6.
Flow Systems XIV
o Express concentration as a function of conversion
b c d
A B C D
a a a
Liquid Phase Gas Phase
Flow FB Batch NB N B Batch FB Flow
CB CB CB CB
v V V v
N P T F P T
No phase change V V0 T 0 v v0 T 0
N T0 P T0 FT0 P T0
v v0 V V0
N B N T0 P T0 FB FT0 P T0
b CB CB
CB CA0 B X N T V0 P0 T FT v0 P0 T
a
N B P T0 F P T0
yA0 CB CT0 CB CT0 B
N T P0 T FT P0 T
d c b
δ 1 No phase change (or semipermeable membrane
a a a b
P y P CA0 B X
CT0 0 , CA0 A0 0 P T
v v0 (1 X ) 0 , CB a P T0
RT0 RT0 P T0 1 X P0 T
Mar/17 2011 Spring 1
4. Isothermal Reactor Design
o Objectives
- Describe the CRE algorithm that allows the reader
to solve chemical reaction engineering problems
through logic rather than memorization.
- Size batch reactors, semibatch reactors, CSTRs,
PFRs, PBRs, membrane reactors, and microreactors
for isothermal operation given the rate law and feed
conditions.
- Account for the effects of pressure drop on
conversion in packed bed tubular reactors and in
packed bed spherical reactors
1. Algorithm for Isothermal Reactor Design I
o Isothermal reactor design algorithm for conversion
1) V dN j
F j 0 F j rj dV
dt
2) Apply mole balance to reactor type
3) Is -rA = f(X) given? ⇒ Then evaluate the equation
4) If not, determine the rate law in terms of conc.
5) Use stoichiometry to express conc. as a function of
conversion
6) Combine step 4) & 5) to obtain -rA = f(X)
7) Consider volume change
8) Combine 4) ~ 7) and solve ODE (Polymath)
Mar/24 2011 Spring 3
1. Algorithm for Isothermal Reactor Design II
Mar/24 2011 Spring 4
1. Algorithm for Isothermal Reactor Design III
Mar/24 2011 Spring 5
2. Applications/Examples of the CRE Algorithm I
Gas Phase
Elementary Additional Information
Reaction
Only A fed P0 = 8.2 atm
T0 = 500 K CA0 = 0.2 mol/dm3
2A → B
k = 0.5 dm3/mol·s vo = 2.5 dm3/s
Solve for X = 0.9 for A is limiting
Mar/24 2011 Spring 6
2. Applications/Examples of the CRE Algorithm II
Reactor Mole Balance Rate Law Stoichiometry
X dX
Batch t N A0 rA kC2 Gas:
0 rAV A
V = V0
FA0 X Gas:
CSTR V rA kCA2
rA T =T0, P =P0
X dX
PFR V FA0 rA kCA2 Gas:
0 rA T =T0, P =P0
Mar/24 2011 Spring 7
2. Applications/Examples of the CRE Algorithm III
Reactor Stoichiometry 2
Per mole A ? N A N A0 (1 X )
CA
Batch V V0
CA0 (1 X )
Per mole A
CSTR A → ½B FA FA0 (1 X )
ε = 1.0(1- ½) = -0.5 CA
v v0 (1 εX )
Per mole A (1 X )
PFR A → ½B CA0
(1 εX )
ε = 1.0(1- ½) = -0.5
Mar/24 2011 Spring 8
2. Applications/Examples of the CRE Algorithm IV
Reactor Stoichiometry 3
1
N A0 ( X )
Batch NB 2 CA0 X
CB
V V0 2
1
CSTR FA0 ( X)
FB 2
CB
v v0 (1 εX )
CA0 X
PFR 2(1 εX )
Mar/24 2011 Spring 9
2. Applications/Examples of the CRE Algorithm V
Reactor Combine Integration
1 X 1 1 X
Batch t
kCA0 0 (1 X ) 2
dX t
kCA0 (1 X )
FA0 X (1 0.5 X ) 2
CSTR V 2ε(1 ε) ln(1 X )
kCA02
(1 X ) 2 FA0
V 2 (1 ε) X
2
kCA0 ε X
2
X (1 0.5 X )
2 1 X
FA0
2 0
PFR V 2
dX
kCA0 (1 X )
Mar/24 2011 Spring 10
2. Applications/Examples of the CRE Algorithm VI
Reactor Evaluate For X = 0.9
Batch kCA0 = (0.5)(0.2) t = 90 s
= 0.1 s-1
kC2A0 = (0.5)(0.2)2
= 0.02mol/dm3·s V = 680.6 dm3
CSTR FA0 = CA0·v0 τ = V/v0 = 272.3 s
= (0.2)(2.5) = 0.5 mol/s
V = 90.7 dm3
PFR τ = V/v0 = 36.3 s
Mar/24 2011 Spring 11
3. Design of CSTRs I
o Single CSTR 1
- Design equation FA0 X
V
( rA ) exit
- Substitute FA0 = v0CA0 v0CA0 X
V
rA
- Space time τ V CA0 X
τ
v0 rA
- 1st order rxn assume CA0 X 1 X
τ
rA k 1 X
- Rearranging X
τk
1 τk
Mar/24 2011 Spring 12
3. Design of CSTRs II
o Single CSTR 2
- CA = CA0(1 - X) CA0
CA
1 τk
- Damköhler number ⇒ dimensionless number
• quick estimate of the degree on conversion
achieved by continuous reactors
rA0V Rate of rxn at entrance
Da
FA0 Entering flow rate of A
A rxn rate
A convection rate
Mar/24 2011 Spring 13
3. Design of CSTRs III
o Single CSTR 3
- Damköhler number for a 1st order irrev. rxn
rA0V kCA0V
Da τk
FA0 v0CA0
- Damköhler number for a 2nd order irrev. rxn
rA0V kCA0
2
V
Da τkCA0
FA0 v0CA0
- Rule of thumb
• if Da < 0.1, then X < 0.1
• if Da > 10, then X > 0.9
☞ 1st order rxn, X = Da/(1 + Da)
Mar/24 2011 Spring 14
3. Design of CSTRs IV
o CSTR in series 1
- 1st order irrev. rxn with no change in volumetric flow
rate, effluent of the first reactor C A0
C A1
1 τ1k1
- For 2nd reactor FA1 FA2 v0 (CA1 CA2 )
V2
rA2 k2 CA2
- Solving for CA2 C
CA1 CA0
1 τ 2 k 2 (1 τ 2 k2 )( 1 τ1 k1 )
A2
- For n CSTRs in series C CA0 CA0
(1 k ) (1 Da ) n
An n
- n tank in series 1 1
X 1 1
Mar/24 2011(1 τk )
Spring
n
(1 Da ) n 15
3. Design of CSTRs V
o CSTR in series 2
CA0
rAn kCAn k
(1 τk ) n
Mar/24 2011 Spring 16
3. Design of CSTRs VI
o CSTR in parallel 1
- One large reactor of volume V
o 2nd order reactor in a CSTR
FA0 X FA0 X v0CA0 X
V
rA kCA 2 2
kCA0 (1 X ) 2
- Dividing by v0 τ V X
v0 kCA0 (1 X ) 2
- For conversion X
(1 2Da) - 1 4Da Ex 4-2,
X
2Da p 163
Mar/24 2011 Spring 17
4. Tubular Reactors I
o Design equation
- Differential form dX
FA0 rA
• Q or ∆P dV
- Integral form X dX
V FA0
• no Q or ∆P 0 r
A
o 2nd order reactor in a PFR 1
X dX
V FA0
0 kC 2
A
Mar/24 2011 Spring 18
4. Tubular Reactors II
o 2nd order reactor in a PFR 2
- Liquid phase reaction (v = v0)
• combining MB & rate law dX kCA2
dV FA0
• conc. of A, C C (1 X )
A A0
• combining
v0 X
x
FA0 dx
V 2
kCA0 0 (1 X ) kCA0 1 X
2
• solving for X
τkCA0 Da 2
X
1 τkCA0 1 Da 2
Mar/24 2011 Spring 19
4. Tubular Reactors III
o 2nd order reactor in a PFR 3
- Gas phase reaction (T = T0, P = P0)
• conc. of A, C A CA0 1 X
1 εX
• combining
X (1 ε X )
V FA0 dX
0 kCA0 (1 X ) 2
v0 X (1 ε X ) 2
V 2
kCA0 0 (1 X ) 2
dX
v0 ( 1 ε ) 2
X
V 2ε (1 ε ) ln( 1 X ) ε X
2
kCA0 1 X
Mar/24 2011 Spring 20
5. Pressure Drop in Reactors I
o Pressure Drop and the Rate Law
- In PBR in terms of catalyst weight
dX gram moles
FA0 rA
dW gram catalyst min
• rate equation, r kC2
A A
• stoichiometry CA0 (1 X ) P T0
CA
1 ε X P0 T
• isothermal dX CA0 (1 X )
2
P
2
FA0 k
dW 1 ε X P0
2 2
dX kCA0 CA0 (1 X ) P dX
1 εX F1 ( X , P )
dW v0 P0 dW
Mar/24 2011 Spring 21
5. Pressure Drop in Reactors II
o Flow through a Packed Bed
- Ergun equation
dP G 1 150(1 )μ
1.75G
dz ρ g c D p
3
Dp
• pressure drop in packed bed
dP P0 T FT
β0
dz P T0 FT 0
P0 T
1 εX
dP
β0
dz P T0
G(1 ) 150(1 )
β0 3
1.75G
ρ 0 g c DP DP
Mar/24 2011 Spring 22
