L7b-1
Review: Fixed-Volume CSTR Start-Up
Isothermal (unusual, but simple case), well-mixed CSTR
Unsteady state: concentrations vary with time & accumulation is non-zero
Goal: Determine the time required to reach steady-state operation and
CA as a function of time
CA0u0
u0CA
In - Out + Generation = Accumulation
FA0

FA


rA V
dNA
dt
moles A in CSTR
D wrt time while
in unsteady state
Use concentration rather than conversion in the balance eqs
dC A
rA  kCA
C A0  CA  rA  
dt
dCA Integrate to find CA (t) while CSTR of 1st
 CA0  CA  kCA  
order rxn is in unsteady-state:
dt


 CA0 
 t 1k  
1

e
 CA
 1  k 


Copyright © 2014, Prof. M. L. Kraft (mlkraft@illinois.edu). All rights reserved.
L7b-2
Review: Time to Reach Steady-State


C A0
At steady state,  CA0 
 t 1k  

 CAS
1 e
 CA


1  k
t is large and:  1  k 
0


In the unsteady state,  CA0 
 CA0 
 t s 1k  

 0.99 
 1 e

when CA = 0.99CAS:
1

k



 1  k 
4.6

 t time to reach 99% (CA = 0.99CAS) of
steady-state concentration in terms of k
1  k
When k is very small
t  4.6
(slow rxn), 1>>k: s
When k is very big
(fast rxn), 1<<k

63% of the steady-state
concentration is achieved at: 1  k
ts 
4.6
k
CA = 0.63CAS

99% of the steady-state
4.6
concentration is achieved at:
1  k
CA  0.99CAS
Copyright © 2014, Prof. M. L. Kraft (mlkraft@illinois.edu). All rights reserved.
Review: Enhanced Yield in SemiBatch Reactor
F
Semi-batch
L7b-3
D
FB
A
V0 + u0t
A+B
A+B
→P
V0
V0 + u0t
A+B⇌
C+D
V0 - u0t
Vf
Scenario 1 shown in blue. Scenario 2 shown in red.
Scenario 1: Enhance selectivity of desired product over undesired side product
Higher concentrations of A favor formation of the desired product
Higher concentrations of B favor formation of the undesired side product
Scenario 2: Improve the product yield obtained from a reversible reaction
A l  B l
C l  D  g
Allowing D(g) to bubble out of solution pushes equilibrium towards completion
Copyright © 2014, Prof. M. L. Kraft (mlkraft@illinois.edu). All rights reserved.
CBu0
Review: Mole Balance on A for
Semi-Batch Reactor
L7b-4
Goal: Find how CA D with time (assume reactor is well-mixed)
In - Out + Generation =
Accumulation
dNA
 0  0  rA V 
dt
Use whatever units are most convenient (NA, CA, XA, etc)
V0 + u0t
Convert NA to CA using:
NA
dCA V  r V  V dCA  C dV
 CA  NA  CA V
A
A
 rA V 
dt
dt
V
dt
Reactor volume balance:
In - Out + Generation = Accumulation
d   V  u = u0
dV
 u0 
0u0  0  0 
 V0  u0t  V
  0
dt
dt
 rA V  V
dCA
 C Au0
dt
Rearrange to get in
terms of dCA/dt
 rA 
CAu0 dCA

V
dt
Copyright © 2014, Prof. M. L. Kraft (mlkraft@illinois.edu). All rights reserved.
Review: Mole Balance on B in
Semi-Batch Reactor
C u
L7b-5
Goal: Find how CB D with time (assume reactor is well-mixed)
B 0
Mole balance on B:
In - Out + Generation =
FB0

V0 + u0t

0 + rB V
dNB
 rB V  FB0
dt

Accumulation
dNB
dt
NB  CB V
dC
d
dV
 CB V   rB V  CB0u0  CB  V B  rB V  CB0u0
dt
dt
dt
dV
u0 
Substitute
dt
dC
CBu0  V B  rB V  CB0u0 Rearrange to get in terms of dCB/dt
dt

u0  CB0  CB 
dCB
 rB 
dt
V
Balance on B
Copyright © 2014, Prof. M. L. Kraft (mlkraft@illinois.edu). All rights reserved.
L7b-6
CBu0
Review: Semi-Batch Mole
Balances in Terms of NA
Goal: Find how NA & NB D with time (reactor is well-mixed)
In - Out + Generation = Accumulation

0
0

rA V

dNA
dt
dNA
 rA V
dt
V0 + u0t
N
C
Substitute: -rA = kACACB and V0  u0 t  V and
V
NANB
dNA
dNA
NANB
then rA  k

r
V



k
A
2
dt
dt
V0  u0 t
V  u t
0
0
NB comes from basic mole balance:
dNB
NANB
dNB



k
 FB0
 rA V  FB0
dt
V0  u0 t
dt
The design eq in terms of XA can be messy. Sometimes it gives a single
equation when using Nj or Cj gives multiple reactor designs
Copyright © 2014, Prof. M. L. Kraft (mlkraft@illinois.edu). All rights reserved.
X=0
P0 = 20 atm
PBR, 1000 kg cat
X=?
P = ? atm
2A→B -rA = kCA2
α = 0.0008/kg
k=0.1 dm6/mol∙min∙kg cat at 300 K
FA0 = 10 mol min
1.
Mole balance
2.
Rate law
dX A
 r 'A
dW
rA  kC A 2
FA0
3. Stoichiometry (put CA C A
in terms of X)
4. Combine
C A0 1  X A   P 

 
1   X A  P0 
dX A

dW

5. Relate P/P0 to W
L7b-7
What conversion and P is
measured at the outlet of the
PBR? The rxn is isothermal at 300
K, assume ideal gas behavior, and
the feed contains pure A (g).
dX A
dW

k C A02
 1  XA 2  P 2
 
 P0 
1   X A 
2
2
kCA0 1  X A   P 

u0 1   X A 2  P0 
FA 0
2
dP
  T   P0 
   
 1   XA 
dW
2  T0   P P0  
1
Copyright © 2014, Prof. M. L. Kraft (mlkraft@illinois.edu). All rights reserved.
X=0
P0 = 20 atm
PBR, 1000 kg cat
X=?
P = ? atm
2A→B -rA = kCA2
α = 0.0008/kg
k=0.1 dm6/mol∙min∙kg cat at 300 K
FA0 = 10 mol min
4. Combine
5. Relate P/P0 to W
L7b-8
What conversion and P is
measured at the outlet of the
PBR? The rxn is isothermal at 300
K, assume ideal gas behavior, and
the feed contains pure A (g).
dX A kCA0 1  X A   P 


dW
u0 1   X A 2  P0 
dP
  P0 
 
 1   X A 
dW
2  P P0  
2
2
Simultaneously solve dXA/dW and dP/dW (or dy/dW) using Polymath
First, need to determine , CA0, & u0.
NTf  NT0
1 2

 
   0.5
NT0
2
What is CA0?
P0
20atm
mol
 C A0  0.813
 CA0 
RT0

dm3
dm3  atm 
 0.082
  300K 
mol  K 

FA0
10mol min
dm3
u0 

 u0  12.3
3
C A0
min
0.813mol dm
P0 V0  NT0RT0 
Copyright © 2014, Prof. M. L. Kraft (mlkraft@illinois.edu). All rights reserved.
X=0
P0 = 20 atm
PBR, 1000 kg cat
X=?
P = ? atm
2A→B -rA = kCA2
α = 0.0008/kg
k=0.1 dm6/mol∙min∙kg cat at 300 K
FA0 = 10 mol min
L7b-10
What conversion and P is
measured at the outlet of the
PBR? The rxn is isothermal at 300
K, assume ideal gas behavior, and
the feed contains pure A (g).
2
dP
  P0 
dX A kCA0 1  X A   P 
 


 1   X A 
 
2
dW
2  P P0  
dW
u0 1   X A   P0 
Simultaneously solve dXA/dW and dP/dW (or dy/dW) using Polymath
3
mol
dm
  0.5
C A0  0.813
u0  12.3
dm3
min
2
Copyright © 2014, Prof. M. L. Kraft (mlkraft@illinois.edu). All rights reserved.
X=0
P0 = 20 atm
PBR, 1000 kg cat
X=?
P = ? atm
2A→B -rA = kCA2
α = 0.0008/kg
k=0.1 dm6/mol∙min∙kg cat at 300 K
FA0 = 10 mol min
L7b-11
What conversion and P is
measured at the outlet of the
PBR? The rxn is isothermal at 300
K, assume ideal gas behavior, and
the feed contains pure A (g).
2
dP
  P0 
dX A kCA0 1  X A   P 
 


 1   X A 
 
2
dW
2  P P0  
dW
u0 1   X A   P0 
Simultaneously solve dXA/dW and dP/dW (or dy/dW) using Polymath
3
mol
dm
  0.5
C A0  0.813
u0  12.3
dm3
min
2
Copyright © 2014, Prof. M. L. Kraft (mlkraft@illinois.edu). All rights reserved.
What conversion and P is L7b-12
measured at the outlet of the
PBR? The rxn is isothermal at
300 K, assume ideal gas
behavior, and the feed contains
pure A (g).
P = 14.28 atm
X = 0.93
Copyright © 2014, Prof. M. L. Kraft (mlkraft@illinois.edu). All rights reserved.
What conversion can be achieved in a L7b-13
fluidized CSTR with the same catalyst weight
and P0 = P (ideal gas behavior, pure A feed)?
F X
CSTR design eq: W  A0 A
r 'A
FA0 X A
W
kC A0 1  X A 
Use info from PBR to determine FA0, CA0 & k
dX A r ' A
dX A kC A
C 1  XA   P   T0 
NT  NT0 1  1






0
CA  A0
  
dW
FA0
dW
FA0
NT0
1
1   XA  P0   T 
Do not plug in P and P0 that occurred in PBR
0
1
yet! Use Ergun eq to get P/P0 as a function of
Use PBR expt
W, plug into design eq & integrate over W!
P
 1   W parameters to
Isothermal and =0. Ergun eq for P/P0 becomes:
solve for α
P0
9atm
 1   1000kg
20atm
 0.2025  1  1000kg
  0.0008 kg1
P
 0.0008 
C

C
1

X

C

C
1

X
1

W
Plug into CA: A

A0 
A 
A
A0 
A


 kg 
 P0 
Copyright © 2014, Prof. M. L. Kraft (mlkraft@illinois.edu). All rights reserved.
L7b-14
What conversion can be achieved in a
fluidized CSTR with the same catalyst weight
and P0 = P (ideal gas behavior, pure A feed)?
F X
CSTR design eq: W  A0 A
r 'A
FA0 X A
W
kC A0 1  X A 
Use info from PBR to determine FA0, CA0 & k
dX A kC A
 0.0008 

CA  CA0 1  XA  1  
W

dW
FA0
 kg 

 0.0008  
k  CA0 1  X A  1  
W  Rearrange


 kg  
dX A



dW
FA0
Plug CA into PBR
design eq:
dX A kC A0 
 0.0008   Integrate so that we can


W
 1  X A  1  


dW
FA0 
 kg   get values of unknowns
Copyright © 2014, Prof. M. L. Kraft (mlkraft@illinois.edu). All rights reserved.
L7b-15
What conversion can be achieved in a
fluidized CSTR with the same catalyst weight
and P0 = P (ideal gas behavior, pure A feed)?
F X
CSTR design eq: W  A0 A
r 'A
FA0 X A
W
kC A0 1  X A 
dX A kC A 0 
 0.0008  
PBR design eq :

W
 1  X A  1  


dW
FA0 
 kg  
X A dX
kC A0 W 
 0.0008  
A
 

W  dW
  1  

FA 0 0 
 kg  
0 1  X A 
1000kg
3 


   0.0008
 1  kC A0 
2
2  
 ln 

1

1

W
   
 F
 3  0.0008 kg    
1

X
kg
 

A
A0 
 

 0

1000kg
3 


1
1
   0.0008
2  

 kC A 0  2  
 ln 

1

1

W
  

 
 

1

0.141
F
3
0.000
8
kg
kg


 
 
A0   

 0

Copyright © 2014, Prof. M. L. Kraft (mlkraft@illinois.edu). All rights reserved.
L7b-16
What conversion can be achieved in a
fluidized CSTR with the same catalyst weight
and P0 = P (ideal gas behavior, pure A feed)?
F X
CSTR design eq: W  A0 A
r 'A
FA0 X A
W
kC A0 1  X A 
dX A kC A 0 
 0.0008  
PBR design eq :

W
 1  X A  1  


dW
FA0 
 kg  
1000kg
3 


1
   0.0008
2  
 1  kC A0  2  
 ln 
1

1

W








 
kg
 0.859  FA0  3   0.0008 kg   
 


 0
3
3 



kC A 0 
2
 0.0008
2
  0.0008
 0.152 
833.3 kg 1  1 
1000kg     833.3 kg 1  1 
0k g   

FA0
kg
kg
 

 

 




3

kC A0

kC A0 
833.3 kg 1  0.0894 
 0.152 
833.3 kg 1  1  0.8  2   833.3 kg 1  1  0.152  F
FA 0 
A0



kC A0
0.152 
 758.8kg 
FA0
 2.0  10 4 kg1 
kC A0
FA0
Plug this value into
the CSTR eq
Copyright © 2014, Prof. M. L. Kraft (mlkraft@illinois.edu). All rights reserved.
L7b-17
What conversion can be achieved in a
fluidized CSTR with the same catalyst weight
and P0 = P (ideal gas behavior, pure A feed)?
F X
CSTR design eq: W  A0 A
r 'A
FA0 X A
W
kC A 0 1  X A 
2.0  10 4 kg1 
kC A0
FA0
XA
1000kg 
2.0  10 4 kg1 1  X A 
1
0.2 
XA
 0.2  0.2X A  X A  0.2  1.2XA  0.17  XA
1  XA 
Conversion in fluidized CSTR, no pressure drop
Copyright © 2014, Prof. M. L. Kraft (mlkraft@illinois.edu). All rights reserved.
A + 2B → C Elementary rxn, feed is a stoichiometric mixture
Fluidized CSTR, isothermal, isobaric, ideal, gas-phase reaction
P0= 6 atm; T = 443K & u0 = 50 dm3/min
k  53
mol
3
kg cat  min atm
L7b-18
at 300K with E=80 kJ/mol, elementary rxn
How many kg of catalyst is required to achieve XA = 0.8?
Fluidized
CSTR
design eq:
F X
C u X
W  A0 A  W  A0 0 A
r 'A
r 'A
1. What is CA0?
CA0  y A0CT0
y A0 =
Known: u0 and XA
Unknown: CA0 & -r’A
NA0
NT0
NT0
P
CT0 

V
RT
Feed is a stoichiometric mixture
1
1
y A0 =

→ 1 part A, 2 parts B
1 2 3
6atm
 P   C   1
mol
C A0  y A0 
A0  


C

0.055
A0
3
3
 RT 
atm  dm3 
dm
 0.082
 443K
mol  K 

Copyright © 2014, Prof. M. L. Kraft (mlkraft@illinois.edu). All rights reserved.
L7b-19
A + 2B → C
Elementary rxn, feed is a stoichiometric mixture
Fluidized CSTR, isothermal, isobaric, ideal, gas-phase reaction
P0= 6 atm; T = 443K & u0 = 50 dm3/min
k  53
mol
3
kg cat  min atm
at 300K with E=80 kJ/mol, elementary rxn
How many kg of catalyst is required to achieve XA = 0.8?
F X
C u X
W  A0 A  W  A0 0 A
r 'A
r 'A
2. What is –r’A? Units on k are:
r 'A  kPA PB2
Ci 
Known: u0, XA, & CA0 (0.055 mol/dm3)
Unknown: -r’A
mol
kg cat  min atm3
2a. What is PA?
Ni
P
 i
V RT
Substitute for Cj & Cj0
Express rate law in terms
of partial pressure, not Cj

C j0  j   j X A
For ideal, isobaric,
Cj 
isothermal rxn:
1   XA
 Pj0 

  j   j XA
Pj
RT



RT
1   XA



Pj0  j   j X A
 Pj 
1   XA
Copyright © 2014, Prof. M. L. Kraft (mlkraft@illinois.edu). All rights reserved.


L7b-20
A + 2B → C
Elementary rxn, feed is a stoichiometric mixture
Fluidized CSTR, isothermal, isobaric, ideal, gas-phase reaction
P0= 6 atm; T = 443K & u0 = 50 dm3/min
k  53
mol
3
kg cat  min atm
at 300K with E=80 kJ/mol, elementary rxn
How many kg of catalyst is required to achieve XA = 0.8?
F X
C u X
W  A0 A  W  A0 0 A
r 'A
r 'A
Known: u0, XA, & CA0 (0.055 mol/dm3)
Unknown: -r’A
2 Units on k necessitate expressing rate law in
2. What is –r’A? r 'A  kPAPB
terms of partial pressure, not Cj
2a. What is PA?
A=-1
A=1
PA0  y A0PT0
Pj 


Pj0  j   j X A

1   XA
NT  NT0
  y A0
NT0
J 
Fj0
FA0

C j0u0
CA0u0

C j0
CA0
1
3

y j0
y A0
   y A0    (1  2  1)    
 1
 PA0    6atm  PA0  2atm
3
2
3
2atm 1  X A 
PA 
1   2 3  XA
Copyright © 2014, Prof. M. L. Kraft (mlkraft@illinois.edu). All rights reserved.
A + 2B → C
Elementary rxn, feed is a stoichiometric mixture
Fluidized CSTR, isothermal, isobaric, ideal, gas-phase reaction
P0= 6 atm; T = 443K & u0 = 50 dm3/min
k  53
mol
3
kg cat  min atm
L7b-21
at 300K with E=80 kJ/mol, elementary rxn
How many kg of catalyst is required to achieve XA = 0.8?
F X
C u X
W  A0 A  W  A0 0 A
r 'A
r 'A
2. What is –r’A? r 'A  kPAPB
2b. What is PB?

 
PA0  j   j X A
PB 
1   XA

2
3
2
Known: u0, XA, & CA0 (0.055 mol/dm3)
Unknown: -r’A
PA 
B=-2
 PB 
2atm 1  X A 
1   2 3  XA
B 
J 
Fj0
FA0

C j0
CA0

FB0 2
 2
FA0 1
2atm  2  2X A 
4atm 1  X A 
 PB 
1   2 3  XA
1   2 3  XA
Copyright © 2014, Prof. M. L. Kraft (mlkraft@illinois.edu). All rights reserved.
y j0
y A0
A + 2B → C
Elementary rxn, feed is a stoichiometric mixture
Fluidized CSTR, isothermal, isobaric, ideal, gas-phase reaction
P0= 6 atm; T = 443K & u0 = 50 dm3/min
k  53
mol
3
L7b-22
at 300K with E=80 kJ/mol, elementary rxn
kg cat  min atm
How many kg of catalyst is required to achieve XA = 0.8?
FA0 XA
CA0u0 XA
W
W
r 'A
r 'A
2
2. What is –r’A? r 'A  kPAPB
2c. What is k at 443K?
 k 443K
Known: u0, XA, & CA0 (0.055 mol/dm3)
Unknown: -r’A
PB 
4atm 1  X A 
1   2 3  XA
k 443K  k 300K e
PA 
2atm 1  X A 
1   2 3  XA
 E  1 1 
   
 R  T1 T2 
1 
 80000J mol  1




  8.314J molK  300K 443K 
e

mol
 53 
 kgcat  min atm3 


 k 443K  1.663  106
mol
kgcat  min atm3
Copyright © 2014, Prof. M. L. Kraft (mlkraft@illinois.edu). All rights reserved.
A + 2B → C
Elementary rxn, feed is a stoichiometric mixture
Fluidized CSTR, isothermal, isobaric, ideal, gas-phase reaction
P0= 6 atm; T = 443K & u0 = 50 dm3/min
k  53
mol
3
L7b-23
at 300K with E=80 kJ/mol, elementary rxn
kg cat  min atm
How many kg of catalyst is required to achieve XA = 0.8?
FA0 XA
CA0u0 XA
W
W
r 'A
r 'A
2
2. What is –r’A? r 'A  kPA PB
Known: u0, XA, & CA0 (0.055 mol/dm3)
Unknown: -r’A
4atm 1  X A 
1   2 3  XA
mol
6
PB 
k 443K  1.663  10
PA 
2atm 1  X A 
1   2 3  XA
kgcat  min atm3

  2atm 1  X A    4atm 1  X A  
mol
6
r 'A   1.663  10


3   1   2 3  X

1

2
3
X


kgcat

min

at
m
A 
A 


 1  X 3 


mol
A

r ' A   1.663  106
32atm3 

3
3

 1   2 3  X  
kgcat  min atm 

A


Copyright © 2014, Prof. M. L. Kraft (mlkraft@illinois.edu). All rights reserved.
2
A + 2B → C
Elementary rxn, feed is a stoichiometric mixture
Fluidized CSTR, isothermal, isobaric, ideal, gas-phase reaction
P0= 6 atm; T = 443K & u0 = 50 dm3/min
k  53
mol
3
kg cat  min atm
at 300K with E=80 kJ/mol, elementary rxn
How many kg of catalyst is required to achieve XA = 0.8?
C A0u0 X A
CA0 =0.055 mol/dm3
W
r 'A
 1  X 3 


mol
A

r ' A   1.663  106
32atm3 

3
3

 1   2 3  X  
kgcat  min atm 

A


mol   dm3 

50
 0.8
 0.055
3  
mi
n

dm  

W
 1  0.8 3 


mol
6

32atm3 
 1.663  10

3
3
 1   2 3  0.8  
kgcat  min atm 



W  5.24  10 7 kg cat
Copyright © 2014, Prof. M. L. Kraft (mlkraft@illinois.edu). All rights reserved.
L7b-24