Multiphase Chemical Reactor Engineering by 5wuVvO

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




   Fluidized Bed Reactor         Fixed Bed Reactor




Slurry Bubble Column Reactor   Trickle Column Reactor
Batch Reactor
         Fixed Bed Rector




Fixed Bed Reactor that converts sulfur in diesel fuel to H2S
      Fluidized Bed Reactor




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




         Stirring Apparatus
      Straight Through Transport
                Reactor

                                                              Riser
Settling
Hopper




Standpipe


            The reactor is 3.5 m in diameter and 38 m tall.
            Sasol/Sastech PT Limited
Slurry Phase Distillate Reactor
Packed Bed Reactor
CSTR
            Hand holes for charging
            reactor



             Connection for heating
             or cooling jacket




       Agitator
           Plug Flow Model
         CA,out
                             H
                          t
                             V

                         C A ,in  C A ,out
                     Particle surrounding by fluid of
                      essential constant concentration,
         CA,in        CA,m

Gas + solids
Batch Mix Flow: Charge Reactor

                Residence time distribution

                Particle stays in the reactor for
                 certain length of time
Countercurrent Flow

        If solids are moving plug flow and
         we have constant flow composition
          CA
        Residence time of solids:
                              H
                           t
                              V

        Heat Effects !!
          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=∞
               Non-Reacting

1.   Small particles  highly conductive particles

2.   Small particles  volumetric reaction
           1) Small Particles:
        Highly Conductive Particles

                 Particle initially at uniform T = Tp
      Tp         At t = 0, we drop it into our furnace




Fluid at Tf
                     Energy Balance
                                                      dH
                    Qconvection  Qradiation       m
                                                       dt
 Heat in by convection and radiation = change in enthalpy of particle

                                                                   d C pTp 
     2
         
4R hcv T f  Tp    m T  T             w
                                               4      4
                                                      p     m      dt
Where,
             Area of sphere = 4πR2
             Hcv = convection coefficient
             σ = Stefan-Boltzman constant
             Єm = emissivity of the particle (wall has Є = 1)
      Energy Balance

      hr   m
                        T w
                            4
                                 Tp4   
                        T F     Tp    

hcv  hr   TF  Tp  
                                   m p C p dTp
                                       A     dt

    Can solve this equation to get Tp =f(t)
                        Find hcv
   Have film: ∆H Tf ≠ Tp
   Use mass transfer analogy to get hcv

           hcvd p                                1    1
                     Nu  2  0.6 Rep Pr        2    3

             kf

                       Vd p            Cp        
               Rep           ; Pr          ; 
                                      kf         
                      2. Small Particles:
                     Volumetric Reaction
   Small such that no internal
    gradients

Heat generated by reaction = Heat transferred to surrounding


      V p  rAv    H r   hAp Tp  T f                   
Steady State:




      Volume of
      particle  Rate of                                      Exothermic Rxn:
                               H r    rAv    R 
                reaction

           T              
                                                             -∆Hr = (+)

                p    Tf                                   -rAv = (+)
                                        h           3
         3. Large Particles:
Possible Internal Particle Gradients
   We have to solve the conduction equation
   Non reacting particle: the conduction equation for sphere:
                 1   2 T               T
                      r ke      C p ,s
                r r 
                  2
                            r            t
 Surface :

                                     
                                            Ke = effective thermoconductivity
    dT
 ke              h T f  Tp                      within the particle
    dr                         r R         ∂T/∂r = 0 at steady state
         r R

                        Heat transferred into particle
Heat conducted into
                        Note: accommodate radiation in the
particle at r =Rp
                        definition of h if that is the case
               Boundary Conditions
                     T
                        0
Symmetry condition   r    r 0



Initial condition    t  0;Tp  Tp ,0 ;T ( r )  Tp

Internal gradient    Tr  0  Tr  R0
                     Tr   r 0
                                  Tr   r  R0


External gradient     T f  Tr          r R
              Reacting Systems
   General equation for volumetric reactions
    (Reaction in porous particles)

   Recall continuity equation:
    continuity for A
     Solve (1), (2), (3) Together

 Continuity for A
  C A 1   2 C A 




                                                    Coupled through the
      2  r De      rAv                   (1)
   t  r r     r 




                                                    reaction rate
 Energy balance
              T 1   2 T 
 C p 1     2  r ke   rAv  H r    (2)
               t r r  r 
 kr C A CSn   rAv 
      m


  k r T C C   rAv 
         m   n                                (3)
         A   S
                         In Steady State
   Showed that for steady conditions:


                        ke
                            dT
                                De
                                    dCA
                                         H r 
                            dr       dr
         Integrate at r = 0, r = R

         For sphere

              Tr 0  TS   C A,s  C A ,r 0   H r 
                             De
                             ke
                                     TS  Tr   r R
                         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.
Shrinking Core: Non-Isothermal

   Heat generated at reaction front, not throughout the volume
     1   2 T               T
          r ke      C p ,s                                              Tc

    r r 
      2
                r            t                                                Ts

   In Steady State,                                               rc

    k e   2 T                                                       r             Tf
          r     0
    r r  r 
      2
                                                               R

   Solve

    T  Tc
            
                          1
                           rc   1
                                 r      ;   dT   Tc  Ts
                                                 1 1 2
    Tc  Ts            
                       1
                       R         r1c       dr   
                                                  R  rc r 
T Conditions

Tc  T r  r
           c


Ts  T r  R
T  T r r
Boundary Condition 1: r = rc

Heat is generated = Heat conducted out through product layer
       Area



akr CS ,0C A ,c  H r   ke
                                dT
                                dr          r  rc

           akr CS ,0C A ,c  H r   1 1 
TC  TS                               
                                      R r 
                     ke                  c 
Boundary Condition 2: r = R

 Heat arriving by conduction    =   Heat removed for
 from within particle               convection



                            hTS  T f 
               dT
           ke
               dr   r R

                                ke                      
         TS  T f  TC  TS          1               
                                         1
                                               
                                hR  R R  r1c
                                                         
                                                           
                 Can be obtained
                                      Bi-1
                 from B.C. 1
                             Solution
   Combine equations and eliminate TS to get Tc-Tf




               TC  T f
                                   akr CS ,0C A,c  H r  rc2
        1  1  1  1
                 
        ke  rc R  hR 2
     Recall from Isothermal SC
               Model
                    De 
                                2
C A ,c
                   akr CS ,0 rc 
C A, f          De  1            1 1
           1             1       
            ak C r  r  Bi  R
               r s ,0 c  c        m 


             Substitute CA,c into (Tc –Tf) equation
                        Tc - Tf

    TC  T f                       C A , f  H r 
                    
1 1 1 1               1 1 1        1        1
      2                  
                            r R  ak C r 2  k R 2
ke  rc R  hR
                      De  c      r S ,0 c  m




  Conduction   Convection   Diffusion in    Reaction   Mass
                            Product Layer              Transfer
     Can Heat Transfer Control the Rate
       in Endo- and Exothermal Rxn?
     Consider CA,c ≈ CA,f; initially rapid reaction

a)    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.

b)    Exothermic
      initial rapid reaction and with poor Q, TC will 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.
Fixed Bed Reactor
                           Fixed Bed Reactor

              Solids take part in reaction  unsteady state or semi-batch mode
              Over some time, solids either replaced or regenerated
                       CA,out
Regeneration
                                                                      Breakthrough
                                                                      curve




                                                     CA,out/CA,in
                       1                   2




                           CA,in                                       t
          Isothermal Reaction:
           Plug Flow Reactor
   Plug flow of fluid – no radial gradients, and no
    axial dispersion

   Constant density with position

   Superficial velocity remains constant
Plug Flow Model



z + dz                        CA,f + dCA,f


    z                         CA,f




                                     U0 
                                                
                                             Vgas m 2 / s   
     U0 (m/s) superficial velocity
                                                      
                                               Axs m 2
                        Mass Balance
                Input – Output – Reaction = Accumulation


            
U 0C A, f  U 0 C A, f                                  
                           dC A, f    rAv   dz     C A, f  z 
                                                          t
 Divide by ∂z and take the limits as ∂z  0


                     C A , f         C A , f
                               U0               rAv  0
                       t                 z

  ε is void fraction in bed
                                                              Void fraction
For first order reaction, fluid only:

                            1 dN A
                                      kv 1   C A , f
                  mol
        rAv  3                       ''

             m reactor  s  Vr dt
For steady state:
                                                             Volume of reactor
                                 C A , f
                                            0
                                   t
Therefore,
                      dC A , f
                U0                kv' 1   C A , f  0
                                    '

                         dz
Conversion as a function of Height
 Integrating with CA,f = CA,f,in at z = 0




                        C A, f                kv' 1    
                                                 '
       X A  1                      1  exp 
                                                          z
                                                            
                       C A, f ,in                 U0       



          Note 1: Same equation as for catalytic reactor with 1st order reaction
          Note 2: Can be used in pseudo-homogeneous reaction
                Balance on Solid
   aA (fluid) + S (solid)  Products
   Input – Output – Reaction = Accumulation
   Over increment of dz: input = 0, output =0

                                  Cs
            rsv  z  1     z
                                   t

            Volume fraction of solid =
                                             mol
                 m3 of solid
                                         m3 of solid · s
             m3 of reactor volume
Balance on Solid

            C s
    1     rsv  0
             t


    rav   a   rsv 

    C s   rAv
                  0
     t a 1   
                  Solve These Equations

= 0 (In quasi steady state, we ignore the
  accumulation of A in gas)

    C A , f            C A , f               C A, fa1    Cs
              U0                  rAv  0                       0
       t                  z                    z     U 0 t

                                                     C 'A , f   Cs '
                                                                       0
                                                       z         t
            Cs   rAv
                         0                         C'A , f  f z ,t 
             t a1   
                                                     C s  f  z ,t 
                                                       '
a)   Shrinking Core Model
b)   Uniform reaction in porous particle, zero order in fluid
c)   Uniform reaction, 1st order in fluid and in solid
d)   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)
e)   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)
         a) In Shrinking Core Model

                    rAv  akv 1   C A ,c CS ,o
   Recall that            De                                  For SCM
C A ,c               akvCS ,o rc2                                                  3
                                                                          rc 
C A, f     
           1 
                 De  1 
                           1 
                                  1 1                         Cs  Cs ,o  
            ak C r  r  Bi  R                                         R
               r S ,o c  c      m 


                              rc R 3
   Solid Phase
                          rc2     kv C A ,c rc  0                       Solve

                              t
                                                                          CA,f = f(z)
                                  3                                       rc = f(z,t)
   Liquid Phase           C A , f
                     U0               akv 1   Cs ,oC A ,c  0
                            z
          Conversion vs Time

    t=0              t>0




z
Overall Conversion of Solid

                   3
              rc 
           L

             R  dz
           0     
                              L
                           1
                             3  c
1 X s        L
                                 3
                                r dz
                          LR 0
                dz
               0
Height Vs time (Graphical)
                 Unreacted
      z/L        bed depth




                             Reaction
All CA has                   zone
been
reacted


                                        Completely
                                        reacted


                                                     t/
             Particles at bed
             entrance are
             completed reacted
b) Uniform Reaction in Porous Particle
       and Zero Order in Fluid
                                                  CS
                                         1 X S 
      k 1  X S 
dX S                            where             C0
 dt                                                    1
                                              dX S        dCS
                                                      CS ,0

               U0       C A , f
                                 kCS  0
             a1    z
             CS
                   kCS  0
              t
c) Uniform Reaction and 1st order in
         Fluid and in solid

    rAv  akv 1   C ACS
    rAv  akv 1    CS


        C A , f
   U0               akv 1    C A ,s CS  0
          z
    C A ,s
             akv 1    C A ,s CS  0
     t
     Non-Isothermal Packed Bed
              Reactor
   For mass continuity  did balance on fluid and
    on solid
   For energy balance, we do balance on each
    phase
    Non-Isothermal Packed Bed
             Reactor
    Assumptions:
    1) Adiabatic reaction – no heat lost through shell to
       surroundings (no radial temperature gradients) q
       =0
    2) Biλ is small – uniform T within particle (an
       exothermic reaction Tp > Tg)
    3) Plug flow of gas and use Tref =0 for enthalpy
       calculations
    4) Assume an average density can be used (ρg =
       constant)
           Modeling

                                q =0



Tf + dTf               z + dz
   Tf                   z




           Tf,0
                           kg 
                  U0
                         G 2   U0g
                          m s
Moving Bed Reactor
     Solids in
         Vs
                               Gas out



                        ∆z



                 U0

                               Gas in




                  Solids out
      Moving Bed Reactor (MBR)
   Steady state reactor where solids moving at near their packed
    bed voidage
   Counter or co-current operation
   Solid usually move downward (vertical shaft reactor or
    furnace)
   Voidage is near that of a packed bed
      Slightly above random loose-packed voidage
   Solids move mainly in a plug floe, but region near wall have a
    velocity distribution
             Advantages of MBR
   True counter-current flow
   Uniform residence time (essentially plug flow)
   Reasonable ∆P
   Throughput variable
   Generally larger particle dp > 2-3 mm
   Difficulties coping with wide size distribution of
    particles (fines tend to block up the void spaces)

								
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