1. Dept of Chemical and Biomolecular Engineering CN2125E Heat and Mass Transfer Dr. Tong Yen Wah, E5-03-15, 6516-8467 chetyw@nus.edu.sg (Mass Transfer, Radiation)

2. Course Outline Week 9-12: Mass Transfer –Week 9: Steady-state Diffusion (WWWR Ch 26) –Week 10: Unsteady-state Diffusion (WWWR Ch 27) –Week 11: Convective Mass Transfer (WWWR Ch 28) Week 13: Radiation Heat Transfer (WWWR Ch 23, ID Ch 12-13)

3. HW/Tutorial Week #9 WWWR Chapters 26, ID Chapter 14 Tutorial #9 WWWR # 26.17 & 26.27 To be discussed on March 24, 2015. By either volunteer or class list.

4. Molecular Diffusion General differential equation One-dimensional mass transfer without reaction

5. Unimolecular Diffusion Diffusivity of gas can be measured in an Arnold diffusion cell

6. Assuming –Steady state, no reaction, and diffusion in z- direction only We get And since B is a stagnant gas,

7. Thus, for constant molar flux of A, when N B,z = 0, –with boundary conditions: at z = z 1, y A = y A1 at z = z 2, y A = y A2 Integrating and solving for N A,z

8. since the log-mean average of B is we get This is a steady-state diffusion of one gas through a second stagnant gas;

9. For film theory, we assume laminar film of constant thickness , then, z 2 – z 1 =  and But we know So, the film coefficient is then

10. To determine concentration profile, if isothermal and isobaric, integrated twice, we get

11. –with boundary conditions: at z = z 1, y A = y A1 at z = z 2, y A = y A2 So, the concentration profile is:

12. Example 1

17. Pseudo-Steady-State Diffusion When there is a slow depletion of source or sink for mass transfer Consider the Arnold diffusion cell, when liquid is evaporated, the surface moves, at any instant, molar flux is

18. Molar flux is also the amount of A leaving Under pseudo-steady-state conditions, which integrated from t=0 to t=t, z=z t0 to z=z t becomes

19. Example 2

27. Equimolar Counterdiffusion Flux of one gaseous component is equal to but in the opposite direction of the second gaseous component Again, for steady-state, no reaction, in the z-direction, the molar flux is

28. In equimolar counterdiffusion, N A,z = -N B,z Integrated at z = z 1, c A = c A1 and at z = z 2, c A = c A2 to: Or in terms of partial pressure,

29. The concentration profile is described by Integrated twice to With boundary conditions at z = z 1, c A = c A1 and at z = z 2, c A = c A2 becomes a linear concentration profile:

30. Systems with Reaction When there is diffusion of a species together with its disappearance/appearance through a chemical reaction Homogeneous reaction occurs throughout a phase uniformly Heterogeneous reaction occurs at the boundary or in a restricted region of a phase

31. Diffusion with heterogeneous first order reaction with varying area: –With both diffusion and reaction, the process can be diffusion controlled or reaction controlled. –Example: burning of coal particles –steady state, one-dimensional, heterogeneous

32. –3C (s) + 2.5 O 2 (g)  2 CO 2 (g) + CO (g) –Along diffusion path, R O2 = 0, then the general mass transfer equation reduces from –to –For oxygen,

33. –From the stoichiometry of the reaction, –We simplify Fick’s equation in terms of oxygen only, –which reduces to

34. –The boundary conditions are: at r = R, y O2 = 0 and at r = , y O2 = 0.21, –Integrating the equation to: –The oxygen transferred across the cross- sectional area is then:

35. –Using a pseudo-steady-state approach to calculate carbon mass-transfer output rate of carbon: accumulation rate of carbon: input rate of carbon = 0 Thus, the carbon balance is

36. –Rearranging and integrating from t = 0 to t = , R = R i to R = R f, we get –For heterogeneous reactions, the reaction rate is

37. –If the reaction is only C (s) + O 2 (g)  CO 2 (g) and if the reaction is not instantaneous, then for a first-order reaction, at the surface, then,

38. –Combining diffusion with reaction process, we get

39. Example 3

43. Diffusion with homogeneous first-order reaction: –Example: a layer of absorbing liquid, with surface film of composition A and thickness , assume concentration of A is small in the film, and the reaction of A is

44. –Assuming one-direction, steady-state, the mass transfer equation reduces from to with the general solution

45. –With the boundary conditions: at z = 0, c A = c A0 and at z = , c A = 0, –At the liquid surface, flux is calculated by differentiating the above and evaluating at z=0,

46. –Thus, –Comparing to absorption without reaction, the second term is called the Hatta number. –As reaction rate increases, the bottom term approaches 1.0, thus

47. –Comparing with we see that k c is proportional to D AB to ½ power. This is the Penetration Theory model, where a molecule will disappear by reaction after absorption of a short distance.

48. Example 4