Ghasem D. Najafpour. Biochemical Engineering and Biotechnology
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(E.1.4)
(E.1.5)
The BOD removal is 80%
(E.1.6)
(E.1.7)
(E.1.8)
(E.1.9)
Now we can increase the BOD removal efficiency with decreasing organic loading. What
would be the amount of oxygen required per kg BOD removal, with 90% removal efficiency?
Given data for constant coefficient a 0.5 and b 0.3
(E.1.10)
Given BOD loading
(E.1.11)
(E.1.12)
In fact, the more organic load there is, the more oxygen is needed in the treatment process.
1
05 04 09 03 04
X
C
t
dO
d
8 kg of O /kg MLSS.day
2
2
.... .
CC
C
C
C
CC
SO S
SO
S
SO
SSO
09 1 09
01
..
.
FC
VX
SO
33
33
(m /day) kg BOD/m
(m ) kg MLSS/m
0.4
()
()
11
X
C
t
a
F
VX
CCb
dO
d
2
SO S
Ê
Ë
Á
ˆ
¯
˜
()
1
04 05508 035 052
X
C
t
dO
d
6
kg O
kg MLSS day
22
..(.). .
CC
SSO
02.
08 1.
C
C
S
SO
80
100
1
CC
C
C
C
SO S
SO
S
SO
Ê
Ë
Á
ˆ
¯
˜
1
1
X
C
t
aC
VX
C
C
dO
d
2
SO S
SO
m
Ê
Ë
Á
ˆ
¯
˜
kg BOD
kg MLSS.day
0. BOD loading 8
314 BIOCHEMICAL ENGINEERING AND BIOTECHNOLOGY
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Example 2
A bioreactor of V 20 m
3
working volume produces penicillin. What sugar concentration will
you choose if OTR is not limited?
Hint: if you aerate a bioreactor, the power consumption is less than a non-aerated biore-
actor and the specific sugar consumption rate is 1 kgkg
1
cellh
1
.
Data given
D
t
2.4 m Three sets of impellers
D
i
0.8 m M 1 mPa.s
N
i
2.5 rps P 1200 kgm
3
Number of blades 8, turbine Aeration rate 1 vvm
P
g
/P 0.4 OTR 6 10
3
kgm
3
Specific O
2
uptake 0.63 mmol O
2
kg
1
cell y
m
0.5 h
1
Solution
Let us find the Reynolds number
When flow is turbulent, the power number is obtained from Figure 6.6, Chapter 6, using Re
from power curves.
N
P
P
no
6
Ungassed power 5 3 sets of impellers 15 hp
Correction factor for scale-up using non-geometry similarity,
Gassed power, P
g
0.4 (15) 6hpfor three sets of impellers using aeration
f
DD H D
DD H D
DD
H
c
()( )
()( )
.
.
**
ti Li
ti Li
ti
L
//
//
/
/
3442
33
125
3
DD
HD
3
20 2 4 4 4 42
2
()()/(.)(/).
*
Li
/
P
Pg
ND
P
ND
g
no
ii
ii
5 hp
6
6
6 1200 2 5 0 8
981
35
35
35
◊
r
r
..
.
Re
ND
r
m
i
2
2
4
1200 0 8 2 5
01
19 10
..
.
.
BIOPROCESS SCALE-UP 315
Ch013.qxd 11/21/2006 9:51 AM Page 315
The mass transfer coefficient for a turbulent regime is
The oxygen transfer rate is
Biomass concentration is calculated.
What is your desired sugar concentration?
To answer the above question, we truly need to know the yield of biomass on substrate.
Based on availability of oxygen, there is a good chance of sufficient biomass being gener-
ated, therefore nutrients and carbon sources must be available to carry on the process.
Example 3
Let us scale-up a small fermenter with volume (V) of 0.3 m
3
. The scale-up factor is 200
fold. The large fermenter has a volume (V
2
) of 60 m
3
. The working volume is 60% of nom-
inal volume, that is
V
L
0.18 m
3
x
610
208 10
3
3
4
.
0 kg cells m
3
2 08 10 0 65 10
33
..
mole O
kg cell
32 kg
kmol
2
q
O
3
2
131
2
kg
m
1
32
kg O kg cell m s
610
3
Ê
Ë
Á
ˆ
¯
˜
OTR xq
O2
kg O m s
2
410
43
1
OTR K a C C
L
() ( )
*
ll
6 10 6 10 0
23
Ka
L
210
6
20
44 6 10
3
06
0 667 2 1
Ê
Ë
Á
ˆ
¯
˜
.
.
(.) s
V
S
3
22
21
20 m /min
/4 2.4 m
4.4 m/min 7 10 m s
()()
316 BIOCHEMICAL ENGINEERING AND BIOTECHNOLOGY
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Aeration rate 1 vvm
Density, r 1200 kgm
3
Oxygen transfer rate, OTR 0.25 kmolm
3
h
1
H
L
/D
t
1.2
Two sets of impellers, turbine and flat blade.
The usual procedures for scaling up are summarised as following:
1. Reynolds number.
2. Power consumption per unit volume of liquid.
3. Tip velocity of an impeller, N
i
D
i.
4. Liquid circulation time, mixing time.
5. Volumetric oxygen transfer coefficient.
Solution
Large reactor:
and
H
L2
1.2 3.36 4.03 m
D
i
2 m336 3 11./ .
D
V
t
36 m
2
2
13
13
03
60
03
3
..
.
/
/
Ê
Ë
Á
ˆ
¯
˜
Ê
Ë
Á
ˆ
¯
˜
H
L
1 m
1
12 0576 069 .. .
DD
it
92 m13 01/.
D
t
m
018
03
0 576
13
.
.
.
/
Ê
Ë
Á
ˆ
¯
˜
VDD
1
2
4
12 01
tt
3
8 m(. ) .
BIOPROCESS SCALE-UP 317
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Air flow rate
Gas supply velocity,
For turbulent flow the power number is about 6 and for two sets of impellers is going to be
doubled.
If you have access to the graph P
g
/P that is good, read gassed power to ungassed power ratio
off of the graph.
Read aeration number (Na) from Figure 6.7, Chapter 6.
Now we make use of Miller’s correlation for gassed power calculations:
5
P
PND
F
i
g
05
1
2
11
3
056
043
.
.
.
Ê
Ë
Á
ˆ
¯
˜
N
Pg
ND
PN N
p
i
1
3
hp
1
1
35
11
3
5
12 1200 0 192 9 81 0 383
r
(. )/ . .
Re N
p
6212
PV
g
8 hp 322 322 018 05....
K
P
V
U
v
g
S
0 0318 1 174
095
067
..
.
.
Ê
Ë
Á
ˆ
¯
˜
K
v
3
31 1
0.25 kmol/m h
0.213 atm
1.174 kmol m h atm
OTR K P
v
O
2
P
H
O
L
1 atm 1
10.3
2
0.213 atm
2
Ê
Ë
Á
ˆ
¯
˜
U
S
.45 m/h
018 60
4 0 576
41
2
.
(/)(. )
F
1
101
. 8 m /min 3 10 m /s
333
318 BIOCHEMICAL ENGINEERING AND BIOTECHNOLOGY
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P
1
0.383N
1
3
0.383(116)
3
2.75 hp
P
g
/P 0.58/2.75 0.2 l
Power input constant, geometric similarities
Constant impeller tip velocity:
N
1
D
i1
N
2
D
i2
P
g
/V 0.58/0.18 3.22 hp
Example 4: Filter and Air Filtration
A batch production of penicillin of 40 m
3
capacity is required to supply sterile air through
bioreactor at 1 volume of air per volume of culture per min (1 vvm). Incoming air contains
P 275
36
018
5.
.
Ê
Ë
Á
ˆ
¯
˜
50 hp
P
g
3
36 m 116 hp
058
018
.
.
()
Ê
Ë
Á
ˆ
¯
˜
NN
D
D
21
116
0 192
112
i1
i2
20 rpm
Ê
Ë
Á
ˆ
¯
˜
Ê
Ë
Á
ˆ
¯
˜
.
.
NN
V
V
D
D
21
2
1
13 53
13
116
60
03
0 192
1
Ê
Ë
Á
ˆ
¯
˜
Ê
Ë
Á
ˆ
¯
˜
Ê
Ë
Á
ˆ
¯
˜
//
/
.
.
.
i1
i2
112
3
53
Ê
Ë
Á
ˆ
¯
˜
/
5 rpm
rrND
V
ND
V
N
N
V
V
D
D
1
3
1
5
1
2
3
2
5
2
2
1
3
2
1
1
5
ii i
i2
Ê
Ë
Á
ˆ
¯
˜
Ê
Ë
Á
ˆ
¯
˜
Ê
Ë
Á
ˆ
¯
˜
058 05
0 383
310
11
2
1
32
1
3056
043
1
..
(. )( )
()
.
.
NND
N
i1
3
6
Ê
Ë
Á
ˆ
¯
˜
rrpm
BIOPROCESS SCALE-UP 319
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3000 bacteria per cubic metre of air, for 100 hours’ operation. Calculate the filter depth, if
the penetration of bacteria is 1 in 1 million.
Solution
Assuming:
D
filter
60 cm based on availability
Air flow rate,
Microbial load 3000 cellsm
3
24000 3000 7.2 10
8
cells
Filter cross-sectional area (4)(0.6)
2
0.0283 m
2
Removal of organisms may follow the following exponential format of equation
(E.4.1)
where L is the length of filter and k is rate constant for bed materials 40 m
1
and/or
84 m
1
. The length of filter is based on filter materials: for large k, a shorter length of
filter was obtained.
Example 5
In a batch fermentation of ethanol, kinetic data were collected as product formed. The data
are shown in Table E.5.1. The data will be used to design a continuous bioreactor (CSTR)
with a 100 l working volume.
L
k
L
1
72 10
1
40
72 10 08
1
84
7 2 10 0
14 14
14
ln( . ) ln( . ) .
ln( . )
5 m
..41 m
N
N
1
2
72 10
.
8
66
cells
1 cell in 10 cells 10 cells
NN
N
N
kL
kL
kt
to
e
ln
ln
.
1
2
8
6
72 10
10
Air velocity
Vol. flow rate
mh
1
A
2400
0 0283
8488
.
F
air
33
40 m 1 vvm 60 min/h 2400 m h
()()( )
1
320 BIOCHEMICAL ENGINEERING AND BIOTECHNOLOGY
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(a) If the feed rate, F, is 50
3
m
3
h
1
, calculate the ethanol concentration in the product
stream. Keep in mind there was no ethanol in the feed stream.
(b) If the product yield is 45%, what would the sugar concentration in the feed stream be?
Experimental data are shown in Table E.5.1.
The data are plotted in Figure E.5.1.
BIOPROCESS SCALE-UP 321
TABLE E.5.1. Ethanol production in batch fermentation
with respect to inoculation time
Time, h Ethanol, gl
1
0 0.0
10 2.0
15 4.0
18 6.2
21 10.0
27 16.0
33 30.0
42 56.5
48 75.7
51 85.0
54 96.0
57 103.5
60 107.3
66 110.5
72 113.0
FIG. E.5.1. Ethanol concentration profile with respect to fermentation time in batch mode of operation.
Time, h
0 1020304050607080
Ethanol production, g/l
0
20
40
60
80
100
120
d
C
p
/d
t
=3.3 g.l
-1
.hr
-1
Ch013.qxd 11/21/2006 9:51 AM Page 321
Use material balance for ethanol production:
(E.5.1)
At steady-state condition the product concentration remained constant, dC
p
/dt 0.
Initially there was no ethanol in feed stream, C
po
0; When there is no accumulation, the
balance equation is reduced to:
(E.5.2)
(E.5.3)
(E.5.4)
There is no wall temperature effect,
(E.5.5)
From plotted data
(E.5.6)
(E.5.7)
Ethanol concentration in the product stream is 66 gl
1
.
S
C
Y
0
66
045
14
P
P/S
1
7g l
.
Y
P/S
045.
C
p
3
3
3
0.1 m
510
6 kg m
Ê
Ë
Á
ˆ
¯
˜
(.)33
330
5
6
V
F
r
C
t
V
F
F
V
C
p
p
p
d
d
Ê
Ë
Á
ˆ
¯
˜
d
d
3g l h
P
11
C
t
3.
m
m
w
Ê
Ë
Á
ˆ
¯
˜
014
1
.
C
V
F
r
PP
rDC
F
V
C
pp p
DC r()
PP
d
d
)
p
po p p
C
t
DC C r(
322 BIOCHEMICAL ENGINEERING AND BIOTECHNOLOGY
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Example 6: Vinegar Production in a Jacketed Bioreactor
Ethanol is oxidised to acetic acid in the production of vinegar using Acetobacter. The reac-
tion is exothermic:
(E.6.1)
The biochemical reaction rate followed the Monod rate model with a Monod rate constant
of k
S
6.2 10
6
gcm
3
and a specific growth rate of
max
6.67 10
7
gcm
3
s
1
.
Design the bioreactor with a suitable heat transfer area.
Given data
N 500 rpm 30000 rph T 32 C 90 F
D
i
0.75 ft k 0.356 Btu.h
1
ft
1
F
1
P 62.4 lbmft
3
C
p
1 Btulb
1
F
1
M 1cp 2.42 lbmft
1
h
1
Solution
(E.6.2)
K
s
6.2 10
6
gcm
3
and
max
6.67 10
7
gcm
3
s
1
Figure E.6.1 shows the jacketed bioreactor.
r
S
KS
S
n
max
S
C H OH O CH COOH H O kcal mol
25 2 3 2 rxn
1
Acetobacter
HæÆææææ 119
BIOPROCESS SCALE-UP 323
w
D
i
L
B
Cooling H
2
O, out
Cooling H
2
O, in
FIG. E.6.1. Diagram of jacketed vessel with impeller location.
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