CHAPTER 3

MATERIAL BALANCESMATERIAL BALANCES

Objectives:• Understand the terms: system,

surroundings, boundary and process in thermodynamics;

• Be able to identify open and closed systems, and batch, semi batch, fed-batch and continuous processes;

• Understand the difference between steady state and equilibrium;

• Be able to write appropriate equations for conservation of mass for processes with and without reaction;

• Be able to solve simple mass-balance

problems with and without reaction; and

• Be able to apply stoichiometric principles for macroscopic analysis of cell growth and product formation.

Overview• One of the simplest concepts in process

engineering is a material or mass balance.

• Because mass in biological systems is conversed at all the times, the law of conservation of mass provides the theoretical framework for material balances.

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 come out to what goes in.

3.1 Thermodynamic Preliminaries

• Thermodynamics is a fundamental branch of science dealing with the properties matter.

• Thermodynamic principles are useful in setting up material balances.

System and Process

• In thermodynamics, a system consists of any matter identified for investigation.

Surroundings

System boundary

System

• The system boundary may be real and tangible, such as wall of a beaker fermenter, or imaginary.

• If the boundary does not allow mass to pass from system to surrounding and vice versa, the system is a closed system with a constant mass.

• Conversely, a system able to exchange mass with surroundings is an open system.

Types of Process

Batchoperates in a

closed system

semi-batch allows either input or output of mass,

but not both.

fed-batchallows input of material

to the system but not output

Continuousallows matter to flow

in and out of the system

In batch cultures, a bioreactor is filled with fresh medium and then inoculated.

At the end of the fermentation,the contents are removed for down stream processing othe reactor is then cleanedosterilized and orefilled for the next fermentation

Batch process

Fed batch

With fed batch cultures, fresh media is added to the bioreactor without continuous removal.

When the reactor volume is full, the fermenter is emptied, either partially or completely, and the process restarted

Continuous process

• With continuous cultures– fresh media is continuously added into the

bioreactor and at the same time – bioreactor fluid is continuously removed.

• The cells thus continuously propagate on the fresh medium entering the reactor and at same time, products, metabolic waste products and cells are removed in the effluent.

• Continuous culture reactors need to be shut down less frequently than batch systems. Cells can also be immobilized in the reactor to maximize their retention and thus increase productivity.

• Microbial populations can be maintained in a state of exponential growth over a long period of time by using a system of continuous culture.

Steady state and Equilibrium

•all properties of a system, such as temperature, pressure, concentration, volume, mass, etc do not vary with time. •does not change with time.

steady state

• According to this definition of steady state, batch, fed batch and semi-batch processes cannot operate under steady state conditions.

• Mass of the system is either increasing or decreasing with time during fed-batch and semi batch process.

• In batch processes, even though the total mass in constant, changes occurring inside the system cause the system properties to vary with time.

• Such processes are called transient or unsteady-state processes.

• 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 operation conditions.

• Steady state is an important and useful concept in engineering analysis.

• However, it is often confused with another thermodynamic term, equilibrium.

• A system at equilibrium is one in which all opposing forces are exactly counter-balanced so that the properties of the system do not change with time.

• From the experience we know the system tend to approach an equilibrium condition when they are isolated from their surrounding.

• A equilibrium there is no net change in either the system or the universe.

• Equilibrium implies that there is no net driving force for change.

• The energy of the system is at minimum and in rough terms, the system is ‘static’, ‘unmoving’ or ‘inert’.

• For example, when liquid and vapour are equilibrium in a closed vessel, although there may be constant exchange of molecule between the phases, there is no net change in either the system or the surroundings.

3.2 Law of Conversation of Mass• Mass is conversed in ordinary chemical and

physical processes.

• Consider the system ,operating as a continuous process with input and output streams containing glucose.

Mi Mo

(kg/h glucose) (kg/h glucose)

System

• The mass flow rate of glucose into the system is Mi kg h-1 and the mass flow rate out Mo kg h-1. If Mi and Mo are different there four possible explanations:

Measurement of Mi and Mo are wrong

The system has a leak allowing glucose to enter or escape undetected Glucose is consumed or generated by chemical reaction within the systemGlucose accumulate within the system

• If we assume that the measurement are correct and there are no leaks, the different 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:

• Equation 3.1

• The accumulation term in the above equation can be either positive or negative.

• Negative accumulation represents depletion of pre-existing reserves.

system

within

daccumulate

mass

system

within

consumed

mass

-

system

within

generated

mass

boundaries

system

through

out mass

-

boundaries

system

through

in mass

• Eq. (3.1) is known as general mass-balance equation.

• The mass referred to in the equation can be total mass, mass of a particular molecular or atomic species, or biomass.

Types of Material Balance

Two Types

Continuous processes it is usual to collect information about the process referring

to a particular instant in time.

Amount of mass entering and leaving the system

are specified using a flow rate

Can be used directly Eq. (3.1)

Called a differential balance.

Batch and semi batch processes

Information about these systems is usually collected over a period of time rather than at a particular instant.

Each term of the mass-balance equation in this case is a quantity

of mass, not a rate.

Called an integral balance.

• Two types of balances may be written: – Differential balances, or balances that indicate

what is happening in a system at an instant in time. Each term of the balance equation is a rate (rate of input, rate of generation etc.) and has units of the balanced quantity unit divided by a time unit (people/yr, g SO2 /s, barrels/day).

– This is the type of balance usually applied to a continuous process.

– Integral balances, or balances that describe what happens between two instants of time. Each term of the equation is an amount of the balanced quantity and has the correspond in g unit (people, g SO2, barrels).

– This type of balance is usually applied to a batch process with the two instants of time being the moment after the input takes place and the moment before the product is withdrawn.

• The following rules may be used to simplify the material balance equation: – If the balanced quantity is total mass, set

generation = 0 and consumption = 0. – If the balanced substance is a nonreactive species

(neither a reactant nor a product), set generation = 0 and consumption = 0.

– If a system is at steady state, set accumulation = 0, regardless of what is being balanced. By definition, in a steady-state system nothing can change with time, including the amount of the balanced quantity.

Simplification of the General Mass-Balance Equation

• If a continuous process is at steady state, the accumulation term on the right-hand side must be zero.

• This follows from the definition of steady state: because all properties of the system, including its mass, must be unchanging with time, a system at steady state cannot accumulate mass.

• Eq. (3.1) becomes:

mass in + mass generated = mass out + mass consumed (Eq. 3.2)

• Eq.(3.2) is called the general steady-state mass-balance equation.

• Eq. (3.2) also applies over the entire duration of batch and fed-batch processes» 'mass out in this case is the total mass harvested from the system so that at the end of the process there is no accumulation.

• 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 Eq (3.1) and (3.2) are zero.

• When reaction does not occur, Eq. (3.2) can be further simplified to:

mass in = mass out (3.3)

Example 3.1: General mass-balance equationA continuous process is set up for treatment of wastewater. Each day, 105 kg cellulose and 103 kg bacteria enter in the feed stream, while 104 kg cellulose and 1.5 x 104 kg bacteria leave in the effluent. The rate of cellulose digestion by the bacteria is 7 x 104 kg. The rate of bacteria growth is 2 x 104kg d-1; the rate of cell death by lysis is 5 x 102kgd-1. Write balance for cellulose and bacteria in the system.

Solution Cellulose is not generated by the process, only consumed. Using a basis of 1 day, the cellulose balance in kg from Eq. (3.1) is:

(105 – 104 + 0 - 7 x 104) = accumulationTherefore » 2 x 104 kg cellulose accumulates in the system each day.

Performing the same balance for bacteria:(103 – 1.5 x 104 + 2 x 104 – 5 x 102) = accumulationTherefore 5.5 x 103 kg bacterial accumulates in the system each day.

Material Balances on a Continuous Distillation Process

• One thousand kilograms per hour of a mixture of benzene (B) and toluene (T) containing 50% benzene by mass is separated by distillation into two fractions. The mass flow rate of benzene in the top stream is 450 kg B/h and that of toluene in the bottom stream is 475 kg T/h. The operation is at steady state. Write balances on benzene and toluene to calculate the unknown component flow rates in the output streams.

Integral Balances on Batch Processes

• Ammonia is produced from nitrogen and hydrogen in a batch reactor.

• At time t = 0 there are no mol of NH3 in the reactor, and at a later time tf the reaction terminates and the contents of the reactor, which include nf mol of ammonia, are withdrawn.

• Between to and tf no ammonia enters or leaves through the reactor boundaries, so the general balance equation (Equa tion 3.1) is simply generation = accumulation.

• Moreover, the quantity of ammonia that builds up (accumulates) in the reactor between to and tf is simply nf - no, the final amount minus the initial amount.

• The same reason may be applied to any substance participating in a batch process to obtain

• accumulation = final output - initial input (by definition) = generation - consumption ( Equation 3.1)

• Equating these two expressions for the accumulation yields

• This equation is identical to Eq. 3.2 for continuous steady-state processes, except that in this case the input and output terms denote the initial and final amounts of the balanced substance rather than flow rates of the balanced substance in continuous feed and product streams.

initial input + generation = final output + consumption (3.3)

3.3 Procedure for material-balance calculations• The first step in material-balance calculations

is to understand the problem. Certain information is available about a process; the task is to calculate unknown quantities.

• Material balances should be carried out in an organised manner; this makes the solution easy to follow

The guidelines

• Draw a clear process flow diagram showing all relevant information.A simple box diagram showing all streams entering or leaving the system allows information about a process to be organised and summarised in a convenient way. All given quantitative information should be shown on the diagram.

• Select a set of units and state it clearly.

• Select a basis for the calculation and state in clearly. In approaching mass-balance problem it is helpful to focus on a specific quantity of material entering or leaving the system. Continuous processes at steady state we usually base the calculation on the amount of material entering or leaving the system within a specified period of time.

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.

• State all assumptions applied to the problem.

• Identify which components of the system, if any, are involved in reaction.This is necessary for determining which mass balance equation (3.2) or (3.3) is appropriate. The simpler Eq. (3.3) can be applied to molecular species which are neither reactants nor products of reaction.

Example 3.2 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 gmol min-1 dry oxygen gas. All the water is evaporated. The outflowing gas is found to contain 1 % (w/w) water. Draw and label the flow sheet for this process.

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:

Unknown flow rates are represented with symbols. As shown in Figure, the flow rate of dry air is denoted D g min-1 and the flow rate of humid, oxygen-rich air is H g min-1. The water content in the humid air is shown as 1 mass%.

3.4 Stoichiometry of growth and product formation

• The law of conservation of mass has been used to determine unknown quantities entering or leaving bioprocesses.

• For mass balances with reaction, the stoichiometry of conversion must be known before the mass balance can be solved.

• When cell growth occurs, cells are a product of reaction and must be represented in the reaction equation.

• In this section we will discuss how reaction equations for growth and product synthesis are formulated.

• Metabolic stoichiometry has many applications in bioprocessing: as well as in mass and energy balances, it can be used to compare theoretical and actual product yields, check the consistency of experimental fermentation data, and formulate nutrient medium.

Growth Stoichiometry and Elemental Balances

• Despite its complexity and the thousands of intracellular reactions involved, cell growth obeys the law of conservation of matter.

• All atoms of carbon, hydrogen, oxygen, nitrogen and other elements consumed during growth are incorporated into new cells or excreted as products.

• Confining our attention to those compounds taken up or produced in significant quantity, if the only extracellular products formed are CO2 and H2O, we can write the following equation for aerobic cell growth:

CwHxOyNz + aO2 + bHgOhNi cCHαOβNδ + dCO2 + eH2O

(Eq. 3.4)

• In Eq. 3.4, CwHxOyNz is the chemical formula for the substrate (e.g. for glucose w= 6, x= 12, y= 6 and z= 0) HgOhNi is the chemical formula for the nitrogen sourceCHαOβNδ is the chemical 'formula' for dry biomass. a, b, c, d and e are stoichiometric coefficients.

• The equation is written on the basis of one mole of substrate; therefore a moles O2 are consumed and d moles CO2 are formed per mole substrate reacted, etc.

• As illustrated in Figure 3.1, the equation represents a macroscopic view of metabolism; it ignores the detailed structure of the system and considers only those components which have net interchange with the environment.

Figure 3.1: Conversion of substrate, oxygen and nitrogen for cell growth

• In Eq. (3.4), biomass is represented by the formula CwHxOyNz

• There is no fundamental objection to having a molecular formula for cells, even if it is not widely applied in biology.

• The formula is a reflection of the biomass composition.

• As shown in Table 3.1, microorganisms such as Escherichia coli contain a wide range of elements; however 90-95% of biomass can be accounted for by four major elements: C, H, O and N.

• Eq. (3.4) is not complete unless the stoichiometric coefficients a, b, c, d and e are known.

Table 3.1 Elemental composition of Escherichia coli bacteria

• Once a formula for biomass is obtained, these coefficients can be evaluated using normal procedures for balancing equations, i.e. elemental balances and solution of simultaneous equations.

• C balance: w =c+d (3.5) • H balance: x + bg =cα+2e (3.6) • O balance: y + 2a+bh =cβ +2d+e (3.7) • N balance: z + bi = cδ (3.8)

• Notice that we have five unknown coefficients (a, b, c, d and e) but only four balance equations. This means that additional information is required before the equations can be solved. Usually this information is obtained from experiments. A use ful measurable parameter is the respiratory quotient (RQ):

• When an experimental value of RQ is available, Eqs (3.5) to (3.9) can be solved to determine the stoichiometric coefficients.

• The results, however, are sensitive to small errors in RQ, which must be measured very accurately.

a

d

consumed O moles

produced CO moles RQ Quotient,y Respirator

2

2

• When Eq. (3.4) is completed, the quantities of substrate, nitrogen and oxygen required for production of biomass can be determined directly.

Example 3.3 Stoichiometric coefficients for cel growth

Production of single-cell protein from hexadecane is described by the following reaction equation:

C16H34 + aO2 + bNH3 cCH1.66O0.27N0.20 + dCO2 + eH2O

where CH1.66O0.27N0.20 represents the biomass. If respiratory quotient,RQ = 0.43, determine the stoichiometric coefficients.

C balance: 16 = c+ d (1)

H balance: 34+3 b=1.66c+2e (2)

O balance: 2 a=0.27 c+ 2 d+ e (3)

N balance: b=0.20 c (4)

RQ: 0.43 = d/a (5)

b is already written simply as a function of c in (4); let us try expressing the other variables solely in terms of c. From (1):

d= 16- c. (6)

From (5):

)7.(326.243.0

dd

a

Combining (6) and (7) gives an expressionfor a in terms of c only: a = 2.326 (16 -c)a = 37.22 - 2.326 c (8)

Substituting (4) into (2) gives: 34 + 3 (0.20 c) = 1.66 c+ 2 e 34 = 1.06 c+ 2 ee=17 - 0.53c

Substituting (8), (6) and (9) into (3) gives:

2 (37.22 - 2.326 c) = 0.27 c+ 2 (16 - c) + (17 - 0.53 c) 25.44 = 2.39 cc= 10.64

Using this result for c in (8), (4), (6) and (9) gives:

a = 12.48b = 2.13d = 5.37e = 11.36

Check that these coefficient values satisfy Eq. (1)-(5). The complete reaction equation is:

C16H34 + 12.5O2 + 2.13NH3 10.6CH1.66O0.27N0.20 +5.37CO2 + 11.4H2O

3.5 Recycle, purges and bypasses

• We have performed mass balances on simple single-unit processes. However, steady-state systems incorporating re cycle, by-pass and purge streams are common in bioprocess industries.

Figure 3.2: Flow sheet for process with (a)recycle (b)by-pass and (c) purge streams

• As an example, consider the system use cells as catalysts in fermentation processes.

• It is often advantageous to recycle biomass from spent fermentation broth.

• Cell recycle requires a separation device, such as a centrifuge or gravity settling tank, to provide a concentrated recycle stream under aseptic conditions.

Figure 3.3: Fermenter with recycle

Figure 3.4: System boundaries for cell-recycle system

• System I represents the overall recycle process; only the fresh feed and final product streams cross this system boundary.

• 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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