lecture 8 - university of michiganessen/html/powerpoints/lecture_notes/lec8/... · 2011-02-01 ·...
TRANSCRIPT
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
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
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
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
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
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
( )
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
+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:
( )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
( ) 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
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
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
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
PBR
13
2/1
2
2
)W1(y
)W1(y
dWdy
1y0WWhen
y2dWdy
0For
α−=
α−=
α−=
==
α−=
=ε
W
( ) 21W1y α−=
P 1
14
CA = CA0 1− X( ) PP0
CA 2
W
ΔP
No ΔP
15
No ΔP
−rA = kCA2
-rA 3
ΔP
W 16
No ΔP
X 4
W
ΔP
17
0
00 T
TPP)X1( ε+υ=υ
18
PPy,TT 0
0 ==
y)X1(1f 0
ε+=
υυ=
No ΔP
5
W
ΔP
1.0
υ
19
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
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 =
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
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
23
( ) ( )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
25
Example A + B → 2C
26
Example A + B → 2C
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
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
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
30
Example 2: Gas Phase Reaction in PBR for δ ≠ 0
31
Example 2: Gas Phase Reaction in PBR for δ ≠ 0
32
T = T0
Engineering Analysis
33
Engineering Analysis
34
Engineering Analysis
35
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
Mole Balance
Rate Laws
Stoichiometry
Isothermal Design
Heat Effects
37
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
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
39
End of Lecture 8
40