Chemical Reaction Engineering 1
Lecture 9
Reactor Design for Multiple Reactions
Ch 6: Reactor Design for Multiple
Reactions
• Usually, more than one reaction occurs within a chemical reactor
• Minimization of undesired side reactions that occur with the desired
reaction contributes to the economic success of a chemical plant
• Goal: determine the reactor conditions and configuration that
maximizes product formation
• Reactor design for multiple reactions
• Parallel reactions
• Series reactions
• Independent reactions
• More complex reactions
• Use of selectivity factor to select the proper reactor that minimizes
unwanted side reactions
Classification of Multiple Reactions
1) Parallel or competing reactions
B
k1
A
k2
C
2) Series reactions
Desired product
k1
A
k2
B
C
Desired product
3) Independent reactions
A
k1
B
4) Complex reactions
C
Crude oil cracking
k2
D
k
1 C+D
A + B ⎯⎯→
k
2 →E
A + C ⎯⎯⎯
Parallel Reactions
Purpose: maximizing the desired product in parallel reactions
kD
D
(desired)
rD = kDCA1CB 1
→ rD =
−ED
AD e RT CA1CB 1
→ rU =
−EU
AUe RT CA 2 CB 2
A+B
kU
k (T) =
U (undesired)
−E
AeRT
rU = kUCA 2 CB 2
Rate of disappearance of A: −rA = rD + rU
−rA =
−EU
−ED
ADe RT C A1CB 1 + AUe RT C A2 CB 2
Define the instantaneous rate selectivity, SD/U
SD U =
SD U =
(
rate of formation of D rD
=
rate of formation of U rU
)
sD U =
AD e
A Ue
−(ED −EU )
AD
e RT
kD
CA1− 2 CB 1− 2 → SD U =
kU
AU
− ED
RT
− EU
RT
(
C A 1CB1
C A 2 CB2
)
CA1− 2 CB 1− 2
Goal: Maximize SD/U to maximize production of the desired product
Maximizing SD/U for Parallel
Reactions: Temperature Control
SD U =
−(ED −EU )
AD
e RT
AU
(CA1−2 ) CB1−2
What reactor conditions and configuration maximizes the selectivity?
Start with temperature (affects k):
b) If ED < EU
a) If ED > EU
−(ED −EU )
ED − EU
 0 → e RT
RT
1
Specific rate of desired reaction
kD increases more rapidly with
increasing T
Use higher temperature to favor
desired product formation
−(ED −EU )
ED − EU
 0 → e RT
RT
1
Specific rate of desired reaction kD
increases less rapidly with
increasing T
Use lower T to favor desired
product formation (not so low that
the reaction rate is tiny)
Maximizing SD/U for Parallel
Reactions: Concentration
kD
D
SD U =
A+B
kU
U
−(ED −EU )
AD
e RT
AU
(
)
CA1−2 CB 1− 2
What reactor conditions and configuration maximizes the selectivity?
Now evaluate concentration:
b) 1  2 → 1 − 2  0
a) 1  2 → 1 − 2  0
CA1−2
CA1−2
→ Use large CA
→ Use small CA
c) 1  2 → 1 − 2  0
d) 1  2 → 1 − 2  0
CB1− 2
CB1− 2
→ Use large CB
→ Use small CB
How do these concentration requirements affect reactor selection?
Concentration Requirements &
Reactor
Selection
k
D
D
A+B
kU
How do concentration requirements play
into reactor selection?
CA0u0
CB0u0
U
PFR
PFR (or PBR): concentration is
high at the inlet & progressively
drops to the outlet concentration
CA(t)
CB(t)
Batch:
concentration is
high at t=0 &
progressively drops
with increasing time
CBu0
CA
CAu0
CBu0
CSTR:
concentration is
always at its
lowest value
(that at outlet)
Semi-batch: concentration
of one reactant (A as
shown) is high at t=0 &
progressively drops with
increasing time, whereas
concentration of B can be
kept low at all times
kD
A+B
kU
D
High CA favors undesired
High
C
favors
desired
A
1  2
1  2
product formation
product formation
(keep CA low)
U
1  2
High CB
favors
desired
product
formation
1   2
High CB
favors
undesired
product
formation
(keep CB
low)
Batch reactor
When CA & CB are low (end time
or position), all rxns will be slow
PFR/PBR
High P for gas-phase rxn, do not
add inert gas (dilutes reactants)
PFR/PBR
Side streams feed low CA
CA
Semi-batch
reactor slowly feed
←High CB
A to large amt of B
CA
CA
CA
CSTRs in
series
CA0u0
CB0u0
CAu0
CBu0
CSTR
PFR/PBR w/ side streams feeding
low CB
CB
Semi-batch
←High CA
PFR/PBR
reactor, slowly
PFR/PBR
w/ high
feed B to large amount of A
recycle
CB
CB
CB
CSTRs in
• Dilute feed with inerts that are
series
easily separated from product
B consumed before leaving CSTRn • Low P if gas phase
Different Types of Selectivity
instantaneous rate selectivity, SD/U
SD U =
rate of formation of D rD
=
rate of formation of U rU
overall rate selectivity, SD U
F
Exit molar flow rate of desired product
SD U = D =
FU Exit molar flow rate of undesired product
N
Final moles of desired product
SD U = D =
NU Final moles of undesired product
instantaneous yield, YD
(at any point or time in reactor)
YD =
r
rate of formation of D
= D
rate of consumption of A −rA
overall yield, YD
FD
flow YD =
FA0 − FA
Evaluated
at outlet
batch YD =
ND
NA0 − NA
Evaluated
at tfinal
Series (Consecutive) Reactions
A
k1
k2
U
D
(desired) (undesired)
Spacetime t for a flow reactor
Time is the key factor here!!!
Real time t for a batch reactor
To maximize the production of D, use:
CSTRs in series
Batch
or
PFR/PBR
or
n
and carefully select the time (batch) or spacetime (flow)
Concentrations in Series Reactions (PFR)
A
k1
B
k2
C
-rA = k1CA
rB,net = k1CA – k2CB
How does CA depend on t?
dFA
dC A
= −k1C A → u0
= −k1C A → C A = C A0e −k1t
dV
dV
How does CB depend on t?
(
)
dFB
dCB
= k1C A − k 2CB → u0
= k1 C A0e −k1t − k 2CB
dV
dV
→
(
)
Substitute
(
dCB
dCB
= k1 CA0e−k1t − k 2CB →
+ k 2CB = k1 CA0e−k1t
dt
dt
)
V
u0
=t
Concentrations in Series Reactions
→
Use integrating
factor
(
dCB
+ k 2CB = k1 CA0e−k1t
dt
)
𝐼. 𝐹. = exp න𝑘2 𝑑τ = exp 𝑘2 τ
→
(
d CBek 2t
dt
at τ = 0, CB=0
) =k C
1
(
e
A0
k 2 −k1)t
 e−k1t − e−k 2t
→ CB = k1C A0 
 k 2 − k1




CC = CA0 − CA − CB
CA0
CC =
k 2 1 − e−k1 t − k1 1 − e−k2 t
k 2 − k1
Reactions in Series: Cj & Yield
CA = CA0e−k1t
B
C
A
 e−k1t − e−k 2t
CB = k1CA0 
 k 2 − k1




CC = CA0 − CA − CB
topt
What is the optimal
t?
The reactor V (for a given u0) and t that maximizes CB occurs when dCB/dt=0
(
)
dCB  k1CA0 
−k1t
−k 2t
=
−
k
e
+
k
e
=0
1
2

dt  k 2 − k1 
t opt =
k
1
ln 1
k1 − k 2 k 2
V
= t so Vopt = u0t opt
u0
Case 2: CSTR Series Reactions
A→B→C
What is the optimal t ?
1) Mole Balances
A:
FA0 − FA + rAV = 0
C A0 v0 − C A v0 + rAV = 0
C A0 − C A + rAt = 0
B:
0 − v0 C B + rBV = 0
− C B + rBt = 0
15
Case 2: CSTR Series Reactions
2) Rate Laws -rA = k1CA
rB,net = k1CA – k2CB
3) Combine
CA 0 - CA - k1CAt = 0
CA 0
CA =
1+ k1t
-CB + ( k1CA - k 2CB )t = 0
16
k1CAt
CB =
1+ k2t
k1CA 0t
CB =
(1+ k2t )(1+ k1t )
Case 2: CSTR Series Reactions
A→B→C
Find
t that gives maximum concentration of B
k1C A0t
CB =
(1 + k2t )(1 + k1t )
dCB
=0
dt
17

t max
1
=
k1k 2
Part 2:
Algorithm for Solution to
Complex Reactions
 Complex Reactions:
A + 2B → C
A + 3C → D
 Example A: Liquid Phase PFR
 Example B: Liquid Phase CSTR
 Example C: Gas Phase PFR
 Example D: Semibatch Reactor
18
New things for multiple reactions are:
1. Number Every Reaction
2. Mole Balance on every species
3. Rate Laws
(a) Net Rates of Reaction for every species
N
rA =  riA
i =1
(b) Rate Laws for every reaction
r1 A = − k1 AC AC B2
r2C = − k 2C C A2 CC3
(c) Relative Rates of Reaction for every reaction
For a given reaction i: (i) aiA+biB →ciC+diD:
19
riA
riB
riC riD
=
=
=
− ai − bi ci
di
Reactor Mole Balance Summary
Reactor Type
Gas Phase
Batch
dN A
= rAV
dt
Semibatch
20
Liquid Phase
dC A
= rA
dt
dN A
= rAV
dt
u 0C A
dC A
= rA −
dt
V
dN B
= rBV + FB 0
dt
u0 CB 0 − CB 
dC B
= rB +
dt
V
Reactor Mole Balance Summary
Reactor Type
21
Gas Phase
Liquid Phase
(C A0 − C A )
CSTR
FA0 − FA
V=
− rA
V = u0
PFR
dFA
= rA
dV
dC A
u0
= rA
dV
PBR
dFA
= rA
dW
dC A
u0
= rA
dW
− rA
Note: The reaction rates in the above mole balances are net rates.
Stoichiometry
22
Batch
Flow
NB
CB =
V
NT P0 T0
V = V0
NT 0 P T
FB
CB =
u
FT P0 T0
u = u0
FT 0 P T
N B NT 0 P T0
CB =
NT V0 P0 T
FB FT 0 P T0
CB =
FT u0 P0 T
N B P T0
CB = CT 0
NT P0 T
FB P T0
CB = CT 0
FT P0 T
Stoichiometry
Concentration of Gas:
𝐹𝐵 𝑃 𝑇0
𝐶𝐵 = 𝐶𝑇0
𝐹𝑇 𝑃0 𝑇
𝐹𝑇 =𝐹𝐴 + 𝐹𝐵 + 𝐹𝐶 + 𝐹𝐷
Isobaric and isothermal systems
Note: For constant volume systmes (i.e., liquid phase),
concentrations are simply defined as:
Flow:
23
FA
CA =
u0
NA
Batch: C A =
V0
Gas Phase
Multiple Reactions
ODE solver can solve
all the coupled
equations. The most
important step is to
define the rnet properly.
24
Example A: Liquid Phase PFR
The complex liquid phase reactions follow
elementary rate laws:
2
−
r
=
k
C
C
(1) A + 2 B → C
1A
1A A B
NOTE: The specific reaction rate k1A is defined with
respect to species A.
(2) 3C + 2 A → D
− r2C = k2C CC3 C A2
NOTE: The specific reaction rate k2C is defined with
respect to species C.
25
Example A: Liquid Phase PFR
Complex Reactions
(1) A + 2 B → C
(2) 3C + 2 A → D
Plot the overall rate
selectivity (SD/U) as a function
of the reactor volume.
1) Mole Balance on each and every species
26
dFA
(1)
= rA
dV
dFB
(2)
= rB
dV
dFC
(3)
= rC
dV
dFD
(4)
= rD
dV
Example A: Liquid Phase PFR
2) Rate Laws:
Net Rates
(5) 𝑟𝐴 = 𝑟1𝐴 + 𝑟2𝐴
(6) 𝑟𝐶 = 𝑟1𝐶 + 𝑟2𝐶
(9) r1 A = −k1 AC AC B2
Rate Laws
(10) r2C = −k 2C C A2 CC3
Relative Rates
Reaction 1
r1 A r1B r1C
=
=
−1 −2 1
(11) r1B = 2r1 A
27
(12) r1C = −r1 A
(7) 𝑟𝐵 = 𝑟1𝐵 + 0
(8) 𝑟𝐷 = 0 + 𝑟2𝐷
Example A: Liquid Phase PFR
Relative Rates
Reaction 2
r2 A r2C r2 D
=
=
−2 −3 1
2
(13) r2 A = r2C
3
r2C
(14) r2 D = −
3
2
rA = −k1 AC AC − k 2C C A2 CC3
3
rB = −2k1 AC AC B2
2
B
rC = k1 AC AC B − k 2C C A2 CC3
28
k 2C 2 3
rD =
C A CC
3
Example A: Liquid Phase PFR
3) Stoichiometry
Liquid
(15) 𝐶𝐴 = 𝐹𝐴 Τ𝜐0
(16) 𝐶𝐵 = 𝐹𝐵 Τ𝜐0
(17) 𝐶𝐶 = 𝐹𝐶 Τ𝜐0
(18) 𝐶𝐷 = 𝐹𝐷 Τ𝜐0
(19) 𝑆ሚ𝐶 Τ𝐷 = 𝑖𝑓 𝑉 > 0.00001 then
29
𝐹𝐶
𝐹𝐷
else 0
Example A: Liquid Phase PFR
4) Parameters
(20) 𝑘1𝐴 = 10
(21) 𝑘2𝐶 = 20
(22) 𝑉𝑓 = 2500
(23) 𝐹𝐴0 = 200
(24) 𝐹𝐵0 = 200
(25) 𝜐0 = 100
With the input variables, the ODE solver can solve
all the coupled equations.
30
Example B: Liquid Phase CSTR
Same reactions, rate laws, and rate constants as
Example A
(1) A + 2 B → C
− r1 A = k1 AC AC
2
B
NOTE: The specific reaction rate k1A is defined with
respect to species A.
(2) 3C + 2 A → D
− r2C = k2C CC3 C A2
NOTE: The specific reaction rate k2C is defined with
respect to species C.
31
Example B: Liquid Phase CSTR
The complex liquid phase reactions take place in
a 2,500 dm3 CSTR. The feed is equal molar in A
and B with FA0=200 mol/min, the volumetric flow
rate is 100 dm3/min and the reaction volume is 50
dm3.
Find the concentrations of A, B, C and D existing
in the reactor along with the existing selectivity.
Plot FA, FB, FC, FD and SC/D as a function of V
32
Example B: Liquid Phase CSTR
(1) A + 2B →C
(2) 2A + 3C → D
r1 A = −k1 AC AC B2
r2C = −k 2C C A2CC3
1) Mole Balance
(1) A
(2) B
(3) C
(4) D
33
u0C A0 − u0C A + rAV = 0
u0CB 0 − u0CB + rBV = 0
0 − u0CC + rCV = 0
0 − u0CD + rDV = 0
Example B: Liquid Phase CSTR
2) Rate Laws: (5)-(14) same as PFR
3) Stoichiometry: (15)-(18)
same as Liquid Phase PFR
(19) SC / D
FC
u0CC
=
=
FD + 0.0001 u0CD + 0.0001
4) Parameters:
k1 A , k 2C , C A0 , C B 0 , V , u0
34
For CSTR case, it can be solved analytically!
Example C: Gas Phase PFR, with ΔP
Same reactions, rate laws, and rate constants as
Example A:
(1) A + 2 B → C
− r1 A = k1 AC AC B2
NOTE: The specific reaction rate k1A is defined with respect to
species A.
(2) 3C + 2 A → D
− r2C = k2C CC3 C A2
NOTE: The specific reaction rate k2C is defined with respect to
species C.
35
Example C: Gas Phase PFR, with ΔP
1) Mole Balance
(1)
(2)
dFA
= rA
dV
dFB
= rB
dV
dFC
(3)
= rC
dV
dFD
(4)
= rD
dV
2) Rate Laws: (5)-(14) same as CSTR
36
Example C: Gas Phase PFR, with ΔP
3) Stoichiometry:
Gas: Isothermal T = T0
FA
𝑃
(15) C A = CT 0
y (16) CB = CT 0
y=
𝑃0
FT
FC
(17) CC = CT 0
y (18) C D = CT 0
FT
(19) FT = FA + FB + FC + FD
Packed Bed with Pressure Drop
37
dy
  FT  T 
 FT




=− 
=−



dW
2 y  FT 0  T0 
2 y FT 0
FB
y
FT
FD
y
FT
Example C: Gas Phase PFR, with ΔP
4) Selectivity
 FC 
FC
S=
= if (V  0.00001) then   else (0) (20)
FD
 FD 
38
Example D : Liquid Phase Semibatch
The complex liquid phase reactions take place
in a semibatch reactor where A is fed to B with
FA0= 3 mol/min. The volumetric flow rate is 10
dm3/min and the initial reactor volume is 1,000
dm3.
The maximum volume is 2,000 dm3 and
CA0=0.3 mol/dm3 and CB0=0.2 mol/dm3. Plot CA,
CB, CC, CD and SS/D as a function of time.
39
Example E: Liquid Phase Semibatch
(1) A + 2B →C
(2) 2A + 3C → D FA0
1) Mole Balances:
40
dN A
dt
dN B
dt
dN C
dt
dN D
dt
B
= rAV + FA0
N A0 = 0
= rBV
N B 0 = C B 0V0 = 2.000
= rCV
NC 0 = 0
= rDV
N D0 = 0
Example E: Liquid Phase Semibatch
2) Rate Laws: (5)-(14)
Net Rate, Rate Laws and relative rate – are the same
as Liquid and Gas Phase PFR and Liquid Phase CSTR
V = V0 + v0t (15)
NA
CA =
V
NC
CC =
V
(16)
(18)
NB
(17 )
CB =
V
ND
(19)
CD =
V
3) Selectivity and Parameters:
SC / D
41
 NC 
 else (0) (20)
= if (t  0.0001) then 
 ND 
u0 = 10 dm3 min V0 = 100dm3
FA 0 = 3 mol min