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Lecture 4. Chemical Reaction Engineering (CRE) is the field that studies the rates and mechanisms of chemical reactions and the design of the reactors in which they take place. Today’s lecture. Block 1 Mole Balances Size CSTRs and PFRs given – r A = f(X ) Block 2 Rate Laws - PowerPoint PPT Presentation
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Chemical Reaction Engineering (CRE) is the field that studies the rates and mechanisms of
chemical reactions and the design of the reactors in which they take place.
Lecture 4
Today’s lectureBlock 1
Mole Balances Size CSTRs and PFRs given –rA=f(X)
Block 2Rate LawsReaction OrdersArrhenius Equation
Block 3Stoichiometry Stoichiometric TableDefinitions of ConcentrationCalculate the Equilibrium Conversion, Xe
2
Reactor Mole Balances in terms of conversionReactor Differential Algebraic Integral
A
0A
r
XFV
CSTR
A0A rdV
dXF
X
0 A0A r
dXFVPFR
Vrdt
dXN A0A
Vr
dXNt
X
0 A0A
Batch
X
t
A0A rdW
dXF
X
0 A0A r
dXFWPBR
X
W3
dDcCbBaA
d
r
c
r
b
r
a
r DCBA
Da
dC
a
cB
a
bA
Last LectureRelative Rates of Reaction
4
BAA CkCr
βα
β
α
OrderRection Overall
Bin order
Ain order
Last LectureRate Laws - Power Law Model
5
C3BA2 A reactor follows an elementary rate law if the reaction orders just happens to agree with the stoichiometric coefficients for the reaction as written.e.g. If the above reaction follows an elementary rate law
2nd order in A, 1st order in B, overall third order
B2AAA CCkr
Last LectureArrhenius Equation
k Ae E RT
k is the specific reaction rate (constant) and is given by the Arrhenius Equation.
Where:
T k A
T 0 k 0
A1013
k
T6
These topics build upon one another
Mole Balance Rate Laws Stoichiometry
Reaction Engineering
7
How to find
rA f X
rA g Ci Step 1: Rate Law
Ci h X Step 2: Stoichiometry
rA f X Step 3: Combine to get
8
We shall set up Stoichiometry Tables using species A as our basis of calculation in the following reaction. We will use the stochiometric tables to express the concentration as a function of conversion. We will combine Ci = f(X) with the appropriate rate law to obtain -rA = f(X).
Da
dC
a
cB
a
bA
A is the limiting Reactant.
9
NA NA 0 NA 0X
NB NB 0 b
aNA 0 NA 0
NB 0
NA 0
b
aX
For every mole of A that react, b/a moles of B react. Therefore moles of B remaining:
Let ΘB = NB0/NA0
Then:
NB NA 0 B b
aX
NC NC 0 c
aNA 0X NA 0 C
c
aX
10
Species
Symbol
Initial Change
Remaining
Batch System Stoichiometry Table
B B NB0=NA0ΘB -b/aNA0X
NB=NA0(ΘB-b/aX)
A A NA0 -NA0X NA=NA0(1-X)
Inert I NI0=NA0ΘI ---------- NI=NA0ΘI
FT0 NT=NT0+δNA0X
Where: 0A
0i
0A
0i
00A
00i
0A
0ii y
y
C
C
C
C
N
N
1a
b
a
c
a
dan
d
C C NC0=NA0ΘC +c/aNA0X
NC=NA0(ΘC+c/aX)
D D ND0=NA0ΘD +d/aNA0X
ND=NA0(ΘD+d/aX)
11δ = change in total number of mol per mol A reacted
Constant Volume BatchNote: If the reaction occurs in the liquid phase
orif a gas phase reaction occurs in a rigid (e.g.
steel) batch reactor
V V0Then
CA NAV
NA 0 1 X
V0
CA 0 1 X
CB NBV
NA 0
V0
B b
aX
CA 0 B
b
aX
etc.12
Suppose
€
−rA = kACA2CB
Batch: 0VV
Stoichiometry
13
€
−rA = kACA 02 1− X( )
2ΘB −
b
aX
⎛
⎝ ⎜
⎞
⎠ ⎟
Equimolar feed:
B 1
Stoichiometric feed:
B b
a
Constant Volume Batch Reactor (BR)
rA f X and we have
if
rA kACA2CB then
rA CA 03 1 X 2 B
b
aX
Constant Volume Batch
14
Consider the following elementary reaction with KC=20 dm3/mol and CA0=0.2 mol/dm3. Xe’ for both a batch reactor and a flow reactor.
Calculating the equilibrium conversion for gas phase reaction,Xe
C
B2AAA K
CCkr
BA2
15
BR Example
Step 1:
dX
dt
rAVNA 0
Lmol 2.0C 0A
KC 20 L mol
Step 2: rate law, BB2AAA CkCkr
Calculate Xe
B
AC k
kK
C
B2AAA K
CCkr
16
BR Example
Symbol
Initial Change
Remaining
B 0 ½ NA0X NA0 X/2
A NA0 -NA0X NA0(1-X)
Totals:NT0=NA
0
NT=NA0 -NA0 X/2
@ equilibrium: -rA=0 C
Be2Ae K
CC0
Ke CBeCAe
2
CAe NAeV
CA 0 1 Xe
CBe CA 0
Xe217
Calculate Xe
BR Example
Species Initial Change Remaining
A NA0 -NA0X NA=NA0(1-X)
B 0 +NA0X/2 NB=NA0X/2
NT0=NA0 NT=NA0-NA0X/2
Solution:
2Ae
BeC C
CK
C
Be2AeAA K
CCk0rAt equilibrium
0VV 2/BA Stoichiomet
ryConstant volumeBatch
Calculating the equilibrium conversion for gas phase reaction
18
BR Example
2e0A
e2
e0A
e0A
eX1C2
X
X1C2
XC
K
82.0202
X1
XCK2 2
e
e0Ae
€
Xeb = 0.703
BR Example Xeb
19
A A FA0 -FA0X FA=FA0(1-X)
Species
Symbol
Reactor Feed
Change
Reactor Effluent
B B FB0=FA0ΘB -b/aFA0X
FB=FA0(ΘB-b/aX)
i Fi0FA 0
Ci00
CA 00
Ci0CA 0
y i0yA 0
Where:
Flow System Stochiometric Table
20
Species
Symbol
Reactor Feed
Change
Reactor Effluent
0A
0i
0A
0i
00A
00i
0A
0ii y
y
C
C
C
C
F
F
Where:
Flow System Stochiometric Table
Inert I FI0=A0ΘI ---------- FI=FA0ΘI
FT0 FT=FT0+δFA0X
C C FC0=FA0ΘC +c/aFA0X
FC=FA0(ΘC+c/aX)
D D FD0=FA0ΘD +d/aFA0X
FD=FA0(ΘD+d/aX)
1a
b
a
c
a
dan
d
A
A
FCConcentration – Flow System
21
Species Symbol Reactor Feed Change Reactor Effluent
A A FA0 -FA0X FA=FA0(1-X)
B B FB0=FA0ΘB -b/aFA0X FB=FA0(ΘB-b/aX)
C C FC0=FA0ΘC +c/aFA0X FC=FA0(ΘC+c/aX)
D D FD0=FA0ΘD +d/aFA0X FD=FA0(ΘD+d/aX)
Inert I FI0=FA0ΘI ---------- FI=FA0ΘI
FT0 FT=FT0+δFA0X
0A
0i
0A
0i
00A
00i
0A
0ii y
y
C
C
C
C
F
F
1a
b
a
c
a
dWhere: and
A
A
FCConcentration – Flow System
Flow System Stochiometric Table
22
A
A
FCConcentration Flow System:
0Liquid Phase Flow System:
CA FA
FA 0 1 X
0
CA 0 1 X
CB NB
NA 0
0
B b
aX
CA 0 B
b
aX
Flow Liquid Phase
etc.
23
We will consider CA and CB for gas phase reactions in the next lecture
Mole Balance
Rate Laws
Stoichiometry
Isothermal Design
Heat Effects
24
End of Lecture 4
25
