Test bank for critical care nursing a holistic approach 11th edition morton f...
Lec4_PDF.pdf
1. 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 4
3. Reactor Differential Algebraic Integral
V
FA 0X
rA
CSTR
FA0
dX
dV
rA
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
FA 0
dX
dW
rA
X
A
A
r
dX
F
W
0
0
PBR
X
W
3
Reactor Mole Balances Summary
in terms of conversion, X
Review Lecture 2
8. 8
i
A C
g
r
Step 1: Rate Law
X
h
Ci
Step 2: Stoichiometry
X
f
rA
Step 3: Combine to get
Two steps to get
rA f X
Review Lecture 2
9. Building Block 2: Rate Laws
9
Power Law Model:
B
A
A C
kC
r
β
α
β
α
Order
Rection
Overall
B
in
order
A
in
order
C
B
A 3
2
A reactor follows an elementary rate law if the reaction
orders just happens to agree with the stoichiometric
coefficients for the reaction as written.
e.g. If the above reaction follows an elementary rate law
2nd order in A, 1st order in B, overall third order
B
A
A
A C
C
k
r 2
Review Lecture 3
10. Arrhenius Equation
10
E = Activation energy (cal/mol)
R = Gas constant (cal/mol*K)
T = Temperature (K)
A = Frequency factor (same units as rate constant k)
(units of A, and k, depend on overall reaction order)
RT
E
Ae
k
T k A
T 0 k 0
A 1013
T
k
Review Lecture 3
12. Algorithm
12
i
A C
g
r
Step 1: Rate Law
X
h
Ci
Step 2: Stoichiometry
X
f
rA
Step 3: Combine to get
How to find
X
f
rA
Review Lecture 3
13. Building Block 3: Stoichiometry
13
We shall set up Stoichiometry Tables using species
A as our basis of calculation in the following
reaction. We will use the stoichiometric tables to
express the concentration as a function of
conversion. We will combine Ci = f(X) with the
appropriate rate law to obtain -rA = f(X).
A is the limiting reactant.
D
a
d
C
a
c
B
a
b
A
Chapter 4
14. Stoichiometry
14
NA NA0 NA0X
NB = NB0 -
b
a
NA0X = NA0
NB0
NA0
-
b
a
X
æ
è
ç
ö
ø
÷
For every mole of A that reacts, b/a moles of B react. Therefore
moles of B remaining:
Let ΘB = NB0/NA0
Then:
X
a
b
N
N B
A
B 0
NC NC 0
c
a
NA 0X NA 0 C
c
a
X
Chapter 4
15. Batch System - Stoichiometry Table
15
Species Symbol Initial Change Remaining
B B NB0=NA0ΘB -b/aNA0X NB=NA0(ΘB-b/aX)
A A NA0 -NA0X NA=NA0(1-X)
Inert I NI0=NA0ΘI ---------- NI=NA0ΘI
FT0 NT=NT0+δNA0X
Where:
0
0
0
0
0
0
0
0
0
0
A
i
A
i
A
i
A
i
i
y
y
C
C
C
C
N
N
1
a
b
a
c
a
d
and
C C NC0=NA0ΘC +c/aNA0X NC=NA0(ΘC+c/aX)
D D ND0=NA0ΘD +d/aNA0X ND=NA0(ΘD+d/aX)
δ = change in total number of mol per mol A reacted
Chapter 4
16. Stoichiometry Constant Volume Batch
16
Note: If the reaction occurs in the liquid phase
or
if a gas phase reaction occurs in a rigid (e.g. steel)
batch reactor
V V0
Then
CA
NA
V
NA 0 1 X
V0
CA 0 1 X
CB
NB
V
NA 0
V0
B
b
a
X
CA 0 B
b
a
X
etc.
Chapter 4
17. Stoichiometry Constant Volume Batch
17
Suppose
rA kACA
2
CB
Batch: 0
V
V
-rA = kACA0
3
1- X
( )
2
QB -
b
a
X
æ
è
ç
ö
ø
÷
Equimolar feed: 1
B
Stoichiometric feed:
a
b
B
Chapter 4
18. Stoichiometry Constant Volume Batch
18
X
f
rA
and we have
If B
A
A
A C
C
k
r 2
, then
X
a
b
X
C
r B
A
A
2
3
0 1 Constant Volume Batch
A
r
1
X
Chapter 4
19. Batch Reactor - Example
19
Consider the following elementary reaction with
KC=20 dm3/mol and CA0=0.2 mol/dm3.
Find Xe for both a batch reactor and a flow reactor.
Calculate the equilibrium conversion for gas phase
reaction, Xe .
C
B
A
A
A
K
C
C
k
r
2
B
A
2
Chapter 4
20. Batch Reactor - Example
20
Step 1:
0
A
A
N
V
r
dt
dX
3
0 2
.
0 dm
mol
CA
mol
dm
KC
3
20
Step 2: rate law: B
B
2
A
A
A C
k
C
k
r
Calculate Xe
B
A
C
k
k
K
C
B
2
A
A
A
K
C
C
k
r
Chapter 4
21. Batch Reactor - Example
21
Symbol Initial Change Remaining
B 0 ½ NA0X NA0 X/2
A NA0 -NA0X NA0(1-X)
Totals: NT0=NA0 NT=NA0 -NA0 X/2
@ equilibrium: -rA=0
C
Be
Ae
K
C
C
2
0
Ke
CBe
CAe
2 CAe
NAe
V
CA0 1 Xe
CBe CA0
Xe
2
Chapter 4
22. Batch Reactor - Example
22
Species Initial Change Remaining
A NA0 -NA0X NA=NA0(1-X)
B 0 +NA0X/2 NB=NA0X/2
NT0=NA0 NT=NA0-NA0X/2
Solution:
2
Ae
Be
C
C
C
K
C
Be
2
Ae
A
A
K
C
C
k
0
r
At equilibrium
0
V
V
2
/
B
A
Stoichiometry:
Constant Volume:
Batch
Chapter 4
23. Batch Reactor - Example
23
2
e
0
A
e
2
e
0
A
e
0
A
e
X
1
C
2
X
X
1
C
2
X
C
K
8
2
.
0
20
2
X
1
X
C
K
2 2
e
e
0
A
e
Xeb 0.703
Chapter 4
24. Flow System – Stoichiometry Table
24
A A FA0 -FA0X FA=FA0(1-X)
Species Symbol Reactor Feed Change Reactor Effluent
B B FB0=FA0ΘB -b/aFA0X FB=FA0(ΘB-b/aX)
i
Fi0
FA 0
Ci00
CA 00
Ci0
CA 0
yi0
yA 0
Where:
Chapter 4
25. Flow System – Stoichiometry Table
25
Species Symbol Reactor Feed Change Reactor Effluent
0
A
0
i
0
A
0
i
0
0
A
0
0
i
0
A
0
i
i
y
y
C
C
C
C
F
F
Where:
Inert I FI0=A0ΘI ---------- FI=FA0ΘI
FT0 FT=FT0+δFA0X
C C FC0=FA0ΘC +c/aFA0X FC=FA0(ΘC+c/aX)
D D FD0=FA0ΘD +d/aFA0X FD=FA0(ΘD+d/aX)
1
a
b
a
c
a
d
and
A
A
F
C
Concentration – Flow System
Chapter 4
26. 26
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
A
0
i
0
A
0
i
0
0
A
0
0
i
0
A
0
i
i
y
y
C
C
C
C
F
F
1
a
b
a
c
a
d
Where: and
A
A
F
C
Concentration – Flow System
Flow System – Stoichiometry Table
Chapter 4
27. 27
Stoichiometry
A
A
F
C
Concentration Flow System:
0
Liquid Phase Flow System:
CA
FA
FA 0 1 X
0
CA 0 1 X
CB
NB
NA0
0
B
b
a
X
CA 0 B
b
a
X
Flow Liquid Phase
etc.
We will consider CA and CB for gas phase
reactions in the next lecture
Chapter 4
