Environmental Engineering Reference
In-Depth Information
The enthalpy introduced by the flow to a slice of the reactor is
X
+
X
φ
n
,
i
φ
n
,
i
H
i
T
re
ðÞ
c
p
,
i
T
ð
−
T
ref
Þ
ð
Eq
:
6
:
31
Þ
The enthalpy that leaves the slice is
X
+
X
φ
n
,
i
φ
n
,
i
H
i
T
re
ðÞ
c
p
,
i
T+dT
ð
−
T
ref
Þ
ð
Eq
:
6
:
32
Þ
The enthalpy generated in the slice due to the reaction of component A is
dV
ð
−
Δ
r
H
Þ −
ð
R
A
X
A
,T
ð Þ
Þ
ð
Eq
:
6
:
33
Þ
If the reactor operates in steady state, there is no accumulation of heat in the slice.
The energy balance then reads
Rate of accumulation = rate of supply
−
rate of release + rate of production
+ heating
=
cooling
)
0=
X
+
X
X
φ
n
,
i
φ
n
,
i
H
i
T
re
ðÞ
c
p
,
i
T
ð
−
T
ref
Þ−
φ
n
,
i
H
i
T
re
ðÞ
+dV
+
X
+d
Q
φ
n
,
i
c
p
,
i
T+dT
ð
−
T
ref
Þ
−
Δ
r
H
Þ −
ð
R
A
X
A
,T
ð Þ
Þ
ð
Eq
:
6
:
34
Þ
This balance neglects any shaft work introduced to the system, for instance, the work
done by the turbine that pumps the flow. Simplifying the balance, we obtain
X
+d
Q
φ
n
,
i
c
p
,
i
dT = dV
ð
−
Δ
r
H
Þ −
R
A
X
A
,T
ð
ð Þ
Þ
ð
Eq
:
6
:
35
Þ
Dividing by dV and rearranging gives
+
d
Q
dV
dV
=
−
Δ
r
H
ð
Þ −
ð
R
A
X
A
,T
ð Þ
Þ
dT
X
ð
Eq
:
6
:
36
Þ
φ
n
,
i
c
p
,
i
Q
=
hA
T
f
−
The heat removed from the system by the coolant,
ð Þ
T
, can be expressed as
d
Q
=
h
T
f
−
=
:
:
ð Þ
T
d
A
=
h
4
ðÞ
d
ð Þ
T
f
−
T
dV
ð
Eq
6
37
Þ
where d is the reactor diameter,
h
is the heat transfer coefficient, and T
f
is the temper-
ature of the coolant. Substituting the expression for d
Q
in Equation (6.36), we obtain
the final expression for the temperature gradient inside the reactor:
dT
dV
=
ð
−
Δ
r
H
Þ −
ð
R
A
X
A
,T
ð Þ
Þ
+
h
4
ðÞ
=
d
ð Þ
T
f
−
T
X
ð
Eq
:
6
:
38
Þ
φ
n
,
i
c
p
,
i
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