Biomedical Engineering Reference
In-Depth Information
Example 5-5. The elementary reaction
A
þ
R
/ 2
R
with a rate constant
k
¼
0.02 L/mol/s is to be carried out isothermally in flow reactors. We
wish to process 120 L/min of a feed containing 5 mol/L of A to the highest conversion
possible in the reactor system consisting of one 100-L PFR and one 100-L CSTR, connected
as you wish and any feed arrangement.
1. Sketch your recommended design and feed arrangement;
2. Determine the final conversion.
Solution. To begin with the solution, let us first examine the reaction stoichiometry and
kinetics. The reaction system can be considered as constant density: isothermal, no molar
change. Stoichiometry leads to
C
A
þ C
R
¼ C
A
0
þ C
R
0
¼ C
A
0
(E5-5.1)
The reaction rate is given by
r
A
¼ n
A
r ¼kC
A
C
R
¼kC
A
ðC
A
0
C
A
Þ
(E5-5.2)
1. Equation
(E5-5.2)
shows that when
C
R
¼
0, the reaction rate is zero. That is, a PFR with
a fresh feed would not convert any reactant to the product. The CSTR is thus placed in the
front, followed by the PFR.
Figure E5-5.1
shows a sketch of the reactor system.
2. Mole balance of A around the first reactor, CSTR, as shown in
Fig. E5-5.1
leads to
QC
A
0
QC
A
1
þ r
A
1
V
1
¼ 0
(E5-5.3)
Substituting
Eqn (E5-5.2) into (E5-5.3)
, we obtain
QC
A
0
QC
A
1
kC
A
1
ðC
A
0
C
A
1
ÞV
1
¼ 0
(E5-5.3)
which is a quadratic equation on
C
A1
that can be solved to give
q
ðQ þ kC
A
0
V
1
Þ
2
Q þ kC
A
0
V
1
4QkC
A
0
V
1
Q
þ
kC
A
0
V
1
ð
Q
kC
A
0
V
1
Þ
2kV
1
C
A
1
¼
¼
2kV
1
(E5-5.4)
C
A0
Q
CSTR
V
1
PFR
C
Ae
,
C
Re
Q
C
A1,
C
R1
Q
V
2
FIGURE E5-5.1
A sketch of the reactor system with one CSTR in front, followed by one PFR, in series.
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