01
[Problem] In cultivation cultivation of of baker’s yeast in a stirred and aerated tank, lethal agents are added to the fermentation medium to kill the organisms immediately. immediately. Increase in dissolved oxygen (DO) concentration upon addition of lethal agents is followed with a DO analyzer and a recorder. Using the following following data, determine determine the oxygen transfer coefficient (k La) for the fermentor. Saturation DO concentration is 9 mg/L. Time (min)
1
2
2.5
3
4
5
DO (mg/L)
1
3
4
5
6.5
7.2
Solution: dC L dt
k L a (C C L ) *
d (C * C L )
(C * C L )
k L adt
ln(C * C L ) k L at Constant
Plot ln(C * C L ) versus t , slope = k La. Time (min) 1 2 2.5 3 4 5 * * C = 9 mg/L
DO (mg/L)
ln(C * C L )
1 3 4 5 6.5 7.2
2.1 1.8 1.6 1.4 0.92 0.59
2.4 2.2 2.0 1.8 )
L
1.6
C *
The oxygen
1.4
C ( n l
1.2
transfer coefficient (k La) for the fermentor is 1 0.39 min .
Slope k L a
0.39 min 1
1.0 0.8 0.6 0.4 0
1
2
3
Time (min)
4
5
6
02
[Problem] A batch fermentation is conducted at 30C. Experiments 1 with sodium sulfite oxidation indicate that k La = 438 h . The solubility of oxygen from air in the fermentation broth is 7.3 mg/L. The culture has a doubling time, in exponential growth, of 30 min, and oxygen yield coefficient Y X/O of 0.6 g cells/g O2. 2
(a) At what biomass level is C L predicted to become zero? (b) In reality, C L does not become zero. C L, which follows the Monod form: max
Taking K O
2
Rather, becomes a function of C L
K O 2 C L
= 0.05 mmol/L, estimate the cell concentration when C L
becomes 5% saturation. Solution:
(a)
dX dt
ln 2 t d
X , ln
0.693 0 .5 h
X X 0
t , ln
2 X 0 X 0
t d
1.386 h 1 max
Material balance on dissolved oxygen: k L a (C * C L ) When C L = 0, X
Y X/O2
k L aC *
0.6 1.386
1 Y X/O2
X
( 438)(7.3 10 3 ) 1.4 g/L
(b) Material balance on dissolved oxygen: k L a (C * C L )
X Y X/O2
max
C L K O 2 C L
438 [(7.3 7.3 0.05) 10 3 ]
X
0 .6
(1.386)
7.3 0.05 (0.05 32) ( 7.3 0.05)
X = 7.1 g/L #
03
[Problem] Escherichia coli has a maximum respiration rate, qO
2 , max
, of
about 240 mg O2/g cell-h. It is desired to achieve a cell mass of 20 g 1 cell/L. The k La is 120 h in a 1000-L fermentor (800 L working volume). A gas stream enriched in oxygen is used (i.e., 80% O2) which * gives a value of C = 28 mg/L. If oxygen becomes limiting, growth and respiration slow according to: qO 2
qO 2 , max C L
0.2 mg/L C L
where C L is the dissolved oxygen concentration in the fermentor. is C L when the cell mass is at 20 g/L?
What
Solution:
Oxygen transfer rate = oxygen uptake rate
k L a (C C L ) qO 2 X *
qO 2 , max C L X
0.2 C L
120( 28 C L )
(3360 120C L )(0.2 C L ) 4800C L
240 C L 20 0.2 C L
672 3360C L 24C L 120C L2 4800C L
120C L2 1464C L 672 0
1464 (1464 ) 2 4 120 672 1464 1570 C L 2 120 240
C L = 0.44 mg/L When the cell mass is at 20 g/L, C L is 0.44 mg/L. #
04 [Problem] A cylindrical tank (1.22-m diameter) is filled with water to an operating level equal to the tank diameter. The tank is agitated with a 0.36-m diameter, flat six-blade disk turbine, which has a power number of 6. The impeller rotational speed is 150 rpm. The air enters through an open-ended tube situated below the impeller and its volumetric flow 3 rate is 0.0217 m /s at 1.08 atm and 25 C. Estimate the power 3 requirement and k La. Water has a density of 997 kg/m and a viscosity 4 of 8.9 10 kg/m-s at 25C. Use the following correlation equations to estimate the desired data. D log10 192 i P0 Dt Pg
4.38
NDi2
0.115
1.96 Di / Dt
Di N 2 g
Q 3 NDi
0.7
P k La = 0.002 g (vs )0.2 V L
Solution: N =
150
= 2.5 rev/s
60
2
Re
NDi
(997 )(2.5)(0.36) 2 8.9 10
4
Power number N p = 6 = D log10 192 i P0 Dt Pg
4.38
P0 3
5
N Di
NDi2
0.36 = 192 1.22
3.6 10 5
0.115
P0 1.96 Di / Dt
Di N 2 g
4.38
(3.6 10 )
0.115
P0 = 565 W
Q 3 ND i
(0.36)(2.5) 9.8
1.960.36
2
5
(997)(2.5) 3 (0.36) 5
1.22
0.0217 = 3 ( 2 . 5 )( 0 . 36 )
0.317
0.7
Pg P0
0.482
P k La = 0.002 g (vs )0.2 V L
Pg = 0.482(565) = 272 W
272 = 0.002 2 (1.22) (1.22) 4 1
(where k La in s ,
Pg V L
3
0.7
0.0217 2 (1.22) 4
in W/m , vs in m/s)
0.2
= 0.0356 s
1
10401 Advanced Chemical Engineering: Part III Homework 2
(2016/1/5 submit)
[Q1] Candida utilis grows aerobically on ethanol, with no other products apart from cell mass and CO2 produced. The observed yield coefficients Y X/S,obs (that is, the yield coefficient not taking account of maintenance), at various specific growth rates were found as the follows: 1 (h )
0.05
0.1
0.15
0.2
0.25
0.3
Y X/S,obs (g cell/g ethanol)
0.48
0.54
0.56
0.57
0.57
0.58
Find the true yield coefficient Y X/S and the maintenance coefficient m. Solution:
Material balance on substrate:
1 dS X dt
1 Y X/S,obs
1
versus
dt
1 dX Y X/S dt
mX
1 1 dX m m Y X/S X dt Y X/S
Plot
dS
dS dX
1
Y X/S, obs
1 Y X/S
m 1 dS dt 1 X X dt dX Y X/S
m
:
0.05
0.1
0.15
0.2
0.25
0.3
1/
20
10
6.7
5
4
3.3
Y X/S,obs
0.48
0.54
0.56
0.57
0.57
0.58
1/Y X/S,obs
2.1
1.9
1.8
1.8
1.8
1.7
2.2 2.1
Slope = m
) l l e c g 2.0 / l o n a h t 1.9 e g ( s b o , S / X
=0.0215 g ethanol/g cell-h Intercept =
1.8
1 Y X/S
= 1.675
or
Y /
1
1.7
Y X/S = 0.60 g cell/g ethanol
1.6 0
5
10
15
20
25
1/ (h)
#
1
[Q2] Escherichia coli is cultivated on glycerol at a specific growth rate of 0.1 h . The cell yield coefficient is 0.45 g cell/g glycerol, and the final cell concentration is 60 g/L. The heats of combustion of glycerol and the cell are 18 and 20 kJ/g, respectively. Estimate the maximal rate of heat evolution per liter of broth. Solution: 1
H S =
18 =
1
Y X/H
Y X/S + H X Y X/S
0.45 + 20 0.45
Y X/H
1 Y X/H
= 20 kJ/g cell
The maximal rate of heat evolution per liter of broth = X (1/Y X/H) = (0.1)(60)(20) = 120 kJ/L-h
[Q3] The following data was obtained for Saccharomyces cerevisae on limiting glucose at three dilution rates, where the feed concentration of glucose was 17.4 g/L. Dilution rate (h 1)
Cell dry weight (g/L)
Glucose (g/L)
Ethanol (g/L)
0.15
8.1
0.03
3.55
0.22
6.2
0.03
4.95
0.28
4.2
0.20
5.30
If cell growth follows the Monod model, estimate max and K S. Solution:
Monod model: D max
0.03(
max
0.22)
0.22 K S
S (
max
D
K S S
K S
S (
max
D)
D 1
K S does not vary with dilution rate. K S
S
The data of D = 0.22 and 0.28 h result in: 0.2( max 0.28)
D)
0.28
0.03(0.294 0.22) 0.22
max =
0.294 h
0.01 g/L
1
[Q4] Suppose you have a microorganism that obeys the Monod equation with max = 1 0.7 h and K S = 5 g/L. The cell yield Y X/S is 0.65 g cell/g substrate. You want to cultivate this microorganism with chemostat. The substrate concentration of the inlet stream is 85 g/L, and the outlet concentration must be 5 g/L. What flow rates should be if you use: (a) one 1,000 L-fermentor; (b) two 1,000 L-fermentors in series; and (c) one 2,000 L-fermentor? Solution:
(a) Use one 1,000 L-fermentor
max
S
K S S
0.7 5 55
0.35 h 1
F DV V 0.35 1000 350 L/h
(b) Use two 1,000 L-fermentors in series D1
F V 1
F
1,000
D2 D
In the first fermentor, X 1 Y X/S ( S 0 S 1 ) 0.65 (85 S 1 ) 1
max
S 1
K S S 1
D
0.7 S 1 5 S 1
In the second fermentor, 2
X 2
2
D2 2
max
X D1 1 X 2
S 2
K S S 2
0.7 5 55
0.65 ( S 1 5) 1.86 D( S 1 5)
0.35
0.35 D 1
0.35
0.35 h 1
0.7 S 1 5 S 1
X 1 D 1.86 D ( S 1 5) 1.86( S 1 5)
X 1
0.65 (85 S 1 ) 1.86( S 1 5)
0.65( S 1 5)(S 1 5) 1.3S 1 ( S 1 5) 0.65( S 1 85)(S 1 5)
0.65S 12 16.3 1.3S 12 6.5S 1 0.65S 12 52 S 1 276
1.3S 12 58.5S 1 260 0 D
D
Y X/S ( S 1 S 2 )
0.7 S 1 5 S 1
0.7 51.6 5 51.6
S 1
0.64 h 1
65 3422 1352
2 1.3
51.6 g/L
F D 1,000 640 L/h
(c) Use one 2,000 L-fermentor
max
S
K S S
0.7 5 55
0.35 h 1
F DV V 0.35 2000 700 L/h
#
