DESIGN OF CATALYST REACTOR WITH DEACTIVATION
Download as PPTX, PDF9 likes3,094 views
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.
DESIGN OF CATALYST REACTOR WITH DEACTIVATION
- 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 1/8/20182 CHEMICAL ENGINEERING DEPARTMENT,DSCE
- 3. FIG :- FIXED BED REACTORS 1/8/20183 CHEMICAL ENGINEERING DEPARTMENT,DSCE
- 4. FIG :- FLUIDISED BED REACTOR 1/8/20184 CHEMICAL ENGINEERING DEPARTMENT,DSCE
- 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 CHEMICAL ENGINEERING 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 1/8/201810
- 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, 1/8/201812 CHEMICAL ENGINEERING DEPARTMENT,DSCE
- 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. 1/8/201814 CHEMICAL ENGINEERING DEPARTMENT,DSCE
- 15. There are three possible reactor policies for a batch of deactivating catalyst. 1/8/201815 CHEMICAL ENGINEERING 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 CHEMICAL ENGINEERING DEPARTMENT,DSCE
- 17. Asas In the regime of strong resistance to pore diffusion we must change T with time such that 1/8/201817 CHEMICAL ENGINEERING DEPARTMENT,DSCE