a b c 1 D AB L k c
….. (112)
d e 2 D AB L
f
….. (113)
i g A
….. (114)
g h 3 D AB L
147
Table Variable
Symbol
Dimensions
characteristic length
L
L
fluid density
ρ
M/L3
fluid viscosity
μ
M/Lt
buoyant force
g Δρ A
mass diffusivity mass-transfer coefficient
D AB
k c
M/L2 t 2 L2 /t L/t
148
Solving for the dimensionless groups, we obtain π 1
π
and
2
k c L
Sh
D AB ρ D AB
3
μ
1 Sc
L 3 g A
….. (115) ….. (116) ….. (117)
D AB
With the multiplication of 2 and 3, we obtain a dimensionless parameter analogous to the Grashof number in heat transfer by natural convection 3 L g D AB A 2 3 D AB
L 3 g A 2
Gr AB
….. (118)
149
The result of the dimensional analysis of mass transfer by natural convection indicates that a correlating relation could be of the form,
Sh Gr AB , Sc
….. (119)
150
A n al o g i es am o n g M as s , H eat an d M o m en t u m Tr an s f er
Analogies among mass, heat and momentum transfer have their origin either in the mathematical description of the effects or in the physical parameters used for quantitative description.
To explore those analogies, it could be understood that the diffusion of mass and conduction of heat obey very similar equations. In particular, diffusion in one dimension is described by the Fick’s Law as
J A D AB
d C A
….. (120)
d z
Similarly, heat conduction is described by Fourier’s law as
d T q k d z where k is the thermal conductivity
….. (121) 151
M as s , H eat an d M o m en t u m Tr an s f e r The similar equation describing momentum transfer as given by Newton’s law is
d d z
….. (122)
where is the momentum flux (or shear stress) and is the viscosity of fluid.
It has become conventional to draw an analogy among mass, heat and momentum transfer. Each process uses a simple law combined with a mass or energy or momentum balance.
The analogies are useful in understanding the transfer phenomena and as a satisfactory means for predicting behaviour of systems for which limited quantitative data are available. 152
M as s , H eat an d M o m en t u m Tr an s f e r The similarity among the transfer phenomena and accordingly the existence of the analogies require that the following five conditions exist within the system
The physical properties are constant
There is no mass or energy produced within the system. This implies that there is no chemical reaction within the system
There is no emission or absorption of radiant energy.
There is no viscous dissipation of energy.
The velocity profile is not affected by the mass transfer. This implies there should be a low rate of mass transfer. 153
R ey n o l d s A n a lo g y The first recognition of the analogous behaviour of mass, heat and momentum transfer was reported by Osborne Reynolds in 1874. Although his analogy is limited in application, it served as the base for seeking better analogies. Reynolds postulated that the mechanisms for transfer of momentum, energy and mass are identical. Accordingly,
k c
h C p
f 2
….. (123) 154
R ey n o l d s A n a lo g y Here h is heat transfer coefficient f is friction factor is velocity of free stream The
Reynolds analogy is interesting because it suggests a very simple relation between different transport phenomena.
This
relation is found to be accurate when Prandtl and Schmidt numbers are equal to one.
This
is applicable for mass transfer by means of turbulent eddies in gases. In this situation, we can estimate mass transfer coefficients from heat transfer coefficients or from friction factors. 155
C h i l t o n -C o l b u r n a n al o g i es Chilton and Colburn, using experimental data, sought modifications to the Reynold’s analogy that would not have the restrictions that Prandtl and Schmidt numbers must be equal to one. They defined a j factor for mass transfer as j D
k c
Sc 2 3
….. (124)
The analogous j factor for heat transfer is j H St Pr 2 3
where St is Stanton number =
Nu h Re Pr v C p
….. (125) ….. (126)
Based on data collected in both laminar and turbulent flow regimes, they found
j D j H
f 2
….. (127) 156
This analogy is valid for gases and liquids within the range of 0.6 < Sc < 2500 and 0.6 < Pr < 100. The Chilton-Colburn analogy has been observed to hold for many different geometries for example, flow over flat plates, flow in pipes, and flow around cylinders.
157
Example A stream of air at 100 kPa pressure and 300 K is flowing on the top surface of a thin flat sheet of solid naphthalene of length 0.2 m with a velocity of 20 m/sec. The other data are: Mass diffusivity of naphthalene vapor in air = 6 * 10 –6 m 2/sec Kinematic viscosity of air = 1.5 * 10 –5 m 2.sc Concentration of naphthalene at the air-solid naphthalene interface = 1 * 10 –5 kmol/m3 Calculate:
the the avera verag ge mass ass tran transf sfer er coef coeffficie icien nt over ver the the fla flat plat plate e the rate of loss of naphthalene from the surface per unit width
Note: For heat transfer over a flat plate, convective heat transfer coefficient for laminar flow can be calc alculated by the equation. 12
Nu 0.664 ReL
Pr 1 3
You may use analogy between mass and heat transfer
158
Solution Give Given: n: Corr Correl elat atio ion n for for heat heat tran transf sfer er
Nu 0.664
12 Re L
Pr
13
The anal analog ogou ouss rel relatio ation n for for mass mass tra transf nsfer is
Sh 0.664
12 ReL
where Sh = Sherwood number = kL/D AB Re L = Reynolds number = L/ Sc = Schmidt number = / ( D AB) k = ove overall all mass ass tran transf sfer er coe coeffici ficie ent L = length of sheet D AB = diffusivity of A in B = velocity of air = viscosity of air = density of air, and kinema ematic tic visc viscos osit ityy of air air. / = kin
Sc
13
159
Subst ubstititut utiing for for the the know known n quan quantitititie es in equ equatio ation n (1) (1)
k 0.2
6 * 10 6
0.2 20 0.664 5 1.5 * 10
12
1.5 * 10 5 6 * 10 6
13
k = 0.014 m/s Rate of loss of naphthalene = k (C Ai – C A) = 0.014 (1 * 10 –5 – 0) = 1.4024 * 10 –7 kmol/m 2 s Rate of loss per meter width = (1.4024 * 10 –7) (0.2) = 2.8048 * 10 –8 kmol/m.s = 0.101 gmol/m.hr . 160
C o n v e c t i v e M as s Tr a n s f e r C o r r e l at i o n s
Extensive data have been obtained for the transfer of mass between a moving fluid and certain shapes, such as flat plates, spheres and cylinders.
The techniques include sublimation of a solid, vapourization of a liquid into a moving stream of air and the dissolution of a solid into water.
These data have been correlated in terms of dimensionless parameters and the equations obtained are used to estimate the mass transfer coefficients in other moving fluids and geometrically similar surfaces.
161
Flat Plate From the experimental measurements of rate of evaporation from a liquid surface or from the sublimation rate of a volatile solid surface into a controlled air-stream, several correlations are available. These correlation have been found to satisfy the equations obtained by theoretical analysis on boundary layers,
Sh 0.664 ReL1 2Sc 1 3 laminar Re L 3 * 10 5 ….. (128)
Sh 0.036 ReL0.8 Sc 1 3 turbulent Re L 3 * 10 5 ….. (129)
Using the definition of j factor for mass transfer in equation (128) and (129) we obtain
j D 0.664 ReL 1 2 laminar Re L 3 * 10 5 ….. (130) 0.2
J D 0.037 ReL
turbulent Re L 3 * 10 5
….. (131)
These equations may be used if the Schmidt number is in the
162
Sing le Sph ere Correlations for mass transfer from single spheres are represented as addition of terms representing transfer by purely molecular diffusion and transfer by forced convection, in the form ….. (132) Sh Sh C Re m Sc n o
Where C, m and n are constants, the value of n is normally taken as 1/3 For very low Reynold’s number, the Sherwood number should approach a value of 2. Therefore the generalized equation becomes
Sh 2 C Re
m
Sc
13
….. (133)
163
Sing le Sph ere For mass transfer into liquid streams, the equation given by Brain and Hales
Sh 4 1.21
2 3 12 Pe AB
….. (134)
correlates the data that are obtained when the mass transfer Peclet number, Pe AB is less than 10,000. This Peclet number is equal to the product of Reynolds and Schmidt numbers (i.e.)
Pe AB Re Sc
….. (135)
For Peclet numbers greater than 10,000, the relation given by Levich is useful 13
Sh 1.01 Pe AB
….. (136) 164
Sing le Sph ere The relation given by Froessling
Sh 2 0.552 Re 1 2 Sc 1 3
….. (137)
correlates the data for mass transfer into gases Reynold’s numbers ranging from 2 to 800 and Schmidt number ranging 0.6 to 2.7. For natural convection mass transfer the relation given by Schultz Sh 2 0.59 Gr AB Sc
14
….. (138)
is useful over the range 2 * 1 0 8 < Gr AB Sc < 1.5 * 10 10
165
Example The mass flux from a 5 cm diameter naphthalene ball placed in stagnant air at 40C and atmospheric pressure, is 1.47 * 10 –3 mol/m 2. sec. Assume the vapor pressure of naphthalene to be 0.15 atm at 40C and negligible bulk concentration of naphthalene in air. If air starts blowing across the surface of naphthalene ball at 3 m/s by what factor will the mass transfer rate increase, all other conditions remaining the same? For spheres :
Sh = 2.0 + 0.6 (Re) 0.5 (Sc)0.33 Where Sh is the Sherwood number and Sc is the Schmids number. The viscosity and density of air are 1.8 * 10 –5 kg/m.s and 1.123 kg/m 3, respectively and the gas constant is 82 06 cm 3 atm/mol.K. 166
Fl o w Th r o u g h P ip e s Mass transfer from the inner wall of a tube to a moving fluid has been studied extensively. Gilliland and Sherwood, based on the study of rate of vapourisation of nine different liquids into air given the correlation
Sh
p B, l m P
0.023 Re
0.83
Sc
0.44
….. (140)
Where p B, lm is the log mean composition of the carrier gas, evaluated between the surface and bulk stream composition. P is the total pressure. This expression has been found to be valid over the range 2000 < Re < 35000 0.6 < Sc < 2.5 167
Linton and Sherwood modified the above relation making it suitable for large ranges of Schmidt number. Their relation is given as
Sh 0.023 Re 0.83 Sc 1 3
….. (141)
and found to be valid for 2000 < Re < 70000 and 1000 < Sc < 2260 Example
A solid disc of benzoic acid 3 cm in diameter is spun at 20 rpm and 25C. Calculate the rate of dissolution in a large volume of water. Diffusivity of benzoic acid in water is 1.0 * 10 –5 cm 2/sec, and solubility is 0.003 g/cc. The following mass transfer correlation is applicable: Sh = 0.62 Re ½ Sc 1/3 2 Where D and is the angular Re speed in radians/time 168
Solution
….. (*1) Dissolution rate = N A S Where N A = mass flux, and S = surface area for mass transfer N A = k c (C As – C A ) ….. (*2) Where C As is the concentration of benzoic acid in water at the surface of the dose. C A is the concentration of benzoic acid in water for an infinite distance from the surface of the disc. Given:
Sh = 0.62 Re ½ Sc 1/3 (i.e.)
1
k c D D AB
D 2 2 0.62
1
3 D AB
….. (*3)
20 * 2 radian sec 60 1 rotation = 2 radian
169
