362
Chapter 9
Certainly more substrate is “lost” due to maintenance, but at the same time a much larger feed
stream has been treated in the system. Relatively speaking the same percentage of substrate is
“lost” (the maintenance loss depends only on
pi).
The stress on the cells caused by rapid pumping
of medium through the recirculation loop and the filter may, however, give an independent
contribution to the maintenance demand of the cell. In that case the recirculation solution may be
at a disadvantage. Otherwise the only negative aspect of cell recirculation is that the pumping
costs may outweigh the savings in capital investment for the reactor. This is treated in Example
9.6.
Example 9.4. Design of cell recirculating system
It is desired to process 1 m
3
h
'1
of a feed with
s/=
4 g L
'1
of the growth-limiting substrate. Assume Monod
kinetics with
Ks=
1 g L
"1
and ^
= 1 h~\
Ya =
0.5 g/gram of substrate. The reactor volume is 500 L. What is
the maximum cell productivity with and without cell recirculation?
a.
x = x
e :
D=\ 1
0.5
=2
h"!, and since
a -Ks
Ay =0.25 and ^„^,=1 h 1, the productivity is zero because
D
/(a + 0
=
0.8
h‘!.
b.
x >
xe:
To avoid washout,
Df
must be smaller than 0.8 h
’1
and, consequently,
/ = l - J ? ( / ? - l ) < « 0 . 4
(
1
)
The maximum productivity is, of course, obtained when 5 = 0 , i.e., when
= s f •
Ysx
= 2 g L
1
and
{Px
)mai ~
Dxe
- 2 - 2 = 4 g L ! or 2 kg h ! in a 500-L reactor. When 5 —> 0, the specific growth rate
tends to zero, and from Eq. (9.39) with a given
D
= 2 h"! one obtains / —» 0, which means that the
reactor cell concentration x —> oc _
This is not a feasible solution.
To illustrate the remaining calculations, choose 5 = 0.309, for which
^
= 0.553 h
'1
and,
consequently,
/ = l - /i( /? -l) = — = 0.276
D
(
2
)
The biomass concentration in the effluent is
xt
= 4 - 0.5 (1 - 0.309) = 1.3 8 2 g L '1
(3)
and
or 1.38 kg h'! . The cell concentration in the reactor is
Px —
Dxt
= 2.764 g L 'h 1
(4)