74
Chapter 3
Ero q* + Ec qc = 0
(3.33)
Here Ec is a (4 x 4) matrix whereas Em
has 4 rows and
N + M
-3 columns. Provided that
detCEJ^O the algebraic equation (3.33) is solved to give
qc = -(Ec)', Emqm
(3.34)
Where (Ec)'1
is the inverse of Ec. The concept of the systematic procedure is illustrated in Example
3.8.
Example 3.8. Anaerobic yeast fermentation with CCb, ethanol and glycerol as metabolic products.
Consider anaerobic fermentation of
S. cerevisiae
where the metabolic products are C02 (c) ethanol (e) and
glycerol (g). Let the carbon source be glucose and the nitrogen source be NH3. The biomass composition is
CH
1
.
6
iO
0
.s
2
N
0
.is. There are two substrates (CH20 and NH3), four products (CH30*, CHwO, C02 and H20) in
addition to the biomass
X.
With 7 reacting species and four constraints one needs to measure 3 rates. The
remaining rates can be calculated.
The measured rates are chosen as qm
=
(q„
and the calculated rates are then qc =
(qt, qn, qt, qw)
r\
1
p
f
2
1.61
4
<ls
+
1
0.52
1
j
X
,0
0.15
0,
q.
0
1
0>
'0>
3
3
2
0
0
0.5
1
0
1
0
1wj
Ee
IV
1
3
-0.3133
9
0
-0.15
0
= ~(Ec)”IE niqm
=
-2
3
-0.6867
7
9
0.45
1
6
/
^ s j
(
1
)
Based on the numbers in the resulting matrix above the yield coefficients in (3.30) are easily calculated:
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