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Bioreactor
design for
fermentation
-
Contents
• Lecture 01 – Bioeconomy,
fermentation and bioreactors
• Lecture 02 – Stoichiometry and
mass balance
• Lecture 03 – Fermentation basics &
Kinetics
• Lecture 04 & 05 – Mixing in stirred
tank for design
• Lecture 06 – Mass transfer for
bioreactor design
• Lecture 07 – Heat transfer for
bioreactor design
• Lecture 08 – Bioreactor design
Stoichiometry
of Growth
and Product
Formation
• For mass balances with reaction the
stoichiometry of conversion must be
known before the mass balance can be
solved.
• Under growth conditions, cells are a
product of reaction and must be
represented in the reaction equation.
• Metabolic stoichiometry has many
applications in bioprocessing: in mass and
energy balances.
1. It can be used to compare theoretical
and actual product yields.
2. Check the consistency of experimental
fermentation data.
3. Formulating nutrient medium.
Growth
Stoichiometry
and
Elemental
Balances
• Cell growth obeys the law of
conservation of matter.
• All atoms of carbon, hydrogen,
oxygen, nitrogen and other
elements consumed during growth
are incorporated into new cells or
excreted as products.
• Confining our attention to those
compounds taken up or produced
in significant quantity, if the only
extracellular products formed are
CO2 H2O and we can write the
following equation for aerobic cell
growth
Stoichiometry of biochemical reaction
with cell growth and product formation
Fermentation
Substrate Cells and product
Substrate
Residual substrate
More new cells
Products
CO2 by respiration
Scenario 1:
Growth
Stoichiometry
and
elemental
balances (e.g.
single-cell
protein)
The stoichiometry of conversion must be known before
the mass balance can be solved
𝑤 𝑥 𝑦 𝑧 2 𝑔 𝑏 𝑖 𝛼 𝛽 𝛿 2 2
(Eq. 1)
𝑢 𝑥 𝑦 𝑧 represents the chemical formula for the
substrate (e.g. for glucose w=6, x=12, y=6 and z=0)
𝑔 𝑏 𝑖 represents the chemical formula for the
nitrogen source
𝛼 𝛽 𝛿 represents the ‘chemical formula’ for dry
biomass (biomass compositions)
− a, b, c, d, e are stoichiometric coefficients
Conversion of substrate, oxygen and
nitrogen for cell growth
Elemental composition of Escherichia
coli bacteria
92%
Elemental composition and degree of
reduction for selected organisms
• Compounds such as vitamins and minerals
taken up during metabolism could be
included; however, since these growth factors
are generally consumed in small quantity we
assume here that their contribution to the
stoichiometry and energetics of reaction can
be neglected.
• Other substrates and products can easily be
added if appropriate.
• Bacteria tend to have slightly higher nitrogen
contents (11-14%) than fungi (6.3-9.0%).
• For a particular species, cell composition
depends also on culture conditions and
substrate utilized, hence the different entries
in Table 4.3 for the same organism.
Elemental
composition
comparison
• The results are remarkably similar
for different cells and conditions;
CH1.8O0.5N0.2 can be used as a
general formula when composition
analysis is not available.
• The average ‘molecular weight’ of
biomass based on C,H, O and N
content is therefore 24.6. although
5-10% residual ash is often added
to account for those elements not
included in the formula.
Element balance:
Five unknown coefficients, but
there are only four equations
Notice that we have five unknown coefficients (a, b, c, d and e) but only four
balance equations.
This means that additional information is required before the equations can be
solved.
Usually this information is obtained from experiments.
One more equation such as RQ is
needed, which could be obtained from
experiment
When Eq above is completed, the quantities of substrate, nitrogen and oxygen
required for production of biomass can be determined directly
Example: Stoichiometric coefficients for
cell growth
Production of single-cell protein from hexadecane is
described by the following reaction equation:
It has been measured that RQ is 0.43 mole CO2
produced per mole O2 consumed. Write the complete
reaction equation and calculate oxygen demand.
Electron
balance
for cell
growth
Definition of available electrons
• Available electrons refers to the number of
electrons available for transfer to oxygen on
combustion of a substrate to CO2, H2O and
nitrogen-containing compounds.
• The number of available electrons found in
organic material is calculated from the
valence of the various elements: 4 for C, 1
for H,-2 for O, 5 for P, and 6 for S. The
number of available electrons for N
depends on the reference state:-3 if
ammonia is the reference, 0 for molecular
nitrogen N2, and 5 for nitrate.
Cont’d
Definition of degree of reduction
• Degree of reduction, is defined as
the number of equivalents of available
electrons in that quantity of material
containing 1 g atom carbon.
• for substrate CwHxOyNz, the number of
available electrons is (4w + x- 2y- 3z).
The degree of reduction for the
substrate, s, is therefore (4w + x - 2y -
3z)/w.
• Degree of reduction for CO2, H2O and
NH3 is zero.
Example: calculate the Degree of
reduction for substrates below:
Ethanol: C2H5OH
Glucose: C6H12O6
Methane: CH4
Glucose is C6H12O6. The degree of reduction = 6 x (+4) +
12 x (+1) + 6 x (-2) = 24
The higher degree of reduction species lower degree of oxidation
Electron balance for cell growth
(cont’d)
• Electrons available for transfer to oxygen are
conserved during metabolism.
• In a balanced growth equation, number of available
electrons is conserved by virtue of the fact that the
amounts of each chemical element are conserved.
Electron balance for cell growth
(cont’d)
RQ=measured
where s and B are the degrees
of reduction of substrate and
biomass, respectively.
This equation
could be used to
get coefficient to
complete
chemical reaction
Biomass yield
• As cells grow there is, as a general approximation, a
linear relationship between the amount of biomass
produced and the amount of substrate consumed.
• This relationship is expressed quantitatively using
the biomass yield, :
Biomass yield (cont’d)
• When is constant throughout growth, its
experimentally determined value can be used to
determine the stoichiometric coefficient c in
chemical reaction
• Biomass yield expressed in terms of the
stoichiometric Eq. above is:
Scenario 2: Stoichiometric coefficients
for cell growth + product formation
where f is the stoichiometric coefficient for product
Product synthesis introduces one extra unknown stoichiometric
coefficient to the equation; thus, an additional relationship
between coefficients is required.
Product Stoichiometry (cont’d)
• Product experimentally-determined yield
coefficient, the product yield
Theoretical oxygen demand
OXYGEN DEMAND IS AN IMPORTANT
PARAMETER IN BIOPROCESSING AS
OXYGEN IS OFTEN THE LIMITING
SUBSTRATE IN AEROBIC
FERMENTATIONS.
IMPORTANT TO DESIGN
BIOREACTOR FOR AIR DEMAND
AND COMPRESSOR SIZING.
Theoretical
oxygen
demand
(Cont’d)
If wen use electron balance, we
can get
Rearrange, we can get
Theoretical
oxygen
demand
(Cont’d)
• It means that if we know which
organism ( B) substrate (w and
s) and product (j and p,) are
involved in cell culture, and the
yields of biomass (c) and product
(f), we can quickly calculate the
oxygen demand.
• Alternatively, we could
determine a by solving for all the
stoichiometric coefficients.
Example 4.7 Stoichiometric coefficients for cell growth
Exercise:
• The chemical reaction equation for respiration of
glucose is
• Candida utilis cells convert glucose to CO2 and H20
during growth. The cell composition is
CH1.8400.55N0.2 plus 5% ash. Yield of biomass from
substrate is 0.5 g g- 1. Ammonia is used as nitrogen
source. What is the oxygen demand with biomass
growth compared to that without biomass grwoth?
Example 4.8 Product yield and oxygen demand
Calculate the stoichiometric coefficients of the following
biological reaction:
Hexadecane: C16H34 + a O2 + b NH3 = c (C4.4H7.3N0.86O1.2) +
d H2O + e CO2
Amount of carbon in 1 mole of substrate = 16 (12) = 192 g
Amount of carbon converted to biomass = 192 (2/3) = 128 g
Then, 128 = c (4.4)(12); c = 2.42
Amount of carbon converted to CO2 = 192 – 128 = 64 g
64 = e (12)
e = 5.33
The nitrogen balance yields
14b = c (0.86)(14)
b = (2.42)(0.86) = 2.085
The hydrogen balance is
34 (1) + 3b = 7.3c + 2d
d = 12.43
The oxygen balance yields
2a(16) = 1.2c(16) + 2e(16) + d(16)
a = 12.427.
Summary
• Stoichiometry for biomass growth
only
• Stoichiometry for biomass growth and
product formation
• Calculation of stoichiometric
coefficients
⁻ Elemental balance
⁻ Electron balance
⁻ Determined biomass and product
yields for stoichiometric coefficients
⁻ Determined respiratory quotient for
stoichiometric coefficients
• Theoretical oxygen demand
MASS &
ENERGY
BALANCE
Case study:
Production of Citric Acid by
Aspergillus niger using Cane
Molasses in a Bioreactor
Reference
1) Pauline M. Doran. (1995). Bioprocess Engineering Principle. Sydney, Australia.
Elsevier Science & Technology Books. ISBN: 0122208552.
2) Electronic Journal of Biotechnology, Vol.5 No.3, Issue of December 15, 2002.
ISSN: 0717-3458 by Universidad Católica de Valparaíso ,Chile.
A laboratory scale stirred fermentor of 15-L
capacity having working volume of 9-L was used
for cultivation process and nutritional analysis. The
strain GCBT7 Aspergillus niger, was found to
enhance citric acid production.
• Volume of fermenter: 15 L
• Working volume: 9 L
• pH value = pH 6.0
• Incubation temperature = 30⁰C
• Raw molasses sugar-mainly sucrose (Substrate): 150 g/L
• Fermentation hours: 144 hours.
• Ammonium nitrate (Nitrogen source) = 0.2%= 2g/L
• Maximum Production citric acid (Product) : 99.56% ± 3.5 g/L
• The dry cell mass, A.niger (Biomass): 18.5 g/L
• (Notes: Assume 100% consumption of sugar and N source).
• Yields: 1) Yx/s (Biomass yield from substrate) = 0.123
2) Yp/s (Product yield from substrate) = 0.664
MASS BALANCE
1) Develop the product stoichiometric equation
-assume extracellular product (citric acid)
CwHxOyNz + aO2 + bHgOhNi cCHαOβNδ + dCO2 +
eH2O + fCjHkOlNm
Substrate (Carbon source-sucrose from molasses)
Nitrogen source (Ammonium nitrate)
Biomass (A.niger)
Product (citric acid)
Stoichiometric coefficient elemental balance:
C balance: w = c + d + fj
H balance: x + bg = cα + 2e + fk
O balance: y + 2a + bh = cβ +2d +e + fl
N balance: z + bi = cδ + fm
CwHxOyNz + aO2 + bHgOhNi
cCHαOβNδ + dCO2 + eH2O + fCjHkOlNm
C12H22O11 + aO2 + bNH4NO3 cCH1.72O0.55N0.17 +
dCO2 + eH2O + fC6H8O7
2) Calculate the stoichiometric coefficient balance
C balance : 12 = c + d + 6f
H balance: 22 + 4b = 1.72c + 2e + 8f
O balance: 11 + 2a +3b = 0.55c + 2d + e + 7f
N balance: 2b = 0.17c
Y x/s = c (Mw cells) / (Mw substrate) **
where Yx/s = 0.123
Mw cells = 24.9 + ash (7.5%)
= 24.9 / (1 – 0.075)
= 26.92 g/mol
Mw substrate (sucrose) = 342 g/mol
0.123 = c (26.92 / 342)
c = 1.56
N balance: 2b = 0.17c
b = (1.56 * 0.17) /2
b = 0.133
Yp/s = 0.664
Yp/s = f (Mw product) / (Mw
substrate) **
Mw citric acid = 192 g/mol
Mw sucrose = 342 g/mol
f = 1.15
C balance : 12 = c + d + 6f
d = 12 -1.56 – 6(1.15)
d = 3.54
H balance: 22 + 4b = 1.72c + 2e +
8f
22 + 4(0.133) = 1.72(1.56) + 2e +
8(1.15)
e = (22.944 – 15.58) / 2
e = 5.32
O balance: 11 + 2a +3b = 0.55c + 2d + e + 7f
a = (16.899 - 11.708) / 2
a = 4.95
C12H22O11 + 4.95O2 + 0.133 NH4NO3
1.56 CH1.72O0.55N0.17 + 3.54CO2 + 5.32H2O +
1.15 C6H8O7
C12H22O11 + 4.95O2 + 0.133 NH4NO3
1.56CH1.72O0.55N0.17 + 3.54CO2 + 5.32H2O +
1.15 C6H8O7
1 mol C12H22O11 produces 1.15 mol C6H8O7
1 mol C12H22O11 produces 1.56 mol CH1.72O0.55N0.17
1 mol C12H22O11 produces 3.54 mol CO2
1 mol C12H22O11 produces 5.32 mol H2O
Mol of 100 tonnes of citric acid = 1 x 108 g / Mw citric acid
= 1 x 108 g / 192 g/mol
= 520 833 moles
Estimation of plant capacity: 100 tonnes citric acid/year
Amount of sucrose consumed:
= (mol sucrose / mol citric acid) * total no. of mol
citric acid) * Mw sucrose
= (1 / 1.15) * 520 833 moles * 342 g/mol
= 154.89 tonnes sucrose/year
Amount of O2 consumed:
= (4.95/1.15) * 520 833 moles * 32 g/mol
= 71.74 tonnes O2 /year
Amount of biomass produced:
= (1.56/1) * 452 895 moles * 26.92 g/mol
= 19.02 tonnes biomass / year
Since the biomass is also a
product side, so, use
sucrose as basis i.e.,
Mol of 154.89 tonnes of
sucrose = 1.55 x 108 g / Mw
sucrose
= 1 x 108 g / 342 g/mol
= 452895 moles)
Amount of Ammonium nitrate consumed:
Mw NH4NO3 = 80 g / mol
= (0.133/1.15) * 520 833 moles * 80 g/mol
= 4.82 tonnes NH4NO3 / year
Amount of CO2 produced:
Mw CO2 = 44 g / mol
=(3.54/1) * 452 895 moles * 44 g/mol
= 70.54 tonnes CO2 / year
Amount of H2O produced:
Mw H2O = 18 g / mol
=(5.32/1) * 452 895 moles * 18 g/mol
= 43.37 tonnes H2O/ year
Mass Balance
Molasses
(1032.60 tonnes/year):
15% sucrose=154.89
tonnes/year
85% H2O in=877.71
tonnes/year
Estimation of plant capacity: 100 tonnes citric acid/year
Air
(341.62 tonnes/year):
21%O2= 71.74 tonnes/year
79% N2= 269.88 tonnes/year
Ammonium nitrate:
0.2 % (2g/L) = 4.82
tonnes/year
Biomass (A.niger):
19.02 tonnes/year
Citric acid:
100 tonnes/year
H2O out
(921.08 tonnes/year):
H2O produced= 43.37
tonnes/year
H2O in=877.71
tonnes/year
Off gas
(340.42 tonnes/year):
CO2= 70.54 tonnes/year
N2= 269.88 tonnes/year
Fermenter
Total Mass Balances (MASSin ≈ MASSout):
Stream Mass In
(tonnes/year)
Mass Out
(tonnes/year)
Sucrose 154.89 0
Ammonium nitrate 4.82 0
O2 71.74 0
N2 269.88 269.88
Biomass, A.niger 0 19.02
Citric acid 0 100.00
CO2 0 70.54
Water 877.71 921.08
Total 1379.04 1380.52 *
Note: * some portions of water lost due to evaporation
E NE RGY
BALANCE
Considering the various quantities of materials
involved, their specific heats, and their
changes in temperature or state.
Heat Management in Bioreactors:
Temperature control essential for optimisation of
biomass production or product formation.
Typical cultivation conditions include:
• Small reactors – large surface area to unit
volume ratio – generally require heat addition.
• Large reactors – small surface area to unit
volume ratio – generally require heat removal.
Growth Temp.
(0C)
Species Min. Opt. Max.
Plant cells --- 25 ---
Animal cells --- 37 ---
E. Coli 10 30-37 45
B. Subtilis 15 30-37 55
S. Cerevisiae 0-5 28-36 40-42
General operating temperature of microbes
Heat Balancing
General energy balance can be applied to a bioreactor.
Qacc= Qmet + Qag + Qaer + Qsen - Qevap - Qhxcr - Qsurr
Where…
Qacc – is the accumulated energy in the system (can be positive
or negative in the case of heat loss)
Qmet – Energy generated by metabolism
Qag – Energy generated by agitation (W)
Qaer – Energy generated by aeration (W)
Qsen – Energy generated by condensation (sensible heat)
Qevap – Heat loss to evaporation
Qhxcr – Heat loss to heat exchanger (can be negative or positive)
Qsurr – Heat loss to surrounding environment
We require steady state conditions in
a fermenter, therefore, in
fermentation we require Qacc=0.
If we ignore heat loss to the
surrounding environment (usually
negligible) we can describe the
heat exchanger duty as:
Qhxcr = Qmet + Qag + Qaer + Qsen – Qevap
Energy
Balance
(study
case)
Assumption:
-no shaft work (impeller), Ws=0 (in this example)
-no evaporation, Mv=0
-heat of reaction, ΔHc at 30 °C is -460 kJ gmol-1 O2 consumed
(for aerobic-consider only O2 combustion,
for anaerobic, you have to find ΔHc for each of the reactants
& products).
- Q accumulation, Qacc = 0
- Negligible sensible Heat change, Qsen = 0
Energy balance equation:
For cell metabolism, the modified energy balance equation is:
–ΔHrxn – MvΔhv – Q – Ws = 0
In this case, since Ws= 0; Mv= 0, therefore:
–ΔHrxn – Q = 0
ΔHrxn is related to the amount of oxygen consumed:
ΔHrxn = (-460 kJ gmol-1) * (71740 kg) * (1000g /1kg)
* (1 gmol/ 32 g)
= -1.03 x 1010 kJ
Since;
–ΔHrxn – Q = 0
Q = +1.03 x 1010 kJ / year
(amount of heat that must be removed from the fermenter
per 100 tonnes citric acid produced )
Data from mass balance
Energy Balance
Molasses
(1032.60 tonnes/year):
15% sucrose=154.89
tonnes/year
85% H2O in=877.71
tonnes/year
Estimation of plant capacity: 100 tonnes citric acid/year
Air
(341.62 tonnes/year):
21%O2= 71.74 tonnes/year
79% N2= 269.88 tonnes/year
Ammonium nitrate:
0.2 % = 4.82
tonnes/year
Biomass (A.niger):
19.02 tonnes/year
Citric acid:
100 tonnes/year
H2O out
(921.08 tonnes/year):
H2O produced= 43.37
tonnes/year
H2O in=877.71
tonnes/year
Off gas
(340.42 tonnes/year):
CO2= 70.54 tonnes/year
N2= 269.88 tonnes/year
Fermenter
30 °C
Q= +1.03 x 1010 kJ
Lesson2 Stoichiometry and mass balance.pdf
Lesson2 Stoichiometry and mass balance.pdf

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Lesson2 Stoichiometry and mass balance.pdf

  • 1. Bioreactor design for fermentation - Contents • Lecture 01 – Bioeconomy, fermentation and bioreactors • Lecture 02 – Stoichiometry and mass balance • Lecture 03 – Fermentation basics & Kinetics • Lecture 04 & 05 – Mixing in stirred tank for design • Lecture 06 – Mass transfer for bioreactor design • Lecture 07 – Heat transfer for bioreactor design • Lecture 08 – Bioreactor design
  • 2. Stoichiometry of Growth and Product Formation • For mass balances with reaction the stoichiometry of conversion must be known before the mass balance can be solved. • Under growth conditions, cells are a product of reaction and must be represented in the reaction equation. • Metabolic stoichiometry has many applications in bioprocessing: in mass and energy balances. 1. It can be used to compare theoretical and actual product yields. 2. Check the consistency of experimental fermentation data. 3. Formulating nutrient medium.
  • 3. Growth Stoichiometry and Elemental Balances • Cell growth obeys the law of conservation of matter. • All atoms of carbon, hydrogen, oxygen, nitrogen and other elements consumed during growth are incorporated into new cells or excreted as products. • Confining our attention to those compounds taken up or produced in significant quantity, if the only extracellular products formed are CO2 H2O and we can write the following equation for aerobic cell growth
  • 4. Stoichiometry of biochemical reaction with cell growth and product formation Fermentation Substrate Cells and product Substrate Residual substrate More new cells Products CO2 by respiration
  • 5. Scenario 1: Growth Stoichiometry and elemental balances (e.g. single-cell protein) The stoichiometry of conversion must be known before the mass balance can be solved 𝑤 𝑥 𝑦 𝑧 2 𝑔 𝑏 𝑖 𝛼 𝛽 𝛿 2 2 (Eq. 1) 𝑢 𝑥 𝑦 𝑧 represents the chemical formula for the substrate (e.g. for glucose w=6, x=12, y=6 and z=0) 𝑔 𝑏 𝑖 represents the chemical formula for the nitrogen source 𝛼 𝛽 𝛿 represents the ‘chemical formula’ for dry biomass (biomass compositions) − a, b, c, d, e are stoichiometric coefficients
  • 6. Conversion of substrate, oxygen and nitrogen for cell growth
  • 7. Elemental composition of Escherichia coli bacteria 92%
  • 8. Elemental composition and degree of reduction for selected organisms
  • 9. • Compounds such as vitamins and minerals taken up during metabolism could be included; however, since these growth factors are generally consumed in small quantity we assume here that their contribution to the stoichiometry and energetics of reaction can be neglected. • Other substrates and products can easily be added if appropriate. • Bacteria tend to have slightly higher nitrogen contents (11-14%) than fungi (6.3-9.0%). • For a particular species, cell composition depends also on culture conditions and substrate utilized, hence the different entries in Table 4.3 for the same organism.
  • 10. Elemental composition comparison • The results are remarkably similar for different cells and conditions; CH1.8O0.5N0.2 can be used as a general formula when composition analysis is not available. • The average ‘molecular weight’ of biomass based on C,H, O and N content is therefore 24.6. although 5-10% residual ash is often added to account for those elements not included in the formula.
  • 11. Element balance: Five unknown coefficients, but there are only four equations Notice that we have five unknown coefficients (a, b, c, d and e) but only four balance equations. This means that additional information is required before the equations can be solved. Usually this information is obtained from experiments.
  • 12. One more equation such as RQ is needed, which could be obtained from experiment When Eq above is completed, the quantities of substrate, nitrogen and oxygen required for production of biomass can be determined directly
  • 13. Example: Stoichiometric coefficients for cell growth Production of single-cell protein from hexadecane is described by the following reaction equation: It has been measured that RQ is 0.43 mole CO2 produced per mole O2 consumed. Write the complete reaction equation and calculate oxygen demand.
  • 14. Electron balance for cell growth Definition of available electrons • Available electrons refers to the number of electrons available for transfer to oxygen on combustion of a substrate to CO2, H2O and nitrogen-containing compounds. • The number of available electrons found in organic material is calculated from the valence of the various elements: 4 for C, 1 for H,-2 for O, 5 for P, and 6 for S. The number of available electrons for N depends on the reference state:-3 if ammonia is the reference, 0 for molecular nitrogen N2, and 5 for nitrate.
  • 15. Cont’d Definition of degree of reduction • Degree of reduction, is defined as the number of equivalents of available electrons in that quantity of material containing 1 g atom carbon. • for substrate CwHxOyNz, the number of available electrons is (4w + x- 2y- 3z). The degree of reduction for the substrate, s, is therefore (4w + x - 2y - 3z)/w. • Degree of reduction for CO2, H2O and NH3 is zero.
  • 16. Example: calculate the Degree of reduction for substrates below: Ethanol: C2H5OH Glucose: C6H12O6 Methane: CH4 Glucose is C6H12O6. The degree of reduction = 6 x (+4) + 12 x (+1) + 6 x (-2) = 24 The higher degree of reduction species lower degree of oxidation
  • 17. Electron balance for cell growth (cont’d) • Electrons available for transfer to oxygen are conserved during metabolism. • In a balanced growth equation, number of available electrons is conserved by virtue of the fact that the amounts of each chemical element are conserved.
  • 18. Electron balance for cell growth (cont’d) RQ=measured where s and B are the degrees of reduction of substrate and biomass, respectively. This equation could be used to get coefficient to complete chemical reaction
  • 19. Biomass yield • As cells grow there is, as a general approximation, a linear relationship between the amount of biomass produced and the amount of substrate consumed. • This relationship is expressed quantitatively using the biomass yield, :
  • 20. Biomass yield (cont’d) • When is constant throughout growth, its experimentally determined value can be used to determine the stoichiometric coefficient c in chemical reaction • Biomass yield expressed in terms of the stoichiometric Eq. above is:
  • 21. Scenario 2: Stoichiometric coefficients for cell growth + product formation where f is the stoichiometric coefficient for product Product synthesis introduces one extra unknown stoichiometric coefficient to the equation; thus, an additional relationship between coefficients is required.
  • 22. Product Stoichiometry (cont’d) • Product experimentally-determined yield coefficient, the product yield
  • 23. Theoretical oxygen demand OXYGEN DEMAND IS AN IMPORTANT PARAMETER IN BIOPROCESSING AS OXYGEN IS OFTEN THE LIMITING SUBSTRATE IN AEROBIC FERMENTATIONS. IMPORTANT TO DESIGN BIOREACTOR FOR AIR DEMAND AND COMPRESSOR SIZING.
  • 24. Theoretical oxygen demand (Cont’d) If wen use electron balance, we can get Rearrange, we can get
  • 25. Theoretical oxygen demand (Cont’d) • It means that if we know which organism ( B) substrate (w and s) and product (j and p,) are involved in cell culture, and the yields of biomass (c) and product (f), we can quickly calculate the oxygen demand. • Alternatively, we could determine a by solving for all the stoichiometric coefficients.
  • 26. Example 4.7 Stoichiometric coefficients for cell growth
  • 27.
  • 28. Exercise: • The chemical reaction equation for respiration of glucose is • Candida utilis cells convert glucose to CO2 and H20 during growth. The cell composition is CH1.8400.55N0.2 plus 5% ash. Yield of biomass from substrate is 0.5 g g- 1. Ammonia is used as nitrogen source. What is the oxygen demand with biomass growth compared to that without biomass grwoth?
  • 29.
  • 30. Example 4.8 Product yield and oxygen demand
  • 31. Calculate the stoichiometric coefficients of the following biological reaction: Hexadecane: C16H34 + a O2 + b NH3 = c (C4.4H7.3N0.86O1.2) + d H2O + e CO2 Amount of carbon in 1 mole of substrate = 16 (12) = 192 g Amount of carbon converted to biomass = 192 (2/3) = 128 g Then, 128 = c (4.4)(12); c = 2.42 Amount of carbon converted to CO2 = 192 – 128 = 64 g 64 = e (12) e = 5.33 The nitrogen balance yields 14b = c (0.86)(14) b = (2.42)(0.86) = 2.085 The hydrogen balance is 34 (1) + 3b = 7.3c + 2d d = 12.43 The oxygen balance yields 2a(16) = 1.2c(16) + 2e(16) + d(16) a = 12.427.
  • 32. Summary • Stoichiometry for biomass growth only • Stoichiometry for biomass growth and product formation • Calculation of stoichiometric coefficients ⁻ Elemental balance ⁻ Electron balance ⁻ Determined biomass and product yields for stoichiometric coefficients ⁻ Determined respiratory quotient for stoichiometric coefficients • Theoretical oxygen demand
  • 33. MASS & ENERGY BALANCE Case study: Production of Citric Acid by Aspergillus niger using Cane Molasses in a Bioreactor Reference 1) Pauline M. Doran. (1995). Bioprocess Engineering Principle. Sydney, Australia. Elsevier Science & Technology Books. ISBN: 0122208552. 2) Electronic Journal of Biotechnology, Vol.5 No.3, Issue of December 15, 2002. ISSN: 0717-3458 by Universidad Católica de Valparaíso ,Chile.
  • 34.
  • 35. A laboratory scale stirred fermentor of 15-L capacity having working volume of 9-L was used for cultivation process and nutritional analysis. The strain GCBT7 Aspergillus niger, was found to enhance citric acid production. • Volume of fermenter: 15 L • Working volume: 9 L • pH value = pH 6.0 • Incubation temperature = 30⁰C • Raw molasses sugar-mainly sucrose (Substrate): 150 g/L • Fermentation hours: 144 hours. • Ammonium nitrate (Nitrogen source) = 0.2%= 2g/L • Maximum Production citric acid (Product) : 99.56% ± 3.5 g/L • The dry cell mass, A.niger (Biomass): 18.5 g/L • (Notes: Assume 100% consumption of sugar and N source). • Yields: 1) Yx/s (Biomass yield from substrate) = 0.123 2) Yp/s (Product yield from substrate) = 0.664
  • 36. MASS BALANCE 1) Develop the product stoichiometric equation -assume extracellular product (citric acid) CwHxOyNz + aO2 + bHgOhNi cCHαOβNδ + dCO2 + eH2O + fCjHkOlNm Substrate (Carbon source-sucrose from molasses) Nitrogen source (Ammonium nitrate) Biomass (A.niger) Product (citric acid)
  • 37. Stoichiometric coefficient elemental balance: C balance: w = c + d + fj H balance: x + bg = cα + 2e + fk O balance: y + 2a + bh = cβ +2d +e + fl N balance: z + bi = cδ + fm CwHxOyNz + aO2 + bHgOhNi cCHαOβNδ + dCO2 + eH2O + fCjHkOlNm
  • 38. C12H22O11 + aO2 + bNH4NO3 cCH1.72O0.55N0.17 + dCO2 + eH2O + fC6H8O7 2) Calculate the stoichiometric coefficient balance C balance : 12 = c + d + 6f H balance: 22 + 4b = 1.72c + 2e + 8f O balance: 11 + 2a +3b = 0.55c + 2d + e + 7f N balance: 2b = 0.17c
  • 39.
  • 40.
  • 41. Y x/s = c (Mw cells) / (Mw substrate) ** where Yx/s = 0.123 Mw cells = 24.9 + ash (7.5%) = 24.9 / (1 – 0.075) = 26.92 g/mol Mw substrate (sucrose) = 342 g/mol 0.123 = c (26.92 / 342) c = 1.56
  • 42. N balance: 2b = 0.17c b = (1.56 * 0.17) /2 b = 0.133 Yp/s = 0.664 Yp/s = f (Mw product) / (Mw substrate) ** Mw citric acid = 192 g/mol Mw sucrose = 342 g/mol f = 1.15
  • 43. C balance : 12 = c + d + 6f d = 12 -1.56 – 6(1.15) d = 3.54 H balance: 22 + 4b = 1.72c + 2e + 8f 22 + 4(0.133) = 1.72(1.56) + 2e + 8(1.15) e = (22.944 – 15.58) / 2 e = 5.32
  • 44. O balance: 11 + 2a +3b = 0.55c + 2d + e + 7f a = (16.899 - 11.708) / 2 a = 4.95 C12H22O11 + 4.95O2 + 0.133 NH4NO3 1.56 CH1.72O0.55N0.17 + 3.54CO2 + 5.32H2O + 1.15 C6H8O7
  • 45. C12H22O11 + 4.95O2 + 0.133 NH4NO3 1.56CH1.72O0.55N0.17 + 3.54CO2 + 5.32H2O + 1.15 C6H8O7 1 mol C12H22O11 produces 1.15 mol C6H8O7 1 mol C12H22O11 produces 1.56 mol CH1.72O0.55N0.17 1 mol C12H22O11 produces 3.54 mol CO2 1 mol C12H22O11 produces 5.32 mol H2O Mol of 100 tonnes of citric acid = 1 x 108 g / Mw citric acid = 1 x 108 g / 192 g/mol = 520 833 moles Estimation of plant capacity: 100 tonnes citric acid/year
  • 46. Amount of sucrose consumed: = (mol sucrose / mol citric acid) * total no. of mol citric acid) * Mw sucrose = (1 / 1.15) * 520 833 moles * 342 g/mol = 154.89 tonnes sucrose/year Amount of O2 consumed: = (4.95/1.15) * 520 833 moles * 32 g/mol = 71.74 tonnes O2 /year Amount of biomass produced: = (1.56/1) * 452 895 moles * 26.92 g/mol = 19.02 tonnes biomass / year Since the biomass is also a product side, so, use sucrose as basis i.e., Mol of 154.89 tonnes of sucrose = 1.55 x 108 g / Mw sucrose = 1 x 108 g / 342 g/mol = 452895 moles)
  • 47. Amount of Ammonium nitrate consumed: Mw NH4NO3 = 80 g / mol = (0.133/1.15) * 520 833 moles * 80 g/mol = 4.82 tonnes NH4NO3 / year Amount of CO2 produced: Mw CO2 = 44 g / mol =(3.54/1) * 452 895 moles * 44 g/mol = 70.54 tonnes CO2 / year Amount of H2O produced: Mw H2O = 18 g / mol =(5.32/1) * 452 895 moles * 18 g/mol = 43.37 tonnes H2O/ year
  • 48. Mass Balance Molasses (1032.60 tonnes/year): 15% sucrose=154.89 tonnes/year 85% H2O in=877.71 tonnes/year Estimation of plant capacity: 100 tonnes citric acid/year Air (341.62 tonnes/year): 21%O2= 71.74 tonnes/year 79% N2= 269.88 tonnes/year Ammonium nitrate: 0.2 % (2g/L) = 4.82 tonnes/year Biomass (A.niger): 19.02 tonnes/year Citric acid: 100 tonnes/year H2O out (921.08 tonnes/year): H2O produced= 43.37 tonnes/year H2O in=877.71 tonnes/year Off gas (340.42 tonnes/year): CO2= 70.54 tonnes/year N2= 269.88 tonnes/year Fermenter
  • 49. Total Mass Balances (MASSin ≈ MASSout): Stream Mass In (tonnes/year) Mass Out (tonnes/year) Sucrose 154.89 0 Ammonium nitrate 4.82 0 O2 71.74 0 N2 269.88 269.88 Biomass, A.niger 0 19.02 Citric acid 0 100.00 CO2 0 70.54 Water 877.71 921.08 Total 1379.04 1380.52 * Note: * some portions of water lost due to evaporation
  • 50. E NE RGY BALANCE Considering the various quantities of materials involved, their specific heats, and their changes in temperature or state. Heat Management in Bioreactors: Temperature control essential for optimisation of biomass production or product formation. Typical cultivation conditions include: • Small reactors – large surface area to unit volume ratio – generally require heat addition. • Large reactors – small surface area to unit volume ratio – generally require heat removal.
  • 51. Growth Temp. (0C) Species Min. Opt. Max. Plant cells --- 25 --- Animal cells --- 37 --- E. Coli 10 30-37 45 B. Subtilis 15 30-37 55 S. Cerevisiae 0-5 28-36 40-42 General operating temperature of microbes
  • 52. Heat Balancing General energy balance can be applied to a bioreactor. Qacc= Qmet + Qag + Qaer + Qsen - Qevap - Qhxcr - Qsurr Where… Qacc – is the accumulated energy in the system (can be positive or negative in the case of heat loss) Qmet – Energy generated by metabolism Qag – Energy generated by agitation (W) Qaer – Energy generated by aeration (W) Qsen – Energy generated by condensation (sensible heat) Qevap – Heat loss to evaporation Qhxcr – Heat loss to heat exchanger (can be negative or positive) Qsurr – Heat loss to surrounding environment
  • 53. We require steady state conditions in a fermenter, therefore, in fermentation we require Qacc=0. If we ignore heat loss to the surrounding environment (usually negligible) we can describe the heat exchanger duty as: Qhxcr = Qmet + Qag + Qaer + Qsen – Qevap
  • 54. Energy Balance (study case) Assumption: -no shaft work (impeller), Ws=0 (in this example) -no evaporation, Mv=0 -heat of reaction, ΔHc at 30 °C is -460 kJ gmol-1 O2 consumed (for aerobic-consider only O2 combustion, for anaerobic, you have to find ΔHc for each of the reactants & products). - Q accumulation, Qacc = 0 - Negligible sensible Heat change, Qsen = 0 Energy balance equation: For cell metabolism, the modified energy balance equation is: –ΔHrxn – MvΔhv – Q – Ws = 0 In this case, since Ws= 0; Mv= 0, therefore: –ΔHrxn – Q = 0
  • 55. ΔHrxn is related to the amount of oxygen consumed: ΔHrxn = (-460 kJ gmol-1) * (71740 kg) * (1000g /1kg) * (1 gmol/ 32 g) = -1.03 x 1010 kJ Since; –ΔHrxn – Q = 0 Q = +1.03 x 1010 kJ / year (amount of heat that must be removed from the fermenter per 100 tonnes citric acid produced ) Data from mass balance
  • 56. Energy Balance Molasses (1032.60 tonnes/year): 15% sucrose=154.89 tonnes/year 85% H2O in=877.71 tonnes/year Estimation of plant capacity: 100 tonnes citric acid/year Air (341.62 tonnes/year): 21%O2= 71.74 tonnes/year 79% N2= 269.88 tonnes/year Ammonium nitrate: 0.2 % = 4.82 tonnes/year Biomass (A.niger): 19.02 tonnes/year Citric acid: 100 tonnes/year H2O out (921.08 tonnes/year): H2O produced= 43.37 tonnes/year H2O in=877.71 tonnes/year Off gas (340.42 tonnes/year): CO2= 70.54 tonnes/year N2= 269.88 tonnes/year Fermenter 30 °C Q= +1.03 x 1010 kJ