Handout for 2_25 Quiz
22
Reactor Differential Algebraic Integral A A r X F V 0 CSTR A A r dV dX F 0 X A A r dX F V 0 0 PFR 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 0 0 PBR X W 1
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Chemical Reaction Quiz Material
Transcript of Handout for 2_25 Quiz
Reactor
Differential
Algebraic
Integral
A
A
r
XFV
0
CSTR
AA rdV
dXF 0
X
A
Ar
dXFV
0
0PFR
Vrdt
dXN AA 0
0
0
X
A
AVr
dXNtBatch
X
t
AA rdW
dXF 0
X
A
Ar
dXFW
0
0PBR
X
W 1
How to find
rA f X
rA g Ci Step 1: Rate Law
Ci h X Step 2: Stoichiometry
rA f X Step 3: Combine to get
2
These topics build upon one another
Mole Balance Rate Laws Stoichiometry
3
Species Symbol Reactor Feed Change Reactor Effluent
A A FA0 -FA0X FA=FA0(1-X)
B B FB0=FA0ΘB -b/aFA0X FB=FA0(ΘB-b/aX)
C C FC0=FA0ΘC +c/aFA0X FC=FA0(ΘC+c/aX)
D D FD0=FA0ΘD +d/aFA0X FD=FA0(ΘD+d/aX)
Inert I FI0=FA0ΘI ---------- FI=FA0ΘI
FT0 FT=FT0+δFA0X
0
0
0
0
00
00
0
0
A
i
A
i
A
i
A
ii
y
y
C
C
C
C
F
F
1
a
b
a
c
a
dWhere: and
A
A
FC Concentration – Flow System
4
A
A
FC Concentration Flow System:
0 Liquid Phase Flow System:
XC
XFFC A
AAA
1
10
0
0
X
a
bCX
a
b
V
N
V
NC BAB
ABB 0
0
0
Flow Liquid Phase
etc.
5
BAA CkCr If the rate of reaction were
X
a
bXCr BAA 1
2
0then we would have
XfrA This gives us
A
A
r
F
0
X6
We obtain:
0
0
0
0T
T
P
P
F
F
T
T
Combining the compressibility factor equation of
state with Z = Z0
000
000
00
TT
TT
T
T
CF
CF
TRZ
PC
ZRT
PC
Stoichiometry:
7
T
T
P
PF
T
T
P
P
F
F
FFC T
T
AAA
0
00
00
0
0
0
T
T
P
P
F
FCC
T
BTB
0
0
0
000 TT FC
8
XFFF ATT 00 The total molar flow rate is:
P
P
T
T
F
XFF
T
AT 0
00
000
P
P
T
TX
F
F
T
A 0
00
00 1
P
P
T
TXyA
0
0
00 1
P
P
T
TX 0
0
0 1
Substituting FT gives:
9
Gas Phase Flow System:
Concentration Flow System:
0
00
0
0
0
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
0
0 1
0
0
0
0
0
0
0
11
P
P
T
T
X
Xa
bC
P
P
T
TX
Xa
bF
FC
BABA
BB
10
2
0
0
2
011
1
T
T
P
P
X
Xa
b
X
XCkr
B
AAA
If –rA=kCACB
This gives us
FA0/-rA
X 11
12
Consider the following elementary reaction with KC=20 dm3/mol
and CA0=0.2 mol/dm3.
Calculate Equilibrium Conversion or both a batch reactor (Xeb)
and a flow reactor (Xef).
Xef
C
BAAA
K
CCkr
2
BA2
13
2AB
B2
1A
Xef ?
Rate law:
C
BAAA
K
CCkr 2
Xeb 0.703
14
Solution:
Species Fed Change Remaining
A FA0 -FA0X FA=FA0(1-X)
B 0 +FA0X/2 FB=FA0X/2
FT0=FA0 FT=FA0-FA0X/2
15
A FA0 -FA0X FA=FA0(1-X)
B 0 FA0X/2 FB=FA0X/2
Stoichiometry: Gas isothermal T=T0, isobaric P=P0
V V0 1X
CA FA0 1 X V0 1X
CA0 1 X 1X
CB FA 0 X 2
V0 1X CA 0 1 X 2 1X
16
Pure A yA0=1, CA0=yA0P0/RT0, CA0=P0/RT0
C
AAAA
KX
XC
X
XCkr
121
1 0
2
0
201
12
e
eeAC
X
XXCK
2
11
2
110
Ay
@ eq: -rA=0
17
82.020223
3
0
dm
mol
mol
dmCK AC
yA0 11
21
1
2
8 Xe 0.5Xe
2
1 2Xe Xe2
8.5Xe217Xe 8 0
Xef 0.757Flow: Recall
Xeb 0.70Batch: 18
19
20
21
22
