Ghasem D. Najafpour. Biochemical Engineering and Biotechnology
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The diameter of small tank is 47.4 cm. Since the impeller is one third of the tank diameter,
D
i
16 cm
(E.2.5)
Power is defined for a well-agitated vessel with a mechanical stirrer; then read power num-
ber for turbulent flow from Figure 6.6, Chapter 6:
(E.2.6)
Calculate power for a small-scale vessel:
(E.2.7)
The power for large vessel is based on scale factor, that is calculated as following:
P
Large
125 1.96 245 W (E.2.8)
For mass transfer calculation
K
L
a constant
V
S
100 l D
i
16 cm D
t
47.43 cm
ND
i
0.5
(E.2.9)
When Re 80,000 for turbulent flow, the power number versus Re, we can read
P
no
6
P
Pg
ND
NO
ii
r
35
N
i
5 rps 1 .5 rpm
05
016
312 87
.
.
.
P
Small
3
0 kg/m
6 W 6
100 3 125 0 16
981
19
35
()(.)(.)
.
.
N
05
016
.
.
ND
i
0.5
P
Pg
ND
no.
C
i
r
3
5
6
Re
i
3
1 g/cm 50 cm/s
0.01
80,000 Tubulent flow
()()()16
304 BIOCHEMICAL ENGINEERING AND BIOTECHNOLOGY
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(E.2.10)
Scale-up factor 125
(E.2.11)
The power required for a large vessel is:
P
L
19.6 125 2450 W (E.2.12)
The power per unit volume of small bioreactor is
P/V 19.6 /0.1 196 W.m
3
(E.2.13)
The volume of a large vessel is
V
L
2450/196 12.5 m
3
D
iL
(0.84 m)
(E.2.14)
We have understood the following concepts are true; our calculations also support the fact
that power per unit volume is constant.
ND
i
constant N
3
D
i
2
constant
V
L
12.5 V
s
DDD
t,L t
L
i,L
2.515 m, 0.84 m
4
12 5
3
.
44
2 515 12
3
3
D
t
3
5 m(. ) .
D
D
iL
iS
Ê
Ë
Á
ˆ
¯
˜
2
25
P
P
ND
ND
s
L
ii
iiL
s
()
()
35
35
1
125
196
01
19 6
3
.
.
.
W
m
P
small
6 W
6 1000 3 125 0 16
981
19
35
..
.
.
BIOPROCESS SCALE-UP 305
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P
no
6
Power 0.3 hp (E.2.15)
Example 3: Power Calculation
A 20 m
3
working volume of bioreactor is used for production of penicillin. What is the initial
substrate concentration, S
o
, that you choose when there is a limitation in sufficient oxygen
transfer rate and there are no limiting reactants?
Bioreactor Characteristics and Given Data
D
tank
2.4 m Turbine
D
i
0.8 m µ
broth
1 mPa.s
N
i
2.3 rps P 1.2 r
H
2
O
Number of blades 8 Aeration rate 1 vvm
P
g
/P 0.4
If you aerate a bioreactor, power consumption is much less than a non-aerated bioreactor.
Oxygen transfer rate, OTR: 6 10
3
kgm
3
Oxygen uptake rate, OUR: 0.65 mmol O
2
kg
1
cell
Specific growth rate, y
max
: 0.5 h
1
Specific sugar consumption rate 1.0 kgkg
1
cellh
1
The mass transfer coefficient is calculated by the following correlation
(E.3.1)
Ka
P
V
V
L
()
.
.
s
g
S
13
06
0 667
210
Ê
Ë
Á
ˆ
¯
˜
Re
ND
Li
ii
kg
m.s
,
()(.)()
.
.
r
m
2
2
5
1000 0 8 1
0 001
62 10
NN
D
D
LS
S
L
ii
i,
i,
1 rp
s
,,
/
/
(. )
.
.
Ê
Ë
Á
ˆ
¯
˜
Ê
Ë
Á
ˆ
¯
˜
23
23
3 125
016
08
306 BIOCHEMICAL ENGINEERING AND BIOTECHNOLOGY
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where P
g
/V is the power per unit volume (hpm
3
) and V
S
is the gas superficial velocity
(cmmin
1
).
Solution
The flow regime is turbulent as the Reynolds number is large.
Read power number N
p
from the graph P
no.
versus Re: N
p
6
Correction factor for non-geometrical similarity,
The power input for three sets of impellers in a single shaft is:
P (5.04 hp)(1.25)(three sets of impellers) 19 hp
(E.3.2)
F
g
20 m
3
.min
1
1 vvm V
L
0.333 m
3
s
1
Given gassed power is 40% of ungassed power,
Aeration rate
a
ii
N
F
ND
g
3
(
()
.HD
Li
*
3
2
/)
20 m
2.4
4
/0.8
442
()
.
.
DD
ti
*
/
24
08
3
f
DD H D
DD H D
c
ti Li
ti Li
/) /)
/) /)
((
((
.
.
**
3442
33
125
N
Pg
ND
PNNDg
p
c
i
pic
/
/
3758
r
r
3
5
3
5
35
6 1200 2 5 0 8 9 81
(.) (.) .
kkg.m/s hp
745.7 kg.m/s
5.04 hp
N
Dv
DND
Re
r
m
r
m
iii
()
(.)(.)
.
.
1200 0 8 2 3
01
192 10
2
4
BIOPROCESS SCALE-UP 307
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P
g
/P 0.4 P
g
0.4 19 7.6 hp
(E.3.3)
The mass transfer coefficient is calculated.
(E.3.4)
With an assumption of oxygen concentration at the interface, equilibrium with liquid phase
is 6 ppm, the oxygen transfer rate is calculated.
(E.3.5)
Cell Balance
(E.3.6)
C
S
3
kg/m
10 3
05
20 6
.
.
.
19 375 0
05
10
.
.
.
max
xx
q
CC
S0
S
Sugar to cell
SS
m
x
002
208 10
10
3
.
.
.3 kg/m
3
OURq
O2
2
kg O /kg cell.s
208 10
3
.
OTR xq x
O
2
(. )( )065 10 32 10
33
OTR K a C C
L
((.)().
*
OO
3
2
) 02 kg/m s
2
3 57 6 10 0
3
Ka
L
2 10 4 42 3
3
06
0 667
7.6 754.7 W/hp
20
57 s
1
Ê
Ë
Á
ˆ
¯
˜
.
.
(. ) .
V
S
3
22
2
20 m /min
4
2.4 m
4.42 m/min 7.4 10 m/
s
=
()
N
F
ND
a
g
ii
33
0 333
25 08
026
.
.(.)
.
308 BIOCHEMICAL ENGINEERING AND BIOTECHNOLOGY
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Example 4: Scale-up Calculations
Calculate mass transfer, gas hold up, gassed and ungassed power for the fermenter with the
given data:
r
broth
1200 kgm
3
m 0.002 N.s.m
2
N
i
90 rpm
D
t
4m D
i
2m w
b
0.4 m H
L
6.5 m
Air is sparged with 0.4vvm, it is equipped with two sets of impellers and a flat-blade tur-
bine with four baffles.
Calculate: (a) power, P; (b) Power, P
g
; (c) K
L
a; (d) gas holdup.
Solution
N 90 rpm/60 1.5 rps
Read N
p
from Figure 6.6, Chapter 6, power versus Re, for a turbulent regime, N
p
6
Correction factor for non-geometrical similarity,
(E.4.1)
P 2 0.8 105 168 hp
The gas flow rate is
F
g
33
7 m /min 0.5445 m /s04 4
4
6 5 32 6
2
...
(
)
Ê
Ë
Á
ˆ
¯
˜
(
)
f
DD H D
DD H D
c
=
()( )
()( )
.
.
**
ti Li
ti Li
//
//
2325
33
085
6
981
1 5 2 1200
10
35
P
P
.
(.)()( )
5 hp
N
ND
Re
i
2
2
6
1 5 2 1200
0 002
36 10
r
m
.
.
.
() ( )
.
.
*
DD H D
ti
*
Li
//2
65
2
325
BIOPROCESS SCALE-UP 309
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The aeration number is
Reading gassed power to ungassed power ratio from the plot for the ratio of P
g
/P, defined
in Figure 6.6, Chapter 6.
The gassed power is always less than the ungassed power.
P
g
0.74P 0.74(168) 124.3 hp
The mass transfer coefficient for turbulent flow is:
(E.4.2)
where the gas and liquid velocities are
Example 5
Calculate mass transfer coefficient in a 60 m
3
fermenter with a gas and liquid interfacial
area of a 0.3 m
2
m
3
, given r
broth
1200 kgm
3
. The small reactor has working volume
of 0.18 m
3
, 1 vvm aeration rate. Oxygen transfer rate (OTR) is 0.25kmolm
3
h
1
. There
are two sets of impellers, and flat-blade turbine types of impeller were used, H
L
1.2D
t
.
Find the exact specifications of a large fermenter.
VDDD
1
23
018
4
12 03 .(.).
tt t
VV
S
3
2
L
3
5 m /s
m
m
0 544
4
46526
.
.
Ka
L
210
7W/hp
3
124 3 754
26
054
4
58
06 0667
.. .
.
..
Ê
Ë
Á
ˆ
¯
˜
Ê
Ë
Á
ˆ
¯
˜
662 s
1
Ka
P
V
V
L
210
3
g
L
S
Ê
Ë
Á
ˆ
¯
˜
06
0 667
.
.
P
P
P
P
g
a
074.
N
F
ND
a
g
ii
33
0 5445
15 2
0 045
.
.()
.
310 BIOCHEMICAL ENGINEERING AND BIOTECHNOLOGY
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The diameter of the tank is
Also, the impeller diameter is
The liquid media height is
H
L
1.2 0.576 0.691 m
The large tank diameter is
and the impeller for the large tank is
H
L2
1.2 3.36 4.03 m
The air flow rate is
Gas superficial velocity,
Oxygen partial pressure:
H:D 3:1 6:1
The column diameter is in the range 0.1 D 7.5 m.
OTR K P K
LL
O
3
3
2
0.25 kmol/m hr
0.213 atm
1.174 kmol/m h at
1
mm
P
Hm
L
O
2
1 atm 1
10.3 in
atm 0.21 0.213 atm
Ê
Ë
Á
ˆ
¯
˜
U
s
1.45 m/h
018 60
4 0 576
4
2
.
(/)(. )
F
1
3
10 310
.18 m /min m s
331
D
i2
/2 m3363 11..
D
V
t2
6 m
2
13
13
03
60
03
33
..
.
/
/
Ê
Ë
Á
ˆ
¯
˜
Ê
Ë
Á
ˆ
¯
˜
DD
it
/2 cm
1
3
0 576 3 19..
D
t
76 m
018
03
05
13
.
.
.
/
Ê
Ë
Á
ˆ
¯
˜
BIOPROCESS SCALE-UP 311
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The mass transfer for freely rising gas bubbles is governed by conservation of mass and
momentum balance. When the distance, velocity and concentration of substances provide the
major driving forces for fluid motion, three dimensionless numbers explain the mass transfer.
The dimensionless numbers are the Schmidt number, Grashof number and Sherwood num-
ber. Schmidt number relates viscous forces with mass diffusivity. Grashof number deals
with free convection and buoyant forces. The Rayleigh number results from the multiplica-
tion of the Schmidt number and the Grashof number. Finally, Sherwood number is the ratio
of conductive forces and mass diffusivity. Based on analogy in mass transfer, the Sherwood
number is similar to the Nusselt number in heat transfer. The Sherwood number can be cor-
related with Schmidt and Grashof numbers for calculating mass transfer coefficients in a
mathematical expression.
(E.5.1)
(E.5.2)
(E.5.3)
And Rayleigh number
(E.5.4)
Calderbank’s correlation for turbulent flow aeration shows mass transfer is proportional to
D
i
2
and Re
3/4
(E.5.5)
(E.5.6)
13.5 DYNAMIC MODEL AND OXYGEN TRANSFER RATE IN
ACTIVATED SLUDGE
The oxygen requirements for an activated sludge system with aeration and agitation for
complete mixing are shown by mathematical model as the rate of oxygen transferred is
U
Q
A
g
g
Sh Sc Re 013
13 34
.
//
Gr Sc Ra
Dg
D
3
2
r
m
lO
Gr
Dg
3
2
18
r
m
l
Sh
kD
D
=
l
O
2
Sc
D
=
n
O
2
312 BIOCHEMICAL ENGINEERING AND BIOTECHNOLOGY
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equal to the difference of inlet and outlet concentration plus the oxygen utilised for gener-
ation of biomass:
(13.5.1)
where V is aeration tank volume in m
3
, is oxygen accumulation rate in kg O
2
m
3
day
1
, a and b are empirical constants (kg O
2
kg
1
BOD) and (kg O
2
kg
1
MLSSday
1
),
respectively. Also is the flow rate of fresh wastewater (m
3
.day
1
), MLSS stands for
mixed liquor suspended solids, C
S
is the BOD in effluent from aeration tank (kg BODm
3
),
C
So
is BOD in fresh wastewater and X is the sludge concentration in the aeration tank (kg
MLSSm
3
).
Example 1
In a specific activated sludge plant, the organic load is carried out at 0.8 kg BOD per kg
MLSSday
1
with an 80% BOD removal efficiency. Values for the above mathematical
model are as follows.
Fresh feed flow rate 1m
3
h
1
, initial BOD concentration is 20,000 ppm and V is 10 m
3
.
The yield of biomass on substrate is 0.5 gg
1
BOD.
a 0.55 kg O
2
kg
1
BOD
b 0.35 kg O
2
kg
1
MLSSday
1
What is the oxygen requirement per kg BOD removal?
Solution
Making oxygen balance for the aeration tank, the dynamic transfer rate of oxygen is obtained:
(E.1.1)
Dividing the oxygen balance by the volume of the aeration tank and concentration of bio-
mass, XV, the above equation will be simplified as follows:
(E.1.2)
(E.1.3)
or
m C
VX
SO
3
3
3
3
m
day
kg BOD
m
m
kg MLSS
m
Ê
Ë
Á
ˆ
¯
˜
Ê
Ë
Á
ˆ
¯
˜
Ê
Ë
Á
ˆ
¯
˜
()
1
2
X
C
t
a
VX
CCb
d
d
O
SO S
m
()
V
C
t
aCCbVX
d
d
O
SO S
2
.( )m
m
d/d
O
Ct
2
V
C
t
aC C bVX
d
d
O
SO S
2
m()
BIOPROCESS SCALE-UP 313
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