lecture 8 - university of michiganelements/5e/powerpoints/2013lectures/lec8... · 2019-08-07 ·...
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
Lecture 8 – Tuesday 2/5/2013
2
� Block 1: Mole Balances � Block 2: Rate Laws � Block 3: Stoichiometry � Block 4: Combine
� Pressure Drop � Liquid Phase Reactions � Gas Phase Reactions
� Engineering Analysis of Pressure Drop
Gas Phase Flow System:
Concentration Flow System:
( )
( )
( )( ) 0
00
0
00
0
11
1
1PP
TT
XXC
PP
TTX
XFFC AAAA εευυ +
−=
+
−==
υA
AFC =
( )PP
TTX 0
00 1 ευυ +=
( ) ( ) 0
00
0
00
0
11 PP
TT
X
XabC
PP
TTX
XabF
FCBABA
BB εευυ +
⎟⎠
⎞⎜⎝
⎛ −Θ=
+
⎟⎠
⎞⎜⎝
⎛ −Θ==
Pressure Drop in PBRs
3
Pressure Drop in PBRs
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
Mole Balances Must use the differential form of the mole balance to separate variables:
Rate Laws Second order in A and irreversible:
ʹ′−= AA rdWdXF 0
2AA kCr =ʹ′−
€
CA =FAυ
=CA01− X( )1+εX( )
PP0T0T
Stoichiometry
€
CA =CA01− X( )1+εX( )
PP0
Isothermal, T=T0
( )( )
2
02
2
0
20
11
⎟⎟⎠
⎞⎜⎜⎝
⎛
+
−=
PP
XX
FkC
dWdX
A
A
εCombine:
Need to find (P/P0) as a function of W (or V if you have a PFR)
5
Pressure Drop in PBRs
( )
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
+−
⎟⎟⎠
⎞⎜⎜⎝
⎛ −−= !"#
$!$"# TURBULENT
LAMINAR
ppc
GDDg
GdzdP 75.111501
3
µφφφ
ρErgun Equation:
6 0
00 )1(
TT
PPXευυ +=
0
0
00 T
TPP
FF
T
Tυυ =
υυ
ρρ
υρρυ
00
00
0
=
=
= mm !!Constant mass flow:
Pressure Drop in PBRs
( )00
03
0
75.111501
T
T
ppc FF
TT
PPG
DDgG
dzdP
⎥⎥⎦
⎤
⎢⎢⎣
⎡+
−⎟⎟⎠
⎞⎜⎜⎝
⎛ −−=
µφφφ
ρ
T
T
FF
TT
PP 00
00ρρ =Variable Density
( )⎥⎥⎦
⎤
⎢⎢⎣
⎡+
−⎟⎟⎠
⎞⎜⎜⎝
⎛ −= G
DDgG
ppc
75.1115013
00
µφφφ
ρβLet
7
Pressure Drop in PBRs
( ) 00
00
1 T
T
cc FF
TT
PP
AdWdP
ρφβ−
−=
( ) ccbc zAzAW ρφρ −== 1Catalyst Weight
( ) 0
0 112
PA cc ρφβ
α−
=Let 8
Pressure Drop in PBRs
€
ρb = bulk density
€
ρc = solid catalyst density
€
φ = porosity (a.k.a., void fraction)
Where
(1 ) solid fractionφ− =
We will use this form for single reactions:
( )( )
( )XTT
PPdWPPd
εα
+−= 112 00
0
dpdW
= −α2p
TT0
FTFT 0
p = PP0
dpdW
= −α2p
TT01+εX( )
dpdW
= −α2p
1+εX( ) Isothermal case
9
Pressure Drop in PBRs
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 Balances, Rate Laws and Stoichiometry.
dXdW
=kCA0
2 1− X( )2
FA0 1+εX( )2p2
( )PXfdWdX ,= ( )PXf
dWdP ,=
dpdW
= f p,X( )and or
10
Pressure Drop in PBRs
11
For ε = 0dpdW
=−α2p(1+εX)
When W = 0 p =1dp2 = −α dW
p2 = (1−αW )p = (1−αW )1/2
Packed Bed Reactors
Example 1: Gas Phase Reaction in PBR for =0
18
Gas Phase reaction in PBR with δ = 0 (Analytical Solution)
A + B à 2C Repeat the previous one with equimolar feed of A and B and: kA = 1.5dm6/mol/kg/min α = 0.0099 kg-1
Find X at 100 kg
00 BA CC =
0AC
0BC?=X
Example 1: Gas Phase Reaction in PBR for =0
19
1) Mole Balance 0
'
A
A
Fr
dWdX −
=
2) Rate Law BAA CkCr =− '
3) Stoichiometry CA =CA0 1− X( ) p
CB =CA0 1− X( ) p
Example 1: Gas Phase Reaction in PBR for =0
20
dpdW
= −α2p
2pdp = −αdW
p2 =1−αW
p = 1−αW( )1 2
−rA = kCA02 1− X( )2 p2 = kCA0
2 1− X( )2 1−αW( )
( ) ( )0
220 11
A
A
FWXkC
dWdX α−−
=
0=W p =1,
4) Combine
Example 1: Gas Phase Reaction in PBR for =0
21
( )( )dWW
FkC
XdX
A
A α−=−
11 0
20
2
⎟⎟⎠
⎞⎜⎜⎝
⎛−=
− 21
2
0
20 WW
FkC
XX
A
A α
XXWWXW ==== , ,0 ,0
( )( )0.., 75.0
6.0==
=
αeidroppressurewithoutXdroppressurewithX
Example 2: Gas Phase Reaction in PBR for ≠0
22
The reaction A + 2B à C is carried out in a packed bed reactor in which there is pressure drop. The feed is stoichiometric in A and B. Plot the conversion and pressure ratio y = P/P0 as a function of catalyst weight up to 100 kg.
Additional Information kA = 6 dm9/mol2/kg/min α = 0.02 kg-1
Example 2: Gas Phase Reaction in PBR for ≠0
23
A + 2B à C
1) Mole Balance 0A
A
Fr
dWdX ʹ′−
=
2) Rate Law 2BAA CkCr =ʹ′−
3) Stoichiometry: Gas, Isothermal
( )PPX 0
0 1 ευυ +=
CA =CA01− X( )1+εX( )
p
Example 2: Gas Phase Reaction in PBR for ≠0
24
4) CB =CA0ΘB − 2X( )1+εX( )
p
5) dpdW
= −α2p
1+εX( )
6) f = υυ0
=1+εX( )p
Initial values: W=0, X=0, p=1 Final values: W=100 Combine with Polymath. If δ≠0, polymath must be used to solve.
7)
02.0,6,2,232]2[
31]211[
00
0
====
−=−=−−=
α
ε
kFC
y
AA
A
Pressure Drop Engineering Analysis
29
𝜌↓0 =𝑀𝑊∗𝐶↓𝑇0 = 𝑀𝑊∗ 𝑃↓0 /𝑅𝑇↓0
𝛼= 2𝑅𝑇↓0 /𝐴↓𝐶 𝜌↓𝐶 𝑔↓𝐶 𝑃↓0↑2 𝐷↓𝑃 𝜙↑3 𝑀𝑊 𝐺[150(1−𝜙)𝜇/𝐷↓𝑃 +1.75𝐺]
𝛼≈(1/𝑃↓0 )↑2
Pressure Drop - Summary
34
� Pressure Drop � Liquid Phase Reactions
� Pressure Drop does not affect concentrations in liquid phase reactions.
� Gas Phase Reactions � Epsilon does not equal to zero
d(P)/d(W)=… Polymath will combine with d(X)/d(W) =… for you
� Epsilon = 0 and isothermal P=f(W) Combine then separate variables (X,W) and integrate
� Engineering Analysis of Pressure Drop
Pressure Change – Molar Flow Rate
35
( ) cc
0
0
0T
T0
1ATT
PP
FF
dWdP
ρϕ−ρ
β−=
dpdW
= −β0
FTFT 0
TT0
pP0Ac 1−φ( )ρc ( ) CC0
0
1AP2
ρϕ−
β=α
dydW
= −α2p
FTFT 0
TT0
Use for heat effects, multiple rxns
( )X1FF
0T
T ε+= Isothermal: T = T0 dXdW
= −α2p
1+εX( )
Example 1: Gas Phase Reaction in PBR for =0
36
A + B à 2C
Case 1:
Case 2: 0AC
0BC??
XP=
=
001
6
,0099.0,min
5.1 AB CCkgkgmoldmk ==
⋅⋅= −α
?,?,100 === PXkgW
?,?,21,2 01021 ==== PXPPDD PP