Transcript Plug Flow Model - Biodiesel Projects

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
Slurry Bubble Column Reactor
Batch Reactor
Fixed Bed Reactor
Trickle Column 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

CA,in
Gas + solids
Particle surrounding by fluid of
essential constant concentration,
CA,m
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

Tp
Fluid at Tf

Particle initially at uniform T = Tp
At t = 0, we drop it into our furnace
Energy Balance
Qconvection  Qradiation
dH
m
dt
Heat in by convection and radiation = change in enthalpy of particle


4R hcv T f  Tp    m T  T
2
4
w
4
p
  m
Where,
Area of sphere = 4πR2
Hcv = convection coefficient
σ = Stefan-Boltzman constant
Єm = emissivity of the particle (wall has Є = 1)
d C pTp 
dt
Energy Balance
hr   m
T
T
4
w
 T p4
F
 Tp
hcv  hr  TF  Tp  


m pC p dTp
A
Can solve this equation to get Tp =f(t)
dt
Find hcv


Have film: ∆H Tf ≠ Tp
Use mass transfer analogy to get hcv
hcvd p
kf
1
2
 Nu  2  0.6 Rep Pr
Rep 
Vd p

; Pr 
Cp
kf

; 

1
3
2. Small Particles:
Volumetric Reaction

Small such that no internal
gradients
Heat generated by reaction = Heat transferred to surrounding
Vp  rAv   Hr   hAp Tp  Tf 
Steady State:
Volume of
particle
Rate of
reaction
T
p
 Tf

 H r    rAv   R 


h
 
3
Exothermic Rxn:
-∆Hr = (+)
-rAv = (+)
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
2
r r 
r 
t
Surface:
dT
ke
dr

 h T f  Tp
r R
Heat conducted into
particle at r =Rp
rR

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
Boundary Conditions
Symmetry condition
T
0
r
r 0
Initial condition
t  0;Tp  Tp ,0 ;T ( r )  Tp
Internal gradient
Tr  0  Tr  R0
 Tr
r  R0
T f  Tr
r R
Tr
External gradient
r 0
Reacting Systems

General equation for volumetric reactions
(Reaction in porous particles)

Recall continuity equation:
continuity for A
Solve (1), (2), (3) Together
C A 1   2 C A 

 2  r De
  rAv
t
r r 
r 
(1)
Energy balance
T 1   2 T 
C p 1     2  r ke   rAv  H r 
t r r 
r 
kr C AmCSn   rAv 
kr T C C   rAv 
m
A
n
S
(2)
(3)
Coupled through the
reaction rate
Continuity for A
In Steady State

Showed that for steady conditions:
dT
dC A
 H r 
 ke
 De
dr
dr
Integrate at r = 0, r = R
For sphere
De
Tr 0  TS   C A,s  C A,r 0   H r 
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
2
r r 
r 
t

Tc
Ts
In Steady State,
rc
ke   2 T 
r
0
2
r r  r 

Solve
T  Tc

Tc  Ts


1
rc
 1r
1
R
 r1c
;

r
R
Tc  Ts
dT
 1 1 2
dr
R  rc r


Tf
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
dT
akr CS ,0C A ,c  H r   ke
dr
r  rc
 akr CS ,0C A ,c  H r   1 1 
TC  TS 
   
ke
 R rc 
Boundary Condition 2: r = R
Heat arriving by conduction
from within particle
dT
 ke
dr
=
Heat removed for
convection
 hTS  T f 
r R
 ke  1
TS  T f  TC  TS  
 hR  R R1  r1c

Can be obtained
from B.C. 1
Bi-1





Solution

Combine equations and eliminate TS to get Tc-Tf
TC  T f
1  1  1   1
ke  rc R  hR2
 akr CS ,0C A ,c  H r   rc2
Recall from Isothermal SC
Model
C A ,c
C A, f
 De 

2
akr CS ,0 rc 



De  1 
1 1
1 
  1 


 ak C r  r
r s ,0 c  c
 Bim  R

Substitute CA,c into (Tc –Tf) equation
Tc - Tf
TC  T f
1 1 1 1
    2
ke  rc R  hR
Conduction

C A, f  H r 
1 1 1
1
1
   

2
De  rc R  akr CS ,0 rc km R 2
Convection
Diffusion in
Product Layer
Reaction
Mass
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
Breakthrough
curve
1
2
CA,in
CA,out/CA,in
Regeneration
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
z
CA,f + dCA,f
CA,f
U0 (m/s) superficial velocity
U0 

Vgas m2 / s
 
Axs m2

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
t
ε is void fraction in bed
U0
C A , f
z
 rAv  0
Void fraction
For first order reaction, fluid only:
mol

 1 dNA
''
rAv  3

 kv 1   C A, f

 m reactor s  Vr dt
For steady state:
C A , f
t
Volume of reactor
0
Therefore,
U0
dCA , f
dz
 kv'' 1   C A , f  0
Conversion as a function of Height
Integrating with CA,f = CA,f,in at z = 0
X A  1
C A, f
C A, f ,in
 kv'' 1    
 1  exp 
z 
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 =
m3 of solid
m3 of reactor volume
mol
m3 of solid · s
Balance on Solid
Cs
1     rsv  0
t

 rav   a   rsv 
Cs
rAv

0
t a1   
Solve These Equations
= 0 (In quasi steady state, we ignore the
accumulation of A in gas)

C A , f
t
U0
C A , f
z
 rAv  0
a1    Cs

0
z
U 0 t
C A, f
C 'A , f
Cs
rAv

0
t a1   
Cs'

0
z
t
C 'A , f  f z ,t 
Cs'  f z ,t 
a)
b)
c)
d)
e)
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)
a) In Shrinking Core Model
rAv  akv 1   CA,cCS ,o
De
akv CS ,o rc2


De  1 
1 1
1 

 ak C r  r  1  Bi  R
r S ,o c  c
m 


Recall that
C A ,c
C A, f
Solid Phase
3

r
R
rc2 c  kvC A ,c rc  0
t
3
Liquid Phase
C A, f
U0
z
For SCM
 rc 
Cs  Cs ,o  
R
 akv 1   Cs ,oC A ,c  0
3
Solve
CA,f = f(z)
rc = f(z,t)
Conversion vs Time
t=0
z
t>0
Overall Conversion of Solid
3
 rc 
  R  dz
0
L
1 X s 
L
 dz
0
L
1
3

r dz
3  c
LR 0
Height Vs time (Graphical)
z/L
All CA has
been
reacted
Unreacted
bed depth
Reaction
zone
Completely
reacted
Particles at bed
entrance are
completed reacted
t/
b) Uniform Reaction in Porous Particle
and Zero Order in Fluid
dX S
 k 1  X S 
dt
where
CS
1 X S 
C0
C A , f
U0

 kCS  0
a1    z
CS
 kCS  0
t
1
 dX S 
dCS
CS ,0
c) Uniform Reaction and 1st order in
Fluid and in solid
rAv  akv 1   C AC S
rAv  akv 1    C S
U0
C A , f
z
 akv 1    C A ,s C S  0
C A ,s
 akv 1    C A ,s C S  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
U0
 kg 
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)