PFR design. Accounting for
pressure drop
Chemical Reaction Engineering I
Aug 2011 - Dec 2011
Dept. Chem. Engg., IIT-Madras
Overview
• Notation
• PFR design equation (mass balance)
• Pressure drop equation
• Accounts for change in number of moles (due to reaction)
• Does not consider phase change
• Methodology, with examples
• Liquid phase reaction: Calculations are simple
• Gas phase reaction: Calculations are more involved
Notation
•
Usual notation applies.
– FA = molar flow rate of A, FT = total molar flow rate
– V = volume of reactor, Q = volumetric flow rate
– D = diameter and A = cross-sectional area of PFR, L = length of PFR, z =
distance (between 0 and L)
– T = temperature, R = universal gas constant, P = pressure, PA = partial pressure
of A
– x = conversion, e = fractional increase in number of moles for 100% conversion,
for a given feed conditions.
– k = rate constant, CA = concentration of A
–
r = density, m = viscosity
– ff = friction factor, Re = Reynolds number
– <Vav> = average velocity of the fluid in the PFR. Used sparingly.
–
Subscript “in” indicates inlet. E.g. FA-in is molar flow rate of A at the inlet.
Definitions and formulas
The following definitions are of use here:
Definition of ‘x’
Definition of ‘e’
Definition of CA
FA  FAin (1  x)
FT  FT in (1  e x)
FA
CA 
Q
For gas phase only:
From ideal gas law
Q
At constant temperature
FT RT FT in (1  e x) RT

P
P
FT in (1  e x) RT (1  e x) PinQin
Q

P
P
Formulas
For liquid as well as gas phase
Re 
D Vav r
m

4Qr
 Dm
32 f f r Q
dP

dz
 2 D5
ff 

2
16
if Re < 2100
Re
1

 6.9  
12.96  log10 

Re



2
if Re > 10,000
You don’t need to memorize these formulas
PFR design equation
• Steady state conditions
• For a first order reaction
FAin
dFA
dx
 FAin
 rA
dV
dV
FAin
dx
 kC A
dV
FAin (1  x)
FA
dx
k
k
dV
Q
Q
• At constant temperature
dx
(1  x)
k (1  x) P
k

dV
Q
PinQin (1  e x)
dx
A k (1  x) P

dz PinQin (1  e x)
PFR design equation
• For any other order of reaction also, write the equation such
that dx/dz = f(x,P). i.e. The only unknowns on the RHS must
be x and P
– i.e. RHS must not have Q or CA. These are not known yet.
– But Qin and CA-in are known and hence RHS may have these terms.
dx
A k (1  x) P
(1  x) P


,
dz PinQin (1  e x)
(1  e x)
where  
Ak
PinQin
• For an ‘n’th order reaction
FAnin1 (1  x) n P n
dx
(1  x) n P n
 kA n n

n
n
dz
Pin Qin 1  e x 
1  e x 
FAnin1
where   kA n n
Pin Qin
Pressure Drop Equation
• Under steady state conditions, Reynolds number is a constant
– Even though local velocity changes
– This is because density also changes with location
– We assume that the viscosity of the medium remains the same,
even when the reaction occurs
• Under steady state conditions, (rQ) = constant
Re 
• Using Re, Calculate the friction factor ff.
• Write pressure drop equation
32 f f rQ 2
32 f f  rQ  Q
32 f f  rinQin  Q
dP



 2 D5
 2 D5
 2 D5
32 f f  rinQin  Q
32 f f  rinQin  PinQin (1  e x)
dP



dz
 2 D5
 2 D5
P
dP
(1  e x)
 
dz
dz
where  
P
32 f f  rinQin  PinQin
 2 D5
4Qr
 Dm
Solution
•
Solve both equations
simultaneously
•
Initial conditions:
– At z =0, P = Pin
– At z = 0, x = 0
•
Special case:
– When e =0, solve the first
equation and find ‘P’. Then
substitute for ‘P’ in the second
equation and solve for ‘x’
dP
(1  e x)
 
dz
P
32 f f  rinQin  PinQin
where  
 2 D5
FAnin1 (1  x) n P n
dx
(1  x) n P n
 kA n n

n
n
dz
Pin Qin 1  e x 
1  e x 
FAnin1
where   kA n n
Pin Qin
Example
• Consider a gas phase reaction under isothermal conditions.
(Isomerization)
• A B
• Pin = 10 atm, Q = 0.005 m3/s, T = 300 K, Pure A is fed, k =
0.1 lit/s, Molecular weight M = 60 g/gmol, Viscosity of the
gas = 10-5 Pa-s
• What should be the length of the PFR if it is constructed
(a) using a 2 cm pipe and the conversion desired is 10%? (b)
using a 1.5 cm pipe and the conversion desired is 10% and (c)
using a 1.5 cm pipe and the conversion desired is 20%
Solution
•
e= 0. This simplifies the equations
dP
(1  e x)
 
dz
P
32 f f  rinQin  PinQin
where  
 2 D5
dP
1
 
dz
P
P2  Pin2  2 z
R=0.08206 atm-lit/(gmol-K)
rin 
Pin M
 24.0558 kg/m3
RT
Vavin  15.9155 m / s
Re  7.65 105
f  0.003
FAin  2.0047 mol/s
  1.847 109 Pa2 / m
  6.2832 109 1/ ( Pa m)
Dia = 2 cm
•
Conversion and Pressure vs distance
Conversion
0.8
0.6
0.4
d = 2 cm
e=0
0.2
0
0
10
20
30
40
50
60
70
80
90
100
40
50
60
70
80
90
100
Pressure / Pa
5
10
x 10
9
d = 2 cm
e=0
8
7
0
10
20
30
PFR length / m
Dia = 2 cm, expanded
•
Conversion and Pressure vs distance
Conversion
0.2
0.15
0.1
d = 2 cm
e=0
0.05
0
0
2
4
6
8
10
12
14
16
18
20
8
10
12
14
16
18
20
Pressure / Pa
5
10
x 10
9.9
9.8
d = 2 cm
e=0
9.7
9.6
0
2
4
6
PFR length / m
Dia = 1.5 cm
• 10% conversion is possible, but 20% is not
Conversion and Pressure vs distance
Conversion
0.2
0.15
0.1
d = 1.5 cm
e=0
0.05
0
0
10
20
30
40
50
60
70
30
40
50
60
70
Pressure / Pa
5
10
x 10
5
d = 1.5 cm
e=0
0
0
10
20
PFR length / m
2nd order reaction, k = 0.01
• Coupled equations give correct answer
Conversion and Pressure vs distance
Conversion
0.4
Conversion and Pressure vs distance
0.5
Pressure / Pa
1
2
40
60
80
100
120
140
160
180
x 10
3
4
5
6
7
8
9
10
4
5
6
7
8
9
10
20
40
60
x 10
5
d = 1 cm
e=0
0
d = 2 cm
e=0
6
0
10
200
0
1
2
3
PFR length / m
80
100
120
140
160
180
200
PFR length / m
Conversion and Pressure vs distance
0.4
Conversion
20
8
4
0
5
0
5
10
d = 1 cm
e=0
0.1
0.3
0.2
d = 1 cm
e = -0.5
0.1
0
0
2
4
6
8
10
12
6
8
10
12
5
10
Pressure / Pa
0
0.2
0
d = 2 cm
e=0
Pressure / Pa
Conversion
1
0.3
x 10
5
d = 1 cm
e = -0.5
0
0
2
4
PFR length / m