Biomedical Engineering Reference
In-Depth Information
There are two solutions:
C
A1
¼
C
A0
and
Q
kV
1
C
A
1
¼
(E5-5.5)
Clearly, this second solution is the one desired. The first solution is the trivial solution where
no product was present prior to the reactor start-up. Substitute the numbers into
Eqn (E5-5.5)
,
we obtain
Q
kV
1
¼
120=60
0:02 100
C
A
1
¼
mol
=
L
¼
1 mol
=
L
Mole balance of A in a differential volume along the axis of the PFR (
V
,
V
þ
d
V
) leads to
d
ðQC
A
Þþr
A
d
V ¼ 0
(E5-5.6)
Since
Q
is constant, substituting
Eqn (E5-5.2)
into
Eqn (E5-5.6)
and rearranging, we obtain
1
C
A
þ
Q
d
C
A
C
A
0
Q
d
C
A
1
C
A
0
C
A
k
d
V ¼
C
A
ðC
A
0
C
A
Þ
¼
(E5-5.7)
Integrating
Eqn (E5-5.7)
around the PFR (or the second reactor in
Fig. E5-5.1
),
1
C
A
þ
Q
d
C
A
C
A
0
Z
V
2
Z
C
Ae
1
C
A
0
C
A
k
d
V ¼
(E5-5.8)
0
C
A
1
we obtain
Q
C
A
0
ln
C
A
1
ð
C
A
0
C
Ae
Þ
C
Ae
ðC
A
0
C
A
1
Þ
kV
2
¼
(E5-5.9)
Equation
(E5-5.9)
can be rearranged to give
C
A
0
exp
kV
2
C
A
0
Q
C
Ae
¼
(E5-5.10)
C
A
0
C
A
1
C
A
1
1 þ
Substituting in the numbers, we have
C
A
0
5
exp
kV
2
C
A
0
Q
¼
exp
0:02 100 5
120=60
C
Ae
¼
mol
=
L
C
A
0
C
A
1
C
A
1
5
1
1
1 þ
1 þ
¼ 0:008408
mol
=
L
The final conversion is thus
QC
A
0
QC
Ae
QC
A
0
C
A
0
C
Ae
C
A
0
5
0
:
008408
5
f
Ae
¼
¼
¼
¼ 0:9983
The conversion is at 99.83%.
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