unit8 avn vtu

8/11/2019 Unit8 Avn Vtu

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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=


10/14

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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