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UNIT 8
STOICHIOMETRY OF BIOPROCESS
The growth of biomass with time is given by
cellsmoreproductslarextracellucellssubstrate ++
The rate of microbial growth is characterized by specific growth rate
dt
dX
X
1net=
net= specific growth rate
X = cell concentration
t = time
net =g - kd where, g=gross specific growth rate,
kd= death rate of cells , if kd=0, net =g
Monod eq: applied for balanced growth, i.e. composition of biomass remains constant and
specific rate of production of each component of culture is equal to .
rZ = z, where Z cellular component(ex: protein, RNA, polysaccharide), rZis volumetric rate of
production of Z, z is concentration of Z
Where m = max specific growth rate,
when S>>Ks, g =m
when S
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Limitations of Monods eq:
applied only for balanced
At extremely low substra
YIELD
True or stoichiometric or
/(mass or moles of reacta
Observed or apparent yie
reactant consumed)
YIELD COEFFICIENT
YX/S = mass or moles o
consumed
YP/S = mass or moles o
YP/X = mass or moles o
YX/O = mass or moles
YCO2/S = mass or mole
RQ = moles of CO2 for
growth
te concentration cannot be applied
theoretical yield = (total mass or mole of produ
nt used to form that particular product)
ld = (mass or moles of product present) /(total
f biomass produced per unit mass or moles of s
product formed per unit mass or moles of subs
f product formed per unit mass or moles of bio
f biomass formed per unit mass or moles of ox
s of CO2 formed per unit mass or moles of subs
ed per mole of O2 consumed
ct formed)
ass or moles of
bstrate
rate consumed
ass formed
gen consumed
trate consumed
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YATP = mass or moles
Ykcal = mass or moles of
Maintenance coefficient :
X
m)dt/ds(m =
Types of maintenance coefficien
D
ATPm
MATP/X
Y
1
APATP/X
Y
1+=
D
2Om
M 2O/XY
1
AP 2O/XY
1+=
D= dilution factor
ELEMENTAL BALANCES (o
C balance: w = c+d
f biomass formed per unit mass or moles of A
iomass formed per Kcal of heat evolved in fer
specific rate of substrate uptake for cellular cel
ts:
)1(
)2(
nly biomass is produced without product for
P formed
entation
l activities.
mation)
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H balance: x+bg = c+2e
O balance: y+2a+bh = c+2d+e
N balance: z+bi=c
RQ=d/a
5 eqs. and 5 unknowns (a,b,c,d,e) solving gives 5 unknowns
Degree of reduction (): Number of equivalents of available electrons per gram atom Carbon.
Available electrons are those that would be transferred to oxygen upon oxidation of a compound
to CO2, H2O, NH3.
for C =4, H=1, N=-3, O=-2, P=5, S=6
Ex: For CH4: [1(4)+4(1)] /1 =8
C6H12O6: [6(4)+12(1)+6(-2)] / 6 =24/6 = 4
C2H5OH: [2(4)+6(1)+1(-2)]/2 = 12/2=6
High degree of reduction means low degree of oxidation CH4> C2H5OH > C6H12O6
wS-4a=c B+fj p-------- (1)
a=1/4 (w S-c B- f j p)
From (1),
Maximum value of stoichiometric coefficient c (all electrons are used for biomass synthesis) is
Sw
pfj
Sw
Bc
Sw
a41
+
+
=
2OtotranferredelectronSw
a4
=
biomasstotranferredelectronSw
Bc=
producttotranferredelectronSw
pfj=
Sw
BcB
=
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& Cmax is given by
Maximum value of stoichiometric coefficient f (no biomass forms) is
Q1. Production of single cell protein from hexadecane is given by the following reaction:
CH1.66O0.27N0.2 represents biomass , If RQ = 0.43,determine the stoichiometric coefficients.
Solution: C balance: 16 = c+d -----(1)
H balance: 34+ 3b = 1.66 c + 2e -----(2)
2a = 0.27c+2d+e ----------(3)
b= 0.2 c ------------(4)
RQ = d/a ------------(5)
Solving (1)-(5),
a=12.48
b= 2.13
c= 10.64
d=5.37
e=11.36
Q2. The respiration of glucose is given by:
O2
eH2
dCO2.0
N27.0
O66.1
cCH2aO34
H16
C +++
B
Sw
maxc
=
Pj
Sw
maxf
=
O2H62CO62O66O12H6C ++
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Candida utiliscells convert glucose to CO2and H2O during growth. The cell composition is
CH1.84O0.55N0.2plus 5% ash. Yield of biomass from substrate is 0.5 g/g. NH3 is used as nitrogen
source.
(a)
Find oxygen demand with growth and without growth.
(b)
if ethanol is used as substrate to produce
(c)
Cells of same composition as above, compare maximum possible biomass yields from
ethanol and glucose.
(d)
Solution: (a) The cell composition is CH1.84O0.55N0.2plus 5% ash. This means 95% of
total weight = 25.44
(e)Mwt. of biomass = 25.44/0.95= 26.78
(f)
B= CH1.84O0.55N0.2= (4x1)+ (1x1.84)-(2x0.55) (3x0.2) =4.14
(g)
s1 = C6H12O6= 4, s 2= C2H5OH = 6
(h)
Yxs = 0.5 g/g =(0.5/26.78)/(1/180)
(i)
= 3.36 gmol/gmol
(j)
a=1/4(wS-cB- f jp), fjp =0 as no products are produced
(k)
=1/4[6(4)-3.36(4.14)] =2.52
(l)
Oxygen demand with growth/without growth = (2.52/6)x100 = 42%
(m)
(b)
C6H12O6as substrate:
Cmax(C6H12O6) = 6(4)/4.14 = 5.8 gmol/gmol
YXS max= (5.8x 26.78)/(1x180)= 0.86 g/g
C2H5OH as substrate:
Cmax(C2H5OH)= 2(6)/4.14 = 2.9 gmol/gmol
B
Sw
maxc
=
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YXS max= (2.9x 26.78)/(1x46)= 1.69 g/g
(Yxs)based on C2H5OH/ (Yxs)based on C6H12O6= 1.69/2.9 = 1.96 2
Q3. Assume that the experimental measurements for a certain organism have shown that the cells
can convert 2/3 of substrate carbon to biomass. (a) Calculate the stoichiometric coefficients offollowing reactions:
(b) Calculate the yield coefficients YX/S, YX/O2 for both reactions
For C16H34, C present = 16x12=192 g
C converted to biomass = 2/3(192) = 128 g
C balance: 128 = c(4.4)(12), c= 2.42
C converted to CO2= 192-128 = 128 g
64 = e(12), e=5.33
N balance: 14 b = c(0.86) (14), b = 2.085
H balance 34(1)+3b =7.3c+2d, d=11.29
O balance: 2a(16)=1.2c(16)+2e(16)+d(16)
a=12.427
For C6H12O6 : C present = 72 g
C converted to biomass = 2/3(72) = 48 g
48=4.4c(12), c=0.909
C converted to CO2 = 72-48=24g
24=12 e, e=2
N balance= 14b = 0.86c(14), c=0.782
H balance: 12+3b=7.3c+2d
d = 3.854
The oxygen balance yields
2eCOO2dH2.1N27.0O66.1cCH3bNH2aO6O12H6C ++++
2eCOO2dH2.1N27.0O66.1cCH3bNH2aO34H16C ++++
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6 (16) + 2(16)a = 1
a = 1.473
b, for hexadecane,
(YX/S)hexadecane> (YX/S)glucose
(YX/O2)glucose> (YX/S)hexadecane
Classification of microbial prod
.2(16)c + 2(16)e + 16d
cts
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1) Growth associated: speci
ex: production of a constitutive e
2) Non growth associated p
zero. qP= = constant
Ex: secondary metabolite produc
3) Mixed growth associate
g +
ex: lactic acid fermentation, a
Luedeking-Piret model
In the eq: qP= g+ ,
If =0, qP= , product is only
If =0, qP= gproduct is only
ic rate of product formation specific growth
nzyme
oduct: takes place in stationary phase and whe
tion of penicillin
product: takes place in slow growth and statio
d xanthan gum production
ongrowth associated
growth associated
rate
growth rate is
ary phases. qP=
8/11/2019 Unit8 Avn Vtu
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A strain of mold was grown in a
Time (h)
0
9
16
23
30
34
36
40
a. Calculate the maximum
b. Calculate the apparent gr
c. C. What maximum ce
with the same size inocul
d. Solution A plot of ln X
MODELS WITH GROWTH IN
batch culture on glucose and the following data
Cell concentration (g/l) Glucose conc
1.25
2.45
5.1
10.5
22
33
37.5
41
100
97
90.4
76.9
48.1
20.6
9.38
0.63
et specific growth rate
owth yield
ll concentration could one except if 50 g of glu
um?
versus t yields a slope of 0.1 h.
IBITORS
were obtained.
entration (g/l)
ose were used
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Substrate Inhibition
The microbial gro
by the substrate
Competitive Subst
Non competitive substrate inhibi
Or if Ks
Product Inhibition
Competitive product inhibition:
Noncompetitive product inhibiti
Example of Non-competitive pr
Ethanol fermentati
th rate at higher substrate concentrations is sai
ate inhibition :
tion:
, then:
n :
duct inhibition
n from glucose
to be inhibited
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Where, Pmis the product concen
where Kpis the product i
Inhibition by tox
Competitive inhibition:
Noncompetitive inhibitio
Uncompetitive inhibition
The net specific rate exp
Where is the death rate
HEAT GENERATION
ration at which growth stops, or
hibition constant.
ic Compounds
n :
:
ession in the presence of death has the followin
constant (h-1)
BY MICROBIAL GROWTH
g form:
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About 40% to 50% energy that i
energy (ATP) during aerobic me
actively growing cells, requires l
growth.
The heat generate during microbof the substrate and of cellular m
utilization of substrate is present
to the sum of the metabolic heat
The above equation is re
The total rate of heat evo
Where VLis the liquid v
the final electron acceptor
stored in a carbon and energy source is conver
tabolism and the rest of the energy is released a
ess maintenance, where the heat evolution is di
ial growth is being calculated using the heat ofaterial. A schematic of an enthalpy balance for
d below as figure. The heat of combustion the
and the heat of combustion of the cellular mate
rranged as to yield
lution in a batch fermentation is
lume (I) and X is the cell concentration (g/l).Si
ted to biological
heat The
ectly related to
he combustionmicrobial
ubstrate is equal
ial.
ce the oxygen is
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Here QGRis in units of kcal/h, while QO2is in millimoles of Q2/h
Metabolic heat released during fermentation can be removed by circulating cooling water
through a cooling coil or cooling jacket in the fermenter. Often, temperature control (adequate
heat removal) is an important limitation on reactor design. The ability to estimate heat-removal
requirements is essential for proper reactor design.
Reference Books
1)
Shuler and Kargi (2004). Bioprocess Engineering:Basic Concepts, 2nded. Prentice Hall.
2)
Doran P.M. (2005). Bioprocess Engineering Principles, 1sted. Academic Press
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