Dr. R. Nagarajan Professor Dept of Chemical Engineering IIT Madras

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

Mass Transport: Illustrative Problems

SOLUTION TO THE PROBLEM

SOLUTION Catalytic Converter

SOLUTION • 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 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 Mmix and Mco are close hence we will assume

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

SOLUTION therefore therefore

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

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

SOLUTION where We estimate at which If then

SOLUTION 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 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 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 Moreover, hence, and

SOLUTION 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 where Mixing cup avg temp at duct outlet ?

SOLUTION Again, we see that Moreover,

SOLUTION therefore and

SOLUTION Therefore,

SOLUTION 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 • Pressure Drop • We have: • Therefore

SOLUTION But Therefore

SOLUTION From overall momentum balance:

Dr. R. Nagarajan Professor Dept of Chemical Engineering IIT Madras