Design of Reactors Presentation Module Useful

Design of Reactors_Module9

1. DESIGN OF REACTORS MODULE 9

2. Types of Reactors Batch No flow of material in or out of reactor Changes with time Fed- Batch Either an inflow or an outflow of material but not both Changes with time Continuous Flow in and out of reactor Continuous Stirred Tank Reactor (CSTR) Plug Flow Reactor (PFR) Steady State Operation

3. Mass Balance on Reactive System In - out + gen - cons = accumulation A mass balance for the system is Where N A is the mass of “A” inside the system.

4. Generalized Design Equation for Reactors In - out + gen - cons = accumulation

5. Batch Reactor Generalized Design Equation No flow into or out of the reactor, then, F A = F A0 = 0 Good mixing, constant volume or

6. Fed Batch Reactor Reactor Design Equation No outflow F A = 0 Good Mixing r A dV term out of the integral

7. Continuous Stirred Tank Reactor rate of flow in = rate of flow out F A = v C A and F A0 = v C A0 v = volumetric flow rate (volume/time)

8. CSTR Contd ……. General Reactor Design Equation Assume Steady State Well Mixed So or

9. Plug Flow Reactor (PFR) Tubular Reactor Pipe through which fluid flows and reacts. Poor mixing Difficult to control temperature variations. An advantage is the simplicity of construction.

10. PFR Design Equation Design Equation Examine a small volume element ( D V) with length D y and the same radius as the entire pipe. If the element is small, then spatial variations in r A are negligible, and Flow of A into Element Flow of A out of Element Assumption of “good mixing” applies only to the small volume element

11. PFR Design Equation If volume element is very small, then assume steady state with no changes in the concentration of A. Simplify design equation to: r A is a function of position y, down the length of the pipe and reactant concentration The volume of an element is the product of the length and cross-sectional area, D V = A D y Design Equation becomes:

12. PFR Design Equation take the limit where the size of a volume element becomes infinitesimally small or because ∆ y A = V, This is the Design Equation for a PFR

13. Mole Balance Reactors

14. Mole Balance Table

15. Conversion The basis of calculation is always the limiting reactant. Consider the general equation The conversion X of species A in a reaction is equal to the number of moles of A reacted per mole of A fed, ie

16. Conversion Maximum value of conversion: irreversible reactions X = 1 reversible reactions X= X e

17. Design Equations Batch Reactor: Differential Design Equation Integral Design Equation

18. Design Equations CSTR: Design Equation

19. Design Equations Plug Flow Reactor: Differential Design Equation Integral Design Equation

20. Design Equations Packed Bed Reactor: Differential Design Equation Integral Design Equation

21. Isothermal Reactor Design Design Procedure: Find Mole balance design equation Find Rate law Find Stoichiometric equation Combine above three to get final equation Evaluate

22. Isothermal Reactor- liquid phase The elementary liquid phase reaction 2A B is carried out isothermally in a CSTR. Pure A enters at a volumetric flow rate of 25 dm 3 /s and at a concentration of 0.2 mol/dm 3 . What CSTR volume is necessary to achieve a 90% conversion when k = 10 dm 3 /(mol*s) ? Solution Mole Balance Rate Law

23. Solution contd …. Stoichiometry Combining

24. Solution contd …. Evaluating at x = 0.9 V = 1125 dm 3 Calculation of space Time:

25. Reactor Design – Gas Phase Reaction Gas Phase Elementary Reaction : 2A B P = 8.2 atm T = 500 K C A0 = 0.2 mol/dm 3 k = 0.5 dm 3 /mol-s v o = 2.5 dm 3 /s X = 90%

26. Solution Design For Batch Reactor:

27. Solution Continued…. Design For CSTR: Mole Balance: Rate Law: Stoichiometry : Combining:

28. Solution Continued…. Evaluation:

29. Soln Contd …Design for PFR Design For PFR: Mole Balance: Rate Law: Stoichiometry : Combining:

30. Contd …Design for PFR Integrating : Evaluating :

31. Reactor Design- Reversible Reaction systems Given that the system is gas phase and isothermal, determine the reactor volume when X = 0.8 X e . Solution : equilibrium constant, K C , for this reaction K C = C Be / C 2 Ae

32. Solution Contd ….

33. Solution Contd …. X = 0.8X e = 0.711

34. Selectivity and Yield

35. Parallel Reactions

36. Reactor Sizing Involves determination of reactor volume to achieve a given conversion or determine the conversion that can be achieved in a given reactor type and size Given - r A as a function of conversion,-r A =f(X), one can size any type of reactor to find - r A = f(X), construct Levenspiel plot Plot F Ao /- r A as a function of X The volume of a CSTR and the volume of a PFR can be represented as the shaded areas in the Levenspiel Plots

37. Reactor Sizing

38. Reactor Sizing Using the Ideal Gas Law to Calculate C A0 and F A0 The entering molar flow rate for a gas is where C A0 = entering concentration, mol/dm 3 y A0 = entering mole fraction of A P = entering total pressure, e.g., kPa P A0 = y A0 P = entering partial pressure of A, e.g., kPa T = entering temperature, K v = volumetric flow rate

39. Reactor Sizing A gas of pure A at 830 kPa (8.2 atm ) enters a reactor with a volumetric flow rate, v , of 2 dm 3 /s at 500 K. Calculate the entering concentration of A, C A0 , and the entering molar flow rate, F A0 . F A0 = 2 * 1* 830 / (8.314 * 500) = 0.399 mol/s C A0 = 0.399/2 = 0.199 mol/dm 3

40. Reactor Sizing Levenspiel Plots in Terms of Concentrations For a plug-flow reactor , Writing the above equation in terms of concentration,C A