Biomedical Engineering Reference
In-Depth Information
where
W
s
is the shaft work exported out of the reaction mixture, which is the negative of the
stirrer output, and
Q
is the heat transfer into the reaction mixture. That is,
Q ¼ UA
H
T
c
T
(5.96)
where
U
is the overall heat-transfer coefficient and
A
H
is the total heat transfer area.
Substituting into
Eqn (5.95)
, we obtain
X
F
j0
H
j
H
j0
N
S
þ V
X
N
R
r
i
DH
R
i
¼ UA
H
ðT
c
TÞ W
s
(5.97)
j¼1
i¼1
Thus, the energy balance equation can be solved simultaneously with the mole balance equa-
tion to determine the reactor output or reactor volume requirement.
For a steady PFR,
Eqn (3.134)
is reduced to
X
N
S
X
N
R
d
Q
d
W
s
F
j0
dH
j
þ
r
i
DH
Ri
d
V ¼
(5.98)
j¼1
i¼1
W
s
¼
Usually, the shaft work is not present or negligible in a PFR,
0. The heat transferred into
the reactor can be written in the same manner as for a CSTR:
d
Q ¼ Ua
T
c
T
d
V
(5.99)
where
a
is the specific heat transfer area, which equals the heat transfer area divided by the
reactor volume. Thus, the energy balance equation for a steady PFR is given by
X
N
S
X
N
R
F
j0
C
Pj
dT þ
r
i
DH
R
i
d
V ¼ UaðT
c
TÞ
d
V
(5.100)
j¼1
i¼1
which is a differential equation, just like the mole balance equation for PFR.
Example 5-9. The exothermic gas phase reaction:
k
f
k
b
%
þ
r ¼ k
f
p
A
k
b
p
B
p
C
A
B
C
is carried out in a CSTR operating at 50 bar. Calculate the maximum rate if the reactor oper-
ates at a 22% conversion of A with a pure A in the feed. Additional data are given below
k
f
¼ 0:435 exp
E
f
RT
mol
s
1
$m
3
$bar
1
;
E
f
¼ 20
kJ
=
mol
k
b
¼ 147 exp
E
b
RT
mol
s
1
$m
3
$bar
2
;
E
b
¼ 60
kJ
=
mol
Solution: For nonisothermal reactors, temperature plays an important role in reactor perfor-
mance. In a CSTR, the temperature is uniform in the reactor and thus is the same as the final
(output) temperature. To find the maximum reaction rate, we perform:
d
r
d
T
¼ p
A
d
k
f
d
k
b
d
T
0 ¼
d
T
p
B
p
C
(E5-9.1)
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