Environmental Engineering Reference
In-Depth Information
the same as the concentration everywhere within the reactor at all times (Figure 6.1).
In other words, the feed is instantaneously mixed in the reactor. The mole balance for
a pollutant
j
in the CSTR is
d
N
j
d
t
=
F
j
0
−
F
j
+
Vv
j
r
.
(6.5)
In many cases, the above equation can be further simplified by assuming a
steady
state
such that d
N
j
/
d
t
0, that is, the rate of accumulation is zero. If we also stipulate
that the rate of the reaction is uniform throughout the reactor, we obtain the following
equation:
=
F
j
=
F
j
0
+
Vv
j
r
.
(6.6)
Thus, the volume of a CSTR is
(F
j
−
F
j
0
)
v
j
r
V
=
.
(6.7)
Let us consider the special case where we have a
constant
volumetric feed flow
rate
Q
0
and an effluent volumetric flow rate of
Q
e
.If
C
A0
and
C
A
represent the
concentration ofA in the feed and the effluent, respectively, we can write
F
j
0
=
Q
0
C
j
0
and
F
j
=
QC
j
. Then, we have the following equation for the reactor volume:
(QC
j
−
Q
0
C
j
0
)
v
j
r
V
=
.
(6.8)
6.1.1.3
Plug-Flow Reactor (PFR) or Tubular Reactor
Another type of a continuous-flow reactor is the
tubular reactor
, where the reaction
occurs such that radial concentration gradient is nonexistent, but a gradient exists in
the axial direction. This is called
plug flow
. The flow of the fluid is one in which
no fluid element is mixed with any other either behind or ahead of it. The reactants
are continually consumed as they flow along the length of the reactor and an axial
concentration gradient develops. Therefore, the reaction rate is a function of the axial
direction. A polluted river that flows as a narrow channel can be considered to be in
plug flow.
To analyze the tubular reactor, we consider a section of the cylindrical pipe of
length
Δ
z
as shown in Figure 6.3. If the rate is constant within the small-volume
Δ
=
A
c
Δ
element
z
and we assume steady state, the following mass balance on
compound
j
for the volume element results:
V
F
j
(z)
−
F
j
(z
+ Δ
z)
+
A
c
d
zv
j
r
=
0.
(6.9)
Since
Δ
V
=
A
c
Δ
z
, where
A
c
is the cross-sectional area, we can write the above
equation as
F
j
(z
+ Δ
z)
−
F
j
(z)
A
c
d
z
=
v
j
r
.
(6.10)
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