bioprocess technology unit i

7/29/2019 Bioprocess Technology Unit I

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Bioprocess Technology


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Introduction

Mass balances provide a very powerful tool in

engineering analysis.

Many complex situations are simplified by looking

at the movement of mass and equating what comes

out to what goes in.

Questions such as:

CO2

concentration in the off-gas.

Fraction of substrate consumed not converted or

converted to product.

Amounts of reactants needed to produce (x) grams

of roduct s .


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Needed oxygen for a fermentation process.

All these problems are answered by mass balances

work.

In this chapter will explain

How the law of conservation of mass is applied to

atoms, molecular species and total mass, Set up formal techniques for solving material-

balance problems with and without reaction.

Aspects of metabolic stoichiometry are alsodiscussed for calculation of nutrient and oxygen

requirements during fermentation processes.


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Thermodynamic Preliminaries

Thermodynamics is a branch of science dealingwith

the properties of matter.

Thermodynamic principles are useful in setting up

material balances.

System and Process

In thermodynamics, asystem consists of any matter

identifiedfor investigation.

(Figure 4.1), the system is set apart from the

surroundings, which are the remainder of the

universe, by a system boundary.


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The system boundary may be real and tangible, such

as the walls of a beaker or fermenter, or imaginary.

If the boundary does not allow mass to pass from

system to surroundings and vice versa, the system is

a closedsystem with constant mass.

Conversely, a system able to exchange mass with its

surroundings is an open system.


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A process causes changes in the system or

surroundings.

To describe processes.

(i) A batch process operates in a closed system.

All materials are added to the system at the start of

the process; the system is then closed and productsremoved only when the process is complete.

(ii) A semi-batchprocess allows either input or

output of mass, but not both.

(iii) A fed-batchprocess allows input of material to

the system but not output.


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(iv) A continuous process allows matter to flow in

and out of the system.

If rates of mass input and output are equal,

continuous processes can be operated indefinitely.

Steady State and Equilibrium

If all properties of a system, such as temperature,pressure, concentration, volume, mass, etc. do not

vary with time, the process is said to be atsteady

state.

Thus, if we monitor any variable of a steady-state

system, its value will be unchanging

with time.


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Steady State and Equilibrium

If all properties of a system, such as temperature,

pressure, concentration, volume, mass, etc. do not varywith time, the process is said to be at steady state.

Batch, fedbatch and semi-batch processes cannot

operate under steady-state conditions. Mass of the system is either increasing or decreasing

with time.

Even though the total mass is constant, changesoccurring inside the system cause the system properties

to vary with time.Transient or unsteady-state processes.


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Continuous processes may be either steady state or

transient.

It is usual to run continuous processes as close to

steady state as possible;

However, unsteady-state conditions will exist during

start-up and for some time after any change in

operating conditions.

Steady state differs from equilibrium?


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Law of Conservation of Mass Mass is conserved in ordinary chemical and physical

processes. Consider the system of Figure 4.2 operating as a

continuous process with input and output streams

containing glucose. The mass flow rate of glucose into the system is Mi

kg h-1; the mass flow rate out is Mo kg h-1. If Mi Mo

are different there are four possible explanations:


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(i) Measurements of Mi and Mo are wrong;

(ii) the system has a leak allowing glucose to enter

or escape undetected;

(iii) glucose is consumed or generated by chemical

reaction within the system; or

(iv) glucose accumulates within the system. If we assume that the measurements are correct and

there are no leaks, the difference between Mi and Mo

must be due to consumption or generation by

reaction, and/or accumulation.

A mass balance for the system can be written in a

general way to account for these possibilities:


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The accumulation term in the above equation can be

either positive or negative; negative accumulationrepresents depletion of pre-existing reserves.

Eq. (4.1) is known as thegeneral mass-balance

equation. The mass referred to in the equation can be total

mass, mass of a particular molecular or atomic

species, or biomass.


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Types of Material Balance The general mass-balance equation (4.1) can be applied.

For continuous processes at particular instant of time

amounts of mass entering and leaving the system are

specified using flow rates,

e.g. molasses enters the system at a rate of 50 lb h- 1; atthe same instant in time, fermentation broth leaves at a

rate of 20 lb h-1.

The two quantities can be used directly in Eq. (4.1) asthe input and output terms.

A mass balance based on rates is called a differential

balance.


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For batch and semibatch processes.

Information is collected over a period of time rather.

E.g., 100 kg substrate is added to the reactor; after 3days' incubation, 45 kg product is recovered.

Each term of the mass-balance equation in this case

is a quantity of mass, not a rate. Integral balance.


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If reaction does not occur in the system, or if the

mass balance is applied to a substance that is neither

a reactant nor product of reaction, the generation and

consumption terms in Eqs (4.1) and (4.2) are zero.

At steady state, for balances on total mass or atomic

species or when reaction does not occur, Eq. (4.2)

can be further simplified to: mass in = mass out. (4.3)


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P d F M t i l B l


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Procedure For Material-Balance

Calculations

(i)Draw a clear process flow diagram showing all relevantinformation.

A simple box diagram showing all streams entering or

leaving the system allows information about a process to be

organized and summarized in a convenient way.

All given quantitative information should be shown on the

diagram.

Note that the variables of interest in material balances aremasses, mass flow rates and mass compositions; if

information about particular streams is given using volume

or molar quantities, mass flow rates and compositions

should be calculated before labeling the flow sheet.

(ii) S l f i d i l l


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(ii) Select a set of units and state it clearly.

All quantities are expressed using consistent units.

Units must also be indicated on process diagrams. (iii) Select a basis for the calculation and state it

clearly.

Focus on a specific quantity of material entering orleaving the system.

For continuous processes at steady state we usually

base the calculation on the amount of materialentering or leaving the system within a specified

period of time.

.

F b h i b h i i i


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For batch or semi-batch processes, it is convenient

to use either the total amount of material fed to the

system or the amount withdrawn at the end.

(iv) State all assumptions applied to the problem.

To solve, you will need to apply some 'engineering'

judgments.

Real-life situations are complex, and there will be

times when one or more assumptions are required

before you can proceed with calculations.

The details omitted can be assumed, provided your

assumptions are reasonable.

M ki i h i i


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Making assumptions when an assumption is

permissible and what constitutes a reasonable

assumption is one of the marks of a skilled engineer.

When you make assumptions it is important that you

state them exactly.

Differential mass balances on continuous processes

are performed under steady state conditions; we can

assume that mass flow rate and compositions do not

change with time and the accumulation term of Eq.

(4.1) is zero.

Another assumption is that the system under

investigation does not leak.

( ) Id if hi h f h if


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(v) Identify which components of the system, if any,

are involved in reaction.

This is necessary for determining which mass

balance equation (4.2) or (4.3), is appropriate.

The simpler Eq. (4.3) can be applied to molecular

species which are neither reactants nor products of

reaction.

E l 4 2 S tti fl h t


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Example 4.2 Setting up a flow sheet

Humid air enriched with oxygen is prepared for a

gluconic acid fermentation.

The air is prepared in a special humidifying

chamber.

1.5 L h- 1 liquid water enters the chamber at the same

time as dry air and 15 g mol min- 1 dry oxygen gas.

All the water is evaporated.

The out flowing gas is found to contain 1% (w/w)water.

Draw and label the flow sheet for this process.

S l i


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Solution:

Let us choose units of g and min for this process; the

information provided is first converted to mass flow

rates in these units.

The density of water is taken to be 103 g L-1;

therefore:

As the molecular weight of O2 is 32:


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U k fl t t d ith b l


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Unknown flow rates are represented with symbols.

As shown in Figure 4E2.1, the flow rate of dry air is

denoted D g min- 1 and the flow rate of humid,

oxygen-rich air is Hg min- 1.

The water content in the humid air is shown as l

mass%.


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Material-Balance Worked Examples

Example 4.4 Batch mixing

Corn-steep liquor contains 2.5 % invert sugars and

50% water; the rest can be considered solids.

Beet molasses containing 50% sucrose, 1% invert

sugars, 18% water and the remainder solids, is

mixed with corn-steep liquor in a mixing tank.

Water is added to produce a diluted sugar mixture

containing 2% (w/w) invert sugars.

125 kg corn-steep liquor and 45 kg molasses are fed

into the tank.


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(ii) System boundary


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(ii) System boundary.

The system boundary is indicated in Figure 4E4.1.

2. Analyse (i)Assumptions.

No leaks

No inversion of sucrose to reducing sugars, or anyother reaction

(ii)Extra data.

No extra data are required. (iii)Basis.

125 kg corn-steep liquor.

(iv) Compounds involved in reaction


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(iv) Compounds involved in reaction.

No compounds are involved in reaction.

(v)Mass-balance equation. The appropriate mass-balance equation is Eq. (4.3):

mass in = mass out.

3. Calculate (i) Calculation table.

Table 4E4.1 shows all given quantities in kg.

Rows and columns on each side of the table havebeen completed as much as possible from the

information provided.

Total is denoted P.


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Using the result from (2)


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Using the result from (2)

79.35 kg H2O in = H2O out.

.'. H20 out= 79.35 kg. These results allow the mass-balance table to be

completed, as shown in Table 4E4.2.

(iii) Check the results. All columns and rows of Table 4E4.2 add up

correctly.


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Finalise


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Finalise

(i) The specific questions.

The water required is 8.75 kg. The sucrose concentration in

the product mixture is:

(22.5/178.75)*100= 12.6%

(ii)Answers.

(a) 8.75 kg water is required. (b) The product mixture contains 13% sucrose.

Material balances on reactive systems are slightly more

complicated than Examples 4.3 and 4.4. To solve problemswith reaction, stoichiometric relationships must be used in

conjunction with mass-balance equations.


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(a) What minimum amount of ethanol is required?


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(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 Figure4E5.1.


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(v) Mass-balance equations


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(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.


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If 2 kg acetic acid represents 12 mass% of the


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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.

If we assume complete conversion of ethanol, 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.


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N2 balance


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N2balance

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.


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H2O balance


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2O 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.


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Therefore, the total molar quantity of off-gas is 0.1667


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kgmol. The off-gas composition is:

(ii)Answers.

Quantities are expressed in kg h-1

rather than kg to reflectthe 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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In mass-balance problems we assume that all


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p

oxygen required by the stoichiometric equation is

immediately available to the cells.

Sometimes it is not possible to solve for unknown

quantities in mass balances until near the end of the

calculation.

In such cases, symbols for various componentsrather than numerical values must be used in the

balance equations.

This is illustrated in the integral mass-balance ofExample 4.6 which analyses batch culture of

growing cells for production of xanthan gum.

Depending on which quantities are known and what


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p g q

information is sought, analysis of more than one

system may be required before the flow rates and

compositions of all streams are known.

Mass balances with recycle, by-pass or purge

usually involve longer calculations than for simple

processes, but are not more difficult conceptually.


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As illustrated in Figure 4.6, the equation represents a


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macroscopic view of metabolism; it ignores the

detailed structure of the system and considers only

those components which have net interchange withthe environment.

Compounds such as vitamins and minerals taken up during


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metabolism could be included; however, since these growth

factors are generally consumed in small quantity we assume

here that their contribution to the stoichiometry andenergetics 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 differententries in Table 4.3 for the same organism.

CH1.8 O0.5N0.2 can be used as a general formula when

composition analysis is not available.


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Example 4.7 Stoichiometric coefficients for cell


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growth


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Although elemental balances are useful, the

f i (4 4) bl


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presence of water in Eq. (4.4) causes some problems

in practical application.

Because water is usually present in great excess andchanges in water concentration are inconvenient to

measure or experimentally verify, H and O balances

can present difficulties. Instead, a useful principle is conservation of

reducing power or available electrons, which can be

applied to determine quantitative relationshipsbetween substrates and products.

An electron balance shows how available electrons

from the substrate are distributed in reaction.

Electron Balances


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Electron Balances Available electrons: number of electrons available for transfer to

oxygen on combustion of a substance to CO2, H2O and nitrogen-

containing compounds.

In organic material, it 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.

For N, it depends on the reference state:-3 if ammonia is the reference,

0 for molecular nitrogen N2, and 5 for nitrate.

The reference state for cell growth is usually chosen to be the same as

the nitrogen source in the medium.

It is assumed for convenience that ammonia is used as nitrogen

source; this can easily be changed if other nitrogen sources are

employed.

Degree o f reduction, is the number of equivalents of

available electrons in that quantity of material containing 1


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available electrons in that quantity of material containing 1

g atom carbon.

Therefore, for substrate CwHxOyN z, the number of availableelectrons is (4w + x - 2y - 3z).

The degree of reduction for the substrate, Ys, is therefore

(4w + x - 2y - 3z)/w.

Degrees of reduction relative to NH3 and N2 for several

biological compounds are given in Table B.2 in Appendix

B.

Degree of reduction for CO2, H2O and NH3 is zero.


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Note that the available-electron balance is not

i d d t f th l t t f l t l


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independent of the complete set of elemental

balances; if the stoichiometric equation is balanced

in terms of each element including H and O, theelectron balance is implicitly satisfied.


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One electron balance, two elemental balances and

d tit till i d t


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one measured quantity are still inadequate

information for solution of five unknown

coefficients; another experimental quantity isrequired.

Cells grow, there is, a linear relationship between

the amount of biomass produced and the amount ofsubstrate consumed.

Biomass yield, Yxs:

Factors influences biomass yield; medium

iti t f th b d it


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composition, nature of the carbon and nitrogen

sources, pH and temperature.

Biomass yield is greater in aerobic than in anaerobiccultures; choice of electron acceptor, e.g. O2, nitrate

or sulphate, can also have a significant effect.

When Yxs is constant throughout growth, itsexperimentally-determined value can be used to

determine the stoichiometric coefficient c in Eq.

(4.4). Eq. (4.11) expressed in terms of the stoichiometric

Eq. (4.4) is:


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where MW is molecular weight and 'MW cells'

means the biomass formula-weight plus any residual

ash.

Before applying measured values of Yxs and Eq.

(4.12) to evaluate c, we must be sure that the

experimental culture system is well represented by

the stoichiometric equation.

E. g., we must be sure that substrate is not used to

synthesize extracellular products other than CO2 &

H O.

One complication with real cultures is that some

f ti f b t t d i l d f


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fraction of substrate consumed is always used for

maintenance activities such as (maintenance of

membrane potential and internal pH, turnover ofcellular components and cell motility).

These metabolic functions require substrate but do

not necessarily produce cell biomass, CO2 and H2Oin the way described by Eq. (4.4).


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Theoretical Oxygen Demand

O i f h li i i b i bi


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Oxygen is often the limiting substrate in aerobic

fermentations.

Oxygen demand is represented by the stoichiometric

coefficient a in Eqs (4.4) and (4.13).

Oxygen requirement is related directly to theelectrons available for transfer to oxygen.

Oxygen demand can be derived from an appropriateelectron balance


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electron balance.

When product synthesis occurs as represented

by Eq. (4.13), the electron balance is:

where yp is the degree of reduction of the product.

Rearranging gives:


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Maximum Possible Yield

F E (4 15) th f ti l ll ti f il bl


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From Eq. (4.15) the fractional allocation of available

electrons in the substrate can be written as:

is the fraction of available electrons transferred

from substrate to oxygen.

is the fraction of available electrons transferred tobiomass.

is the fraction of available electrons transferred

to product.


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Cmax can be converted to a biomass yield with massunits using Eq (4 12)


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units using Eq. (4.12).

If we do not know the stoichiometry of growth, we

can quickly calculate an upper limit for biomass

yield from the molecular formulae for substrate andproduct.

If the composition of the cells is unknown, YB can

be taken as 4.2 corresponding to the averagebiomass formula CH1.8O0.5N0.2.


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Eq. (4.20) allows "us to quickly calculate an upperlimit for product yield from the molecular formulae


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limit for product yield from the molecular formulae

for substrate and product.

Example 4.8 Product yield and oxygen demand

The chemical reaction equation for respiration of

glucose is:

Candida utilis cells convert glucose to CO 2 andH2Oduring growth.

The cell composition is CH1.84O0.55N0.2plus 5% ash.

Yield of biomass from substrate is 0.5 g g-1.

Ammonia is used as nitrogen source.


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(b) Maximum possible biomass yield is given by Eq.(4 19)


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(4.19).

Using the data above, for glucose:

Converting this to a mass basis:

For ethanol:

And


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Therefore, on a mass basis, the maximum possible

amount of biomass produced per gram ethanol

consumed is roughly twice that per gram glucose

consumed.

This result is consistent with the data in Table 4.4.


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In Fig 4.5; at least four different system boundaries canbe defined.


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System I represents the overall recycle process; only the

fresh feed and final product streams cross this systemboundary.

In addition, separate material balances can be

performed over each process unit: the mixer, the

fermenter and the settler.

Other system boundaries could also be defined; for

example, we could group the mixer and fermenter, or

settler and fermenter, together.

Material balances with recycle involve carrying out

individual mass-balance calculations for each

designated system.


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At the end of this Chapter you should:

(i) understand the terms: system, surroundings, boundary


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( ) y , g , y

and process in thermodynamics;

(ii) be able to identify openand closed systems, and batch,semibatch, fed-batch and continuous processes;,

(iii) understand the difference between steady state and

equilibrium;

(iv) be able to write appropriate equations for conservation

of mass for processes with and without reaction;

(v) be able to solve simple mass-balance problems with and

without reaction; and (vi) be able to apply stoichiometric principles for

macroscopic

bioprocess technology unit i

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