L7-1
Review: Liquid Phase Reaction in PFR
Be able to do these 4 steps,
integrate & solve for V for ANY
ORDER RXN
LIQUID PHASE: Ci ≠ f(P) → no pressure drop
2A → B -rA = kCA2 2nd order reaction rate
Calculate volume required to get a conversion of XA in a PFR
dX A rA

 Mole balance
dV
FA0
 Rate law
rA  kCA 2
 Stoichiometry (put CA in
terms of X)
CA  CA0 (1  X A )
dX A

dV
 Combine


FA0
k CA0
2

XA

0
V
dX A
1  XA 
2
  dV
See Appendix A for integrals
frequently used in reactor design
0




k CA02 1  X A 
FA0
k CA02
2
FA0

 XA
 1 X

A

V

Liquid-phase 2nd order reaction in PFR
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
L7-2
Review: Liquid Phase Reaction in PBR
LIQUID PHASE: Ci ≠ f(P) → no pressure drop
Be able to do these 4 steps, integrate
& solve for V for ANY ORDER RXN
2A → B -r’A = kCA2 2nd order reaction rate
Calculate catalyst weight required to get a conversion of XA in a PBR
dX A r 'A

 Mole balance
dW
FA0
r 'A  kCA 2
 Rate law
CA  CA0 (1  X A )
 Stoichiometry (put CA in
terms of X)
dX A

dW
 Combine


FA0
k CA0
2

XA
dX A
0
1  XA 

W
2
  dW 
0



k CA02 1  X A 
FA0
k C A02
2
FA0

 XA 
 1 X   W

A 
Liquid-phase 2nd order reaction in PBR
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
L7-3
Review: Isobaric, Isothermal, Ideal
Gas-Phase Rxns in Tubular Reactors
Gas-phase reactions are usually carried out in tubular reactors (PFRs & PBRs)
• Plug flow: no radial variations in concentration, temperature, & ∴ -rA
• No stirring element, so flow must be turbulent
FA0
FA
C j0   jCA0 XA
C j0   jCA0 XA  P   T0  Z0 
 Cj 
GAS PHASE: C j 
  

1   XA
1   XA
 P0   T  Z 
1
1
1
Stoichiometry for basis species A:
CA0 1  XA 
CA0  CA0 XA
CA 
 CA 
1   XA
1   XA
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
Review: Effect of  on u and XA

L7-4
NTf  NT0 Change in total # moles at X A  1

NT0
total moles fed
: expansion factor, the fraction of change in V per mol A reacted
u0: volumetric flow rate
u varies if gas phase & moles product ≠ u  u 1   X  Z  T   P0 
  
0
A 
moles reactant, or if a DP, DT, or DZ occurs
Z
 0  T0   P 
No DP, DT, or DZ occurs, but moles product ≠ moles reactant → u  u0 1   X A 
•  = 0 (mol product = mol reactants): u  u0: constant volumetric flow rate as XA
↑ < 0 (mol product < mol reactants): u < u0 volumetric flow rate ↓ as XA ↑
Q1: For an irreversible gas-phase reaction, how does the residence time and
XA change when  < 0?
a)They don’t
b)The residence time is longer & XA increases
c)The residence time is longer & XA decreases
d)The residence time is shorter, & XA decreases
e)The residence time is shorter & XA increases
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
Review: Effect of  on u and XA

L7-5
NTf  NT0 Change in total # moles at X A  1

NT0
total moles fed
: expansion factor, the fraction of change in V per mol A reacted
u0: volumetric flow rate
u varies if gas phase & moles product ≠ u  u 1   X  Z  T   P0 
  
0
A 
moles reactant, or if a DP, DT, or DZ occurs
Z
 0  T0   P 
No DP, DT, or DZ occurs, but moles product ≠ moles reactant →
u  u0 1   X A 
•  = 0 (mol product = mol reactants): u  u0: constant volumetric flow rate as XA
increases
•  < 0 (mol product < mol reactants): u < u0 volumetric flow rate decreases as
XA increases
• Longer residence time than when u  u0
• Higher conversion per volume of reactor (weight of catalyst) than if u  u0
•  > 0 (mol product > mol reactants): u > u0 with increasing XA
• Shorter residence time than when u  u0
• Lower conversion per volume of reactor (weight of catalyst) than if u  u0
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
L7-6
Review: Isobaric, Isothermal, Ideal Rxn
in PFR
GAS PHASE:
Be able to do these 5 steps, & solve for V
for ANY ORDER RXN
Ci = f() → no DP, DT, or DZ
2A → B
-rA = kCA2 2nd order reaction rate
Calculate PFR volume required to get a conversion of XA
dX A rA

 Mole balance
dV
FA0
rA  kCA 2
 Rate law
CA0 1  X A 

1   XA
 Stoichiometry (put CA in
terms of X)
CA
 Combine
dX A k  CA0  1  X A 

dV
1   XA 2 FA0
2
V
V

FA0
k CA0

FA0
k CA02

2

XA

0
1   XA 2 dX
A
2
1  XA 
2
Integral A-7 in appendix

 1   2 X
A
2 1    ln 1  X A    2 X A  
 1  XA






Gas-phase 2nd
order rxn in
PFR no DP,
DT, or DZ
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
L7-7
Review: Pressure Drop in PBRs
GAS PHASE: A → B -r’A = kCA2 2nd order reaction rate
Calculate dXA/dW for an isothermal ideal gas phase reaction with DP
dX A
FA0
 r 'A
 Mole balance
dW
 Rate law
r 'A  kCA 2
CA 
 Stoichiometry (put CA in
terms of X)
 Combine
 Relate P/P0 to W
(Ergun equation)
dX A

dW

CA0 1  X A   P 
 
1   X A  P0 
k CA02
FA 0
 1  XA 2  P 2
1   XA 
2
 
 P0 
dP
  T   P0 
   
 1   X A 
dW
2  T0   P P0  
Ergun Equation can be simplified
by using y=P/P0 and T=T0:
dy

  1   X A 
dW
2y
Simultaneously solve dXA/dW and dP/dW (or dy/dW) using Polymath
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
L7-8
Review: Ergun Equation
dP
  T   P0 
Calculates pressure drop in a packed bed.
   
1


X



A
This equation can be simplified to:
dW
2  T0   P P0  
Differential form of Ergun equation
for pressure drop in PBR:
y
P
P0

T
dy

  1   X A   
dW
2y
 T0 
NTf  NT0
  y A0
NT0
1    : fraction of solid in bed =

20
A c c 1    P0
volume of solid
total bed volume
AC: cross-sectional area
C: particle density
: constant for each reactor, calculated using a complex
equation that depends on properties of bed (gas density,
particle size, gas viscosity, void volume in bed, etc)
: constant dependant on the packing in the bed
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
L7-9
L7: Unsteady-State Isothermal
Reactor Operation: CSTR Start-Up
and Semi-Batch Reactors
CBu0
Semi-batch
A
A+B
V0
V0 + u0t
Vf
start
time t
end
• Time required to reach steady-state after CSTR start-up
• Predicting concentration and conversion as a function of time
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
L7-10
Start-Up of a Fixed-Volume CSTR
Isothermal (unusual, but simple case), well-mixed CSTR
Unsteady state: concentrations vary with time & accumulation is non-zero
Goal: Determine the time necessary to reach steady-state operation
u0CA
CA0u0
In - Out + Generation = Accumulation
moles A in CSTR
dNA changes with
FA0  F A  rA V 
dt time until steady
state is reached
Use concentration rather than conversion in the balance eqs
dNA
CA0u0  C A u0  rA V 
dt

C A0u0  C Au0  rA V dNA  1 

 
V
dt  V 
Divide by V to convert dNA to dCA

V
u0

C A0


CA

 rA 
dC A
dt
Multiply
by 
dC A
C A0  CA  rA  
dt
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
L7-11
CSTR Start-Up: 1st Order Reaction
dC A
dC A
Combine C

C

k
C



A0
A
A
dt rA  kCA
dt
Integrate this eq to find CA (t) while 1st order rxn in CSTR is at unsteady-state:
C A0  CA  rA  
Bring variables to
one side & factor
dC A 1
  C A0  CA 1  k  
dt

 CA0

dCA  1 

   1  k  
 CA 
dt
 
 1  k 

Put like variables
with their integrals
  CA0

t 1  k 
 CA  

dCA
1  k 
1  k 


dt
 C




 ln
0
t

A0  C
0
0
C



A0  0
1  k  A
 1  k 



1k  t

CA
 1
e 
 C

 t 1k  
C A0
  A0  1  e 
 CA
 1  k 
1  k 
CA


Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
L7-12
CSTR Start-Up: 1st Order Reaction
dCA We integrated this eq to find CA (t) while
CA0  CA  kCA  
dt CSTR of 1st order rxn is in unsteady-state:
At steady state,  CA0  1  e t1k    C  C A0  C
AS
A
1

k

t is large and:  1  k 
0


Is this consistent with steady C  C  kC    dCA No accumulation
A0
A
A
state balance eq for CSTR?
dt at steady state
Yes, same!
0
C A0
 C A0  C A  kC A  0 
 C AS Goal: combine start-up and SS eqs
1  k 
to estimate time to reach SS (ts)


In the unsteady state,  CA0 
 CA 0 
 t s 1k  

1

e

0.99

 1  k 
when CA = 0.99CAS:
1

k





Solve for ts to determine time to reach 99% of steady-state concentration
 t 1k  
 t 1k  
 t 1k  
 1 e s
 0.99  0.01  e s 
 ln 0.01  ln e s 

 1  k
 4.6   t s 
 



4.6
 ts

1  k




time to reach 99% of steady-state
concentration in terms of k
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
L7-13
CSTR Start-Up: 1st Order Reaction
 C



1

e
C

 1  k  


A0
 t 1k 
A
In the unsteady state, the time to
reach CA = 0.99CAS is:
When k is very small
t  4.6
(slow rxn), 1>>k: s
t  4.6

1  k
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
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
Better Selectivity in a Semi-Batch
Reactor
L7-14
To enhance selectivity of desired product over side product
2
kP
A  B 
P rP  kpCA CB
Desired product P
2
kS
A  B 
 S rS  kSCA CB
Undesired side product S
Instantaneous selectivity, SP/S, is the ratio of the relative
rates*:
rP kPCA 2CB kP CA

SP/S 

2
k S CB
rS k SCA CB
Higher concentrations of A favor formation of the desired product P
Higher concentrations of B favor formation of the undesired side product S
To maximize the formation of the desired product:
Slowly feed B into the reactor containing A
Commonly used in bioreactors, when the enzyme is inhibited by excess substrate
*We’ll look at this concept of instantaneous selectivity in more detail in L9
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
L7-15
Semi-Batch Reactor Design Equation
CBu0
V0 + u0t
Do a mole balance on A since it does not enter or leave
the reactor (assume the reactor is well-mixed)
In - Out + Generation = Accumulation
FA0

0

dNA
F A  rA V 
dt
dNA
0  rA V 
dt
Use whatever units are most convenient (NA, CA, XA, etc)
Convert NA to CA using:
NA
 CA  NA  CA V
V
dC A
dV
 rA V  V
 CA
dt
dt
dC A V
 rA V 
dt
2 parts: how CA changes with
t and how V changes with t
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
L7-16
Semi-Batch Reactor Design Equation
CBu0
Do a mole balance on A since it does not enter or leave
the reactor (assume the reactor is well-mixed)
In - Out + Generation =
0

 rA V  V
V0 + u0t
0

rA V  t 
dC A
dV
 CA
dt
dt

Accumulation
dNA
dt
2 parts: how CA changes with t
and how V changes with t
Reactor volume at any time can be found with a mole balance
In - Out + Generation = Accumulation
d   V  u = u0
dV
 u0 
0u0  0  0 
 V0  u0t  V
  0
dt
dt
Substitute:  rA V  V
dC A
 C Au0
dt
dC A
 rA V  C Au0  V
dt
 rA 
Rearrange to get in terms of dCA/dt
CAu0 dCA

V
dt
Balance on A
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
L7-17
Semi-Batch Reactor Design Equation
CBu0
Mole Balance on B
In - Out + Generation =
FB0

V0 + u0t

0

dNB
 rB V  FB0
dt


dNB
 rB V  FB0
dt
dVCB
d
  CB V   rB V  FB0
dt
dt
dCB
dV
 CB
V
 rB V  CB0u0
dt
dt
 CBu0  V

rB V
Accumulation
dNB
dt
dCB
 rB V  CB0u0
dt
u0 
dV
dt
Substitute
Rearrange to get in terms of dCB/dt
u0  CB0  CB 
dCB
Balance on B

 rB 
dt
V
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
L7-18
Semi-Batch Reactor Design Equation:
in Terms of NA
C u
B 0
In - Out + Generation =
0

0

dNA
 rA V
dt
V0 + u0t
rA V

Accumulation
dNA
dt
Substitute V  V0  u0t
Reactor design eq. provided
that rA is a function of NA
NA NB
 NA  NB

r


k

r


k
-rA = kACACB
A
A
 V  V


 V  u t 2
dNA
 rA  V0  u0 t 
dt
0
0
dNA
NA NB
dNA



k

 rA V
dt
V0  u0 t
dt
dNB
NANB
dNB
 k
 FB0
 rA V  FB0 
NB comes from BMB:
dt
V

u
t
dt
0
0
The design eq in terms of XA can be messy. Sometimes it gives a single
equation when using Nj or Cj gives multiple reactor design equations.
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
L7-19
Improving Yields of Reversible Rxns with
Semi-Batch Reactors
Semi-batch
A+B
V0
FD
A+B⇌C+D
V0 - u0t
Vf
To improve product yield in a reversible reaction: A  l   B  l 
C l   D  g
• Start with A(l) and B(l) in the reactor
• D(g) bubbles out of the liquid phase, pushing the equilibrium to the right
and forcing the reaction to go to completion
Boil off water to
produce high
MW polymer
Common industrial reaction:
+
n
n
+
nylon
H2O
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
L7-20
Improving Yields of Reversible Rxns with
Semi-Batch Reactors
Semi-batch
A+B
V0
FD
A+B⇌C+D
V0 - u0t
Vf
To improve product yield in a reversible reaction: A  l   B  l 
C l   D  g
• Start with A(l) and B(l) in the reactor
• D(g) bubbles out of the liquid phase, pushing the equilibrium to the right
and forcing the reaction to go to completion
How do we account for the loss of product D in the material balance?
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
L7-21
Loss of Mass in Semi-Batch Reactor
A l  B l
D(g)
C l   D  g  elementary rxn
u = u0    0
Overall Mass balance:
In - Out + Generation = Accumulation
g 
dm
m  
u0  m 
0 
 gas leaving reactor
dt
 time 
m
m
↑want in terms of dV/dt
 V
V

u0 m dm  1 
m dV
V0 + u0t



u


Divide mass balance by 
0




dt   

dt
Relate ṁ to a rate:
rA 
moles
volume  time
From stoichiometry, rD = -rA
Next, convert mass
units to:
time
Conversion 2:
Conversion 1: rA  V  
massD
molesD 
 molesD MWD   massD
MWD
r V MWD 
dV
 u0  A
dt

rA  V  MW 
moles
time
mass
m
time
Substitute for ṁ
One of the diff. eq. that are simultaneously solved
(by Polymath)
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.