Design of Fermentation Processes
415
supply cheap methanol for an SCP production of 50,000 tons year
1
{16 273 tons h
' 1
based on 3000 h y r
1
on
stream).
Laboratory tests show that the biomass composition CH
1
.gO
0
.
5
N
0
,;>
and biomass yield on methanol
Ya
= 9/16
kg kg
"1
are approximately independent of the growth rate. NH
3
is the nitrogen source, and only C0
2
and H20
are produced besides the biomass.
One literature source states the following cell growth kinetics at 55 °C:
qx = —mszl
-----£l
— x
(kg of cells nT
3
IT1)
(
1
)
s +
Ks sY
+ K {
s
is the methanol concentration (kg m 3), and
Sj
is the oxygen concentration (PM).
Ks
= 0.0832 kg m"3, and
Kj =
2 PM = 2 x 10
'3
moles of 0 2/ m3.
jim
irc
= 0.9 h"1.
a.
Calculate the maximum steady-state productivity of cells in a stirred tank reactor fed with
S/=
50
kg m
"3
sterile methanol. It is assumed that the oxygen concentration
s} » K;.
Calculate the
corresponding minimum reactor volume for production of 16 2/3 tons of biomass h 1. What is the
0
2
requirement to sustain the production rate calculate above? Calculate the minimum value of
kta
necessary to transfer the required 0
2
to the liquid from air (21% oxygen). The saturation
concentration ^ of oxygen in the liquid is 910 PM at 55 °C. Is it possible to transfer the required
quantity of oxygen to the liquid from air using commercially available gas dispersers (spargers,
agitators, etc.)?
b.
Assume that
kjct
= 0.2 s
1
is the highest mass transfer coefficient which can be obtained in the
stirred tank reactor. With this value of
k,a
and the full expression (1) for the rate of cell production
calculate corresponding values of
qx, s
and
V
for different values of
s,.
The
s
value calculated in
question (a) seems to have a special significance. Why? Can s, = 0.2 PM be used? Is there
something wrong with the model (1)—or perhaps with some of the assumptions? See Section
9.1.4 for an explanation.
c.
A much better design than the monstrous stirred tank of (b) can be devised: Pure oxygen is used
rather than air, the reactor volume necessary to produce 16 2/3 tons of biomass h
' 1
is broken up
into a number of smaller reactors, and the design of the individual reactor is improved. Assume
that each reactor has a medium volume of 30 m3. The reactor is designed as a tube formed into a
loop equipped with a number of highly effective static mixers through which the liquid is pumped
with a circulation time of 30 s. Pure oxygen enters near the top of the loop and follows the liquid
through the loop while being constantly redispersed into small bubbles by the mixers. At the very
top of the loop (just over the gas inlet), the spent gas leaves the reactor to enter the head space. In
this apparatus, one can achieve
ksa
= 0.45 s
'1
as an average for the total circulated liquid volume
when the liquid circulation time is 30 s, corresponding to a superficial liquid velocity of 1 m s'1.
The liquid-phase methanol concentration
s
is 1 kg m '\ and
sf
= 50 kg m'3.
s,
is taken to be 20 PM at all points in the loop (is this assumption critical?).
Calculate the biomass productivity in each 30 m
3
reactor when (for economic reasons) 95% of the
oxygen is to be taken up by the bioreaction during a single passage through the loop. How many
reactors must be constructed to reach the productivity of 16 2/3 tons h 1?
d.
How would you modify the reactor design if the feed is methane and oxygen rather than methanol
and oxygen. The inlet feed is pure oxygen and >99 % methane. The reaction stoichiometry is that
of problem 9.4, and data for saturation concentrations of CH
4
is taken either from Example 9.7 or
from tables.