MASS & ENERGY BALANCE
Case study:
Production of citric acid by Aspergillus
niger using cane molasses in a
fermentor
Electronic Journal of Biotechnology ISSN: 0717-3458 Vol.5 No.3, Issue
of December 15, 2002 2002 by Universidad Católica de Valparaíso -Chile Received May 16, 2002 / Accepted November 5, 2002
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.
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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)
Substrate (Carbon source-sucrose from molasses)
CwHxOyNz + aO2 + bHgOhNi
eH2O + fCjHkOlNm
Product (citric acid)
Biomass (A.niger)
cCHαOβNδ + dCO2 +
Nitrogen source (Ammonium nitrate)
CwHxOyNz + aO2 + bHgOhNi
cCHαOβNδ + dCO2 + eH2O + fCjHkOlNm
Stoichiometric coefficient 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
C12H22O11 + aO2 + bNH4NO3
dCO2 + eH2O + fC6H8O7
cCH1.72O0.55N0.17 +
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
Estimation of plant capacity: 100 tonnes citric acid/year
Mol of 100 tonnes of citric acid = 1 x 108 g / Mw citric acid
= 1 x 108 g / 192 g/mol
= 520 833 moles
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
Estimation of plant capacity: 100 tonnes citric acid/year
Off gas
(340.42 tonnes/year):
CO2= 70.54 tonnes/year
N2= 269.88 tonnes/year
Molasses
(1032.60 tonnes/year):
15% sucrose=154.89
tonnes/year
85% H2O in=877.71
tonnes/year
Biomass (A.niger):
19.02 tonnes/year
Fermenter
Ammonium nitrate:
0.2 % (2g/L) = 4.82
tonnes/year
Air
(341.62 tonnes/year):
21%O2= 71.74 tonnes/year
79% N2= 269.88 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
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
877.71
921.08
1379.04
1380.52 *
Water
Total
Note: * some portions of water lost due to evaporation
ENERGY 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.
General operating temperature of microbes
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
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, there 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)
Data from mass balance
= -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 )
Energy Balance
Estimation of plant capacity: 100 tonnes citric acid/year
Off gas
(340.42 tonnes/year):
CO2= 70.54 tonnes/year
N2= 269.88 tonnes/year
Molasses
(1032.60 tonnes/year):
15% sucrose=154.89
tonnes/year
85% H2O in=877.71
tonnes/year
Q= +1.03 x 1010 kJ
Biomass (A.niger):
19.02 tonnes/year
Fermenter
30 °C
Ammonium nitrate:
0.2 % = 4.82
tonnes/year
Air
(341.62 tonnes/year):
21%O2= 71.74 tonnes/year
79% N2= 269.88 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
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