Transcript Entrained Bed Reactor

Entrained Bed Reactor
Quak Foo Lee
Department of Chemical and Biological Engineering
Entrained Bed Reactor

Entrained bed, pneumatic transport reactor

Riser reactor, raining bed reactor
 Dilute suspension of solids and usually reactors have large
L/D ratio
 Near plug flow
 If feed particles are all of the same size, them all have same
residence time
Entrained Bed Reactor

Advantage is one can control the residence time

Small particle and co-current flow

Simplified analysis
 Flat velocity profile for gas
 Plug flow of solids
 Isothermal
 Constant fluid properties
Reactor Configurations
Gas + Solid
Down flow
D
Up flow
L
Gas + Solid
Riser Reactor
Upward particle velocity
U p  U0  U S
Superficial
gas velocity
Settling velocity of
particle
Gas + Solids
U S  Ut
Terminal velocity of a single particle in an unbounded fluid
Note: Concentration of solids is low
Terminal Velocity
 4d p  s   g g 
Ut  

 g CD


CD  f Re p ,  
1/ 2
Reynolds Number Region
Strokes Region
Re p  1
Ut 
d g  s   g 
2
p
18

d gp
2
p
18
1  Re p  500
 4  s   g  g 
Ut  d p 

g 
 225

2
2
1/ 3
For Higher Solid Concentrations
U s  Ut  e
U s  Ut 
4.11 0.64  
2.17
Example

If we considered U0 ~Up, and we have 1 mol of gas reacting
with 1 mol of solid of Mwt = 100 g/mol and p = 2 g/cm3
Vgas
Vsolids
22,400cm3 (STP) 400


1
 100g 
 2 g / cm3 


1
400
  0.9975
 1
1  


For uniform gas composition
Plug flow of solids
H Xs1  U p Ri   Ri 
H Xs1  U 0  U t Ri  Ri 
Where,  = time for complete reaction
For Strokes Region and SCM

With no Mass Transfer (MT) of Product Layer Diffusion (PLD)
Resistance:
R

k r C A, f
U0R
4 g R 3
H Xs1 

k r C A, f 18  k r C A, f
1
H Xs1 
k r C A, f

4 g 3 
U 0 R  18  R 


Critical Point
Goes through a maximum
at some critical R
Hcrit
Hmax
HXs=1
Rcrit
R
Find Hcrit and Rcrit

Differentiate HXs=1
1
H Xs1 
k r C A, f
dH
1

dR k r C A, f

4 g 3 
U 0 R  18  R 



2 g 2 
U 0  3   Rcrit   0


1/ 2
 3 U 0 

Rcrit  
 2 g 
Find Hcrit and Rcrit

Substitute Rcrit into HXs=1 to get Hcrit
H crit
7

9k r C A, f
1/ 2
3  
3/ 2
U
0
 2 g 


Note: To transport largest particles through the system:
U 0  U t ( Rmax )
Within the Strokes Region
4  g  2
 Rmax
U 0  
18   
Substitute U0 into Hcrit to
get
H crit
3
 g  Rmax

 0.0998
   k r C A, f
Up flow
Notes

Comparison indicates Rcrit ~ 0.77 Rmax

One would do a similar calculation for any
expression for τ and also can use more general
equation for CD.
CA, f

Note: for CA,f to be constant, e.g. at some
over
reactor height, we need small gas conversion (or
large molar excess of species A in gas phase).
Down Flow – Raining Bed
Gas + Solid
U p  U0 Us
U p  U0 Ut
For some simplified case:
H Xs1  U p Ri  Ri 
At U 0  Ut Ri 
U p  U0  U s  2Ut Rmax 
For chemical reaction controlling:
3
 g  Rmax

H Xs1  0.444
Down flow
   k r C A, f
Recall
3
 g  Rmax

 0.0998
   k r C A, f
Up flow
H crit
Down flow
 g  R

H Xs1  0.444
   k r C A, f
Ratio
3
max
H crit downflow 0.444

 4.4
H crit upflow
0.0998