Transcript 01 - rshanthini

CP302 Separation Process Principles
Mass Transfer - Set 1
Course content of
Mass transfer section
L
T
A
Diffusion
Theory of interface mass transfer
Mass transfer coefficients, overall coefficients
and transfer units
Application in absorption, extraction and
adsorption
Concept of continuous contacting equipment
Simultaneous heat and mass transfer in gasliquid contacting, and solids drying
CSK
08
02
05
06
01
05
Prof. R. Shanthini
21 Feb 2013
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CP302 Separation Process Principles
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
4. Chapter 10 of R.P. Singh and D.R. Heldman
Introduction to Food Engineering
3rd edition, Academic Press
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Modes of mass transfer
Mass transfer could occur by the following three ways:
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. It cannot therefore 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.
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Modes of mass transfer
Stirring the water
with a spoon
creates forced
convection.
That helps the
sugar molecules to
transfer to the bulk
water much faster.
Diffusion
(slower)
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Convection
(faster)
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Objectives of the slides that follow:
Understanding diffusion in one dimension
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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.
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Diffusion
Solvent B
Solute A
concentration of A is the same everywhere
Diffusion is the macroscopic result of random thermal
motion on a micoscopic scale (Brownian motion).
It occurs even when there is no concentration gradient
(but there will be no net flux).
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A
B
Liquids A and B are separated from each other.
Separation removed.
A
B
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
DIFFUSION. Equilibrium is reached
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Examples of Diffusion
A diffusion problem that occurs in the field of microelectronics is
the oxidation of silicon according to the reaction Si + O2
SiO2.
When a slab of the material is exposed to gaseous oxygen, the
oxygen undergoes a first-order reaction to produce a layer of the
oxide. The task is to predict the thickness d of the very slowlygrowing oxide layer as a function of time t.
Oxygen
CA = CA0 at Z = 0
Silicon oxide layer
CA = CAδ at Z = δ
Silicon
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Examples of Diffusion
If an insect flight muscle contains
tracheal tubules which allow air to
diffuse into all parts of the muscle,
and the tracheal tubules make up
20% of the volume of the muscle,
how large can the muscle be?
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Objectives of the slides that follow:
Mathematical modelling of steady-state
one dimensional diffusive mass transfer
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Fick’s First Law of Diffusion
For mass transfer occurring only in z-direction
JA = - DAB
dCA
dz
(1)
Mixture
of A & B
CA
What is JA?
JA
CA + dCA
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dz
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Fick’s First Law of Diffusion
JA = - DAB
dCA
dz
concentration gradient of
A in z-direction
Unit: mass (or 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
Unit: mass (or moles) per area per time
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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.
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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 the Fick’s law and integrating gives the following:
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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)
(3)
DAB is given as 0.687 x 10-4 m2/s
(z2 – z1) is given as 0.2 m
(CA2 – CA1) = ?
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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
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CA =
pA
RT
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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
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Summary: Modelling diffusion in z-direction
JA = - DAB
z1
CA1
JA
(CA2 – CA1)
(z2 – z1)
z2
CA2
= - DAB
z
(pA2 – pA1)
RT(z2 – z1)
CA1 and pA1 at z1 and
CA2 and pA2 at z2
remain unchanged with
time (steady state).
DAB is constant
Longitudinal flow:
Flow area perpendicular to the flow direction is a constant.
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Objectives of the slides that follow:
Derivation of DAB = DBA under certain conditions
DAB: diffusivity of A in B
DBA: diffusivity of B in A
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Equimolar counter-diffusion in gases
Consider steady-state diffusion in an ideal mixture of 2 ideal
gases A & B at constant total pressure and temperature.
Molar diffusive flux of A in B:
JA = - DAB
Molar diffusive flux of B in A:
JB = - DBA
dCA
dz
dCB
dz
(4)
(5)
JA and JB : molar diffusive flux of A and B, respectively (moles/area.time)
CA and CB : concentration of A and B, respectively (moles/volume)
DAB and DBA : diffusivities of A in B and of B in A, respectively
z : distance in the direction of transfer
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Since the total pressure remains constant, there is no net mass
transfer. That is,
(6)
JA + JB = 0
For an ideal gas mixture at constant pressure,
CA + CB = pA/RT + pB/RT = P/RT = constant
Therefore,
dCA + dCB = 0
(7)
Substituting (6) and (7) in (4) and (5), we get
DAB = DBA = D (say)
Therefore, (4) & (5) give
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JA = - D
JB = - D
dCA
dz
dCB
dz
(8a)
(8b)
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This is known as equimolar counter diffusion.
This describes the mass transfer arising solely from the
random motion of the molecules (i.e., only diffusion)
It is applicable to stationary medium or a fluid in
streamline flow.
JA = - D
Prof. R. Shanthini
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JB = - D
dCA
dz
dCB
dz
(8a)
(8b)
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Objectives of the slides that follow:
Mathematical modelling of steady-state
one dimensional convective mass transfer
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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.
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Molar diffusive flux of A in B:
JA = - DAB
dCA
dz
(4)
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)
(9)
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)
(10)
The velocity of the bulk motion can be given by
(NA + NB) (mol/m2.s)
vbulk (m/s) =
(CT) (mol/m3)
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Total concentration
(11)
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vA,convection =
vA,diffusion + vbulk
Multiplying the above by CA, we get
CA vA,convection =
CA vA,diffusion + CA vbulk
Using equations (9) to (11) in the above, we get
NA = JA + CA (NA + NB)
CT
(12)
Substituting JA from equation (4) in (12), we get
NA = -DAB
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dCA
dz
+ CA
(NA + NB)
CT
(13)
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Let us introduce partial pressure pA into (13) as follows:
CA =
nA
V
pA
=
RT
(14a)
CT =
nT
V
P
=
RT
(14b)
Total number of moles
Total pressure
Using (14a) and (14b), equation (13) can be written as
NA = -
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DAB
dpA
RT
dz
pA (N + N )
A
B
+
P
(15)
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Let us introduce molar fractions xA into (13) as follows:
NA
xA =
(NA + NB)
=
CA
CT
(16)
Using (16), equation (13) can be written as
NA = -CT DAB dxA
dz
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+ xA (NA + NB)
(17)
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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
(NA + NB)
+
CT
(13)
In terms of partial pressures (using pA = CART and P = CTRT):
DAB
NA = RT
dpA
dz
pA (N + N )
A
B
+
P
In terms of molar fraction of A (using xA = CA /CT):
NA = -CT DAB dxA + xA (NA + NB)
dz
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(15)
(17)
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A diffusing through stagnant, non-diffusing B
Air (B)
2
z2 – z1
1
Liquid
Benzene
(A)
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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.
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Therefore, NB = 0
Substituting NB = 0 in equation (15), 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 =
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dpA
dz
pA2
DAB
RT
⌠
⌡
pA1
dpA
(1 - pA/P)
DAB P
P - pA2
ln
RT(z2 – z1)
P – pA1
(18)
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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 (18) is therefore written as follows:
NA =
DAB P
(pA1 - pA2 )
RT(z2 – z1) pB,LM
(19)
Equation (19) is the most used form.
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Using xA = CA /CT, pA = CART and P = CTRT,
equation (18) can be converted to the following:
NA =
DAB CT ln
(z2 – z1)
1 - xA2
1 – xA1
(20)
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 (20) becomes the following:
DAB CT
NA = (xA1 - xA2 )
(z2 – z1) xB,LM
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(21)
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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
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