lecture 8 - university of michiganessen/html/powerpoints/lecture_notes/lec8/... · 2011-02-01 ·...
TRANSCRIPT
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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 8
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Today’s lecture � Block 1: Mole Balances on PFRs and PBR
� Must Use the Differential Form � Block 2: Rate Laws � Block 3: Stoichiometry
� Pressure Drop: Liquid Phase Reactions: Pressure Drop does not affect the concentrations in liquid phase rx.
Gas Phase Reactions: Epsilon not Equal to Zero
d(P)/d(W)=.. Polymath will combine with d(X)/f(W)=..for you
Epsilon = 0 and Isothermal P=f(W) Combine then Separate Variables (X,W) and Integrate 2
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Reactor Mole Balances in terms of conversion
Reactor
Differential
Algebraic
Integral
A
0A
rXFV
−=CSTR
A0A rdVdXF −= ∫ −=
X
0 A0A rdXFVPFR
VrdtdXN A0A −=
VrdXNt
X
0 A0A ∫ −=Batch
X
t
A0A rdWdXF ′−= ∫ ′−
=X
0 A0A rdXFWPBR
X
W3
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Gas Phase Flow System:
Concentration Flow System:
( )( )
( )( ) 0
00A
0
00
0AAA P
PTT
X1X1C
PP
TTX1
X1FFCε+−=
ε+υ
−=υ
=
υ= A
AFC
( )PP
TTX1 0
00 ε+υ=υ
( ) ( ) 0
0B0A
0
00
B0AB
B PP
TT
X1
XabC
PP
TTX1
XabF
FCε+
⎟⎠⎞⎜
⎝⎛ −Θ
=ε+υ
⎟⎠⎞⎜
⎝⎛ −Θ
=υ
=
4
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Note: Pressure drop does NOT affect liquid phase reactions
Sample Question: Analyze the following second order gas phase reaction that occurs isothermally in a PBR:
AàB
′−= A0A rdWdXF
Mole Balance: Must use the differential form of the mole balance to separate variables:
2AA kCr =′−Second order in A and irreversible:
Rate Law:
Pressure Drop in Packed Bed Reactors
5
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CA =FAυ
=CA01− X( )1+εX( )
PP0T0T
Stoichiometry:
CA = CA01− X( )1+εX( )
PP0
Isothermal, T=T0
( )( )
2
02
2
0A
20A
PP
X1X1
FkC
dWdX
⎟⎟⎠
⎞⎜⎜⎝
⎛ε+−=Combine:
Need to find (P/P0) as a function of W (or V if you have a PFR)
Pressure Drop in Packed Bed Reactors
6
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( )
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
+µφ−⎟⎟⎠
⎞⎜⎜⎝
⎛φφ−
ρ−=
TURBULENT
LAMINAR
p3
pc
G75.1D11501
DgG
dzdPErgun Equation:
Pressure Drop in Packed Bed Reactors
7 0
00 T
TPP)X1( ε+υ=υ0
0
0T
T0 T
TPP
FFυ=υ
υυρ=ρ
υρ=ρυ=
00
00
0mm Constant mass flow:
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( )0T
T
0
0
p3
pc0 FF
TT
PPG75.1
D11501
DgG
dzdP
⎥⎥⎦
⎤
⎢⎢⎣
⎡+µφ−
⎟⎟⎠
⎞⎜⎜⎝
⎛φφ−
ρ−=
T
0T0
00 F
FTT
PPρ=ρVariable Density
( )⎥⎥⎦
⎤
⎢⎢⎣
⎡+µφ−
⎟⎟⎠
⎞⎜⎜⎝
⎛φφ−
ρ=β G75.1
D11501
DgG
p3
pc00Let
Pressure Drop in Packed Bed Reactors
8
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( ) 0T
T
0
0
cc
0
FF
TT
PP
1AdWdP
ρφ−β−=
( ) ccbc 1zAzAW ρφ−=ρ=Catalyst Weight
ρb = bulk density
ρc = solid catalyst density
φ = porosity (a.k.a., void fraction)
Where
( ) 0cc
0
P1
1A2
ρφ−β=αLet
Pressure Drop in Packed Bed Reactors
9
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We will use this form for single reactions: ( )
( ) ( )X1TT
PP1
2dWPPd
00
0 ε+α−=
0T
T
0 FF
TT
y2dWdy α−=
0PPy =
( )X1TT
y2dWdy
0
ε+α−=
( )X1y2dW
dy ε+α−= Isothermal case
Pressure Drop in Packed Bed Reactors
10
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The two expressions are coupled ordinary differential equations. We can only solve them simultaneously using an ODE solver such as Polymath. For the special case of isothermal operation and epsilon = 0, we can obtain an analytical solution.
Polymath will combine the mole balance, rate law and stoichiometry.
( )( )
22
0A
220A y
X1FX1kC
dWdX
ε+−=
( )P,XfdWdX = ( )P,Xf
dWdP = ( )X,yf
dWdy =and or
Pressure Drop in Packed Bed Reactors
11
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PBR
1) Mole Balance: 0A
A
Fr
dWdX −=
2) Rate Law: ( )( )
22
20AA y
X1X1kCr ⎥
⎦
⎤⎢⎣
⎡ε+−=−
AàB
( )( )
( )( )yX1
X1CPP
X1X1CC 0A
00AA ε+
−=⎟⎟⎠
⎞⎜⎜⎝
⎛ε+−=
12
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PBR
13
2/1
2
2
)W1(y
)W1(y
dWdy
1y0WWhen
y2dWdy
0For
α−=
α−=
α−=
==
α−=
=ε
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W
( ) 21W1y α−=
P 1
14
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CA = CA0 1− X( ) PP0
CA 2
W
ΔP
No ΔP
15
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No ΔP
−rA = kCA2
-rA 3
ΔP
W 16
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No ΔP
X 4
W
ΔP
17
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0
00 T
TPP)X1( ε+υ=υ
18
PPy,TT 0
0 ==
y)X1(1f 0
ε+=
υυ=
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No ΔP
5
W
ΔP
1.0
υ
19
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Example 1: Gas Phase Reaction in PBR for δ = 0 Gas Phase Reaction in PBR with δ = 0 (Polymath Solution) A + B à 2C
Repeat the previous one with equil molar feed of A and B and kA = 1.5dm9/mol2/kg/min α = 0.0099 kg-1
Find X at 100 kg
20
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A + B à 2C
min kg moldm5.1k
6
= 1kg 0099.0 −=α
kg 100W = ?X = ?P =
1PP D2D = 0102 P21P =Case 2:
Example 1: Gas Phase Reaction in PBR for δ = 0
21
Case 1:
?X = ?P =
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1) Mole Balance: 0A
A
F'r
dWdX −=
2) Rate Law: BAA CkC'r =−
3) ( )yX1CC 0AA −=
4) ( )yX1CC 0AB −=
0W = 1y =
Example 1: Gas Phase Reaction in PBR for δ = 0
22
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y2dWdy α−=5) dWydy2 α−=
W1y2 α−=
( ) 21W1y α−=
( ) ( ) ( )W1X1kCyX1kCr 220A
2220AA α−−=−=−
( ) ( )0A
220A
FW1X1kC
dWdX α−−=
Example 1: Gas Phase Reaction in PBR for δ = 0
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( ) ( )dWW1FkC
X1dX
0A
20A
2 α−=−
⎟⎟⎠
⎞⎜⎜⎝
⎛ α−=− 2
WWFkC
X1X 2
0A
20A
XX ,WW ,0X ,0W ====
( )( )0.e.i,droppressurewithout 75.0X
droppressurewith 6.0X
=α=
=
Example 1: Gas Phase Reaction in PBR for δ = 0
24
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25
Example A + B → 2C
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26
Example A + B → 2C
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Example 2: Gas Phase Reaction in PBR for δ ≠ 0 Polymath Solution A + 2B à C
is carried out in a packed bed reactor in which there is pressure drop.The fed is stoichiometric in A and B.
Plot the conversion and pressure ratio y = P/P0 as a function of catalyst weight upto 100 kg.
Additional Information kA = 6dm9/mol2/kg/min α = 0.02 kg-1
27
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A + 2B à C
1) Mole Balance: 0A
A
F'r
dWdX −=
2) Rate Law: 2BAA CkC'r =−
3) Stoichiometry: Gas, Isothermal
( )PPX1VV 0
0 ε+=
( )( )yX1
X1CC 0AA ε+−=
Example 2: Gas Phase Reaction in PBR for δ ≠ 0
28
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4) CB = CA0!B ! 2X( )1+ !X( ) y
5) ( )X1y2dW
dy ε+α−=
6) f = !!0
=1+ "X( )y
7) 02.0 6k 2F 2C 32
0A0A =α===−=ε
Initial values: W=0, X=0, y=1 à W=100
Combine with Polymath.
If δ≠0, polymath must be used to solve.
Example 2: Gas Phase Reaction in PBR for δ ≠ 0
29
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Example 2: Gas Phase Reaction in PBR for δ ≠ 0
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Example 2: Gas Phase Reaction in PBR for δ ≠ 0
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T = T0
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Engineering Analysis
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Engineering Analysis
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Engineering Analysis
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Pressure Change – Molar Flow Rate
( ) cc
0
0
0T
T0
1ATT
PP
FF
dWdP
ρϕ−ρ
β−=
( ) cc0
00T
T0
1AyPTT
FF
dWdy
ρϕ−
β−= ( ) CC0
0
1AP2
ρϕ−β=α
00T
T
TT
FF
y2dWdy α−= Use for heat effects, multiple rxns
( )X1FF
0T
T ε+= Isothermal: T = T0 ( )X1y2dW
dX ε+α−=36
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Mole Balance
Rate Laws
Stoichiometry
Isothermal Design
Heat Effects
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Gas Phase Flow System:
Concentration Flow System:
( )( )
( )( ) 0
00A
0
00
0AAA P
PTT
X1X1C
PP
TTX1
X1FFCε+−=
ε+υ
−=υ
=
υ= A
AFC
( )PP
TTX1 0
00 ε+υ=υ
( ) ( ) 0
0B0A
0
00
B0AB
B PP
TT
X1
XabC
PP
TTX1
XabF
FCε+
⎟⎠⎞⎜
⎝⎛ −Θ
=ε+υ
⎟⎠⎞⎜
⎝⎛ −Θ
=υ
=
38
Gas Phase Reaction with Pressure Drop
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A + 2B à C
1) Mole Balance:
dXdW
= ! " r AFA0
2) Rate Law:
! " r A = kCACB2
3) Stoichiometry: Gas, Isothermal
! = !0 1+ "X( ) P0P
CA = CA01!X( )1+ "X( )
y
Example 2: Gas Phase Reaction in PBR for δ ≠ 0
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End of Lecture 8
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