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Chemical Reaction Engineering (CRE)is the field that studies the rates and
mechanisms of chemical reactions andthe design of the reactors in which they
take place.
Lecture 4
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Lecture 4 Tuesday 1/18/2011Block 1
Size CSTRs and PFRs given r A=f(X)Block 2
Reaction Orders Arrhenius Equation
Block 3
Stoichiometric TableDefinitions of ConcentrationCalculate the Equilibrium Conversion, X e
2
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Reactor Differential Algebraic Integral
A
A
r
X F V 0CSTR
A A r dV dX F 0
X
A A r dX F V 0
0PFR
V r
dt
dX N A A 0
0
0
X
A
A V r dX
N t Batch
X
t
A A r dW
dX F 0
X
A A
r
dX F W
00
PBR
X
W3
ReviewLecture 2
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F A 0
r A
X
ReviewLecture 2
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F A 0r A
Are a = Volume of PF
V 0
X 1 F A 0
r A
dX
X 1
ReviewLecture 2
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6
reactor firsttofedAof molesi pointtoupreactedAof moles
X i
Only valid if there are no side streams
ReviewLecture 2
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7
ReviewLecture 2
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Two steps to get
Step 1: Rate Law
Step 2: Stoichiometry
Step 3: Combine to get
r A f X
r A
g C i
C i h X
r A f X
ReviewLecture 2
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BAA CkCr
Order RectionOverall
Binorder
AinorderPower Law Model
9
C3BA2 A reactor follows an elementary rate law if the
reaction orders just happens to agree with thestoichiometric coefficients for the reaction as written.e.g. If the above reaction follows an elementary ratelaw B
2AAA CCk r
ReviewLecture 3
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E = Activation energy (cal/mol)R = Gas constant (cal/mol*K)T = Temperature (K)
A = Frequency factor (same units as rate constant k)(units of A, and k, depend on overall reaction order)
10
RT E Aek
T k AT 0 k 0
A 10 13
T
k
ReviewLecture 3
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These topics build upon one another
Mole Balance Rate Laws Stoichiometry
11
ReviewLecture 3
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r A f X
r A g C i Step 1: Rate Law
C i h X Step 2: Stoichiometry
r A f X Step 3: Combine to get
12
ReviewLecture 3
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We shall set up Stoichiometry Tables using species A as our basis of calculation in the followingreaction. We will use the stochiometric tables toexpress the concentration as a function ofconversion. We will combine C i = f(X) with theappropriate rate law to obtain -r A = f(X).D
a
dC
a
cB
a
bA
A is the limiting Reactant.
13
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N A N A 0 N A 0 X
N B N B 0 b
a N A 0 N A 0 N B 0 N A 0
b
a X
For every mole of A that react, b /a moles of B react.Therefore moles of B remaining:
Let B = N B0 /N A0
Then:
N B N A 0 B ba X
N C N C 0 c
a N A 0 X N A 0 C
c
a X
14
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Species
Symbol
Initial Change Remaining
B B N B0 =N A0 B -b/aN A0 X N B=N A0 ( B-b/aX)
A A N A0 -N A0 X N A=N A0 (1-X)
Inert I N I0 =N A0 I ---------- N I =N A0 I
F T0 N T =N T0 +N A0 X
Wher e: 0
0
0
0
00
00
0
0
A
i
A
i
A
i
A
ii y
yC C
C C
N N
1
ab
ac
ad and
C C N C0 =N A0 C +c/aN A0 X N C =N A0 ( C +c/aX )D D N D0 =N A0 D +d/aN A0
XN D=N A0 ( D+d/aX)
15 = change in total number of mol per mol Areacted
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Note: If the reaction occurs in the liquid phaseor
if a gas phase reaction occurs in a rigid (e.g. steel)
batch reactorV V 0Then
C A N A
V N A 0 1 X
V 0C A 0 1 X
C B N B
V N A 0
V 0 B
ba
X
C A 0 B ba
X
etc.16
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Suppose r A k AC A2
C B
Batch: 0VV
17
r A k AC A 0 2 1 X 2
B ba X
Equimolar feed: B 1
Stoichiometric feed: B ba
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r A f X and we have
if r A k A C A2 C
B then
r A C A 03
1 X 2 B ba
X
Constant Volume Batch
18
Ar
1
X
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Consider the following elementary reaction withKC=20 dm 3/mol and C A0 =0.2 mol/dm 3.Xe for both a batch reactor and a flow reactor.
Calculating the equilibrium conversionfor gas phase reaction,X e
C
B2AAA K
CCk r
BA2
19
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Step 1: dX dt
r AV N A 0
30 2.0 dmmol C A
mol dm K C 3 20
Step 2: rate law, BB2AAA Ck Ck r
Calculate X e
B
AC
k
k K
C
B2AAA K
CCk r
20
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Symbol
Initial Change Remaining
B 0 N A0 X N A0 X/2
A N A0 -N A0 X N A0 (1-X)
Totals: N T0=N A0 NT=N A0 -N A0 X/2
@ equilibrium: -r A=0C
Be2Ae K
CC0
K e C BeC Ae
2 C Ae N Ae
V C A 0 1 X e
C Be C A 0 X e
221
Calculate X e
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Species Initial Change Remaining A N A0 -N A0 X N A=N A0 (1-X)B 0 +N A0 X/2 N B=N A0 X/2
NT0=N A0 NT=N A0 -N A0 X/2
S o l u t i o n :
2Ae
BeC C
CK
C
Be2AeAA K
CCk 0r At equilibrium
0VV 2/BAStoichiometry
ConstantvolumeBatch
Calculating the equilibrium conversion
for gas phase reaction
22
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2e0Ae
2e0A
e0A
eX1C2
X
X1C2
XC
K
82.0202
X1
XCK 2 2
e
e0Ae
X eb 0.703
23
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A A F A0 -F A0 X F A=F A0 (1-X)
Species
Symbol
ReactorFeed
Change
ReactorEffluent
B B F B0=F A0 B -b/aF A0 X
FB=F A0 ( B-b/aX)
i F i0
F A 0
C i 0 0
C A 0 0
C i0
C A 0
yi 0
y A 0Where:
Flow System Stochiometric Table
24
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Species Symbol ReactorFeed Change ReactorEffluent
0A
0i
0A
0i
00A
00i
0A
0ii y
yCC
CC
FFWhere:
Inert I F I0= A0 I ---------- F I=F A0 I FT0 F T=F T0+F A0 X
C C F C0 =F A0 C +c/aF A0 X F C=F A0 ( C+c/aX)
D D F D0=F A0 D +d/aF A0 X F D=F A0 ( D+d/aX)
1a
bac
adand
A
A
F
CConcentration Flow System
25
Flow System Stochiometric Table
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Species Symbol Reactor Feed Change Reactor Effluent
A A F A0 -F A0 X F A=F A0 (1-X)
B B F B0=F A0 B -b/aF A0 X F B=F A0 ( B-b/aX)
C C F C0 =F A0 C +c/aF A0 X F C=F A0 ( C+c/aX)
D D F D0=F A0 D +d/aF A0 X F D=F A0 ( D+d/aX)
Inert I F I0=F A0 I ---------- F I=F A0 I
FT0 F T=F T0+F A0 X
0A
0i
0A
0i
00A
00i
0A
0ii y
yCC
CC
FF
1a
bac
adWhere: and
AA
FCConcentration Flow System
26
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AA
FCConcentration Flow System:
0Liquid Phase Flow System:
C A F A
F A 0 1 X
0 C A 0 1 X
C B N B
N A 0
0
B ba
X
C A 0 B ba
X
Flow Liquid Phase
etc.
27
We will consider C A and C B for gas phasereactions in the next lecture
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Mole Balance
Rate LawsStoichiometry
Isothermal Design
Heat Effects
28
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End of Lecture 4
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