Transcript 02 - Rshanthini.com

CP302 Separation Process Principles
Mass Transfer - Set 2
Prof. R. Shanthini
26 Feb 2013
1
Example 6.2.2 from Ref. 1 (from Set 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
26 Feb 2013
2
Solution: The set-up of Example 6.2.2 is shown in the figure.
Assuming steady state, equation (19) applies.
NA =
DAB P
(pA1 - pA2 )
RT(z2 – z1) pB,LM
2
(19)
where
pB,LM =
Air (B)
(pA1 – pA2 )
ln[(P - pA2 )/ (P - pA1 )]
Data provided are the following:
Prof. R. Shanthini
26 Feb 2013
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)
3
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
26 Feb 2013
4
Example 6.2.3 from Ref. 1
Diffusion in a tube with change in
path length:
Diffusion of water vapour in a narrow
tube is occurring as in Example 6.2.2
under the same conditions. However,
as shown in the figure, at a given
time t, the level is z from the top. As
diffusion proceeds the level drops
slowly. Drive the equation for the
time tF for the level to drop from a
starting point z0 m at t = 0 to zF at t =
tF s as shown.
Air (B)
2
z
Z
zFF 0
z
1
Water (A)
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26 Feb 2013
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Solution: Since the level drops
very slowly, we assume
pseudo-steady-state condition.
Therefore, equation (19)
applies in which (z2 – z1) must
be replaced by z.
NA =
DAB P
RTz pB,LM
Air (B)
2
z
Z
zFF 0
(pA1 - pA2 )
z
(22)
1
Water (A)
Prof. R. Shanthini
26 Feb 2013
6
Suppose that the level reduce by dz in dt time.
Mass balance yields the following:
(NA x A x dt) x MA = ρA x (A x dz)
where A is the cross-sectional
area of the tube, MA is the
molecular mass of water and
ρA is the density of water.
Air (B)
(23)
2
z
Z
zFF 0
z
1
dz
Water (A)
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26 Feb 2013
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Combining equations (22) and (23), we get
MA DAB P (p - p ) dt
ρA dz =
A1
A2
RTz pB,LM
(24)
Rearranging (24), we get
zF
ρ ⌠z dz =
⌡
A
tF
⌠
⌡
MA DAB P (p - p ) dt
A1
A2
RT pB,LM
z0
0
Is it okay to take
pA1 and pA2 as
constants?
(zF2 – z02)
MA DAB P (p - p ) t
ρA
=
A1
A2
F
2
RT pB,LM
tF =
Prof. R. Shanthini
26 Feb 2013
ρA RT pB,LM (zF2 – z02)
2 MA DAB P (pA1 - pA2 )
(25)
This above equation is used to
experimentally determine diffusivity DAB
8
Example 6.2.4 from Ref. 1
Evaporation of a Naphthalene sphere:
A sphere of naphthalene having a radius
of 2.0 mm is suspended in a large volume
of still air at 318 K and 1.01325 x 105 Pa
(1 atm). The surface temperature of the
naphthalene can be assumed to be at 318
K and its vapour pressure at 318 K is
0.555 mm Hg. The DAB of naphthalene in
air at 318 K is 6.92 x 10-6 m2/s. Calculate
the rate of evaporation of naphthalene
from the surface.
If the radius of the sphere decreases
slowly with time, drive the equation for the
time taken (tF) for the sphere to evaporate
completely.
Prof. R. Shanthini
26 Feb 2013
pA2
dr
pA1
r
r1
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Solution: All this time the mass transfer of A is taken as NA (in
moles/m2.s) and it is assumed to remain constant for systems at
steady-state or pseudo steady-state while carrying out the
integrations concerned. For the system considered here, we
cannot do that since the area across which the mass transfer
occurs vary as the radius changes.
p
A2
Therefore, we use the following definition:
Mass transfer
per area per
time
NA =
Mass transfer per time
nA
dr
(26)
A
pA1
r
Area of mass transfer
For the given system, equation (26) gives
nA
(27)
NA =
4πr2
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26 Feb 2013
r1
10
Since the given system is a case of A diffusing through
stagnant, non-diffusing B, we could start from equation (15).
DAB dpA
pA (N + N )
NA = +
A
B
RT
dz
P
Setting NB = 0 and rearranging the above gives:
NA = -
DAB
RT
pA2
dpA
(1 - pA/P) dz
Using equation (26) and replacing dz by dr, the
above could be written as follows:
NA =
(15)
nA
DAB
=2
4πr
RT
dpA
(1 - pA/P) dr
dr
pA1
r
r1
which can be rearranged to give
Prof. R. Shanthini
26 Feb 2013
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nA
4π
dr
r2
=-
DAB
RT
dpA
(1 - pA/P)
Integrating the above between r1 and some point r2
(which is a large distance away) gives the following:
r2
nA
4π
⌠
⌡
pA2
dr
=-
r2
r1
nA
4π
1
r1
DAB
RT
dp
⌠
⌡ (1 - p /P)
A
pA1
-
Prof. R. Shanthini
26 Feb 2013
pA2
1
r2
=
DABP
RT
A
P - pA2
P – pA1
dr
pA1
r
r1
(28)
12
Since r2 >> r1, 1/ r2 is much smaller than 1/ r1 in equation (28),
and therefore neglected. So, equation (28) gives the following:
nA
=
4πr1
DABP
RT
P - pA2
P – pA1
pA2
Using equation (27) and the definition of pB,LM,
the above could be written as
NA =
nA
4πr1
2
=
DABP
RTr1
pA1 - pA2
pB,LM
Equation (29) gives the rate of
evaporation of naphthalene from the
surface at r1 radius.
Prof. R. Shanthini
26 Feb 2013
(29)
dr
pA1
r
r1
13
Data provided are the following:
DAB = 6.92 x 10-6 m2/s;
P = 1 atm;
T = 318 K;
r1 = 2 mm = 2/1000 m;
pA1 = 0.555/760 atm (saturated vapour pressure);
pA2 = 0 atm (no naphthalene vapour far away from the sphere)
Substituting the data provided, we get the following:
(0.555/760 – 0 )
pB,LM =
ln[(1 - 0 )/ (1 – 0.555/760 )]
NA =
= 0.9996 atm
(6.92x10-6 m2/s)(1.01325x105 Pa)(0.555/760 - 0) atm
(8314 J/kmol.K) (318 K) (2/1000 m) (0.9996 atm)
Prof. R. Shanthini
26 Feb 2013
= 9.686 x 10-8 kmol/m2.s
14
Now, we need to drive the equation for the
time taken (tF) for the sphere to evaporate
completely. That is, the radius of the sphere is
r1 at t = 0 and it is zero at t = tF.
Since the radius of the sphere is said to
decrease slowly, we assume pseudo-steadystate condition. Therefore, equation (29)
applies in which r1 is replaced by  as follows:
nA
NA = 4π2 =
DABP
RT
pA1 - pA2
pB,LM
pA2
pA1
(30)

d
r1
Sphere is enlarged.
Prof. R. Shanthini
26 Feb 2013
15
Suppose that the surface reduce by d in dt
time. Mass balance yields the following:
(NA x A x dt) x MA = ρA x (A x (-d))
(31)
where A is the surface area of the sphere at
radius , MA is the molecular mass of the
sphere and ρA is the density of the sphere.
pA2
pA1

d
r1
Sphere is enlarged.
Prof. R. Shanthini
26 Feb 2013
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Combining equations (30) and (31), we get
MA DAB P (p - p ) dt
- ρA d =
RT pB,LM A1 A2
Rearranging and integrating, we get
0
⌠
⌡
- ρA
pA2
tF
⌠
⌡
MA DAB P (p - p ) dt
d =
A1
A2
RT pB,LM
r1
0
Is it okay to assume that pA1 and pA2
remain constants as the radius of the
sphere () reduces from r1 to 0?
pA1

d
r1
Sphere is enlarged.
Prof. R. Shanthini
26 Feb 2013
17
Integration gives the following results:
ρA
(r12 – 0)
2
tF =
MA DAB P (p - p ) t
=
A1
A2
F
RT pB,LM
ρA r12 RT pB,LM
(32)
pA2
2 MA DAB P (pA1 - pA2 )
The above gives the time taken for
the sphere to vanish completely at a
slow rate of evaporation.
pA1

d
r1
Sphere is enlarged.
Prof. R. Shanthini
26 Feb 2013
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Objectives of the slides that follow:
Estimating diffusivities
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26 Feb 2013
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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
26 Feb 2013
20
Diffusivity of gases
An example at 1 atm and 298 K:
Prof. R. Shanthini
26 Feb 2013
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
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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
26 Feb 2013
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Binary Gas Diffusivity
Prof. R. Shanthini
26 Feb 2013
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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
26 Feb 2013
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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
26 Feb 2013
25
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
26 Feb 2013
26
Objectives of the slides that follow:
Mathematical modelling of steady-state
one dimensional diffusive mass transfer
in solids
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26 Feb 2013
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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
26 Feb 2013
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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
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26 Feb 2013
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Diffusion in solids following Fick’s Law
Start with equation (13) from Set 1:
NA = -DAB dCA
dz
CA (N + N )
A
B
+
CT
(13)
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
26 Feb 2013
(33)
30
Diffusion through a slab
Applying equation (33) for steady-state diffusion through a solid
slab, we get
NA = DAB (CA1 - CA2)
z2 - z1
(34)
where NA and DAB are taken as constants.
Drive (34) and compare it with
heat conduction equivalent.
CA1
CA2
z2-z1
Prof. R. Shanthini
26 Feb 2013
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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
26 Feb 2013
32
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
26 Feb 2013
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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
26 Feb 2013
34
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
26 Feb 2013
35
Diffusion through a cylinder wall
Applying equation (33) 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
(35)
Area of mass transfer
nA =
2πL DAB(CA1 - CA2)
ln(r2 / r1)
Drive (36) and compare it with
heat conduction equivalent.
Prof. R. Shanthini
26 Feb 2013
(36)
CA2
CA1
r2
r
r1
36
Diffusion through a spherical shell
Applying equation (33) 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
(37)
Area of mass transfer
nA = 4πr1r2 DAB(CA1 - CA2)
(r2 - r1)
Drive (38) and compare it with
heat conduction equivalent.
Prof. R. Shanthini
26 Feb 2013
(38)
CA2
CA1
r2
r
r1
37
Diffusion in porous solids
Read Section 6.5C (page 445 to 446) of Ref. 1
Prof. R. Shanthini
26 Feb 2013
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