1. Mr Zulkarnain, Mr Huzairy & Mr Fahrurrazi
MASS & ENERGY BALANCE
Case study:
Production of Citric Acid by Aspergillus niger using Cane Molasses in a Bioreactor
By:
Mr Zulkarnain, Mr Huzairy & Mr Fahrurrazi

3. 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): 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

4. 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)
Biomass (A.niger)
Nitrogen source (Ammonium nitrate)
Product (citric acid)

5. H balance: x + bg = cα + 2e + fk
CwHxOyNz + aO2 + bHgOhNi cCHαOβNδ + dCO2 + eH2O + fCjHkOlNm
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

6. C12H22O11 + aO2 + bNH4NO3 cCH1.72O0.55N0.17 + dCO2 + eH2O + fC6H8O7
2) Calculate the stoichiometric coefficient balance
C balance : 12 = c + d + 6f
H balance: b = 1.72c + 2e + 8f
O balance: a +3b = 0.55c + 2d + e + 7f
N balance: 2b = 0.17c

9. Y x/s = c (Mw cells) / (Mw substrate) ** where Yx/s = 0.123
Mw cells = ash (7.5%)
= 24.9 / (1 – 0.075)
= g/mol
Mw substrate (sucrose) = 342 g/mol
0.123 = c (26.92 / 342)
c = 1.56

10. Yp/s = f (Mw product) / (Mw substrate) ** Mw citric acid = 192 g/mol
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

11. C balance : 12 = c + d + 6f d = 12 -1. 56 – 6(1. 15) d = 3
C balance : 12 = c + d + 6f d = – 6(1.15) d = 3.54 H balance: b = 1.72c + 2e + 8f (0.133) = 1.72(1.56) + 2e + 8(1.15) e = ( – 15.58) / 2 e = 5.32

12. O balance: 11 + 2a +3b = 0. 55c + 2d + e + 7f a = (16. 899 - 11
O balance: a +3b = 0.55c + 2d + e + 7f a = ( ) / 2 a = 4.95 C12H22O O NH4NO CH1.72O0.55N CO H2O C6H8O7

13. C12H22O O NH4NO3
1.56CH1.72O0.55N CO H2O +
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
= moles
Estimation of plant capacity: 100 tonnes citric acid/year

14. Amount of sucrose consumed:
= (mol sucrose / mol citric acid) * total no. of mol citric acid) * Mw sucrose
= (1 / 1.15) * moles * 342 g/mol
= tonnes sucrose/year
Amount of O2 consumed:
= (4.95/1.15) * moles * 32 g/mol
= tonnes O2 /year
Amount of biomass produced:
= (1.56/1) * moles * g/mol
= tonnes biomass / year
Since the biomass is also a product side, so, use sucrose as basis i.e.,
Mol of tonnes of sucrose = 1.55 x 108 g / Mw sucrose
= 1 x 108 g / 342 g/mol
= moles)

15. Amount of Ammonium nitrate consumed: Mw NH4NO3 = 80 g / mol = (0.133/1.15) * moles * 80 g/mol = 4.82 tonnes NH4NO3 / year Amount of CO2 produced: Mw CO2 = 44 g / mol =(3.54/1) * moles * 44 g/mol = tonnes CO2 / year Amount of H2O produced: Mw H2O = 18 g / mol =(5.32/1) * moles * 18 g/mol = tonnes H2O/ year

16. Mass Balance Estimation of plant capacity: 100 tonnes citric acid/year
Off gas
( tonnes/year):
CO2= tonnes/year
N2= tonnes/year
Molasses
( tonnes/year):
15% sucrose= tonnes/year
85% H2O in= tonnes/year
Biomass (A.niger):
19.02 tonnes/year
Fermenter
Citric acid:
100 tonnes/year
Ammonium nitrate:
0.2 % (2g/L) = 4.82 tonnes/year
H2O out
( tonnes/year):
H2O produced= tonnes/year
H2O in= tonnes/year
Air
( tonnes/year):
21%O2= tonnes/year
79% N2= tonnes/year

17. Mass Out (tonnes/year)
Total Mass Balances (MASSin ≈ MASSout):
Stream
Mass In (tonnes/year)
Mass Out (tonnes/year)
Sucrose
154.89
Ammonium nitrate
4.82 O2
71.74 N2
269.88
Biomass, A.niger
19.02
Citric acid
100.00
CO2
70.54
Water
877.71
921.08
Total *
Note: * some portions of water lost due to evaporation

18. 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.

19. 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 55
S. Cerevisiae
0-5
28-36
40-42

20. Qacc= Qmet + Qag + Qaer + Qsen - Qevap - Qhxcr - Qsurr
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

21. 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

22. 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

23. ΔHrxn is related to the amount of oxygen consumed:
ΔHrxn = (-460 kJ gmol-1) * (71740 kg) * (1000g /1kg) * (1 gmol/ 32 g)
= x kJ
Since;
–ΔHrxn – Q = 0
Q = x kJ / year
(amount of heat that must be removed from the fermenter per 100 tonnes citric acid produced )
Data from mass balance

24. Energy Balance
Estimation of plant capacity: 100 tonnes citric acid/year
Off gas
( tonnes/year):
CO2= tonnes/year
N2= tonnes/year
Q= x kJ
Molasses
( tonnes/year):
15% sucrose= tonnes/year
85% H2O in= tonnes/year
Biomass (A.niger):
19.02 tonnes/year
Fermenter
30 °C
Citric acid:
100 tonnes/year
Ammonium nitrate:
0.2 % = 4.82 tonnes/year
H2O out
( tonnes/year):
H2O produced= tonnes/year
H2O in= tonnes/year
Air
( tonnes/year):
21%O2= tonnes/year
79% N2= tonnes/year

27. Reference
1) Pauline M. Doran. (1995). Bioprocess Engineering Principle. Sydney, Australia. Elsevier Science & Technology Books. ISBN: ) Electronic Journal of Biotechnology, Vol.5 No.3, Issue of December 15, ISSN: by Universidad Católica de Valparaíso ,Chile.

28. THANK YOU