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Analogy of Mass, Heat and Momentum Transport
General molecular transport equations
ππ§ = βπΏπΞ
ππ§
Momentum Heat Mass
ππ§π₯ = β (π
π)
π(ππ£π₯)
ππ§
ππ§π΄
= βπΌπ(ππΆππ)
ππ§ π½π΄π§
β = βπ·π΄π΅π(ππ΄)
ππ§
Turbulent diffusion equation
Momentum Heat Mass
ππ§π₯ = β (π
π+ ππ‘)
π(ππ£π₯)
ππ§
ππ§π΄
= β(πΌ + πΌπ‘)π(ππΆππ)
ππ§
π½π΄π§β
= β(π·π΄π΅ + ππ)π(ππ΄)
ππ§
Diffusion
Diffusion results from random motions of two types: the random motion of
molecules in a fluid, and the random eddies which arise in turbulent flow.
Diffusion from the random molecular motion is termed molecular diffusion;
diffusion which results from turbulent eddies is called turbulent diffusion or
eddy diffusion.
Why diffusion occurs?
Initially, there are solute molecules (A) on the left side of a barrier and
none on the right. The barrier is removed, and the solute (A) diffuses into B
to fill the whole container.
Diffusion of molecules is due to concentration gradient.
Momentum
diffusivity (
π
π) (
π2
π )
Thermal diffusivity πΌ (π2
π )
Molecular diffusivity π·π΄π΅ (π2
π )
Molecular diffusivity depends on pressure, temperature, and composition of the system.
Diffusivities of gases at low density are almost composition independent, increase with the temperature and vary inversely with
pressure (Table 6.2-1 CJG).
Liquid and solid diffusivities are strongly concentration dependent and increase with temperature.
General range of diffusivities:
Gases: 5 Γ 10 β6 β 1 Γ 10-5 m2 / s
Liquids: 10 β6 β 10-9 m2 / s
Solids: 5 Γ 10 β14 β 1 Γ 10-10 m2 / s
In the absence of experimental data, semi-theoretical expressions have been
developed which give approximate values of molecular diffusivities.
Turbulent (eddy) momentum diffusivity ππ‘ (π2
π )
Turbulent (eddy) thermal diffusivity πΌπ‘ (π2
π )
Turbulent (eddy) mas diffusivity ππ (π2
π )
Fickβs Law for Molecular Diffusion
π½π΄π§β = βππ·π΄π΅
π(π₯π΄)
ππ§
where c is the total concentration of A and B in (kg mol A + B)/m3, and π₯π΄ is the mole fraction of A in the mixture A and B.
For constant concentration (c),
ππ΄ = ππ₯π΄
πππ΄ = π(ππ₯π΄) = ππ(π₯π΄)
π½π΄π§β = βπ·π΄π΅
π(ππ΄)
ππ§
For constant molar flux, the above equation can be integrated as follows to
give,
π½π΄π§β β« ππ§
π§2
π§1
= βπ·π΄π΅ β« π(ππ΄)ππ΄2
ππ΄1
π½π΄π§β = β
π·π΄π΅(ππ΄1 β ππ΄2)
(π§1 β π§2)
Since the concentration is related to partial pressure,
ππ΄ =ππ΄π π
π½π΄π§β = β
π·π΄π΅(ππ΄1 β ππ΄2)
π π(π§1 β π§2)
π½π΄π§β =
π·π΄π΅(π§2 β π§1)
1
π π(ππ΄1 β ππ΄2)
8314
T 298 K
P 1.013E+05 Pa
D_AB1 6.870E-05 m2/s
p_A1 6.080E+04 Pa
p_A2 2.027E+04 Pa
dZ 0.2 m
J_AZ1 5.6192E-06 (kg mol A)/(s.m2)
T_1 298 K
D_AB1 6.87E-05 Pa
T_2 398 K
D_AB2 1.14E-04 Pa
T 398 K
P 1.013E+05 Pa
D_AB2 1.140E-04 m2/s
p_A1 6.080E+04 Pa
p_A2 2.027E+04 Pa
dZ 0.2 m
J_AZ2 6.9812E-06 (kg mol A)/(s.m2)
Change 24.2 %
Gas constants: R_
Example 6.1-1: Molecular diffusion of Helium in Nitrogen
(m3.Pa)/(kg mol . K)
Helium-N2 @ 298 K
Helium-N2 @ 398 K
Diffusivity Helium-N2 @ 398 K
Molecular Diffusion in Gases
Equi-molar Counter Diffusion in Gasses
Consider diffusion that occurs in a tube connecting two tanks containing a
binary gas mixture of species A and B. If both tanks as well as the connecting
tube are at a uniform pressure and temperature, the total molar concentration
would be uniform throughout the tanks and the connecting tube. If π₯π΄1 , the mole fraction of A in tank 1, is larger than π₯π΄2 , the mole fraction of A in tank 2, A would diffuse from tank 1 to tank 2 through the connecting tube,
while B would diffuse from tank 2 to tank 1 through the same connecting
tube.
Because the temperature and pressure are uniform, the molar flux of
A from tank 1 to tank 2 through the connecting tube must be the
same as the molar flux of B from tank 2 to tank 1.
π½π΄π§β = βπ½π΅π§
β
[βπ·π΄π΅π(ππ΄)
ππ§] = β [βπ·π΅π΄
π(ππ΅)
ππ§]
Since, π = ππ΄ + ππ΅ = ππππ π‘πππ‘, therefore,
π = ππ΄ + ππ΅ = ππππ π‘πππ‘.
Differentiating
0 = πππ΄ + πππ΅
Substituting πππ΅ = βπππ΄ gives,
[βπ·π΄π΅π(ππ΄)
ππ§] = β [βπ·π΅π΄
π(ππ΅)
ππ§] = β [βπ·π΅π΄ (β
π(ππ΄)
ππ§)] = [βπ·π΅π΄
π(ππ΄)
ππ§]
[βπ·π΄π΅π(ππ΄)
ππ§] = [βπ·π΅π΄
π(ππ΄)
ππ§]
π·π΄π΅ = π·π΅π΄
Conclusion For a binary gas mixture of A and B. the diffusion coefficient
for A diffusing in B is same as gas B diffusing in A.
Molecular diffusivity is independent of concentration.
T 298 K
P 1.01E+05 Pa
D_AB1 2.30E-05 m2/s
p_A1 1.01E+04 Pa
p_A2 5.07E+03 Pa
dZ 0.1 m
J_AZ 4.6973E-07 (kg mol A)/(s.m2)
T 298 K
P 1.01E+05 Pa
D_AB1 2.30E-05 m2/s
p_B1 9.12E+04 Pa
p_B2 9.63E+04 Pa
dZ 0.1 m
J_BZ -4.697E-07 (kg mol A)/(s.m2)
Example 6.2-1: Equimolar Counter Diffusion
Diffusion of Gases A and B with Convective Flow
Diffusion velocity (m/s) of A = π£π΄π
Therefore, molar diffusion flux,
π½π΄π§β
ππ πππ π΄
π β π2= ππ΄π£π΄π
ππ πππ π΄
π3π
π
The velocity of A relative to the stationary point is the sum of the diffusion velocity
(π£π΄π) and the average velocity (π£π , molar average velocity of the whole fluid relative to a stationary point). Mathematically,
π£π΄ = π£π΄π + π£π
ππ΄π£π΄ = ππ΄π£π΄π + ππ΄π£π
Total convective flux of the whole stream relative to the stationary point:
ππ£π = π = ππ΄ + ππ΅; π£π =ππ΄ + ππ΅
π
Therefore,
ππ΄ = βππ·π΄π΅π(π₯π΄)
ππ§+
ππ΄π
(ππ΄ + ππ΅)
ππ΅ = βππ·π΅π΄π(π₯π΅)
ππ§+
ππ΅π
(ππ΄ + ππ΅)
Total flux of A relative to stationary point
ππ΄π£π΄(= ππ΄)
Diffusion flux relative to moving fluid
ππ΄π£π΄π(=, π½π΄π§β )
Convective flux of A relative to stationary
point ππ΄π£π = +
Total flux of
A, ππ΄ Diffusion flux, π½π΄π§
β
Convective flux of A,
ππ΄ (ππ΄ + ππ΅
π)
= +
Special Case for A Diffusing through Stagnant Film of B
Example 1: The benzene vapor (A) diffuses through the air (B) in the tube. The air is
insoluble in benzene liquid, there is no movement in air (ππ΅ = 0).
Example 2: The ammonia vapor (A) diffuses through the air (B) in the tube and gets
absorbed in water. The air is slightly soluble in water, there is no
movement in air (ππ΅ = 0).
Therefore,
ππ΄ = βππ·π΄π΅π(π₯π΄)
ππ§+
ππ΄π
(ππ΄ + 0)
Since,
π =π
π π; ππ΄ = π₯π΄π;
ππ΄π
=ππ΄π
Therefore, for constant pressure
ππ΄ = βπ·π΄π΅π π
π(ππ΄)
ππ§+
ππ΄π
ππ΄
ππ΄ (1 βππ΄π
) = βπ·π΄π΅π π
π(ππ΄)
ππ§
π΅π¨ = βπ«π¨π©π·
πΉπ»
π
(π· β ππ¨)
π (ππ¨)
π π
ππ΄ β« ππ§π2
π1
= βπ·π΄π΅π
π πβ«
πππ΄(π β ππ΄)
ππ΄2
ππ΄1
ππ΄ =π·π΄π΅
(π§2 β π§1)
π
π πln
π β ππ΄2π β ππ΄1
Since, the total pressure is the sum of partial pressures, i.e., π = ππ΄1 + ππ΅1 = ππ΄2 +ππ΅2
Therefore,
ππ΄ =π·π΄π΅
(π§2 β π§1)
π
π πππ
ππ΅2ππ΅1
But,
ππ΅π =ππ΅2 β ππ΅1
ππππ΅2ππ΅1
=ππ΄1 β ππ΄2
ππππ΅2ππ΅1
ππππ΅2ππ΅1
=ππ΄1 β ππ΄2
ππ΅π
Therefore,
π΅π¨ =π«π¨π©
(ππ β ππ)
π·
πΉπ»
ππ¨π β ππ¨πππ©π΄
π΅π¨ =π«π¨π©
(ππ β ππ)
π
πΉπ»
π·
ππ©π΄(ππ¨π β ππ¨π)
π±π¨πβ =
π«π¨π©(ππ β ππ)
π
πΉπ»(ππ¨π β ππ¨π)
EXAMPLE 6.2-2: 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 dry) is 1.01325 Γ 105 Pa (1.0 atm) and the temperature is 293 K (20Β°C). Water evaporates and diffuses through the air in the tube and the diffusion path (π§2 β π§1) is 0.1524 m long. The diagram is similar to Fig. 6.2-2a. Calculate the rate of evaporation at steady state. The diffusivity of water vapor at 293 K and 1 atm pressure is 0.250Γ10-4 m2/s. Assume that the system is isothermal.
Solution:
ππ΄1 = 2.341 Γ 103 Pa (Vapor pressure of water at 20Β°C from Appendix A.2)
ππ΄2 = 0 (Assuming dry air, i.e. no water vapor)
ππ΅1 = P β pAl = 1.01325 Γ 105 β 2.341 Γ 103 = 98984
ππ΅2 = P β pA2 = 1.01325 Γ 105 β 0 = 1.01325 Γ 105 Pa
ππ΅π =ππ΅2 β ππ΅1
ππππ΅2ππ΅1
=ππ΄1 β ππ΄2
ππππ΅2ππ΅1
= 100149.4 ππ
When,
ππ΅1 β ππ΅2; ππ΅π β ππ΅1 + ππ΅2
2= 100154.5 ππ
ππ΄ =π·π΄π΅
(π§2 β π§1)
π
π π
ππ΄1 β ππ΄2ππ΅π
=0.25 Γ 10β4
0.1524
1.01325 Γ 105
8314 Γ 293
(2.341 Γ 103 β 0)
100149.4= 1.595 Γ 10β7
ππ πππ
π2 β π
Question: Water at 20Β°C is flowing in a covered irrigation ditch below ground. There is a vent line 30 mm inside diameter and 1.0 m long to the outside atmosphere at 20Β°C. The percent relative humidity in Riyadh under present weather conditions is about 10%. As a result, the partial pressure of the water vapor in the outside air can be taken as 234 Pa. Determine the molar flux of water vapor in (ππ πππ π2 β π β )
(Data: Use the diffusivity data from Table 6.2-1. You may need to change its value to the required temperature if needed. Vapor pressure of water vapor at 20Β°C = 2340 Pa)
Solution:
π = 293 πΎ; π = 1.01325 Γ 105 ππ; π·π΄π΅= 2.6 Γ 10β5 π2 π β @298 πΎ;
π·π΄π΅2π·π΄π΅1
= (π2π1
)1.75
; π·π΄π΅2 = 2.6 Γ 10β5 (
293
298)
1.75
= π. ππ Γ ππβπ ππ πβ
ππ΄1 = 2340 Pa (Vapor pressure of water vapor at 20Β°C) ππ΄2 = 234 (10% Relative Humidity) ππ΅1 = P β ππ΄1 = 101.325 Γ 10
3 β 2.34 Γ 103 = 98,985 Pa ππ΅2 = P β ππ΄2 = 101325 β 234 = 101,091 Pa
ππ΅π =ππ΅2 β ππ΅1
ππππ΅2ππ΅1
=ππ΄1 β ππ΄2
ππππ΅2ππ΅1
= 100,034 ππ
ππ΄ =π·π΄π΅
(π§2 β π§1)
π
π π
ππ΄1 β ππ΄2ππ΅π
=2.52 Γ 10β5
1.0
1.01325 Γ 105
8314 Γ 293
(2340 β 234)
100,034
= π. ππ Γ ππβπππ πππ
ππ β π
Question 1 (33 pts): Water in the bottom of a narrow metal tube is held at a temperature of 303 K. The total pressure of air (assumed dry) is 1.01325 x 105 Pa (1.0 atm) and the temperature is 303 K. Water evaporates and diffuses through the air in the tube and the diffusion path (z2 - z1) is 0.2 m long. The tube diameter is 10 mm. The diagram is similar to Fig. 6.2-2a. The vapor pressure of water vapors at 303 K is 4242 Pa. The experimental value of the diffusion coefficient at 298 K is 2.6Γ10-5 m2/s.
I. Determine the molecular diffusion coefficient at 303 K.
π·π΄π΅2π·π΄π΅1
= (π2π1
)1.75
; π·π΄π΅2 = 2.6 Γ 10β5 (
303
298)
1.75
= π. ππ Γ ππβπ ππ πβ
7
II. Determine pBM in Pa.
ππ΅π =ππ΅2 β ππ΅1
ππππ΅2ππ΅1
=ππ΄1 β ππ΄2
ππππ΅2ππ΅1
=4242 β 0
ππ101325 β 0
101325 β 4242
= ππ, ππππ·π
5
III. Calculate the rate of evaporation (NA) at steady state in ππ πππ π2 β π β and ππ π2 β π β
ππ΄ =π·π΄π΅
(π§2 β π§1)
π
π π
ππ΄1 β ππ΄2ππ΅π
=2.68 Γ 10β5
0.2
1.01325 Γ 105
8314 Γ 303
(4242 β 0)
99,189= π. π Γ ππβπ
ππ πππ
ππ β π
= ππ. π Γ ππβπππ
ππ β π
13
IV. Calculate the steady state rate of evaporation (NA) if the air is humid (not dry) and the percentage relative humidity, i.e. 100(ππ΄ ππ΄
πβ ) = 30%, where ππ΄ is the partial pressure of the water vapor in the air and ππ΄
π is the vapor pressure.
ππ΄2 =30
100ππ΄
π = 0.3 Γ 4242 = 1273ππ
ππ΅π =4242 β 1273
ππ101325 β 1273101325 β 4242
= 98,560 ππ
ππ΄ =π·π΄π΅
(π§2 β π§1)
π
π π
ππ΄1 β ππ΄2ππ΅π
=2.68 Γ 10β5
0.2
1.01325 Γ 105
8314 Γ 303
(4242 β 1273)
98,560= π. ππ Γ ππβπ
ππ πππ
ππ β π
Note: There is almost 30% decrease in the flux due to a 30% decrease in driving force since the change in the ππ΅π is negligible.
8
Interface Fall for A Diffusing through Stagnant Film of B
The initial and final heights are
π§0, π§πΉ respectively.
If the level drops dz m in dt s, and
the total rate of diffusion due to
evaporation (in kg mol per s) will be
ππ΄ = ππ΄ Γ π΄πππ
ππ΄ Γ π΄πππ
=ππ΄ππ΄
ππ§
ππ‘Γ π΄πππ
π·π΄π΅π§
π
π π
ππ΄1 β ππ΄2ππ΅π
=ππ΄ππ΄
ππ§
ππ‘
π·π΄π΅π
π π
ππ΄1 β ππ΄2ππ΅π
β« ππ‘ =π‘πΉ
0
ππ΄ππ΄
β« π§ππ§π§πΉ
π§0
π‘πΉ =ππ΄ππ΄
(π§πΉ2βπ§0
2)
2π·π΄π΅(
π
π π
ππ΄1 β ππ΄2ππ΅π
)βπ
The above equation can be used to experimentally determine the molecular diffusivity
π·π΄π΅.
(π§πΉ2βπ§0
2) = (π
π π
ππ΄1 β ππ΄2ππ΅π
) (ππ΄ππ΄
) (2π·π΄π΅)π‘πΉ
Diffusion through a Varying Cross-Sectional Area
Case (A): Diffusion from a sphere:
Example:
Evaporation of a drop of liquid
Evaporation of a ball of
naphthalene
Diffusion of nutrients from a
spherical micro-organism
Consider a sphere of A of radius π1 in an
infinite medium of gas B. Component A at partial pressure ππ΄1 at the surface is
diffusing in the surrounding stagnant medium (B), where ππ΄2 = 0 at large distance away.
If the area changes, the flux (ππ΄ππ πππ
π2π ) will also change, but (ππ΄
ππ πππ
π ) will remain
constant. For spherical geometry,
ππ΄ =ππ΄
4ππ2== βπ·π΄π΅
π
π π
1
(π β ππ΄)
πππ΄ππ
ππ΄4π
β«ππ
π2
π2
π1
= βπ·π΄π΅π
π πβ«
1
(π β ππ΄)πππ΄
ππ΄2
ππ΄1
ππ΄4π
(1
π1β
1
π2) = π·π΄π΅
π
π πln
π β ππ΄2π β ππ΄1
For π2 β« π1, 1 π2 β 0β , gives
ππ΄4ππ1
= π·π΄π΅π
π πln
π β ππ΄2π β ππ΄1
ππ΄ =ππ΄
4ππ12 =
π·π΄π΅π1
π
π π[ln
π β ππ΄2π β ππ΄1
] =π·π΄π΅π1
π
π π[ππ΄1 β ππ΄2
ππ΅π]
For ππ΄1 βͺ π β ππ΅π β π . Therefore,
ππ΄ =π·π΄π΅π1
1
π π(ππ΄1 β ππ΄2)
Since π = π π πβ for liquids
ππ΄ =π·π΄π΅π1
(ππ΄1 β ππ΄2)
For a tube of constant cross-sectional area, the total time of evaporation for the change
in level from π§0 to π§πΉ:
π‘πΉ =ππ΄ππ΄
(π§πΉ2βπ§0
2)
2π·π΄π΅(
π
π π
ππ΄1 β ππ΄2ππ΅π
)βπ
For a sphere, for the complete evaporation of initial radius π1:
π‘πΉ =ππ΄ππ΄
(π12)
2π·π΄π΅(
π
π π
ππ΄1 β ππ΄2ππ΅π
)βπ
Time to Completely Evaporate a Sphere
A drop of liquid toluene is kept at a uniform temperature of 25.9 Β°C and is suspended
in air by a fine wire. The initial radius π0 = 2ππ. The vapor pressure of toluence at
25.9 Β°C is ππ΄1 = 3.84 πππ and the density of liquid toluene is 866 kg/m3. Calculate the
time in seconds for complete evaporation, i.e. ππΉ = 0 ππ.
π‘πΉ =ππ΄ππ΄
(π02 β ππΉ
2)
2π·π΄π΅(
π
π π
ππ΄1 β ππ΄2ππ΅π
)βπ
=ππ΄ππ΄
(π02 β ππΉ
2)
2π·π΄π΅
π π
π
ππ΅πππ΄1 β ππ΄2
Question 2 (33 pts):
Water drop (spherical) is suspended in still air (assumed dry) by a fine wire
at 303K at 1.01325 x 105 Pa (1.0 atm). Its initial radius was r0 = 4 mm. The
vapor pressure of water at 303 K is ππ΄0 = 4242 ππ and the density of water
is 995.71 kg/m3. Note that
Conditions in this problem are same as in Question 1
π΄πππ, π΄ = 4ππ2; ππππ’ππ, π = (4 3β )ππ3; πππ π = ππ
The time of evaporation can be computed using,
π‘πΉ =ππ΄ππ΄
(π02 β ππΉ
2)
2π·π΄π΅
π π
π
ππ΅πππ΄1 β ππ΄2
I. Calculate the time in seconds for its complete evaporation (rF = 0 mm).
π‘πΉ =ππ΄ππ΄
(π02 β ππΉ
2)
2π·π΄π΅
π π
π
ππ΅πππ΄1 β ππ΄2
=995.71
18
(0.0042 β 0)
2 β 2.68 Γ 10β58314 Γ 303
101325
99,189
4242 β 0= ππππ π
12
II. Calculate the time in second required for the evaporation of half of the total initial mass
of the water drop
πππ π Fπππ π i
=ππππ’ππFππππ’ππi
=πF
3
πi3 = 0.5; πF
3 = 0.5πi3; ππΉ = 3.175 ππ
π‘πΉ =ππ΄ππ΄
(π02 β ππΉ
2)
2π·π΄π΅
π π
π
ππ΅πππ΄1 β ππ΄2
=995.71
18
(0.0042 β 0.0031752 )
2 β 2.68 Γ 10β58314 Γ 303
101325
99,189
4242 β 0
= ππππ π
10
III. How much time in seconds will be required for its complete evaporation (rF = 0 mm) if
initial radius was r0 = 2 mm. (Detailed calculations not required).
Since π‘πΉ β π02 decreasing the initial radius by (1/2) will cause a (1/4) in the time of evaporation,
i.e. (9599/4) = 2400 s.
5
IV. Calculate the time in seconds for its complete evaporation when P = 0.1 atm = 1.01325 x
104 Pa
π‘πΉ =ππ΄ππ΄
(π02 β ππΉ
2)
2π·π΄π΅
π π
π
ππ΅πππ΄1 β ππ΄2
=995.71
18
(0.0042 β 0 )
2 β 26.8 Γ 10β58314 Γ 303
10132.5
π, πππ
4242 β 0= ππππ
Note: (π·π΄π΅π)is independent of pressure, but (ππ΅π) will change.
6
Case (B): Diffusion through a tapered conduit:
The radius at any location is related to length as
π = (π2 β π1π§2 β π§1
) π§ + π1
and area
π΄ = ππ2 = π [(π2 β π1π§2 β π§1
) π§ + π1]2
Therefore,
ππ΄ =ππ΄
ππ2== βπ·π΄π΅
π
π π
1
(π β ππ΄)
πππ΄ππ
ππ΄ =ππ΄π
β«ππ§
[(π2 β π1π§2 β π§1
) π§ + π1]2
π2
π1
= βπ·π΄π΅π
π πβ«
1
(π β ππ΄)πππ΄
ππ΄2
ππ΄1
Diffusion Coefficients for Gases
Experimental determination of diffusion coefficients
Evaporation of pure liquid in a narrow tube with a gas passed over the top
Evaporation of a sphere for vapors of solids such as naphthalene, iodine, and
benzoic acid in a gas
Two-bulb method, where sampling for π2(π‘) with the help of following relations
can be used to obtain diffusion coefficient
πππ£ β π2
πππ£ β π20 = exp [β
π·π΄π΅(π1 + π2)
(πΏ π΄β )(π2π1)π‘]
where,
π1 + π2 = π10 + π2
0
πππ£ =π1π1
0 + π2π20
π1 + π2
Experimental Diffusion Coefficients of Gases at P = 1 atm (CJG: Table 6.2-1)
Prediction of diffusion coefficients
For a binary pair of gases, the ChapmanβEnskog correlation can be used (Eq. 6.2-44 in
C. J. Geankoplis):
π·π΄π΅ = 1.8583 Γ 10β7
ππ΄π΅2 Ξ©π·,π΄π΅
(π3 2β
π) (
1
ππ΄+
1
ππ΅)
1 2β
where, A and B are two kinds of molecules present in the gaseous mixture, π·π΄π΅is the
diffusivity (m2/s), T is the absolute temperature (K), M is the molar mass (kg/ kg mol),
P is the pressure (atm), and π is the βaverage collision diameterβ, Ξ© is a temperature-
dependent collision integral. Prediction accuracy of the above equation is about 8% up
to about 1000 K.
Another simpler correlation that can be used is (Eq. 6.2-45 in C. J. Geankoplis):
π·π΄π΅ = 1.0 Γ 10β7
[(β π£π΄)1 3β + (β π£π΅)
1 3β ]2(
π3.5 2β
π) (
1
ππ΄+
1
ππ΅)
1 2β
where β π£π΄=sum of structural volume increments (Table 6.2-2, CJG).
It is clear from the above equation that temperature and pressure dependence of the
gas diffusivity is given by:
π·π΄π΅ β (π3.5 2β
π)
π·π΄π΅2π·π΄π΅1
= (π2π1
)1.75
π·π΄π΅2π·π΄π΅1
= (π1π2
)
Prediction accuracy of Fuller et al. Equation
Schmidt number for gases
πππ = π πβ
π·π΄π΅=
ππππππ‘π’π πππππ’π ππ£ππ‘π¦
πππ π πππππ’π ππ£ππ‘π¦
For gases, πππ = 0.5β 2.0
For liquid, , πππ = 100β 10,000
Recommended