Biomedical Engineering Reference
In-Depth Information
which yields
C
0
B
0
C
B
Q Q
0
¼ F
A
0
f
A
(E5-7.13)
To find
Q
0
, we refer to
Fig. E5-7.2
for clarity. Mass balance of B for the feed stream to the PFR
lead to:
ðQ
0
Q
A
ÞC
0
B
0
¼ Q
0
Z
C
B
(E5-7.14)
Q
A
C
B
C
0
B
0
C
B
Q
0
¼ Q
A
þ
(E5-7.15)
Thus,
F
A
0
F
A
0
Q ¼ Q
0
þ
C
0
B
0
C
B
f
A
¼
C
0
B
0
C
B
ð1 þ f
A
Þ
(E5-7.16)
or
F
A
F
A
0
ð1 f
A
Þ
F
A
0
C
0
B
0
C
B
ð1 þ f
A
Þ
1
f
A
1 þ f
A
¼ðC
0
B
0
C
B
Þ
C
A
¼
Q
¼
(E5-7.17)
Having established how the concentration changes with the conversion, we can then employ
the selectivity to solve the problem.
1
f
A
1 þ f
A
k
1
ðC
0
B
0
C
B
Þ
k
1
C
A
k
1
C
A
þ k
2
C
1=2
s
R
=
A
¼
¼
1 f
A
1 þ f
A
þ k
2
C
1=2
k
1
ðC
0
B
0
C
B
Þ
B
B
(E5-7.18)
k
1
ðC
0
B
0
C
B
Þð1 f
A
Þ
k
1
ðC
0
B
0
C
B
Þþk
2
C
1=2
h
k
1
ðC
0
B
0
C
B
Þk
2
C
1=
B
i
f
A
¼
B
19
ð
1
f
A
Þ
21 17f
A
¼
which can be integrated to yield the overall selectivity:
f
Ae
Z
DF
A
due to the formation of R
DF
A
j
total
s
R
=
A
d
f
A
19
1
f
A
21 17f
A
f
Ae
Z
1
f
Ae
0
S
R
=
A
¼
¼
f
Ae
0
¼
d
f
A
0
f
Ae
1
d
f
A
Z
19
17f
Ae
4
21 17f
A
¼
0
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