pressure drop
Post on 09-Dec-2015
36 Views
Preview:
DESCRIPTION
TRANSCRIPT
Chemical Reaction Engineering (CRE) is thefield that studies the rates and mechanisms of
chemical reactions and the design of the reactors inwhich they take place.
Lecture 5b
Lecture 5b
2
Block 1: Mole Balances Block 2: Rate Laws Block 3: Stoichiometry Block 4: Combine
Pressure Drop
Concentration Flow System:
( )( )
( )( ) 0
00
0
00
0
1
1
1
1
P
P
T
T
X
XC
P
P
T
TX
XFFC AAA
A +−=
+
−==
A
A
FC =
( )P
P
T
TX 0
00 1 +=
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:
Mole BalancesMust use the differential form of the mole balance toseparate variables:
Rate LawsSecond order in A and irreversible:
€
CA =FA
=CA 01− X( )1+X( )
PP0T0T
Stoichiometry
€
CA = CA 01− X( )1+X( )
PP0
Isothermal, T=T0
5
Pressure Drop in PBRs
( )
+−
−−= TURBULENT
LAMINAR
ppc
GDDg
G
dz
dP75.1
115013
Ergun Equation:
60
00 )1(
T
T
P
PX +=
Constant mass flow:
Pressure Drop in PBRs
T
T
F
F
T
T
P
P 00
00 =Variable Density
( )
+−
−= GDDg
G
ppc
75.111501
30
0
Let
7
Pressure Drop in PBRs
( ) 00
00
1 T
T
cc F
F
T
T
P
P
AdW
dP
−−=
( ) ccbc zAzAW −== 1Catalyst Weight
( ) 0
0 1
1
2
PA cc −
=Let8
Pressure Drop in PBRs
We will use this form for single reactions:
( )( ) ( )X
T
T
PPdW
PPd +−= 11
2 00
0
002 T
T
F
F
T
T
ydW
dy −=0P
Py =
Isothermal case
9
Pressure Drop in PBRs
The two expressions are coupled ordinary differentialequations. We can only solve them simultaneously usingan ODE solver such as Polymath. For the special case ofisothermal operation and epsilon = 0, we can obtain ananalytical solution.
Polymath will combine the Mole Balances, Rate Lawsand Stoichiometry.10
Pressure Drop in PBRs
11
Packed Bed Reactors
Example 4-4 Calculating Pressure Dropin a Packed Bed
12
Calculate the pressure drop in a 60 ft length of 1 1/2-in.schedule 40 pipe packed with catalyst pellets 114-h. indiameter when 104.4 lb/h of gas is passing through the bed.The temperature is constant along the length of pipe at260°C. The void fraction is 45% and the properties of thegas are similar to those of air at this temperature. Theentering pressure is 10 atm.
13
W
P
1
14
Pressure Drop in a PBR
15
Concentration Profile in a PBR2
€
No ∆P
16
Reaction Rate in a PBR3
€
No ∆P
2
2 2
0
(1 )A A
Pr kC k X
P
− = = −
-rA
€
∆P
W
17
Conversion in a PBR4
€
No ∆P
X
W
€
∆P
18
Flow Rate in a PBR5
€
No∆P
W
€
∆P
10=f
19
0TT =
0 1
(1 )f
X y
= =+
P
Py 0=
Example 4.5: Effect of Pressure Dropon the Conversion
20
21
22
1. Mass transfer (diffusion) of the reactant(s) (e.g., species A) from the bulk fluid to theexternal surface of the catalyst pellet2. Diffusion of the reactant from the pore mouth through the catalyst pores to theimmediate vicinity of the internal catalytic surface3. Adsorption of reactant A onto the catalyst surface4. Reaction on the surface of the catalyst (e.g., A B)5.Desorption of the products (e.g., B) from the surface6. Diffusion of the products from the interior of the pellet to the pore mouth at theexternal surface.7. Mass transfer of the products from the external pellet surface to the bulk fluid
Finding the optimum particle diameter
23
Example 4-6: Calculating X in a Reactorwith Pressure Drop
24
Calculate the catalyst weight necessary to achieve 60% conversion when ethyleneoxide is to be made by the vapor-phase catalytic oxidation of ethylene with air.
Ethylene and oxygen are fed in stoichiometric proportions to a packed-bed reactoroperated isothermally at 260°C. Ethylene is fed at a rate of 0.30 Ib mol/s at apressure of 10 atm. It is proposed to use 10 banks of 1 f -in.-diameter schedule 40tubes packed with catalyst with 100 tubes per bank. Consequently, the molar flowrate to each tube is to be 3 X Ib mol/s. The properties of the reacting fluid are tobe considered identical to those of air at this temperature and pressure. Thedensity of the a -in.-catalyst particles is 120 lb/ft3 and the bed void fraction is0.45. The rate law is:
25
26
Example 1:Gas Phase Reaction in PBR for =0
27
Gas Phase reaction in PBR with δ = 0 (AnalyticalSolution)
Repeat the previous one with equimolar feed of Aand B and:
kA = 1.5dm6/mol/kg/minα = 0.0099 kg-1
Find X at 100 kg
00 BA CC =
0AC
0BC
?=X
Example 1:Gas Phase Reaction in PBR for =0
28
1) Mole Balance0
'
A
A
F
r
dW
dX −=
2) Rate Law BAA CkCr =− '
( )yXCC AB −= 10
Example 1:Gas Phase Reaction in PBR for =0
29
ydW
dy
2
−= , dWydy −=2
Wy −= 12
( ) ( )0
220 11
A
A
F
WXkC
dW
dX −−=
0=W 1=y,
4) Combine
Example 1:Gas Phase Reaction in PBR for =0
30
( ) ( )dWWF
kC
X
dX
A
A −=−
11 0
20
2
( )( )0..,75.0
6.0
===
eidroppressurewithoutX
droppressurewithX
Example 2:Gas Phase Reaction in PBR for ≠0
31
The reaction
is carried out in a packed bed reactor in which there ispressure drop. The feed is stoichiometric in A and B.
Plot the conversion and pressure ratio y = P/P0 as afunction of catalyst weight up to 100 kg.
Additional InformationkA = 6 dm9/mol2/kg/minα = 0.02 kg-1
Example 2:Gas Phase Reaction in PBR for ≠0
32
A + 2B C
1) Mole Balance0A
A
F
r
dW
dX ′−=
3) Stoichiometry: Gas, Isothermal
( )P
PX 0
0 1 +=
Example 2:Gas Phase Reaction in PBR for ≠0
33
4)
5) ( )XydW
dy +−= 12
Initial values: W=0, X=0, y=1Final values: W=100Combine with Polymath.If δ≠0, polymath must be used to solve.
7)
02.0,6,2,23
2]2[
3
1]211[
00
0
====
−=−=−−=
kFC
y
AA
A
Example 2:Gas Phase Reaction in PBR for ≠0
34
Example 2:Gas Phase Reaction in PBR for ≠0
35
36
T = T0
Pressure DropEngineering Analysis
38
= ∗ = ∗= 2 150 1 − + 1.75≈ 1
Pressure DropEngineering Analysis
39
Pressure DropEngineering Analysis
40
Mole Balance
Rate Laws
Stoichiometry
Isothermal Design
Heat Effects
41
End of Lecture 5b
42
top related