Ghasem D. Najafpour. Biochemical Engineering and Biotechnology
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244 BIOCHEMICAL ENGINEERING AND BIOTECHNOLOGY
Solution
C balance: 16 c d $ d 16 c
H balance: 34 3b 1.66c 2e
O balance: 2a 0.27c 2d e
N balance: b 0.2c
RQ: 0.43 d/a $ d 0.43a
d 0.43a 16 c
Substituting and Rearranging:
34 0.6c 1.66c 2e
2a 0.27c 2(16 c) e 2a 0.27c 32 2c e
74.4 4.65c 2c 0.27c 32 e
2 (e 2.92c 42.4) c 10.63
$
(2e 1.06c 34) d 5.37
a 12.49
b 2.13
e 2 (12.49) 0.27 (10.63) 2 (5.37) 11.37
Therefore the exact stoichiometric coefficients for the above reaction are:
9.6 EMBDEN–MEYERHOFF–PARNAS PATHWAY
The metabolic pathway for bacterial sugar fermentation proceeds through the Embden–
Meyerhof–Paranas (EMP) pathway. The pathway involves many catalysed enzyme reactions
which start with glucose, a six-carbon carbohydrate, and end with two moles of three carbon
intermediates, pyruvate. The end pyruvate may go to lactate or be converted to acetyl CoA
for the tricarboxylic acid (TCA) cycle. The fermentation pathways from pyruvate and the
resulting end products are shown in Figures 9.7 and 9.8.
The overall reaction of glucose catabolism to lactate and acetate fermentation from 2 moles
of glucose yields 2 moles of lactic acid, 3 moles of acetic acid and 5 moles of ATP, as shown
below:
(9.6.1)
Reactions involve several enzymes, which have to follow in sequence for lactic acid and
alcohol fermentation. This is known as the glucose catabolism pathway, with emphasis on
energetic and energy carrier molecules such as ATP, ADP, NAD
and NADH. In this pathway
the six-carbon substrate yields two three-carbon intermediates, each of which passes through
a sequence of reactions to the stable end product of pyruvic acid.
2C H O 2 lactic acid 3 acetic acid 5 ATP
6126
C H 12.5O 2.13NH 10.63CH O N 5.37CO 11.4H O
16 34 2 3 1.66 0.27 0.2 2 2
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MATERIAL AND ELEMENTAL BALANCE 245
Example 1
There was an old and outdated fermenter in our school warehouse. The dean of the school
wished the equipment to be used for undergraduate students to learn about bioprocess con-
cepts. The technician ran the equipment was experimentally but he forgot to weigh the glu-
cose carbon source and did not analyse for ethanol. The product stream was analysed as
follows: lactic acid 10 moles, acetic acid 5 moles, carbon dioxide 15 moles and hydrogen
10 moles. It was expected that sugar was oxidised, then converted to pyruvic acid, then pro-
ceeded though the Embden–Meyerhof–Parnas pathway. No other products except ethanol
were formed. The problem was adjusted by pH control and restarting the process without
any contamination. How many moles of ethanol were in the waste product stream? The
overall biological reaction is summarised as:
(E. 1.1)
abcC H O H O CH CH OH 10CH CH(OH)COOH
5CH COOH 15CO 10
6126 2 3 2 3
32
HH
2
↔
Glucose
Glucose-6-
phosphate
Fructose-6-
phosphate
Fructose-1, 6-
bisphosphate
Dihydroxyacetone
phosphate
Glyceraldehyde-3-
Phosphate
1,3 Bisphosphoglycerate
3 Phosphoglycerate
Phosphoenlolpyruvate
Pyruvate
FIG. 9.7. Glyconeogenesis in the Embden–Meyerhoff–Parnas pathway.
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246 BIOCHEMICAL ENGINEERING AND BIOTECHNOLOGY
Carbon balance: 6a 2c 30 10 15
Hydrogen balance: 12a 2b 6c 60 20 20
Oxygen balance: 6a b c 30 10 30
Multiplying the first equation with a negative sign gives:
62 65
2
ab
ab685
再
12 2 3 6 3 55 100
12 2 6 55 140
ab a
aba
再
2c 6a 55
12 2 2 100
670
abc
abc
Ï
Ì
Ô
Ó
Ô
6255
12 2 2 100
670
ac
abc
abc
Ï
Ì
Ô
Ó
Ô
2Glucose
2Glucose-6-phosphate
Fructose- 6-phosphate
Er ythrase-4- phosphate
Acetyl phosphate
Fructose-6- phosphate
Heptose phosphate +
triose phosphate
Acetic Acid
2 Acetyl phosphate
2 Acetic Acid
2 Glyceraldehyde-3-phosphate
2 Pyruvic Acid
2 Lactic Acid
2ATP
2ADP
ADP
ATP
2ADP
2ATP
2NADH
2
2NAD
2NADH
2
2NAD
4ADP + 2P
i
4ATP
FIG. 9.8. Fermentation of glucose to acetate or lactate via the Embden–Meyerhoff–Parnas pathway.
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MATERIAL AND ELEMENTAL BALANCE 247
Solving the above equation for b, yields:
4b 2
b 5
a 12.5
c 10
The defined equation is:
Example 2
Assume that the cells can convert 67% of carbon source to biomass. Hexadecane and glu-
cose are used as carbon sources. Calculate the stoichiometric coefficients of the following
reactions:
(E. 2.1)
Solution
Figure 9.9 shows the flow sheet for stoichiometry and material balance in a cell. According
to the problem, two thirds of the carbon source goes to biomass.
Therefore:
b
1
2.424
2
3
16 4 4
1
() . b
CH O NH (C H N O ) CO HO
16 34 1 2 2 3 1 4.4 7.3 0.86 1.2 2 2 3 2
aa b b b
12.5C H O 5H O 10CH CH OH 10CH CH(OH)COOH
5CH COOH 15CO
6126 2 3 2 3
32
10H
2
c
a
655
2
CELL
Substrate
C
w
H
x
O
y
N
z
a O
2
N source
b H
g
O
h
N
i
Biomass
c CH
O
N
d CO
2
e H
2
O
FIG. 9.9. Flow sheet for stoichiometry and material balance in a cell.
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248 BIOCHEMICAL ENGINEERING AND BIOTECHNOLOGY
The reminder of carbon source, one third, goes to carbon dioxide
Nitrogen balance: a
2
0.86 b
1
2.085
H
2
balance: 34 3a
2
7.3 b
1
2b
3
34 3(2.085) 7.3(2.424) 2b
3
b
3
11.28
O
2
balance: 2a
1
1.2b
1
2b
2
b
3
1.2(2.424) 2(5.333) 11.28
a
1
12.43
Glucose oxidation to SCP:
(E. 2.2)
a
1
147
.
2 1 2 0 909 2 2 3 85 6
1
a
.(. ) .
O balance: 2 1.2 2
2
6
1123
abbb
b
3
385
.
N balance: 0.86 0.78
H balance: 12 3 7.3 2
2
2
ab
abb
21
113
12 3(0.78) 7.3 (0.909) 2b
3
1
3
620
21
() .b
b
1
0 909
.
C balance:
2
3
644
1
() . b
CH O O NH (C H N O ) CO HO
6 12 6 2 3 4.4 7.3 0.86 1.2 2 2
aa b b b
12 1 2 3
1
3
16 5 333
2
() .b
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Cells are able to utilise chemical energy very efficiently. Like any actual process, energy
retained in the substrate is released as heat.
Example 3
From the above biomass production, calculate the yield coefficients. Based on definition,
the yield of biomass on glucose means how much biomass is produced per gram of glucose
utilised in a cell. The yield of biomass on glucose is calculated as follows:
(E. 3.1)
The yield of biomass produced per gram of oxygen consumed is:
Cells utilise energy from substrate in an efficient way, like a real chemical process. Some
of the energy is released as heat. The generated metabolic heat has to be removed from
the bioreactor; the temperature must be controlled for optimal cell growth. Also, the liber-
ated heat is proportional to cell growth. The yield factor Y
is defined as grams of cell pro-
duced per kilocalorie of heat evolved. This is analogous to yield of biomass to substrate and
heat combustion of substrate and cell, H
s
, and H
c
, respectively. The following equation
relates the yield factor to the yield of biomass and the net heat evolved resulting from cell
growth.
(E. 3.2)
Given heat of combustion of sugar:
H
S
673
180
374
kcal/mol
kcal/g biomass.
Y
Y
HY H
XS
XS
=
/
s/c
Y
X/O
1.76 g cells/g oxygen
2
83 03
2 1 47 16
.
.
Y
XS/
0.46 g cell/g glucose
83 03
180
.
Y
XS/
090944 12 73 086 14 12 16
180
.(. . . . )
Y
X/S
Biomass
Glucose
MATERIAL AND ELEMENTAL BALANCE 249
Ch009.qxd 11/16/2006 12:50 PM Page 249
Heat of combustion of cell:
H
C
5.8 kcal/g dry cell mass
Now substitute the values into (E. 3.2) to obtain the yield factor.
The yield of biomass to substrate for hexadecane is obtained.
The molecular mass of C
16
H
34
is 226.
H
S
is the possible heat evolved from oxidation of hexadecane.
Example 4
Citric Acid Production in Solid State Fermentation
Citric acid is manufactured by submerged cultures of Aspergillus niger in a batch reactor
operated at 30 C. For an incubation period of 2 days, 2500kg glucose and 860 kg oxygen
were consumed to produced 1500 kg citric acid, 500 kg biomass and other by-products.
Ammonia was used as nitrogen source. Power input to the system by mechanical agitation
was about 15 kW. Approximately 100 kg of water evaporated during the 2 day operation.
Estimate the energy for cooling requirements.
Solution
Figure 9.10 shows the flow diagram for production of citric acid in solid-state fermentation.
The basis of calculation is 1500 kg citric acid.
H
evap., H
2
O
at 30 C, steam table 2430.7 kJ/kg
Heat of reaction at 30 C 460 kJ/(mol O
2
used)
Y
0 988
11 2 0 988 5 8
.
.. .
0.18 g cell/g glucose
Y
X/O 2
0.56 g cell/g O
2
221 4
12 43 32
.
.
Y
XS/
g cell/g glucose
2 424 91 34
226
0 988
.(.)
.
Y
046
374 046 58
043
.
...
. kcal/g biomass
250 BIOCHEMICAL ENGINEERING AND BIOTECHNOLOGY
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MATERIAL AND ELEMENTAL BALANCE 251
Fermentation reaction:
(E. 4.1)
Mass is balanced at the beginning and the end of process. The evaporated water in solid-state
fermentation may not be required for energy balance.
The first law of thermodynamics is valid for the energy balance.
H
rxn
– m
H
2
O
H
evap
dQ – dW (E. 4.2)
H
rxn
(460)(861 kg)(1000 g/kg)(1 gmol/32 g) 1.24 10
7
kJ
Heat lost by vaporisation of 100kg H
2
O
m
v
H
evap
(100 kg)(2430.7) 2.43 10
5
kJ
Mechanical work:
W
s
(15 kW)(kJ/s/1 kW)(2 days)(24 hours/day) 2.59 10
6
kJ
Q 1.24 10
7
2.43 10
5
2.59 10
6
1.47 10
7
kJ (removed heat)
REFERENCES
1. Aiba, S., Humphrey, A. E. and Millis, N. F., “Biochemical Engineering”, 2nd edn. Academic Press, New York,
1973.
2. Baily, J.E. and Ollis, D.F., “Biochemical Engineering Fundamentals”, 2nd edn. McGraw-Hill, New York,
1986.
3. Blanch, H.W. and Clark, S.D., “Biochemical Engineering”. Marcel Dekker, New York, 1996.
4. Doran, P.M., “Bioprocess Engineering Principles”. Academic Press, New York, 1995.
5. Shuler, M.L. and Kargi, F., “Bioprocess Engineering, Basic Concepts”. Prentice-Hall, New Jersey, 1992.
6. Ghose, T.K., “Bioprocess Computation in Biotechnology”, vol. 1. Ellis Horwood Series in Biochemistry and
Biotechnology, New York, 1990.
Mass Mass H O
beginning end 2 vap
Glucose NH O Biomass CO H O Citric acid
32 2 2
Q = ?
NH
3
+ 860 kg O
2
1500 kg Citric acid
500 kg Biomass
100 kg H
2
O vapour
2500 kg glucose
Fermenter
30 °C
FIG. 9.10. Flow sheet for citric acid in solid-state fermentation.
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CHAPTER 10
Application of Fermentation Processes
10.1 INTRODUCTION
In World War I, Germany needed to synthesise glycerol for manufacturing explosives.
It was found that glycerol was generated in alcoholic fermentation. Nearberg discov-
ered that the addition of sodium bisulphite to the fermentation broth was favoured and
enhanced the production of glycerol at the expense of ethanol. German scientists quickly
developed industrial-scale fermentation with a yield capacity of 1000 tons of glycerol per
month.
1,2
Ethanol is an essential chemical which is used as a raw material for a vast range of appli-
cations including chemicals, fuel (bioethanol), beverages, pharmaceuticals and cosmetics.
The feedstock for ethanol generally comes from renewable sources such as starch (from
wheat, barley, maize, potato, cassava, sweet potato, etc.) and molasses or syrups originat-
ing from sugar beet or sugar cane, etc. Bioethanol is produced by biological fermentation
technology. The fermentation product stream is processed with subsequent enrichment by
distillation/rectification and dehydration.
Bioethanol is becoming a viable solution as a source of renewable energy, as it is a
non-fossil fuel. It may originate from renewable agricultural sources, resulting in cleaner
combustion without any emissions to the air.
10.2 PRODUCTION OF ETHANOL BY FERMENTATION
Carbohydrates obtained from grains, potatoes or molasses are fermented by yeasts to
produce ethanol in the production of beer, alcohols and distilled spirits. Fermentation of
sugar using Saccharomyces cerevisiae produces ethanol under anaerobic conditions. The
batch fermentation system is affected by high substrate and product inhibition. Glucose
concentration plays the major role in increasing the concentration of ethanol and cell
growth rate in the fermentation broth. Cell density, ethanol concentration and glucose
concentration are measured. Industrial grade sugars and molasses are used for the produc-
tion of bioethanol. Today, alcohol technologies are well developed. Many extraction and
purification techniques have enhanced ethanol production and the process is considered
economically feasible.
252
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APPLICATION OF FERMENTATION PROCESSES 253
10.3 BENEFITS FROM BIOETHANOL FUEL
Alcoholic fermentation, ethanol production, has been best known for a few decades by
S. cerevisiae. Many obligate aerobic fungi, such as common moulds of the genera Aspergillus,
Fusarium and Mucor are also well known for their ability to produce ethanol.
2
The bene-
fits are:
• renewable resources
• cleaner environment due to cleaner combustion
• lower net carbon dioxide emissions
• expanded market opportunities in agriculture
• less dependence on crude oil
10.4 STOICHIOMETRY OF BIOCHEMICAL REACTION
The following biochemical reaction represents sugar fermentation. Ethanol is the end prod-
uct, which may be used as a useful bioprocess chemical.
(10.4.1)
There are two popular microorganisms that can produce high concentrations of alcohol.
Their tolerance to high concentrations of ethanol and substrates are stated in the litera-
ture.
3–5
The most common are Zymomonas mobilis and Saccharomyces cerevisiae.
10.5 OPTICAL CELL DENSITY
Cell growth is defined with cell density. Cell concentration is an indication of viability of
microorganism. Measure the optical cell density of Saccharomyces cerevisiae at a wave-
length of 580 nm. Other wavelengths such as 600nm or less may also be used, but one must
be consistent. Draw a growth curve based on incubation time and cell dry weight. A stan-
dard calibration curve is needed before any actual experiment. Generate a calibration curve
to relate the absorbance with cell dry weight. The usual rules of operating a spectropho-
tometer apply here, as well. For example, the accuracy of the method is greatest when the
absorbance readings are in the range 0.1–1. For a given culture sample, a good spectro-
photometer should yield a linear relation between the number of cells and the absorbance.
However, optical density is also a function of cell morphology such as size and shape,
because the amount of transmitted or scattered light depends strongly on these factors.
Consequently, an independent calibration curve is required for each condition in accurate
research work, as the cell size and shape depend on the specific growth rate and the nutrient
composition. As a rule of thumb, an optical density of one unit corresponds to approximately
1 g⭈l
⫺1
of dry cell. This is also commonly referred to as the turbidity measurement.
CH O CHOH CO
Glucose
Yeast fermentation
6126 25 2
22æÆæææææ ⫹
Ethanol carbon dioxide
180 g/mol 46 g/mol 44 g/mol
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