Pressure drop in
Packed Bed Reactors
Chemical Reaction Engineering I
Aug 2011 - Dec 2011
Dept. Chem. Engg., IIT-Madras
Overview
• Notation
• PFR design equation (mass balance)
• Pressure drop equation for packed bed
•
•
•
•
Modified for catalytic reactions
Accounts for change in number of moles (due to reaction)
Does not consider phase change
Methodology, with examples
Basics
•
Usual notation applies.
•
Reaction is usually written as rA , in mol/vol/time
•
For catalytic reactions, it is written as
• rA’, in mol/g-catalyst/time
• Rate law is written as
rA'  kCAn
• Where the rate constant k is in appropriate units
• Remember that CA is in mol/vol
•
Packed bed: The PFR equation will be modified. Instead of dV, we will use dW
dFA
 rA'
dW
Design Equation
For constant temperature and with ideal gas law
FAin
F 
dx
k A 
dW
Q
n
F 1  x 

k
n
Ain
Pn
PinnQinn 1   x 
 n 1
dx kFAin
 n n
dW
Pin Qin
n
n
1

x
P
 
n
1   x 
n
Pressure Drop Equation
Ergun equation, written in terms of mass flow rate

dP
G  1    150 1    

 1.75G 
 3 
dz
 Dp    
Dp

Need not
memorize
this
Superficial mass velocity G =  Vsup
Porosity 
Diameter of particles is Dp
 is a variable. It should be written in terms of X and P
Similarly, z must be written in terms of W
Pressure drop equation
• Mass flow rate is a constant
 Q   inQin 
inQin 


Q
• For ideal gas, constant temperature, no phase
change
inQin  P
in  P




 PinQin  1   x   Pin  1   x 
  Pin  1   x 
dP
G  1    150 1    

 1.75G 
 3 
dz
D p    
Dp
  in  P
Pressure drop equation
• Relate ‘W’ to ‘z’
• Weight of catalyst = Volume of catalyst * density
of catalyst
• W = Az * (1-) * c.
• dW = A (1-) * c * dz
• For ideal gas, constant temperature, no phase
change
  Pin  1   x 
dP
1
G  1    150 1    

 1.75G 
 3 
dW
 A1    c  Dp     Dp
  in  P
1  x

dP
 
dW
P
Simplifications
• If there is no change in volume due to reaction (if  = 0)
• Pressure equation can be solved
P2  P02  2W
• If parameter  is small, and X is also small, then we can
neglect that in the pressure drop equation, but not in the
design equation
Examples
Pin = 5e5; %Pa;
T = 400 ; %Kelvin
Qin = 0.01 ; % m3/s
M = 50 ; % molecular weight, g/gmol
dia = 0.1; % diameter in m
mu = 2e-5; % viscosity in Pa-s
k = 0.001; % rate constant with correct units. lit/g-catalyst/s, for
example
n = 1; %order of reaction
epsilon = -0.2; % fractional change in volume for 100% conversion ,
for the given feed
phi = 0.3 ; % Void fraction, volume of void/total volume
RhoC = 2000 * 1000; % catalyst density, in g/m^3
W_end = 1000; % in g
Dp = 0.005; % diameter of particle, in m
Solution
Rho in – 7.5174 kg/m3
Fain = 1.5035 gmol/s
Area = 0.0079 m2
Va-in=1.2732 m/s
G = 9.5715 kg/s
Alpha = 5.155 E6, Pa2/g catalystt
Beta = 1.6 e-9 units
Results
Conversion and Pressure vs Catalyst Weight
Conversion
0.8
0.6
0.4
d = 10 cm
 = -0.2
0.2
0
0
100
200
300
400
500
600
700
800
900
1000
400
500
600
700
800
900
1000
Pressure / Pa
5
5
x 10
4.95
d = 10 cm
 = -0.2
4.9
0
100
200
300
Catalyst Weight / g
Solution
If dia is changed from 10 cm to 5 cm
Conversion and Pressure vs Catalyst Weight
Conversion
0.06
0.04
d = 5 cm
 = -0.2
0.02
0
0
50
100
150
200
250
300
350
400
150
200
250
300
350
400
5
Pressure / Pa
•
6
x 10
4
d = 5 cm
 = -0.2
2
0
0
50
100
Catalyst Weight / g