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Lecture 5. 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 5 - Thursday 1/20/2011. Block 1: Mole Balances Block 2: Rate Laws Block 3: Stoichiometry - 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 5
Lecture 5 - Thursday 1/20/2011Block 1: Mole Balances Block 2: Rate LawsBlock 3: Stoichiometry
Stoichiometric Table: FlowDefinitions of Concentration: FlowGas Phase Volumetric Flow rateCalculate the Equilibrium Conversion Xe
2
Reactor Mole Balance SummaryIn terms of conversionReactor Differential Algebraic Integral
A
A
rXFV
0CSTR
AA rdVdXF 0
X
AA r
dXFV0
0PFR
VrdtdXN AA 0
0
0
X
AA Vr
dXNtBatchX
t
AA rdWdXF 0
X
AA r
dXFW0
0PBRX
W3
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
4
Algorithm
Reaction Engineering
These topics build upon one another
Mole Balance Rate Laws Stoichiometry
5
Species Symbol Reactor Feed Change Reactor EffluentA 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
0
0
0
0
00
00
0
0
A
i
A
i
A
i
A
ii y
yCC
CC
FF
1
ab
ac
ad
Where: and
A
AFC Concentration – Flow System
Flow System Stochiometric Table
6
A
AFC Concentration Flow System:
0 Liquid Phase Flow System:
XCXFFC AAA
A
110
0
0
X
abCX
ab
VN
VNC BAB
ABB 0
0
0
Flow Liquid Phase
etc.7
Liquid Systems
Stoichiometry
BAA CkCr If the rate of reaction were
X
abXCr BAA 12
0then we would have
XfrA This gives us
A
A
rF
0
X8
Liquid Systems
For Gas Phase Flow Systems
We obtain:
0
0
00 T
TPP
FF
T
T
Combining the compressibility factor equation of state with Z = Z0
000
000
00
TT
TT
T
T
CFCF
TRZPC
ZRTPC
Stoichiometry:
9
Stoichiometry
TT
PPF
TT
PP
FFFFC T
T
AAA
0
00
00
0
00
TT
PP
FFCCT
BTB
0
00
000 TT FC
10
For Gas Phase Flow SystemsStoichiometry
XFFF ATT 00 The total molar flow rate is:
PP
TT
FXFF
T
AT 0
00
000
PP
TTX
FF
T
A 0
00
00 1
PP
TTXyA 0
000 1
PP
TTX 0
00 1
Substituting FT gives:
11
StoichiometryFor Gas Phase Flow Systems
Gas Phase Flow System:
Concentration Flow System:
0
00
0
00
0
11
1
1PP
TT
XXC
PP
TTX
XFFC AAAA
A
AFC
PP
TTX 0
00 1
0
00
0
00
0
11 PP
TT
X
XabC
PP
TTX
XabF
FCBABA
BB
12
For Gas Phase Flow Systems
2
0
0
20 11
1TT
PP
X
Xab
XXCkr
B
AAA
If –rA=kCACB
This gives usFA0/-rA
X13
For Gas Phase Flow Systems
14
For Gas Phase Flow Systems
Consider the following elementary reaction with KC=20 dm3/mol and CA0=0.2 mol/dm3. Calculate Equilibrium Conversion or both a batch reactor (Xeb) and a flow reactor (Xef).
Example: Calculating the equilibrium conversion for gas phase reaction in a flow reactor, Xef
C
BAAA K
CCkr 2
BA2
15
Gas Flow Example Xef
2A B
B21A
X ef ?
Rate law:
C
BAAA KCCkr 2
X eb 0.703
16
Solution:
Species
Fed Change Remaining
A FA0 -FA0X FA=FA0(1-X)B 0 +FA0X/2 FB=FA0X/2
FT0=FA0 FT=FA0-FA0X/2
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Gas Flow Example Xef
A FA0 -FA0X FA=FA0(1-X)B 0 FA0X/2 FB=FA0X/2
Stoichiometry: Gas isothermal T=T0, isobaric P=P0
V V0 1X
CA FA 0 1 X V0 1X
CA 0 1 X 1X
CB FA 0 X 2V0 1X
CA 0 1 X 2 1X 18
Gas Flow Example Xef
Pure A yA0=1, CA0=yA0P0/RT0, CA0=P0/RT0
C
AAAA KX
XCXXCkr
1211 0
2
0
20 1
12e
eeAC X
XXCK
211
2110
Ay
@ eq: -rA=0
19
Gas Flow Example Xef
82.02022 3
3
0
dmmol
moldmCK AC
yA 0 112
1
12
8 X e 0.5X e
2
1 2X e X e2
8.5Xe2 17Xe 8 0
Xef 0.757Flow: Recall
Xeb 0.70Batch:20
Gas Flow Example Xef
21
22
23
End of Lecture 5
24
