The document discusses the design of catalyst reactors accounting for catalyst deactivation. It begins by introducing fixed bed and fluidized bed reactors. It then discusses criteria for selecting between these reactors, including catalyst deactivation behavior and reaction conditions. The document goes on to provide steps for designing catalyst reactors and single adiabatic packed bed reactors. It also discusses models for fluidized bed reactors and approaches for designing reactors to account for catalyst deactivation over time.

1. SEMINAR ON :
DESIGN OF CATALYST
REACTOR
WITH DEACTIVATION
CATALYTIC REACTION
ENGINEERING (HCE23)
DONE BY:
KISHAN KASUNDRA
1/8/20181
CHEMICAL ENGINEERING
DEPARTMENT,DSCE

2. INTRODUCTION:-
Reactant gas can be made to contact solid catalyst in many
ways, and each has its specific advantages and
disadvantage below figures illustrate a number of the
contacting patterns.
These may be divided into two broad types
1) The fixed bed reactors
2) The fluidized-bed reactors
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3. FIG :- FIXED BED REACTORS
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4. FIG :- FLUIDISED BED REACTOR
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5. COMPARISION BETWEEN FIXED BED AND
FLUIDISED BED REACTOR:-
FIXED BED REACTOR FLUIDIZED BED REACTOR
In fixed-bed reactors, the catalyst
remains essentially stationary until it is to
be reactivated or discarded.
In fluidized-bed reactors, the catalyst
creates bed which is not stationary.
Effective temperature control of large
fixed beds can be difficult.
Temperature control of fluidized bed is
easier than fixed bed.
Fixed beds cannot use very small sizes
of catalyst because of plugging and high
pressure drop.
Fluidized beds are well able to use small
size particles.
These types of reactors give low cost
and minimal maintenance.
It requires maintenance frequently.
It is not easier to replace the catalyst. It is easier to regenerate or replacement
of the catalyst.
Fixed beds are favored for reactions that
involve high pressure.
Fluidized beds are favored for strongly
exothermic or endothermic reactions.
1/8/20185
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DEPARTMENT,DSCE

6. SELECTION CRITERIA FOR CATALYST
REACTORS:-
Catalyst deactivation behavior and regeneration policy:
a. Fixed beds are favored if the life of the catalyst is longer than about 3 months.
b. Fluidized or slurry beds are favored in process situations that involve rapid
deactivation and the need to regenerate the catalyst.
Reaction conditions:
a. Fixed beds are favored for high pressures;
b. Fluid or slurry beds are favored for strongly exothermic or endothermic reactions
as they facilitates effective heat transfer and temperature control.
Catalyst strength and attrition resistance:
a. Fixed-bed catalysts must be strong enough to avoid being crushed at reactor
bottom.
b. In fluidized or slurry bed reactors, attrition of just a few percent of the total catalyst
charge per day leads to uneconomical loss of catalyst.
Process economics:
a. Capital cost depends on the complexity of design, cost of materials, reactor
fabrication, and catalyst cost 1/8/2018
CHEMICAL ENGINEERING
DEPARTMENT,DSCE

7. DESIGN OF CATALYST REACTOR:-
A practical sequence to achieve the optimal combination of catalyst and reactor
is as follows:
1. For various candidate reactor types that seem promising, to specify the
reactor size (including the amount of catalyst needed), the concentration,
temperature profiles, the quantity of heat that must be added or removed, and
the rate of deactivation.
2. Choose among the candidates, using these criteria:
(a) Minimize the volume of catalyst, and therefore reactor size, required. For
instance, in the case of an irreversible first-order reaction, a tubular reactor
requires a lesser volume than does either a slurry reactor or a fluidized bed
reactor.
(b) Provide for efficient heat transfer when dealing with strongly exothermic or
endothermic reactions. Slurry or fluid-bed reactors are attractive under these
circumstances.
(c) Also for strongly exothermic or endothermic reactors, consider reactors or at
least catalyst trays in series, with inter-stage heating or cooling.
(d) In situations in which the catalyst becomes deactivated rapidly (in seconds1/8/20187
CHEMICAL ENGINEERING
DEPARTMENT,DSCE

8. 3. Choose the catalyst type and form that will:
(a) Maximize activity, selectivity and stability.
(b) Minimize pressure drop and maximize access of reactants to the porous catalyst
interior.
(c) Be compatible with the reactor- design needs; for instance, with high thermal
conductivity for highly endothermic or exothermic reactions, and with sufficient
mechanical strength so that catalyst at the bottom of the reactor can withstand the
full weight of material above it.
4. Choose the reactor-catalyst combinations that will minimize capital cost and
overall production costs. In this connection, be aware that the price of a catalyst is
usually a relatively minor consideration in comparison to its activity, selectivity, and
stability.
1/8/20188
CHEMICAL ENGINEERING
DEPARTMENT,DSCE

9. DESIGN OF SINGLE ADIABATIC PACKED BED
REACTOR:-
Find out conversion for different temperature for process and also find rate of
reaction and plot a graph of conversion vs temperature.
Now, find out slope of adiabatic line is;
Slope = Cp/-∆Hr
Tabulate XA , -rA’ and 1/(-rA’).
Then, plot 1/(-rA’) vs XA and find the area under the curve.
Then amount of catalyst needed W = FA0 X area
Where, FA0 = Feed rate
Now, find out heat duty at the reactor and after the reactor the temperature of all
flowing streams and recommend the design for reaction.
1/8/20189
CHEMICAL ENGINEERING
DEPARTMENT,DSCE

10. DESIGN OF FLUIDIZED BED REACTOR:-
In our development we make two questionable assumptions:
1) We ignore the flow of gas through the cloud since the cloud volume is very small for
fast bubbles.
2) We ignore the flow of gas, either up or down, through the emulsion since this flow is
much smaller than the flow through the bubbles.
Assume first order reaction; Let the reaction be
A → R, -rA’’’ = k’’’ CA , mol/m3s.s
Then for any slice of bed we have
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11. Here, these five resistances in series-parallel, eliminating cloud and emulsion
concentrations, and integrating from the bottom to the top of the bed gives,
Where,
fb = volume of solid in bubble/volume of bed = 0.001~0.01 (rough approx from
experiment)
k’’’ = effective rate constant for fluidized bed
δ = bed fraction in bubbles, m3 bubbles/m3 bed
Kbc = interchange volume between bubble and cloud/volume of bubble
fc = volume of solids in cloud and wake/volume of bed
fe = volume of the solids in the rest of the emulsion/volume of bed
Kce = interchange volume between cloud and emulsion/volume of bubble
ftotal = volume of solids
HBFB = W/A ρ (1-εf)
εf = fraction of void in the bed under bubbling fluidized bed conditions
1/8/2018

12. We also find that the average gas composition seen by the solids is,
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13. DESIGN OF DEACTIVATION OF CATALYST:-
The rate at which the catalyst pellet deactivates can be written as
(Deactivation rate) = f(main stream temperature). f(main stream concentration).
f(present state of catalyst pellets)
for deactivation which in general is dependent on the concentration of gas phase
species
Where d is called the order of deactivation, m measures the concentration
dependency and Ed is the activation energy or temperature dependency of the
deactivation.
1/8/201813
CHEMICAL ENGINEERING
DEPARTMENT,DSCE

14. If deactivating catalyst the conversion change slowly with time, then the average
conversion during a run can be found by calculating the steady-state conversion at
various times and summing over time. In symbols, then,
There are two important and real problems with deactivating catalysts.
The operational problem: how to best operate a reactor during a run. Since
temperature is the most important variable affecting reaction and deactivation this
problem reduces to finding the best temperature progression during the run.
The regeneration problem: when to stop a run and either discard or regenerate
the catalyst. This problem is easy to treat once the first problem has been solved
for a range of run times and final catalyst activities.
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15. There are three possible reactor policies for a batch of deactivating catalyst.
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DEPARTMENT,DSCE

16. There are two performance equation which can be useful to solve for average
conversion,
From figure we can see that for case K, a decreases with time, and so does XA
For case L, a decreases with time so that the feed rate must also be lowered.
Nearly always we find case M is optimum. And above equation becomes,
So k’aCA0, should remain constant as the temperature is raised. This means that in
the regime of no pore diffusion resistance we must change T with time such that1/8/201816
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17. Asas
In the regime of strong resistance to pore diffusion we must change T with time
such that
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18. THANK YOU!!
1/8/201818
CHEMICAL ENGINEERING
DEPARTMENT,DSCE