Multiphase Chemical Reactor Engineering

1. Multiphase Chemical Reactor Engineering Quak Foo Lee Ph.D. Candidate Chemical and Biological Engineering The University of British Columbia

2. Different Types of Reactor Fluidized Bed Reactor Fixed Bed Reactor Slurry Bubble Column Reactor Batch Reactor Trickle Column Reactor

3. Fixed Bed Rector Fixed Bed Reactor that converts sulfur in diesel fuel to H2S

4. Fluidized Bed Reactor Fluidized Bed Reactor using H2SO4 as a catalyst to bond butanes and iso-butanes to make high octane gas

5. Batch Reactor Stirring Apparatus

6. Straight Through Transport Reactor Riser Settling Hopper Standpipe The reactor is 3.5 m in diameter and 38 m tall. Sasol/Sastech PT Limited

7. Slurry Phase Distillate Reactor

8. Packed Bed Reactor

9. CSTR Hand holes for charging reactor Connection for heating or cooling jacket Agitator

10. Plug Flow Model • Particle surrounding by fluid of essential constant concentration, CA,m CA,out CA,in Gas + solids

11. Residence time distribution Particle stays in the reactor for certain length of time Batch Mix Flow: Charge Reactor

12. If solids are moving plug flow and we have constant flow composition  Residence time of solids: Heat Effects !! Countercurrent Flow

13. Heat Effects on Reactions of Single Particles • Normally (developed) dealing with exothermic and endothermic reaction. • If reaction occurs at a rate such that the heat absorbed (endothermic) or generated (for exothermic) can’t be transferred rapidly enough, then non-isothermal effects become important: The particle T ≠ the fluid T • For exothermic reaction, Tp will increase and the rate of reaction will increase above that expected for the isothermal case. • Two conditions: • i) Film ∆T (external ∆T) Tf (bulk fluid) ≠ Tp (particle) • ii) Intraparticle ∆T (internal ∆T) Tr=Rp ≠ Tr=∞

14. Non-Reacting • Small particles  highly conductive particles • Small particles  volumetric reaction

15. Small Particles: Highly Conductive Particles • Particle initially at uniform T = Tp • At t = 0, we drop it into our furnace Tp Fluid at Tf

16. Energy Balance Heat in by convection and radiation = change in enthalpy of particle Where, Area of sphere = 4πR2 Hcv = convection coefficient σ = Stefan-Boltzman constant Єm = emissivity of the particle (wallhas Є = 1)

17. Energy Balance Can solve this equation to get Tp =f(t)

18. Find hcv • Have film: ∆H Tf ≠ Tp • Use mass transfer analogy to get hcv

19. 2. Small Particles: Volumetric Reaction • Small such that no internal gradients Heat generated by reaction = Heat transferred to surrounding Steady State: Volume of particle Rate of reaction Exothermic Rxn: -∆Hr = (+) -rAv = (+)

20. 3. Large Particles:Possible Internal Particle Gradients • We have to solve the conduction equation • Non reacting particle: the conduction equation for sphere: Ke = effective thermoconductivity within the particle ∂T/∂r = 0 at steady state Heat transferred into particle Note: accommodate radiation in the definition of h if that is the case Heat conducted into particle at r =Rp

21. Boundary Conditions Symmetry condition Initial condition Internal gradient External gradient

22. Reacting Systems • General equation for volumetric reactions (Reaction in porous particles) • Recall continuity equation: continuity for A

23. Solve (1), (2), (3) Together Continuity for A (1) (2) (3) Coupled through the reaction rate Energy balance

24. In Steady State • Showed that for steady conditions: Integrate at r = 0, r = R For sphere

25. Some Notes • If we know CA,s (surface concentration) and CA,r=0 (CA within pellet at r = 0), we can calculate temperature gradient, previous equation tell us either we need or don’t need to worry about T gradient within particle. • Where isothermal (approach) approximation can be used and where internal T gradients must be considered. • Volumetric reaction for porous particles, heat is generated in a volume.

26. Shrinking Core: Non-Isothermal • Heat generated at reaction front, not throughout the volume • In Steady State, • Solve Tc Ts rc r Tf R

27. T Conditions

28. Boundary Condition 1: r = rc Heat is generated = Heat conducted out through product layer Area

29. Boundary Condition 2: r = R Heat arriving by conduction = Heat removed for from within particle convection Can be obtained from B.C. 1 Bi-1

30. Solution • Combine equations and eliminate TS to get Tc-Tf

31. Recall from Isothermal SC Model Substitute CA,c into (Tc –Tf) equation

32. Tc - Tf Conduction Convection Reaction Mass Transfer Diffusion in Product Layer

34. Can Heat Transfer Control the Rate in Endo- and Exothermal Rxn? • Consider CA,c≈CA,f; initially rapid reaction • Endothermic with poor heat transfer, heat will be consumed in reaction, and if can’t transfer heat in, TC will drop  reaction rate ↓ markedly and rate of reaction become the slow step occurring at a rate dictated by the flow of heat. • Exothermic initial rapid reaction and with poor Q, TCwill increased, then rate of reaction ↑ and eventually reach point where gaseous reactant can’t be transferred fast enough (external mass transfer or diffusion). Hence rate is limited.

35. Fixed Bed Reactor

36. CA,out Regeneration 1 2 CA,in Fixed Bed Reactor • Solids take part in reaction  unsteady state or semi-batch mode • Over some time, solids either replaced or regenerated Breakthrough curve CA,out/CA,in t

37. Isothermal Reaction:Plug Flow Reactor • Plug flow of fluid – no radial gradients, and no axial dispersion • Constant density with position • Superficial velocity remains constant

38. Plug Flow Model z + dz CA,f + dCA,f z CA,f U0 (m/s) superficial velocity

39. Mass Balance Input – Output – Reaction = Accumulation Divide by ∂z and take the limits as ∂z  0 ε is void fraction in bed

40. Void fraction For first order reaction, fluid only: For steady state: Volume of reactor Therefore,

41. Conversion as a function of Height Integrating with CA,f = CA,f,in at z = 0 Note 1: Same equation as for catalytic reactor with 1st order reaction Note 2: Can be used in pseudo-homogeneous reaction

42. Balance on Solid • aA (fluid) + S (solid)  Products • Input – Output – Reaction = Accumulation • Over increment of dz: input = 0, output =0 Volume fraction of solid = m3 of solid m3 of reactor volume mol m3 of solid · s

43. Balance on Solid

44. Solve These Equations = 0 (In quasi steady state, we ignore the accumulation of A in gas) Substitute rAv

45. Shrinking Core Model • Uniform reaction in porous particle, zero order in fluid • Uniform reaction, 1st order in fluid and in solid • Park et al., “An Unsteady State Analysis of Packed Bed Reactors for Gas-Solid Reactions”, J. Chem. Eng. Of Japan, 17(3):269-274 (1984) • Evans et al., “Application of a Porous Pellet Model to Fixed, Moving and Fluid Bed Gas-Solid Reactors”, Ind. Eng. Chem. Proc. Des. 13(2):146-155 (1974)

46. a) In Shrinking Core Model Recall that For SCM Solid Phase Liquid Phase Solve CA,f = f(z) rc = f(z,t)

47. Conversion vs Time t = 0 t > 0 z

48. Overall Conversion of Solid

49. t/ Height Vs time (Graphical) Unreacted bed depth z/L Reaction zone All CA has been reacted Completely reacted Particles at bed entrance are completed reacted

50. b) Uniform Reaction in Porous Particleand Zero Order in Fluid where