Mass Transfer
447
Example 10.6 Mass transfer to a single cell
The small dimension of the cell ensures that the lower limit of 2 for the Sherwood number can be applied.
For transport of oxygen from the bulk liquid to a spherical cell of diameter 2 pm we find:
D r
~
2 1
k } = 2
^
d„„
(
1
)
This gives
k} =
2-2.1-10 9m V
~2-10~6m
= 2.1 10 3m s'
(
2
)
The specific interfacial area per unit biomass is given by
6
acell= -r-r*
------x------
(3)
d cell
( ! -
W )PcelI
where w is the fractional water content. With
w
= 0.7 and a density of the cell of 106 g m'3, the volumetric
mass transfer coefficient for the cell is
kjO
cell
6
k,
d « n ( l ~ W )PcH t
6-2.1-10“3m-s_1
2 • 10"6 m( 1 - 0.7) 106 g DW ■
m “3
= 0.021 m V s'
(4)
With a bulk phase oxygen concentration equal to 60% of the saturation value, the maxium specific oxygen
transport rate with air sparged through the reactor can be calculated assuming a concentration of oxygen at
the cell surface very close to 0. This gives:
^
! a c e ll(CO M lk ~ C0 ,surface
) *
^ i a ce!lC 0 ,bu!k
( \$ )
and
= 0.021 • 0.6 • 0.2095 • 1.22 *
10“3 = 3.23 • 10”3 mol O, (g DW)’1
s'1
(6)
Thus, mass transfer to the cells is very rapid compared with the oxygen requirement inside the cell (see
Example 10.1), and it is therefore not necessary to consider this process in a model. Even if the mass
transfer coefficient should be much lower than that used in Eq. (1), the maximum mass transfer rate from
the bulk liquid to the cell is still very rapid compared to the cellular oxygen requirements._______________