Continuous Cultivation - Bioprocess Engineering: Kinetics, Biosystems, Sustainability and Reactor Design - page 620

a chemostat. A thermophilic organism can be selected from a natural population by oper-
ating a chemostat at an elevated temperature (e.g. 50 e 60 C). Selection in chemostats also
presents significant problems in the culture of cells containing recombinant DNA. The
most productive cells often grow more slowly and are displaced by less productive cells.
We will discuss this problem in more detail in Chapter 16.

12.2. CHOOSING THE CULTIVATION METHOD

One of the first decisions is whether to use a batch or continuous cultivation scheme.
Although a simple batch and CSTR represent two extremes, consideration of these two
extreme alternatives will clarify some important issues in reactor selection. First to consider
is the productivity. The simplest case is for the production of cell mass or a primary product.
For a batch reactor, four distinct phases are present: lag phase, maximum growth phase, har-
vesting, and preparation for a new batch (e.g. reactor cleaning, sterilizing, and filling). Let us
define t P as the sum of the times required for the lag phase, harvesting, and preparation. The
value for t P will vary with size of the equipment and the nature of the fermentation but is
normally in the range of several hours (3 e 10 h). Thus, the total time to complete a batch cycle,
t B , assuming growth rate is not limited by nutrients,

m net ln X

1

t B ¼

X 0 þ t P

(12.25)

where k d ), X is the

maximal attainable cell concentration and X 0 is the cell concentration at inoculation. In
arriving at Eqn (12.25) , we have assumed that the fermentation will not go beyond the
maximum growth rate regime.
The total amount of cell mass produced comes from knowing the total amount of growth
extent-limiting nutrient present and its yield factor:

m net is the specific growth rate in maximum growth regime ( m net ¼ m max

X X 0 ¼ YF X = S S 0 (12.26)

The rate of cell mass production in one batch cycle ( P XB )is

YF X = S S 0
m 1
net

YF X = S S 0

P XB ¼ X X 0

¼

X 0 þ t P ¼

(12.27)

ln X

1 þ YF X = S S 0
X 0

t B 1

m net ln

þ t P

The maximum productivity of a chemostat is found by differentiating D X with respect to D
and setting d

ðDXÞ
d

¼ 0 . From Eqns (12.19) and (12.23) ,

D

r

K S
S 0 þ K S

2

P XC ¼ðDXÞ opt ¼ YF X = S ðm max k d Þ 1

2
4

3
5 p

K S ðS 0 þ K S Þ

S 0

r

K S
S 0 þ K S

(12.28)

k d
m max k d

1 þ