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.

THANKS FOR YOUR

ATTENTION

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04/03/2018