Microbial Kinetics - SNUwemt.snu.ac.kr/lecture 2013-2/env/2013 Ch 3 Microbial...

CHAPTER 3.

Microbial Kinetics

3.Microbial Kinetics

• Microorganisms fuel their lives by performing redox reaction that 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 oftheir redox reactions to exploit chemical resources in their environment.

• Engineers want to take advantage of these microbially catalyzed reactionsbecause 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), etc.

3.Microbial Kinetics

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

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

First : Active microorganisms catalyze the pollutant removing reactions.So 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 energy - and electron (e-) - generating primary substrates.So the rate of biomass production is proportional to the utilizationrate 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.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 substrate

.

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

]1.3[1 ∧

SKS

dtdX

Xsyn

a

asyn

syn

aX

t S

∧

K

3.1 Monod equation

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

= concentration of active biomass(MxL-3)

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

= Time (T)

= maximum specific growth rate(T-1)

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

]1.3[)1(∧

SKS

dtdX

X syna

asyn

3.1 Monod equation

[1.13]SK

SvvM

m Michaelis-Menten equation :

- developed in 1913- theory of enzyme action and kinetics

• 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.)

Compare eq. 3.1 with eq. 1.13

: Reaction velocity

: Substrate concentration givingone-half of the maximum velocity

vMK

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

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).

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

Xdecay

a

adec

b

dec


= endogenous-decay coefficient(T-1)

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

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

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

]2.3[-1 bdtdX

Xdecay

a

adec

]3.3[-1 bfdtdX

X ddecay

a

aresp

respdecinert μμμ -=

]4.3[)1()(11 bfbfbdtdX

XdtdX

X ddinert

a

a

i

a

df

3.1 Endogenous Decay

: inert biomass concentration(MxL-3)iX

: fraction of active biomass that is biodegradable

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

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

]5.3[-1 ∧

bSK

SdtdX

X decsyna

a

3.1 Net specific growth rate

Figure3.1 schematic of how the synthesis and net specific growth rates depend on the substrate concentration.At S=20K, = o.95

Maximum specific growth rate

3.1 Net specific growth rate

syn ∧

If S =o, then b

bSK

Sdecsyn

0

-∧

• The ultimate interest is to remove substrate

• Substrate utilization and biomass growth are connected by

]6.3[∧

aut XSKSqr

utr∧

q

]7.3[∧∧

Yq

3.1 Rate of substrate utilization

= rate of substrate utilization (MsL-3T-1)

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

]8.3[-∧

aanet bXXSKSqYr

netr

3.1 Net rate of active-biomass growth

aanet bXXSK

SXar - ∧

]5.3[-∧

bSK

S

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

The net rate of cell growth :

]7.3[=∧∧

Yqμ(T-1) = (MsMx

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

•

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

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

. with connected is netr

]9.3[-∧

bSKSqY

Xr

a

net

]10.3[)-(∧

mSKSqY


Some 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].

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

= b/y

]9.3[-∧

bSKSqY

Xr

a

net

)/(

)-(∧

mXdtdSY

mSKSqY

a


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

0

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

Ybm

XdtdS

a

/

- The energy supplied through substrate utilization is just equal to m, the maintenance ene

Ybm

XdtdS

a

/

, X a = const.

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

0

3.2 PARAMETER VALUES

3.2 PARAMETER VALUES---Y

• H2-Oxidizing sulfate reducers:

• Aerobic heterotrophs:

• Denitrifying heterotrophs:

Next slide

3.2 PARAMETER VALUES---Y

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

conc. from inoculum until the time of harvesting.- The true yield is estimated from

- The batch technique is adequate for rapidly growing cells, but can create errors when the cells grow slowly so that biomass decay can not be neglected.

SXY /-

• Experimental estimation of Y

3.2 PARAMETER VALUES -- q

]13.3[)07.1( )-(∧∧ R

RTT

TT qq

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

∧∧T

T qq

]11.3[/ 0∧∧

ee fqq

CatdVSSeqeaboutqe0-

∧

20 -/1

function ure temperatis ∧

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, where 1 g VSS represents 1g of biomass.

∧

q

/273.0//8/ 0∧∧

dgVSSBODgdVSSBODgfqq LLee

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

3.2 PARAMETER VALUES--q

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

b < 0.05 /d for slower-growing species

)-()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

• K is the Monod half-maximum-rate concentration most variable and least predictable parameter Its value can be affected by the substrate`s affinity for transport

or metabolic enzymes mass-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

]8.3[-∧

aanet bXXSKSqYr

3.3 Basic Mass Balances

Q = constant feed flow rateSo = feed substrate concentration

Chemostat: a completely mixed reactor, uniform and steady concentrationsof active cell, substrate, inert biomass and any other constituents.

• Active biomass

• Substrate

1430 .QXVX aa

1530 0 .SSQVrut

• Active biomass

1430 .QXVX aa

1630 .QXVbXVXSKSq̂Y aaa

1831

1.

QVb

QVq̂Y

QVb

KS

93 .bSKSq̂YXr anet


• Substrate

1530 0 .SSQVrut

1730 0 .SSQVXSKSq̂

a

1931

10 .

QVb

SSYX a

63.XSKSq̂r aut

163.fromVXSKSq̂

YQXVbX

aaa


• HRT (Hydraulic detention time)

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

20.3/)(det QVTtimeentionhydraulic

2131 .V/QD)T(ratedilution

22.31 biomassactiveofrateproductionsystemtheinbiomassactive

3.3 HRT & SRT

]5.3[-1 ∧

bSK

SdtdX

X decsyna

a

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

2231 .biomassactiveofrateproductionsystemtheinbiomassactive

3.3 SRT ( )

• SRT with quantitative parameters

in chemostat at steady state 23.31DQ

VQXVX

a

a

1831

1.

QVb

QVq̂Y

QVb

KS

1931

10 .

QVb

SSYX a

2431

1.

bq̂Yb

KS

2531

0

.bSSYXa

3.3 SRT ( )

QV

• S and Xa are controlled by SRT( )

3.3 SRT ( )

24.31ˆ

1

bqY

bKS

2531

0

.bSSYXa

26.3ˆ0

0min

bKbqYSSK

0,:)1 0 aXSSsmallveryat

min: washout

3.3 SRT ( )

2431

1.

bq̂Yb

KS

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

min : value at which washout begins

min

.)(,/ constVQthenQV

]27.3[ˆ1

/ˆ1/

ˆ][

0

0

0

0

limmin

lim

lim

bqY

SbKbqYSK

bKbqYSSK

o

o

s

s

3.3 SRT ( )

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

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

26.3ˆ0

0min

bKbqYSSK

decreases with increasing S0min

limmin ][

3.3 SRT ( )2) For all > ,

S declines monotonically with increasing

min

min

.)(,/ constVQthenQV

3.3 SRT (Smin ),3) For very large S approaches another key limiting value, S min .

24.31ˆ

1

bqY

bKS

bqYbK

bqYb

K

bqYb

KS

x

x

ˆ

/1ˆ/1

lim

1ˆ1

limmin

.)(,/ constVQthenQV

3.3 SRT (Smin)

• 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 .

- Biomass will not accumulate or will gradually disappear.

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

]9.3[-∧

bSKSqY

Xr

a

net

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

because Q(S0 – S) increases as becomes larger. - 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, Xa would approach zero (eq 3.25)

min

2531

0

.bSSYX a

3.3 Microbiological Safety Factor

min5) 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 substrate removal (S0 – S), biomass production (= Q Xa), reactor volume (V)and other factors.

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

- Typical safety factor = 5 ~ 100 .

min

3.4 Mass Balances on Inert Biomass and Volatile Solids

• Steady-state mass balance on inert biomass

Formation rate of inert biomassfrom active biomass decay 29310 0 .XXQVbXf iiad

30.310 bfXXX daii

iX

0iX

QV

= concentration of inert biomass= input concentration of inert biomass

• 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 ( )that we cannot differentiate easily from inactive biomass.

VbXf ad1

0iX

3.4 Mass Balances on inert Biomass and Volatile Solides

30.310 bfXXX daii

0iX

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

From 3.25 and 25.31

0

bSSYX a

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

, =

bbf

SSYXX dii

1100

3.4 Maximum of inert biomass ( ) iX

minmin00max

00

00max

lim

lim

lim

3.3)1(

)1(11

)1(1

SSFigfromfdSSYXX

fd

b

SSYX

bfdbSSYXX

ii

i

ii

bbf

SSYXX dii

1100

great accumulation of inert biomass

3.4 Mass Balances on inert Biomass and Volatile Solides

- Xi increases monotonically from Xi0 to a maximum Xi

max

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

)1(min00max fdSSYXX ii

Xv = Xi + Xa

Xv: volatile suspended solids concentration(VSS)

]31.3[1

])1(1[)(

))1(1(

)1(

00

0

0

bbfdSSY

X

bfdXX

XabfdXXX

i

ai

aiv

3.4 Xv: volatile suspended solids concentration (VSS)

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

- if x< xmin, Xv = Xi

0, S = S0

3.4 Xv: volatile suspended solids concentration (VSS)

uitionbySSYX

bbfSSY

X

bfXXXXfrom

ni

di

daaiv

int)(

1])1(1[)(

)1(]31.3[

00

00

0

Yn: net yield or observed yield (Yobs)

]32.3[1

)1(1

x

xdn b

bfYY

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

]33.3[1

)1(10

x

xdss b

bfff

Net accumulation of biomass fromSynthesis and decay.

2531

0

.bSSYXa

3.4 Yn: net yield or observed yield (Yobs)

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 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 : 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)SMP = UAP + BAP

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

BAP ( biomass-associated products )• non-growth associated• related to decay and lysis of cell• release 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 :

rUAP = - k1 rut

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

-1)

- BAP formation kinetics :

rBAP = k2 Xa

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

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

LL

Lt

The BOD exertion equation :

]42.3[}5exp{1:5 dkBODBOD L

original substrate 0.23/d 0.68 active biomass 0.1/d 0.40 SMP 0.03/d 0.14

k BOD5:BODL ratio

3.5 Soluble Microbial ProductsExample 3.3 Effluent BOD5

3.6 Nutrients and Electron Acceptors

• 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.

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

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

N = 14g N/113g cell VSS = 0.124g N/g VSS < C5H7O2N >P = 0.2 * N =0.025 g P/g VSS

]43.3[1

)1(1

x

xdutn

utnn

bbfYr

Ynrr

3.6 How to determine nutrient requirements

]32.3[]1

)1(1[x

xdn b

bfYY

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

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

Y = (MxMs-1)

- Overall mass balance

]44.3[0 0 VrQCQC nnn

]45.3[0 nnn rCC

3.6 Nutrient requirements

nC

0nC : Influent concentration of nutrient ( Mn L-3)

: Effluent concentration of nutrient ( Mn L-3)

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

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

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

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

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

a = 1g O2 /g COD for oxygena = 0.35g NO3

- - N/g COD for NO3- - N

]46.3[42.1 00 QXgVSSgCODQSinputsOD v

]47.3[42.1)( QXgVSSgCODSMPQQSoutputsOD v

]48.3[)(42.1 00vva

a XXSMPSSQtS

3.6 Use rate of Electron Acceptors

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

See table 2.1

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

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)

]49.3[0aaa

a RSSQtS

3.6 Amoumt of e- acceptors to be supplied

aR

Acceptor consumption rate

3.7 Input Active Biomass

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

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

• 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.

]50.3[ˆ

0 0 VbXVXSKSqYQXQX aaaa

1630 .QXVbXVXSKSq̂Y aaa

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


]51.3[1ˆ

1

xx

x

bqYbKS

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

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

x

24.31ˆ

1

bqY

bKS

]52.3[01

aa

ax QXQX

VX

but the is now redefined.

]50.3[ˆ


•

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

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

contains some biomass.

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

0aX

0aX

0aX

0aX

3.7 HOW does affect S and washout ?

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

]53.3[1

1)( 0

x

xa b

SSYX

]54.3[)1(0 adaii bXfXXX

]55.3[))1(1(0 bfXXXXX daiaiv


]50.3[ˆ


1730 0 .SSQVXSKSq̂

a

For biomass,

For substrate,

))1(1(0bfdXXX aiv

For inert, 29310 0 .XXQVbXf iiad

]31.3[))1(1(0bfdXXX aiv When Xa

0 = 0,

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

• θx/θ ratio represents the build up of active biomass by the input. Rearranging Eq. 3.52,

- Xa0 < Xa: θx > 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

]53.3[1

1)( 0

x

xa b

SSYX

]56.3[/1

10

aa

x

XX


bqYbK

bqYb

KSx

ˆ

/1ˆ/1

limmin

bqYb

K

bqYb

KS

/1ˆ/11ˆ

1

• Organic substrate of particles (large polymers)

- Organic substrates that originate as particles or large polymer take important application in environmental biotechnology

• Effect of hydrolysis

- Before the bacteria can begin the oxidation reaction, the particles or large polymers must be hydrolyzed to smaller molecules

3.8 Hydrolysis of Particulate and Polymeric Substrates

- Hydrolysis kinetics is not settled, because the hydrolytic enzymes are not necessarily associated with or proportional to the active biomass

- Extracellular enzymes catalyze these hydrolysis reaction

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

• Hydrolysis of kinetics

- Reliable approach is a first-order relationship

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

* rhyd: rate of accumulation of particulate substrate due to hydrolysis(MsL-3T-1)

]57.3[phydhyd Skr

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

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

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

khyd = Xa ( = specific hydrolysis rate coefficient)'hydk

'hydk


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

]59.3[1

0

hyd

pp k

SS

]58.3[)(0 0 VSkPSQ phydpp

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

]14.3[)(0 0 SSQVrut


• The steady-state mass balance on soluble substrate

]60.3[//ˆ

)(0 0 QVSkQVXSKSqSS phyda

]17.3[)(ˆ

0 0 SSQVXSKSq

a

* S0 effectively is increased by QVSk phyd /

- The steady-state mass balance on substrate

- The steady-state mass balance on soluble substrate


• Other constituents of particulate substrate also are conserved during hydrolysis

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

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

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

]61.3[phydnhydn Skr


▣ Example 3.6


1

Y of casein (pp. 170)

]53.3[1

1)( 0

x

xa b

SSYX

]54.3[)1(0 adii bXfXX

]'55.3[]55.3[ paivaiv SXXXXXX

2

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

3

The biodegradable fraction = 0.8 (pp. 171)


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