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ERT 316
ISOTHERMAL REACTOR DESIGN
PART 1
Copyright Cheng 05
START
1. The general mole equation
2. Design Equations: • Batch • CSTR • PFR
3. Is –rA=f(X) given?
4. Determine the rate law in terms of the concentration
of the reacting species
5. Use Stoichiometry to express concentration as a function of conversion • Liquid phase or Gas phase • Constant Volume Batch • Constant P and T
Evaluate the algebraic (CSTR) or integral (PFR)
equations
6. Combine steps 4 and 5 to obtain –rA=f(X)
END
YES
NO
Algorithm for Isothermal Reactor
4.1 Design Algorithm for Isothermal Reactors
• To design an isothermal reactors, the following sequence is
highly recommended.
4.1 Design Algorithm for Isothermal Reactors
To carry out the evaluation, the following method can
be used:
Graphically (Chapter 2 plot)
Numerical (Quadrature Formulas Chapter 2 and
Appendix A4)
Analytical (Integral Tables)
Software (Polymath)
Copyright Cheng 05
Algorithm for isothermal
Reactor (PFR) reactor
volume for 1st order gas-
phase reaction
Scale up of Batch Reactor to
the design of CSTR
- scale up of a pilot plant - costly
- Instead - build microplant (laboratory
bench scale)
- example: analyze data from a
laboratory batch reactor-determine
specific reaction rate, k - use it to
design full-scale flow reactor
Batch Operation
A B
Step 1: Write the mole balance
Step 2: Write the rate law
Step 3: Derive concentration term from stoichiometry
Vrdt
dXN AA 0
- Calculation of time taken to achieve a given conversion X
2
AA kCr Example for second order
reaction
Batch Operation
A B
Step 4: Combine equation from step 1,2,3
Step 5: Evaluate
2
0 1 XkCdt
dXA
- Calculation of time taken to achieve a given conversion X
t X
A X
dX
kCdt
0 0
2
0 1
1
This is the reaction time or tR
Algorithm to estimate reaction time
Mole Balance
Rate Law First order Second order
Stoichiometry
Combine
Evaluate
(integrate)
Vrdt
dXN A
R
A 0
2
AA kCr AA kCr
2
0 1 XkCdt
dXA
R
Xkdt
dX
R
1
Batch Operation
A B
To reach 90% conversion in a constant-volume batch reactor scales:
if k = 10-4 s-1
- Calculation of time taken to achieve a given conversion X
For first order
4.2 Design of CSTR
Step 1: Write the mole balance of CSTR
Design equation for CSTR is
If volumetric flow rate does not change with the
reaction, (i.e. v = v0), then
exitA
AO
exitA
AO
r
XCv
r
XFV
)()(
0
A
A
A
A
r
XC
vr
XCv
v
V
0
0
00
0
where is the space time
Step 2: Write the rate law
For 1st order irreversible reaction,
Step 3: Derive concentration in terms of conversion
(from stoichiometry)
Step 4: Combine eq from step 1,2, 3
Rearranging;
4.2 Design of CSTR
k
kX
1
AA kCr
X
X
k 1
1
is often referred to
as Damköhler number
(for 1st order)
k
Step 2: Write the rate law
For 1st order irreversible reaction,
Step 3: Derive concentration in terms of conversion
(from stoichiometry)
Step 4: Combine eq from step 1,2, 3
Rearranging;
4.2 Design of CSTR
k
CC A
A
1
0
k
kX
1
AA kCr
X
X
k 1
1
Damköhler number
• Is the ratio of the rate of reaction of A to the rate of convective transport of A at the entrance to the reactor.
rate of reaction at entrance
entering flow rate of A
• For first order irreversible reaction;
• For second order irreversible reaction;
0
0
A
A
F
VrDa
kCv
VkC
F
VrDa
A
A
A
A
00
0
0
0
0
00
2
0
0
0A
A
A
A
A kCCv
VkC
F
VrDa
How to estimate degree of conversion for a CSTR?
By using Damkohler number,
0
0
A
A
F
VrDa
If Da 0.1, X < 0.1 If Da 10, X > 0.9
If first degree order, Da = k If second degree order, Da =kCA0
4.2 Design of CSTR (for first order)
For CSTRs in series, conversion as a function of
the number of tanks in series:
For CSTRs in parallel, conversion is:
nkX
)1(
11
k
kX
1Just like a single
CSTR
Example: Producing 200 Million
Pounds per Year in a CSTR It is desired to produce 200 million pounds per
year of ethylene glycol (EG). The reactor is to
be operated isothermally. A 1lb mol/ft3 solution
of ethylene oxide (EO) in water is fed to the
reactor shown in figure 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 800-gal reactors were arranged in parallel,
what is the corresponding conversion?
(c) If 800-gal reactors were arranged in series,
what is the corresponding conversion?
CBA catalyst
Example: Producing 200 Million
Pounds per Year in a CSTR
Extract the given information:
FC = 2 x 108 lbm/yr x 1 yr/365 days x 1day/24 h x 1hr/60 min x 1lbmol/62lbm
= 6.137 lbmol/min
From reaction stoichiometry,
FC = FA0X
min67.7
8.0
137.60
lbmol
X
FF C
A
Example: Producing 200 Million
Pounds per Year in a CSTR
STEP 1: Design equation of CSTR
STEP 2: Rate Law
STEP 3: Stoichiometry (Liquid phase, v = v0 )
STEP 4: Combining;
exitA
AO
r
XFV
)(
AA kCr
)1()(
00
Xk
Xv
r
XFV
exitA
A
Example: Producing 200 Million
Pounds per Year in a CSTR
STEP 5: Evaluate
The entering volumetric flowrate of stream A, with CA01 = 1 lb mol/ft3 before mixing is;
From the problem statement,
Thus, the total entering volumetric flow rate of liquid is
Substituting all the values to calculate volume of reactor;
min67.7
/1
min/67.7 3
3
01
00
ft
ftlbmol
lbmol
C
Fv
A
AA
00 AB vv
min34.1567.767.7
3
000
ftvvv BA
galft
ft
Xk
Xv
r
XFV
exitA
A 14763.1978.01min311.0
8.0
min34.15
)1()(
3
1
3
00
v0VV
Example: Producing 200 Million
Pounds per Year in a CSTR
b) CSTR in parallel.
Rearranging the equation of volume in part a)
)1(
0
Xk
XvV
)1(0 Xk
X
v
V
)1( Xk
X
)1( X
Xk
)1( k
kX
)1( Da
Da
Example: Producing 200 Million
Pounds per Year in a CSTR
b) CSTR in parallel.
)1( k
kX
)1( Da
Da
min94.13min/67.7
1
48.7
1800
3
3
0
ftgal
ftgal
v
V
34.4min311.0min94.13 1 kDa
81.0)34.41(
34.4
X
Example: Producing 200 Million
Pounds per Year in a CSTR
c) CSTR in series
min97.6min/34.15
1
48.7
1800
3
3
0
ftgal
ftgal
v
V
167.2min311.0min97.6 1 kDa
nkX
)1(
11
90.0
167.21
11
2
X
4.3 PFR Assume no dispersion and no radial gradients in either
temperature, velocity, or concentration and in the absence of
pressure drop or heat exchange.
STEP 1: Write the mole balance of PFR:
STEP 2: Write the rate law
Eg: For second order,
X
A
Ar
dXFV
0
0
2
AA kCr
4.3 PFR STEP 3: Write concentration in terms of conversion
(from stoichiometry)
XCC AA 10
For liquid phase
For gas phase
x
XCC AA
1
10
4.3 PFR STEP 4: Combine all the equations
Rearranging,
X
AA
A
X
X
kC
v
X
dX
kC
FV
0 0
0
22
0
0
11For liquid phase
For gas phase
X
A
A dXX
X
kC
FV
0
2
2
2
0
0
1
1
X
XXX
kC
v
A 1
)1()1ln()1(2
22
0
0
2
2
0
0
11 Da
Da
kC
kCX
A
A
Design a PFR: summary
X
X
kC
Cv
X
dX
kC
FV
A
AX
A
A
112
0
00
0 22
0
0
• mole balance
• rate laws
• Stoichiometry
• combination
AA rdV
dXF 0
2
AA kCr
)1(0 XCC AA
or
2
2
0
0
11 Da
Da
kC
kCX
A
A
No pressure drop
No heat exchange
X
A
Ar
dXFV
00
Damköhler number for 2nd-order reaction
In case of 2nd order rxn, liquid phase, isothermal
Design a PFR: summary
X
A
A dXXkC
XFV
0 22
0
2
01
1
• mole balance
• rate laws
• Stoichiometry
• combination
AA rdV
dXF 0
2
AA kCr
)1(
)1(
)1(
)1(
)1(0
0
0
0 X
XC
Xv
XF
Xv
F
v
FC A
AAAA
No pressure drop
No heat exchange
X
A
Ar
dXFV
00
X
XXX
kC
vV
A 1
1)1ln()1(2
2
2
0
0
In case of 2nd order rxn, gas phase, isothermal
Pressure Drop In Reactors
Isothermal Reactor Design – Part 2
Design a PBR
2
0
2
0 1
1
P
P
X
X
v
kC
dW
dX AO
• mole balance
• rate laws
• stoichiometry
• combination
AA rdW
dXF 0
2
AA kCr
T
T
P
P
X
XCC AA
0
0
0)1(
)1(
PXFdW
dX,1
In case of 2nd order rxn, gas phase, isothermal
Need to relate pressure
drop to catalyst weight
(in order to determine
conversion)
= pressure (kPa) = inlet pressure (kPa) = temperature (K) = inlet temperature (K)
= porosity = volume of void = void fraction = volume of solid
total bed volume total bed volume
= cross sectional area (m2)
=diameter of particle in the bed, ft (m)
=viscosity of gas passing through the bed, (kg/m.s)
=length down the packed bed of pipe, ft (m)
=superficial velocity = volumetric flow ÷ cross-sectional area of pipe (m/s)
=gas density (kg/m3) = solid density (kg/m3) = inlet gas density
= = superficial mass velocity, (kg/m2.s)
Design a PBR
P
1
pD
G
cA
z
u
Ergun equation
c0
0P 0TT
XPP
P
T
T
dW
dP
1
2 0
0
0
G
DDg
G
ppc
75.111501
3
0
0
0
0
1
2
PA cc
cg = 32.174lbmft/s2lbf
(conversion factor)
For isothermal operation, we have two sets of equation with
two unknowns, X & P
(1) (2)
Special case: if ε=0, an analytical solution to second equation is
obtained as follows
Design a PBR
2
00 1
1
P
P
X
X
v
kC
dW
dX AO
X
PP
P
T
T
dW
dP
1
2 0
0
0
21
0
1 WP
P Used only when ε=0
Design a PBR
2
0
2
0 1
1
P
P
X
X
v
kC
dW
dX AO
• mole balance
• rate laws
• stoichiometry
• Combination
• Solve
AA rdW
dXF 0
2
AA kCr
T
T
P
P
X
XCC AA
0
0
0)1(
)1(
In case of 2nd order rxn, gas phase, isothermal
WXv
kC
dW
dX AO 112
0
When ε=0
21
1
0 WW
X
X
kC
v
AO
By integration;
Solving for conversion gives:
Solving for catalyst weight,
Design a PBR
21
1
0 WW
X
X
kC
v
AO
211
21
0
0
0
0
W
v
WkC
W
v
WkC
XA
A
21
00 1//211 XXkCvW A
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
↑ W, ↓P, ↓CA
↑ ΔP, ↓P, ↓CA, ↓-rA
↑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
.
hcatkgkmol
mk
612
(a) Conversion for ΔP = 0
α = 0 thus,
[Volume of catalyst] x [catalyst density]
211
21
0
0
0
0
W
v
WkC
W
v
WkC
XA
A
Effect of pressure drop on the conversion profile
0
0
0
0
1v
WkC
v
WkC
XA
A
ccLAW 1
32 /1923200013.045.01 mkgmm
kg5.27
6.4/15.7
15.27/1.0
123
36
0
0
hmkgmkmol
hcatkgkmol
m
v
WkCA
82.06.41
6.4
X
(b) Conversion with pressure drop
211
21
0
0
0
0
W
v
WkC
W
v
WkC
XA
A
Effect of pressure drop on the conversion profile
0
0
1
2
PA ccc
693.049.06.41
49.06.4
X
kPamkgm
mkPa
101345.01/19230013.0
/8.25232
1037.0 kg
59.0
2
5.27037.01
21
1
kgkgW
(c) Conversion when catalyst diameter were doubled.
(increase by a factor of 2, )
From , thus
Effect of pressure drop on the conversion profile
dominant
3
0
2
0
175.1
pcDg
G
pD
10
0
0
1
2
PA ccc
0
pD
10
122 pp DD
G
DDg
G
ppc
75.111501
3
0
0
(c) Conversion when catalyst diameter were doubled.
Thus,
Conversion increases from 0.693 to 0.774 by increasing catalyst diameter by a factor of 2.
Effect of pressure drop on the conversion profile
122 pp DD
11
12 0185.02
1037.0
2
1
kgkg
D
D
p
p
774.0
2
5.270185.016.41
2
5.270185.016.4
kg
kg
X
Increasing particle size
decrease the pressure
drop parameter, increase
conversion & reaction
rate