Analogy of Mass, Heat and Momentum Transport...Momentum Heat Mass 𝜏𝑧𝑥=−(𝜇 + ) ( 𝑥) 𝑧 𝐴 =−(𝛼+𝛼 ) ( 𝑃𝑇) 𝑧 ⋆ =−( + 𝑀) ( ) Diffusion Diffusion

Analogy of Mass, Heat and Momentum Transport

General molecular transport equations

𝜓𝑧 = −𝛿𝑑Γ

𝑑𝑧

Momentum Heat Mass

𝜏𝑧𝑥 = − (𝜇

𝜌)

𝑑(𝜌𝑣𝑥)

𝑑𝑧

𝑞𝑧𝐴

= −𝛼𝑑(𝜌𝐶𝑃𝑇)

𝑑𝑧 𝐽𝐴𝑧

⋆ = −𝐷𝐴𝐵𝑑(𝑐𝐴)

𝑑𝑧

Turbulent diffusion equation

Momentum Heat Mass

𝜏𝑧𝑥 = − (𝜇

𝜌+ 𝜀𝑡)

𝑑(𝜌𝑣𝑥)

𝑑𝑧

𝑞𝑧𝐴

= −(𝛼 + 𝛼𝑡)𝑑(𝜌𝐶𝑃𝑇)

𝑑𝑧

𝐽𝐴𝑧⋆

= −(𝐷𝐴𝐵 + 𝜀𝑀)𝑑(𝑐𝐴)

𝑑𝑧

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 (

𝜇

𝜌) (

𝑚2

𝑠)

Thermal diffusivity 𝛼 (𝑚2

𝑠)

Molecular diffusivity 𝐷𝐴𝐵 (𝑚2

𝑠)

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 × 10 –6 – 1 × 10-5 m2 / s

Liquids: 10 –6 – 10-9 m2 / s

Solids: 5 × 10 –14 – 1 × 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 𝜀𝑡 (𝑚2

𝑠)

Turbulent (eddy) thermal diffusivity 𝛼𝑡 (𝑚2

𝑠)

Turbulent (eddy) mas diffusivity 𝜀𝑀 (𝑚2

𝑠)

Fick’s Law for Molecular Diffusion

𝐽𝐴𝑧⋆ = −𝑐𝐷𝐴𝐵

𝑑(𝑥𝐴)

𝑑𝑧

where c is the total concentration of A and B in (kg mol A + B)/m3, and 𝑥𝐴 is the mole fraction of A in the mixture A and B.

For constant concentration (c),

𝑐𝐴 = 𝑐𝑥𝐴

𝑑𝑐𝐴 = 𝑑(𝑐𝑥𝐴) = 𝑐𝑑(𝑥𝐴)

𝐽𝐴𝑧⋆ = −𝐷𝐴𝐵

𝑑(𝑐𝐴)

𝑑𝑧

For constant molar flux, the above equation can be integrated as follows to

give,

𝐽𝐴𝑧⋆ ∫ 𝑑𝑧

𝑧2

𝑧1

= −𝐷𝐴𝐵 ∫ 𝑑(𝑐𝐴)𝑐𝐴2

𝑐𝐴1

𝐽𝐴𝑧⋆ = −

𝐷𝐴𝐵(𝑐𝐴1 − 𝑐𝐴2)

(𝑧1 − 𝑧2)

Since the concentration is related to partial pressure,

𝑐𝐴 =𝑝𝐴𝑅𝑇

𝐽𝐴𝑧⋆ = −

𝐷𝐴𝐵(𝑝𝐴1 − 𝑝𝐴2)

𝑅𝑇(𝑧1 − 𝑧2)

𝐽𝐴𝑧⋆ =

𝐷𝐴𝐵(𝑧2 − 𝑧1)

1

𝑅𝑇(𝑝𝐴1 − 𝑝𝐴2)

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 𝑥𝐴1 , the mole fraction of A in tank 1, is larger than 𝑥𝐴2 , 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.

𝐽𝐴𝑧⋆ = −𝐽𝐵𝑧

⋆

[−𝐷𝐴𝐵𝑑(𝑐𝐴)

𝑑𝑧] = − [−𝐷𝐵𝐴

𝑑(𝑐𝐵)

𝑑𝑧]

Since, 𝑃 = 𝑝𝐴 + 𝑝𝐵 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, therefore,

𝑐 = 𝑐𝐴 + 𝑐𝐵 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡.

Differentiating

0 = 𝑑𝑐𝐴 + 𝑑𝑐𝐵

Substituting 𝑑𝑐𝐵 = −𝑑𝑐𝐴 gives,


𝑑𝑧] = − [−𝐷𝐵𝐴

𝑑(𝑐𝐵)

𝑑𝑧] = − [−𝐷𝐵𝐴 (−

𝑑(𝑐𝐴)

𝑑𝑧)] = [−𝐷𝐵𝐴

𝑑(𝑐𝐴)

𝑑𝑧]


𝑑𝑧] = [−𝐷𝐵𝐴

𝑑(𝑐𝐴)

𝑑𝑧]

𝐷𝐴𝐵 = 𝐷𝐵𝐴

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 = 𝑣𝐴𝑑

Therefore, molar diffusion flux,

𝐽𝐴𝑧⋆

𝑘𝑔 𝑚𝑜𝑙 𝐴

𝑠 ∙ 𝑚2= 𝑐𝐴𝑣𝐴𝑑

𝑘𝑔 𝑚𝑜𝑙 𝐴

𝑚3𝑚

𝑠

The velocity of A relative to the stationary point is the sum of the diffusion velocity

(𝑣𝐴𝑑) and the average velocity (𝑣𝑀 , molar average velocity of the whole fluid relative to a stationary point). Mathematically,

𝑣𝐴 = 𝑣𝐴𝑑 + 𝑣𝑀

𝑐𝐴𝑣𝐴 = 𝑐𝐴𝑣𝐴𝑑 + 𝑐𝐴𝑣𝑀

Total convective flux of the whole stream relative to the stationary point:

𝑐𝑣𝑀 = 𝑁 = 𝑁𝐴 + 𝑁𝐵; 𝑣𝑀 =𝑁𝐴 + 𝑁𝐵

𝑐

Therefore,

𝑁𝐴 = −𝑐𝐷𝐴𝐵𝑑(𝑥𝐴)

𝑑𝑧+

𝑐𝐴𝑐

(𝑁𝐴 + 𝑁𝐵)

𝑁𝐵 = −𝑐𝐷𝐵𝐴𝑑(𝑥𝐵)

𝑑𝑧+

𝑐𝐵𝑐

(𝑁𝐴 + 𝑁𝐵)

Total flux of A relative to stationary point

𝑐𝐴𝑣𝐴(= 𝑁𝐴)

Diffusion flux relative to moving fluid

𝑐𝐴𝑣𝐴𝑑(=, 𝐽𝐴𝑧⋆ )

Convective flux of A relative to stationary

point 𝑐𝐴𝑣𝑀 = +

Total flux of

A, 𝑁𝐴 Diffusion flux, 𝐽𝐴𝑧

⋆

Convective flux of A,

𝑐𝐴 (𝑁𝐴 + 𝑁𝐵

𝑐)

= +

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 (𝑁𝐵 = 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 (𝑁𝐵 = 0).

Therefore,

𝑁𝐴 = −𝑐𝐷𝐴𝐵𝑑(𝑥𝐴)

𝑑𝑧+

𝑐𝐴𝑐

(𝑁𝐴 + 0)

Since,

𝑐 =𝑃

𝑅𝑇; 𝑝𝐴 = 𝑥𝐴𝑃;

𝑐𝐴𝑐

=𝑝𝐴𝑃

Therefore, for constant pressure

𝑁𝐴 = −𝐷𝐴𝐵𝑅𝑇

𝑑(𝑝𝐴)

𝑑𝑧+

𝑝𝐴𝑃

𝑁𝐴

𝑁𝐴 (1 −𝑝𝐴𝑃

) = −𝐷𝐴𝐵𝑅𝑇

𝑑(𝑝𝐴)

𝑑𝑧

𝑵𝑨 = −𝑫𝑨𝑩𝑷

𝑹𝑻

𝟏

(𝑷 − 𝒑𝑨)

𝒅(𝒑𝑨)

𝒅𝒛

𝑁𝐴 ∫ 𝑑𝑧𝑍2

𝑍1

= −𝐷𝐴𝐵𝑃

𝑅𝑇∫

𝑑𝑝𝐴(𝑃 − 𝑝𝐴)

𝑝𝐴2

𝑝𝐴1

𝑁𝐴 =𝐷𝐴𝐵

(𝑧2 − 𝑧1)

𝑃

𝑅𝑇ln

𝑃 − 𝑝𝐴2𝑃 − 𝑝𝐴1

Since, the total pressure is the sum of partial pressures, i.e., 𝑃 = 𝑝𝐴1 + 𝑝𝐵1 = 𝑝𝐴2 +𝑝𝐵2

Therefore,


(𝑧2 − 𝑧1)

𝑃

𝑅𝑇𝑙𝑛

𝑝𝐵2𝑝𝐵1

But,

𝑝𝐵𝑀 =𝑝𝐵2 − 𝑝𝐵1

𝑙𝑛𝑝𝐵2𝑝𝐵1

=𝑝𝐴1 − 𝑝𝐴2




𝑝𝐵𝑀

Therefore,

𝑵𝑨 =𝑫𝑨𝑩

(𝒛𝟐 − 𝒛𝟏)

𝑷

𝑹𝑻

𝒑𝑨𝟏 − 𝒑𝑨𝟐𝒑𝑩𝑴

𝑵𝑨 =𝑫𝑨𝑩

(𝒛𝟐 − 𝒛𝟏)

𝟏

𝑹𝑻

𝑷

𝒑𝑩𝑴(𝒑𝑨𝟏 − 𝒑𝑨𝟐)

𝑱𝑨𝒛⋆ =

𝑫𝑨𝑩(𝒛𝟐 − 𝒛𝟏)

𝟏

𝑹𝑻(𝒑𝑨𝟏 − 𝒑𝑨𝟐)

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 × 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. 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 0.250×10-4 m2/s. Assume that the system is isothermal.

Solution:

𝑝𝐴1 = 2.341 × 103 Pa (Vapor pressure of water at 20°C from Appendix A.2)

𝑝𝐴2 = 0 (Assuming dry air, i.e. no water vapor)

𝑝𝐵1 = P – pAl = 1.01325 × 105 – 2.341 × 103 = 98984

𝑝𝐵2 = P – pA2 = 1.01325 × 105 – 0 = 1.01325 × 105 Pa





= 100149.4 𝑃𝑎

When,

𝑝𝐵1 ≈ 𝑝𝐵2; 𝑝𝐵𝑀 ≅𝑝𝐵1 + 𝑝𝐵2

2= 100154.5 𝑃𝑎


(𝑧2 − 𝑧1)

𝑃

𝑅𝑇

𝑝𝐴1 − 𝑝𝐴2𝑝𝐵𝑀

=0.25 × 10−4

0.1524

1.01325 × 105

8314 × 293

(2.341 × 103 − 0)

100149.4= 1.595 × 10−7

𝑘𝑔 𝑚𝑜𝑙

𝑚2 ∙ 𝑠

Question: Water at 20°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 20°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 (𝑘𝑔 𝑚𝑜𝑙 𝑚2 ∙ 𝑠⁄ )

(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 20°C = 2340 Pa)

Solution:

𝑇 = 293 𝐾; 𝑃 = 1.01325 × 105 𝑃𝑎; 𝐷𝐴𝐵= 2.6 × 10−5 𝑚2 𝑠⁄ @298 𝐾;

𝐷𝐴𝐵2𝐷𝐴𝐵1

= (𝑇2𝑇1

)1.75

; 𝐷𝐴𝐵2 = 2.6 × 10−5 (

293

298)

1.75

= 𝟐. 𝟓𝟐 × 𝟏𝟎−𝟓 𝒎𝟐 𝒔⁄

𝑝𝐴1 = 2340 Pa (Vapor pressure of water vapor at 20°C) 𝑝𝐴2 = 234 (10% Relative Humidity) 𝑝𝐵1 = P – 𝑝𝐴1 = 101.325 × 10

3 – 2.34 × 103 = 98,985 Pa 𝑝𝐵2 = P – 𝑝𝐴2 = 101325 – 234 = 101,091 Pa





= 100,034 𝑃𝑎


(𝑧2 − 𝑧1)

𝑃

𝑅𝑇


=2.52 × 10−5

1.0

1.01325 × 105

8314 × 293

(2340 − 234)

100,034

= 𝟐. 𝟐𝟏 × 𝟏𝟎−𝟖𝒌𝒈 𝒎𝒐𝒍

𝒎𝟐 ∙ 𝒔

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 2.6×10-5 m2/s.

I. Determine the molecular diffusion coefficient at 303 K.


= (𝑇2𝑇1

)1.75

; 𝐷𝐴𝐵2 = 2.6 × 10−5 (

303

298)

1.75

= 𝟐. 𝟔𝟖 × 𝟏𝟎−𝟓 𝒎𝟐 𝒔⁄

7

II. Determine pBM in Pa.





=4242 − 0

𝑙𝑛101325 − 0

101325 − 4242

= 𝟗𝟗, 𝟏𝟖𝟗𝑷𝒂

5

III. Calculate the rate of evaporation (NA) at steady state in 𝑘𝑔 𝑚𝑜𝑙 𝑚2 ∙ 𝑠⁄ and 𝑘𝑔 𝑚2 ∙ 𝑠⁄


(𝑧2 − 𝑧1)

𝑃

𝑅𝑇


=2.68 × 10−5

0.2

1.01325 × 105

8314 × 303

(4242 − 0)

99,189= 𝟐. 𝟑 × 𝟏𝟎−𝟕

𝒌𝒈 𝒎𝒐𝒍

𝒎𝟐 ∙ 𝒔

= 𝟒𝟏. 𝟒 × 𝟏𝟎−𝟕𝒌𝒈

𝒎𝟐 ∙ 𝒔

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(𝑝𝐴 𝑝𝐴

𝑜⁄ ) = 30%, where 𝑝𝐴 is the partial pressure of the water vapor in the air and 𝑝𝐴

𝑜 is the vapor pressure.

𝑝𝐴2 =30

100𝑝𝐴

𝑜 = 0.3 × 4242 = 1273𝑃𝑎

𝑝𝐵𝑀 =4242 − 1273

𝑙𝑛101325 − 1273101325 − 4242

= 98,560 𝑃𝑎


(𝑧2 − 𝑧1)

𝑃

𝑅𝑇


=2.68 × 10−5

0.2

1.01325 × 105

8314 × 303

(4242 − 1273)

98,560= 𝟏. 𝟔𝟐 × 𝟏𝟎−𝟕

𝒌𝒈 𝒎𝒐𝒍

𝒎𝟐 ∙ 𝒔

Note: There is almost 30% decrease in the flux due to a 30% decrease in driving force since the change in the 𝑝𝐵𝑀 is negligible.

8

Interface Fall for A Diffusing through Stagnant Film of B

The initial and final heights are

𝑧0, 𝑧𝐹 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

𝑁𝐴 = 𝑁𝐴 × 𝐴𝑟𝑒𝑎

𝑁𝐴 × 𝐴𝑟𝑒𝑎

=𝜌𝐴𝑀𝐴

𝑑𝑧

𝑑𝑡× 𝐴𝑟𝑒𝑎

𝐷𝐴𝐵𝑧

𝑃

𝑅𝑇


=𝜌𝐴𝑀𝐴

𝑑𝑧

𝑑𝑡

𝐷𝐴𝐵𝑃

𝑅𝑇


∫ 𝑑𝑡 =𝑡𝐹

0

𝜌𝐴𝑀𝐴

∫ 𝑧𝑑𝑧𝑧𝐹

𝑧0

𝑡𝐹 =𝜌𝐴𝑀𝐴

(𝑧𝐹2−𝑧0

2)

2𝐷𝐴𝐵(

𝑃

𝑅𝑇


)−𝟏

The above equation can be used to experimentally determine the molecular diffusivity

𝐷𝐴𝐵.

(𝑧𝐹2−𝑧0

2) = (𝑃

𝑅𝑇


) (𝑀𝐴𝜌𝐴

) (2𝐷𝐴𝐵)𝑡𝐹

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 𝑟1 in an

infinite medium of gas B. Component A at partial pressure 𝑝𝐴1 at the surface is

diffusing in the surrounding stagnant medium (B), where 𝑝𝐴2 = 0 at large distance away.

If the area changes, the flux (𝑁𝐴𝑘𝑔 𝑚𝑜𝑙

𝑚2𝑠) will also change, but (𝑁𝐴

𝑘𝑔 𝑚𝑜𝑙

𝑠) will remain

constant. For spherical geometry,

𝑁𝐴 =𝑁𝐴

4𝜋𝑟2== −𝐷𝐴𝐵

𝑃

𝑅𝑇

1

(𝑃 − 𝑝𝐴)

𝑑𝑝𝐴𝑑𝑟

𝑁𝐴4𝜋

∫𝑑𝑟

𝑟2

𝑟2

𝑟1


𝑅𝑇∫

1

(𝑃 − 𝑝𝐴)𝑑𝑝𝐴

𝑝𝐴2

𝑝𝐴1

𝑁𝐴4𝜋

(1

𝑟1−

1

𝑟2) = 𝐷𝐴𝐵

𝑃

𝑅𝑇ln


For 𝑟2 ≫ 𝑟1, 1 𝑟2 ≅ 0⁄ , gives

𝑁𝐴4𝜋𝑟1

= 𝐷𝐴𝐵𝑃

𝑅𝑇ln


𝑁𝐴 =𝑁𝐴

4𝜋𝑟12 =

𝐷𝐴𝐵𝑟1

𝑃

𝑅𝑇[ln


] =𝐷𝐴𝐵𝑟1

𝑃

𝑅𝑇[𝑝𝐴1 − 𝑝𝐴2

𝑝𝐵𝑀]

For 𝑝𝐴1 ≪ 𝑃 ⇒ 𝑝𝐵𝑀 ≅ 𝑃 . Therefore,

𝑁𝐴 =𝐷𝐴𝐵𝑟1

1

𝑅𝑇(𝑝𝐴1 − 𝑝𝐴2)

Since 𝑐 = 𝑃 𝑅𝑇⁄ for liquids

𝑁𝐴 =𝐷𝐴𝐵𝑟1

(𝑐𝐴1 − 𝑐𝐴2)

For a tube of constant cross-sectional area, the total time of evaporation for the change

in level from 𝑧0 to 𝑧𝐹:


(𝑧𝐹2−𝑧0

2)

2𝐷𝐴𝐵(

𝑃

𝑅𝑇


)−𝟏

For a sphere, for the complete evaporation of initial radius 𝑟1:


(𝑟12)

2𝐷𝐴𝐵(

𝑃

𝑅𝑇


)−𝟏

Time to Completely Evaporate a Sphere

A drop of liquid toluene is kept at a uniform temperature of 25.9 °C and is suspended

in air by a fine wire. The initial radius 𝑟0 = 2𝑚𝑚. The vapor pressure of toluence at

25.9 °C is 𝑝𝐴1 = 3.84 𝑘𝑃𝑎 and the density of liquid toluene is 866 kg/m3. Calculate the

time in seconds for complete evaporation, i.e. 𝑟𝐹 = 0 𝑚𝑚.


(𝑟02 − 𝑟𝐹

2)

2𝐷𝐴𝐵(

𝑃

𝑅𝑇


)−𝟏

=𝜌𝐴𝑀𝐴

(𝑟02 − 𝑟𝐹

2)

2𝐷𝐴𝐵

𝑅𝑇

𝑃

𝑝𝐵𝑀𝑝𝐴1 − 𝑝𝐴2

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 𝑝𝐴0 = 4242 𝑃𝑎 and the density of water

is 995.71 kg/m3. Note that

Conditions in this problem are same as in Question 1

𝐴𝑟𝑒𝑎, 𝐴 = 4𝜋𝑟2; 𝑉𝑜𝑙𝑢𝑚𝑒, 𝑉 = (4 3⁄ )𝜋𝑟3; 𝑀𝑎𝑠𝑠 = 𝜌𝑉

The time of evaporation can be computed using,


(𝑟02 − 𝑟𝐹

2)

2𝐷𝐴𝐵

𝑅𝑇

𝑃


I. Calculate the time in seconds for its complete evaporation (rF = 0 mm).


(𝑟02 − 𝑟𝐹

2)

2𝐷𝐴𝐵

𝑅𝑇

𝑃


=995.71

18

(0.0042 − 0)

2 ∙ 2.68 × 10−58314 × 303

101325

99,189

4242 − 0= 𝟗𝟓𝟗𝟗 𝒔

12

II. Calculate the time in second required for the evaporation of half of the total initial mass

of the water drop

𝑀𝑎𝑠𝑠F𝑀𝑎𝑠𝑠i

=𝑉𝑜𝑙𝑢𝑚𝑒F𝑉𝑜𝑙𝑢𝑚𝑒i

=𝑟F

3

𝑟i3 = 0.5; 𝑟F

3 = 0.5𝑟i3; 𝑟𝐹 = 3.175 𝑚𝑚


(𝑟02 − 𝑟𝐹

2)

2𝐷𝐴𝐵

𝑅𝑇

𝑃


=995.71

18

(0.0042 − 0.0031752 )

2 ∙ 2.68 × 10−58314 × 303

101325

99,189

4242 − 0

= 𝟑𝟓𝟓𝟏 𝒔

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 𝑡𝐹 ∝ 𝑟02 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


(𝑟02 − 𝑟𝐹

2)

2𝐷𝐴𝐵

𝑅𝑇

𝑃


=995.71

18

(0.0042 − 0 )

2 ∙ 26.8 × 10−58314 × 303

10132.5

𝟕, 𝟖𝟐𝟏

4242 − 0= 𝟕𝟓𝟕𝒔

Note: (𝐷𝐴𝐵𝑃)is independent of pressure, but (𝑝𝐵𝑀) will change.

6

Case (B): Diffusion through a tapered conduit:

The radius at any location is related to length as

𝑟 = (𝑟2 − 𝑟1𝑧2 − 𝑧1

) 𝑧 + 𝑟1

and area

𝐴 = 𝜋𝑟2 = 𝜋 [(𝑟2 − 𝑟1𝑧2 − 𝑧1

) 𝑧 + 𝑟1]2

Therefore,

𝑁𝐴 =𝑁𝐴

𝜋𝑟2== −𝐷𝐴𝐵

𝑃

𝑅𝑇

1

(𝑃 − 𝑝𝐴)

𝑑𝑝𝐴𝑑𝑟

𝑁𝐴 =𝑁𝐴𝜋

∫𝑑𝑧

[(𝑟2 − 𝑟1𝑧2 − 𝑧1

) 𝑧 + 𝑟1]2

𝑟2

𝑟1


𝑅𝑇∫

1

(𝑃 − 𝑝𝐴)𝑑𝑝𝐴

𝑝𝐴2

𝑝𝐴1

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 𝑐2(𝑡) with the help of following relations

can be used to obtain diffusion coefficient

𝑐𝑎𝑣 − 𝑐2

𝑐𝑎𝑣 − 𝑐20 = exp [−

𝐷𝐴𝐵(𝑉1 + 𝑉2)

(𝐿 𝐴⁄ )(𝑉2𝑉1)𝑡]

where,

𝑐1 + 𝑐2 = 𝑐10 + 𝑐2

0

𝑐𝑎𝑣 =𝑉1𝑐1

0 + 𝑉2𝑐20

𝑉1 + 𝑉2

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 Chapman–Enskog correlation can be used (Eq. 6.2-44 in

C. J. Geankoplis):

𝐷𝐴𝐵 = 1.8583 × 10−7

𝜎𝐴𝐵2 Ω𝐷,𝐴𝐵

(𝑇3 2⁄

𝑃) (

1

𝑀𝐴+

1

𝑀𝐵)

1 2⁄

where, A and B are two kinds of molecules present in the gaseous mixture, 𝐷𝐴𝐵is the

diffusivity (m2/s), T is the absolute temperature (K), M is the molar mass (kg/ kg mol),

P is the pressure (atm), and 𝜎 is the ‘average collision diameter’, Ω 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):

𝐷𝐴𝐵 = 1.0 × 10−7

[(∑ 𝑣𝐴)1 3⁄ + (∑ 𝑣𝐵)

1 3⁄ ]2(

𝑇3.5 2⁄

𝑃) (

1

𝑀𝐴+

1

𝑀𝐵)

1 2⁄

where ∑ 𝑣𝐴=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:

𝐷𝐴𝐵 ∝ (𝑇3.5 2⁄

𝑃)


= (𝑇2𝑇1

)1.75


= (𝑃1𝑃2

)

Prediction accuracy of Fuller et al. Equation

Schmidt number for gases

𝑁𝑆𝑐 = 𝜇 𝜌⁄

𝐷𝐴𝐵=

𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑣𝑖𝑡𝑦

𝑚𝑎𝑠𝑠 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑣𝑖𝑡𝑦

For gases, 𝑁𝑆𝑐 = 0.5– 2.0

For liquid, , 𝑁𝑆𝑐 = 100– 10,000

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