Transcript PM3125_Lectures_1to5..

PM3125: Lectures 1 to 5
Content:
Mass transfer: concept and theory
uploaded at http://www.rshanthini.com/PM3125.htm
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Reference books used for ppts
1. C.J. Geankoplis
Transport Processes and Separation Process Principles
4th edition, Prentice-Hall India
2. J.D. Seader and E.J. Henley
Separation Process Principles
2nd edition, John Wiley & Sons, Inc.
3. J.M. Coulson and J.F. Richardson
Chemical Engineering, Volume 1
5th edition, Butterworth-Heinemann
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Modes of mass transfer
Mass transfer could occur by the following three ways:
Diffusion is caused by concentration gradient.
Advection is caused by moving fluid. (It cannot therefore
happen in solids.)
Convection is the net transport caused by both diffusion
and advection. (It occurs only in fluids.)
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Stirring the water
with a spoon
creates forced
convection.
That helps the
sugar molecules to
transfer to the bulk
water much faster.
Diffusion
(slower)
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Convection
(faster)
Diffusion
Solvent B
Solute A
concentration of A
is high
concentration of A
is low
Mass transfer by diffusion occurs when a component
in a stationary solid or fluid goes from one point to
another driven by a concentration gradient of the
component.
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Example of diffusion mass transfer
At the surface of the lung:
Air
Blood
Oxygen
High oxygen concentration
Low carbon dioxide concentration
Low oxygen concentration
High carbon dioxide concentration
Carbon dioxide
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27 Feb & 05 Mar 2012
Fick’s First Law of Diffusion
JA = - DAB
for mass transfer
in z-direction only
dCA
dz
(1)
A&B
CA
JA
Prof. R. Shanthini
27 Feb & 05 Mar 2012
CA + dCA
dz
JA = - DAB
dCA
dz
concentration
gradient of A
in z-direction
(mass/moles per
volume per distance)
diffusion coefficient
(or diffusivity) of A in B
diffusion flux of A in relation to the
bulk motion in z-direction
(mass/moles per area per time)
Prof. R. Shanthini
27 Feb & 05 Mar 2012
What is the unit of diffusivity?
Unit and Scale of Diffusivity
For dissolved matter in water:
D ≈ 10-5 cm2/s
For gases in air at 1 atm and at room temperature:
D ≈ 0.1 to 0.01 cm2/s
Diffusivity depends on the type of solute, type of
solvent, temperature, pressure, solution phase
(gas, liquid or solid) and other characteristics.
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Example 6.1.1 from Ref. 1
Molecular diffusion of Helium in Nitrogen: A mixture of He
and N2 gas is contained in a pipe (0.2 m long) at 298 K and 1
atm total pressure which is constant throughout. The partial
pressure of He is 0.60 atm at one end of the pipe, and it is
0.20 atm at the other end. Calculate the flux of He at steady
state if DAB of He-N2 mixture is 0.687 x 10-4 m2/s.
Solution:
Use Fick’s law of diffusion given by equation (1) as
JA = - DAB
dCA
dz
Rearranging equation (1) and integrating gives the following:
Prof. R. Shanthini
27 Feb & 05 Mar 2012
z2
⌠
J
⌡ A
z1
CA2
D
dz = -⌠
dC
AB
A
⌡
(2)
CA1
At steady state, diffusion flux is constant.
Diffusivity is taken as constant.
Therefore, equation (2) gives
JA(z2 – z1) = - DAB (CA2 – CA1)
DAB is given as 0.687 x 10-4 m2/s
(z2 – z1) is given as 0.2 m
(CA2 – CA1) = ?
Prof. R. Shanthini
27 Feb & 05 Mar 2012
(3)
Even though CA is not given at points 1 and 2, partial pressures
are given. We could relate partial pressure to concentration as
follows:
CA =
nA
V
pA V = nA RT
Number of moles of A
Total volume
Absolute temperature
Gas constant
Partial pressure of A
Combining the above we get
Prof. R. Shanthini
27 Feb & 05 Mar 2012
CA =
pA
RT
Equation (3) can therefore be written as
JA(z2 – z1) = - DAB
(pA2 – pA1)
RT
which gives the flux as
JA = - DAB
(pA2 – pA1)
RT(z2 – z1)
JA = - (0.687x10-4 m2/s)
(0.6 – 0.2) x 1.01325 x 105 Pa
(8314 J/kmol.K) x (298 K) x (0.20–0) m
JA = 5.63 x 10-6 kmol/m2.s
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Diffusion of gases A & B plus convection
Diffusion is the net transport of substances in a stationary
solid or fluid under a concentration gradient.
Advection is the net transport of substances by the moving
fluid, and so cannot happen in solids. It does not include
transport of substances by simple diffusion.
Convection is the net transport of substances caused by both
advective transport and diffusive transport in fluids.
JA is the diffusive flux described by Fick’s law, and we have
already studied about it.
Let us use NA to denote the total flux by convection (which is
diffusion plus advection.
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Molar diffusive flux of A in B:
JA = - DAB
dCA
dz
(1)
The velocity of the above diffusive flux of A in B can be given by
JA (mol/m2.s)
vA,diffusion (m/s) =
CA (mol/m3)
(4)
The velocity of the net flux of A in B can be given by
vA,convection (m/s) =
NA (mol/m2.s)
CA (mol/m3)
(5)
The velocity of the bulk motion can be given by
(NA + NB) (mol/m2.s)
vbulk (m/s) =
(CT) (mol/m3)
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Total concentration
(6)
Ignore the derivation, if you wish
vA,convection =
vA,diffusion + vbulk
Multiplying the above by CA, we get
CA vA,convection =
CA vA,diffusion + CA vbulk
Using equations (4) to (6) in the above, we get
NA = JA + CA (NA + NB)
CT
(7)
Substituting JA from equation (1) in (7), we get
NA = -DAB
Prof. R. Shanthini
27 Feb & 05 Mar 2012
dCA
dz
+ CA
(NA + NB)
CT
(8)
Ignore the derivation, if you wish
Let us introduce partial pressure pA into (8) as follows:
CA =
nA
V
pA
=
RT
(9a)
CT =
nT
V
P
=
RT
(9b)
Total number of moles
Total pressure
Using (9a) and (9b), equation (8) can be written as
NA = -
DAB
dpA
RT
dz
Prof. R. Shanthini
27 Feb & 05 Mar 2012
pA (N + N )
A
B
+
P
(10)
Ignore the derivation, if you wish
Let us introduce molar fractions xA into (8) as follows:
NA
xA =
(NA + NB)
=
CA
CT
(11)
Using (11), equation (8) can be written as
NA = -CT DAB dxA
dz
Prof. R. Shanthini
27 Feb & 05 Mar 2012
+ xA (NA + NB)
(12)
Ignore the derivation, if you wish
Diffusion of gases A & B plus convection:
Summary equations for (one dimensional) flow in z direction
In terms of concentration of A:
NA = -DAB dCA
dz
CA (N + N )
A
B
+
CT
(8)
Total concentration
In terms of partial pressures (using pA = CART and P = CTRT):
DAB dpA
pA
(NA + NB)
(10)
NA = +
dz
P
RT
Total pressure
In terms of molar fraction of A (using xA = CA /CT):
NA = -CT DAB dxA + xA (NA + NB)
(12)
dz
NA
CA
=
Molar fraction xA =
Prof. R. Shanthini
(NA + NB)
CT
27 Feb & 05 Mar 2012
A diffusing through stagnant, non-diffusing B
Air (B)
2
z2 – z1
1
Liquid
Benzene
(A)
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Evaporation of a pure liquid (A) is
at the bottom of a narrow tube.
Large amount of inert or nondiffusing air (B) is passed over
the top.
Vapour A diffuses through B in
the tube.
The boundary at the liquid
surface (at point 1) is
impermeable to B, since B is
insoluble in liquid A.
Hence, B cannot diffuse into or
away from the surface.
Therefore, NB = 0
Substituting NB = 0 in equation (10), we get
NA = -
DAB
RT
dpA
dz
pA (N + 0)
A
+
P
Rearranging and integrating
NA (1 - pA/P) = -
DAB
RT
z2
N ⌠dz = ⌡
A
z1
NA =
Prof. R. Shanthini
27 Feb & 05 Mar 2012
dpA
dz
pA2
DAB
RT
⌠
⌡
pA1
dpA
(1 - pA/P)
DAB P
P - pA2
ln
RT(z2 – z1)
P – pA1
(13)
Introduce the log mean value of inert B as follows:
pB,LM =
=
(pB2 – pB1 )
ln(pB2 /pB1 )
=
(P – pA2 ) – (P – pA1 )
ln[(P - pA2 )/ (P - pA1 )]
(pA1 – pA2 )
ln[(P - pA2 )/ (P - pA1 )]
Equation (13) is therefore written as follows:
NA =
DAB P
(pA1 - pA2 )
RT(z2 – z1) pB,LM
(14)
Equation (14) is the most used form.
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Using xA = CA /CT, pA = CART and P = CTRT,
equation (13) can be converted to the following:
NA =
DAB CT ln
(z2 – z1)
1 - xA2
1 – xA1
(15)
Introduce the log mean value of inert B as follows:
xB,LM =
=
(xB2 – xB1 )
ln(xB2 /xB1 )
=
(1 – xA2 ) – (1 – xA1 )
ln[(1 - xA2 )/ (1 - xA1 )]
(xA1 – xA2 )
ln[(1 - xA2 )/ (1 - xA1 )]
Therefore, equation (15) becomes the following:
DAB CT
NA = (xA1 - xA2 )
(z2 – z1) xB,LM
Prof. R. Shanthini
27 Feb & 05 Mar 2012
(16)
Example 6.2.2 from Ref. 1
Diffusion of water through stagnant, non-diffusing air:
Water in the bottom of a narrow metal tube is held at a constant
temperature of 293 K. The total pressure of air (assumed to be
dry) is 1 atm and the temperature is 293 K. Water evaporates
and diffuses through the air in the tube, and the diffusion path is
0.1524 m long. Calculate the rate of evaporation at steady state.
The diffusivity of water vapour at 1 atm and 293 K is 0.250 x 10-4
m2/s. Assume that the vapour pressure of water at 293 K is
0.0231 atm.
Answer: 1.595 x 10-7 kmol/m2.s
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Solution: The set-up of Example 6.2.2 is shown in the figure.
Assuming steady state, equation (14) applies.
NA =
DAB P
(pA1 - pA2 )
RT(z2 – z1) pB,LM
2
(14)
where
pB,LM =
Air (B)
(pA1 – pA2 )
ln[(P - pA2 )/ (P - pA1 )]
Data provided are the following:
z2 – z1
1
DAB = 0.250 x 10-4 m2/s;
Water (A)
P = 1 atm; T = 293 K;
z2 – z1 = 0.1524 m;
pA1 = 0.0231 atm (saturated vapour pressure);
pA2 = 0 atm (water vapour is carried away by air at point 2)
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Substituting the data provided in the equations given, we get
the following:
(0.0231 – 0 )
pB,LM =
ln[(1 - 0 )/ (1 – 0.0231 )]
NA =
= 0.988 atm
(0.250x10-4 m2/s)(1x1.01325x105 Pa)(0.0231 - 0) atm
(8314 J/kmol.K) (293 K) (0.1524 m) (0.988 atm)
= 1.595 x 10-7 kmol/m2.s
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Estimating Diffusivity
Diffusivities for different systems could be estimated using
the empirical equations provided in the following slides as
well as those provided in other reference texts available
in the library and other sources.
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Diffusivity of gases
An example at 1 atm and 298 K:
Prof. R. Shanthini
27 Feb & 05 Mar 2012
System
H2-NH3
Diffusivity (cm2/s)
0.783
H2-CH4
Ar-CH4
He-CH4
He-N2
0.726
0.202
0.675
0.687
Air-H2O
Air-C2H5OH
Air-benzene
0.260
0.135
0.0962
Binary Gas Diffusivity
DAB
P
Mi
T
Vi
- diffusivity in cm2/s
- absolute pressure in atm
- molecular weight
- temperature in K
- sum of the diffusion volume for component i
DAB is proportional to 1/P and T1.75
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Binary Gas Diffusivity
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Diffusivity in Liquids
For very large spherical molecules (A) of 1000 molecular
weight or greater diffusing in a liquid solvent (B) of small
molecules:
DAB =
DAB
T
μ
VA
9.96 x 10-12 T
μ VA1/3
applicable
for biological
solutes such
as proteins
- diffusivity in cm2/s
- temperature in K
- viscosity of solution in kg/m s
- solute molar volume at its normal boiling point
in m3/kmol
DAB is proportional to 1/μ and T
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Diffusivity in Liquids
For smaller molecules (A) diffusing in a dilute liquid solution of
solvent (B):
DAB =
1.173 x 10-12 (Φ MB)1/2 T
μB VA0.6
applicable
for biological
solutes
DAB - diffusivity in cm2/s
MB - molecular weight of solvent B
T - temperature in K
μ - viscosity of solvent B in kg/m s
VA - solute molar volume at its normal boiling point in m3/kmol
Φ - association parameter of the solvent, which 2.6 for water,
1.9 for methanol, 1.5 for ethanol, and so on
DAB is proportional to 1/μB and T
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Diffusivity of Electrolytes in Liquids
For smaller molecules (A) diffusing in a dilute liquid solution of
solvent (B):
8.928 x 10-10 T (1/n+ + 1/n-)
DoAB =
(1/λ+ + 1/ λ-)
DoAB is diffusivity in cm2/s
n+ is the valence of cation
n- is the valence of anion
λ+ and λ- are the limiting ionic conductances in very dilute
solutions
T is 298.2 when using the above at 25oC
DAB is proportional to T
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Diffusion in solids
Diffusion in solids are occurring at a very slow rate.
In gas:
DAB = 0.1 cm2/s
Time taken 2.09 h
In liquid:
DAB = 10-5 cm2/s
Time taken
In solid:
DAB = 10-9 cm2/s
Time taken
Prof. R. Shanthini
27 Feb & 05 Mar 2012
2.39 year
239 centuries
Diffusion in solids
Diffusion in solids are occurring at a very slow rate.
However, mass transfer in solids are very important.
Examples:
Leaching of metal ores
Drying of timber, and foods
Diffusion and catalytic reaction in solid catalysts
Separation of fluids by membranes
Treatment of metal at high temperature by gases.
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Diffusion in solids
Diffusion in solids occur in two different ways:
- Diffusion following Fick’s law (does not depend on
the structure of the solid)
- Diffusion in porous solids where the actual structure
and void channels are important
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Diffusion in solids following Fick’s Law
Start with equation (8):
NA = -DAB dCA
dz
CA (N + N )
A
B
+
CT
Bulk term is set to zero in solids
Therefore, the following equation will be used to describe
the process:
NA = -DAB dCA
dz
Prof. R. Shanthini
27 Feb & 05 Mar 2012
(17)
(8)
Diffusion through a slab
Applying equation (17) for steady-state diffusion through a solid
slab, we get
NA = DAB (CA1 - CA2)
z2 - z1
(18)
where NA and DAB are taken as constants.
CA1
CA2
Similar to heat conduction.
Prof. R. Shanthini
27 Feb & 05 Mar 2012
z2-z1
Relating the concentration and solubility
The solubility of a solute gas in a solid is usually expressed by
the notation S.
Unit used in general is the following:
m3 solute at STP
m3 solid . atm partial pressure of solute
Relationship between concentration and solubility:
S pA
CA =
kmol solute /m3 solid
22.414
where pA is in atm
STP of 0oC and 1 atm
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Relating the concentration and permeability
The permeability of a solute gas (A) in a solid is usually
expressed by the notation PM. in m3 solute at STP (0oC and 1
atm) diffusing per second per m2 cross-sectional area through a
solid 1 m thick under a pressure difference of 1 atm.
Unit used in general is the following:
m3 solute at STP . 1 m thick solid
s . m2 cross-sectional area . atm pressure difference
Relationship between concentration and permeability:
PM = DAB S
where DAB is in m2/s and S is in m3/m3.atm
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Example 6.5.1 from Ref. 1
Diffusion of H2 through Neoprene membrane:
The gas hydrogen at 17oC and 0.010 atm partial pressure is
diffusing through a membrane on vulcanized neoprene rubber
0.5 mm thick. The pressure of H2 on the other side of neoprene
is zero. Calculate the steady-state flux, assuming that the only
resistance to diffusion is in the membrane. The solubility S of H2
gas in neoprene at 17oC is 0.051 m3 (at STP of 0oC and 1
atm)/m3 solid. atm and the diffusivity DAB is 1.03 x 10-10 m2/s at
17oC.
Answer: 4.69 x 10-12 kmol H2/m2.s
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Example 6.5.2 from Ref. 1
Diffusion through a packging film using permeability:
A polythene film 0.00015 m (0.15 mm) thick is being considered
for use in packaging a pharmaceutical product at 30oC. If the
partial pressure of O2 outside the package is 0.21 atm and inside
it is 0.01 atm, calculate the diffusion flux of O2 at steady state.
Assume that the resistances to diffusion outside the film and
inside are negligible compared to the resistance of the film.
Permeability of O2 in polythene at 303 K is 4.17 x 10-12 m3 solute
(STP)/(s.m2.atm.m).
Answer: 2.480 x 10-12 kmol O2/m2.s
Would you prefer nylon to polythene? Permeability of O2 in nylon
at 303 K is 0.029 x 10-12 m3 solute (STP)/(s.m2.atm.m). Support
your answer.
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Diffusion through a cylinder wall
Applying equation (17) for steady-state diffusion through a
cylinder wall of inner radius r1 and outer radius r2 and length L in
the radial direction outward, we get
Mass transfer
per area per
time
N =
A
Mass transfer per time
nA
dCA
2 π r L = -DAB dr
(19)
Area of mass transfer
nA =
2πL DAB(CA1 - CA2)
ln(r2 / r1)
Similar to heat conduction.
Prof. R. Shanthini
27 Feb & 05 Mar 2012
(20)
CA2
CA1
r2
r
r1
Diffusion through a spherical shell
Applying equation (17) for steady-state diffusion through a
spherical shell of inner radius r1 and outer radius r2 in the radial
direction outward, we get
Mass transfer
per area per
time
N =
A
Mass transfer per time
nA
4 π r2
= -DAB
dCA
dr
(21)
Area of mass transfer
nA = 4πr1r2 DAB(CA1 - CA2)
(r2 - r1)
(22)
r2
Similar to heat conduction.
Prof. R. Shanthini
27 Feb & 05 Mar 2012
CA2
CA1
r
r1
Microscopic (or Fick’s Law) approach:
JA = - DAB
dCA
dz
(1)
good for diffusion dominated problems
Macroscopic (or mass transfer coefficient) approach:
NA = - k ΔCA
(50)
where k is known as the mass transfer coefficient
good for convection dominated problems
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Mass Transfer Coefficient Approach
NA = kc ΔCA
= kc (CA1 – CA2 )
(51)
kc is the liquid-phase mass-transfer coefficient
based on a concentration driving force.
CA1
A&B
What is the unit of kc?
NA
Prof. R. Shanthini
27 Feb & 05 Mar 2012
CA2
Mass Transfer Coefficient Approach
= kc (CA1 – CA2 )
NA = kc ΔCA
(51)
Using the following relationships between concentrations and
partial pressures:
CA1 = pA1 / RT;
CA2 = pA2 / RT
Equation (51) can be written as
NA = kc (pA1 – pA2) / RT
where kp = kc / RT
= kp (pA1 – pA2)
(52)
(53)
kp is a gas-phase mass-transfer coefficient
based on a partial-pressure driving force.
Prof. R. Shanthini
27 Feb & 05 Mar 2012
What is the unit of kp?
Models for mass transfer between phases
Mass transfer between phases across the following interfaces
are of great interest in separation processes:
- gas/liquid interface
- liquid/liquid interface
Such interfaces are found in the following separation
processes:
- absorption
- distillation
- extraction
- stripping
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Models for mass transfer at a fluid-fluid interface
Theoretical models used to describe mass transfer between a
fluid and such an interface:
- Film Theory
- Penetration Theory
- Surface-Renewal Theory
- Film Penetration Theory
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Film Theory
Entire resistance to mass transfer in a given turbulent phase
is in a thin, stagnant region of that phase at the interface,
called a film.
For the system shown, gas is
taken as pure component A,
which diffuses into nonvolatile
liquid B.
In reality, there may be mass
transfer resistances in both
liquid and gas phases. So we
need to add a gas film in which
gas is stagnant.
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Liquid
film
pA
Bulk
liquid
CAi
Gas
CAb
z=0
z=δL
Mass transport
Two Film Theory
There are two stagnant films (on either side of the fluid-fluid
interface).
Each film presents a resistance to mass transfer.
Concentrations in the two fluid at the interface are assumed to
be in phase equilibrium.
Liquid
film
Gas
film
Gas
phase
pAb
Liquid
phase
pAi
CAi
CAb
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Mass transport
Two Film Theory
Gas
film
Interface
Gas
phase
pAb
Interface
Liquid
film
Liquid
phase
Gas
phase
pAb
Liquid
phase
pAi
pAi
CAi
CAi
CAb
Mass transport
Concentration gradients for
the film theory
Prof. R. Shanthini
27 Feb & 05 Mar 2012
CAb
Mass transport
More realistic concentration
gradients
Two Film Theory applied at steady-state
Mass transfer in the gas phase:
NA = kp (pAb – pAi)
(52)
Mass transfer in the liquid phase:
NA = kc (CAi – CAb )
(51)
Phase equilibrium is assumed at
the gas-liquid interface.
Applying Henry’s law,
pAi = HA CAi
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Liquid
film
Gas
film
Gas
phase
pAb
Liquid
phase
pAi
CAi
CAb
(53)
Mass transport, NA
Henry’s Law
pAi = HA CAi at equilibrium,
where HA is Henry’s constant for A
Note that pAi is the gas phase pressure
and CAi is the liquid phase concentration.
Liquid
film
Gas
film
pAb
Unit of H:
[Pressure]/[concentration] = [ bar / (kg.m3) ]
Prof. R. Shanthini
27 Feb & 05 Mar 2012
pAi
CAi
CAb
Two Film Theory applied at steady-state
We know the bulk concentration and partial pressure.
We do not know the interface concentration and partial pressure.
Therefore, we eliminate pAi and
CAi from (51), (52) and (53) by
combining them appropriately.
Liquid
film
Gas
film
Gas
phase
pAb
Liquid
phase
pAi
CAi
CAb
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Mass transport, NA
Two Film Theory applied at steady-state
From (52):
pAi = pAb - NA
kp
(54)
From (51):
CAi = CAb + NA
kc
(55)
Substituting the above in (53) and rearranging:
NA =
pAb - HA CAb
(56)
HA / kc + 1 / kp
The above expression is based on gas-phase and
liquid-phase mass transfer coefficients.
Let us now introduce overall gas-phase and
overall liquid-phase mass transfer coefficients.
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Introducing overall gas-phase mass transfer
coefficient:
Let’s start from (56).
Introduce the following imaginary gas-phase partial pressure:
pA* ≡ HA CAb
(57)
where pA* is a partial pressure that would have been in
equilibrium with the concentration of A in the bulk liquid.
Introduce an overall gas-phase mass-transfer coefficient (KG) as
1
KG
≡
1
kp
+
HA
kc
(58)
Combining (56), (57) and (58):
Prof. R. Shanthini
27 Feb & 05 Mar 2012
NA = KG (pAb - pA* )
(59)
Introducing overall liquid-phase mass transfer
coefficient:
Once again, let’s start from (56).
Introduce the following imaginary liquid-phase concentration:
pAb ≡ HA CA*
(60)
where CA* is a concentration that would have been in
equilibrium with the partial pressure of A in the bulk gas.
Introduce an overall liquid-phase mass-transfer coefficient (KL) as
1
KL
≡
1
+
HAkp
1
kc
(61)
Combining (56), (60) and (61):
Prof. R. Shanthini
27 Feb & 05 Mar 2012
NA = KL (CA* - CAb)
(62)
Gas-Liquid Equilibrium Partitioning Curve showing the
locations of p*A and C*A
pA
pAb
pAb = HACA*
pAi
pAi = HA CAi
p A*
pA* = HA CAb
CAb
Prof. R. Shanthini
27 Feb & 05 Mar 2012
CAi
CA*
CA
Summary:
NA = KL (CA* - CAb)
(62)
= KG (pAb - pA*)
(59)
CA* = pAb / HA
(60)
pA* = HA CAb
(57)
where
1
KG
=
HA
KL
Prof. R. Shanthini
27 Feb & 05 Mar 2012
=
1
kp
+
HA
kc
(58 and 61)
Example 3.20 from Ref. 2 (modified)
Sulfur dioxide (A) is absorbed into water in a packed
column. At a certain location, the bulk conditions are 50oC,
2 atm, yAb = 0.085, and xAb = 0.001. Equilibrium data for
SO2 between air and water at 50oC are the following:
pA (atm)
0.0382
0.0606
0.1092
CA (kmol/m3)
0.03126
0.04697 0.07823
0.1700
0.10949
Experimental values of the mass transfer coefficients are
kc = 0.18 m/h and kp = 0.040 kmol/h.m2.kPa.
Compute the mass-transfer flux by assuming an average
Henry’s law constant and a negligible bulk flow.
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Solution:
Data provided:
T = 273oC + 50oC = 323 K;
PT = 2 atm;
yAb = 0.085;
xAb = 0.001;
kc = 0.18 m/h;
kp = 0.040 kmol/h.m2.kPa
0.2
HA = 1.4652 atm.m3/kmol
slope of
the curve
HA = 161.61
kPa.m3/kmol
pA (atm)
0.16
0.12
0.08
0.04
0
0.02
Prof. R. Shanthini
27 Feb & 05 Mar 2012
y = 1.4652x
R2 = 0.9759
0.04
0.06
0.08
CA (kmol/m3)
0.1
0.12
Equations to be used:
NA = KL (CA* - CAb)
(62)
= KG (pAb - pA*)
(59)
CA* = pAb / HA
(60)
pA* = HA CAb
(57)
where
1
KG
=
HA
KL
Prof. R. Shanthini
27 Feb & 05 Mar 2012
=
1
kp
+
HA
kc
(58 and 61)
Calculation of overall mass transfer coefficients:
1
KG
=
1
kp
1
=
h.m2.kPa/kmol
0.040
HA
161.61 kPa.m3/kmol
kc
=
HA
=
KL
1
+
kp
0.18 m/h
HA
kc
(58 and 61)
= 25 h.m2.kPa/kmol
= 897 h.m2.kPa/kmol
KG = 1/(25 + 897) = 1/922 = 0.001085 kmol/h.m2.kPa
KL = HA KG = 161.61/922 = 0.175 m/h
Prof. R. Shanthini
27 Feb & 05 Mar 2012
NA = KL (CA* - CAb)
(62) is used to calculate NA
CA* = pAb / HA = yAb PT / HA
= 0.085 x 2 atm / 1.4652 atm.m3/kmol = 0.1160 kmol/m3
CAb = xAb CT = 0.001 CT
CT = concentration of water (assumed)
= 1000 kg/m3 = 1000/18 kmol/m3 = 55.56 kmol/m3
CAb = 0.001 x 55.56 kmol/m3 = 0.05556 kmol/m3
NA = (0.175 m/h) (0.1160 - 0.05556) kmol/m3
= 0.01058 kmol/m2.h
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Alternatively,
NA = KG (pAb - pA*)
(59) is used to calculate NA
pA* = CAb HA = xAb CT HA
= 0.001 x 55.56 x 161.61 kPa = 8.978 kPa
pAb = yAb PT = 0.085 x 2 x 1.013 x 100 kPa = 17.221 kPa
NA = (1/922 h.m2.kPa/kmol) (17.221 - 8.978) kPa
= 0.00894 kmol/m2.h
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Summary: Two Film Theory applied at steady-state
NA = kp (pAb – pAi) = kc (CAi – CAb ) = KG (pAb - pA*) = KL (CA* - CAb)
(52)
(51)
pAi = HA CAi
(59)
(53)
pAb = HA CA*
(60)
pA* = HA CAb
(57)
1
KG
=
HA
KL
=
(62)
1
kp
+
HA
kc
Liquid
film
Gas
film
Liquid
phase
pAb
Gas
phase
pAi
CAi
(58 and 61)
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Mass transport, NA
CAb
Summary equations with mole fractions
NA = ky (yAb – yAi) = kx (xAi – xAb )
= Ky (yAb - yA*) = Kx (xA* - xAb)
(63)
yAi = KA xAi
(64)
yAb = KA xA*
(65)
yA* = KA xAb
(66)
1
Ky
=
KA
Kx
=
Prof. R. Shanthini
27 Feb & 05 Mar 2012
1
ky
+
KA
kx
Liquid
film
Gas
film
Liquid
phase
yAb
(67)
Gas
phase
yAi
xAi
xAb
Mass transport, NA
Notations used:
xAb
yAb
xAi
yAi
xA*
yA*
kx
ky
Kx
Ky
KA
: liquid-phase mole fraction of A in the bulk liquid
: gas-phase mole fraction of A in the bulk gas
: liquid-phase mole fraction of A at the interface
: gas-phase mole fraction of A at the interface
: liquid-phase mole fraction of A which would have been in
equilibrium with yAb
: gas-phase mole fraction of A which would have been in
equilibrium with xAb
: liquid-phase mass-transfer coefficient
: gas-phase mass-transfer coefficient
: overall liquid-phase mass-transfer coefficient
: overall gas-phase mass-transfer coefficient
: vapour-liquid equilibrium ratio (or equilibrium distribution
coefficient)
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Gas-liquid equilibrium ratio (KA) curve
yA
yAb
yAb = KAxA*
yAi
yAi = KA xAi
yA*
yA* = KA xAb
xAb
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27 Feb & 05 Mar 2012
xAi
x A*
xA
Gas & Liquid-side Resistances in Interfacial Mass Transfer
1
KG
1
=
kp
+
H
kc
fG = fraction of gas-side resistance
1/kp
=
1/KG
1
KL
=
1
H kp
1/kp
=
1/kp + H/kc
+
=
kc
kc + H kp
1
kc
fL = fraction of liquid-side resistance
1/kc
=
1/KL
Prof. R. Shanthini
27 Feb & 05 Mar 2012
1/kc
=
=
1/Hkp + 1/kc
kp
kp + kc/H
Gas & Liquid-side Resistances in Interfacial Mass Transfer
If fG > fL, use the overall gas-side mass transfer
coefficient and the overall gas-side driving force.
If fL > fG use the overall liquid-side mass transfer
coefficient and the overall liquid-side driving force.
Prof. R. Shanthini
27 Feb & 05 Mar 2012
1
=
KG
HA
=
KL
1
+
kp
HA
kc
(58 and 61)
The above is also written with the following notations:
1
KOG
=
Prof. R. Shanthini
27 Feb & 05 Mar 2012
H
KOL
=
1
KG
+
H
KL
Other Driving Forces
Mass transfer is driven by concentration gradient as well as
by pressure gradient as we have just seen.
In pharmaceutical sciences, we also must consider mass
transfer driven by electric potential gradient (as in the
transport of ions) and temperature gradient.
Transport Processes in Pharmaceutical Systems (Drugs
and the Pharmaceutical Sciences, vol. 102), edited by G.L.
Amidon, P.I. Lee, and E.M. Topp (Nov 1999)
Encyclopedia of Pharmaceutical Technology (Hardcover)
by James Swarbrick (Author)
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Example 1
An exhaust stream from a containing 3 mole% acetone and
90 mole% air is fed to a mass transfer column in which the
acetone is stripped by a countercurrent, falling 293 K water
stream. The tower is operated at a total pressure of
1.013x105 Pa. If a combination of Raoult-Dalton equilibrium
relation may be used to determine the distribution of acetone
between the air and aqueous phases, determine
(a) The mole fraction of acetone within the aqueous phase
which would be in equilibrium with the 3 mole% acetone gas
mixture, and
(b) The mole fraction of acetone in the gas phase which
would be in equilibrium with 20 ppm acetone in the aqueous
phase.
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Example 1 worked out
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27 Feb & 05 Mar 2012
Example 2
The Henry’s law constant for oxygen dissolved in water is
4.06x109 Pa/(mole of oxygen per total mole of solution) at 293
K. Determine the solution concentration of oxygen in water
which is exposed to dry air at 1.013x105 Pa and 293 K.
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Example 2 worked out
Henry’s law can be expressed in terms of the mole fraction
units by
pA = H’ xA
where H’ is 4.06x109 Pa/(mol of oxygen/total mol of solution).
Dry air contains 21 mole percent oxygen. By Dalton’s law
pA = yA P = (0.21)(1.013x105 Pa)
= 2.13 x 104 Pa
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Example 3
In an experiment study of the absorption of ammonia by
water in a wetted-wall column, the overall mass-transfer
coefficient, KG, was found to be 2.74 x 10-9 kgmol/m2.s.Pa.
At one point in the column, the gas phase contained 8
mole% ammonia and the liquid-phase concentration was
0.064 kgmole ammonia/m3 of solution. The tower operated at
293 K and 1.013x105 Pa. At that temperautre, the Henry’s
law constant is 1.358x103 Pa/(kgmol/m3).
If 85% of the total resistance to mass transfer is encountered
in the gas phase, determine the individual film mass-transfer
coefficients and the interfacial compositions.
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Example 3 worked out
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27 Feb & 05 Mar 2012
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Example 4
A wastewater stream is introduced to the top of a masstransfer tower where it flows countercurrent to an air
stream. At one point in the tower, the wastewater stream
contains 10-3 mole A/m3 and the air is essentially free of
any A. At the operation conditions within the tower, the film
mass-transfer coefficients are KL = 5x10-4
kmole/m2.s.(kmole/m3) and KG = 0.01 kmole/m2.s.atm.
The concentrations are in the henry’s law region where
pA,i = H CA,i with H =1 0 atm/(kmole/m3). Determine the
following:
(a) The overall mass flux of A
(b) The overall mass-transfer coefficients, KOL and KOG.
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Example 4 worked out
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Prof. R. Shanthini
27 Feb & 05 Mar 2012
Prof. R. Shanthini
27 Feb & 05 Mar 2012