Analogy of Mass, Heat and Momentum Transport...Momentum Heat Mass ๐๐ง๐ฅ=โ(๐ + ) ( ๐ฅ)...
Transcript of Analogy of Mass, Heat and Momentum Transport...Momentum Heat Mass ๐๐ง๐ฅ=โ(๐ + ) ( ๐ฅ)...
-
Analogy of Mass, Heat and Momentum Transport
General molecular transport equations
๐๐ง = โ๐ฟ๐ฮ
๐๐ง
Momentum Heat Mass
๐๐ง๐ฅ = โ (๐
๐)
๐(๐๐ฃ๐ฅ)
๐๐ง
๐๐ง๐ด
= โ๐ผ๐(๐๐ถ๐๐)
๐๐ง ๐ฝ๐ด๐ง
โ = โ๐ท๐ด๐ต๐(๐๐ด)
๐๐ง
Turbulent diffusion equation
Momentum Heat Mass
๐๐ง๐ฅ = โ (๐
๐+ ๐๐ก)
๐(๐๐ฃ๐ฅ)
๐๐ง
๐๐ง๐ด
= โ(๐ผ + ๐ผ๐ก)๐(๐๐ถ๐๐)
๐๐ง
๐ฝ๐ด๐งโ
= โ(๐ท๐ด๐ต + ๐๐)๐(๐๐ด)
๐๐ง
-
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 (
๐
๐) (
๐2
๐ )
Thermal diffusivity ๐ผ (๐2
๐ )
Molecular diffusivity ๐ท๐ด๐ต (๐2
๐ )
-
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 ร 10 โ6 โ 1 ร 10-5 m2 / s
Liquids: 10 โ6 โ 10-9 m2 / s
Solids: 5 ร 10 โ14 โ 1 ร 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 ๐๐ก (๐2
๐ )
Turbulent (eddy) thermal diffusivity ๐ผ๐ก (๐2
๐ )
Turbulent (eddy) mas diffusivity ๐๐ (๐2
๐ )
-
Fickโs Law for Molecular Diffusion
๐ฝ๐ด๐งโ = โ๐๐ท๐ด๐ต
๐(๐ฅ๐ด)
๐๐ง
where c is the total concentration of A and B in (kg mol A + B)/m3, and ๐ฅ๐ด is the mole fraction of A in the mixture A and B.
For constant concentration (c),
๐๐ด = ๐๐ฅ๐ด
๐๐๐ด = ๐(๐๐ฅ๐ด) = ๐๐(๐ฅ๐ด)
๐ฝ๐ด๐งโ = โ๐ท๐ด๐ต
๐(๐๐ด)
๐๐ง
For constant molar flux, the above equation can be integrated as follows to
give,
๐ฝ๐ด๐งโ โซ ๐๐ง
๐ง2
๐ง1
= โ๐ท๐ด๐ต โซ ๐(๐๐ด)๐๐ด2
๐๐ด1
๐ฝ๐ด๐งโ = โ
๐ท๐ด๐ต(๐๐ด1 โ ๐๐ด2)
(๐ง1 โ ๐ง2)
Since the concentration is related to partial pressure,
๐๐ด =๐๐ด๐ ๐
๐ฝ๐ด๐งโ = โ
๐ท๐ด๐ต(๐๐ด1 โ ๐๐ด2)
๐ ๐(๐ง1 โ ๐ง2)
๐ฝ๐ด๐งโ =
๐ท๐ด๐ต(๐ง2 โ ๐ง1)
1
๐ ๐(๐๐ด1 โ ๐๐ด2)
-
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 ๐ฅ๐ด1 , the mole fraction of A in tank 1, is larger than ๐ฅ๐ด2 , 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.
๐ฝ๐ด๐งโ = โ๐ฝ๐ต๐ง
โ
[โ๐ท๐ด๐ต๐(๐๐ด)
๐๐ง] = โ [โ๐ท๐ต๐ด
๐(๐๐ต)
๐๐ง]
Since, ๐ = ๐๐ด + ๐๐ต = ๐๐๐๐ ๐ก๐๐๐ก, therefore,
๐ = ๐๐ด + ๐๐ต = ๐๐๐๐ ๐ก๐๐๐ก.
Differentiating
0 = ๐๐๐ด + ๐๐๐ต
Substituting ๐๐๐ต = โ๐๐๐ด gives,
[โ๐ท๐ด๐ต๐(๐๐ด)
๐๐ง] = โ [โ๐ท๐ต๐ด
๐(๐๐ต)
๐๐ง] = โ [โ๐ท๐ต๐ด (โ
๐(๐๐ด)
๐๐ง)] = [โ๐ท๐ต๐ด
๐(๐๐ด)
๐๐ง]
[โ๐ท๐ด๐ต๐(๐๐ด)
๐๐ง] = [โ๐ท๐ต๐ด
๐(๐๐ด)
๐๐ง]
๐ท๐ด๐ต = ๐ท๐ต๐ด
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 = ๐ฃ๐ด๐
Therefore, molar diffusion flux,
๐ฝ๐ด๐งโ
๐๐ ๐๐๐ ๐ด
๐ โ ๐2= ๐๐ด๐ฃ๐ด๐
๐๐ ๐๐๐ ๐ด
๐3๐
๐
The velocity of A relative to the stationary point is the sum of the diffusion velocity
(๐ฃ๐ด๐) and the average velocity (๐ฃ๐ , molar average velocity of the whole fluid relative to a stationary point). Mathematically,
๐ฃ๐ด = ๐ฃ๐ด๐ + ๐ฃ๐
๐๐ด๐ฃ๐ด = ๐๐ด๐ฃ๐ด๐ + ๐๐ด๐ฃ๐
Total convective flux of the whole stream relative to the stationary point:
๐๐ฃ๐ = ๐ = ๐๐ด + ๐๐ต; ๐ฃ๐ =๐๐ด + ๐๐ต
๐
Therefore,
๐๐ด = โ๐๐ท๐ด๐ต๐(๐ฅ๐ด)
๐๐ง+
๐๐ด๐
(๐๐ด + ๐๐ต)
๐๐ต = โ๐๐ท๐ต๐ด๐(๐ฅ๐ต)
๐๐ง+
๐๐ต๐
(๐๐ด + ๐๐ต)
Total flux of A relative to stationary point
๐๐ด๐ฃ๐ด(= ๐๐ด)
Diffusion flux relative to moving fluid
๐๐ด๐ฃ๐ด๐(=, ๐ฝ๐ด๐งโ )
Convective flux of A relative to stationary
point ๐๐ด๐ฃ๐ = +
Total flux of
A, ๐๐ด Diffusion flux, ๐ฝ๐ด๐ง
โ
Convective flux of A,
๐๐ด (๐๐ด + ๐๐ต
๐)
= +
-
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 (๐๐ต = 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 (๐๐ต = 0).
Therefore,
๐๐ด = โ๐๐ท๐ด๐ต๐(๐ฅ๐ด)
๐๐ง+
๐๐ด๐
(๐๐ด + 0)
Since,
๐ =๐
๐ ๐; ๐๐ด = ๐ฅ๐ด๐;
๐๐ด๐
=๐๐ด๐
Therefore, for constant pressure
๐๐ด = โ๐ท๐ด๐ต๐ ๐
๐(๐๐ด)
๐๐ง+
๐๐ด๐
๐๐ด
๐๐ด (1 โ๐๐ด๐
) = โ๐ท๐ด๐ต๐ ๐
๐(๐๐ด)
๐๐ง
-
๐ต๐จ = โ๐ซ๐จ๐ฉ๐ท
๐น๐ป
๐
(๐ท โ ๐๐จ)
๐ (๐๐จ)
๐ ๐
๐๐ด โซ ๐๐ง๐2
๐1
= โ๐ท๐ด๐ต๐
๐ ๐โซ
๐๐๐ด(๐ โ ๐๐ด)
๐๐ด2
๐๐ด1
๐๐ด =๐ท๐ด๐ต
(๐ง2 โ ๐ง1)
๐
๐ ๐ln
๐ โ ๐๐ด2๐ โ ๐๐ด1
Since, the total pressure is the sum of partial pressures, i.e., ๐ = ๐๐ด1 + ๐๐ต1 = ๐๐ด2 +๐๐ต2
Therefore,
๐๐ด =๐ท๐ด๐ต
(๐ง2 โ ๐ง1)
๐
๐ ๐๐๐
๐๐ต2๐๐ต1
But,
๐๐ต๐ =๐๐ต2 โ ๐๐ต1
๐๐๐๐ต2๐๐ต1
=๐๐ด1 โ ๐๐ด2
๐๐๐๐ต2๐๐ต1
๐๐๐๐ต2๐๐ต1
=๐๐ด1 โ ๐๐ด2
๐๐ต๐
Therefore,
๐ต๐จ =๐ซ๐จ๐ฉ
(๐๐ โ ๐๐)
๐ท
๐น๐ป
๐๐จ๐ โ ๐๐จ๐๐๐ฉ๐ด
-
๐ต๐จ =๐ซ๐จ๐ฉ
(๐๐ โ ๐๐)
๐
๐น๐ป
๐ท
๐๐ฉ๐ด(๐๐จ๐ โ ๐๐จ๐)
๐ฑ๐จ๐โ =
๐ซ๐จ๐ฉ(๐๐ โ ๐๐)
๐
๐น๐ป(๐๐จ๐ โ ๐๐จ๐)
-
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 ร 105 Pa (1.0 atm) and the temperature is 293 K (20ยฐC). Water evaporates and diffuses through the air in the tube and the diffusion path (๐ง2 โ ๐ง1) 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 0.250ร10-4 m2/s. Assume that the system is isothermal.
Solution:
๐๐ด1 = 2.341 ร 103 Pa (Vapor pressure of water at 20ยฐC from Appendix A.2)
๐๐ด2 = 0 (Assuming dry air, i.e. no water vapor)
๐๐ต1 = P โ pAl = 1.01325 ร 105 โ 2.341 ร 103 = 98984
๐๐ต2 = P โ pA2 = 1.01325 ร 105 โ 0 = 1.01325 ร 105 Pa
๐๐ต๐ =๐๐ต2 โ ๐๐ต1
๐๐๐๐ต2๐๐ต1
=๐๐ด1 โ ๐๐ด2
๐๐๐๐ต2๐๐ต1
= 100149.4 ๐๐
When,
๐๐ต1 โ ๐๐ต2; ๐๐ต๐ โ ๐๐ต1 + ๐๐ต2
2= 100154.5 ๐๐
๐๐ด =๐ท๐ด๐ต
(๐ง2 โ ๐ง1)
๐
๐ ๐
๐๐ด1 โ ๐๐ด2๐๐ต๐
=0.25 ร 10โ4
0.1524
1.01325 ร 105
8314 ร 293
(2.341 ร 103 โ 0)
100149.4= 1.595 ร 10โ7
๐๐ ๐๐๐
๐2 โ ๐
-
Question: Water at 20ยฐC 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 20ยฐC. 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 (๐๐ ๐๐๐ ๐2 โ ๐ โ )
(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 20ยฐC = 2340 Pa)
Solution:
๐ = 293 ๐พ; ๐ = 1.01325 ร 105 ๐๐; ๐ท๐ด๐ต= 2.6 ร 10โ5 ๐2 ๐ โ @298 ๐พ;
๐ท๐ด๐ต2๐ท๐ด๐ต1
= (๐2๐1
)1.75
; ๐ท๐ด๐ต2 = 2.6 ร 10โ5 (
293
298)
1.75
= ๐. ๐๐ ร ๐๐โ๐ ๐๐ ๐โ
๐๐ด1 = 2340 Pa (Vapor pressure of water vapor at 20ยฐC) ๐๐ด2 = 234 (10% Relative Humidity) ๐๐ต1 = P โ ๐๐ด1 = 101.325 ร 10
3 โ 2.34 ร 103 = 98,985 Pa ๐๐ต2 = P โ ๐๐ด2 = 101325 โ 234 = 101,091 Pa
๐๐ต๐ =๐๐ต2 โ ๐๐ต1
๐๐๐๐ต2๐๐ต1
=๐๐ด1 โ ๐๐ด2
๐๐๐๐ต2๐๐ต1
= 100,034 ๐๐
๐๐ด =๐ท๐ด๐ต
(๐ง2 โ ๐ง1)
๐
๐ ๐
๐๐ด1 โ ๐๐ด2๐๐ต๐
=2.52 ร 10โ5
1.0
1.01325 ร 105
8314 ร 293
(2340 โ 234)
100,034
= ๐. ๐๐ ร ๐๐โ๐๐๐ ๐๐๐
๐๐ โ ๐
-
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 2.6ร10-5 m2/s.
I. Determine the molecular diffusion coefficient at 303 K.
๐ท๐ด๐ต2๐ท๐ด๐ต1
= (๐2๐1
)1.75
; ๐ท๐ด๐ต2 = 2.6 ร 10โ5 (
303
298)
1.75
= ๐. ๐๐ ร ๐๐โ๐ ๐๐ ๐โ
7
II. Determine pBM in Pa.
๐๐ต๐ =๐๐ต2 โ ๐๐ต1
๐๐๐๐ต2๐๐ต1
=๐๐ด1 โ ๐๐ด2
๐๐๐๐ต2๐๐ต1
=4242 โ 0
๐๐101325 โ 0
101325 โ 4242
= ๐๐, ๐๐๐๐ท๐
5
III. Calculate the rate of evaporation (NA) at steady state in ๐๐ ๐๐๐ ๐2 โ ๐ โ and ๐๐ ๐2 โ ๐ โ
๐๐ด =๐ท๐ด๐ต
(๐ง2 โ ๐ง1)
๐
๐ ๐
๐๐ด1 โ ๐๐ด2๐๐ต๐
=2.68 ร 10โ5
0.2
1.01325 ร 105
8314 ร 303
(4242 โ 0)
99,189= ๐. ๐ ร ๐๐โ๐
๐๐ ๐๐๐
๐๐ โ ๐
= ๐๐. ๐ ร ๐๐โ๐๐๐
๐๐ โ ๐
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(๐๐ด ๐๐ด
๐โ ) = 30%, where ๐๐ด is the partial pressure of the water vapor in the air and ๐๐ด
๐ is the vapor pressure.
๐๐ด2 =30
100๐๐ด
๐ = 0.3 ร 4242 = 1273๐๐
๐๐ต๐ =4242 โ 1273
๐๐101325 โ 1273101325 โ 4242
= 98,560 ๐๐
๐๐ด =๐ท๐ด๐ต
(๐ง2 โ ๐ง1)
๐
๐ ๐
๐๐ด1 โ ๐๐ด2๐๐ต๐
=2.68 ร 10โ5
0.2
1.01325 ร 105
8314 ร 303
(4242 โ 1273)
98,560= ๐. ๐๐ ร ๐๐โ๐
๐๐ ๐๐๐
๐๐ โ ๐
Note: There is almost 30% decrease in the flux due to a 30% decrease in driving force since the change in the ๐๐ต๐ is negligible.
8
-
Interface Fall for A Diffusing through Stagnant Film of B
The initial and final heights are
๐ง0, ๐ง๐น 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
๐๐ด = ๐๐ด ร ๐ด๐๐๐
๐๐ด ร ๐ด๐๐๐
=๐๐ด๐๐ด
๐๐ง
๐๐กร ๐ด๐๐๐
๐ท๐ด๐ต๐ง
๐
๐ ๐
๐๐ด1 โ ๐๐ด2๐๐ต๐
=๐๐ด๐๐ด
๐๐ง
๐๐ก
๐ท๐ด๐ต๐
๐ ๐
๐๐ด1 โ ๐๐ด2๐๐ต๐
โซ ๐๐ก =๐ก๐น
0
๐๐ด๐๐ด
โซ ๐ง๐๐ง๐ง๐น
๐ง0
๐ก๐น =๐๐ด๐๐ด
(๐ง๐น2โ๐ง0
2)
2๐ท๐ด๐ต(
๐
๐ ๐
๐๐ด1 โ ๐๐ด2๐๐ต๐
)โ๐
The above equation can be used to experimentally determine the molecular diffusivity
๐ท๐ด๐ต.
(๐ง๐น2โ๐ง0
2) = (๐
๐ ๐
๐๐ด1 โ ๐๐ด2๐๐ต๐
) (๐๐ด๐๐ด
) (2๐ท๐ด๐ต)๐ก๐น
-
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 ๐1 in an
infinite medium of gas B. Component A at partial pressure ๐๐ด1 at the surface is
diffusing in the surrounding stagnant medium (B), where ๐๐ด2 = 0 at large distance away.
If the area changes, the flux (๐๐ด๐๐ ๐๐๐
๐2๐ ) will also change, but (๐๐ด
๐๐ ๐๐๐
๐ ) will remain
constant. For spherical geometry,
๐๐ด =๐๐ด
4๐๐2== โ๐ท๐ด๐ต
๐
๐ ๐
1
(๐ โ ๐๐ด)
๐๐๐ด๐๐
๐๐ด4๐
โซ๐๐
๐2
๐2
๐1
= โ๐ท๐ด๐ต๐
๐ ๐โซ
1
(๐ โ ๐๐ด)๐๐๐ด
๐๐ด2
๐๐ด1
๐๐ด4๐
(1
๐1โ
1
๐2) = ๐ท๐ด๐ต
๐
๐ ๐ln
๐ โ ๐๐ด2๐ โ ๐๐ด1
For ๐2 โซ ๐1, 1 ๐2 โ 0โ , gives
๐๐ด4๐๐1
= ๐ท๐ด๐ต๐
๐ ๐ln
๐ โ ๐๐ด2๐ โ ๐๐ด1
๐๐ด =๐๐ด
4๐๐12 =
๐ท๐ด๐ต๐1
๐
๐ ๐[ln
๐ โ ๐๐ด2๐ โ ๐๐ด1
] =๐ท๐ด๐ต๐1
๐
๐ ๐[๐๐ด1 โ ๐๐ด2
๐๐ต๐]
-
For ๐๐ด1 โช ๐ โ ๐๐ต๐ โ ๐ . Therefore,
๐๐ด =๐ท๐ด๐ต๐1
1
๐ ๐(๐๐ด1 โ ๐๐ด2)
Since ๐ = ๐ ๐ ๐โ for liquids
๐๐ด =๐ท๐ด๐ต๐1
(๐๐ด1 โ ๐๐ด2)
For a tube of constant cross-sectional area, the total time of evaporation for the change
in level from ๐ง0 to ๐ง๐น:
๐ก๐น =๐๐ด๐๐ด
(๐ง๐น2โ๐ง0
2)
2๐ท๐ด๐ต(
๐
๐ ๐
๐๐ด1 โ ๐๐ด2๐๐ต๐
)โ๐
For a sphere, for the complete evaporation of initial radius ๐1:
๐ก๐น =๐๐ด๐๐ด
(๐12)
2๐ท๐ด๐ต(
๐
๐ ๐
๐๐ด1 โ ๐๐ด2๐๐ต๐
)โ๐
-
Time to Completely Evaporate a Sphere
A drop of liquid toluene is kept at a uniform temperature of 25.9 ยฐC and is suspended
in air by a fine wire. The initial radius ๐0 = 2๐๐. The vapor pressure of toluence at
25.9 ยฐC is ๐๐ด1 = 3.84 ๐๐๐ and the density of liquid toluene is 866 kg/m3. Calculate the
time in seconds for complete evaporation, i.e. ๐๐น = 0 ๐๐.
๐ก๐น =๐๐ด๐๐ด
(๐02 โ ๐๐น
2)
2๐ท๐ด๐ต(
๐
๐ ๐
๐๐ด1 โ ๐๐ด2๐๐ต๐
)โ๐
=๐๐ด๐๐ด
(๐02 โ ๐๐น
2)
2๐ท๐ด๐ต
๐ ๐
๐
๐๐ต๐๐๐ด1 โ ๐๐ด2
-
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 ๐๐ด0 = 4242 ๐๐ and the density of water
is 995.71 kg/m3. Note that
Conditions in this problem are same as in Question 1
๐ด๐๐๐, ๐ด = 4๐๐2; ๐๐๐๐ข๐๐, ๐ = (4 3โ )๐๐3; ๐๐๐ ๐ = ๐๐
The time of evaporation can be computed using,
๐ก๐น =๐๐ด๐๐ด
(๐02 โ ๐๐น
2)
2๐ท๐ด๐ต
๐ ๐
๐
๐๐ต๐๐๐ด1 โ ๐๐ด2
I. Calculate the time in seconds for its complete evaporation (rF = 0 mm).
๐ก๐น =๐๐ด๐๐ด
(๐02 โ ๐๐น
2)
2๐ท๐ด๐ต
๐ ๐
๐
๐๐ต๐๐๐ด1 โ ๐๐ด2
=995.71
18
(0.0042 โ 0)
2 โ 2.68 ร 10โ58314 ร 303
101325
99,189
4242 โ 0= ๐๐๐๐ ๐
12
II. Calculate the time in second required for the evaporation of half of the total initial mass
of the water drop
๐๐๐ ๐ F๐๐๐ ๐ i
=๐๐๐๐ข๐๐F๐๐๐๐ข๐๐i
=๐F
3
๐i3 = 0.5; ๐F
3 = 0.5๐i3; ๐๐น = 3.175 ๐๐
๐ก๐น =๐๐ด๐๐ด
(๐02 โ ๐๐น
2)
2๐ท๐ด๐ต
๐ ๐
๐
๐๐ต๐๐๐ด1 โ ๐๐ด2
=995.71
18
(0.0042 โ 0.0031752 )
2 โ 2.68 ร 10โ58314 ร 303
101325
99,189
4242 โ 0
= ๐๐๐๐ ๐
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 ๐ก๐น โ ๐02 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
๐ก๐น =๐๐ด๐๐ด
(๐02 โ ๐๐น
2)
2๐ท๐ด๐ต
๐ ๐
๐
๐๐ต๐๐๐ด1 โ ๐๐ด2
=995.71
18
(0.0042 โ 0 )
2 โ 26.8 ร 10โ58314 ร 303
10132.5
๐, ๐๐๐
4242 โ 0= ๐๐๐๐
Note: (๐ท๐ด๐ต๐)is independent of pressure, but (๐๐ต๐) will change.
6
-
Case (B): Diffusion through a tapered conduit:
The radius at any location is related to length as
๐ = (๐2 โ ๐1๐ง2 โ ๐ง1
) ๐ง + ๐1
and area
๐ด = ๐๐2 = ๐ [(๐2 โ ๐1๐ง2 โ ๐ง1
) ๐ง + ๐1]2
Therefore,
๐๐ด =๐๐ด
๐๐2== โ๐ท๐ด๐ต
๐
๐ ๐
1
(๐ โ ๐๐ด)
๐๐๐ด๐๐
๐๐ด =๐๐ด๐
โซ๐๐ง
[(๐2 โ ๐1๐ง2 โ ๐ง1
) ๐ง + ๐1]2
๐2
๐1
= โ๐ท๐ด๐ต๐
๐ ๐โซ
1
(๐ โ ๐๐ด)๐๐๐ด
๐๐ด2
๐๐ด1
-
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 ๐2(๐ก) with the help of following relations
can be used to obtain diffusion coefficient
๐๐๐ฃ โ ๐2
๐๐๐ฃ โ ๐20 = exp [โ
๐ท๐ด๐ต(๐1 + ๐2)
(๐ฟ ๐ดโ )(๐2๐1)๐ก]
where,
๐1 + ๐2 = ๐10 + ๐2
0
๐๐๐ฃ =๐1๐1
0 + ๐2๐20
๐1 + ๐2
-
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 ChapmanโEnskog correlation can be used (Eq. 6.2-44 in
C. J. Geankoplis):
๐ท๐ด๐ต = 1.8583 ร 10โ7
๐๐ด๐ต2 ฮฉ๐ท,๐ด๐ต
(๐3 2โ
๐) (
1
๐๐ด+
1
๐๐ต)
1 2โ
where, A and B are two kinds of molecules present in the gaseous mixture, ๐ท๐ด๐ตis the
diffusivity (m2/s), T is the absolute temperature (K), M is the molar mass (kg/ kg mol),
P is the pressure (atm), and ๐ is the โaverage collision diameterโ, ฮฉ 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):
๐ท๐ด๐ต = 1.0 ร 10โ7
[(โ ๐ฃ๐ด)1 3โ + (โ ๐ฃ๐ต)
1 3โ ]2(
๐3.5 2โ
๐) (
1
๐๐ด+
1
๐๐ต)
1 2โ
where โ ๐ฃ๐ด=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:
๐ท๐ด๐ต โ (๐3.5 2โ
๐)
๐ท๐ด๐ต2๐ท๐ด๐ต1
= (๐2๐1
)1.75
๐ท๐ด๐ต2๐ท๐ด๐ต1
= (๐1๐2
)
-
Prediction accuracy of Fuller et al. Equation
-
Schmidt number for gases
๐๐๐ = ๐ ๐โ
๐ท๐ด๐ต=
๐๐๐๐๐๐ก๐ข๐ ๐๐๐๐๐ข๐ ๐๐ฃ๐๐ก๐ฆ
๐๐๐ ๐ ๐๐๐๐๐ข๐ ๐๐ฃ๐๐ก๐ฆ
For gases, ๐๐๐ = 0.5โ 2.0
For liquid, , ๐๐๐ = 100โ 10,000