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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 4R 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 hTS 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, fa1 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 a1 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 a1 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 U0g 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)