CHAPTER 3CHAPTER 3.

Microbial KineticsMicrobial Kinetics

3.Microbial Kinetics

• Microorganisms fuel their lives by performing redox reaction that generate the energy and reducing power needed to maintainthat generate the energy and reducing power needed to maintainand construct themselves.

• Because redox reactions are nearly always very slow unless catalyzed, microorganisms produce enzyme catalysts that increase the kinetics ofmicroorganisms produce enzyme catalysts that increase the kinetics oftheir redox reactions to exploit chemical resources in their environment.

• Engineers want to take advantage of these microbially catalyzed reactionsg g y ybecause the chemical resources of the microorganisms usually are the pollutants that the engineers must control :

Organics (BOD, e- donor), NH4+ (e- donor, nutrient),

NO3- (e- acceptor nutrient) PO4

3- (nutrient) etcNO3 (e acceptor, nutrient), PO4 (nutrient), etc.

3.Microbial Kinetics

• Engineers who employ microorganisms for pollution controlmust recognize two interrelated principles (connections betweeng p p (

the active biomass (the catalyst) and the primary substances):

First : Active microorganisms catalyze the pollutant removing reactions.S th t f ll t t l d d th t ti fSo the rate of pollutant removal depends on the concentration ofthe catalyst, or the active biomass.

Second : The active biomass is grown and sustained through the utilizationof its energ and electron (e ) generating primar s bstratesof its energy - and electron (e-) - generating primary substrates.So the rate of biomass production is proportional to the utilizationrate of the primary substrates.rate of the primary substrates.

3.1 Basic Rate Expressions

• Mass – balance modeling is based on i) the active biomass and ii) the primary substrate that limits the growth rate of the biomass.

• In the vast majority of cases the rate limiting substrate is e- donor• In the vast majority of cases, the rate limiting substrate is e donor.So the term substrate now refers to the primary e- donor substrate.

• Jacques Monod : Nobel prize winner, director of Pasteur Institute.

Monod equation : relationship most frequently used to represent bacterial growth kinetics developed in the 1940s.

: His original work related the specific growth rate of fast – growing bacteria to the concentration of a rate limiting e- donor substratelimiting, e donor substrate

.

3.1 Monod equationMonod eq : largely empirical, but wide spread applied for microbial systems

1 ∧ SdX ⎟⎞

⎜⎛

]1.3[1 ∧

SKS

dtdX

Xsyn

a

asyn +

=⎟⎟⎠

⎞⎜⎜⎝

⎛= μμ

synμ

aX

= specific growth rate due to synthesis(T-1)

= concentration of active biomass(MxL-3)a

tS

concentration of active biomass(MxL )

= concentration of the rate limiting substrate (M L-3)

= Time (T)

S

μ∧

= concentration of the rate-limiting substrate (MsL-3)

= maximum specific growth rate(T-1)μK

g ( )

= substrate concentration giving one-half the maximum rate(M L-3)(MsL )

3.1 Monod equation

]13[)1(∧ SdX

Compare eq. 3.1 with eq. 1.13

]1.3[)1(SK

SdtdX

X syna

asyn +

== μμ

[1.13]SK

SvvM

m += Michaelis-Menten equation :

- developed in 1913th f ti d ki ti- theory of enzyme action and kinetics

: Reaction velocity

: Substrate concentration giving

vMK

• Microorganisms are “bags full of enzymes” so that

one-half of the maximum velocity

• Microorganisms are bags full of enzymes so that it is not surprising that the growth rate of microorganisms (Monod eq.)is related to the reactions of the catalysts that mediate many reactions(Michaelis and Menten eq.)

3.1 Basic Rate Expressions

Endogenous decay:1) Environmental engineers study more slowly growing bacteria that has an energy ) g y y g g gy

demand for maintenance.2) Active biomass has an energy demand for maintenance of cell functions ( motility repair and re-synthesis osmotic regulation transport and heat loss)( motility, repair and re synthesis, osmotic regulation, transport and heat loss).

3) Flow of energy and electrons are required to meet maintenance needs.4) The cell oxidize themselves to meet maintenance- energy needs.

]2.3[-1 bdtdX

Xa

dec =⎟⎟⎠

⎞⎜⎜⎝

⎛=μ

dtXdecaya⎟⎠

⎜⎝

b = endogenous-decay coefficient(T-1)b

decμ

g y ( )

= specific growth rate due to decay(T-1)

3.1 Endogenous Decay

]2.3[-1 bdtdX

Xdecay

a

adec =⎟⎟

⎠

⎞⎜⎜⎝

⎛=μ

• Oxidation decay rate : Although most of the decayed biomassis oxidized, a small fraction accumulates as inert biomass

y

]3.3[-1 bfdtdX

X ddecay

a

aresp =⎟⎟

⎠

⎞⎜⎜⎝

⎛=μ

• The rate at which active biomass is converted to inert biomass

df : fraction of active biomass that is biodegradable

The rate at which active biomass is converted to inert biomass

respdecinert μμμ -=11 dXdX ⎟

⎞⎜⎛

]4.3[)1()(11 bfbfbdtdX

XdtdX

X ddinert

a

a

i

a

−−=−−−=⎟⎟⎠

⎞⎜⎜⎝

⎛=−

: inert biomass concentration(M L-3)iX : inert biomass concentration(MxL 3)iX- inert biomass formed from active biomass decay-inert biomass represents a fraction of only the organic portion of active biomass

3.1 Net specific growth rate

The net specific growth rate of active biomass(μ) is the sum of g ( )new growth and decay

1 ∧ SdX ]5.3[-1 ∧

bSK

SdtdX

X decsyna

a +=+== μμμμ

3.1 Net specific growth rate

Maximum specific growth rate

Fig re3 1 schematic of ho the s nthesisFigure3.1 schematic of how the synthesis and net specific growth rates depend on the substrate concentration.concentration.At S=20K, = o.95synμ μ

∧

S∧

If S =o, then b

bSK

Sdecsyn

−=+

=+=

0

-∧

μμμμ

3.1 Rate of substrate utilization• The ultimate interest is to remove substrate

]63[∧

XSq ]6.3[aut XSK

qr+

−=

utr = rate of substrate utilization (M L-3T-1)ut

∧

q

rate of substrate utilization (MsL T )

= maximum specific rate of substrate utilization (MsMx-1T-1)

• Substrate utilization and biomass growth are connected by

∧∧

]7.3[Yq=μ

3.1 Net rate of active-biomass growth

]7.3[=∧∧

Yqμ

The net rate of cell growth :

(T-1) = (MsMx-1T-1) (MxMs

-1) = (T-1)

]5.3[-∧

bSK

S+

= μμ

aanet bXXSK

SXar - ∧

+== μμ

]8.3[-∧

aanet bXXSKSqYr+

=S

netr = the net rate of active-biomass growth (MxL-3T-1)

3.1 Basic Rate Expressions

• . with connected is μnetr∧

Sqr ]9.3[- bSKSqY

Xr

a

net

+==μ

Some prefer to think of cell maintenance as being a shunting ofSome prefer to think of cell maintenance as being a shunting of substrate - derived electrons and energy directly for maintenance. This is expressed by [3.10].

]10.3[)-(∧

mSKSqY+

=μ

p y [

m = maintenance-utilization rate of substrate (MsMx-1 T-1)

SK +

When systems go to steady state, there is no difference in the two approach,and b= Ym

3.1 Basic Rate Expressions

]9.3[-∧

bSKSqY

Xr

a

net

+==μ

m = maintenance-utilization rate of substrate (MsMx-1 T-1)

= b/y

a

)/(

)-(∧

mdtdSY

mSKSqY

−−

=

+=μ

)( mX

Ya

=

bdtdS<

− /At starvation state, the substrate utilization rate is less than m

0<

At steady state, the net specific growth rate of cell (μ), or the net yield (Yn) is zero. Y

mXa

=< 0<μ

0=μ

bmdtdS==

− /

, X a = const.

- Any growth of biomass (Ym) only supplements the decay of biomass (b).So apparently the concentration of biomass does not change

Ym

Xa

So apparently the concentration of biomass does not change.- The energy supplied through substrate utilization is just equal to m, the maintenance ene

3.2 PARAMETER VALUES

3.2 PARAMETER VALUES---Y• Aerobic heterotrophs:

• Denitrifying heterotrophs:

• H2-Oxidizing sulfate reducers:

Next slide

3.2 PARAMETER VALUES---Y• Experimental estimation of Y

- A small Inoculum is grown to exponential phase and harvestedfrom batch growth. Measure the changes in biomass and substrate

f i l til th ti f h ticonc. from inoculum until the time of harvesting.- The true yield is estimated from

/

- The batch technique is adequate for rapidly growing cells,

SXY ΔΔ= /-

q q p y g g ,but can create errors when the cells grow slowly so that biomass decay can not be neglected.

3.2 PARAMETER VALUES -- q

is controlled largely by e- flow to the electron acceptor.For 20 oC the maximum flow to the energy reaction is about 1 e- eq / gVSS d

∧

q

0∧

For 20 oC, the maximum flow to the energy reaction is about 1 e eq / gVSS – d, where 1 g VSS represents 1g of biomass.

]11.3[/ 0∧∧

fqq =

CatdVSSeqeaboutqe0- 20-/1=

]11.3[/ ee fqq

∧∧

For example (Table 3.1), if fs0 = 0.7, then fe0 = 0.3. So

functionuretemperatis∧

q

/273.0//8/ 0∧∧

dgVSSBODgdVSSBODgfqq LLee −=−==

)(∧∧ RTT

]12.3[)07.1(= )20-(20

∧∧T

T qqfunctionure temperatis q

]13.3[)07.1( )-( R

RTT

TT qq =

3.2 PARAMETER VALUES--q

b is depends on species type and temperature.b = 0 1- 0 3 /d for aerobic heterotrophsb = 0.1- 0.3 /d for aerobic heterotrophs

b < 0.05 /d for slower-growing species

R )-()07.1(R

RTT

TT bb =

Biodegradable fraction (fd) is quite reproducible and has a value near 0.8 for a wide range of microorganisms

3.2 PARAMETER VALUES---k

]8.3[-∧

aanet bXXSKSqYr = aanet SK+

• K is the Monod half-maximum-rate concentration most variable and least predictable parameterIts value can be affected by the substrate`s affinity for transportor metabolic enzymes

mass-transport resistances (approaching of substrate andmass-transport resistances (approaching of substrate and microorganisms to each other) ignored for suspended growth are lumped into the Monod kinetics by an increase in K

3.3 Basic Mass BalancesChemostat: a completely mixed reactor, uniform and steady concentrations

of active cell, substrate, inert biomass and any other constituents.

• Active biomass

[ ]1430 .QXVX aa −= μ

• Substrate

( ) [ ]( ) [ ]1530 0 .SSQVrut −+=

Q = constant feed flow rateQ constant feed flow rateSo = feed substrate concentration

3.3 Basic Mass Balances

• Active biomass

[ ]1430 .QXVX aa −= μ

[ ]Sq̂

[ ]93 .bSKSq̂YXr anet −+

==μ

[ ]1630 .QXVbXVXSKSqY aaa −−+

=

1QVb ⎟⎟⎠

⎞⎜⎜⎝

⎛+

[ ]1831

.

QVb

QVq̂Y

QKS

⎟⎟⎞

⎜⎜⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛+−⎟⎟

⎠

⎞⎜⎜⎝

⎛

⎟⎠

⎜⎝=

QQq ⎟

⎠⎜⎝

⎟⎠

⎜⎝

⎟⎠

⎜⎝

3.3 Basic Mass Balances

• Substrate

( ) [ ]1530 0 .SSQVrut −+=

( ) [ ]0Sq̂

[ ]63.XSKSq̂r aut +

−=

( ) [ ]1730 0 .SSQVXSKSq

a −++

−=

[ ]163.fromVXSq̂QXVbXa

aa =+

( ) [ ]10

[ ]fVSKY a+

( ) [ ]1931

10 .

QVb

SSYX a⎟⎟⎠

⎞⎜⎜⎝

⎛+

−=

Q ⎠⎝

3.3 HRT & SRT

• HRT (Hydraulic detention time)

[ ]20.3/)(det QVTtimeentionhydraulic ==θ

[ ]2131 V/QD)T(tdil ti −

SRT (Solids retention time) = MCRT (Mean cell residence time)

[ ]2131 .V/QD)T(ratedilution ==

• SRT (Solids retention time) = MCRT (Mean cell residence time)= Sludge age

[ ]1systemtheinbiomassactive [ ]22.31−== μθχ biomassactiveofrateproductionsystemtheinbiomassactive

]5.3[-1 ∧

bSK

SdtdX

X decsyna

a +=+== μμμμ

3.3 SRT ( )χθ

[ ]1systemtheinbiomassactive [ ]2231 .biomassactiveofrateproductionsystemtheinbiomassactive −== μθχ

• SRT : Physiological status of the system: Information about the specific growth rate of the microorganisms

3.3 SRT ( )χθ

• SRT with quantitative parameters

in chemostat at steady state[ ]23.31VVXa ==== θθ θθ = in chemostat at steady state[ ]23.3DQQXa

θθχ

⎞⎛

θθ χ =

[ ]1831

1.

QVb

QVq̂Y

QVb

KS

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛+−⎟⎟

⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛+

= ( ) [ ]2431

1.

bq̂Yb

KSχχ

χ

θθθ+−

+=

VQQ ⎟⎠

⎜⎝ ⎠⎝⎠⎝

χθ== θQV

( ) [ ]1931

10 .Vb

SSYX a⎟⎟⎞

⎜⎜⎛

−= [ ]2531

0

.bSSYXa ⎟⎟⎠

⎞⎜⎜⎝

⎛

+−

=θ1

Qb ⎟⎟

⎠⎜⎜⎝

+ 1 b ⎟⎠

⎜⎝ + χθ

3.3 SRT ( )χθ

• S and Xa are controlled by SRT( )χθ

( ) [ ]24.31ˆ

1 χ

θθθbqY

bKS

++

= ( )1 χχ θθ bqY +−

[ ]2531

0

.bSSYXa ⎟⎟⎠

⎞⎜⎜⎝

⎛

+−

=θ1 b ⎟

⎠⎜⎝ + χθ

3.3 SRT ( )minχθ

0,:)1 0 == aXSSsmallveryat χθ

min: χχ θθ <washout

minχθ : value at which washout beginsχθ

• by letting S = S0 in eq 3.24 and solving for χθ

( ) [ ]2431

1.

bq̂Yb

KSχχ

χ

θθθ+−

+=

( ) [ ]26.3ˆ0

0min

bKbqYSSK−−

+=χθ ( ) bKbqYS

.)(,/ constVQthenQV =↓↑=== χχ θθθ

3.3 SRT ( )limmin ][ χθ

( ) [ ]26.3ˆ0

0min

bKbYSSK +

=χθ ( )0 bKbqYS −−χ

decreases with increasing S0minχθ

( )ˆ][ 0

0

limmin lim bKbqYS

SK +=χθ ( )

( ) /ˆ1/

0

0

lim SbKbqYSK

bKbqYS

o

o

s

s

−−+

=

−−

∞→

∞→

( )

]27.3[ˆ1

/

bqY

SbKbqYs

−=

∞→

limmin ][ χθ defines an absolute minimum (or maximum ) boundary

for having steady-state biomass.χθ μ

for having steady state biomass. It is a fundamental delimiter of a biological process.

3.3 SRT ( )minχθ

2) For all > , S declines monotonically with increasing

minχθχθ

χθ

.)(,/ constVQthenQV =↓↑=== χχ θθθ

3.3 SRT (Smin ),χθ3) For very large S approaches another key limiting value, S min .

1 θb( ) [ ]24.31ˆ

1

χχ

χ

θθθbqY

bKS

+−+

=

( )bqYb

KSx +−

+=

∞→ 1ˆ1

limminχ

θ θθθ( )

bqYb

K

bqY

−−+

=

+

/1ˆ/1

lim

1

χ

χχ

θθ

θθ

bqYbK

bqYx

=

∞→

ˆ

/1 χθ θ

bqY −

.)(,/ constVQthenQV =↓↑=== χχ θθθ

3.3 SRT (Smin)

• S is the minimum concentration capable of supporting steady state biomass• Smin is the minimum concentration capable of supporting steady-state biomass.

- If S < Smin , the cells net specific growth rate is negative.

- In eq 3.9, S 0, then - b .

∧

μ

]9.3[-∧

bSKSqY

Xr

a

net

+==μ

- Biomass will not accumulate or will gradually disappear.

- Therefore, steady-state biomass can be sustained only when S > Smin .

3.3 SRT (Xa variation)4) When > , Xa rises initially,

because Q(S0 – S) increases as becomes larger. However X reaches a maximum and declines

minχθχθ

χθ- However, Xa reaches a maximum and declines

as decay becomes dominant (Q(S0 – S) decreases) for large (small Q)

- If were to extend to infinity X would approach zero (eq 3 25)

χθθ- If were to extend to infinity, Xa would approach zero (eq 3.25)χθ

[ ]2531

0

.bSSYX a ⎟⎟⎠

⎞⎜⎜⎝

⎛

+−

=χθ ⎠⎝

3.3 Microbiological Safety Factor

minχθχθ5) We can reduce S from So to Smin as we increase from to infinity.

- The exact value of we pick depends on a balancing of χθsubstrate removal (S0 – S), biomass production (= Q Xa), reactor volume (V)and other factors.

- In practice, engineers often specify a microbiological safety factor, p , g p y g y ,defined by / .

- Typical safety factor = 5 ~ 100

χθ minχθ

Typical safety factor 5 100 .

3.4 Mass Balances on Inert Biomass and Volatile Solids

0X

• Because some fraction of newly synthesized biomass is refractory to self-oxidation, endogenous respiration leads to the accumulation of inactive biomass ( ).

• Real influents often contain refractory volatile suspended solids ( )( ) VbXf ad−1

iXReal influents often contain refractory volatile suspended solids ( )that we cannot differentiate easily from inactive biomass.

• Steady-state mass balance on inert biomass

Formation rate of inert biomass( ) ( ) [ ]0

from active biomass decay( ) ( ) [ ]29310 0 .XXQVbXf iiad −+−=

iX = concentration of inert biomass= input concentration of inert biomass0

iX

( ) [ ]

QV=θ

( ) [ ]30.310 θbfXXX daii −+=

3.4 Mass Balances on inert Biomass and Volatile Solides

( ) [ ]30.310 θbfXXX daii −+=

inert biomass formed from active biomass decayInfluent inert

- inert biomass formed from active biomass decay-inert biomass represents a fraction of only the organic portion of active biomass

0iX

⎞⎛

By introducing , we are relaxing the original requirement that only substrate enters the influent.

From 3.25 and [ ]25.31

0

⎟⎟⎠

⎞⎜⎜⎝

⎛

+−

=χθbSSYX a,χθθ =

( )( ) χθbfSSYXX d−−+=

100 ( )χθb

SSYXX ii ++

1

3.4 Maximum of inert biomass ( ) iX

( )( ) θbfd−100 ( )( )χ

χ

θθ

bbf

SSYXX dii +

−+=1

100

00max li )1( bfdSSYXX ⎟

⎞⎜⎛ −

+ θ0max lim )1(1

bfdb

YXX ii

⎟⎞

⎜⎛

⋅⋅−⎟⎟⎠

⎜⎜⎝ +

+=∞→χ

χχθ

θθ

00lim )1(

11 fdSSYX i −⎟⎟⎟⎟⎞

⎜⎜⎜⎜⎛

−+=

∞→θ

( )00max lim33)1(

1

SSFigfromfdSSYXX

b

=−−+=

⎟⎠

⎜⎝

+⋅

∞→χ

χ

θ

θ

( ) minmin lim3.3)1( SSFigfromfdSSYXX ii =−−+=∞→χθ

3.4 Mass Balances on inert Biomass and Volatile Solides

( ) )1(min00max fdSSYXX ii −−+=

great accumulation of inert biomass

X increases monotonicall from X 0 to a ma im m X max- Xi increases monotonically from Xi0 to a maximum Xi

max

- Operation at a large θ results in greater accumulation of inert biomass

3.4 Xv: volatile suspended solids concentration (VSS)

Xv = Xi + Xa

Xv: volatile suspended solids concentration(VSS)

))1(1(

)1(0

0χ

θ

θ

bfdXX

XabfdXXX

i

aiv

−++=

+−+=

]31.3[1

])1(1[)(

))1(1(0

0 χ

χ

θθ

θ

bbfdSSY

X

bfdXX

i

ai

+−+⋅−

+=

++

1 χθb+

3.4 Xv: volatile suspended solids concentration (VSS)

Xv generally follow the trend of X but it does not equal zero;- Xv generally follow the trend of Xa, but it does not equal zero;(When Xa goes to zero, Xv equals Xi )

- if θ < θ min X = X 0 S = S0- if θx< θx , Xv = Xi , S = S

3.4 Yn: net yield or observed yield (Yobs)

−++= bfXXXXfrom

daaiv )1(]31.3[

0χθ

Net accumulation of biomass from

−+⋅−+=

bbfSSY

X di 1

])1(1[)( 00 χ

θθ

Net accumulation of biomass fromSynthesis and decay.

⎞⎛

><−+=

+

uitionbySSYX

b

ni

i

int)(

100

χθ [ ]2531

0

.bSSYXa ⎟⎟⎠

⎞⎜⎜⎝

⎛

+−

=χθ

Yn: net yield or observed yield (Yobs)

)1(1 bf θ+ ]32.3[1

)1(1

x

xdn b

bfYYθ

θ⋅+

⋅⋅−+⋅=∴

[3.32] is parallel to the relationship between fs and fs0:

]333[)1(10 xd bfff θ⋅⋅−+=∴ ]33.3[

1 xss bff

θ⋅+⋅=∴

3.5 Soluble Microbial Products

SMP(soluble microbial products)

• Much of the soluble organic matter in the effluent from a biological reactor is of microbial origin.

• It is produced by the microorganisms as they degrade the organic substrate in the influent to the reactor.

• The major evidence for this phenomenon comes from experiments ;

“single soluble substrates of known composition were fed to microbial culturessingle soluble substrates of known composition were fed to microbial cultures and the resulting organic compounds in the effluent were examined forthe presence of the influent substrate.”

“The bulk of the effluent organic matter was not the original substrate and was of high molecular weight, suggesting that it was of microbial origine.”

• It is called “ Soluble Microbial Products (SMP)”.

3.5 Soluble Microbial ProductsSMP: Biodegradable, although some at a very low rate

: Moderate formula weights(100s to 1000s): The majority of the effluent COD and BOD j y: They can complex metals, foul membranes and cause color or foaming: They are thought to arise from two processes:

1) growth associated (UAP) , 2) non-growth associated (BAP)1) growth associated (UAP) , 2) non growth associated (BAP)SMP = UAP + BAP

UAP ( substrate-utilization-associated products )• growth associated• they are generated from substrate utilization and biomass growththey are generated from substrate utilization and biomass growth• they are not intermediates of catabolic pathways

BAP ( biomass-associated products )BAP ( biomass associated products )• non-growth associated• related to decay and lysis of cell• release of soluble cellular constituents through lysis andrelease of soluble cellular constituents through lysis and solubilization of particulate cellular components

3.5 SMP vs. EPS

- SMP School focuses on i) SMP, ii) active biomass and iii) inert biomass,but does not include EPS.

- EPS School focuses on i) EPS (soluble and bound) and ii) active biomass,but does not include SMP.

- Compared to other aspects of biochemical operations, little research has been done on the production and fate of soluble microbial products. So little is known about the characteristics of UAP and BAP.

3.5 UAP and BAP formation kinetics

- UAP formation kinetics :

r k rrUAP = - k1 rut

rUAP = rate of UAP-formation (MpL-3T-1)UAP ( )k1 = UAP- formation coefficient (MpMs

-1)

f- BAP formation kinetics :

rBAP = k2 XrBAP = k2 Xa

rBAP = rate of BAP-formation (MpL-3T-1)1k2 = BAP- formation coefficient (MpMx-1)

3.5 Soluble Microbial Products

The BOD exertion equation :

Example 3.3 Effluent BOD5

kt}-exp{BOD-BOD [3.41]kt})-exp{-(1BODBOD

LL

Lt

==

]42.3[}5exp{1:5 dkBODBOD L ⋅−−=

original substrate 0.23/d 0.68

k BOD5:BODL ratio

gactive biomass 0.1/d 0.40 SMP 0.03/d 0.14

3.6 Nutrients and Electron Acceptors

Bi t h l i l P t id ffi i t t i t d• Biotechnological Process must provide sufficient nutrients and electron acceptors to support biomass growth and energy generation.

• Nutrients, being elements comprising the physical structure of the cells, are needed in proportion to the net production of biomass.

• Active and inert biomass contain nutrients, as long as they are produced microbiologically.

• e- acceptor is consumed in proportion to e- donor utilization multiplied by the sum of exogenous and endogenous flows of e- to the terminal acceptor.

3.6 How to determine nutrient requirements

- Nutrients are needed in proportion to the net production of biomass (rut Yn)

- In chemostat model, rate of nutrients consumption (rn)

utnn Ynrr γ=

, p ( n)

Y = (MxMs-1)

]43.3[1

)1(1

x

xdutn b

bfYrθ

θγ+−+

=

rn = rate of nutrient consumption (MnL-3T-1) (negative value)

γ = the stochiometric ratio of nutrient mass to VSS for the biomass [M M -1]γn the stochiometric ratio of nutrient mass to VSS for the biomass [MnMx ]

γN = 14g N/113g cell VSS = 0.124g N/g VSS < C5H7O2N >γP = 0 2 * γN =0 025 g P/g VSSγP = 0.2 γN =0.025 g P/g VSS

]323[])1(1[ xd bfYY θ⋅⋅−+ ]32.3[]1

)([x

xdn b

fYYθ⋅+

⋅=

3.6 Nutrient requirements- Overall mass balance

]44.3[0 0 VrQCQC nnn +−=

]453[0 θrCC += ]45.3[θnnn rCC +=

0C : Influent concentration of nutrient ( M L-3)

nCnC : Influent concentration of nutrient ( Mn L )

: Effluent concentration of nutrient ( Mn L-3)

* The supply of nutrients must be supplemented if is negative in eq 3 45C

nr : rate of nutrient consumption (MnL-3T-1) (negative value)

* The supply of nutrients must be supplemented if is negative in eq 3.45.nC

3.6 Use rate of Electron Acceptors

• Use rate of electron acceptor [ΔSa/ Δ t (MaT-1)]

mass balance on electron equivalents expressed as oxygen demand (OD)- mass balance on electron equivalents expressed as oxygen demand (OD)

]46.3[42.1 00 QXgVSSgCODQSinputsOD v+= See table 2.1g

]47.3[42.1)( QXgVSSgCODSMPQQSoutputsOD v++=

- the acceptor consumption as O2 equivalents , ΔSa/ Δ t = OD inputs – OD outputs

[ ] ]48.3[)(42.1 00vva

a XXSMPSSQtS

−+−−=ΔΔ

γ

γa = 1g O2 /g COD for oxygenγa = 0.35g NO3

- - N/g COD for NO3- - N

tΔ

γa 3 3

O2 (32g) + 4 e- 2 O-2

NO3- (14g) + 5 e- N 0

14g NO3 –N /5e- / 32g O2/4e- = 0.35 g NO3 –N / g O2

3.6 Amoumt of e- acceptors to be supplied

- the acceptor can be supplied in the influent flow or by other means,such as aeration to provide oxygen ( ).aRp yg ( )

[ ] ]49.3[0aaa

a RSSQtS

+−=ΔΔ

a

[ ] aaatΔ

Ra: the required mass rate of acceptor supply (M T-1)

Acceptor consumption rate

Ra: the required mass rate of acceptor supply (Ma T-1)Sa

0 : Influent concentration of e- acceptor ( Ma L-3) Sa : effluent concentration of e- acceptor ( Ma L-3)

3.7 Input Active Biomass

- Three circumstances for inputs of active biomass in degradation of the substrate :the substrate :

• When processes are operated in series, the downstream process often receives biomass from the upstream process.p p

• Microorganisms may be discharged in a waste stream or grown in the sewers.

• Bioaugmentation is the deliberate addition of microorganisms to improve performance.

- When active biomass is input, the steady-state mass balance for active biomass (3.16), must be modified.

Sq̂

]503[ˆ

0 0 VbXVXSqYQXQX +=

[ ]1630 .QXVbXVXSKSqY aaa −−+

=

]50.3[0 VbXVXSK

YQXQX aaaa −+

+−=

3.7 Input Active BiomassˆS

• Eq 3 50 can be solved for S when active biomass is input

]50.3[ˆ

0 0 VbXVXSKSqYQXQX aaaa −+

+−=

( ) ]51.3[ˆ

1 xbKS θ+=

• Eq 3.50 can be solved for S when active biomass is input,

( ) ][1ˆ xx bqY θθ +−

• The solution for s (eq 3.51) is exactly the same as eq 3.24,

( ) [ ]24.31ˆ

1

χχ

χ

θθθbqY

bKS

+−+

=

xθbut the is now redefined.

- When the SRT is redefined as the reciprocal of the net specific growth rate,

1 VX ]52.3[01

aa

ax QXQX

VX−

== −μθ

0aX3.7 HOW does affect S and washout ?

•

- = 0, washout occurs for θx of about 0.6 d.

- As Increases, complete washout is eliminated, because the reactor always

0aX

0aX

contains some biomass.

- Increasing , also makes S lower, and the effect is most dramatic near washout.

a

0aX

3.7 Input Active Biomass

]50.3[ˆ

0 0 VbXVXSKSqYQXQX aaaa −+

+−=For biomass,

( ) [ ]1730 0 .SSQVXSKSq̂

a −++

−=For substrate,

For inert, ( ) ( ) [ ]29310 0 .XXQVbXf iiad −+−=

• Due to the change in definition of θx, solutions of the mass balances differ.

))1(1(0χθbfdXXX aiv −++=

]53.3[1

1)( 0⎥⎦

⎤⎢⎣

⎡+

−=x

xa b

SSYXθθ

θ

]54.3[)1(0 θadaii bXfXXX −+=

]55.3[))1(1(0 θbfXXXXX daiaiv −++=+=

]31.3[))1(1(0χθbfdXXX aiv −++=When Xa

0 = 0,

3.7 Input Active Biomass• Due to the change in definition of θx, solutions of the mass balances differ.

]53.3[1

1)( 0⎥⎤

⎢⎡

−= xa b

SSYXθθ

θ

• θx/θ ratio represents the build up of active biomass by the input.

][1

)( ⎥⎦

⎢⎣ + x

a b θθ

Rearranging Eq. 3.52,

]56.3[1x =θ

- Xa0 < Xa: θx > θ

]56.3[/1 0

aa XX−θ

- Xa0 > Xa: θx is negative.

The net biomass decay which makes the steady-state μ negative and sustain S < Smin. If θx is negative, S < Smin from eqs below

bqYb

KS−−

+=

/1ˆ/1

limmin

χ

χ

θ θθ

( )bqYb

KS+−

+=

χχ

χ

θθθ

1ˆ1

bqYbK

bqx

−=

∞→

ˆ

/ χθ θ( )

bqYb

K

q

−−+

=χ

χ

χχ

θθ

/1ˆ/1

3.8 Hydrolysis of Particulate and Polymeric Substrates

• Organic substrate of particles (large polymers)

O i b t t th t i i t ti l l l t k- Organic substrates that originate as particles or large polymer take important application in environmental biotechnology

• Effect of hydrolysis

B f th b t i b i th id ti ti th ti l- Before the bacteria can begin the oxidation reaction, the particles or large polymers must be hydrolyzed to smaller molecules

Extracellular enzymes catalyze these hydrolysis reaction

- Hydrolysis kinetics is not settled, because the hydrolytic enzymes t il i t d ith ti l t th ti bi

- Extracellular enzymes catalyze these hydrolysis reaction

are not necessarily associated with or proportional to the active biomass

-level of hydrolytic enzymes is not established, and measurement i ith i lis neither simple

3.8 Hydrolysis of Particulate and Polymeric Substrates

• Hydrolysis of kinetics

- Reliable approach is a first-order relationship

]573[Skr −=

* rhyd: rate of accumulation of particulate substrate due to hydrolysis

]57.3[phydhyd Skr −=

(MsL-3T-1)* Sp : concentration of the particulate (or large-polymer) substrate (MsL-3)

* k : first order hydrolysis rate coefficient (T-1)

- khyd is proportional to the concentration of hydrolytic enzymes as well as to the intrinsic hydrolysis kinetics of the enzymes

khyd : first-order hydrolysis rate coefficient (T 1)

to the intrinsic hydrolysis kinetics of the enzymes

- some researchers include the active biomass concentration as part of khyd

k X ( ifi h d l i t ffi i t)'k 'kkhyd = Xa ( = specific hydrolysis rate coefficient)hydk hydk

3.8 Hydrolysis of Particulate and Polymeric Substrates

- When Equation 3.57 is used , the steady-state mass balance on particulate substrate

]14.3[)(0 0 SSQVrut −+=

]59.3[1

0

θp

p kS

S+

=]58.3[)(0 0 VSkPSQ phydpp −−=1 θhydk+

pypp

- represent the liquid detention time and should not be substituted byθ xθ x

3.8 Hydrolysis of Particulate and Polymeric Substrates

• The steady-state mass balance on soluble substrate

]173[)(ˆ

0 0 SSQVXSq−+−=

- The steady-state mass balance on substrate

]17.3[)(0 SSQVXSK a +

+

]603[//ˆ

)(0 0 QVSkQVXSqSS

- The steady-state mass balance on soluble substrate

]60.3[//)(0 0 QVSkQVXSKSqSS phyda ++

−−=

* S0 effectively is increased by QVSk phyd /

3.8 Hydrolysis of Particulate and Polymeric Substrates

• Other constituents of particulate substrate also are conserved d i h d l iduring hydrolysis

- Good examples of particulate substrate : the nutrient nitrogen, phosphate, sulfur

]61.3[phydnhydn Skr γ=

* rhydn: rate of accumulation of a soluble form of nutrient n by hydrolysis(M L-3T-1)(MnL T )

nγ* : stoichiometric ratio of nutrient n in the particulate substrate( )1−snMM

3.8 Hydrolysis of Particulate and Polymeric Substrates

▣ Example 3.6

3.8 Hydrolysis of Particulate and Polymeric SubstratesY of casein (pp. 170)

]53.3[1

1)( 0

θθ

θx

xa b

SSYX ⎥⎦

⎤⎢⎣

⎡+

−=x ⎦⎣

]54.3[)1(0 θadii bXfXX −+=

1

]'55.3[]55.3[ paivaiv SXXXXXX ++=→+=

COD’/weight = the conversion factor (pp 129 181)

2

COD’/weight = the conversion factor (pp. 129, 181)

3

The biodegradable fraction = 0.8 (pp. 171)