64
Chapter 3
- (4 *
1 + 0 ■
0.0164) + (6 ■
0.510 + 0'0.275 + 4.67 - 0.077 + 4.18 • 0.137) = - 0.0078 * 0
(1)
The redox balance (1) is very nearly satisfied. Previously it was shown that the carbon balance is
satisfied, and the combined tests confirm to a high degree of certainty that not only have the right
compounds of the net reaction been found, but they are also determined in the right ratios._____________
Example 3.2. Aerobic growth with ammonia as nitrogen source
Let the stoichiometry of a general, aerobic process with one product
P
be
-S-YNH,
-
YO
.,
+Y„X + YC02+YP =
0
(1)
S
is the carbon- and energy substrate C /fr|
with degree of reduction K„
X
is biomass
CH x Ox N Xi
with degree of reduction
\
and
P
is a product
0P
i
with degree of reduction Kp. With the above
overall reaction stoichiometry the carbon, nitrogen and degree of reduction balances are:
Y„ + YK+YV =
1
X!
+ JQ
, fjp - X! - (-4) Tjo = 0
Equations (2) and (4) are solved to obtain
YK
and
in terms of
YK
and
Yx
_ {4 -kpRQ)Yso+kp- ks
1SX
K P ~ K X
(
2
)
(3)
(4)
(5)
(
6
)
Where
Let the substrate be glucose and ethanol the product. For a standard biomass with
\
= 4.20:
(4 -
6RQ)Y!0
+ 2
Y
(4 .2 0 * 6 -4 )r,„ -0 -2 0
1.80
*
1.80
(7)
(
8
)
In industrial yeast fermentation both
YK
and RQ are usually monitored more or less continuously, and
Eq. (8) is used to calculate the biomass yield on glucose and the ethanol yield on glucose. Thus for RQ =