CH2015: HEAT AND MASS TRANSFR
L12: DIFFUSION MASS TRANSFER
Dr. Dilhara Sethunga
dilharap@uom.lk|0778589964
27 December 2022
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Diffusion in Gasses
Equimolar Counter Diffusion
Diffusion of A through stagnant, non diffusing B
Diffusion through varying cross-sectional area
Diffusion coefficient for gasses
2
Equimolar counter diffusion in
Gasses
• This is the situation in which the components A and
B both diffuses at equal rates but in opposite
directions.
NA= -NB
𝐶𝐴
𝜕𝐶𝐴
𝑁𝐴 = (𝑁𝐴 + 𝑁𝐵 ) − 𝐷𝐴𝐵
𝐶
𝜕𝑧
𝜕𝐶𝐴
𝑁𝐴 = −𝐷𝐴𝐵
𝜕𝑧
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Equimolar Counter diffusion in
Gasses
Carbon particle burns in air
Here the carbon particle is surrounded by an air film
through which the molecules of O2 diffuse and reach
the surface of the particle to sustain combustion. If a
O2 molecule diffuses to the surface a CO2 molecule is
formed (provided that CO2 is the only product of
combustion) which diffuses through the air film at
steady state. Therefore, O2 and CO2 undergo
“equimolar counter diffusion”.
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Equimolar Counter diffusion in
Gasses
Industrial Example: Distillation
The vapor rising through a distillation column remains
in intimate contact with the down-flowing liquid.
Exchange of mass occurs between the phases-the more
volatile component moves from the liquid to the vapor
phase and the less volatile one gets transported from
vapor to the liquid phase.
If the column is perfectly insulated against the heat loss
and the molar heats of vaporization of the components
are equal, the mass exchange between the phases will
occur in equimolar counter diffusion mode.
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Equimolar counter diffusion in
Gasses - Examples
Ammonia gas (A) is diffusing through a uniform tube
0.10 m long containing N2 gas (B) at 1.0132 x 105 Pa
pressure and 298 K. At point 1, pA1 = 1.013 x 104 Pa
and at point 2, pA2 = 0.507 x 104 Pa. The diffusivity
DAB = 0.230 x 10-4 m2/s.
(a) Calculate the flux NA at steady state
(b) Repeat for NB
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Counter diffusion in Gasses Examples
An aqueous solution of methanol is being separated by distillation in a column.
Methanol (A), which is the more volatile component, moves from the liquid phase
to the vapor phase while water (B), the less volatile component, gets transported
in the opposite direction. At a section of the column the vapor phase contains
0.76 mole fraction methanol, and the liquid has 0.6 mole fraction of it. The
temperature is 71.2 0C and the total pressure is essentially atmospheric. The
diffusional resistance offered is equivalent to that of a stagnant vapor film of 1
mm thickness. If the latent heat of vaporization of methanol is 274.6 kcal/kg and
that of water is 557.7 kcal/kg at the given temperature, calculate the flux of
methanol and that of water vapor.
Given: if the mole fraction of methanol in the solution is 0.6, its mole fraction in
the equilibrium vapor would be 0.825. The vapor-phase mutual diffusivity
DAB=1.816 x 10-5 m2/s
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Diffusion of A through Stagnant,
Non-Diffusive B
Industrial Example
Dry air is needed for burning sulfur in Sulfuric Acid
Manufacturing Plant. Dry air is produced in a packed tower
by contacting air with concentrated sulfuric acid. The
moisture (A) in the air (B) diffuses through the air layer
(film) into the concentrated sulfuric acid. (air is the source
of moisture and the concentrated sulfuric acid is the sink
for moisture). Since air is virtually insoluble with Sulfuric
acid, it can be considered as non-diffusive with respect to
moisture.
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Diffusion of A through Stagnant,
Non-Diffusive B
General Mass Transfer Equation
𝑁𝐴 =
𝐷𝐴𝐵 𝑃
𝑙𝑛
𝑅𝑇𝑧
𝑃 − 𝑝𝐴𝑧
𝑃 − 𝑝𝐴0
or
𝑁𝐴 =
𝐷𝐴𝐵 𝑃(𝑝𝐴0 − 𝑝𝐴𝑧 )
𝑅𝑇𝑧𝑝𝐵,𝐿𝑀
Where;
𝑝𝐵,𝐿𝑀 =
𝑝𝐵𝑧 − 𝑝𝐵0
𝑝
𝑙𝑛 𝐵𝑧
𝑝𝐵0
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Diffusion of A through Stagnant,
Non-Diffusive B
Example
Water in the bottom of a narrow metal tube is held at a
constant temperature of 293 K. The total pressure of air
(assumed dry) is 1.01325 x 105 Pa (1.0 atm) and the
temperature is 293 K (20 °C). Water evaporates and diffuses
through the air in the tube, and the diffusion path z2-z1 is
0.1524 m long. Calculate the rate of evaporation at steady
state. The diffusivity of water vapor at 293 K and 1 am
pressure is 0.250 x 10-4 m2/s. Assume that the system is
isothermal.
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Diffusion of A through Stagnant,
Non-Diffusive B
Example
Ammonia (A) diffuses through a stagnant layer of air (B), 1
cm thick, at 25 0C and 1 atm total pressure. The partial
pressures of ammonia on the two sides of the air layer are
0.9 atm and 0.1 atm respectively. Air is non-diffusing.
Calculate,
(a) molar flux of ammonia
(b) prepare the plots of partial pressure distributions of
ammonia and air along the diffusion path.
Given DAB=0.214 cm2/s
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Diffusion of A through Stagnant,
Non-Diffusive B
Example
A test tube, 1.5 cm in diameter and 12 cm tall, is partly filled with a
solution of alkaline pyrogallate. The depth of the empty space above
the solution is 5 cm. The temperature is 25 0C and the total pressure is
1 atm. Air may be assumed to contain 21% O2 and 79% N2. The
diffusivity of O2 in N2 at the given condition is 0.21 cm2/s.
a) Calculate the rate of absorption of oxygen from air in the solution
at steady state if air flows gently over the open end of the test tube.
Make plots of the distribution of partial pressures of the gasses along
the diffusion path.
b) Calculate the partial pressure gradient of oxygen midway in the
diffusion path
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Diffusion of through Varying Cross
Sectional Area
Diffusion in Spherical Geometry
Example: Evaporation of liquid droplet, burn carbon particle in
air, Catalytic Reactions in spherical geometry, etc..
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Diffusion of through Varying Cross
Sectional Area
General mass transfer equation
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Diffusion of through Varying Cross
Sectional Area
Example:
Calculate the time required for the sublimation of 3g of
naphthalene from a naphthalene ball of mass 4g kept suspension
in a large volume of stagnant air at 45 0C and 1.013 bar pressure.
Diffusivity of naphthalene in air under the given conditions is
6.92×10-6 m2/s, its density is 1140 kg/m3 and its sublimation
pressure at 45 0C is 0.8654 mmHg.
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Diffusion of through Varying Cross
Sectional Area
Example:
A 1 cm diameter spherical pellet of a strongly
basic oxide is held at the center of a closed vessel
of 5 litre volume. Initially the vessel is evacuated
and then filled with an equimolar mixture of H2S
and N2 at 25 °C and 1 atm total pressure. As the
molecules of H2S reach the surface of the pellet,
they get absorbed instantaneously so that the
concentration of H2S at the surface remains zero
at all time. Diffusion of H2S occurs through a
stagnant film of estimated thickness of 4 mm
surrounding the pellet. The bulk of the gas may
be assumed to have a uniform composition at any
time. Calculate the time of absorption of 95% of
the H2S gas. The temperature remains at 25°C and
the diffusivity of H2S in nitrogen is 1.73 x 10-5
m2/s at this temperature and 1 atm pressure.
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Diffusion of through Varying Cross
Sectional Area
Diffusion in Cylindrical Geometry
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Diffusion of through Varying Cross
Sectional Area
Diffusion in Cylindrical Geometry
Example: Sublimation of a 20 cm long naphthalene cylinder of
mass 10 g occurs in a large volume of air in a room at 45°C and
1 atm pressure. The vapor diffuses through a stagnant film of air
of thickness 3 mm surrounding the cylinder. The air beyond the
film is well-mixed. Calculate the time required for sublimation of
half of the naphthalene. Loss of mass because of sublimation
from the flat ends may be neglected (as their combined area is
considerably smaller than that of the cylindrical surface). The
data given in above may be used.
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Diffusion Coefficient
The units of diffusion coefficient is m2/s. In general,
the smaller the component (species) size the higher the
diffusion coefficients.
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Diffusion Coefficient
Empirical Equations
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Diffusion Coefficient
Empirical Equations
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Diffusion Coefficient
In general,
Diffusion Coefficient is a function of temperature and
pressure. The relation is;
𝑇𝑛
𝐷∝
𝑃
Example: The diffusivity of H2S in nitrogen is 1.73 x 10 -5 m2/s
at 25°C temperature and 1 atm pressure. Calculate the diffusivity
of H2S in nitrogen at 50 0C and 3 atm.
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Diffusion in Liquids
𝑁𝐴 =
𝐶𝐴
𝐶
(𝑁𝐴 + 𝑁𝐵 ) − 𝐷𝐴𝐵
𝜌
𝐶=
𝑀
𝑑𝐶𝐴
𝑑𝑧
𝑎𝑣
𝜌: density of the solution
𝑀:Molecular weight of the solution
𝜌
:
𝑀 𝑎𝑣
Average molar concentration of the solution
Above equation can be re arranged as:
𝑁𝐴 = (𝑁𝐴 + 𝑁𝐵 )𝑥𝐴 − 𝐷𝐴𝐵
𝜌
𝑀
𝑑𝑥𝐴
𝑎𝑣 𝑑𝑧
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Diffusion in Liquids
Case a: for the diffusion of A through non-diffusing B;
𝜌
𝐷𝐴𝐵 𝑀
𝑎𝑣
𝑁𝐴 =
𝑥𝐴0 − 𝑥𝐴𝑧
𝑧𝑥𝐵,𝐿𝑀
Where 𝑥𝐵,𝐿𝑀 =
𝑥𝐵𝑧 −𝑥𝐵0
𝑥
ln 𝑥𝐵𝑧
𝐵0
Case b: for equimolar counter diffusion of A and B
𝑁𝐴 =
𝐷𝐴𝐵
𝜌
𝑀
𝑧
𝑎𝑣
𝑥𝐴0 − 𝑥𝐴𝑧
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Diffusion in Liquids
Exercise:
In a slurry reactor the gas A is sparged in an agitated
suspension of catalyst particles, 1 mm in average diameter, in a liquid (B).
The gas dissolves in the liquid and is transported to the surface of the
catalyst particles where it undergoes an instantaneous reaction. In a
particular case the concentration of A in the liquid is 1 kmol/m3, the rate
of reaction is 3.15 ×10-6 kmol/m2.s based on the external surface area of
the catalyst particles, and the diffusivity of A in the liquid is 7×10-10 m2/s.
If the diffusion of dissolved A to the catalyst surface occurs through a
stagnant film surrounding a particle, calculate the thickness of the liquidfilm.
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Mass Transfer Coefficient
Any solution obtained in above discussion can be written in the
form of following equation.
𝑁𝐴 = 𝑘 𝐶𝐴2 − 𝐶𝐴1
NA- The molar flux
CA2 - concentration of species A at point 2
CA1 - concentration of species A at point 1
kthe mass transfer coefficient.
k is depending on,
1. the type of the medium which the mass transfer occurs (gas or liquid)
2. the driving force (concentration, partial pressure, mole fraction)
3. mass transport of A through non diffusing B or counter diffusion
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Mass Transfer Coefficient
Different Mass Transfer Coefficients
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Mass Transfer Coefficient
Notations
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Mass Transfer Coefficient
Notations
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Mass Transfer Coefficient
Conversions between different mass transfer
coefficients
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Diffusion through Solids
𝑁𝐴 = −𝐷𝐴
𝜕𝐶𝐴
𝜕𝑛
Diffusion through a Solid Slab with Constant Area
𝑧
𝐶𝐴2
න 𝑁𝐴 𝑑𝑧 = − න
0
𝐷𝐴 𝑑𝐶𝐴
𝐶𝐴1
𝑁𝐴 = 𝐷𝐴
(𝐶𝐴1 − 𝐶𝐴2 )
𝑧
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Diffusion through Solids
𝑁𝐴 = −𝐷𝐴
𝜕𝐶𝐴
𝜕𝑛
Diffusion through a Hollow Cylinder
𝐴𝐿𝑀 𝐷𝐴 𝐶𝐴1 − 𝐶𝐴2
𝑊=
𝑡
𝑡 = 𝑟𝑜 − 𝑟𝑖
𝐴𝐿𝑀 = 2𝜋𝑙
𝑟𝑜 − 𝑟𝑖
𝑟
𝑙𝑛 𝑟𝑜
𝑖
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Diffusion through Porous Solids
Practical Examples: Catalytic reactions inside porous solids,
adsorption or membrane (porous membranes) separation
Diffusion mechanism: 1. Molecular diffusion (d/λ>20, d=pore
diameter, λ: mean free path: mean free path is the average distance
travelled by a moving particle between successive impacts
(collisions))
𝐶𝐴
𝑑𝐶𝐴
𝑁𝐴 =
𝑁 + 𝑁𝐵 − 𝐷𝐴𝐵,𝑒𝑓𝑓
𝐶 𝐴
𝑑𝑧
𝐷𝐴𝐵,𝑒𝑓𝑓 𝑁𝐴 𝐶
𝑁𝐴 =
𝑙𝑛
𝑁𝐴 + 𝑁𝐵 𝑧
𝐶𝑁𝐴
− 𝐶𝐴2
𝑁𝐴 + 𝑁𝐵
𝐶𝑁𝐴
− 𝐶𝐴1
𝑁𝐴 + 𝑁𝐵
At steady-state NA and NB and hence NA+NB are constant. Assumed that
flow through a constant cross-sectional area
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Diffusion through Porous Solids
Practical Examples: Catalytic reactions inside porous solids,
adsorption or membrane (porous membranes) separation
Diffusion mechanism: 2. Knudsen diffusion (d/λ<0.2)
𝐷𝐾,𝐴 𝑝𝐴1 − 𝑝𝐴2
𝑁𝐴 =
𝑙
𝐷𝐾,𝐴 : Knudsen diffusivity
𝑙: length of the pore
𝑝𝐴𝑖 : partial pressure of the diffusing
substance A at each side
Knudsen Diffusivity can be calculated by;
𝐷𝐾,𝐴
𝑑
=
3
8𝑔𝑅𝑇
𝜋𝑀𝐴
1/2
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Diffusion through Porous Solids
Practical Examples: Catalytic reactions inside porous solids,
adsorption or membrane (porous membranes) separation
Diffusion mechanism: 3. If 0.2< d/λ<20, both molecular and
Knudsen diffusion take place
𝑇ℎ𝑒 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡𝑇𝑜𝑡𝑎𝑙 𝑀𝑜𝑙𝑎𝑟 𝐷𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛 =
𝐶𝐷𝐴𝐵,𝑒𝑓𝑓
𝑁𝐴 =
ln
𝑁𝐴 + 𝑁𝐵 𝑧
𝑑𝐶𝐴
𝑑𝐶𝐴
+
𝑑𝑧 𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟
𝑑𝑧 𝐾𝑛𝑢𝑑𝑠𝑜𝑛
𝐶𝐴2 −
𝑁𝐴 𝐶 𝐷𝐴𝐵,𝑒𝑓𝑓 + 𝐷𝐴,𝑒𝑓𝑓
𝑁𝐴 + 𝑁𝐵 𝐷𝐴𝐵,𝑒𝑓𝑓 𝐷𝐴,𝑒𝑓𝑓
𝐶𝐴1 −
𝑁𝐴 𝐶 𝐷𝐴𝐵,𝑒𝑓𝑓 + 𝐷𝐴,𝑒𝑓𝑓
𝑁𝐴 + 𝑁𝐵 𝐷𝐴𝐵,𝑒𝑓𝑓 𝐷𝐴,𝑒𝑓𝑓
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Diffusion in Crystalline Solids
• Interstitial mechanism
• Vacancy mechanism
• Interstitialcy mechanism
• Crowded-ion mechanism
• Diffusion along grain boundaries
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