1. Dr. R. Nagarajan Professor Dept of Chemical Engineering IIT Madras Advanced Transport Phenomena Module 6 Lecture 29 1 Mass Transport: Illustrative Problems

2. 2

3. SOLUTION TO THE PROBLEM 3

4. 4 SOLUTION Catalytic Converter

5. a. Mechanism of CO(g) transport to the wall If Re < 2100 (see below),transport to the wall is by Fick diffusion of CO(g) through the prevailing mixture. SOLUTION 5

6. Therefore Analogous heat transfer diffusivity is for gas mixture b. Discuss whether the Mass transfer Analogy Conditions(M A C) and Heat transfer Analogy Conditions (H A C) are met; implications ?  Since M mix and M co are close hence we will assume SOLUTION 6

7. c. Sc for the mixture : Now: and: SOLUTION 7

8. therefore SOLUTION 8

9. d. L=? We will need Re Now: therefore (laminar-flow regime) SOLUTION 9

10. For a square channel and (used below). If then the mass-transfer analogy is: SOLUTION 10

11. where We estimate at which If then SOLUTION 11

12. therefore Tentatively, assume F (entrance =1).Then: that is, (at which F (entrance) is indeed ). Solving for L gives: L =8.3 cm ( needed to give 95 % CO-Conversion). SOLUTION 12

13. e. Discuss underlying assumptions, e.g., fully developed flow? nearly constant thermo physical properties? no homogeneous chemical reaction? “ diffusion-controlled” surface reaction? f. If the catalyst were “ poisoned,” it would not be able to maintain. This would cause to exceed 8.3 cm. If catalyst were completely deactivated, then and, of course, SOLUTION 13

14. g. If the heat of combustion is 67.8 kcal/mole CO, how much heat is delivered to the catalyst channel per unit time? Overall CO balance gives the CO-consumption rate/channel: where SOLUTION 14

15. Moreover, hence, and SOLUTION 15

16. Therefore  The “sensible” heat transfer required to keep the wall at 500 K can be calculated from a heat balance on the 8.3 cm-long duct- i.e., once we calculate, we have : SOLUTION 16

17. where Mixing cup avg temp at duct outlet ? SOLUTION 17

18. Again, we see that Moreover, SOLUTION 18

19. therefore and SOLUTION 19

20. Therefore, SOLUTION 20

21. h. “Quasi-Steady” Application of These Results? Note that: and hence, if the characteristic period of the unsteadiness >> 8.2 ms, the previous results can be used at each flow condition. SOLUTION 21

22. i.Pressure Drop We have: Therefore SOLUTION 22

23. But Therefore SOLUTION 23

24. From overall momentum balance: SOLUTION 24