Post on 14-Apr-2018
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MASS TRANSFER 1
CLB 20804
MASS TRANSFER
BASIC PRINCIPLES AND APPLICATIONS
CHAPTER 1
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Topic Outcomes
At the end of Chapter 1, you should:
Define and explain the basic concept of mass transfer.
DeriveFicks Law equation.
Calculate mass transfer rate based on unimolecular
diffusion and equimolar counterdiffusion.
Explain the Inter phase mass transfer.
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Introduction toBasic Principles of
diffusion andapplications
Principles of DiffusionFicks
Law
Application tounimolecular diffusion
and equimolarcounterdiffusion
What are in this chapter?
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Introduction
Mass transfer refers to mass in transit due to a speciesconcentration gradient in a mixture.
When a system contains two or more components whoseconcentration vary from point to point , there is a naturaltendency for mass to be transferred.
Minimizing the concentration differences within the systemand moving it towards equilibrium.
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Principles of Mass transfer
Mass transfer by ordinary molecular diffusion occurs becauseof a concentration difference(driving force for transfer).
The mass transfer rate is proportional to the area normal to
the direction of mass transfer. Not to the volume. So, the rateexpressed as a flux.
Mass transfer stops when the concentration is uniform.Mechanism of mass transfer involves both molecular diffusion
and convection.
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Molecular Diffusion
Net transport of molecules from a region of higher concentration to aregion of lower concentration byrandom molecular motion.
matter wants to "get away" from the other similar matter and go to aplace where there is open space.
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A B
A B
Liquids A and B are separated from each other.
Partition removed.
A goes from high concentration of A to low
concentration of A.
B goes from high concentration of B to low
concentration of B.
Molecules of A and B are uniformly distributed
everywhere in the vessel purely due to the
DIFFUSION.
Molecular Diffusion
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xamp e:Molecular Diffusion
Sprayed air freshener.
drop of liquid dye
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Example of Mass Transfer
At the surface of the lung:
Air Blood
Oxygen
Carbon dioxide
High oxygen concentration
Low carbon dioxide concentrationLow oxygen concentration
High carbon dioxide concentration
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Properties of Mixtures
Mass transfer always involves mixtures.
Consequently, we must account for the variation ofphysical properties which normally exist in a given
system. In order to understand the future discussions, let
us first consider definitions and relations whichare often used to explain the role of components
within a mixture.
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Chemical
Composition
Moles andMolecular
Weight
Mass andMole Fractions
AverageMolecular
Weight
Concentration
Properties of Mixtures
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Definitions
Definitions:iC Molarconcentration of species i. 3kmol/m:i Mass density (kg/m3) of species i.
:iM Molecularweight (kg/kmol) of species i.
i i iC M
*:iJ Molarflux of species i due to diffusion. 2kmol/s m Transport of i relative to molar average velocity (v*) of mixture.
:iN Absolute molar flux of species i. 2kmol/s m Transport of i relative to a fixed reference frame.
:ij Mass flux of species i due to diffusion. 2kg/s m Transport of i relative to mass-average velocity (v) of mixture.
Transport of i relative to a fixed reference frame.
:ix Mole fraction of species i / .i ix C C
:im Mass fraction of species i / .i im
Absolute mass flux of species i. 2
kg/s m:in
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Concentrations
Molar concentration of species A:(liquids,solids) ,(gases)
Where is molecular weight of species A.
Mole fraction:
)(
solids)&(liquids
gases
c
cy
c
cx
A
A
A
A
RT
p
V
n
Mc AA
A
AA
AM
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For a gaseous mixture that obeys the ideal gas law,mole fraction, can be written in terms ofpressures
For gases, P
p
RTP
RTpy AAA
Ay
Daltons
law
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General Ficks Law Equation of diffusion
Ficks Law
dz
dcDJ AABAz
*
The basic empirical relation to estimate the rate of moleculardiffussion:
Molar Flux
Units -
mol/(s.m2)
Diffusivity
(constant)
Units -cm2/s
alternative form
dz
dyCDJ AABA
JAzis the molar flux of A by ordinary molecular
diffusion relative to the molar average velocity of the
mixture in positive zdirection, DAB is the mutual
diffusion coefficient of A in B, and d cA/d z is the
c o n c e n t r a t i o n g r a d i e n t o f A , w h i c h i s
negative in the direction of ordinary molecular diffusion.
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Check your understanding-Theory of diffusion
Think of the last time that you washed the dishes. You placed the firstgreasy plate into the water and the dishwater got a thin film of oil on thetop of it. Assume that there is no oil at the top of the sink yet. Find thediffusion flux, J of oil droplets through the water to the top surface. Thesink is 18 cm deep, and the concentration of oil on the plate is 0.1mol/cm3. Given the diffusivity is 7x10-7 cm2/s.
6.1-1-Geankoplis(413)
Given : D = 7x10-7 cm2/s.
Find : Calculate the flux.
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dC = conc. at the top of the sink conc. of oil on the plate.
The concentration at the top of the sink = 0The concentration of oil on the plate = 0.1 mol/cm3.
dC = 0 0.1 = -0.1 mol/cm3
dx = the depth of the sink = 18 cm
Check your understanding-Theory of diffusion
J = 3.89 x 10-9 mol / (cm2.s)
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1.4 Diffusion Through Moving Bulk Fluid
Movement of A is now due to 2 contributions:
Molecular diffusion fluxes :
Fluxes resulting from bulk flow= CAV (kg-mole/m3. m/s)
Concentration of A at any point inthe mixture = CA (kg-mole/m
3)
(1)
Consider abulk fluid of binary mixture A and B moving in the z-directionas shown, with an average bulk fluid velocity V m/s, as shown in the Figurebelow:
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1.4.1 Diffusion Through MovingBulk Fluid
Molar flux of A :
Similarly for comp. B:
Total molar flux of A and B:
VCJN BBB
NNN BA
VCJN AAA (2)
(3)
(4)
Also, N = C V
Where, C = total molar concentrationV= average molar velocity(m/s)
(5)
Substitute eqn (4) into eqn (5) :
C
NNV
BA
(6)
1 4 1 s on ro g o ng
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1.4.1 us on roug ov ngBulk Fluid
Substitute eqn (1) & (6) into eqn (2) for comp. A: In term mole fraction:
In term concentration
)( BAAA
ABA NNydz
dyCDN
(7)
NA Mass Transfer Relative to a fixed position
J*AzMass Transfer Relative to a mass or molar average velocity
1 5 Stea y State Equimo ar Counter
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1.5 Stea y-State Equimo ar Counter-
Diffusion For binary mixture A and B,
When gas system is closed and are connected by straight tube, and at
constant pressure and temperature.
Molar fluxes A and B areequal, but opposite in direction
Total pressure is constant throughout.
Thus, diffussion flux are also equal but opposite in direction
0 BA NNN
Diffusivity, DAB = DBA
(1)
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Mass transfer in the connecting tube is equimolar counterdiffusion by molecular
1.5 Steady-State Equimolar Counter-
Diffusion
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1.5.1 Steady-State Equimolar Counter-
Diffusion Refer to eqn diffusion through moving bulk fluid:
in term mole fract ion
Integrated at z= z1, cA= cA1 and at z= z2, cA= cA2 to:
after simplify and rearrangement, the equation becomes :
)( BAAA
ABA NNydz
dycDN
0
dzAdyD
ABcJN AA
2
1
Ay
Ay dz
Ady
cD
A
J AB
21)( 12 AC
AC
DA
Jzz
AB
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1.5.2 Steady-State Equimolar Counter-
Diffusion of Ideal Gas Mixture:
Consider 2-component gas mixture (A and B)
Ideal Gas Law:
P = Total pressuren = Total moles of gas
Component-A:
Where;
pA = partial pressure of A
nA = moles of A
nRTPv
RTnVP AA
1 5 2 Steady State Equimolar Counter
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1.5.2 Steady-State Equimolar Counter-Diffusion of Ideal Gas Mixture:
Concentration of A:
Differentiating with respect to distance z
Replacing into the original Fick's Law
Integrated Eqn(3) over a diffusional path from z2 to z1 &pA1 to PA2
RT
p
V
nC AAA
RTdz
dp
dz
dC AA 1
dz
Dp
RT
DJ AABA
(1)
(2)
(3)
(4)
2
1
2
1
.
PA
pA
AAB
z
z
A dpRT
DdzJ
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1.5.2 Steady-State Equimolar Counter-Diffusion of Ideal Gas Mixture:
where :
pA1 = partial pressure of A at point 1
pA2 = partial pressure of A at point 2:
(5) 21)( 12
AAAB
A ppzzRT
DJ
Rearrange equation:
Diffusion flux for components A
Diffusion flux for component-B
)(
)(
12
21
zzRT
ppDJ BBABB
St d St t Eq i l C t
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Steady-State Equimolar Counter-Diffusion
21)( 12
AAAB
A ppzzRT
DJ
Or, In terms of partial pressure
In terms of concentration
21)( 12 AC
AC
DA
Jzz
AB
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Ammonia gas (A) diffusing through a uniform tube 0.10 m longcontaining N2 gas (B) at 1.0132x105 Pa pressure and 298 K. At point 1,
pA1=1.013x104 Pa and point pA2= 0.507x10
4 Pa. The diffusivity DAB =
0.230 x 10-4 m2/s. R = 8314 m3.Pa/kg-mole.K
(a) Calculate the diffusion flux JA at steady-state.
(b) Repeat for calculate the flux JB.
Check your understanding
Given:Total pressure PT = 1.0132 x 10
5 Pa(constant)
Temperature T = 298 KDAB = 0.230 x 10-4 m2/s
R = 8314 m3.Pa/kg-mole.KAt point 1, pA1 = 1.013 x 104 PaAt point 2, pA2 = 0.507 x 104 PaDiffusion path = ( z2 - z1 ) = 0.1 m
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Component-A is diffusing from Point 1 to Point 2, as its partial
pressure is higher at point 1.
Solution:
)()(
12
21
zzRTppDJ AAABA
(a) Calculate the diffusion flux JA at steady-state.
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(b) Calculate the diffussion flux JB.
Use the Dalton's Law of partial pressures to determine the partial pressures
ofcomponent-B at points 1 and 2:
PT = pA+ pBAt point 1:pB1 = PT - pA1PB1 = 1.0132 x 105 - 1.013 x 104 = 91,190 PaAt point 2:
pB2 = PT - pA2PB2 = 1.0132 x 105 - 0.507 x 104 = 96,250 Pa
Component-B is diffusing in the opposite direction to component-A:from Point 2 to Point 1, as the partial pressure for B at point 2 is higher.
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Flux for component-B is the same as the flux for component-A, butwith a negative sign, indicating that it is in the opposite direction.
Calculate the flux of B in A using:
)(
)(
12
21
zzRT
ppDJ BBABB
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1.5.2 Steady-State Equimolar Counter-Diffusion ofIdeal Gas Mixture:
Rate of Diffusion= (molar flux) x (Surface area)
= JAx S
Where ;Rate of Diffusion(unit: kmole/s)
JA= Diffusion flux of component A (kmole/m2 .s)
S= Surface area(unit m2 )
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1.6 Steady-State One componentDiffusions
Opposite of equimolar counterdiffusion.
One component diffuses, while the other remains stagnant(nonmoving).
Fig 1.6, point 0 is saturated with component A.
while a stream of pure B flows past the end of the tube removing any A that
has reached point z.
In this situation, the concentration gradient along the tube is exponential
Fig 1.6
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1.6 Steady-State One componentDiffusions
Example:
Diffusion of Components liquid benzene (A) through stagnant (nonmoving)
component air (B).
Evaporation of a pure (A) at the bottom of a narrowtube.
Where a large amount of nondiffussing air (B) ispassed over the top.
Benzene vapor (A) diffuses through the air (B) in thetube.
Point 1 (Boundary at the liquid surface) ~impermeableto air (since air is insoluble in benzene liquid).
Hence, air (B) cannot diffuse into or away from thesurface.
Point 2~ partial pressure PA2=0(since a large volume ofair is passing by. Component B cannot diffuse,
NB=0
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Rearrange eqn (1) to a ficks law form
dz
dy
y
CDN A
A
ABA
1
:(2)
(1)
1.6 Steady-State One component Diffusions
)( BAAA
ABA NNydz
dyCDN
Refer to equation diffusion through moving bulk fluid
Equation simplifies to
0
)( AAAABA NydzdyCDN
dz
dyCDNyN
AABAAA )(
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1.6 Steady-State One component Diffusions
At quasi steady state condition, rearange eqn becomes in integral form;
Upon integration yields
Log mean (LM) of (1-yA) at the two ends of the stagnants layer :
1
2
12 11
lnA
AAB
yy
zzCD
AN
2
1
2
1 1
yA
yAA
A
A
ABz
z y
dy
N
CDdz
Integration rules
(3)
(4)
)1/(1
112
12
AA
AA
LMAyyIn
yyy
(5)
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Rearrange eqn (5)
Subsitute eqn (6)into eqn (4),
LMA
AA
A
A
y
yy
y
yIn
1)1(
1 21
1
2
LMA
AAABA
y
yy
zz
CDN
)1(
21
12
Finally:the rate of diffusion is given by (in mole fraction terms)
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using the Ideal Gas equation,
RT
pC
RT
PC
AA
(3.1)
General equation Diffusion through moving bulk fluid:
Component B cannot diffuse, NB= 0
)( BAAA
ABA NNC
C
dz
dCDN
(1)
)0(
A
AAABA N
C
C
dz
dCDN (2)
(3.2)
1.6.2 Steady-State One component Diffusions
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Differentiate eqn (3.2) with respect to distance z,
Subsitute eqn (4) and (5) into eqn (2)
AAA
ABA N
P
P
dz
dCDN
Divide eqn (3.2) with eqn (3.1);
re-arrange:
dz
dp
RTdz
dC AA 1
P
p
C
CAA
dz
dp
RT
D
P
pN AABAA 1
(4)
(5)
(6)
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Integrating Eqn (6) from point 1 to point 2: the partial pressure of
A changes from pA1 to pA2 :
Upon integration
2
1
2
1 1
pA
pA A
AAB
z
z
A
Pp
dp
RT
DdzN
Integration rules
(7)
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Log mean (LM) of (P-PA) at the two ends of the stagnants
layer :
Rearange eqn (5)
LMA
AAABA
PP
pP
zzRT
PDN
,
1
12 )(
)(
)(2
LMA
AA
A
A
PPPP
pPPPIn
21
1
2
)(
Subsitute eqn (6)into eqn (4),
Finally:
(9)
(10)
)/( 1221
AA
AA
LMA PPPPIn
PP
PP
Rate of diffusion is given by (inpartial pressure terms)
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An open beaker, 6 cm in height, is filled with liquid benzene at 25C towithin 0.5 cm of the top. A gentle breeze of dry air at 25C and 1 atm is
blown by a fan across the mouth of the beaker so that evaporatedbenzene is carried away by convection after it transfers through astagnant air layer in the beaker. The vapor pressure of benzene at 25Cis 0.131 atm. The mutual diffusion coefficient for benzene in air at 25Cand 1 atm is 0.0905 cm2/s. Determine the initial rate of evaporation ofbenzene as a molar flux in mol/cm2.s
Check your understanding
Evaporation of Benzene from abeaker
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Then
scm
molxNA
.1004.1
933.0
131.0
)06.82)(298(5.0
0905.02
6
LMA
AAABA
PP
pP
zzRT
PDN
,
1
12 )(
)(
)(2
Solution:
Let A = benzene, B = air.
Take Zl = 0.
Then Z2 - Zl= Z = 0.5 cm.
One component equation:
933.0)131.01/(01131.0
)/( 12
21
InPP
PPPPIn
PPPP
LMA
AA
AA
LMA
Log mean (LM) of (P-PA) at the two ends of the stagnants layer :
1.7 Diffusion between phases - Phase
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1.7 Diffusion between phases PhaseEquilibrium
When two immiscible phases are in contact:
Example. gas-liquid, two immiscible liquids, it is possible for
diffusing molecules to pass from one phase to the other across
the interphase boundary.
When diffusion between two phases occurs, there are two
factors to take into account;
1 Equilibrium relationships between the two phases
2 rate at which the diffusion takes place
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1.7 Diffusion between phases - PhaseEquilibrium
Referring to Fig 1.7:
If a component, A is passing between phases 1 & 2, there will ultimately be adynamic equilibrium established where the rate of diffusion of A is the same in
both directions.
Depending on the type of system (liquid vapour, liquid liquid etc.), the
equilibrium concentration may often be expressed by simple relationships.
Fig 1.7
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1.7 Diffusion between phases - PhaseEquilibrium
2 simple relationships are : Distribution Coefficients
1.6
Where: XA = concentration of the component A in Phase 1
YA = concentration of the component A in Phase2
K = constant
YA = KXA
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Henry's Law
1.7 Diffusion between phases - PhaseEquilibrium
Where:
P
A= Partial pressure of A in the gas phase
C
A= Concentration of A in the liquid phase
H = constant - called Henry's law constant
Henry's law is a special case of the distribution coefficient for gas-liquid systems.
In the case of systems where one of the phases is a gas or vapour itis common to express the concentration in the gas phase terms ofpartial pressure.
Thus Henry's Law states that
PA = HCA
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1.8 Rate of diffusion between phases -Mass Transfer
The theory discussed so far enables us to calculate the rates ofdiffusion within a single phase and to calculate the concentrationsof a component in two phases when the system is in a state ofequilibrium.
However, many practical problems concern the rate of diffusion
between two phases when the two phases are not in equilibrium.
The rate of mass transfer between two phases is dependant on anumber of factors including:
1. The diffusivity of the diffusing component in the two
phases
2. How far the system is from equilibrium
3. The resistance to transfer across the interface between the
two phases
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Mass Transfer EquationRate of mass transfer is directly proportional to the driving force for
transfer, and the area available for the transfer process to take place,
that is:
Transfer rate transfer area driving forceThe proportional coefficient in this equation is called the masstransfercoefficient, so that:
Transfer rate = mass-transfer coefficient
transfer area driving force
1.8 Rate of diffusion between phases -Mass Transfer
NA = kACA
Where:
NA = mass transfer rate of A across thephase boundary
K = mass transfer coefficient
C = concentration driving force
1 8 Interphase mass transfer
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1.8 Interphase mass transfer
Two-Film Theory of Mass TransferIn order to illustrate the concept of
interphase mass transfer lets considerthe process of transport of a volatile
chemical across the air/water
interphase.
Gas molecules must diffuse from the
main body of the gas phase to thegas-liquid interface, then cross thisinterface into the liquid side, andfinally diffuses from the interface intothe main body of the liquid.
C t ti d i i f
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The concentration driving force is a measure of how far the systemis from equilibrium and may be developed with reference to the
diagram below. (Fig 1.9)
1.9 Concentration driving force
Consider a component, such as oxygen, diffusing from air to water. Assume the
system is at steady state but not at equilibrium.
PA = partial pressure of oxygen in the air and CA is the concentration ofoxygen in the water.
CA* = concentration of oxygen in water that will be in equilibrium with PA
PA* = partial pressure of oxygen in air that will be in equilibrium with CA
(Fig 1.9)
C t ti d i i f
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1.9 Concentration driving force
The rate of oxygen mass transfer between air and water may be
expressed in two ways. One based on partial pressure of oxygen in air and one based on the
concentration of oxygen in water.
NA = KG (PA P*A)
NA = KL (C*A CA)
Where:
PAP
A
*= partial pressure driving force
CACA
*= concentration driving force
KL = mass transfer coefficient based on the liquid phase.
KG = mass transfer coefficient based on the gas phase.
The approriate equilibrium value PA* or CA* may be calculated using Henry's law
Gas Liquid Equilibrium Partitioning Curve
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Gas-Liquid Equilibrium Partitioning Curve
CA,i
PA,i
PA
CA
P*A
C*A
CA
PA
PA,i = H CA,i
P*A= H CA
PA= H C*A
PA* = HCA
PA= HCA*
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What have you learn from thisChapter?