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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.

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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

solution
SOLUTION

Catalytic Converter

solution1
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.

solution2
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
solution3
SOLUTION

c. Sc for the mixture :

Now:

and:

solution4
SOLUTION

therefore

therefore

solution5
SOLUTION

d. L=? We will need Re

Now:

therefore (laminar-flow regime)

solution6
SOLUTION

For a square channel

and (used below).

If then the mass-transfer analogy is:

solution7
SOLUTION

where

We estimate at which

If

then

solution8
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).

solution9
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,

solution10
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

solution11
SOLUTION

Moreover,

hence,

and

solution12
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 :
solution13
SOLUTION

where

Mixing cup avg temp at duct outlet ?

solution14
SOLUTION

Again, we see that

Moreover,

solution15
SOLUTION

therefore

and

solution16
SOLUTION

Therefore,

solution17
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.

solution18
SOLUTION
  • Pressure Drop
  • We have:
  • Therefore
solution19
SOLUTION

But

Therefore

solution20
SOLUTION

From overall momentum balance:

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