Isothermal Reactor Design – Part 2
Pressure Drop In Reactors
Design a PBR
In case of 2nd order rxn, gas phase,
isothermal
• mole balance
dX
FA0
 rA
dW
• rate laws
 rA  kCA2
• stoichiometry
• combination
(1  X ) P T0
C A  C A0
(1  X ) P0 T
dX kCAO  1  X   P 


  
dW
v0  1  X   P0 
2
dX
 F1  X , P 
dW
2
Need to relate pressure
drop to catalyst weight
(in order to determine
conversion)
Design a PBR
Ergun equation
P0 = inlet pressure (kPa) T = temperature (K) T0 = inlet temperature (K)
1   = volume of solid
 = porosity = volume of void = void fraction
P = pressure (kPa)
total bed volume
Ac = cross sectional area (m2)
total bed volume
D p =diameter of particle in the bed, ft (m)
 =viscosity of gas passing through the bed, (kg/m.s)
z =length down the packed bed of pipe, ft (m)
u =superficial velocity = volumetric flow ÷ cross-sectional area of pipe (m/s)
 =gas density (kg/m3)  c = solid density (kg/m3)
G == superficial mass velocity, (kg/m2.s)
 0 = inlet gas density
Design a PBR
 For isothermal operation, we have two sets of equation with
two unknowns, X & P
dX kCAO  1  X


dW
v0  1  X
dP   T  P0 
 P 
1  X  (2)
  
  (1)
dW 2  T0  P P0 
 P0 
2
 Special case: if ε=0, an analytical solution to second equation is
obtained as follows
P
12
 1  W 
P0
Used only when ε=0
Design a PBR
In case of 2nd order rxn, gas phase,
isothermal
• mole balance
dX
FA0
 rA
dW
• rate laws
 rA  kCA2
• stoichiometry
• Combination
• Solve
When ε=0
(1  X ) P T0
C A  C A0
(1  X ) P0 T
dX kCAO  1  X   P 


  
dW
v0  1  X   P0 
2
2
dX kCAO
2
1  X  1  W 

dW
v0
By integration;
v0  X 
 W 

  W 1 

kCAO  1  X 
2 

Design a PBR
v0  X 
 W 

W


1 

kCAO  1  X 
2


Solving for conversion gives:
kCA0W  W 
1 

v0 
2 
X
kC W  W 
1  A 0 1 

v0 
2 
Solving for catalyst weight,
1  1  2v0 / kCA0 X / 1  X 
W

12
 For gas phase reactions, as the pressure drop increases, the
concentration decreases, resulting in a decreased rate of
reaction, hence a lower conversion when compared to a
reactor without a pressure drop.
↑ W, ↓P,
↑ΔP
↑ ΔP, ↓P, ↓CA, ↓-rA
↑ W, ↓P, ↓CA
↑W, ↑X
Effect of pressure drop on the conversion profile
Consider a packed bed column with a second order reaction is taking place in 20
meters of a 1 ½ schedule 40 pipe packed with catalyst.
2A  B + C
The following data are given:
Inlet pressure, P0 = 10 atm=1013 kPa
Entering flowrate, v0 = 7.15 m3/h
Catalyst pellet size, Dp = 0.006 m
Solid catalyst density: ρc = 1923 kg/m3
Cross sectional area of 1 ½ -in schedule 40 pipe: AC =0.0013 m2
Pressure drop parameter, β0 = 25.8 kPa/m
Reactor length, L = 20 m
Void fraction = 45%
(a) Calculate the conversion in the absence of pressure drop.
(b) Calculate the conversion accounting for pressure drop.
(c) What is conversion in part (b) if the catalyst particle diameter were doubled.
The entering concentration of A is 0.1 kmol/m3 and the specific reaction rate is
.
12m6
k
kmol  kg  cat  h
Effect of pressure drop on the conversion profile
kCA0W  W 
1 

v0 
2 
X
kC W  W 
1  A 0 1 

v0 
2 
(a) Conversion for ΔP = 0
α = 0 thus,
W  1  Ac L  c
 
kCA0W
v0
X
kCA0W
1
v0
X 
[Volume of catalyst] x [catalyst density]


 1  0.45 0.0013m2 20m1923kg / m3
 27.5kg


kCA0W
12m6
1



0.1kmol / m3 27.5kg 
  4.6
3
v0
kmol  kg  cat  h
 7.15m / h 
4.6
 0.82
1  4.6
Effect of pressure drop on the conversion profile
(b) Conversion with pressure drop
kCA0W  W 
1 

v0 
2 
X
kC W  W 
1  A 0 1 

v0 
2 
2 0

Ac  c 1  c P0
225.8kPa / m 

0.0013m 2 1923kg / m3 1  0.451013kPa



 0.037kg 1
4.60.49
X 
 0.693
1  4.60.49
0.037kg 1 27.5kg 
1
 1
 0.59
2
2
W
Effect of pressure drop on the conversion profile
(c) Conversion when catalyst diameter were doubled.
(increase by a factor of 2, D p  2 D p )
2
1.75G 2 1   
0 
 0 g c D p 3
2 0
From  
,
Ac  c 1  c P0
1
dominant
1
 0
Dp
 0
1
thus  0
Dp
Effect of pressure drop on the conversion profile
(c) Conversion when catalyst diameter were doubled.
D p2  2 D p1
 2  1
Thus,
D p1
D p2


1
 0.037kg    0.0185kg 1
2
1
 0.018527.5kg  
4.61 

2

  0.774
X
 0.018527.5kg  
1  4.61 

2


Increasing particle size
decrease the pressure
drop parameter, increase
conversion & reaction
rate
Conversion increases from 0.693 to 0.774 by increasing catalyst diameter
by a factor of 2.