140
Chapter 5
YK = ^
+ ° - 1V| ^ = 1 - 0.952(1
-Y „ ) =
1 -
Ya
(6)
~ rs
C onsequently the carbon balance is autom atically satisfied. A lso
Y„
is independent o f the stoichiom etric
coefficient o f C 0 2 in reaction (1) as is easily seen from the tw o relations for (-r0) and (-rs) in (4). These
results w ill be seen to hold, also w hen ethanol is produced.
From Eq. (5) T„ is calculated to be 0.548 C -m ole biom ass (C -m ole g lucose)'1, w hich is exactly the sam e
value as obtained experim entally. This is not strange since
Y„
= 0.425 w as calculated based on a yield
coefficient o f biom ass
= 0.548.
The A T P balance gives:
-
avt +
(2v2 + 0. lv, )P /0 + 2 / 3v2 = 0
(V)
Inserting v, and v, from Eq. (4) gives:
2.1
0Y„P/Q
+ 0.70(1 -1.048(1 -
Yso
))
1 - K .
(
8
)
Van G ulik and H eijnen (1995) have analyzed a large num ber o f aerobic ferm entation data on yeast and they
conclude that the effective P/O ratio is 1.2-1.3. If w e use 1.25 then a is calculated to 2.42 m ole A TP (C-
m ole b io m a ss)'.
C ase B :
D =
0.30 h 1
or D -
0.40 h '1, v3
*
0
In this case the substrate consum ption is:
rs =
1.10v, + v 2 + 1 .5 v 3
(9)
w hile the oxygen consum ption is given by the upper equation in (4). Solution for v, and v2 in term s o f (-r0),
(-rs) and
rc =
v3 yields:
v, =
(0.952(1 - y J - l.4 2 9 r sJ - 0
(10)
v 2
=(l-1.10(0.952(l-r„)-1.429y„)-1.5y„X -rJ)
(11)
w hich gives
Ya =
0.952(1
-Y j-L 4 2 9 Y st
Y
= O-lv, + v 2 + 0 -5 v 3
_ l
Y
_ Y
(
12
)
(13)
A s w as m entioned in the case v3 = 0 the carbon balance autom atically closes and Tsx is independent o f the
stoichiom etric coefficient for C 0 2 in reaction (1).
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