58
Chapter 3
even for growth at different growth conditions. Thus, Fig. 3.4 shows the rates of production of
ethanol (e), C02 (c), biomass (x) and glycerol (g) in a continuous steady state anaerobic
fermentation of
S. cerevisiae
as functions of the consumption of glucose (s) that was the limiting
substrate in all the experiments. A vertical line through any point
{-q%V)
on the abscissa will give
the amount of products synthesized per hour in the reactor of volume
V
at the dilution rate which
corresponds to
(~qtV).
Increasing values of (-?,F) corresponds to increasing values of the dilution
rate
D = v/V.
The four lines very nearly intersect at the origo of the diagram, and by regression the
slopes are calculated to have the following values (all in C-mole per C-mole glucose):
j Z - = Yse=
0.510 ;
Y1C
=
0.275 ;
Ym
-0.137 ;
Ytg
=0.077
(3.22)
-<l*v
Consequently the stoichiometry is
-
CH2 O
+ 0.5
\QCH3Ox,2
+0.275
C 02 +0.137X+ 0.077CHg/30 =
0
(3.23)
Since
Y№
+
YK
+
YH
+ TK = 0.999 all significant carbon containing compounds are accounted for.
The composition of ash free biomass was experimentally determined:
-^CHi 74Oo
.6oNo.12
(3.24)
except for small amounts of S, P and minerals.
Only the carbon containing compounds are included in (3.23). In experiments where the nitrogen
source appears in no products except the biomass the stoichiometric coefficient Tsn is easily found.
Thus, in the experiments of Duboc
et al.
(1998) the nitrogen source was NH3
and
Ysn
= 0.12 1
T„
with the biomass composition in equation (3.24). The yield coefficient of H20, Tsw
is not included
either. It can be found from either an O or an H balance (in mole H20 per C-mole glucose):
YIW
=
1 -
1
- 0.510 - 2 • 0.275 - 0.6 ■
0.137 - 0.077 = 0.0358
Tw = 0.5(2 + 3 • 0.0164 - 3 ■
0.510 - 1.8 • 0.137 - j • 0.077) = 0.0377
The two values are identical to within the experimental error. As mentioned earlier the small
amount of water produced by the overall reaction (3.23) has no perceptible influence on the
molarity of the reactants. Thus, complete conversion of 1
C-mole glucose = 30 g glucose per liter
medium leads to formation of 0.04 mole H20, a negligible amount compared to the 55 moles of
water initially present per liter medium.
From (3.23) the yield coefficients on biomass produced are easily obtained using (3.14) (both on
C-mole per C-mole biomass):
Y
= Y~l
= 7.30 ;
Y
- Y Y
=3.72
IS
SX
»
xe
se XI
(3.25)
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