Design of Reactors Presentation Module Useful

Types of Reactors
 Batch
◦ No flow of material in or out of reactor
◦ Changes with time
 Fed- Batch
◦ Either an inflow or an outflow of material but not both
◦ Changes with time
 Continuous
◦ Flow in and out of reactor
◦ Continuous Stirred Tank Reactor (CSTR)
◦ Plug Flow Reactor (PFR)
◦ Steady State Operation

Mass Balance on Reactive System
 In - out + gen - cons = accumulation
 A mass balance for the system is
Where NA is the mass of “A” inside the system.
GA
Rate of
generation/
consumption
FA0
Rate of flow in
FA
Rate of flow out
System
GA
Rate of
generation/
consumption
FA0
Rate of flow in
FA
Rate of flow out
System
dt
dN
G
F
F A
A
A
A 


0

Generalized Design Equation for
Reactors
 In - out + gen - cons = accumulation
dt
dN
dV
r
F
F A
V
A
A
A 

 
0

Batch Reactor
 Generalized Design Equation
 No flow into or out of the reactor,
then, FA = FA0 = 0
 Good mixing, constant volume
or
dt
dN
dV
r
F
F A
V
A
A
A 

 
0


V
A
A
dV
r
dt
dN
V
r
dt
dN
A
A

 
A
A
A
r
dt
dC
dt
V
N
d



Fed Batch Reactor
 Reactor Design Equation
 No outflow FA = 0
 Good Mixing rA dV term out of
the integral
dt
dN
dV
r
F
F A
V
A
A
A 

 
0
 
dt
V
C
d
dt
dN
V
r
F A
A
A
A





0
  A
A
A
A
r
C
V
F
dt
dC


 1
0

Continuous Stirred Tank
Reactor
 rate of flow in = rate of flow out
 FA = v CA and FA0 = v CA0
 v = volumetric flow rate (volume/time)

CSTR Contd…….
 General Reactor Design Equation
 Assume Steady State
 Well Mixed
 So or
dt
dN
dV
r
F
F A
V
A
A
A 

 
0
0

dt
dNA
A
V
A Vr
dV
r 

0
0 

 A
A
A Vr
F
F
A
A
A
r
F
F
V


 0

Plug Flow Reactor (PFR)
 Tubular Reactor
 Pipe through which fluid flows and
reacts.
 Poor mixing
 Difficult to control temperature

PFR Design Equation
 Design Equation
 Examine a small volume element (DV) with length
Dy and the same radius as the entire pipe.
 If the element is small, then spatial variations in rA are
negligible, and
dt
dN
dV
r
F
F A
V
A
A
A 

 
0
Flow of A
into
Element
Flow of A
out of
Element
V
r
dV
r A
V
A D


Assumption of “good mixing”
applies only to the small
volume element

PFR Design Equation
 If volume element is very small, then assume steady state with no
changes in the concentration of A.
 Simplify design equation to:
 rA is a function of position y, down the length of the pipe and
reactant concentration
 The volume of an element is the product of the length and cross-
sectional area,
DV = A Dy
 Design Equation becomes:
0

dt
dNA
    0

D

D

 V
r
y
y
F
y
F A
A
A
   
A
A
A
Ar
y
y
F
y
y
F







D

D


PFR Design Equation
 take the limit where the size of a volume element
becomes infinitesimally small
 or because ∆y A = V,
 This is the Design Equation for a PFR
A
A
y
Ar
dy
dF


D
lim
0
A
A
r
dV
dF


Design of Reactors Presentation Module Useful

Conversion
 The basis of calculation is always the limiting
reactant.
Consider the general equation
 The conversion X of species A in a reaction is equal
to the number of moles of A reacted per mole of A
fed, ie

Conversion
Maximum value of conversion:
irreversible reactions X = 1
reversible reactions X=Xe

Design Equations
Batch Reactor:
Differential Design Equation
Integral Design Equation

Design Equations
CSTR:
Design Equation

Design Equations
Plug Flow Reactor:

Design Equations
Packed Bed Reactor:

Isothermal Reactor Design
Design Procedure:
1. Find Mole balance design equation
2. Find Rate law
3. Find Stoichiometric equation
4. Combine above three to get final
equation
5. Evaluate

Isothermal Reactor- liquid phase
 The elementary liquid phase reaction 2AB is carried out
isothermally in a CSTR. Pure A enters at a volumetric flow rate of
25 dm3/s and at a concentration of 0.2 mol/dm3. What CSTR
volume is necessary to achieve a 90% conversion when k = 10
dm3/(mol*s)?
Solution
Mole Balance
Rate Law

Solution contd….
Stoichiometry
Combining

Solution contd….
Evaluating at x = 0.9
V = 1125 dm3
Calculation of space Time:

Reactor Design – Gas Phase Reaction
Gas Phase Elementary Reaction :
2AB
P0 = 8.2 atm
T0 = 500 K
CA0 = 0.2 mol/dm3
k = 0.5 dm3/mol-s
vo = 2.5 dm3/s
X = 90%

Solution
Design For Batch Reactor:

Solution Continued….
Design For CSTR:
Mole Balance:
Rate Law:
Stoichiometry:
Combining:

Solution Continued….
Evaluation:

Soln Contd…Design for PFR
Design For PFR:
Mole Balance:
Rate Law:
Stoichiometry:
Combining:

Contd…Design for PFR
Integrating:
Evaluating:

Reactor Design- Reversible Reaction systems
Given that the system is gas phase and isothermal,
determine the reactor volume when X = 0.8 Xe.
Solution :
equilibrium constant, KC, for this reaction
KC = CBe / C2Ae

Solution Contd….
X = 0.8Xe = 0.711

Reactor Sizing
 Involves determination of
◦ reactor volume to achieve a given conversion or
◦ determine the conversion that can be achieved in a given
reactor type and size
 Given -rA as a function of conversion,-rA=f(X), one can size
any type of reactor
 to find -rA= f(X), construct Levenspiel plot
 Plot FAo/-rA as a function of X
 The volume of a CSTR and the volume of a PFR can be
represented as the shaded areas in the Levenspiel Plots

Reactor Sizing
Using the Ideal Gas Law to Calculate CA0 and FA0
The entering molar flow rate for a gas is
where CA0=
entering concentration,
mol/dm3
yA0= entering mole fraction of A
P0=
entering total pressure, e.g.,
kPa
PA0=
yA0P0 = entering partial
pressure of A, e.g., kPa
T0= entering temperature, K
v0= volumetric flow rate

Reactor Sizing
A gas of pure A at 830 kPa (8.2 atm) enters a reactor
with a volumetric flow rate, v0, of 2 dm3/s at 500 K.
Calculate the entering concentration of A, CA0, and the
entering molar flow rate, FA0.
FA0 = 2 * 1* 830 / (8.314 * 500) = 0.399 mol/s
CA0 = 0.399/2 = 0.199 mol/dm3

Reactor Sizing
Levenspiel Plots in Terms of Concentrations
For a plug-flow reactor ,
Writing the above equation in terms of concentration,CA

Design of Reactors Presentation Module Useful

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Design of Reactors Presentation Module Useful