Design of Fermentation Processes
375
jc
0
=4.8- 2.5/3 = 4 gm '
12-2.5+ 30-0.5
.3
s0
= ---------
1
----------= 15 gm
(
8
)
And the exit conditions ares = 3 g m ', x = x
0
+ 0.1(15 - 3) = 5.2 g m'3. Thus
4/15
4
r -
3
1
+
1 + 4/(0.1 -15)
i.e.,
V2
= 3/4 • 0.3985 • 3 = 0.897 m
1
or
V, +V2
=3.397 m l With an extra stream, the minimum of
V, +V2
is
not quite at the point where the chemostat operation is optimized. The cell-free side stream requires a few
more cells to be delivered from the chemostat to operate the plug flow reactor better. The true optimum is
obtained for (x, j) = (4.87, 11.3) gm
3
from the chemostat and
(Vi+ViUm
= 3.3939 m3._________ ;_________
A tubular reactor can operate on a sterile feed if, once inoculated, a portion of the effluent stream is
returned to the inlet. The general situation is shown in Fig. 9.5. The feed to the reactor contains
both cells and substrate, and a portion v* =
vR
of the net stream is returned to the inlet and mixed
with the feed stream. A mass balance at the point where the recycle stream joins the feed stream
gives:
vsf +Rvs = (\ + R)vsl
(9.74)
vxf +Rvx = (
1
+ Ä)vx,
(
9
.
7 5
)
V.
s* X
,
S,
v r = v -R ,
s,
x
Figure 9.5
Schematic of a plug flow reactor with external recycle.
From Eqs. (9.74) and (9.75), the inlet concentrations to the plug flow reactor are obtained as
functions of
sf , xf , s,
and
x.
The true flow through the reactor is v(l +
R),
and Eq. (9.69) can be
applied directly to calculate the reactor volume. In the important case
xf
= 0, one obtains from Eq.
(9.73)
=
(1
+
a f
) ln I — -
a f
In —
v
(
i
+
ä
)
J
U J
f
u
(9.76)
where
cif = Ks
/
sf
as before.