Decomposition Type
Analysis of Continuous Reactors
▪ PFR: Liquid Reactions
For reactant A:
VPFR =
Lecture No. 6.1 A
PLUG FLOW REACTORS (PFR)
or Tubular Reactors (TR)
dFA
FAo r
A
FA
VPFR =
vo
k A C A o n −1
VPFR =
For n=1
A. Constant Volume ( A=0, for liquids)
B. Variable Volume ( A 0, for gases)
For n 1
Ch.E. 2115
Lola Domnina B. Pestaňo, Ph.D.
XA
0
dX A
(1 − X A )n
VPFR =
=0
dCA
−k A CA n
CA
CAo
PFR
vo
− ln (1 − X A )
kA
=
PFR
vo
1
−1
k A ( n − 1) CAo n −1 (1 − X A )n −1
=
1
1
−1
k A ( n − 1) CAo n −1 (1 − X A )n −1
Fj
dFj
Fjo
rj
VPFR = v o
1
k A C A o n −1
VPFR =
PFR
1
VPFR = v o
A
=
XA
0
XA
0
−CAodX A
−k A CA n
dX A
(1 − X A )n
1
− ln (1 − X A )
kA
NOTE: The space time
equations are the same
as the time formulas for
CV Batch Reactors
2
@UST Ch.E. Department
2
Problems
Decomposition Type
▪ PFR: Gaseous Reactions
VPFR =
0
A
Fj
dFj
Fjo
rj
PS
3A
For reactant A:
VPFR =
FA
FAo
dFA
rA
VPFR = v o
VPFR = v o
PFR
=
dCA
−k A CA n
VPFR = v o
−CAodX A
XA
o
NAo (1 − X A )
Vo (1 + A X A )
−k A
VPFR =
CA
CAo
v oCAo
k A CAon
1
k A CAon −1
XA
o
XA
o
1 + A XA
1 − XA
1 + A XA
1 − XA
XA
0
−CAodX A
−k A CA n
For Variable Volume
(Constant Pressure) gas
reactions: V=Vo(1+ AXA)
n
PS
3B
n
dX A
(variable density)
Solved numerically
n
PS
3C
dX A For gas reactions
3
@UST Ch.E. Department
3
3.Repeat Problems 1 and 2 if the decomposition is
now 2nd order and half-life from batch reactor
data is still 30 min for the same CAo in all cases.
4.Given the gas phase elementary reaction: 2A → R
with a feed that is 80%A, 20% inerts. If batch data
at constant pressure gave a half-life of 30 min, what
space time will be needed in a PFR to attain 80%
conversion of A? (Answer: 80.15mins)
5.Repeat 4 if the reaction is first order.
(Answer: 55.65mins)
6.Repeat 4 and 5 to determine the % conversion in
the PFR if the space time is to be 2 hours.
(Answer: 0.8691, 0.98105)
4
@UST Ch.E. Department
4
Problem 4
4.Given the gas phase
elementary reaction:
2A → R with a feed that
is 80%A, 20% inerts. If
batch data at constant
pressure gave a half-life
of 30 min, what space
time will be needed in a
PFR to attain 80%
conversion of A?
(Answer: 80.15mins)
Given:
gas phase elem. reaction:
2A → R n = 2
yAo=0.8 yIo=0.2
Constant Pressure (VV)
Batch reactor: t1/2 = 30 mins
5
@UST ChE Department
Problem 4
Required: PFR: =? if XA=0.8
Required: PFR: =? if XA=0.8
4.Given the gas phase
Solution:
elementary reaction:
0.0292
vol
For
n=2, 2A → P: k A =
2A → R with a feed that
CA mole − min.
PFR:
Gas
Phase
is 80%A, 20% inerts. If
n
batch data at constant
X
1 + A XA
1
dX A
pressure gave a half-life
PFR =
k A CAon −1 o
1 − XA
of 30 min, what space
2
time will be needed in a
X
1 + A XA
1
dX A
PFR to attain 80%
PFR =
k A CAo o
1 − XA
conversion of A?
when XA=0.8 PFR=?
(Answer: 80.15mins)
2
Given:
0.8 1 − 0.4X
1
A
dX A
gas phase elem. reaction:
PFR =
0
0.0292
1 − XA
CAo
2A → R n = 2
CAo
yAo=0.8 yIo=0.2
Constant Pressure (VV)
PFR = 80.15 mins.
Batch reactor: t1/2 = 30 mins
6
Solution:
Detm. kA from Constant
Pressure (or VV) Batch data:
o
For n 1, VVBR decomposition:
t=
1
k A CnAo−1
XA
X A )n −1dX A
(1 − X A )n
(1 +
A
0
V
=
Vo
A
A XA
A
For: 2A → R:
A
=
A
= (1 − 2) / 2 = −0.5
y AO = −0.5(0.8) = −0.4
1
kA =
30 * C Ao
kA =
A
0.5
0
(1 − 0.4X A )dX A
(1 − X A )2
0.0292
vol
CAo mole − min.
5
@UST Ch.E. Department
@UST Ch.E. Department
6
1
At the end of this topic, you should be
able to:
Analysis of Continuous Reactors
▪ Understand the concept of space time ( )
and space velocity (SV) for continuous
reactors.
▪ Use analytical or numerical methods to
solve the steady-state applications for:
▪ Constant Stirred Tank Reactor (CSTR) or
Mixed Flow Reactor (MFR)
▪ Plug Flow Reactor (PFR) or Tubular
Reactor (TR)
▪ Combination of CSTR and PFR
Lecture No. 6A
A. Reactors in Series
B. Space Time
C. Numerical and Analytical Method
Ch.E. 2115
Lola Domnina B. Pestaňo, Ph.D.
@UST Ch.E. Department
@UST Ch.E. Department
1
2
2
Review of Continuous Reactors
Review of Continuous Reactors
MFR or CSTR
TR or PFR
Reactants are continuously fed
to a tube so that what goes in
first also comes out first.
Reaction proceeds across the
reactor length and
concentrations vary from
entrance to exit as there is no
lateral mixing. Radial
concentrations are however
assumed constant.
MFR or CSTR Reactants are
continuously fed to a well
mixed tank and the products
are also withdrawn
continuously. At steady state,
the concentration of any
material in the tank is also
the same as that in the
product stream.
@UST Ch.E. Department
3
3
@UST Ch.E. Department
4
4
MFR or CSTR
Plug Flow Reactor (PFR)
Steady State Flow:
No Spatial Variation:
5
UST Chemical Engineering Dept.
5
@UST Ch.E. Department
6
6
1
Plug Flow Reactor (PFR)
Summary of Mole Balances
FA = C A v
7
UST Chemical Engineering Dept.
7
@UST Ch.E. Department8
8
Reactors in Series
• reactors are connected together so that the exit
stream of one reactor is the feed stream for another
reactor.
FA 0
XA0 = 0
I. Reactors in Series
FA1 , X A1
V1
V2
X Ai =
FA3 , X A3
V3
FA 2 , X A 2
total number of moles of A reacted up to pt. i
mole of A fed into the first reactor
10
@UST Ch.E. Department
9
10
FA 0
V1
XA0 = 0
FA1 , X A1
FA 0
XA0 = 0
FA1 , X A1
V1
V2
V2
FA3 , X A3
V3
Consider the figure above,
X Ai =
FA 2 , X A 2
Entering Molar
rate of A
Considering the individual reactors above,
FA1
Reactor 1: FA1 = FA0 − FA0 X A1
Reactor 2: FA 2 = FA0 − FA0 X A 2
11
FA 2 , X A 2
Consider the CSTR:
total number of moles of A reacted up to pt. i
mole of A fed into the first reactor
Reactor 3: FA3 = FA0 − FA0 X A3
FA3 , X A3
V3
Rearranging,
11
@UST Ch.E. Department
-
Exit Molar
rate of A
+
Rate of
generation of A
=0
-
FA 2
+
rA 2 V2
=0
V2 =
FA1 − FA 2
-rA 2
12
@UST Ch.E. Department
12
2
Space Time ( )
II. Space Time,
Space time, : the time necessary to process one
reactor volume of fluid based on entrance
conditions.
and
=
Space Velocity, SV
V
vo
vo = entering volumetric flow rate
Space velocity (SV): number of reactor
volumes of feed at specified conditions which
can be treated in unit time
SV =
vo
1
=
V
14
@UST Ch.E. Department
13
14
Space Time ( )
Space Time ( )
20 m
20 m
Fluid directly
upstream the reactor
Example:
= 2 min
A space time of 2 minutes means that every 2 min
one reactor volume of feed at specified conditions
is being treated by the reactor.
REACTOR
* space time is also referred to as “holding
time” or “mean residence time”
Example: SV = 5/hr
* considering the figure above: the time it
takes for the fluid to enter the reactor
Thus, a space-velocity of 5 hr-1 means that five
reactor volumes of feed at specified conditions are
being fed into the reactor per hour.
completely is referred to as the space time.
15
@UST Ch.E. Department
15
16
@UST Ch.E. Department
16
Covered Applications
Determination of Reactor Volume and/or
Space time
▪ Single MFR or CSTR
III. Numerical and Analytical
▪ Equivalent PFR
▪ Combination of Reactors
Method
Determination of Over-all Conversion
▪ Single MFR or CSTR
▪ Equivalent PFR
▪ Combination of Reactors
Determination of Reaction Kinetics
▪ Unknown Order and Rate Constant
18
UST Chemical Engineering Dept.
17
18
3
Decomposition Type (aA → P, -rA=kACAn)
Decomposition Type (aA → P, -rA=kACAn)
▪ Single CSTR
▪ Multiple CSTRs in Series
V =
v o ( C jo - C j )
V =
- rj
V=
- rj
=
F jo - F j
C jo - C j
C
=
- rj
jo
= C j - rj
=
F jo X j
Vm =
- rj
v o (C jo -C j )
V=
kC jn
C jo - C j
kC j
C
jo
= C j + kC
Vm =
Xj
=
n
vo Xj
kC jo n-1(1-X j )n
k C jo
n -1
m
(1 - X j ) n
=
v o ( C jm −1 - C jm )
- r jm
m
- r jm
jm − 1
= C
jm
19
@UST Ch.E. Department
19
m
=
=
Fjm −1X jm
-rjm
v o ( C jm −1 - C jm )
k m C jm
n
C jm −1 - C jm
k m C jm
C
jm − 1
τm =
n
= C
Vm =
jm
+ kmC
v o C Ao (X jm - X jm -1 )
k m C jo n (1 - X jm )n
C jo (X jm - X jm -1)
k m C jo n (1 - X jm )n
n
jm
m
20
@UST Ch.E. Department
20
Decomposition Type (aA → P,
Problems
-rA=kACAn)
1. A first order liquid decomposition gave a half-life
of 30 min in a batch reactor. What total space time
will be needed to attain 80% conversion in:
a. One CSTR (Answer: 173.16 mins)
b. A PFR (Answer: 69.67 mins)
c. Two equal sized CSTRs (Answer: 107.02mins,
where XA1 = 0.5528)
d. Equal sized CSTR and PFR in series
(Answer: 81.76mins, where XA1 = 0.4857)
e. Equal sized PFR and CSTR in series
(Answer: 81.76mins, where XA1 = 0.6111)
▪ Single MFR or CSTR
Fjo -Fj
V=
-rj
=
Fjo X j
V =
-rj
v o ( C jo - C j )
- rj
C jo - C j
=
- rj
▪ Multiple MFR or CSTR in Series
Vm =
Fjm −1-Fjm
-rjm
=
Fjo (X jm − X jm −1 )
Vm =
-rjm
v o ( C jm −1 - C jm )
- r jm
C jm −1 - C jm
=
m
- r jm
▪ Single TR or PFR
V=
Fj
dFj
Fjo
rj
V = vo
Cj
dC j
C jo
rj
=
Cj
dC j
C jo
rj
C jm
dC j
C jm−1
rj
=
XA
−C j0 dX j
rj
0
▪ TR or PFR in series
Vm =
Fjm
dFj
Fjm−1
rj
Vm = v o
C jm
dC j
C jm−1
rj
m
=
m
=
X jm
−C j0 dX j
X jm−1
rj
21
@UST Ch.E. Department
21
22
@UST Ch.E. Department
22
Problems
Problems
2. A first order liquid decomposition gave a half-life of
30 min in a batch reactor. What % over-all conversion
is attained in
a. One CSTR, space time of 1 hour(Answer: 58%)
b. A PFR, space time of 1 hour (Answer: 75%)
c. Two equal sized CSTRs, each with a space time of
30 min (Answer: 65.11% where XA1 = 0.41)
d. CSTR and PFR in series, each with a space time of
30 min
(Answer: 70.5% where XA1=0.41)
e. PFR and CSTR in series, each with a space time of
30 min. (Answer:70.5% where XA1 =0.5)
3.Repeat Problems 1 and 2 if the decomposition is
now 2nd order and half-life from batch reactor
data is still 30 min for the same CAo in all cases.
4.Given the gas phase elementary reaction: 2A → R
with a feed that is 80%A, 20% inerts. If batch data
at constant pressure gave a half-life of 30 min, what
space time will be needed in a PFR to attain 80%
conversion of A? (Answer: 80.15mins)
5.Repeat 4 if the reaction is first order.
(Answer: 55.65mins)
6.Repeat 4 and 5 to determine the % conversion in
the PFR if the space time is to be 2 hours.
(Answer: 0.8691, 0.98105)
PS
3A
PS
3B
PS
3C
23
@UST Ch.E. Department
23
- rm j
-rjm
Vm =
C jm −1 - C jm
C
n
j
Fjm −1-Fjm
24
@UST Ch.E. Department
24
4
At the end of this topic, you should be
able to:
▪ Apply the mole balance and its
applications to steady state:
▪ Constant Stirred Tank Reactor
(CSTR) or Mixed Flow Reactor
(MFR)
▪ Plug Flow Reactor (PFR) or
Tubular Reactor (TR)
▪ Combination of CSTR and PFR
▪ Use both graphical and
numerical or analytical methods
to solve the applications
Analysis of Continuous Reactors
Lecture No. 6B
Graphical Method
Ch.E. 2115
Lola Domnina B. Pestaňo, Ph.D.
@UST ChE Department
1
2
Covered Applications
Common Nomenclatures
Determination of Reactor Volume
▪FJo = input molar rate of specie J = voCJo
▪FJf = output molar rate of specie J = vfCJf
▪vo, vf = volumetric feed & exit rates
▪Cjo, CJf = inlet & outlet concentrations
▪-rJ = rate of reaction of J
▪XJ = (FJo – FJf)/FJo = fract. conv. of J
▪τ = Space Time = time to process one
reactor volume of feed = V/vo
▪V = reactor volume
▪θ = residence time = V/vf
▪ Single MFR or CSTR
▪ Equivalent PFR
▪ Combination of Reactors
Determination of Over-all Conversion
▪ Single MFR or CSTR
▪ Equivalent PFR
▪ Combination of Reactors
Determination of Reaction Kinetics
▪ Unknown Order and Rate Constant
3
4
@UST ChE Department
3
@UST ChE Department
4
Solution Methods
Graphical Method
▪ Given the following –rA and CA data:
▪ Graphical
▪
▪
Applicable if data of –rJ and CJ or XJ are
available
Uses:
▪ Levenspiel plots - Plot of 1/-rJ vs CJ or XJ
▪ or Walas Plot - Plot of -rJ vs CJ
CAi
-rAi
▪ Numerical or Algebraic Methods
▪
Applicable if the kinetics of the reaction is
known or unknown, meaning the order,
rate constants are available or unknown
5
@UST ChE Department
5
@UST ChE Department
@UST ChE Department
6
1
Levenspiel Plot
Graphical Method
Yi =
1
−rAi
X Ai =
C A 0 − C Ai
CA0
By Levenspiel
▪ Volume of a CSTR is represented partly
by the area of rectangles bounded by
the plot of 1/-rJ vs XJ, with height equal
to 1/-rJm and width as ΔXJm
Yi
▪ Volume of a PFR is represented partly by
the area under the curve bounded by a
plot of 1/-rJ vs XJ between 0 and XJ.
XAi
7
8
@UST ChE Department
7
@UST ChE Department
8
Solution Methods
Walas Plot
▪ Graphical
▪
▪
Applicable if data of –rJ and CJ or XJ are
available
Uses:
▪ Levenspiel plots - Plot of 1/-rJ vs CJ or XJ
▪ or Walas Plot - Plot of -rJ vs CJ
-rAi
▪ Numerical or Algebraic Methods
▪
Applicable if the kinetics of the reaction is
known or unknown, meaning the order,
rate constants are available or unknown
CAi
9
@UST ChE Department
9
10
@UST ChE Department
10
Graphical Method
Graphical Method
▪ Given the following –rA and CA data:
By Walas (For CSTR only)
CAi
▪ A plot of -rJ vs CJ is used
-rAi
CAo = 0.5
▪ Each CSTR is represented by a straight line
with an x-intercept of CJm-1 and slope of -1/τm
▪ For multiple CSTRs of same size, the slopes of
each line must be the same.
vo = 25
FAo = vo CAo = 12.5
vo = volumetric flow rate
FAo = molar flow rate
11
@UST ChE Department
11
@UST ChE Department
@UST ChE Department
12
2
Problems:
Solution: Problem 2
2. Determine the volume of a PFR to achieve same
conversion in (1)
1. Find the volume of a single CSTR needed to achieve
80% conversion of A.
2. Determine the volume of a PFR to achieve the same
conversion in (1).
3. Suppose 2 equal sized CSTRs are connected in series.
What total volume is needed for the same
conversion?
4. Suppose equal sized CSTR and PFR are connected in
series. What total volume is needed for the same
conversion if CSTR is ahead?
5. Repeat 1-4 if the total volume is to be 60 ft3 and the
final conversion is unknown.
VPFR =
For PFR:
VPFR
=
vo
FA
FAo
dFA
rA
XA
=
XA
−CA0 dX A
rA
0
−CA0 dX A
rA
0
XA
VPFR = FA0
0
VPFR = FA0
0.8
0
1
dX A
−rA
1
dX A
−rA
VPFR = FAo*(Area under the curve)
13
14
@UST ChE Department
13
@UST ChE Department
14
Solution: Problem 2 (A PFR – Numerical Sol’n)
Solution: Problem 2 (A PFR Graphical)
2. Determine the volume of a PFR to achieve same
conversion in (1)
2. Determine the volume of a PFR to achieve same
conversion in (1)
Or: Using Simpson’s Rule with x = 0.40:
Y1 = 2.6 at XA1 = 0.55
From Levenspiel Plot:
𝑥𝑛
Y2 = 7.5 at XA2 = 0.8-0.55 = 0.25
න
XA = 0.8 and FAo = 12.5 lb-mole/hr
VPFR = FA0
0.8
0
𝑓 𝑥 𝑑𝑥 ≈
𝑥0
ℎ
[𝑓 𝑥0 + 4𝑓 𝑥1 + 2𝑓 𝑥2 + 4𝑓 𝑥3 +. . . +4𝑓 𝑥𝑛−1 + 𝑓 𝑥𝑛 ]
3
XA = 0.8 and FAo = 12.5 lb-mole/hr
1
dX A
−rA
VPFR = FA0
0.8
0
x
1
dX A = FA0 A
3
−rA
1
−rA
VPFR = FAo*(Area under the curve)
XA=0
Integral=(2.6x0.55) + (7.5x0.25)] = 3.305
VPFR = (12.5)
VPFR = 12.5(3.25) = 41.31ft 3
1
−rA
+4
1
+2
2
XA=0.2
1
−rA
1
−rA
+4
3
XA=0.4
XA=0.6
5
XA=0.8
0.2
1.176 + 4 (1.887 ) + 2 ( 3.226 ) + 4 (5.556 ) + 12.346
3
VPFR = 41.45 ft 3
15
16
@UST ChE Department
15
1
−rA
+
4
@UST ChE Department
16
Solution: follow-up (2 PFRs instead of a PFR)
Solution: follow-up (2 PFRs instead of a PFR)
What if 2 Plug Flow Reactors were used instead of only
one PFR, calculate the reactor volumes V1 and V2 for
when the intermediate conversion is 40% and the final
conversion is 80%. The entering molar flow rate is the
same as in the previous examples, 12.5 lb-mole/hr.
VPFR = FA0
0.4
0
1
dX A +
−rA
0.8
0.4
What if 2 Plug Flow Reactors were used instead of only
one PFR, calculate the reactor volumes V1 and V2 for
when the intermediate conversion is 40% and the final
conversion is 80%. The entering molar flow rate is the
same as in the previous examples, 12.5 lb-mole/hr.
1
dX A
−rA
VPFR = FA0
For the first PFR: Using Simpson’s Rule with x = 0.2:
VPFR1 = FA0
0.4
0
VPFR1 = (12.5)
1
dX A
−rA
= FA0
xA
3
1
−rA
+4
1
1
−rA
+
2
1
−rA
@UST ChE Department
1
dX A +
−rA
0.8
0.4
1
dX A
−rA
For the 2nd PFR: Using Simpson’s Rule with x = 0.2:
VPFR2 = FA0
0.8
0.4
1
dX A = F x A
A0
−rA
3
3
XA=0
XA=0.2 XA=0.4
0.2
(1.176 ) + 4 (1.887 ) + (3.226 )
VPFR1 = 9.96 ft 3
17
3
@UST ChE Department
17
0.4
0
VPFR2 = 12.5
0.2
3
1
−rA
+4
1
1
−rA
+
2
1
−rA
3
XA=0
XA=0.2 XA=0.4
( 3.226 ) + 4 ( 5.556 ) + (12.346 ) VPFR2 = 31.49 ft 3
VT = 9.96 + 31.49 = 41.45 ft 3
18
@UST ChE Department
18
3
Solution: follow-up (2 PFRs instead of a PFR)
Solution: Problem 3
3. Suppose 2 equal sized CSTRs are connected in series.
What total volume is needed for the same conversion?
What if 2 Plug Flow Reactors were used instead of only
one PFR, calculate the reactor volumes V1 and V2 for
when the intermediate conversion is 40% and the final
conversion is 80%. The entering molar flow rate is the
same as in the previous examples, 12.5 lb-mole/hr.
A PFR:
VPFR = FA0
0.4
1
dX A = FA0
−rA
0.4
0
1
dX A +
−rA
0.8
0.4
xA1
xAo =0
2 PFRs in Series:
0.8
FA1
FAo
1
dX A
−rA
V1
FA2
xA2 = 0.80
V2
Therefore,
Volume of a = Total volume of
single PFR
PFRs in series
VT
V1
V2
19
@UST ChE Department
19
@UST ChE Department
20
For CSTR:
Solution by Walas Plot
For single CSTR: V =
Analysis of Continuous Reactors
FAo -FA
- rA
For CSTRs in series: Vm =
Lecture No. 6.1 B
Graphical Method
Walas Plot
Fjm −1-Fjm
Vm
=
vo
Linearizing:
- rjm = −
1
m
Y
Ch.E. 2115
Lola Domnina B. Pestaňo, Ph.D.
Y = -rjm
X = Cjm
=
- rjm
C jm +
m
=
v o (C jm −1-C jm )
- rjm
C jm −1-C jm
C jm −1
- rjm
Eqn. of straight line
m
x
m
X-intercept =Cjm-1
b
m=−
1
m=−
m
vo
Vm
22
@UST ChE Department
21
22
Demonstration of Walas Method
=
m
-rAi
Problem Set 3F: (Use Levenspiel Plot)
C jo - C j
=
1. Suppose a PFR followed by CSTR are used
whose intermediate conversion is 0.6. What
total volume is needed for the same conversion
(XA=0.8)?
- rj
C jm −1 - C jm
- r jm
2. Suppose a PFR followed by CSTR of equal size
are used. What total volume is needed for the
same conversion(XA=0.8)?
-rA1
m=−
1
1
CAi
@UST ChE Department
3. Suppose 2 PFR of equal size are connected in
series. What total volume is needed for the
same conversion(XA=0.8)?
CA0
CA1
23
PS 3F
Graph
ical
23
@UST Ch.E. Department
24
4
Problem Set 3F: (Use Levenspiel & Walas Plot)
PS 3F
Graph
ical
End.
4. Suppose 3 equal sized CSTRs are used. What
total volume is needed for the same conversion
(XA=0.8)? Use 2 Methods
5. Suppose 2 CSTRs of volumes 45ft3 and 20ft3
respectively are connected in series. What %
conversion is expected? Use 2 Methods
26
@UST ChE Department
25
@UST ChE Department
26
5