Chemical Engineering March 2014 - 44

SPX
Figure 2. This configuration of a combined radialand
axial-impeller system is typical to provide mixing
in an aerobic fermenter
* The fermenter turnaround time
is 25 h (to harvest, clean, sanitize,
fill and inoculate)
* Planned down time is 30 days for
an annual overhaul, plus 15 days
of contingency
* The downstream yield is 95%
Calculations:
1. Fermenter production requirement
= (100,000 ton/yr)/(95%
yield) = 105,000 ton/yr
2. Fermenter broth required =
[(105,000 ton/yr)(2,000 lb/ton)]/
(0.05 ton product/ton broth) =
4,200,000,000 lb broth/yr
3. Fermenter volumetric production
= (4,200 million lb broth/yr)/
[(8.34 lb/gal)(1.02)] = 494,000,000
gal/yr = 1,540,000 gal/d = 64,300
gal/h = 1,070 gal/min
4. Total fermenter capacity requirement
(working volume) =
(64,300 gal/h)(125 h/fermenter
cycle) = 8,000,000 gal
How many fermenters would be
needed to offer 8,000,000 gallons
of net tank capacity? Table 1 offers
some options for the number of fermenters
required versus fermenter
size, using the assumption that the
fermenter height is limited to 60 ft.
Ten-foot-diameter fermenters are
known to be capable of production
rates of 100 mmole/L/h and are
economical, but that size would require
284 fermenters and 110 " seed
trains " (see next section). It is hard
to believe this would be an economical
plant design. If the fermenters
could be 60 ft in diameter, then
there would be eight of them and
four seed trains.
Seed trains: A fermentation
process typically
involves inoculating a
batch of sterile growth
media with a " seed, "
which consists of viable
microbes of the desired
type. A 1-mL vial could
inoculate a 100-mL flask,
which would grow enough
to inoculate a 10-L vessel,
which would grow to
inoculate a 1,000-L tank
and so on. In this way, a
production fermenter requires
a series of smaller
fermenters to produce a
sufficient volume of inoculum. Since
the seed fermenters operate in series,
they are often referred to as a
" seed train. "
Oxygen transfer
A fermenter's oxygen transfer rate
(OTR) is a function of the oxygen
transfer driving force, the surface
area across which the oxygen
flows, and the resistance to oxygen
transfer:
OTR = kL × a(Cbubble-Cliquid)
(1)
where OTR is the oxygen transfer
rate in mmol/h; kL = conductance
(reciprocal of resistance) to oxygen
transfer; a is the surface area of oxygen
transfer in square feet; and C
is the oxygen concentration.
This means that the oxygen transfer
rate can be increased by increasing
kL, a, or the change in C.
The effect of tank height
One of the primary constraints associated
with mass transfer in fermenters
is that bubbles rise only so
fast. No matter how much air is introduced
at the bottom, the bubbles
will rise at a rate dependent on the
bubble size and the liquid density
and viscosity, not on the rate of air
being blown into the tank. The effect
is that increasing airflow increases
the availability of air in
the fermenter. The inventory of air
at any time, the void fraction, displaces
product.
For a very large fermenter with
water-like fermentation broth, the
average-sized air bubbles could
rise at a rate of about 0.6 meters
per second (m/s). That means that
a superficial air velocity of 0.3 m/s
results in a fermenter that is 50%
liquid and 50% air bubbles. That is
not a very productive fermenter.
As the gas bubbles rise, oxygen
is transferred from the air to the
liquid. The average oxygen concentration
in the gas phase goes down
with increasing height.
Consider a representative volume
element of the fermenter that is one
meter per side and one centimeter
tall as in Figure 1. Assume that the
oxygen uptake rate is 100 mmol O2/
L/h throughout the fermenter; the
superficial gas rate is 0.1 m/s (0.1
m3/s per square meter of horizontal
surface); and the bubble rise velocity
for this system is 0.6 m/s. The
maximum fermenter height can be
calculated as follows:
1. Oxygen supplied to the bottom
square meter column element =
[(0.1 m3/s)(1,000 L/m3)(0.209 mol
O2/mol air)]/(24.5 L/mol air at
25°C) = 0.83 mol O2/s
2. The void fraction in a representative
volume element = (0.1 m/s)/
(0.6 m/s) = 0.17
3. The liquid volume in a representative
volume element = (1m)(1m)
(0.01m)(1,000 L/m3)(1 - 0.17) =
8.3 L
4. Oxygen consumed by each volume
element =[(100 mmol O2/L/h)(8.3
L)]/[(1,000 mmol/mol)(3,600 s/h)
= 0.00023 mol O2/s per volume
element
5. The number of volume elements
in column of liquid = 0.83 mol/s)/
(0.00023 mol O2/s/volume element)
= 3,600 elements = 3,600
cm = 36 m = 118 ft
It makes no sense to scale this process
up to a height of above 36 m
because the oxygen is completely
depleted from the sparge air at
that height. Actually, the oxygen
concentration would never drop to
zero, because the oxygen transfer
driving force falls along with the
oxygen concentration, so the top of
the fermenter suffers from diminishing
returns.
In the above calculation it has
been assumed that there is negChemiCal
engineering www.Che.Com marCh 2014 45
http://www.Che.Com
Chemical Engineering March 2014

Table of Contents for the Digital Edition of Chemical Engineering March 2014
