Environmental Engineering Reference
In-Depth Information
Note that in Equation (6.38), the coolant temperature is considered to be constant. This
holds when the coolant is at the boiling point. If the coolant temperature changes along
the length of the reactor, we must add the coolant energy balance. For more informa-
tion, see Froment et al. (2010).
Example 6.4 Calculation of conversion and temperature
profile in a non-isothermal PFR
A first-order exothermic reaction A
!
B is carried out
in a PFR in the
liquid phase.
Data
Enthalpy of reaction (
kmol
−1
−
Δ
r
H
) = 40,000 kJ
10
3
s
−1
Activation energy E
a
= 25,000 kJ
Pre-exponential factor k
0
= 1.2
kmol
−1
Inlet temperature T
0
= 300 K
Molar heat capacity c
p
= 132
K
−1
Superficial velocity of the fluid in the tube
u
= 1.5 m
mol
−1
:
6J
s
−1
m
−3
Initial concentration of A c
A0
= 0.2 kmol
Tube diameter d = 0.05 m
Heat transfer coefficient
h
= 250 W
m
−2
K
−1
Length of the reactor L = 5 m
Universal gas constant R
u
= 8.314 kJ
kmol
−1
K
−1
From a cost analysis, it is concluded that to make a profit the conversion of
A needs to be at least 90%. The operators decide to introduce the flow at room
temperature (300 K). However, B degrades at 500 K and the reaction is exother-
mic, so the operators fear they will have to cool the reactor.
Is it necessary to cool the reactor? If so, what coolant temperature is most suit-
able for the purposes of the company?
Solution
The energy balance Equation (6.38) reads
dV
=
−
Δ
r
H
dT
ð
Þ −
ð
R
A
X
A
,T
ð Þ
Þ
+
h
4
ðÞ
=
d
ð Þ
T
f
−
T
X
φ
n
,
i
c
p
,
i
For simplification, the overall thermal properties of the system will be considered,
so
X
φ
n
c
p
, where < > indicate average specific heat capacity. The
number of moles is conserved due to the stoichiometry of the reaction so
φ
n
h
φ
n
,
i
c
p
,
i
=
.
With V =
S
L, the energy balance then is
c
p
i
=
φ
n
,
A
0
h
c
p
i
dT
dL
=
S
φ
n
,
A
0
c
p
−
Δ
r
H
S
φ
n
,
A
0
c
p
ð
Þ
kc
A0
1
X
ð Þ
−
+
h
4
ðÞ
=
d
ð Þ ð
T
f
−
T
Eq
:
6
:
39
Þ
Search WWH ::
Custom Search