Upload
premkumar164
View
328
Download
8
Embed Size (px)
Citation preview
7/29/2019 Bioprocess Technology Unit I
1/110
Bioprocess Technology
7/29/2019 Bioprocess Technology Unit I
2/110
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 .
7/29/2019 Bioprocess Technology Unit I
3/110
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.
7/29/2019 Bioprocess Technology Unit I
4/110
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.
7/29/2019 Bioprocess Technology Unit I
5/110
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.
7/29/2019 Bioprocess Technology Unit I
6/110
7/29/2019 Bioprocess Technology Unit I
7/110
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.
7/29/2019 Bioprocess Technology Unit I
8/110
(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.
7/29/2019 Bioprocess Technology Unit I
9/110
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.
7/29/2019 Bioprocess Technology Unit I
10/110
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?
7/29/2019 Bioprocess Technology Unit I
11/110
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:
7/29/2019 Bioprocess Technology Unit I
12/110
(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:
7/29/2019 Bioprocess Technology Unit I
13/110
7/29/2019 Bioprocess Technology Unit I
14/110
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.
7/29/2019 Bioprocess Technology Unit I
15/110
7/29/2019 Bioprocess Technology Unit I
16/110
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.
7/29/2019 Bioprocess Technology Unit I
17/110
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.
7/29/2019 Bioprocess Technology Unit I
18/110
7/29/2019 Bioprocess Technology Unit I
19/110
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)
7/29/2019 Bioprocess Technology Unit I
20/110
P d F M t i l B l
7/29/2019 Bioprocess Technology Unit I
21/110
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
7/29/2019 Bioprocess Technology Unit I
22/110
(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
7/29/2019 Bioprocess Technology Unit I
23/110
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
7/29/2019 Bioprocess Technology Unit I
24/110
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
7/29/2019 Bioprocess Technology Unit I
25/110
(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
7/29/2019 Bioprocess Technology Unit I
26/110
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
7/29/2019 Bioprocess Technology Unit I
27/110
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:
7/29/2019 Bioprocess Technology Unit I
28/110
U k fl t t d ith b l
7/29/2019 Bioprocess Technology Unit I
29/110
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%.
7/29/2019 Bioprocess Technology Unit I
30/110
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.
7/29/2019 Bioprocess Technology Unit I
31/110
7/29/2019 Bioprocess Technology Unit I
32/110
(ii) System boundary
7/29/2019 Bioprocess Technology Unit I
33/110
(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
7/29/2019 Bioprocess Technology Unit I
34/110
(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.
7/29/2019 Bioprocess Technology Unit I
35/110
7/29/2019 Bioprocess Technology Unit I
36/110
7/29/2019 Bioprocess Technology Unit I
37/110
Using the result from (2)
7/29/2019 Bioprocess Technology Unit I
38/110
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.
7/29/2019 Bioprocess Technology Unit I
39/110
Finalise
7/29/2019 Bioprocess Technology Unit I
40/110
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.
7/29/2019 Bioprocess Technology Unit I
41/110
(a) What minimum amount of ethanol is required?
7/29/2019 Bioprocess Technology Unit I
42/110
(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.
7/29/2019 Bioprocess Technology Unit I
43/110
7/29/2019 Bioprocess Technology Unit I
44/110
7/29/2019 Bioprocess Technology Unit I
45/110
7/29/2019 Bioprocess Technology Unit I
46/110
(v) Mass-balance equations
7/29/2019 Bioprocess Technology Unit I
47/110
(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.
7/29/2019 Bioprocess Technology Unit I
48/110
If 2 kg acetic acid represents 12 mass% of the
7/29/2019 Bioprocess Technology Unit I
49/110
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.
7/29/2019 Bioprocess Technology Unit I
50/110
N2 balance
7/29/2019 Bioprocess Technology Unit I
51/110
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.
7/29/2019 Bioprocess Technology Unit I
52/110
7/29/2019 Bioprocess Technology Unit I
53/110
H2O balance
7/29/2019 Bioprocess Technology Unit I
54/110
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.
7/29/2019 Bioprocess Technology Unit I
55/110
7/29/2019 Bioprocess Technology Unit I
56/110
Therefore, the total molar quantity of off-gas is 0.1667
7/29/2019 Bioprocess Technology Unit I
57/110
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.
7/29/2019 Bioprocess Technology Unit I
58/110
7/29/2019 Bioprocess Technology Unit I
59/110
7/29/2019 Bioprocess Technology Unit I
60/110
7/29/2019 Bioprocess Technology Unit I
61/110
In mass-balance problems we assume that all
7/29/2019 Bioprocess Technology Unit I
62/110
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
7/29/2019 Bioprocess Technology Unit I
63/110
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.
7/29/2019 Bioprocess Technology Unit I
64/110
7/29/2019 Bioprocess Technology Unit I
65/110
7/29/2019 Bioprocess Technology Unit I
66/110
As illustrated in Figure 4.6, the equation represents a
7/29/2019 Bioprocess Technology Unit I
67/110
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
7/29/2019 Bioprocess Technology Unit I
68/110
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.
7/29/2019 Bioprocess Technology Unit I
69/110
7/29/2019 Bioprocess Technology Unit I
70/110
7/29/2019 Bioprocess Technology Unit I
71/110
7/29/2019 Bioprocess Technology Unit I
72/110
7/29/2019 Bioprocess Technology Unit I
73/110
Example 4.7 Stoichiometric coefficients for cell
7/29/2019 Bioprocess Technology Unit I
74/110
growth
7/29/2019 Bioprocess Technology Unit I
75/110
7/29/2019 Bioprocess Technology Unit I
76/110
7/29/2019 Bioprocess Technology Unit I
77/110
Although elemental balances are useful, the
f i (4 4) bl
7/29/2019 Bioprocess Technology Unit I
78/110
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
7/29/2019 Bioprocess Technology Unit I
79/110
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
7/29/2019 Bioprocess Technology Unit I
80/110
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.
7/29/2019 Bioprocess Technology Unit I
81/110
Note that the available-electron balance is not
i d d t f th l t t f l t l
7/29/2019 Bioprocess Technology Unit I
82/110
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.
7/29/2019 Bioprocess Technology Unit I
83/110
One electron balance, two elemental balances and
d tit till i d t
7/29/2019 Bioprocess Technology Unit I
84/110
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
7/29/2019 Bioprocess Technology Unit I
85/110
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:
7/29/2019 Bioprocess Technology Unit I
86/110
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
7/29/2019 Bioprocess Technology Unit I
87/110
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).
7/29/2019 Bioprocess Technology Unit I
88/110
7/29/2019 Bioprocess Technology Unit I
89/110
7/29/2019 Bioprocess Technology Unit I
90/110
Theoretical Oxygen Demand
O i f h li i i b i bi
7/29/2019 Bioprocess Technology Unit I
91/110
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
7/29/2019 Bioprocess Technology Unit I
92/110
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:
7/29/2019 Bioprocess Technology Unit I
93/110
Maximum Possible Yield
F E (4 15) th f ti l ll ti f il bl
7/29/2019 Bioprocess Technology Unit I
94/110
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.
7/29/2019 Bioprocess Technology Unit I
95/110
Cmax can be converted to a biomass yield with massunits using Eq (4 12)
7/29/2019 Bioprocess Technology Unit I
96/110
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.
7/29/2019 Bioprocess Technology Unit I
97/110
7/29/2019 Bioprocess Technology Unit I
98/110
Eq. (4.20) allows "us to quickly calculate an upperlimit for product yield from the molecular formulae
7/29/2019 Bioprocess Technology Unit I
99/110
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.
7/29/2019 Bioprocess Technology Unit I
100/110
7/29/2019 Bioprocess Technology Unit I
101/110
7/29/2019 Bioprocess Technology Unit I
102/110
(b) Maximum possible biomass yield is given by Eq.(4 19)
7/29/2019 Bioprocess Technology Unit I
103/110
(4.19).
Using the data above, for glucose:
Converting this to a mass basis:
For ethanol:
And
7/29/2019 Bioprocess Technology Unit I
104/110
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.
7/29/2019 Bioprocess Technology Unit I
105/110
7/29/2019 Bioprocess Technology Unit I
106/110
7/29/2019 Bioprocess Technology Unit I
107/110
In Fig 4.5; at least four different system boundaries canbe defined.
7/29/2019 Bioprocess Technology Unit I
108/110
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.
7/29/2019 Bioprocess Technology Unit I
109/110
At the end of this Chapter you should:
(i) understand the terms: system, surroundings, boundary
7/29/2019 Bioprocess Technology Unit I
110/110
( ) 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