Mass Balance
Nur Istianah, ST.,MT.,M.Eng
04/03/2018
04/03/2018
Fermentation
04/03/2018
04/03/2018
Fermentation-Chemical reaction
• Example2. Continuous acetic acid fermentation
• Acetobacter aceti bacteria convert ethanol to acetic acid under aerobic conditions.
• A continuous fermentation process for
vinegar production is proposed using non- viable A. aceti cells immobilized on the
surface of gelatin beads. The production target is 2 kg h - 1 acetic acid;
• However the maximum acetic acid
concentration tolerated by the cells is 12%.
• Air is pumped into the fermenter at a rate of 200 gmol h- 1.
• (a) What minimum amount of ethanol is required?
• (b) What minimum amount of water must be used to dilute the ethanol to avoid acid inhibition?
• (c) What is the composition of the fermenter off-gas?
• Solution:
• 1. Assemble
• (i) Flow sheet.
• The flow sheet for this process is shown in Figure 4E5.1.
• (ii) System boUndary.
• The system boundary is shown in Figure 4E5.1.
• (iii) Write down the reaction equation.
• In the absence of cell growth, maintenance or other metabolism ofsubstrate, the reaction equation is:
• C2H5OH + O2 → CH3COOH + H2O
• (ethanol) (acetic acid)
• Analyse
• (i) Assumptions.
• Steady state
• Inlet air is dry
• Gas volume% = mole%
• No evaporation of ethanol, H20 or acetic acid
• Complete conversion of ethanol
• Ethanol is used by the cells for synthesis of acetic acid only; no side-reactions occur
• Oxygen transfer is sufficiently rapid to meet the demands of the cells.
• Concentration of acetic acid in the product stream is 12%.
• (ii) Extra data.
• Molecular weights: ethanol = 46
• acetic acid = 60,. O2=32,. N2=28,. H2O= 18
• Composition of air: 21% O2, 79% N2.
• (iii) Basis.
• The calculation is based on 2 kg acetic acid leaving the system, or 1 hour.
• (iv) Compounds involved in reaction.
• The compounds involved in reaction are
ethanol, acetic acid,O2 and H2O. N 2 is not involved in reaction.
• (v) Mass-balance equations.
• For ethanol, acetic acid, O2 and H2O, the appropriate mass-balance equation is Eq.
(4.2):
• mass in + mass generated = mass out + mass consumed.
• For total mass and N2, the appropriate mass-balance equation is Eq. (4.3):
mass in = mass out.
• Calculate
• (i) Calculation table.
• The mass-balance table with data provided is shown as Table 4E5.1; the units are kg.
EtOH denotes ethanol; HAc is acetic acid.
• If 2 kg acetic acid represents 12 mass% of the product stream, the total mass of the
product stream must be 2/0.12 = 16.67 kg.
• the only components of the product stream are acetic acid and water; therefore water must account for 88 mass% of the product stream = 14.67 kg.
• In order to represent what is known about the inlet air, some preliminary calculations are needed.
• Therefore, the total mass of air in = 5.768 kg. The masses ofO 2 and N 2 can now be entered in the table, as shown.
• E and W denote the unknown quantities of ethanol and water in the feed stream,
respectively; G represents the total mass of off-gas.
• The question marks in the table show which other quantities must be calculated.
• (ii) Mass-balance and staichiametry calculations.
• As N2 is a tie component, its mass balance is straightforward.
1
2
C2H5OH + O2 → CH3COOH + H2O
Mol
• N2 balance
• 4.424 kg N2 in = N2 out.
• .'. N2 out = 4.424 kg.
• To deduce the other unknowns, we must use stoichiometric analysis as well as mass
balances.
• HAc balance
• 0 kg HAc in + HAc generated = 2 kg HAc out + 0 kg HAc consumed.
• .'. HAc generated = 2 kg.
2
C2H5OH + O2 → CH3COOH + H2O
Mol
4
3
• From reaction stoichiometry, we know that generation of 3.333*10-2 kgmol HAc requires 3.333*10-2 kgmol each of EtOH and O2, and is accompanied by generation of 3.333*10-2 kgmol H2O:
• We can use this information to complete the mass balances for EtOH, O2 and H2O.
• EtOH balance
• EtOH in + 0 kg EtOH generated = 0 kg EtOH out + 1.533 kg EtOH consumed.
• .'. EtOH in = 1.533 kg = E.
• O2 balance
• 1.344 kg O2 in + 0 kg O2 generated = O2 out + 1.067 kg O2 consumed.
• .'. O2 out = 0.277 kg.
• Therefore, summing the O2 and N2 components of the off-gas:
• G= (0.277 + 4.424) kg = 4.701 kg.
• H2O balance
• W kg H2O in + 0.600 kg H2O generated - 14.67 kg H2O out + 0 kg H2O consumed.
• .'. W = 14.07 kg.
• These results allow us to complete the mass-balance table, as shown in Table 4E5.2.
• (iii) Check the results.
• All rows and columns of Table 4E5.2 add up correctly.
• Finalize
• (i) The specific questions.
• The ethanol required is 1.533 kg.
• The water required is 14.07 kg.
• The off-gas contains 0.277 kg O2 and 4.424 kg N2.
• Since gas compositions are normally
expressed using volume or mole%, we must convert these values to moles:
• Therefore, the total molar quantity of off-gas is 0.1667 kgmol. The off-gas composition is:
• (ii) Answers.
• Quantities are expressed in kg h-1 rather than kg to reflect the continuous nature of the process and the basis used for
• calculation.
• (a) 1.5 kg h-1 ethanol is required.
• (b) 14.1 kg h-1 water must be used to dilute the ethanol in the feed stream.
• (c) The composition of the fermenter off-gas is 5.2%
O2 and 94.8% N2.
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04/03/2018
