L7b Pressure drop, CSTR start up and semibatch reactors examples.pptx

L7b-1
Copyright © 2014, Prof. M. L. Kraft (mlkraft@illinois.edu). All rights reserved.
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
moles A in CSTR
D wrt time while
in unsteady state
In Out
- +Generation = Accumulation
A
A0 A A
dN
F F r V
dt
  
CA0u0
u0CA
Use concentration rather than conversion in the balance eqs
 
 
t 1 k
A0
A
C
1 e C
1 k
 

 
 
 
 

 
A
A0 A A
dC
C C r
dt
 
   A A
r kC
 
Integrate to find CA (t) while CSTR of 1st
order rxn is in unsteady-state:
A
A0 A A
dC
C C kC
dt
 
   

L7b-2
Review: Time to Reach Steady-State
 
 
t 1 k
A0
A
C
1 e C
1 k
 

 
 
 
 

 
At steady state,
t is large and: 0
A0
AS
C
C
1 k
 

In the unsteady state,
when CA = 0.99CAS:
 
 
t 1 k
A0 A0
s
C C
1 e 0.99
1 k 1 k
 
 
 
   
  
   
 
   
4.6 t
1 k




time to reach 99% (CA = 0.99CAS) of
steady-state concentration in terms of k
99% of the steady-state
concentration is achieved at: A AS
4.6 C 0.99C
1 k




When k is very small
(slow rxn), 1>>k: s
t 4.6

When k is very big
(fast rxn), 1<<k s
4.6
t
k

63% of the steady-state
concentration is achieved at: 1 k



CA = 0.63CAS

L7b-3
Semi-batch
FB
Review: Enhanced Yield in Semi-
Batch Reactor
V0 Vf
FD
V0 - u0t
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
A+B⇌
C+D
A+B
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
A
A+B
→P
V0 + u0t
V0 + u0t
Scenario 1 shown in blue. Scenario 2 shown in red.

L7b-4
Review: Mole Balance on A for
Semi-Batch Reactor
CBu0
V0 + u0t
In Out
- + Generation = Accumulation
A
A
dN
0 0 r V
dt
   
Use whatever units are most convenient (NA, CA, XA, etc)
A
A A A
N
C N C V
V
   A
A
dC V
r V
dt
 
A
A A
dC dV
r V V C
dt dt
  
Convert NA to CA using:
In Out
Reactor volume balance:
 
0 0
d V
0 0
dt

 u   
u = u0
  0
0
dV
dt
u
  0 0
V t V
u
  
A
A A 0
dC
r V V C
dt
u
  
Rearrange to get in
terms of dCA/dt
A 0 A
A
C dC
r
V dt
u
  
Goal: Find how CA D with time (assume reactor is well-mixed)

L7b-5
Review: Mole Balance on B in
Semi-Batch Reactor
CBu0
V0 + u0t
Mole balance on B:
B
B B0
dN
r V F
dt
  
In Out
B
B0 B
dN
F 0 + r V
dt
 
0
dV
dt
u 
  B
B B B0 0 B B B0 0
dC
d dV
C V r V C C V r V C
dt dt dt
u u
      
 
0 B0 B
B
B
C C
dC
r
dt V
u 
 
Substitute
Balance on B
B B
N C V

Rearrange to get in terms of dCB/dt
B
B 0 B B0 0
dC
C V r V C
dt
u u
  
Goal: Find how CB D with time (assume reactor is well-mixed)

L7b-6
Review: Semi-Batch Mole
Balances in Terms of NA
CBu0
V0 + u0t
A
A
dN
r V
dt

A
A
dN
0 0 r V
dt
  
In Out
0 0
N
V t V and C
V
u
  
A A A B
A
0 0
dN dN N N
r V k
dt dt V t
u
   

NB comes from basic mole balance:
B
A B0
dN
r V F
dt
 
B A B
B0
0 0
dN N N
k F
dt V t
u
   

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
 
A B
A 2
0 0
N N
then r k
V t
u
 

Substitute: -rA = kACACB and
Goal: Find how NA & NB D with time (reactor is well-mixed)

L7b-7
X=0
P0 = 20 atm
PBR, 1000 kg cat
X=?
P = ? atm
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).
2A→B -rA = kCA
2 α = 0.0008/kg
k=0.1 dm6/mol∙min∙kg cat at 300 K
FA0 = 10 mol min
  A
A
A0
dX
F
d
r '
W
1. Mole balance
2. Rate law   2
A A
r kC
 
  
 
 


A0 A
A
A 0
C 1 X P
C
1 X P

3. Stoichiometry (put CA
in terms of X)
4. Combine
   
 
  
 




2 2
2
A0 A
0
A
A
0
2
A
C 1 X P
k
P
1 X
dX
dW F 
 
 
2
2
A
A0
A
2
0 0
A
1 X
kC
dX P
dW P
1 X
u 
  
   
  
5. Relate P/P0 to W
 
 
0
A
0 0
P
dP T
1 X
dW 2 T P P


 
 
  
 
 
  
1

L7b-8
X=0
P0 = 20 atm
PBR, 1000 kg cat
X=?
P = ? atm
2A→B -rA = kCA
2 α = 0.0008/kg
FA0 = 10 mol min
4. Combine
 
 
2
2
A
A0
A
2
0 0
A
1 X
kC
dX P
dW P
1 X
u 
  
   
  
5. Relate P/P0 to W
 
 
0
A
0
P
dP
1 X
dW 2 P P


 
  
 
 
Simultaneously solve dXA/dW and dP/dW (or dy/dW) using Polymath
First, need to determine , CA0, & u0. Tf T0
T0
N N
N



1 2
0.5
2
 

    
What is CA0?
0
0 0 T0 0 A0
0
P
P V N RT C
RT
  
 
A0 3
3
20atm mol
C 0.813
dm
dm atm
0.082 300K
mol K
  
 

 
 

 
A0
0
A0
F
C
u 
3
0
3
10mol min dm
12.3
min
0.813mol dm
u
  

L7b-10
X=0
P0 = 20 atm
PBR, 1000 kg cat
X=?
P = ? atm
2A→B -rA = kCA
2 α = 0.0008/kg
FA0 = 10 mol min
 
 
2
2
A
A0
A
2
0 0
A
1 X
kC
dX P
dW P
1 X
u 
  
   
    
 
0
A
0
P
dP
1 X
dW 2 P P


 
  
 
 
0.5
  
A0 3
mol
C 0.813
dm

3
0
dm
12.3
min
u 

L7b-11
X=0
P0 = 20 atm
PBR, 1000 kg cat
X=?
P = ? atm
2A→B -rA = kCA
2 α = 0.0008/kg
FA0 = 10 mol min
 
 
2
2
A
A0
A
2
0 0
A
1 X
kC
dX P
dW P
1 X
u 
  
   
    
 
0
A
0
P
dP
1 X
dW 2 P P


 
  
 
 
0.5
  
A0 3
mol
C 0.813
dm

3
0
dm
12.3
min
u 

L7b-12
X = 0.93
PBR? The rxn is isothermal at
300 K, assume ideal gas
behavior, and the feed contains
pure A (g).
P = 14.28 atm

L7b-13
What conversion can be achieved in a
fluidized CSTR with the same catalyst weight
and P0 = P (ideal gas behavior, pure A feed)?
A A
A0
dX r '
dW F

  
A0 A 0
A
A 0
C 1 X T
P
C
1 X P T

   
   
  
 
1
T T0
T0
N N 1 1
0
N 1

 
  
0
A0 A
A
F X
CSTR design eq: W
r'


Use info from PBR to determine FA0, CA0 & k
A A
A0
dX kC
dW F
 
   
A A0 A A A0 A
0
P 0.0008
C C 1 X C C 1 X 1 W
P kg
   
     
   
 
 
Isothermal and =0. Ergun eq for P/P0 becomes:
0
P
1 W
P

 
 
A
A
A0
A0
F
k
W
C
X
1 X
 

9atm
1 1000kg
20atm

  0.2025 1 1000kg

   1
0.0008 kg
 

Plug into CA:
Do not plug in P and P0 that occurred in PBR
yet! Use Ergun eq to get P/P0 as a function of
W, plug into design eq & integrate over W!
Use PBR expt
parameters to
solve for α

L7b-14
A0 A
A
F X
CSTR design eq: W
r'


Use info from PBR to determine FA0, CA0 & k
A A
A0
dX kC
dW F

 
A A0 A
0.0008
C C 1 X 1 W
kg
 
    
 
 
A
A
A0
A0
F
k
W
C
X
1 X
 

Plug CA into PBR
design eq:
 
A0
A
A
A0
kC
dX 0.0008
1 X 1 W
dW F kg
 
 
   
 
 
 
 
 
 
A0 A
A
A0
0.0008
k C 1 X 1 W
kg
dX
dW F
 
 
 
 
 
 
 
 
 
Rearrange
Integrate so that we can
get values of unknowns

L7b-15
A0 A
A
F X
CSTR design eq: W
r'


 
A
A
A0
A0
F
k
W
C
X
1 X
 

 
A
A
A
0
A0
kC
dX 0.0008
PBR design eq : 1 X 1 W
dW kg
F
 
 
  
 
 
 
 
 
 
 
 
  
 
   
 
  
 
A0
A
X W
A
A
A
0 0 0
dX 0.00
kC
F
08
1 W dW
1 X kg
 
 
 
   
 
     
   
      
 
 

      
 
 
 
 
1000k
A 2
0
A0
g
3
0
1 2 1 0.0008
ln 1 1 W
1 0.141 3 0.000
kC
F 8 kg kg
 
 
 
 
 
   
 
 
   
 
   
 
 
  
   
 
 
 
 
1000kg
3
2
A
0
A0
A0
1 2 0.0008
ln 1 1 W
1 X 3 0.0008
kC
F kg kg

L7b-16
A0 A
A
F X
CSTR design eq: W
r'


 
A
A
A0
A0
F
k
W
C
X
1 X
 

 
A
A
A
0
A0
kC
dX 0.0008
PBR design eq : 1 X 1 W
dW kg
F
 
 
  
 
 
 
 
 
 
 
 
   
 
     
   
      
 
 
      
 
 
 
 
A0
A0
1000kg
3
2
0
1 2 1 0.0008
ln 1 1 W
0.859 3 0.0008 kg kg
kC
F
 
 A0
A0
kC
F
0.152 758.8kg
 
  
4 1 A0
A0
2.0 10 kg
kC
F
Plug this value into
the CSTR eq
   
 
   
   
 
   
      
   
 
   
   
 
   
   
 
A
3 3
2 2
0
A0
0.0008 0.0008
0.152 833.3 kg 1 1 1000kg 833.3 kg 1 1 0k
k
g
kg
C
F kg
   
 
 
     
 
 
 
 
 
A0
A
3
0
2
0.152 833.3 kg 1 1 0.8 833.3 kg 1 1
kC
F
 
  
 
 
A0
A0
0.152 833.3 kg 1 0.0894
kC
F

L7b-17
A0 A
A
F X
CSTR design eq: W
r'


 
A
A
A
A0
0
F
kC
X
W
1 X


4 1 A0
A0
kC
2.0 10 kg
F
 


 
4
A
A
1
1
2.0 10 k
X
1000kg
1 X
g
 



 
A
A A
A
X
0.2 0.2 0.2X X
1 X
   

A A
0.2 1.2X 0.17 X
   
Conversion in fluidized CSTR, no pressure drop

L7b-18
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
3
mol
k 53 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?
1. What is CA0? A0 A0 T0
C y C
 A0
A0
T0
N
y =
N
T0
T0
N P
C
V RT
 
A0 A A0 0 A
A A
F X C X
W W
r' r'
u
  
 
Known: u0 and XA
Unknown: CA0 & -r’A
Fluidized
CSTR
design eq:
Feed is a stoichiometric mixture
→ 1 part A, 2 parts B A0
1 1
y =
1 2 3


A0 A0
P
C y
RT
 
  
  A0 3
mol
C 0.055
dm
 
A0 3
1 6atm
C
3 atm dm
0.082 443K
mol K
 
   
 
  
 
 

 

L7b-19
P0= 6 atm; T = 443K & u0 = 50 dm3/min
2. What is –r’A?
A0 A A0 0 A
A A
F X C X
W W
r' r'
u
  
 
Known: u0, XA, & CA0 (0.055 mol/dm3)
Unknown: -r’A
  3
mol
Units on k are:
kg cat min atm
Express rate law in terms
of partial pressure, not Cj

 A
2
B
A kP P
r '
3
mol
kg cat min atm

 
2a. What is PA?
 
 


j
0
j
A
A
j j
C
C
X
1 X


For ideal, isobaric,
isothermal rxn:
 
i i
i
N P
C
V RT
Substitute for Cj & Cj0
 
 
 
 
 
 

j j A
A
j0
j
P
P RT
R
X
1 X
T


 
 
 

j0 j j A
j
A
P X
P
1 X



L7b-20
P0= 6 atm; T = 443K & u0 = 50 dm3/min
A0 A A0 0 A
A A
F X C X
W W
r' r'
u
  
 
Unknown: -r’A
Units on k necessitate expressing rate law in
terms of partial pressure, not Cj
3
mol
kg cat min atm

 
 
j0 j j A
j
A
P X
P
1 X


 


2a. What is PA?
A=-1 A=1
j0 j0 0 j0 j0
J
A0 A0 0 A0 A0
F C C y
F C C y
u
u
    
T T0
A0
T0
N N
y
N
 

  A0
1 2
y (1 2 1)
3 3
   
       
A0 A0 T0
P y P
 A0 A0
1
P 6atm P 2atm
3
 
   
 
 
 
 


 A
A
A
2atm 1 X
P
1 2 3 X

 A
2
B
A kP P
r'

L7b-21
P0= 6 atm; T = 443K & u0 = 50 dm3/min
A0 A A0 0 A
A A
F X C X
W W
r' r'
u
  
 
Unknown: -r’A
3
mol
kg cat min atm

 
2b. What is PB? B=-2
j0 j0 j0
J
A0 A0 A0
F C y
F C y
   
2
3
   B0
B
A0
F 2
2
F 1
   
 
 


B
A0 j j A
A
P
P X
1 X


 
 



 A
B
A
4atm 1 X
P
1 2 3 X
 
 

 

B
A
A
2atm 2 2X
1 2 3 X
P
 
 


 A
A
A
2atm 1 X
P
1 2 3 X

 A
2
B
A kP P
r'

L7b-22
P0= 6 atm; T = 443K & u0 = 50 dm3/min
A0 A A0 0 A
A A
F X C X
W W
r' r'
u
  
 
Unknown: -r’A
3
mol
kg cat min atm

 
 
 


 A
B
A
4atm 1 X
P
1 2 3 X
 
 


 A
A
A
2atm 1 X
P
1 2 3 X
2c. What is k at 443K?
 
 

  
  

E 1 1
R T T
1 2
44 0
K 3
3 0 K
k e
k
  

 
 
  
 
 
   
 
 
 
80000J mol 1 1
8.314J mol K 300K 443K
3
443K
mol
53 e
kgcat min atm
k




 6
443K 3
mol
k 1.663 10
kgcat min atm

 A
2
B
A kP P
r'

L7b-23
A0 A A0 0 A
A A
F X C X
W W
r' r'
u
  
 
Unknown: -r’A

 A
2
B
A kP P
r '  
 


 A
B
A
4atm 1 X
P
1 2 3 X
 
 


 A
A
A
2atm 1 X
P
1 2 3 X
 
 
6
443K 3
mol
k 1.663 10
kgcat min atm
 
 
 
 
 


 

 
 

 
  


 


 
 
 A
A
3
2
A
A
A
6 2a 4a
mol
1.663 10
kgcat min at
tm 1 X
1 2 3
tm 1 X
1 2 3 X X
m
r '
 
 
 
 
  
 
 
 
   
  
  


3
6 3 A
A 3 3
A
1 X
mol
r' 1.663 10 32atm
kgcat min atm 1 2 3 X
P0= 6 atm; T = 443K & u0 = 50 dm3/min
3
mol
kg cat min atm

 

L7b-24
3
mol
kg cat min atm

 
 
 
 
 
 
 
  
  
 
  
 

 
   
  
   

3
3
3
6 3
3 3
1 0.8
mol
1.663 10 32atm
k
dm
50
gcat min atm 1 2 3 0.8
mol
0.055
d
W
8
mi
0.
m n
 
 
 
 
  
 
 
 
   
  
  


3
6 3 A
A 3 3
A
1 X
mol
r' 1.663 10 32atm
kgcat min atm 1 2 3 X

 A0 0
A
A
C X
W
r '
u
CA0 =0.055 mol/dm3
7
W 5.24 10 kg cat

 
P0= 6 atm; T = 443K & u0 = 50 dm3/min

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