Phases and Information. Moon phases There are 8 moon phases narcycles/lunarapplet.html
Multi Phases
Transcript of Multi Phases
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Multiphase ChemicalMultiphase Chemical
Reactor EngineeringReactor EngineeringQuak Foo Lee
Ph.D. Candidate
Chemical and BiologicalEngineering
The University of British Columbia
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Different Types of Different Types of
ReactorReactor
Fluidized Bed Reactor
Trickle Column Reactor Slurry Bubble Column Reactor Batch Reactor
Fixed Bed Reactor
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Fixed Bed RectorFixed Bed Rector
Fixed Bed Reactor that converts sulfur in diesel fuel to H2S
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Fluidized Bed ReactorFluidized Bed Reactor
Fluidized Bed Reactor using H2SO4 as a catalyst to bond butanes
and iso-butanes to make high octane gas
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Batch ReactorBatch Reactor
Stirring Apparatus
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Straight ThroughStraight Through
Transport ReactorTransport Reactor
Riser
Standpipe
Settling
Hopper
The reactor is 3.5 m in diameter and 38 m tall.
Sasol/Sastech PT Limited
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Slurry Phase DistillateSlurry Phase Distillate
ReactorReactor
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Packed Bed ReactorPacked Bed Reactor
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CSTRCSTR
Agitator
Connection for heating
or cooling jacket
Hand holes for charging
reactor
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CA,in
CA,out
Gas + solids
Plug Flow ModelPlug Flow Model
ρ V
H t =
out , Ain , A C C ≅
Particle surrounding byParticle surrounding by
fluid of essential constantfluid of essential constantconcentration,concentration, CC A,m A,m
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Countercurrent FlowCountercurrent Flow
If solids are moving plugIf solids are moving plug
flow and we have constantflow and we have constant
flow compositionflow composition
Residence time of solids:Residence time of solids:
Heat Effects !!Heat Effects !!
ρ V
H t =
AC
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Heat Effects on ReactionsHeat Effects on Reactions
of Single Particlesof Single Particles Normally (developed) dealing with exothermic and endothermicNormally (developed) dealing with exothermic and endothermic
reaction.reaction.
If reaction occurs at a rate such that the heat absorbed (endothermic)If reaction occurs at a rate such that the heat absorbed (endothermic)
or generated (for exothermic) can’t be transferred rapidly enough,or generated (for exothermic) can’t be transferred rapidly enough,
then non-isothermal effects become important:then non-isothermal effects become important:The particle T The particle T ≠ the fluid T≠ the fluid T
For exothermic reaction, TFor exothermic reaction, Tpp will increase and the rate of reaction willwill increase and the rate of reaction will
increase above that expected for the isothermal case.increase above that expected for the isothermal case.
Two conditions: Two conditions:
i) Filmi) Film ∆T (external ∆T) T ∆T (external ∆T) Tf f (bulk fluid) ≠ T(bulk fluid) ≠ Tpp (particle)(particle) ii) Intraparticle ∆T (internal ∆T) Tii) Intraparticle ∆T (internal ∆T) Tr=Rpr=Rp ≠ T≠ Tr=∞r=∞
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Non-ReactingNon-Reacting
1.1. Small particlesSmall particles highly conductivehighly conductive
particlesparticles
2.2. Small particlesSmall particles volumetricvolumetric
reactionreaction
1)1)S ll P i lS ll P ti l
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1)1)Small Particles:Small Particles:
Highly ConductiveHighly Conductive
ParticlesParticles Particle initially at uniform TParticle initially at uniform T
= T= Tpp
At t = 0, we drop it into ourAt t = 0, we drop it into ourfurnacefurnace
Fluid at Tf
Tp
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Energy BalanceEnergy Balance
dt
dH mQQ radiationconvection =+
( ) ( )[ ]( )dt
T C d mT T T T h R
p p
pwm p f cv =−∈+− 4424 σ π
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 (wall has Є = 1)
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Energy BalanceEnergy Balance
( )( ) p F
pw
mr T T
T T h
−−
∈=44
σ
( ) ( )dt
dT
A
C mT T hh
p p p
p F r cv
ρ
=−⋅+
Can solve this equation to get Tp =f(t)
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Find hFind hcvcv
Have film:Have film: ∆H T ∆H Tf f ≠ T≠ Tpp
Use mass transfer analogy to get hUse mass transfer analogy to get hcvcv
3
1
2
1
602 Pr Re. Nuk
d h p
f
pcv +==
ρ
µ ν
µ
ν === ;
k
C Pr ;
Vd Re
f
p p
p
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2. Small Particles:2. Small Particles:
Volumetric ReactionVolumetric Reaction Small such that noSmall such that no
internal gradientsinternal gradients
( ) ( ) ( ) f p pr Av p T T hA H r V −=−⋅− ∆
Heat generated by reaction = Heat transferred to surrounding
Steady State:
Volume of particle Rate of
reaction
( ) ( ) ( )
⋅
−⋅−=−
3
R
h
r H T T Avr f p
∆
Exothermic Rxn:
-∆Hr = (+)
-r Av = (+)
3 L P ti l3 L P ti l
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3. Large Particles:3. Large Particles:
Possible Internal ParticlePossible Internal Particle
GradientsGradients We have to solve the conduction equationWe have to solve the conduction equation
Non reacting particle: the conduction equation for sphere:Non reacting particle: the conduction equation for sphere:
t
T
C r
T
k r r r s , pe ∂∂
=
∂∂
∂∂
ρ 2
2
1
( ) Rr p f
Rr e T T hdr
dT
k
:Surface
== −=
Heat conducted into
particle at r =Rp
Heat transferred into particle
Note: accommodate radiation in the
definition of h if that is the case
Ke = effective thermoconductivity
within the particle∂T/∂r = 0 at steady state
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Boundary ConditionsBoundary Conditions
0
0
=
=∂
∂
r r
T
p , p p T )r ( T ;T T ;t === 00
00
00
Rr r r r
r r
T T
RT T
== ≠
=≠=
Symmetry condition
Initial condition
Internal gradient
External gradient Rr r f T T =
≠
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Reacting SystemsReacting Systems
General equation for volumetricGeneral equation for volumetric
reactionsreactions
(Reaction in porous particles)(Reaction in porous particles)
Recall continuity equation:Recall continuity equation:
continuity for Acontinuity for A
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Solve (1), (2), (3)Solve (1), (2), (3)
TogetherTogether
Av A
e A r
r
C Dr
r r t
C −
∂∂
∂∂
=∂∂ 2
2
1ε
( ) ( )
( ) Av
n
S
m
Ar
r Ave p
r C C k
H r r
T k r
r r t
T C
−=
−+
∂∂
∂∂
=∂∂
− ∆ε ρ 2
2
11
Continuity for A
Energy balance
( ) ( ) Av
n
S
m
Ar r C C T k −=
(1)
(2)
(3)
Coupledthroug
hthe
reaction
rate
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In Steady StateIn Steady State
Showed that for steady conditions:Showed that for steady conditions:
( )r A
ee H dr
dC D
dr
dT k ∆−=−
( ) ( ) ( )r r , A s , Ae
e
S r H C C k
D
T T ∆−⋅−=− == 00
Integrate at r = 0, r = R
For sphere
Rr r S T T =
=
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Some NotesSome Notes
If we knowIf we know CC A,s A,s (surface concentration) and(surface concentration) and CC A,r=0 A,r=0 ((CC A A
within pellet at r = 0), we can calculate temperaturewithin pellet at r = 0), we can calculate temperature
gradient, previous equation tell us either we need orgradient, previous equation tell us either we need or
don’t need to worry about T gradient within particle.don’t need to worry about T gradient within particle.
Where isothermal (approach) approximation can beWhere isothermal (approach) approximation can be
used and where internal T gradients must beused and where internal T gradients must be
considered.considered.
Volumetric reaction for porous particles, heat isVolumetric reaction for porous particles, heat is
generated in a volume.generated in a volume.
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Shrinking Core: Non-Shrinking Core: Non-
IsothermalIsothermal
Heat generated at reactionHeat generated at reaction frontfront, not throughout the volume, not throughout the volume
In Steady State,In Steady State,
SolveSolve
t
T C
r
T k r
r r s , pe ∂∂
=
∂∂
∂∂
ρ 2
2
1
02
2
=
∂∂
∂∂
r
T r
r r
k e
( )( ) ( ) 21111
11
r
T T
dr
dT ;
T T
T T
cc
c
r R
sc
r R
r r
sc
c
−−
=−
−=
−−
R
r
r c
Ts
Tf
Tc
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T ConditionsT Conditions
r r
Rr s
r r c
T T
T T
T T c
=
=
=
=
=
=
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Boundary Condition 1:Boundary Condition 1: r r == r r
cc
Heat is generated = Heat conducted out through product layer
Area
( )
( )
−⋅
−−
=−
−=−=
ce
r c , A ,S r
S C
r r
er c , A ,S r
r Rk
H C C ak
T T
dr dT k H C C ak
c
110
0
∆
∆
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Boundary Condition 2:Boundary Condition 2: r r == RR
Heat arriving by conduction = Heat removed for
from within particle convection
( )
( ) ( )
−
−−=−
−=−=
cr R
e
S C f S
f S
Rr
e
RhR
k
T T T T
T T hdr
dT k
11
1
Bi-1Can be obtained
from B.C. 1
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SolutionSolution
Combine equations and eliminateCombine equations and eliminate T T SS to getto get T T cc--T T f f
( ) 2
0
2
1111cr c , A ,S r
ce
f C r H C C ak
hR Rr k
T T ⋅−=+
−
− ∆
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T T cc
-- T T f f
( )
22
0
2
111111111
Rk r C ak Rr D
H C
hR Rr k
T T
mc ,S r ce
r f , A
ce
f C
++
−
−=
+
−
− ∆
Conduction Convection Diffusion in
Product Layer
Reaction Mass
Transfer
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Fixed Bed ReactorFixed Bed Reactor
Solids take part in reactionSolids take part in reaction unsteady state or semi-batch modeunsteady state or semi-batch mode
Over some time, solids either replaced or regeneratedOver some time, solids either replaced or regenerated
1 2
CA,in
CA,out
Regeneration
t
CA,out
/CA,in
Breakthroughcurve
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Isothermal Reaction:Isothermal Reaction:
Plug Flow ReactorPlug Flow Reactor
Plug flow of fluid – no radialPlug flow of fluid – no radial
gradients, and no axial dispersiongradients, and no axial dispersion
Constant density with positionConstant density with position
Superficial velocity remains constantSuperficial velocity remains constant
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( )f , A
' '
v A
r
Av C k dt
dN
V sreactor m
mol r ε −==
⋅
11
3
0→∂
∂
t
C f , A
For first order reaction, fluid only:
For steady state:
Therefore,
( ) 010
=−+ f , A
' '
v
f , AC k
dz
dC U ε
Volume of reactor
Void fraction
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Balance on SolidBalance on Solid
( )
( ) ( )
( )0
1
01
=−
+∂∂
−⋅=−
=+∂∂
−
ε
ε
a
r
t
C
r ar
r t
C
Av s
svav
sv s
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Solve These EquationsSolve These Equations
00 =+∂
∂
+∂
∂ Av
f , A f , A
r z
C
U t
C
ε
( )0
1=
−+
∂∂
ε a
r
t
C Av s
( )0
1
0
=∂
∂−−
∂
∂
t
C
U
a
z
C s f , A ε
= 0 (In quasi steady state, we ignore the
accumulation of A in gas)
S u b s t i t u
t e r A
v
( )
( )t , z f C
t , z f C t
C
z
C
'
s
'
f , A
'
s
'
f , A
=
=
=∂
∂+
∂
∂0
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a)a) Shrinking Core ModelShrinking Core Model
b)b) Uniform reaction in porous particleUniform reaction in porous particle, zero order, zero orderin fluidin fluid
c)c) Uniform reactionUniform reaction, 1, 1stst order in fluid and in solidorder in fluid and in solid
d)d) Park et al., “An Unsteady State Analysis of Park et al., “An Unsteady State Analysis of Packed Bed Reactors for Gas-Solid Reactions”,Packed Bed Reactors for Gas-Solid Reactions”,
J. Chem. Eng. Of Japan, 17(3):269-274 (1984) J. Chem. Eng. Of Japan, 17(3):269-274 (1984)
e)e) Evans et al., “Application of a Porous PelletEvans et al., “Application of a Porous Pellet
Model to Fixed, Moving and Fluid Bed Gas-Model to Fixed, Moving and Fluid Bed Gas-Solid Reactors”, Ind. Eng. Chem. Proc. Des.Solid Reactors”, Ind. Eng. Chem. Proc. Des.13(2):146-155 (1974)13(2):146-155 (1974)
) I Sh i ki C) I Sh i ki C
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a) In Shrinking Corea) In Shrinking Core
ModelModel
( ) o ,S c , Av Av C C ak r ε −= 1
R Bir r C ak
D
r C ak
D
C
C
mcco ,S r
e
co ,S v
e
f , A
c , A
111
11
2
−−
+
=3
= Rr C C c
o , s s
Recall that
03
3
2
=+∂∂ cc , Avc
c r C k R
t
r r
( ) 010
=−+∂
∂c , Ao , sv
f , AC C ak
z
C U ε
Solid Phase
Liquid Phase
For SCM
SolveCA,f = f(z)
r c = f(z,t)
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O ll C i fO ll C i f
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Overall Conversion of Overall Conversion of
SolidSolid
∫
∫
∫ =
=− L
c L
L
c
s dz r LR
dz
dz R
r
X 0
3
3
0
0
3
11
H i ht V tiH i ht V ti
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Height Vs timeHeight Vs time
(Graphical)(Graphical)
z/L
t/τ
All CA has
been
reacted
Particles at bed
entrance are
completed reacted
Unreacted
bed depth
Reaction
zone
Completelyreacted
b) Uniform Reaction in Porousb) Uniform Reaction in Porous
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b) Uniform Reaction in Porousb) Uniform Reaction in Porous
ParticleParticle
and Zero Order in Fluidand Zero Order in Fluid
( )S S X k
dt
dX −= 1
S ,S
S
S S
dC C dX
C
C X
0
0
1
1
=−
=−where
( )0
0
1
0
=+∂∂
=+
∂
∂⋅
−S
S
S
f , A
kC t
C
kC
z
C
a
U
ε
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c) Uniform Reaction and 1c) Uniform Reaction and 1stst order in Fluid and in solidorder in Fluid and in solid
( )
( )
( )
( ) 01
01
1
1
0
=−+∂
∂
=−+
∂
∂
−=
−=
S s , Av
s , A
S s , Av
f , A
S v Av
S Av Av
C C ak t
C
C C ak
z
C U
C ak r
C C ak r
η ε
η ε
η ε
ε
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Non-Isothermal PackedNon-Isothermal Packed
Bed ReactorBed Reactor For mass continuityFor mass continuity did balance ondid balance on
fluid and on solidfluid and on solid For energy balance, we do balanceFor energy balance, we do balance
on each phaseon each phase
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ModelingModeling
Tf + dTf
Tf
z + dz
z
Tf,0
U0
q =0
g U sm
kg G ρ
02=
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Moving Bed ReactorMoving Bed Reactor
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Moving Bed ReactorMoving Bed Reactor
(MBR)(MBR) Steady state reactor where solids moving atSteady state reactor where solids moving at
near their packed bed voidagenear their packed bed voidage
Counter or co-current operationCounter or co-current operation
Solid usually move downward (vertical shaftSolid usually move downward (vertical shaftreactor or furnace)reactor or furnace)
Voidage is near that of a packed bedVoidage is near that of a packed bed
Slightly above random loose-packedSlightly above random loose-packed
voidagevoidage Solids move mainly in a plug floe, but regionSolids move mainly in a plug floe, but region
near wall have a velocity distributionnear wall have a velocity distribution
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Advantages of MBRAdvantages of MBR
True counter-current flow True counter-current flow
Uniform residence time (essentially plug flow)Uniform residence time (essentially plug flow)
ReasonableReasonable ∆P ∆P
Throughput variable Throughput variable Generally larger particle dGenerally larger particle dpp > 2-3 mm> 2-3 mm
Difficulties coping with wide size distribution of Difficulties coping with wide size distribution of
particles (fines tend to block up the voidparticles (fines tend to block up the void
spaces)spaces)