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

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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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Module 6 Lecture 29

Mass Transport: Illustrative Problems

Dr. R. Nagarajan

Professor

Dept of Chemical Engineering

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:

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therefore

therefore

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d. L=? We will need Re

Now:

therefore (laminar-flow regime)

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For a square channel

and (used below).

If then the mass-transfer analogy is:

SOLUTION

where

We estimate at which

If

then

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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

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Moreover,

hence,

and

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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 ?

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Again, we see that

Moreover,

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therefore

and

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Therefore,

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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: