# Ch. 11 Mass transfer principles

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1 Transport of chemical species in solid, liquid, or gas mixture Transport driven by composition gradient, similar to temperature gradients driving heat transport We will look at two mass transport mechanisms, both analogous to heat transport: Diffusion mass transfer Convection mass transfer 1

2 Diffusion mass heat transfer Results from intermolecular collisions that lead to net motion of molecules in direction of decreasing concentration. Analogous to conduction heat transfer Examples: Doping of semiconductor wafers to achieve desired electrical properties Diffusion of carbon atoms during heat treating of steel Leaking of gas through container wall 2

3 Convection mass transfer Combination of diffusion mass transfer combined with bulk fluid motion Examples: Drying of materials such as paper and food products Evaporation of water in power plant cooling tower Chemical reactions and combustion processes determination of distribution of reactants and products 3

4 Convection mass transfer drying process 4

5 Diffusion mass transfer diffusion in a vertical column 5

6 Conservation of chemical species Species concentration may be specified in terms of mass, moles, mass fraction, mole fraction, partial pressure, etc. Begin by considering mass concentration r i, the local density of species i in units of mass/volume 6

7 7

8 Conservation of species i results in: 8

9 In limit of small Dx, Dy Sum terms in Eq over all components results in: 9

10 We know from continuity requirements that: Bulk (mass) averaged velocity components: 10

11 Difference between component velocity and massaveraged velocity = diffusion velocity: 11

12 Diffusion flux defined as: Substituting back into Eq yields: 12

13 Using the continuity equation: We now need mass flux, j i, terms Fick s Law 13

14 Basic equation for binary mass transfer under isothermal, isobaric, steady conditions: Where j x,1 is the mass flux of species 1 in to species 2 per unit area, per unit time (kg/m 2 s or moles/m 2 s) D 12 is the mass diffusion coefficient of species 1 into species 2 (m 2 /s or ft 2 /s) 14

15 Substituting Fick s law: This is the mass-diffusion equation, analogous to the heat diffusion equation: 15

16 Mole fraction of species i: Relation between mass concentration (density) and mole fraction: Molar concentration: 16

17 Steady diffusion through a stationary medium: 17

18 Steady diffusion through a stationary medium: BC s 18

19 Mass diffusivity binary diffusion 19

20 Table 11.2: binary gas mixture diffusivities To find D at other P, T values: 20

21 Table 11.3: gases and organic solutes in water To find D at other T values (2 subscript is for solvent) 21

22 22

23 To solve our differential equation for the mass distribution, we need some boundary conditions and initial conditions (if transient). Initial condition simply specifies distribution of mass at beginning of problem (zero solute, or uniform distribution solute, for example). Boundary conditions more difficult since concentration of species is normally not continuous across an interface 23

24 For example, consider a solid in contact with a gas: Gas Solid y C gas,amb C gas,1 If we want to solve for the distribution of the gas inside the solid, the boundary condition is the concentration of gas on the solid side of the boundary, C gas,1 at y = 0 24

25 Case a) Ideal gas mixture in contact with a liquid phase of one of the mixture components Gas mixture Liquid, species 1 y C gas,1 If we are interested in diffusion of species 1 into gas mixture, we need concentration of species 1 at interface on the gas side 25

26 Boundary condition given as: Partial pressure of species 1 on mixture side is saturation pressure of species 1 at liquid temperature, T To relate to concentration: Gas mixture y P 1 Liquid, species 1 26

27 Case b) Ideal gas mixture in contact with a liquid phase of one of the mixture components Suppose now we are interested in diffusion of species in liquid Boundary condition is Henry s Law: Gas mixture P i Liquid y x i 27

28 H = Henry s constant P i = partial pressure in gas mixture x i = mole fraction of gas in liquid Henry s Law limitations: Dilute solutions (Mole fraction, x i is small) Low pressures, P i < 1 atm 28

29 Case c) Binary liquid mixture with species 1 as solute, in contact with pure substance of species 1 Concentration of species 1 on liquid side assume thermodynamic equilibrium at interface Find solubility data for species 1 into liquid Example from text: Solubility of NaCl in H 2 O at 0 o C: 35.7 grams of NaCl per 100 grams of H 2 O 29

30 Case d) Solid medium in contact with a gas that is soluble in the solid Boundary condition is: Gas mixture Solid y P i n i P 1 = partial pressure in gas mixture C L = molar concentration of gas on solid side of interface (kmol/m 3 ) S = Solubility of species i in solid 30

31 A few solubility values 31

32 Solving convection problems Boundary layers Equivalent form of Newton s law of cooling: 32

33 Heat and mass transfer analogy: Sh = Sherwood number Sc = Schmidt number 33

34 For example: laminar parallel flow over a flat plate in the presence of a concentration gradient Appropriate average Nusselt number: Sherwood number becomes: 34

35 Limit of heat and mass transfer analogy: 35

36 Example 36

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