Analogy of Mass, Heat and Momentum Transport...Momentum Heat Mass ๐œ๐‘ง๐‘ฅ=โˆ’(๐œ‡ + ) ( ๐‘ฅ)...

28
Analogy of Mass, Heat and Momentum Transport General molecular transport equations = โˆ’ ฮ“ Momentum Heat Mass =โˆ’( ) ( ) = โˆ’ ( ) โ‹† = โˆ’ ( ) Turbulent diffusion equation Momentum Heat Mass =โˆ’( + ) ( ) = โˆ’( + ) ( ) โ‹† = โˆ’( + ) ( )

Transcript of Analogy of Mass, Heat and Momentum Transport...Momentum Heat Mass ๐œ๐‘ง๐‘ฅ=โˆ’(๐œ‡ + ) ( ๐‘ฅ)...

  • Analogy of Mass, Heat and Momentum Transport

    General molecular transport equations

    ๐œ“๐‘ง = โˆ’๐›ฟ๐‘‘ฮ“

    ๐‘‘๐‘ง

    Momentum Heat Mass

    ๐œ๐‘ง๐‘ฅ = โˆ’ (๐œ‡

    ๐œŒ)

    ๐‘‘(๐œŒ๐‘ฃ๐‘ฅ)

    ๐‘‘๐‘ง

    ๐‘ž๐‘ง๐ด

    = โˆ’๐›ผ๐‘‘(๐œŒ๐ถ๐‘ƒ๐‘‡)

    ๐‘‘๐‘ง ๐ฝ๐ด๐‘ง

    โ‹† = โˆ’๐ท๐ด๐ต๐‘‘(๐‘๐ด)

    ๐‘‘๐‘ง

    Turbulent diffusion equation

    Momentum Heat Mass

    ๐œ๐‘ง๐‘ฅ = โˆ’ (๐œ‡

    ๐œŒ+ ๐œ€๐‘ก)

    ๐‘‘(๐œŒ๐‘ฃ๐‘ฅ)

    ๐‘‘๐‘ง

    ๐‘ž๐‘ง๐ด

    = โˆ’(๐›ผ + ๐›ผ๐‘ก)๐‘‘(๐œŒ๐ถ๐‘ƒ๐‘‡)

    ๐‘‘๐‘ง

    ๐ฝ๐ด๐‘งโ‹†

    = โˆ’(๐ท๐ด๐ต + ๐œ€๐‘€)๐‘‘(๐‘๐ด)

    ๐‘‘๐‘ง

  • 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œ‡

    ๐œŒ) (

    ๐‘š2

    ๐‘ )

    Thermal diffusivity ๐›ผ (๐‘š2

    ๐‘ )

    Molecular diffusivity ๐ท๐ด๐ต (๐‘š2

    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 โ€“6 โ€“ 1 ร— 10-5 m2 / s

    Liquids: 10 โ€“6 โ€“ 10-9 m2 / s

    Solids: 5 ร— 10 โ€“14 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 ๐œ€๐‘ก (๐‘š2

    ๐‘ )

    Turbulent (eddy) thermal diffusivity ๐›ผ๐‘ก (๐‘š2

    ๐‘ )

    Turbulent (eddy) mas diffusivity ๐œ€๐‘€ (๐‘š2

    ๐‘ )

  • Fickโ€™s Law for Molecular Diffusion

    ๐ฝ๐ด๐‘งโ‹† = โˆ’๐‘๐ท๐ด๐ต

    ๐‘‘(๐‘ฅ๐ด)

    ๐‘‘๐‘ง

    where c is the total concentration of A and B in (kg mol A + B)/m3, and ๐‘ฅ๐ด is the mole fraction of A in the mixture A and B.

    For constant concentration (c),

    ๐‘๐ด = ๐‘๐‘ฅ๐ด

    ๐‘‘๐‘๐ด = ๐‘‘(๐‘๐‘ฅ๐ด) = ๐‘๐‘‘(๐‘ฅ๐ด)

    ๐ฝ๐ด๐‘งโ‹† = โˆ’๐ท๐ด๐ต

    ๐‘‘(๐‘๐ด)

    ๐‘‘๐‘ง

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

    give,

    ๐ฝ๐ด๐‘งโ‹† โˆซ ๐‘‘๐‘ง

    ๐‘ง2

    ๐‘ง1

    = โˆ’๐ท๐ด๐ต โˆซ ๐‘‘(๐‘๐ด)๐‘๐ด2

    ๐‘๐ด1

    ๐ฝ๐ด๐‘งโ‹† = โˆ’

    ๐ท๐ด๐ต(๐‘๐ด1 โˆ’ ๐‘๐ด2)

    (๐‘ง1 โˆ’ ๐‘ง2)

    Since the concentration is related to partial pressure,

    ๐‘๐ด =๐‘๐ด๐‘…๐‘‡

    ๐ฝ๐ด๐‘งโ‹† = โˆ’

    ๐ท๐ด๐ต(๐‘๐ด1 โˆ’ ๐‘๐ด2)

    ๐‘…๐‘‡(๐‘ง1 โˆ’ ๐‘ง2)

    ๐ฝ๐ด๐‘งโ‹† =

    ๐ท๐ด๐ต(๐‘ง2 โˆ’ ๐‘ง1)

    1

    ๐‘…๐‘‡(๐‘๐ด1 โˆ’ ๐‘๐ด2)

  • 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 ๐‘ฅ๐ด1 , the mole fraction of A in tank 1, is larger than ๐‘ฅ๐ด2 , 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.

    ๐ฝ๐ด๐‘งโ‹† = โˆ’๐ฝ๐ต๐‘ง

    โ‹†

    [โˆ’๐ท๐ด๐ต๐‘‘(๐‘๐ด)

    ๐‘‘๐‘ง] = โˆ’ [โˆ’๐ท๐ต๐ด

    ๐‘‘(๐‘๐ต)

    ๐‘‘๐‘ง]

    Since, ๐‘ƒ = ๐‘๐ด + ๐‘๐ต = ๐‘๐‘œ๐‘›๐‘ ๐‘ก๐‘Ž๐‘›๐‘ก, therefore,

    ๐‘ = ๐‘๐ด + ๐‘๐ต = ๐‘๐‘œ๐‘›๐‘ ๐‘ก๐‘Ž๐‘›๐‘ก.

    Differentiating

    0 = ๐‘‘๐‘๐ด + ๐‘‘๐‘๐ต

    Substituting ๐‘‘๐‘๐ต = โˆ’๐‘‘๐‘๐ด gives,

    [โˆ’๐ท๐ด๐ต๐‘‘(๐‘๐ด)

    ๐‘‘๐‘ง] = โˆ’ [โˆ’๐ท๐ต๐ด

    ๐‘‘(๐‘๐ต)

    ๐‘‘๐‘ง] = โˆ’ [โˆ’๐ท๐ต๐ด (โˆ’

    ๐‘‘(๐‘๐ด)

    ๐‘‘๐‘ง)] = [โˆ’๐ท๐ต๐ด

    ๐‘‘(๐‘๐ด)

    ๐‘‘๐‘ง]

    [โˆ’๐ท๐ด๐ต๐‘‘(๐‘๐ด)

    ๐‘‘๐‘ง] = [โˆ’๐ท๐ต๐ด

    ๐‘‘(๐‘๐ด)

    ๐‘‘๐‘ง]

    ๐ท๐ด๐ต = ๐ท๐ต๐ด

    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 = ๐‘ฃ๐ด๐‘‘

    Therefore, molar diffusion flux,

    ๐ฝ๐ด๐‘งโ‹†

    ๐‘˜๐‘” ๐‘š๐‘œ๐‘™ ๐ด

    ๐‘  โˆ™ ๐‘š2= ๐‘๐ด๐‘ฃ๐ด๐‘‘

    ๐‘˜๐‘” ๐‘š๐‘œ๐‘™ ๐ด

    ๐‘š3๐‘š

    ๐‘ 

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

    (๐‘ฃ๐ด๐‘‘) and the average velocity (๐‘ฃ๐‘€ , molar average velocity of the whole fluid relative to a stationary point). Mathematically,

    ๐‘ฃ๐ด = ๐‘ฃ๐ด๐‘‘ + ๐‘ฃ๐‘€

    ๐‘๐ด๐‘ฃ๐ด = ๐‘๐ด๐‘ฃ๐ด๐‘‘ + ๐‘๐ด๐‘ฃ๐‘€

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

    ๐‘๐‘ฃ๐‘€ = ๐‘ = ๐‘๐ด + ๐‘๐ต; ๐‘ฃ๐‘€ =๐‘๐ด + ๐‘๐ต

    ๐‘

    Therefore,

    ๐‘๐ด = โˆ’๐‘๐ท๐ด๐ต๐‘‘(๐‘ฅ๐ด)

    ๐‘‘๐‘ง+

    ๐‘๐ด๐‘

    (๐‘๐ด + ๐‘๐ต)

    ๐‘๐ต = โˆ’๐‘๐ท๐ต๐ด๐‘‘(๐‘ฅ๐ต)

    ๐‘‘๐‘ง+

    ๐‘๐ต๐‘

    (๐‘๐ด + ๐‘๐ต)

    Total flux of A relative to stationary point

    ๐‘๐ด๐‘ฃ๐ด(= ๐‘๐ด)

    Diffusion flux relative to moving fluid

    ๐‘๐ด๐‘ฃ๐ด๐‘‘(=, ๐ฝ๐ด๐‘งโ‹† )

    Convective flux of A relative to stationary

    point ๐‘๐ด๐‘ฃ๐‘€ = +

    Total flux of

    A, ๐‘๐ด Diffusion flux, ๐ฝ๐ด๐‘ง

    โ‹†

    Convective flux of A,

    ๐‘๐ด (๐‘๐ด + ๐‘๐ต

    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 (๐‘๐ต = 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 (๐‘๐ต = 0).

    Therefore,

    ๐‘๐ด = โˆ’๐‘๐ท๐ด๐ต๐‘‘(๐‘ฅ๐ด)

    ๐‘‘๐‘ง+

    ๐‘๐ด๐‘

    (๐‘๐ด + 0)

    Since,

    ๐‘ =๐‘ƒ

    ๐‘…๐‘‡; ๐‘๐ด = ๐‘ฅ๐ด๐‘ƒ;

    ๐‘๐ด๐‘

    =๐‘๐ด๐‘ƒ

    Therefore, for constant pressure

    ๐‘๐ด = โˆ’๐ท๐ด๐ต๐‘…๐‘‡

    ๐‘‘(๐‘๐ด)

    ๐‘‘๐‘ง+

    ๐‘๐ด๐‘ƒ

    ๐‘๐ด

    ๐‘๐ด (1 โˆ’๐‘๐ด๐‘ƒ

    ) = โˆ’๐ท๐ด๐ต๐‘…๐‘‡

    ๐‘‘(๐‘๐ด)

    ๐‘‘๐‘ง

  • ๐‘ต๐‘จ = โˆ’๐‘ซ๐‘จ๐‘ฉ๐‘ท

    ๐‘น๐‘ป

    ๐Ÿ

    (๐‘ท โˆ’ ๐’‘๐‘จ)

    ๐’…(๐’‘๐‘จ)

    ๐’…๐’›

    ๐‘๐ด โˆซ ๐‘‘๐‘ง๐‘2

    ๐‘1

    = โˆ’๐ท๐ด๐ต๐‘ƒ

    ๐‘…๐‘‡โˆซ

    ๐‘‘๐‘๐ด(๐‘ƒ โˆ’ ๐‘๐ด)

    ๐‘๐ด2

    ๐‘๐ด1

    ๐‘๐ด =๐ท๐ด๐ต

    (๐‘ง2 โˆ’ ๐‘ง1)

    ๐‘ƒ

    ๐‘…๐‘‡ln

    ๐‘ƒ โˆ’ ๐‘๐ด2๐‘ƒ โˆ’ ๐‘๐ด1

    Since, the total pressure is the sum of partial pressures, i.e., ๐‘ƒ = ๐‘๐ด1 + ๐‘๐ต1 = ๐‘๐ด2 +๐‘๐ต2

    Therefore,

    ๐‘๐ด =๐ท๐ด๐ต

    (๐‘ง2 โˆ’ ๐‘ง1)

    ๐‘ƒ

    ๐‘…๐‘‡๐‘™๐‘›

    ๐‘๐ต2๐‘๐ต1

    But,

    ๐‘๐ต๐‘€ =๐‘๐ต2 โˆ’ ๐‘๐ต1

    ๐‘™๐‘›๐‘๐ต2๐‘๐ต1

    =๐‘๐ด1 โˆ’ ๐‘๐ด2

    ๐‘™๐‘›๐‘๐ต2๐‘๐ต1

    ๐‘™๐‘›๐‘๐ต2๐‘๐ต1

    =๐‘๐ด1 โˆ’ ๐‘๐ด2

    ๐‘๐ต๐‘€

    Therefore,

    ๐‘ต๐‘จ =๐‘ซ๐‘จ๐‘ฉ

    (๐’›๐Ÿ โˆ’ ๐’›๐Ÿ)

    ๐‘ท

    ๐‘น๐‘ป

    ๐’‘๐‘จ๐Ÿ โˆ’ ๐’‘๐‘จ๐Ÿ๐’‘๐‘ฉ๐‘ด

  • ๐‘ต๐‘จ =๐‘ซ๐‘จ๐‘ฉ

    (๐’›๐Ÿ โˆ’ ๐’›๐Ÿ)

    ๐Ÿ

    ๐‘น๐‘ป

    ๐‘ท

    ๐’‘๐‘ฉ๐‘ด(๐’‘๐‘จ๐Ÿ โˆ’ ๐’‘๐‘จ๐Ÿ)

    ๐‘ฑ๐‘จ๐’›โ‹† =

    ๐‘ซ๐‘จ๐‘ฉ(๐’›๐Ÿ โˆ’ ๐’›๐Ÿ)

    ๐Ÿ

    ๐‘น๐‘ป(๐’‘๐‘จ๐Ÿ โˆ’ ๐’‘๐‘จ๐Ÿ)

  • 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 (๐‘ง2 โˆ’ ๐‘ง1) 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 0.250ร—10-4 m2/s. Assume that the system is isothermal.

    Solution:

    ๐‘๐ด1 = 2.341 ร— 103 Pa (Vapor pressure of water at 20ยฐC from Appendix A.2)

    ๐‘๐ด2 = 0 (Assuming dry air, i.e. no water vapor)

    ๐‘๐ต1 = P โ€“ pAl = 1.01325 ร— 105 โ€“ 2.341 ร— 103 = 98984

    ๐‘๐ต2 = P โ€“ pA2 = 1.01325 ร— 105 โ€“ 0 = 1.01325 ร— 105 Pa

    ๐‘๐ต๐‘€ =๐‘๐ต2 โˆ’ ๐‘๐ต1

    ๐‘™๐‘›๐‘๐ต2๐‘๐ต1

    =๐‘๐ด1 โˆ’ ๐‘๐ด2

    ๐‘™๐‘›๐‘๐ต2๐‘๐ต1

    = 100149.4 ๐‘ƒ๐‘Ž

    When,

    ๐‘๐ต1 โ‰ˆ ๐‘๐ต2; ๐‘๐ต๐‘€ โ‰…๐‘๐ต1 + ๐‘๐ต2

    2= 100154.5 ๐‘ƒ๐‘Ž

    ๐‘๐ด =๐ท๐ด๐ต

    (๐‘ง2 โˆ’ ๐‘ง1)

    ๐‘ƒ

    ๐‘…๐‘‡

    ๐‘๐ด1 โˆ’ ๐‘๐ด2๐‘๐ต๐‘€

    =0.25 ร— 10โˆ’4

    0.1524

    1.01325 ร— 105

    8314 ร— 293

    (2.341 ร— 103 โˆ’ 0)

    100149.4= 1.595 ร— 10โˆ’7

    ๐‘˜๐‘” ๐‘š๐‘œ๐‘™

    ๐‘š2 โˆ™ 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 (๐‘˜๐‘” ๐‘š๐‘œ๐‘™ ๐‘š2 โˆ™ 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 ๐พ; ๐‘ƒ = 1.01325 ร— 105 ๐‘ƒ๐‘Ž; ๐ท๐ด๐ต= 2.6 ร— 10โˆ’5 ๐‘š2 ๐‘ โ„ @298 ๐พ;

    ๐ท๐ด๐ต2๐ท๐ด๐ต1

    = (๐‘‡2๐‘‡1

    )1.75

    ; ๐ท๐ด๐ต2 = 2.6 ร— 10โˆ’5 (

    293

    298)

    1.75

    = ๐Ÿ. ๐Ÿ“๐Ÿ ร— ๐Ÿ๐ŸŽโˆ’๐Ÿ“ ๐’Ž๐Ÿ ๐’”โ„

    ๐‘๐ด1 = 2340 Pa (Vapor pressure of water vapor at 20ยฐC) ๐‘๐ด2 = 234 (10% Relative Humidity) ๐‘๐ต1 = P โ€“ ๐‘๐ด1 = 101.325 ร— 10

    3 โ€“ 2.34 ร— 103 = 98,985 Pa ๐‘๐ต2 = P โ€“ ๐‘๐ด2 = 101325 โ€“ 234 = 101,091 Pa

    ๐‘๐ต๐‘€ =๐‘๐ต2 โˆ’ ๐‘๐ต1

    ๐‘™๐‘›๐‘๐ต2๐‘๐ต1

    =๐‘๐ด1 โˆ’ ๐‘๐ด2

    ๐‘™๐‘›๐‘๐ต2๐‘๐ต1

    = 100,034 ๐‘ƒ๐‘Ž

    ๐‘๐ด =๐ท๐ด๐ต

    (๐‘ง2 โˆ’ ๐‘ง1)

    ๐‘ƒ

    ๐‘…๐‘‡

    ๐‘๐ด1 โˆ’ ๐‘๐ด2๐‘๐ต๐‘€

    =2.52 ร— 10โˆ’5

    1.0

    1.01325 ร— 105

    8314 ร— 293

    (2340 โˆ’ 234)

    100,034

    = ๐Ÿ. ๐Ÿ๐Ÿ ร— ๐Ÿ๐ŸŽโˆ’๐Ÿ–๐’Œ๐’ˆ ๐’Ž๐’๐’

    ๐’Ž๐Ÿ โˆ™ 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 2.6ร—10-5 m2/s.

    I. Determine the molecular diffusion coefficient at 303 K.

    ๐ท๐ด๐ต2๐ท๐ด๐ต1

    = (๐‘‡2๐‘‡1

    )1.75

    ; ๐ท๐ด๐ต2 = 2.6 ร— 10โˆ’5 (

    303

    298)

    1.75

    = ๐Ÿ. ๐Ÿ”๐Ÿ– ร— ๐Ÿ๐ŸŽโˆ’๐Ÿ“ ๐’Ž๐Ÿ ๐’”โ„

    7

    II. Determine pBM in Pa.

    ๐‘๐ต๐‘€ =๐‘๐ต2 โˆ’ ๐‘๐ต1

    ๐‘™๐‘›๐‘๐ต2๐‘๐ต1

    =๐‘๐ด1 โˆ’ ๐‘๐ด2

    ๐‘™๐‘›๐‘๐ต2๐‘๐ต1

    =4242 โˆ’ 0

    ๐‘™๐‘›101325 โˆ’ 0

    101325 โˆ’ 4242

    = ๐Ÿ—๐Ÿ—, ๐Ÿ๐Ÿ–๐Ÿ—๐‘ท๐’‚

    5

    III. Calculate the rate of evaporation (NA) at steady state in ๐‘˜๐‘” ๐‘š๐‘œ๐‘™ ๐‘š2 โˆ™ ๐‘ โ„ and ๐‘˜๐‘” ๐‘š2 โˆ™ ๐‘ โ„

    ๐‘๐ด =๐ท๐ด๐ต

    (๐‘ง2 โˆ’ ๐‘ง1)

    ๐‘ƒ

    ๐‘…๐‘‡

    ๐‘๐ด1 โˆ’ ๐‘๐ด2๐‘๐ต๐‘€

    =2.68 ร— 10โˆ’5

    0.2

    1.01325 ร— 105

    8314 ร— 303

    (4242 โˆ’ 0)

    99,189= ๐Ÿ. ๐Ÿ‘ ร— ๐Ÿ๐ŸŽโˆ’๐Ÿ•

    ๐’Œ๐’ˆ ๐’Ž๐’๐’

    ๐’Ž๐Ÿ โˆ™ ๐’”

    = ๐Ÿ’๐Ÿ. ๐Ÿ’ ร— ๐Ÿ๐ŸŽโˆ’๐Ÿ•๐’Œ๐’ˆ

    ๐’Ž๐Ÿ โˆ™ 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(๐‘๐ด ๐‘๐ด

    ๐‘œโ„ ) = 30%, where ๐‘๐ด is the partial pressure of the water vapor in the air and ๐‘๐ด

    ๐‘œ is the vapor pressure.

    ๐‘๐ด2 =30

    100๐‘๐ด

    ๐‘œ = 0.3 ร— 4242 = 1273๐‘ƒ๐‘Ž

    ๐‘๐ต๐‘€ =4242 โˆ’ 1273

    ๐‘™๐‘›101325 โˆ’ 1273101325 โˆ’ 4242

    = 98,560 ๐‘ƒ๐‘Ž

    ๐‘๐ด =๐ท๐ด๐ต

    (๐‘ง2 โˆ’ ๐‘ง1)

    ๐‘ƒ

    ๐‘…๐‘‡

    ๐‘๐ด1 โˆ’ ๐‘๐ด2๐‘๐ต๐‘€

    =2.68 ร— 10โˆ’5

    0.2

    1.01325 ร— 105

    8314 ร— 303

    (4242 โˆ’ 1273)

    98,560= ๐Ÿ. ๐Ÿ”๐Ÿ ร— ๐Ÿ๐ŸŽโˆ’๐Ÿ•

    ๐’Œ๐’ˆ ๐’Ž๐’๐’

    ๐’Ž๐Ÿ โˆ™ ๐’”

    Note: There is almost 30% decrease in the flux due to a 30% decrease in driving force since the change in the ๐‘๐ต๐‘€ is negligible.

    8

  • Interface Fall for A Diffusing through Stagnant Film of B

    The initial and final heights are

    ๐‘ง0, ๐‘ง๐น 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

    ๐‘๐ด = ๐‘๐ด ร— ๐ด๐‘Ÿ๐‘’๐‘Ž

    ๐‘๐ด ร— ๐ด๐‘Ÿ๐‘’๐‘Ž

    =๐œŒ๐ด๐‘€๐ด

    ๐‘‘๐‘ง

    ๐‘‘๐‘กร— ๐ด๐‘Ÿ๐‘’๐‘Ž

    ๐ท๐ด๐ต๐‘ง

    ๐‘ƒ

    ๐‘…๐‘‡

    ๐‘๐ด1 โˆ’ ๐‘๐ด2๐‘๐ต๐‘€

    =๐œŒ๐ด๐‘€๐ด

    ๐‘‘๐‘ง

    ๐‘‘๐‘ก

    ๐ท๐ด๐ต๐‘ƒ

    ๐‘…๐‘‡

    ๐‘๐ด1 โˆ’ ๐‘๐ด2๐‘๐ต๐‘€

    โˆซ ๐‘‘๐‘ก =๐‘ก๐น

    0

    ๐œŒ๐ด๐‘€๐ด

    โˆซ ๐‘ง๐‘‘๐‘ง๐‘ง๐น

    ๐‘ง0

    ๐‘ก๐น =๐œŒ๐ด๐‘€๐ด

    (๐‘ง๐น2โˆ’๐‘ง0

    2)

    2๐ท๐ด๐ต(

    ๐‘ƒ

    ๐‘…๐‘‡

    ๐‘๐ด1 โˆ’ ๐‘๐ด2๐‘๐ต๐‘€

    )โˆ’๐Ÿ

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

    ๐ท๐ด๐ต.

    (๐‘ง๐น2โˆ’๐‘ง0

    2) = (๐‘ƒ

    ๐‘…๐‘‡

    ๐‘๐ด1 โˆ’ ๐‘๐ด2๐‘๐ต๐‘€

    ) (๐‘€๐ด๐œŒ๐ด

    ) (2๐ท๐ด๐ต)๐‘ก๐น

  • 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 ๐‘Ÿ1 in an

    infinite medium of gas B. Component A at partial pressure ๐‘๐ด1 at the surface is

    diffusing in the surrounding stagnant medium (B), where ๐‘๐ด2 = 0 at large distance away.

    If the area changes, the flux (๐‘๐ด๐‘˜๐‘” ๐‘š๐‘œ๐‘™

    ๐‘š2๐‘ ) will also change, but (๐‘๐ด

    ๐‘˜๐‘” ๐‘š๐‘œ๐‘™

    ๐‘ ) will remain

    constant. For spherical geometry,

    ๐‘๐ด =๐‘๐ด

    4๐œ‹๐‘Ÿ2== โˆ’๐ท๐ด๐ต

    ๐‘ƒ

    ๐‘…๐‘‡

    1

    (๐‘ƒ โˆ’ ๐‘๐ด)

    ๐‘‘๐‘๐ด๐‘‘๐‘Ÿ

    ๐‘๐ด4๐œ‹

    โˆซ๐‘‘๐‘Ÿ

    ๐‘Ÿ2

    ๐‘Ÿ2

    ๐‘Ÿ1

    = โˆ’๐ท๐ด๐ต๐‘ƒ

    ๐‘…๐‘‡โˆซ

    1

    (๐‘ƒ โˆ’ ๐‘๐ด)๐‘‘๐‘๐ด

    ๐‘๐ด2

    ๐‘๐ด1

    ๐‘๐ด4๐œ‹

    (1

    ๐‘Ÿ1โˆ’

    1

    ๐‘Ÿ2) = ๐ท๐ด๐ต

    ๐‘ƒ

    ๐‘…๐‘‡ln

    ๐‘ƒ โˆ’ ๐‘๐ด2๐‘ƒ โˆ’ ๐‘๐ด1

    For ๐‘Ÿ2 โ‰ซ ๐‘Ÿ1, 1 ๐‘Ÿ2 โ‰… 0โ„ , gives

    ๐‘๐ด4๐œ‹๐‘Ÿ1

    = ๐ท๐ด๐ต๐‘ƒ

    ๐‘…๐‘‡ln

    ๐‘ƒ โˆ’ ๐‘๐ด2๐‘ƒ โˆ’ ๐‘๐ด1

    ๐‘๐ด =๐‘๐ด

    4๐œ‹๐‘Ÿ12 =

    ๐ท๐ด๐ต๐‘Ÿ1

    ๐‘ƒ

    ๐‘…๐‘‡[ln

    ๐‘ƒ โˆ’ ๐‘๐ด2๐‘ƒ โˆ’ ๐‘๐ด1

    ] =๐ท๐ด๐ต๐‘Ÿ1

    ๐‘ƒ

    ๐‘…๐‘‡[๐‘๐ด1 โˆ’ ๐‘๐ด2

    ๐‘๐ต๐‘€]

  • For ๐‘๐ด1 โ‰ช ๐‘ƒ โ‡’ ๐‘๐ต๐‘€ โ‰… ๐‘ƒ . Therefore,

    ๐‘๐ด =๐ท๐ด๐ต๐‘Ÿ1

    1

    ๐‘…๐‘‡(๐‘๐ด1 โˆ’ ๐‘๐ด2)

    Since ๐‘ = ๐‘ƒ ๐‘…๐‘‡โ„ for liquids

    ๐‘๐ด =๐ท๐ด๐ต๐‘Ÿ1

    (๐‘๐ด1 โˆ’ ๐‘๐ด2)

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

    in level from ๐‘ง0 to ๐‘ง๐น:

    ๐‘ก๐น =๐œŒ๐ด๐‘€๐ด

    (๐‘ง๐น2โˆ’๐‘ง0

    2)

    2๐ท๐ด๐ต(

    ๐‘ƒ

    ๐‘…๐‘‡

    ๐‘๐ด1 โˆ’ ๐‘๐ด2๐‘๐ต๐‘€

    )โˆ’๐Ÿ

    For a sphere, for the complete evaporation of initial radius ๐‘Ÿ1:

    ๐‘ก๐น =๐œŒ๐ด๐‘€๐ด

    (๐‘Ÿ12)

    2๐ท๐ด๐ต(

    ๐‘ƒ

    ๐‘…๐‘‡

    ๐‘๐ด1 โˆ’ ๐‘๐ด2๐‘๐ต๐‘€

    )โˆ’๐Ÿ

  • 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 ๐‘Ÿ0 = 2๐‘š๐‘š. The vapor pressure of toluence at

    25.9 ยฐC is ๐‘๐ด1 = 3.84 ๐‘˜๐‘ƒ๐‘Ž and the density of liquid toluene is 866 kg/m3. Calculate the

    time in seconds for complete evaporation, i.e. ๐‘Ÿ๐น = 0 ๐‘š๐‘š.

    ๐‘ก๐น =๐œŒ๐ด๐‘€๐ด

    (๐‘Ÿ02 โˆ’ ๐‘Ÿ๐น

    2)

    2๐ท๐ด๐ต(

    ๐‘ƒ

    ๐‘…๐‘‡

    ๐‘๐ด1 โˆ’ ๐‘๐ด2๐‘๐ต๐‘€

    )โˆ’๐Ÿ

    =๐œŒ๐ด๐‘€๐ด

    (๐‘Ÿ02 โˆ’ ๐‘Ÿ๐น

    2)

    2๐ท๐ด๐ต

    ๐‘…๐‘‡

    ๐‘ƒ

    ๐‘๐ต๐‘€๐‘๐ด1 โˆ’ ๐‘๐ด2

  • 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 ๐‘๐ด0 = 4242 ๐‘ƒ๐‘Ž and the density of water

    is 995.71 kg/m3. Note that

    Conditions in this problem are same as in Question 1

    ๐ด๐‘Ÿ๐‘’๐‘Ž, ๐ด = 4๐œ‹๐‘Ÿ2; ๐‘‰๐‘œ๐‘™๐‘ข๐‘š๐‘’, ๐‘‰ = (4 3โ„ )๐œ‹๐‘Ÿ3; ๐‘€๐‘Ž๐‘ ๐‘  = ๐œŒ๐‘‰

    The time of evaporation can be computed using,

    ๐‘ก๐น =๐œŒ๐ด๐‘€๐ด

    (๐‘Ÿ02 โˆ’ ๐‘Ÿ๐น

    2)

    2๐ท๐ด๐ต

    ๐‘…๐‘‡

    ๐‘ƒ

    ๐‘๐ต๐‘€๐‘๐ด1 โˆ’ ๐‘๐ด2

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

    ๐‘ก๐น =๐œŒ๐ด๐‘€๐ด

    (๐‘Ÿ02 โˆ’ ๐‘Ÿ๐น

    2)

    2๐ท๐ด๐ต

    ๐‘…๐‘‡

    ๐‘ƒ

    ๐‘๐ต๐‘€๐‘๐ด1 โˆ’ ๐‘๐ด2

    =995.71

    18

    (0.0042 โˆ’ 0)

    2 โˆ™ 2.68 ร— 10โˆ’58314 ร— 303

    101325

    99,189

    4242 โˆ’ 0= ๐Ÿ—๐Ÿ“๐Ÿ—๐Ÿ— ๐’”

    12

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

    of the water drop

    ๐‘€๐‘Ž๐‘ ๐‘ F๐‘€๐‘Ž๐‘ ๐‘ i

    =๐‘‰๐‘œ๐‘™๐‘ข๐‘š๐‘’F๐‘‰๐‘œ๐‘™๐‘ข๐‘š๐‘’i

    =๐‘ŸF

    3

    ๐‘Ÿi3 = 0.5; ๐‘ŸF

    3 = 0.5๐‘Ÿi3; ๐‘Ÿ๐น = 3.175 ๐‘š๐‘š

    ๐‘ก๐น =๐œŒ๐ด๐‘€๐ด

    (๐‘Ÿ02 โˆ’ ๐‘Ÿ๐น

    2)

    2๐ท๐ด๐ต

    ๐‘…๐‘‡

    ๐‘ƒ

    ๐‘๐ต๐‘€๐‘๐ด1 โˆ’ ๐‘๐ด2

    =995.71

    18

    (0.0042 โˆ’ 0.0031752 )

    2 โˆ™ 2.68 ร— 10โˆ’58314 ร— 303

    101325

    99,189

    4242 โˆ’ 0

    = ๐Ÿ‘๐Ÿ“๐Ÿ“๐Ÿ 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 ๐‘ก๐น โˆ ๐‘Ÿ02 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

    ๐‘ก๐น =๐œŒ๐ด๐‘€๐ด

    (๐‘Ÿ02 โˆ’ ๐‘Ÿ๐น

    2)

    2๐ท๐ด๐ต

    ๐‘…๐‘‡

    ๐‘ƒ

    ๐‘๐ต๐‘€๐‘๐ด1 โˆ’ ๐‘๐ด2

    =995.71

    18

    (0.0042 โˆ’ 0 )

    2 โˆ™ 26.8 ร— 10โˆ’58314 ร— 303

    10132.5

    ๐Ÿ•, ๐Ÿ–๐Ÿ๐Ÿ

    4242 โˆ’ 0= ๐Ÿ•๐Ÿ“๐Ÿ•๐’”

    Note: (๐ท๐ด๐ต๐‘ƒ)is independent of pressure, but (๐‘๐ต๐‘€) will change.

    6

  • Case (B): Diffusion through a tapered conduit:

    The radius at any location is related to length as

    ๐‘Ÿ = (๐‘Ÿ2 โˆ’ ๐‘Ÿ1๐‘ง2 โˆ’ ๐‘ง1

    ) ๐‘ง + ๐‘Ÿ1

    and area

    ๐ด = ๐œ‹๐‘Ÿ2 = ๐œ‹ [(๐‘Ÿ2 โˆ’ ๐‘Ÿ1๐‘ง2 โˆ’ ๐‘ง1

    ) ๐‘ง + ๐‘Ÿ1]2

    Therefore,

    ๐‘๐ด =๐‘๐ด

    ๐œ‹๐‘Ÿ2== โˆ’๐ท๐ด๐ต

    ๐‘ƒ

    ๐‘…๐‘‡

    1

    (๐‘ƒ โˆ’ ๐‘๐ด)

    ๐‘‘๐‘๐ด๐‘‘๐‘Ÿ

    ๐‘๐ด =๐‘๐ด๐œ‹

    โˆซ๐‘‘๐‘ง

    [(๐‘Ÿ2 โˆ’ ๐‘Ÿ1๐‘ง2 โˆ’ ๐‘ง1

    ) ๐‘ง + ๐‘Ÿ1]2

    ๐‘Ÿ2

    ๐‘Ÿ1

    = โˆ’๐ท๐ด๐ต๐‘ƒ

    ๐‘…๐‘‡โˆซ

    1

    (๐‘ƒ โˆ’ ๐‘๐ด)๐‘‘๐‘๐ด

    ๐‘๐ด2

    ๐‘๐ด1

  • 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 ๐‘2(๐‘ก) with the help of following relations

    can be used to obtain diffusion coefficient

    ๐‘๐‘Ž๐‘ฃ โˆ’ ๐‘2

    ๐‘๐‘Ž๐‘ฃ โˆ’ ๐‘20 = exp [โˆ’

    ๐ท๐ด๐ต(๐‘‰1 + ๐‘‰2)

    (๐ฟ ๐ดโ„ )(๐‘‰2๐‘‰1)๐‘ก]

    where,

    ๐‘1 + ๐‘2 = ๐‘10 + ๐‘2

    0

    ๐‘๐‘Ž๐‘ฃ =๐‘‰1๐‘1

    0 + ๐‘‰2๐‘20

    ๐‘‰1 + ๐‘‰2

  • 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 Chapmanโ€“Enskog correlation can be used (Eq. 6.2-44 in

    C. J. Geankoplis):

    ๐ท๐ด๐ต = 1.8583 ร— 10โˆ’7

    ๐œŽ๐ด๐ต2 ฮฉ๐ท,๐ด๐ต

    (๐‘‡3 2โ„

    ๐‘ƒ) (

    1

    ๐‘€๐ด+

    1

    ๐‘€๐ต)

    1 2โ„

    where, A and B are two kinds of molecules present in the gaseous mixture, ๐ท๐ด๐ตis 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œŽ is the โ€˜average 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):

    ๐ท๐ด๐ต = 1.0 ร— 10โˆ’7

    [(โˆ‘ ๐‘ฃ๐ด)1 3โ„ + (โˆ‘ ๐‘ฃ๐ต)

    1 3โ„ ]2(

    ๐‘‡3.5 2โ„

    ๐‘ƒ) (

    1

    ๐‘€๐ด+

    1

    ๐‘€๐ต)

    1 2โ„

    where โˆ‘ ๐‘ฃ๐ด=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:

    ๐ท๐ด๐ต โˆ (๐‘‡3.5 2โ„

    ๐‘ƒ)

    ๐ท๐ด๐ต2๐ท๐ด๐ต1

    = (๐‘‡2๐‘‡1

    )1.75

    ๐ท๐ด๐ต2๐ท๐ด๐ต1

    = (๐‘ƒ1๐‘ƒ2

    )

  • Prediction accuracy of Fuller et al. Equation

  • Schmidt number for gases

    ๐‘๐‘†๐‘ = ๐œ‡ ๐œŒโ„

    ๐ท๐ด๐ต=

    ๐‘š๐‘œ๐‘š๐‘’๐‘›๐‘ก๐‘ข๐‘š ๐‘‘๐‘–๐‘“๐‘“๐‘ข๐‘ ๐‘–๐‘ฃ๐‘–๐‘ก๐‘ฆ

    ๐‘š๐‘Ž๐‘ ๐‘  ๐‘‘๐‘–๐‘“๐‘“๐‘ข๐‘ ๐‘–๐‘ฃ๐‘–๐‘ก๐‘ฆ

    For gases, ๐‘๐‘†๐‘ = 0.5โ€“ 2.0

    For liquid, , ๐‘๐‘†๐‘ = 100โ€“ 10,000