Chapter 4
Isothermal Reactor Design
Chemical Reaction Engineering 1
반응공학 1
Objectives
• Describe the algorithm that allows the reader to solve chemical
reaction engineering problems through logic rather than
memorization.
• Sizing batch reactors, semi-batch reactors, CSTRs, PFRs, and PBRs
for isothermal operation given the rate law and feed conditions.
• Studying a liquid-phase batch reactor to determine the specific
reaction rate constant needed for the design of a CSTR.
• Design of a tubular reactor for a gas-phase reaction.
• Account for the effects of pressure drop on conversion in packed bed
tubular reactors and in packed bed spherical reactors.
• The principles of unsteady operation and semi-batch reactor.
Fig. 4-1
Isothermal
Reaction
Design
Algorithm
for Conversion
4.1 Design structure for isothermal reactors
Algorithm for isothermal reactor design
1. Mole balance and design equation
2. Rate law
3. Stoichiometry
4. Combine
5. Evaluate
We can solve the equations in the
combine step either
A. Graphically (Chapter 2)
B. Numerical (Appendix A4)
C. Analytical (Appendix A1)
D. Software packages (polymath)
Algorithm for Isothermal Reactors
French Menu Analogy
Scale-up of Liquid-Phase Batch Reactor
to the Design of a CSTR
PAST
Laboratory
Experiment
Pilot plant
Operation
Full-scale
Production
High cost of a pilot-plant leads to jump pilot plant operation
FUTURE
Microplant
(Lab-bench-scale unit)
Full-scale
Production
To make this jump successfully requires
a through understanding of the chemical kinetics and transport limitations.
Batch Operation
Algorithm for isothermal reactor design
1. Mole balance and design equation
1  dN A 

  rA
V  dt 
1 dN A d N A / V0  dC A


 rA
V0 dt
dt
dt
2. Rate law
A
 B
3. Stoichiometry
4. Combination
5. Analytical Evaluation


 rA  kCA2
dC A
 kCA2
dt
dC A
kCA2
1

k
CA=CAo(1-X)
dX
CAo
=kCAo2(1-X)2
dt
 dt
t
X dX
1
kCAo 0 (1-X)2 = dt
0


CA
dC A
C A0
C A2

irreversible, 2nd order in A
1 1
1


k  C A C A0

t
 dt
0

t


1
t=
kCAo
(
X
1-X
)
Batch Reaction Times
A
 B
r
dX
 AV
dt
N A0
Mole balance
Rate law
First - order
Second order
 rA  kCA
 rA  kCA2
CA 
Stoichiometry (V=V0)
NA
 C A0 (1  X )
V0
Combine
dX
 k (1  X )
dt
dX
 kCA0 (1  X ) 2
dt
Integration
1
1
t  ln
k 1 X
t
X
kCA0 (1  X )
Batch Reaction Times
1st - order (X  0.9, k  10 -4 s 1 )
tR 
1
1
ln
k 1 X
2nd order (X  0.9, kC A0  10 3 s 1 )
tR 
X
kCA0 (1  X )
1
1
 ln
k 1  0 .9
0.9

kCA0 (1  0.9)
2 .3

k
9

kCA0

2. 3
10
 4 1
s

9
10 3 s 1
 23,000 sec
 9,000 sec
 6.4hr
 2.5hr
Batch Reaction Times (Table 4-2)
The order of magnitude of time
to achieve 90% conversion
For first- and second-order irreversible batch reactions
1st-order
k (s-1)
2nd-order
kCA0 (s-1)
Reaction time
tR
10-4
10-3
Hours
10-2
10-1
Minutes
1
10
Seconds
1,000
10,000
Milliseconds
Reaction Time in Batch Operation (Table 4-3)
1 1
1
t  

k  C A C A0




2nd order
Isothermal
Liquid-phase
Batch reaction
This time is the time t needed to reduce the reactant concentration in a batch reactor
from an initial value CA0 to some specified value CA.
Typical cycle times for a batch polymerization process
tt = tf + te + tR + tc
Activity
1.
2.
3.
4.
Heat to reaction temperature, te
Charge feed to the reactor and agitate, tf
Carry out reaction, tR
Empty and clean reactor, tc
Total cycle time excluding reaction
Time (h)
1.5-3.0
1.0-2.0
Varies (5-60)
0.5-1.0
3.0-6.0
Decreasing the reaction time with a 60-h reaction is a critical problem. As the reaction time is
reduced, it becomes important to use large lines and pumps to achieve rapid transfer and to
utilize efficient sequencing to minimize the cycle time
Example 4-1: Design a Reactor Producing Ethylene Glycol
Design a CSTR to Produce 200 million pounds of ethylene glycol per year by
hydrolyzing ethylene oxide. However, before the design can be carried out , it
is necessary to perform and analyze a batch reactor experiment to determine
the specific reaction rate constant (kA). Since the reaction will be carried out
isothermally, kA will need to be determined only at the reaction temperature of
the CSTR. At high temperature there is a significant by-product formation,
while at temperature below 40oC the reaction does not proceed at a significant
rate; consequently, a temperature of 55oC has been chosen. Since the water is
usually present in excess, its concentration may be considered constant during
the course of the reaction. The reaction is first-order in ethylene oxide.
O
CH2-OH
CH2-CH2 + H2O
A + B
H2SO4
Catalyst
CH2-OH
C
Example 4-1: Determining k from Batch Data
In the lab experiment, 500mL of a 2 M solution (2 kmol/m3) of EO in water was mixed
with 500mL of water containing 0.9 wt % sulfuric acid catalyst. At T=55oC, the CEG was
recorded with time.
Time
(min)
0.0
0.5
1.0
1.5
2.0
3.0
4.0
6.0
10.0
Concentration of EG
(kmol/m3)
0.000
0.145
0.270
0.376
0.467
0.610
0.715
0.848
0.957
Problem Solving Algorithm
Example 4-1 Determining k from Batch Data
A.
B.
C.
Problem statement. Determine the kA D.
Sketch
Identify
C1. Relevant theories
Rate law:  rA  k AC A
dN A
 rAV
Mole balance:
dt
C2. Variables
Dependent: concentrations
Independent: time
C3. Knowns and unknowns
Knowns: CEG = f(t)
Unknowns:
1. CEO = f(t)
2. kA
3. Reactor volume
C4. Inputs and outputs: reactant fed
all at once a batch reactor
C5. Missing information: None
Assumptions and approximations:
Assumptions
1. Well mixed
2. All reactants enter at the same time
3. No side reaction
4. Negligible filling and emptying time
5. Isothermal operation
Approximations
1. Water in excess (CH2O~constant)
E. Specification. The problem is neither
overspecified nor underspecified.
F. Related material. This problem uses the
mole balances developed in Chap. 1
for a batch reactor and the
stoichiometry and rate laws developed
in Chap. 3.
G. Use an Algorithm.
dCA
= -kCA
dt
-
CA dC
A

CA0 CA
= k

t
dt
0
ln
CAo
= kt
CA
CA = CAoe-kt
A + B  C
NC=NAoX=NAo-NA
CC=NC/V=NC/Vo= CAo-CA=CAo- CAoe-kt =CAo(1-e-kt)
Rearranging and taking the logarithm
of both side yields
ln
C A0  C C
 kt
C A0
We see that a plot ln[(CA0-CC)/CA0] as
a function of t will be a straight line with
a slope –k.
k
ln 10
2.3

 0.311 min 1
t 2  t1 8.95  1.55
 rA  (0.311 min 1 )C A
Design of CSTR
Design Equation for a CSTR
Mole balance
V
FA0 X
(rA ) exit
 C  CA 

V  v0  A0
  rA 

V C A0  C A

v0
 rA
the space time
For a 1st-order irreversible reaction, the rate law is
Rate law
Combine
rA  kCA

C A0  C A
kCA
Solving for the effluent concentration of A, we obtain
CA 
C A0
 C A0 (1  X )
1  k
X 
k
1  k
Relationship between
space time and
conversion for a 1storder liquid-phase rxn
Reaction Damköhler number
Da 
rA0V Rate of Reaction at Entrance
" a reaction rate"


FA0
Entering Flow Rate of A
" a convection rate"
The Damkohler is a dimensionless number that can give us a quick
estimate of the degree of conversion that can be achieved in continuousflow reactor.
rA0V kCA0 0V

 k
FA0
v 0 C A0
For 1st-order irreversible reaction
Da 
For 2nd-order irreversible reaction
 rA0V kCA2 0V
Da 

 kCA0
FA0
v0C A0
Da  0.1 will usually give less than 10% conversion.
Da  10.0 will usually give greater than 90% conversion.
Da
k
For first order reaction, X =
=
1 + Da
1 + k
CSTRs in Series
CA1, X1
CA0
v0
CA2, X2
-rA1, V1
-rA2, V2
For first-order irreversible reaction with no volume change (v=v0) is carried out in
two CSTRs placed in series. The effluent concentration of A from reactor 1 is
C A1 
C A0
1  1k1
From a mole balance on reactor 2,
V2 
FA1  FA2 v0 C A1  C A2 

 rA2
k 2 C A2
CSTRs in Series
Solving for CA2, the concentration exiting the second reactor, we get
C A2
C A0
C A1


1   2 k 2 1   2 k 2 1  1k1 
If instead of two CSTRs in series we had n equal-sized CSTRs connected in
series (1 = 2 = … = n = i = (Vi/v0)) operating at the same temperature (k1 =
k2 = … = kn = k), the concentration leaving the last reactor would be
C An 
C A0
1  k n

C A0
1  Da n
The conversion and the rate of disappearance of A for these n tank reactors
in series would be
X  1
1
1  k 
n
X= 1-CA/CAo
1
=1(1+Da)n
 rAn  kCAn 
kCA0
1  k n
Conversion as a Function of Reactors in Series
for different Damkohler numbers for a first-order recation
Da=k=1
Da=k=0.5
Da=k=0.1
Da  1, 90% conversion is achieved in two or three reactors;
thus the cost of adding subsequent reactors might not be justified
Da ~0.1, the conversion continues to increase significantly with each reactor added
CSTRs in Parallel
A balance on any reactor i, gives the individual
reactor volume
 Xi
Vi  FA0i 
  rAi
FA0




1
FA0 i
-rA1, V1, X1
X1  X 2    X n  X
i
 rA1   rA2     rAn   rA
The volume of each individual reactor, Vi, is
related to the total volume, V, of all the reactors,
and similar relationship exists for the total molar
flow rate
FA0
V
Vi 
FA0i 
n
n
-rAi, Vi
n
-rAn, Vn
CSTRs in Parallel
Substituting these values into Eq (4-12) yields
F
FA0i  A0
n
V
Vi 
n
 Xi 

Vi  FA0i 

  rAi 
FA01
FA0
1
FA0i
-r-r
A1, V
, 1V, X1
A1
1
i
V FA 0  X i


n
n   rAi
V




FA0n
-r-r
, ,VVi, i Xi
AiAi
FA 0 X i FA 0 X

 rAi
 rA
The conversion achieved in any one of the reactors in
parallel is identical to what would be achieved if the
reactant were fed in one stream to one large reactor of
volume V
n
-r-rAn, ,VVn, Xn
An
n
A Second-Order Reaction in a CSTR
For a 2nd-order liquid-phase reaction
being carried out in a CSTR, the
combination of the rate law and the
design equation yields
V 
FA0 X FA0 X

 rA
kCA2
We solve the above eq. for X:
X 
(4-14)

For const density v=v0, FA0X=v0(CA0-CA)

C C
V

 A0 2 A
v0
kCA
Using our definition of conversion, we
have
X

(4-15)
2
kCA0 (1  X )
1  2kCA0   1  2kCA0 2  2kCA0 2
2kCA0
1  2kCA0  
1  4kCA0
2kCA0
1  2Da  
1  4Da
(4-16)
2Da
The minus sign must be chosen in the
quadratic equation because X cannot be
greater than 1.
A Second-Order Reaction in a CSTR
0.88
0.67
6
60
Example 4-2: Producing 200 Million Pound/Year in a CSTR
A 1 lb mol/ft3 solution of ethylene oxide (EO) in
water is fed to the reactor together with an equal
volumetric solution of water containing 0.9 wt%
of the catalyst H2SO4. The specific reaction rate
constant is 0.311 min-1.
(a) If 80% conversion is to be achieved, determine
the necessary CSTR volume.
(b) If two 800-gal reactors were arranged in parallel,
what is the corresponding conversion?
(c) If two 800-gal reactors were arranged in series,
what is the corresponding conversion?
Tubular Reactors
- 2nd-order gas-phase reaction
- Turbulent
- No dispersion
- No radial gradients in T, u, or C
Rate law
V  FA 0

X
dX
kCA2
0
For Liquid-Phase Reaction
PLUG-FLOW REACTOR
V  FA 0
PFR mole balance
FA 0

X
dX
must be used when there is a DP or heat
exchange between PFR & the surrounds.
In the absence of DP or heat exchange, the
integral form of the PFR design equation
is used.
V  FA 0

X
0
dX
 rA
dX
 kC (1-X)
v
X
=
kC ( 1-X )
X
( 1-X )
0 kC 2
A
dX
  rA
dV
X
= FAo
0
2
Ao
2
o
Ao
1
V/vo=  = kC
Ao
Da2
kCAo
X=
=
1+kCAo 1 + Da2
(Da2=Damköhler number for a 2nd-order RXN)
For n-th order RXN, Dan=kCAon-1
For Gas-Phase Reaction
V  FA 0

X
0
dX
kCA2
Conversion as a Function of Distance Down the Reactor
v=vo(1+X)
v  (1  0.5 X )v0
the reactant spends more time
v  (1  2 X )v0
the reactant spends less time
ö
V(m3)
The volumetric flow rate decreases with increasing conversion, and the reactant
spends more time in the reactor than reactants that produce no net change in the
total number of moles.
Change in Gas-Phase Volumetric Flow Rate Down the Reactor
v=vo(1+X)
Example 4-3: Determination of a PFR Volume
Determine the PFR volume necessary to produce 300 million pounds of ethylene a year
from cracking a feed stream of pure ethane. The reaction is irreversible and follows an
elementary rate law. We want to achieve 80% conversion of ethane, operating the
reactor isothermally at 1100K at a pressure of 6 atm.
C2H6 (A)  C2H4 (B) + H2 (C)
FB=300x106 lb/year=0.340 lb-mol/sec
FB=FAoX
FAo=FB/X=0.340/0.8=0.425 lb-mol/sec
Pressure Drop in Reactors
 In liquid-phase reaction
- the concentration of reactants is insignificantly affected by even
relatively large change in the total pressure
- ignore the effect of pressure drop on the rate of reaction
when sizing liquid-phase chemical reactors
- that is, pressure drop is ignored for liquid-phase kinetics calculations
 In gas-phase reaction
- the concentration of the reacting species is proportional to
the total pressure
- the effects of pressure drop on the reaction system are a key factor
in the success or failure of the reactor operation
- that is, pressure drop may be very important for gas-phase reactions
Pressure Drop and Rate Law
• For an ideal gas,
Fi
FA0  i  vi X 
Ci 

v v0 (1  X )( P0 / P)(T / T0 )
(4-18)
기상반응에서는 반응 성분
의 농도가 반응압력에 비례
하므로 압력강하에 대한
고려가 필수적이다.
  i  vi X  P To
C i  C A0 

 1  X  P0 T
- determine the ratio P/P0 as a function of V or W
- combine the concentration, rate law, and design equation
- the differential form of the mole balance (design equation) must be used
Pressure Drop and Rate Law
• For example,
- the second order isomerization reaction in a packed-bed reactor
2A  B + C
-the differential form of the mole balance
FA0
- rate law
dX
 rA
dW

gmoles

 g catalyst  min
 rA  kCA2
-from stoichiometry for gas-phase reactions
 1  X  P T0
C A  C A0 

 1  X  P0 T



Pressure drop and the rate law
• Then, the rate law

 1 X
 rA  k C A0 
 1  X

 P T0 


P
T
 0 
2
(4-20)
- the larger the pressure drop from frictional losses, the smaller the reaction rate
• Combining with the mole balance and assuming isothermal operation (T=To)
FA0
dX
 C A0 (1  X ) 
 k

dW
1


X


2
P
 
P 
 0
• Dividing by FA0
dX kCA0

dW
v0
 1  X   P 

  
 1  X   P0 
2
2
2
Pressure Drop and Rate Law
• For isothermal operation (T =T0)
-a function of only conversion and pressure
dX
 f ( X , P)
dW
(4-21)
-Another equation is needed to determine the conversion as a function of
catalyst weight
- that is, we need to relate the pressure drop to the catalyst weight
P  f (W )
Flow Through a Packed Bed
• The majority of gas-phase reactions are catalyzed by passing the reactant
through a packed of catalyst particles
• Ergun equation: to calculate pressure drop in a packed porous bed

dP
G  1   150 (1  )

 1.75G 
 3 
dz
g c D p   
Dp

laminar
(4-22)
turbulent
G=u=superficial mass velocity [g/cm2s]; u=superficial velocity [cm/s]; Dp=diameter of
particle in the bed [cm]; f=porosity=volume of void/total bed volume; 1- f =volume of
solid/total bed volume
• The gas density is the only parameter that varies with pressure on the right-hand
side. So, calculate the pressure drop through the bed laminar turbulent
Flow through a Packed Bed
• Equation of continuity
 0  m 
m
m
o=m
 0 v0  v
- steady state  the mass flow rate at any point is equal to the entering
mass flow rate
• Gas-phase volumetric flow rate
 T  FT
 
T  F
 0  T0
(3-41)
v0
P  T0  FT 0
 0
 
v
P0  T  FT
(4-23)
P
v  v0 0
P
• Then,
  0
Pressure Drop in a Packed Bed Reactor
• then, Ergun equation
dP
G(1  )

dz
0 g c D p  3
• Simplifying
We need
150 (1  )
 P0
 1.75G 

Dp

 P
P0  T  FT
dP
 
  0
dz
P  T0  FT 0
(4-24)

G(1  ) 150 (1  )
0 
 1.75G  (4-25)

3
Dp
0 g c D p  

dP
dW
• The catalyst weight, W
 T  FT
 
T  F
 0  T0
 (1  ) Ac z  c
Volume of
solid
dW
Density of
solid catalyst
 c (1  ) Ac dz
(4-26)
Pressure Drop in a Packed Bed Reactor
• then, Ergun equation
• Simplifying
Let y=P/Po
0
P0  T  FT
dP
 

dW
Ac (1  ) c P  T0  FT 0




(4-27)
P0T
dP
 T F
dy

(1  X )
2y
To
dW
2 T0 PF/ P
0
(4-28)
2 0
Ac  c (1  ) P0
(4-29)
dP
 T P0

dW
2 T0 P / P0

 F

FT  FT 0  FT 0 X  FT 0 1  A0 X 
 FT 0

 FT

F
 T0
FT
 1  X
FT 0
  y A0  
FA0

FT 0
Pressure drop in a packed bed reactor
dP
 T P0
dy

(1+X)(1  X )
2y
dW
2 T0 P / P0
(4-30)
ε < 0, the pressure drop (DP) will be less than ε = 0
ε > 0, the pressure drop (DP) will be greater than ε = 0
• For isothermal operation
dP
 f ( X , P)
dW
and
dX
 f ( X , P)
dW
(4-31)
• The two expressions are coupled ordinary differential equations. We can solve
them simultaneously using an ODE solver such as Polymath.
• For the special case of isothermal operation and ε = 0, we can obtain an
analytical solution.
• Polymath will combine the mole balance, rate law and stoichiometry
Pressure Drop in a Packed Bed Reactor
dP
 T P0
dy

(1+X)(1  X )
2y
dW
2 T0 P / P0
(4-30)
Analytical Solution
If ε = 0 or ε X ≪ 1, we can obtain an analytical solution to Eq. (4-30) for isothermal
operation (i.e., T=T0). For isothermal operation with ε = 0, Eq. (4-30) becomes
dy
dP
 P0

2y
dW
2 ( P / P0 )
Isothermal (T=To)
with ε = 0
dy
2y
dW = - 
At W=0, y=1 (P/Po=1)
y2= 1- w
dy2
dW = - 
Pressure Drop in a Packed Bed Reactor
If ε = 0 or ε X ≪ 1, we can obtain an analytical solution to Eq. (4-30) for isothermal
operation (i.e., T=T0). For isothermal operation with ε = 0, Eq. (4-30) becomes
Pressure ratio
only for ε = 0
y=
P
 1  W
P0
(4-33)
2 0

Ac (1  ) c P0
 (1  ) Ac z  c
(4-26)
2 0 z
P
y=
=f (z)
 1
P0
P0
(4-34)
W
Pressure as a
function of
reactor length, z
Pressure Drop in Pipes
Pressure drop along the length of the pipe
dP
d 2 fG 2
 G

dL
dL
D
P dP
2 fG 2
2 dP
0
G

0
P0 dL
PdL
D
Integrating with P=P0 at L=0, and assuming that f = constant
0
P02  P 2
P 
P 
L
 G 2 0  2 f  ln 0 
2
0 
D
P
Rearranging, we get
P0
 1   pV
P
4 fG 2
p 
0 P0 Ac D
Example 4-4: 1½” schedule 40 x1000-ft L (p=0.018), DP<10%
Analytical Solution for Reaction with Pressure Drop
Conversion as a function of catalyst weight
2nd-order isothermal reaction
A
B
dX
  r A
dW
Gas-phase isothermal with =0
C A  C A0 (1  X )
FA0
dX
kCA2 0
(1  X )
2
 1  W dW
at X  0, W  0 and FA0  C A0 v0
 rA  kCA2
y=

2
Separating variable and Integrating
Rate law:
Stoichiometry:

dX kCA2 0

(1  X ) 2 1  W 1/ 2
dW
FA0
Mole balance:
F A0
Combining
P
P0
P
 1  W
P0
C A  C A0 (1  X )(1  W )1/ 2
v0  X 
 W 

  W 1 

2 
kCA0  1  X 

kCA0W  W 
1 

v0 
2 
X
kC W  W 
1  A0 1 

v0 
2 
(4-38)
Reaction with Pressure Drop
Conversion as a function of catalyst weight
kCA0W  W 
1 

v0 
2 
X
kC W  W 
1  A0 1 

v0 
2 
(4-38)
Catalyst weight for 2nd-order isothermal reaction in PFR with DP
1  1  (2v0 ) / kCA0 X /(1  X )1/ 2
W

(4-39)
Reaction with Pressure Drop
Conversion as a function of catalyst weight
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.
Example 4-5 and Example 4-6
지금까지 배운 지식을 활용하여
적들이 방심하고 있는 사이에
이 예제들은 집에서 풀어 봐야지...
Spherical Packed-Bed Reactors
Spherical Ultraformer Reactor (Amoco) for dehydrogenation reaction such as
Paraffin
Aromatic + 3 H2
 Spherical reactor
- minimize pressure drop
- inexpensive
- the most economical shape for high pressure
Coordinate system and variables
used with a spherical reactor
Synthesizing a Chemical Plant
 Always challenge the assumptions, constraints, and boundaries of
the problem
 The profit from a chemical plant will be the difference between
income from sales and the cost to produce the chemical
Profit = (value of products) – (cost of reactants)
– (operating costs) – (separation costs)
 The operating cost: energy, labor, overhead, and depreciation of
equipment
Production of Ethylene Glycol
402 million
lbC2H6 /yr
1
H2, C2H4
C2H6  C2H4 + H2
2
V=81 ft3, X=0.8
C2H6
O2, C2H4, N2, C2H4O
separator
Ag
C2H4+ ½ O2  C2H4O
6
5
C2H4O
3
4
separator
H2O
C2H4
W=45,440 lb, X=0.6
7
8
H2O, 0.9wt% H2SO4
C2H4O(aq)
9
V=197 ft3, X=0.8
absorber
Cat.
C2H4O + H2O 
CH2OH
CH2OH
200 million
lb EG/yr
Air
Synthesizing a Chemical Plant

Ethylene glycol = $0.38/lb (2x108 lb/yr)
Ethane = $0.04/lb (4x106 lb/yr)
Sulfuric acid = $0.043/lb (2.26x108 lb/yr)
Operating cost = $8x106/yr
 Profit = ($0.38/lb x 2x108 lb/yr) – ($0.04/lb x 4x108 lb/yr)
-($0.043/lb x 2.26x106 lb/yr) – ($8x106/yr)
= $52 million
 How the profit will be affected by conversion, separation, recycle stream,
and operating costs?
Using CA (liquid) and FA (gas)
in the mole balance and rate laws
 More convenient to work in terms of the number of moles or molar
flow rate rather than conversion.
Membrane reactor, multiple reaction, and unsteady state
 Must write a mole balance on each species when molar flow
rates (Fi) and concentrations (Ci) are used as variables
b
c
d
A B 
 C  D
a
a
a
rC rD
rA
rB



a b c
d
 rA  k AC A C B
Liquid Phase
For liquid-phase reaction with no volume change  Concentration is the preferred variable
aA  bB  cC  dD
b
c
d
A B 
C D
a
a
a
Mole balance for liquid-phase reactions
Batch
dC A
 rA
dt
dC B b
 rA
dt
a
CSTR
v 0 (C A 0  C A )
V 
 rA
v 0 (C B 0  C B )
V 
 (b / a ) rA
PFR
dC A
v0
 rA
dV
dC B b
v0
 rA
dV
a
PBR
dC A
v0
 rA
dW
dC B b
v0
 rA
dW
a
Gas Phase
 For gas phase reactions
 need to be expressed in terms of the molar flow rates
dF j
dV
 rj
(1-11)
 Fj
C j  CT 0 
 FT
dy
- FT
=
dW
2y FTo
 P

 P
 0
y
 T0 
 
 T
 
(3-42)
(4-28, T=To)
 Total molar flow rate
n
FT 
F
j 1
j
= FA + FB + FC + FD + FI
Algorithm for Gas Phase Reaction
aA  bB 
 cC  dD
 Mole balances:
Batch
CSTR
PFR
dN A
 rAV
dt
F  FA
V  A0
 rA
dFA
 rA
dV
dN B
 rBV
dt
F  FB
V  B0
 rB
dFB
 rB
dV
dN C
 rCV
dt
F  FC
V  C0
 rC
dFC
 rC
dV
dN D
 rDV
dt
F  FD
V  D0
 rD
dFD
 rD
dV
Algorithm for Gas Phase Reaction
 Rate law:
 rA 
 
k AC A CB
 Stoichiometry:
Relative rate of reaction:
r
rA
r
r
 B  C  D
a b c
d
b
rB  rA
a
Concentration:
Total molar flow rate:
c
rC   rA
y a
d
rD   rA
a
C A  CT 0
FA P T0
FT P0 T
C B  CT 0
FB P T0
FT P0 T
CC  CT 0
FC P T0
FT P0 T
C D  CT 0
FD P T0
FT P0 T
FT  FA  FB  FC  FD
Algorithm for Gas Phase Reaction
 Combine:
dFA
    FA 

 k ACT 0 
dV
 FT 

dFC c
    FA 

 k ACT 0 
dV
a
 FT 
 FB

 FT

 FB

 FT








dFB
b
    FA 

  k ACT 0 
dV
a
 FT 
dFB d
    FA 

 k ACT 0 
dV
a
 FT 


 FB

 FT
 FB

 FT






- Specify the parameter values: kA, CT0, , , T0, a, b, c, d
- Specify the entering number: FA0, FB0, FC0, FD0, and final value: Vfinal
 Use an ODE solver


Microreactors
 Microreactors are used for the production of special chemicals,
combinatorial chemical screening, lab-on-a-chip, and chemical sensors.
Example 4-7: Gas-Phase Reaction in a Microreactor
The gas phase reaction, 2NOCl 2NO + Cl2, is carried out at 425oC and 1641 kPa
(16.2 atm). Pure NoCl is to be fed, and the reaction follows an elementary rate law.
It is desired to produce 20 tons of NO per year in a micro reactor system using a
bank of ten microreactors in parallel.
Each microreactor has 100 channels
with each channel 0.2 mm square and
250 mm in length. Plot the molar flow
rates as a function of volume down the
length of the reactor. The volume of
each channel is 10-5 dm3.
Rate constant and activation energy (given): k=0.29 dm3/mol-sec at 500K with E=24 kcal/mol
To produce 20 tons/year of NO at 85% conversion would require a feed rate of 0.0226 mol/s.
Solution
For one channel,
FAo=22.6 mol/s
FB=19.2 mol/s, X=0.85
2NOCl  2NO + Cl2
2A  2B + C
A  B +1/2C
1. Mole balance:
dF j
dV
dFA
= rA
dV
dFB
= rB
dV
dFC
= rC
dV
 rj
2. Rate law:
-rA=kCA2, k=0.29 dm3/mol-sec
3. Stoichiometry: Gas phase with T=To and P=Po, then v=vo(FT/FTo)
3-1. Relative rate
rB
rA
=
1
-1
rC
=
1/2
rB = -rA, rC= -0.5rA
3-2. Concentration
By applying Equation (3-42)
CA=CTo(FA/FT), CB=CTo(FB/FT), CC=CTo(FC/FT) with FT=FA+FB+FC
4. Combine
5. Evaluate
-rA=kCA2=kCTo2(FA/FT)2
dFA
= -kCTo2(FA/FT)2
dV
dFB
= kCTo2(FA/FT)2
dV
dFC = 0.5kC 2(F /F )2
To
A T
dV
CTo=Po/RT=(1641)/(8.314)(698K)
=0.286 mol/dm3=0.286 mmol/cm3
Use Polymath to solve the ODE
FA
FB
FC
Membrane Reactors
 Membrane reactors can be used to achieve conversions greater
than the original equilibrium value. These higher conversions are the
result of Le Chatelier's Principle; you can remove one of the reaction
products and drive the reaction to the right. To accomplish this, a
membrane that is permeable to that reaction product, but is
impermeable to all other species, is placed around the reacting
mixture
 By having one of the products pass throughout the membrane, we
drive the reaction toward completion
What kinds of membrane reactors are available?
 Membrane reactors are most commonly used when a reaction involves some
form of catalyst, and there are two main types of these membrane reactors: the
inert membrance reactor and the catalytic membrane reactor.
 The inert membrane reactor allows catalyst pellets to flow with the reactants on
the feed side (usually the inside of the membrane). It is known as an IMRCF,
which stands for Inert Membrane Reactor with Catalyst on the Feed side. In this
kind of membrane reactor, the membrane does not participate in the reaction
directly; it simply acts as a barrier to the reactants and some products.
 A catalytic membrane reactor (CMR) has a membrane that has either been
coated with or is made of a material that contains catalyst, which means that the
membrane itself participates in the reaction. Some of the reaction products (those
that are small enough) pass through the membrane and exit the reactor on the
permeate side.
Membrane Reactors
 Inert membrane reactor with catalyst pellets on the feed
side (IMRCF)
C6H12

C6H6 + 3H2
H2 molecule is small enough to diffuse through the small pore of the
membrane while C6H12 and C6H6 cannot..
Membrane Reactors
 Catalytic membrane reactor (CMR)
C6H12

C6H6 + 3H2
Membrane reactors are commonly used in dehydrogenation reactions (e.g.,
dehydrogenation of ethane), where only one of the products (molecular hydrogen) is
small enough to pass through the membrane. This raises the conversion for the
reaction, making the process more economical.
Dehydrogenation Reaction
 According to The DOE, an energy saving of 10 trillion Btu per year could result
from the use of catalytic membrane reactors as replacements for conventional
reactors for dehydrogenation reactions such as the dehydrogenation of
ethylbenzene to styrene.
A 
CH2CH3
B +C
CH=CH
+ H2 (B)
C4H10
C3H8


C4H8 + H2 (B)
C3H6 + H2 (B)
Basic Algorithm for Membrane Reactor (Example 4-8)
DV
RB
P=8.2 atm
FA
T=227oC
F
FA0=10mol/min B
FA
FB
FC
A  B +C
KC=0.05 mol @227oC
membrane
V+DV
RB=kcCB
There are
two “OUT”
terms
1. Mole balance:
for a differential mole balance on B
in the catalytic bed at steady state
for a differential mole balance on A
in the catalytic bed at steady state
FA V  FA V  DV  rA DV  0
dFA
 rA
dV
for a differential mole balance on C
in the catalytic bed at steady state
FB
V
 FB
V  DV
 RB DV  rB DV  0
dFB
 rB  RB
dV
dFC
 rC
dV
Basic Algorithm for Solving Reaction in the Membrane Reactor
2. Rate law:

C C
 rA  k  C A  B C
KC


;


rB  rA ; rC  rA
3. Transport out the sides of the reactor:
RB  kC CB
kc is a transport coefficient. kc=f(membrane & fluid properties, tube diameter…)  constant
4. Stoichiometry:
C A  CT 0
F
FA
F
; C B  CT 0 B ; CC  CT 0 C
FT
FT
FT
FT  FA  FB  FC
 rA  rB  rC
Basic Algorithm for Solving Reaction in the Membrane Reactor
5. Combining and Summarizing:
dFA
 rA ;
dV
 FB
dFB
 rA  k c CT 0 
dV
 FT
 FA
 rA  k c CT 0 
 FT
 CT 0
 
 KC

;

 FB

 FT
dFC
 rA
dV
 FC

 FT



FT  FA  FB  FC
6. Parameter evaluation:
CT0=0.2 mol/L, k=0.7 min-1, KC=0.05 mol/L, kc=0.2 min-1
FA0=10 mol/min, FB0=FC0=0
7. Numerical solution:
Solve with POLYMATH or MATLAB
Effects of Side Stream, RB=kcCB
in a membrane reactor
10
kc=0.20 min-1
FC
FA
FB
FA
5
Conversion
X=(10-4)/10=0.6
FB
FC
0
0
100
200
300
400
500
Reactor volume, V [L]
Large side stream
Little side stream
10
10
FB
kc=20
FC
FA
FA
5
FC
0
0
FB
FA
FB
100
200
kc=0.0022 min-1
min-1
FA
5
FB FC
FC
300
400
Reactor volume, V [L]
500
0
0
100
200
300
400
Reactor volume, V [L]
500
Unsteady-State Operation of Stirred Reactors
 Determine the time to reach steady-state operation
 Predict the concentration and conversion as a function of time
B
C
A
 Semibatch reactor
(b) ammonolysis, chlorination, hydrolysis
(c) acetylation reaction, esterification reaction
A, B
Startup of a CSTR
Time to Reach Steady State for a First-Order Reaction in a CSTR
To determine the time to reach steady-state operation of a CSTR, we begin with the general
mole balance equation applied to a well-mixed CSTR.
FA0  FA  rAV 
dN A
dt
(4-45)
Utilizing the definitions of FA and NA, we have
C A0 v0  C A v  rAV 
d (C AV )
dt
Conversion does not have any meaning in startup because one cannot separate moles
reacted from moles accumulated. Consequently, we must use concentration as the
variable in our balance equation. For liquid-phase reactions V =V0 and for a constant
overflow, v =v0. After dividing by v0 and replacing V/v0 by the space time , we find that
C A0  C A  rA   
dC A
dt
(4-46)
Startup of a CSTR
For a first-order reaction:
rA  kCA
C
dC A 1  k

C A  A0
dt


Solution
CA 
C A0
1  k

t 

1

exp

(
1


k
)







(4-47)
Letting ts be the time necessary to reach 99% of the steady state concentration, CAS:
C AS 
C A0
1  k
Rearranging (4-46) for CA=0.99CAS yields
t S  4 .6

1  k
(4-48)
Startup of a CSTR
Time to reach steady state in an isothermal CSTR

t S  4 .6
1  k
(4-48)
For slow reactions :
t S  4.6
For rapid reactions:
tS 
4.6
k
(4-49)
(4-50)
For most first-order
system, steady state
is achieved in three
to four space times
Semibatch Reactors
A B 
 C
0
B
Semibatch
reactor
volume as a
function of time
0
B
 v0C A  VrA 
A
A
0
- [ out ]

0
rAV 
+ [ gen. ]

rAV (t )
 0 v0
=
[ acc. ]

dN A
dt
d C AV  VdC A C A dV


dt
dt
dt
[ in ]
(4-51)
(4-52)
0
dV
 v0
dt

0

(4-56)
d (V )
dt
for    0
- [ out ]
FB 0 
0

+ [ gen. ]
=
[ acc. ]
rBV (t )

dN B
dt
dN B
 rBV  FB 0
dt
(4-57)
dN B d (VC B )
dC B
dV

 CB
V
dt
dt
dt
dt
- [ out ] + [ gen. ] = [ acc. ]

VdC A
dt
Mole balance on specials B:
Mass balance of all specials:
[ in ]
(4-55)
v
dC A
 rA  0 C A
dt
V
Mole balance on specials A:
[ in ]
V  V0  0t
(4-53)
(4-54)
rBV  FB 0  rBV  v0 C B 0
v (C  C B )
dCB
 rB  0 B 0
dt
V
(4-58)
Example 4-9: Isothermal Semibatch Reactor with 2nd–order Reaction
CNBr + CH3NH2  CH3Br + NCNH2
A + B
 C + D
dN C
 rCV   rAV
dt
dN C
d (CCV )
dCC
dV

V
 CC
dt
dt
dt
dt
Isothermal elementary reaction
in a semibatch reactor
t=0,
CA=0.05 gmol/, CB=0.025 gmol/ℓ,
v0=0.05ℓ/s, k=2.2ℓ/s·mol, V0=5ℓ
V
Mole balance of A, B, C, and D
rA  kCAC B
A
v
dC A
 kCAC B  0 C A
dt
V
B
v
dCB
 kCAC B  0 (C B 0  C B )
dt
V
V  V0  v0 t
dCC
 v0 CC
dt
C
dCC
v
 kCAC B  0 CC
dt
V
D
v
dC D
 kCAC B  0 C D
dt
V
Conversion, X
X 
N A0  N A C A0V0  C AV

N A0
C A0V0
Concentration-time Trajectories
in Semibatch Reactor
0.05
CNBr + CH3NH2  CH3Br + NCNH2
(A)
(B)
(C)
(D)
0.04
Concentration
t=0,
CA=0.05 gmol/, CB=0.025 gmol/ℓ,
v0=0.05ℓ/s, k=2.2ℓ/s·mol, V0=5ℓ
0.03
CA
0.02
CC
0.01
CB
0.00
0
100
200
300
Time
400
500
Reaction Rate-time Trajectories
in Semibatch Reactor
Reaction rate [mole/s•L)
0.0025
CNBr + CH3NH2  CH3Br + NCNH2
(A)
(B)
(C)
(D)
0.0020
t=0,
CA=0.05 gmol/, CB=0.025 gmol/ℓ,
v0=0.05ℓ/s, k=2.2ℓ/s·mol, V0=5ℓ
0.0015
0.00010
0.00005
0.00
0
50
100
150
200
250
Reactor Equations in terms of Conversion
in semibatch reactor
A B  C  D 
The number of moles of A remaining at any time , t
0
0
B
B
number of moles  number of moles  number of moles 

 
 

 of A in the vat    of A in the vat    of A reacted up 

 
 

at time t
initially
to time t

 
 

NA

N A0

N A0 X
A
(4-59)
where X is the mole of A reacted per mole of A
initially in the vat.
The number of moles of B remaining at any time , t
number of moles  number of moles  number of moles  number of moles 

 
 
 

 of B in the vat    of B in the vat    of B added to    of B reacted up 

 
 
 

at time t
initially
the vat
to time t

 
 
 

NB

N Bi

t
F
0
B 0 dt

N A0 X
(4-60)
For a constant molar feed rate
N B  N Bi  FB0t  N A0 X
(4-61)
A
Reactor Equations in terms of Conversion
in semibatch reactor
0
0
B
B
A B  C  D
The concentration of A and B
A
A
CA 
N (1  X )
NA
 A0
V
V0  v0 t
CB 
N  FB 0 t  N A0 X
NB
 Bi
V
V0  v0 t
CC 
NC
N A0 X

V
V0  v0 t
CA 
N A0 X
ND

V
V0  v0 t
A mole balance on specials A:
rAV 
dN A
dt
(4-62)
The number of moles of C and D cab be taken
directly from the stoichiometric table
N C  N Ci  N A0 X
(4-63)
N D  N Di  N A0 X
Combine
The rate law (reversible 2nd-order reaction)

C C
 rA  k  C AC B  C D
KC






(4-66)

dX k 1  X N Bi  FB 0t  N A0 X   N A0 X 2 / K C

dt
V0  v0t
(4-65)
(4-66) can be solved numerically

Equilibrium conversion
in semibatch reactor
0
0
B
B
A
A
For reversible reactions carried out in a
semibatch reactor, the maximum attainable
conversion (i.e., the equilibrium conversion) will
change as the reaction proceeds because more
reactant is continually to the right.
A B
CD
The rate law (reversible 2nd-order reaction)

C C
 rA  k  C AC B  C D
KC

Equilibrium
conversion in a
semibatch reactor




KC 
CCe C De ( N Ce / V )( N De / V )

C AeC Be ( N Ae / V )( N Be / V )
KC 
N Ce N De
N Ae N Be

( N A0 X e )( N A0 X e )
N A0 (1  X e )( FB 0 t  N A0 X e )
N A0 X e2

(1  X e )( FB 0 t  N A0 X e )
N A0
t
K C FB 0
2

X
e
 KC X e 

1 X e





(4-68)
(4-69)
2



FB 0 t 
FB 0 t 
F t
   K C 1 
  4( K C  1) K C B 0
K C 1 
N A0 
N A0 
N A0



Xe 
2( K C  1)
(4-70)
Reactive Distillation

The distillation of chemically reacting mixtures has become increasingly
common in chemical industries.
 Carrying out these two operations, reaction and distillation, simultaneously in a
single unit results in significantly lower capital cost and operating costs.

Reactive distillation is particularly attractive when one of the reaction products
has a lower boiling point, resulting in its volatilization from the reacting liquid
mixture.
Reactive Distillation
 An example of reactive distillation is the production of methyl acetate:

By continually removing the volatile reaction product, methyl acetate, from
the reacting liquid-phase reaction, the reverse reaction is negligible and the
reaction continues to proceed towards completion in the forward direction.
 Reactive distillation is used with reversible, liquid phase reactions.
Suppose a reversible reaction had the following chemical equation :
Reactive Distillation
 For many reversible reactions the equilibrium point lies far to the left and
little product is formed :
 However, if one or more of the products are removed more of the product
will be formed because of Le Chatlier's Principle :
 Removing one or more of the products is one of the principles behind
reactive distillation. The reaction mixture is heated and the product(s) are
boiled off. However, caution must be taken that the reactants won't boil off
before the products.
Homework
P4-11B
P4-12B
P4-13B
P4-14C
Due Date: Next Week