Chemical Reaction Engineering

Lecturer : 郭修伯

Lecture 6

This course focuses on “Nonelementary reaction kinetics”.

Elementary vs. nonelementary

• Elementary: the reaction orders and stoichiometric coefficients are identical.

• Nonelementary reaction kinetics: no direct correspondence between reaction order and stoichiometry.

• Four topics will be introduced in this course:– Pseudo-steady-state hypothesis (PSSH)

– Polymerization

– Enzymatic reactions

– Bioreactors

Nonelementary reaction

• Gas-phase decomposition of azomethane (AZO)

262223)( NHCNCH

when AZO at pressure greater than 1 atm: AZON Cr 2

From experimental observation:

First - order reaction

when AZO at pressure below 50 mmHg: 2

2 AZON Cr Second - order reaction

Why? What happened?

Theory of Lindemann (1922)

• Collision or interaction between molecules forms an activated molecule, [(CH3)2N2]*

262223)( NHCNCH

*223223

1223223 ])[()()()( NCHNCHNCHNCH k

The activation can occur when translational kinetic enery is transferred into energystored in internal degrees of freedom, particularly vibrational degrees of freedom.

The concentration of the active intermediate is very difficult to measure,because AZO* is highly reactive and very short-lived (~ 10-9 s).

Translational kinetic energyCollision, photo, ...etc. Energy stored in internal

degrees of freedom

Active intermediate

• The energy must be absorbed into the chemical bonds where high-amplitude oscillations will lead to bond ruptures, molecular rearrangement, and decomposition.

• The sources of the energy:– photochemical effects of similar phenomena

– molecular collision or interaction.

• Types of active intermediates– free radicals (one or more unpaired electrons, e.g., H•)– ionic intermediates (e.g., carbonium ion)– enzyme-substrate complexes– etc.

*223223

1223223 ])[()()()( NCHNCHNCHNCH k

21* AZOAZOCkr

2232232

223*

223 )()()(])[( NCHNCHNCHNCH k

Deactivated through collision with another molecule

** 2 AZOAZOAZOCCkr

2623*

223 ])[( NHCNCH k ** 3 AZOAZOCkr

*2 3 AZON Ckr

We wanted to know why the orders of reation of rN2 are different at low and high pressures.

Difficult to measure

Actually, these are series reactions (multiple reactions):

q

iijj rr

1

**** 322

1

3

1AZOAZOAZOAZO

iiAZOAZO

CkCCkCkrr

What’s next? CAZO

*

Measurable concentration

relate to

Pseudo-Steady-State Hypothesis (PSSH)

• The active intermediate molecule has a very short lifetime low concentration

the rate of formation = the rate of disappearance

The net rate of formation of the acrive intermadiate is zero:

0* ermediateintactiver

0*** 322

1 AZOAZOAZOAZOAZO

CkCCkCkr

AZO

AZOAZO Ckk

CkC

23

21

*

*2 3 AZON Ckr

AZO

AZON Ckk

Ckkr

23

231

2

AZO

AZON Ckk

Ckkr

23

231

2

At low concentration: 32 kCk AZO 2

12 AZON Ckr

First - order reaction

Second - order reaction

At high concentration: 32 kCk AZO AZOAZON kCCk

kkr

2

312

The reaction is apparent first-order at high azomethane concentrations andapparent second-order at low azomethane concentration

Reaction mechanism

AZO

AZOAZO Ckk

CkC

23

21

*

The active intermediate, AZO*, collides with azomethane, AZO

The active intermediate, AZO*, decomposes spontaneously

The active intermediate, AZO*, is formed from AZO

The Stern-Volmer equation

High-intensity ultrasonic wave applied to water Light

wave compression microsize bubbles

Temperature rising

Generation of intermediatesChemical reaction in the bubbles

The intensity of the light given off, I, is proportional to the rate of reaction ofan activated water molecule formed in the microbubble.

hvOHOH k 23*

2 *2

*2

3)(OHOH

Ckrnsityinte 放出的光

When either carbon disulfide or carbon tetrachloride is added to the water,the intensity of sonoluminescence increases an order-of-magnitude.

聲納冷光

hvCSCS k 24*

2 *2

*2

4)(CSCS

Ckrnsityinte

However, when an aliphatic alcohol, X, is added to the solution, the intensityof sonoluminescence decreases with increasing concentration of alcohol.

Relative intensity I0/I

Alcohol concentration Cx

Sugget a mechanism consistent with experimental observation

XBCAI

I0

XBCAI

I0

)(

1

0 XBAI

I

The active intermediate collides with alcohol

productondeactivatirmediateinteX

Set CX = (X)

X is called a “scavenger” to deactivate the active intermediate清除者

)( 2CSnsityinte Active intermediate was probably formed from CS2

MCSCSM *22

M is a third body (CS2, H2O, etc.)

Purposed mechanisms:

Activation: MCSCSM k *2

12

Deactivation: MCSCSM k 22*

2

Deactivation: XCSCSX k 23*

2

Luminescence: hvCSCS k 24*

2*

24 CSCkI

Using PSSH on CS2*: 0* ermediateintactiver

0)())(())(())(( *24

*23

*2221*

2 CSkCSXkMCSkMCSkr

CS

*2

4 CSCkI

432

214

)()(

))((

kXkMk

MCSkkI

No alcohol (X=0)

42

2140 )(

))((

kMk

MCSkkI

)(1)()(

142

30 XkXkMk

k

I

I

This equation and similar equations involving scavengers are called Stern-Volmer equations.

Chain reaction

• A chain reaction consists of the following sequence:– Initiation

• formation of an active intermediate

– Propagation or chain transfer• interaction of an active intermediate with the reactant or

product to produce another active intermediate

– Termination• deactivation of the active intermediate

PSSH applied to thermal cracking of ethane

The thermal decomposition of ethane to ethylene, methane, butane andhydrogen is believed to proceed in the following sequence:

Initiation: 362 262 CHHC HCk ][ 6211 6262HCkr HCHC

Propagation: 5242

623 HCCHHCCH k ]][[ 62322 62HCCHkr HC

HHCHC k42

352

][ 5233 42 HCkr HC

2524

62 HHCHCH k ]][[ 6244 62HCHkr HC

Termination: 1045

522 HCHC k 25255 ][

5252 HCkr HCHC

(a) Use the PSSH to derive a rate law for the rate of formation of ethylene(b) Compare the PSSH solution in Part (a) to that obtained by solving the complete setof ODE mole balance

The rate of formation of ethylene ][ 52342 HCkr HC

The net rates of reaction of active intermediates CH3•, C2H5•, H• are (PSSH):

052624262

5252525252

5432

5432

HCHCHCHC

HCHCHCHCHC

rrrr

rrrrr

0262623 21 HCHCCH rrr

06242 43 HCHCH rrr

][ 6211 6262HCkr HCHC

]][[ 62322 62HCCHkr HC

][ 5233 42 HCkr HC

]][[ 6244 62HCHkr HC

25255 ][

5252 HCkr HCHC

substitute

2

13

2][

k

kCH

0][]][[ 25256232 HCkHCCHk

Purpose: replace [C2H5•]

21

625

152 ][

2][

HCk

kHC

21

625

13523 ][

2][

42

HCk

kkHCkr HC

The rate of disappearance of ethane ]][[]][[][ 624623262162HCHkHCCHkHCkr HC

The net rates of reaction of active intermediates CH3•, C2H5•, H• are (PSSH):

052624262

5252525252

5432

5432

HCHCHCHC

HCHCHCHCHC

rrrr

rrrrr

0262623 21 HCHCCH rrr

06242 43 HCHCH rrr

][ 6211 6262HCkr HCHC

]][[ 62322 62HCCHkr HC

][ 5233 42 HCkr HC

]][[ 6244 62HCHkr HC

25255 ][

5252 HCkr HCHC

substitute

21

62

21

5

1

4

3 ][2

][

HC

k

k

k

kH 2

1

625

136211 ][

2])[2(

62

HCk

kkHCkkr HC

Purpose: replace [CH3•] and [H•]

2

13

2][

k

kCH

For a constant-volume batch reactor :

21

625

13 ][

242

HCk

kkr HC

21

625

136211 ][

2])[2(

62

HCk

kkHCkkr HC

21

5

1311 ]

2)2(

6262

62

HCHCHC C

k

kkCkk

dt

dC21

5

13 62

422

HCHC C

k

kk

dt

dC

For given initial concentration of C2H6 and temperature,these two equation can be solved simultaneously

We obtain the concentration - time relationship using PSSH

Other methods can also be used…...

1. Mole balances:

C2H6 11 r

dt

dC

CH3•

CH4

C2H5•

C2H4

2. Rate laws for each species:

614212111 CCkCCkCkr

(Batch)

31

62 2CHHC k

5242

623 HCCHHCCH k

HHCHC k42

352

2524

62 HHCHCH k

1045

522 HCHC k

H2

33 r

dt

dC

44 r

dt

dC

55 r

dt

dC

77 r

dt

dC

22 r

dt

dC 122112 2 CCkCkr

2123 CCkr

245614432124 CkCCkCkCCkr

435 Ckr

614436 CCkCkr H• 66 r

dt

dC

C4H10 88 r

dt

dC

6147 CCkr

2458 2

1Ckr

All these O.D.Es can besolved simultaneously

The comparisons of the results obtained from the two methods are shown on page. 351.The two results are identical, indicatingthe validity of the PSSH underthese conditions

Reaction pathways

The second method is more frequently recently used due to the increase in computing power.

The key is to identify which intermediate reactions are important in theoverall sequence in predicting the end products.

The study of reaction pathways

Reaction pathways - smog formation

Nitrogen and oxygen react to form nitric oxide in automobile engines.The NO from automobile exhaust is oxidized to NO2 in the presence of peroxide radicals:

2NOORNOORO Nitrogen dioxide is then decomposed photochemically to give nascent oxygen:

ONOhvNO 2which reacts to form ozone:

32 OOO The ozone then involves in a whole series of reactions with hydro-carbons in the atmosphereto form aldehydes, various free radicals, and other intermediates, which react further toproduce undesirable products in air pollution:

radicalsfreealdehydesolefinO 3

OCHORRCHOCHRRCHO 3

hv OCHR

e.g.HCHOCHCHOCHCHCHCHCHO hv 22233

2

severe eye irritants

OROOR 2 P.354

Finding the reaction mechanism ...

Use both Figure 7-2 and Table 7-2 on page 354.

We will use the same idea to study “Polymerization”

A polymer is a molecule made up of repeating structural (monomer) units.

Polymerization is the process in which monomer units are linked together by chemicalreaction to form long chains. The polymer chains can be linear, branched, or cross-linked.

Polymerization reactions

Step reactions(Condensation reaction)

Chain reactions(Addition reaction)

Require bifunctional or polyfunctional monomers Require an initiator

Copolymers: polymers made up of two or more repeating units

Block

Alternating

Random

Graft

Statistical

SQSQSQ

SSSQQQ

QSSQSQ

QQQQQQ

SSSSfollow certain laws

Step polymerization reactions

• It requires that there is at least a reactive functional group on each end of the monomer

• Example:– has an amine group at one end and a carboxyl group at the other

– common functional groups are -OH, -COOH, -COCl,-NH2

• The molecular weight usually builds up slowly.• It is not meaningful to use conversion of monomer as a measure.

– The reaction will still proceed even though all the monomers has been consumed.

• We measure the progress by the parameter p which is the fraction of functional groups.

COOHCHNH 522 )( amino-caproic acid

Degree of polymerization

0

0

M

MMp

= fraction of functional groups that have been reacted

M = concentration of functional groups

pM

MX n

1

10The number of average degree of polymerization isthe average number of structural units per chain:

The number average molecular weight, is the average molecular weight of a structureunit, times the average number of structral unit per chain, plus the molecular weightof the end group, :

nM

sM nX

egM

egsnn MMXM

Interests to step polymerization

• Conversion of the functional groups

• Degree of polymerization

• Number average molecular weight

• Distribution of chain lengths, n, (i.e. molecular weight, Mn)

Determining the concentration of polymers for step polymerization

Determine the concentration and mole fraction of polymers of chain length j in terms ofinitial concentration of ARB, M0, the concentration of unreacted functional groups M,the propagation constant k and time t.

Set BRAP 1 BRAP 22BRAP jj ...

Reaction

212 PP

321 PPP

431 PPP

422 PPP

21

11

211 2

,2 1

21kP

rrkPr P

PP

21222 2321

PkPrrr PPP

31333 2431

PkPrrr PPP

22

44

224 2

,2 2

42kP

rrkPr P

PP

Rate laws

...

Two ways :A-R-BB-R-A

The net rate of reaction of P1, P2, and P3 for the first 4 reactions are:

31212

11 2221

PkPPkPkPrr P 2

2212

12 222

kPPkPkPrr P

3231213 2223

PkPPkPPkPrr P

Continue...

1

11 21

jjP PkPrr

1jjP Total concentration of functional groups = M

MkPrr P 11 21

Similarly, for j 2

1

1

2j

ijijij MkPPPkr

For a batch reactor,

The mole balance on P1: MkPdt

dP1

1 2 ktM

MM

0

0

1 2

001 1

1

ktM

MP

1

1

2j

ijijij MkPPPkr

MkPkPdt

dPr 2

21

22 2

ktM

MM

0

0

1

ktM

PkMktM

kMdt

dP

020

0

20

2

1

12

1

1

B.C.t = 0, P2=0

ktM

ktM

ktMMP

0

0

2

002 11

1

1

0

0

2

00 11

1

j

j ktM

ktM

ktMMP

generalizing

0

0

M

MMp

120 1 j

j ppMP

Chain polymerization reactions

• Chain polymerization requires an initiator (I) and proceeds by adding one repeating unit a time :

...43

32

21

1

RRM

RRM

RRM

RMI

The molecular weight in a chain usually builds up rapidly once a chain is initiated.

Free-radical polymerization

• The basic steps in free-radical polymerization:– Initiation

• formation of an active intermediate

– Propagation or chain transfer• interaction of an active intermediate with the reactant or

product to produce another active intermediate

– Termination• deactivation of the active intermediate

Initiation• An initiation step is needed to start the polymer chain growth.• It can be achieved by adding a small amount of a chemical

that decomposes easily to form free radicals:

• Initiators can be monofunctional and form the same free radicals, or they can be multifunctional and form different radicals.

• For monofunctional initiators, the reaction sequence between monomer and initiator:

II k 202

1RMI ki

• The propagation sequence between a free radical R1 with a monomer unit is:

• In general:

• The specific reaction rate kp is assumed to be identical for the addition of each monomer to the growing chain.

• The specific reaction rate ki is often taken to be equal to kp.

32

21

RMR

RMRkp

kp

1 jkp

j RMR

Propagation

• The transfer of a radical from a growing polymer chain can occur in the following ways:– Transfer to a monomer

• A live polymer Rj transfers its free radical to the monomer to from the radical R1 and a dead polymer Pj

– Transfer to another species

– Transfer of the radical to the solvent

1RPMR jkm

j

1RPCR jkc

j

1RPSR jks

j

Chain transfer

Chain transter (cont…)

• The species involved in the various chain transfer reactions are all assumed to have the same reactivity as Ri.

• The choice of solvent in which to carry out the polymerization is important.– For example, ks is 10000 times greater in CCl4 than in

benzene.

• The specific reaction rates in chain transfer are all assumed to be independent of the chain length.

Termination

• Termination to form dead polymer occurs primarily by two mechanisms:– Addition (coupling) of two growing polymers:

– Termination by disproportionation:

kjka

kj PRR

kjkd

kj PPRR

Free-radical polymerization reaction

• For example: the polymerization of styrene at 80 C initiated by 2,2-azobisisobutyronitrile:

ismpa kkkkk ~

Addition (termination)

Propagation

Chain transfer (monomer)

Chain transfer (solvent)

Initiation

Initial concentration: 0.01 M for initiator, 3 M for the monomer, and 7 M for the solvent

Rate laws for chain polymerization

• Initiation:– Only a certain fraction f will be successful in initiating

polymer chains. The rate law for the formation of the initiator free radical:

• f is the fraction of initiator free radical successful in initiating chaining and has a typical value in the range 0.2 to 0.7.

– The rate law for the formation of R1 in the initiation step:

)(2 20 IfkrIf

))((1 IMkrr iiR

Rate laws at the initiation step

Using the PSSH for the initiator free radical, I,

)(2 20 IfkrIf ))((1 IMkrr iiR

0))(()(2 20 IMkIfkr ii

)(

)(2)( 20

Mk

IfkI

i

)(2 20 Ifkri

Identical!

Rate laws for R1

• In general:

• The total loss of R1 radicals is found by adding the loss of R1 radicals in each reaction. The rate of disappearance of R1 by termination addition is given by:

• The net rate of disappearance of the free radical R1:

11 jka

j PRR

111

131212

11 ......

jjat

jaaaat

RRkr

RRkRRkRRkRkr

2221

11

111j

jsj

jcj

jmj

jdj

japi RSkRCkRMkRRkRRkMRkrr

initiation + propagation + termination (addition) + termination (disproportionation) + chain transfer

Rate laws for Rj (j 2)

• In general, the net rate of disappearance of live polymer chains with j monomer units (species j, j2)

jsjcjmi

ijdajjpj SRkCRkMRkRRkkRRMkr

11 )()(

We will then use PSSH to solve polymerization problems.

Let represent th total concentration of the radical Rj.

1

*

jjRR

1

2*)(j

tij Rkrr

jsjcjmi

ijdajjpj SRkCRkMRkRRkkRRMkr

11 )()(

2221

11

111j

jsj

jcj

jmj

jdj

japi RSkRCkRMkRRkRRkMRkrr

(j2)

Termination term

PSSH

1

0j

jr

tt

i

k

Ifk

k

rR

)(2 20*

Total free-radical concentration

The net rate of monomer consumption, -rM: )()()( mpiM rrrr

consumption by initiator

consumption by propagation

consumption by monomer chain transferLong Chain approximation (LCA)

ip rr

tpp

jjppM k

IfkMkMRkRMkrr

)(2 20*

1

The rate of disappearance of monomer

The net rate of formation of dead polymer Pj (by addition) is:

1

12

1 jk

kkjkaP RRkr

j

The rate of formation of all dead polymers

t

aa

jPp k

IfkkRkrr

j

)()(

2

1 202*

1

Enzymatically catalyzed reactions

• An enzyme, E, is a protein or proteinlike substance with catalytic properties.

• A substrate, S, is the substance that chemically transformed at an accelerated rate because of the action of the enzyme on it.

• One enzyme can catalyze only one reaction. Unwanted products are easily controlled.

• Enzymes are produced only by living organisms (bacteria, for example).

• Enzymes usually work under mild conditions (pH 4~9, 75 ~ 160 F).

Enzymes

• Most enzymes are named in terms of the reactions they catalyze (***ase), for example:

• urease: the enzyme that catalyzes the decomposition of urea

• tyrosinase: the enzyme that attacks tyrosine

• Three major types of enzyme reactions:– soluble enzyme - insoluble substrate (e.g. laundry detergents)

– insoluble enzyme - soluble substrate (similar to packed catalytic bed rxn)

– soluble enzyme - soluble substrate (e.g. many biological rxns)

our interest

Enzymatically catalyzed rxn example

The proposed mechanisms of the catalytic action of urease which causes urea to decompose into ammonia and carbon dioxide : (Levine and LaCourse, 1967)

1.The enzyme urease reacts with the substrate urea to form an enzyme-substrate complex, E•S:

*22

122 ][ ureaseCONHNHureaseCONHNH k

2.This complex can decompose back to urea and urease:

ureaseCONHNHureaseCONHNH k 222*

22 ][

3.Or, it can react with water to give ammonia, carbon dioxide, and urease:

ureaseCONHOHureaseCONHNH k 233

2*

22 2][

S E E•S

W P

The rate of disappearance of the substrate is: )())(( 21 SEkSEkrS

SEES k 1

SESE k 2

EPWSE k 3

The net rate of formation of the E•S complex is: ))(()())(( 321 SEWkSEkSEkr SE

The enzyme is not consumed by the reactions: )()()( SEEEt PSSH 0 SEr

)()(

))(()(

321

1

WkkSk

SEkSE t

)()(

))()((

321

31

WkkSk

SEWkkr t

S

)()(

))()((

321

31

WkkSk

SEWkkr t

S

Since the reaction of urea and urease is carried outin aqueous solution (water): (W) ~ constant

k’3

m

tS KS

SEkr

)(

))((3

1

23

k

kkKm

This is the form of the “Michaelis-Menten Equation”and Km is call the Michaelis constant

Vmax

(S)

-rS

Vmax

At low substrate concentration: (S) << Km

mS K

SVr

)(max

At high substrate concentration: (S) >> Km

maxVrS

In a special case, when 2maxV

rS 2/max| Vrm s

SK

Vmax/2

Km

Km is equal to the substrate concentration at whichthe rate of reaction is equal to one-half the maximum rate.

Vmax and Km characterise the enzymatic reactions described by Michaelis-Menten kinetics.Vmax is dependent on total enzyme concentration.Km is independent of total enzyme concentration.

)(

)(

)(

3max

max

t

mS

EkV

KS

SVr

Michaelis-Menten equation

Michaelis-Menten kinetics exampleSEES k 1 SESE k 2 EPWSE k 3

Curea (kmol/m3) 0.2 0.02 0.01 0.005 0.002

-rurea (kmol/m3s) 1.08 0.55 0.38 0.2 0.09

mS KS

SVr

)(

)(max

)(

1

)(

)(1

maxmaxmax SV

K

VSV

KS

rmm

S

1/-rs

(1/S) (i.e. 1/Curea)

Slope = 0.02 =maxV

Km

Intercept = 0.75 =max

1

V

urea

ureaS C

Cr

0266.0

33.1

Batch enzymatically catalyzed rxn Artificial kidney design

murea

ureaureaurea KC

CV

dt

dCr

maxA batch reactor in liquid phase:

max

00

max

lnV

CC

C

C

V

Kt ureaurea

urea

uream

)1(0 XCC

tK

XC

K

V

Xt m

urea

m

0max

1

1ln

1

Xt 1

1ln

1

t

X

Intercept = mK

Vmax

Slope =m

urea

K

C 0

Inhibition of enzyme reactions

• The rate of enzyme-catalyzed reactions is affected by pH and inhibitors.

• Three most common types of reversible inhibition:– Competitive

• Substrate and inhibitor are usually similar molecules that compete for the same site on the enzyme.

– Uncompetitive• The inhibitor deactivetes the enzyme-substrate complex, usually by attaching

itself to both the substrate and enzyme molecules of the complex.

– Noncompetitive• Enzymes containing at least two different types of sites. The inhibitor attaches

to only one type of site and the substrate only to the other.

Bioreactors

• Microorganisms and mammalian cells are used to produce a variety of products, such as insulin(胰島素 ), most antibiotics(抗生素 ), and polymers.

• Advantages:– mild reaction conditions– high yields– can catalyze successive steps in a reaction for organisms

contain several enzymes– stereospecific (立體的 ) catalyst (single desired isomer can be formed)

In general, the growth of an aerobic organism follows:

][][][][

...][][][][][

22),,( cellsmoreproductOHCO

sourcephosphatesourceoxygensourcenitrogensourcecarboncellsetcetemperaturpHconditionsmediaculture

ductProcellsMoreSubstrate cell

The rate of this reaction is proportional to the cell concentration andthe reaction is autocatalytic.

• (1) Lag phase– little increase in cell concentration– synthesizing transport proteins for moving the

substrate into the cell– synthesizing enzymes for utilizing the new substrate– beginning the work for replicating the cell’s genetic

material

time

Log cell concentration

1 432

Four phases are included in cell growth:

• (2) Exponential growth phase– the cell’s growth rate is proportional to the cell

concentration– the cells are able to use the nutrients most efficiently

• (3) Stationary phase– the cell reach a minimum biological space where the

lack of one or more nutrients limits cell growth.– many important fermentation products, including

most antibiotics, are produced in the stationary phase

• (4) Dead phase– result of either the toxic by-product and/or the

depletion of nutrient supply

Rate laws of bioreactors

ductProcellsMoreSubstrateCells conditions

The most commonly used expression is the Monod equation for exponential growth:

cg Cr

Cell growth rate Cell concentration

Specific growth rate = ss

s

CK

C

max

Maximum specific growth reaction rateMonod constant

Substrate concentration

ss

scg CK

CCr

max account for inhibition

ss

sc

n

p

pg CK

CC

C

Cr

max

*1

(one model here)

where Cp* = product concentration at which all metabolism ceases and n = experimental constant

Other cell growth rates:

Monod equation

cs

g Ck

Cr

exp1maxTessier equation

)1(max

s

cg

kC

CrMoser equation

The cell death rate is:

cttdd CCkkr )(

Concentration of a substrate toxic to the cell

Specific natural death rate constant

Specific toxic death rate constant

Stoichiometry for cell growth

• Very complex : vary with microorganism / nutrient system; vary with environmental conditions; even more complex when more than one nutrient contributes to cell growth.

• May be simplified as:

ductProcellsMoreSubstrateCells conditions

PYCYS spsccells

//

where Yc/s is the yield coefficient :cellsnewproducetoconsumedsubstrateofmass

formedcellsnewofmassY sc /

where Yp/s is the product coefficient :productformtoconsumedsubstrateofmass

formedproductofmassY sp /

Substrate consumptionProduce new cells

Maintain a cell’s daily activities

timecellsofmass

ceaintenanmforconsumedsubstrateofmassm

typical value = 0.05 h-1

The yield coefficient, Y’c/s , account for substrate consumption for maintenance:

consumedsubstrateofmass

formedcellsnewofmassY sc /

Therefore, the rate of substrate consumption for maintenance: Csm mCr

Product formation

During the growth phase

During the stationary phase

gcpp rYr /

)(/ sspp rYr

The net rate of substrate consumption:

cppsgcss mCrYrYr //

Consumption rate by cell growths

Consumption rate to form product

Consumption rate for maintenance

cppsgcss mCrYrYr //

If the product is formed during the growth phase cgcss mCrYr / gcpp rYr /

If the product is formed during the stationary phase cppsnsn mCrYr /

snsn

csnpp CK

CCkr

where Csn is the concentration of the secondary nutrient

Mass balance on cellsFor a CSTR, a mass balance on the microorganism gives :

Rate of accumulation of cells

VrrvCCvdt

dCV dgcc

c )(00

Rate of cells entering

Rate of cells leaving

Rate of net generation of live cells

Mass balance on substratesFor a CSTR, a mass balance on the substrate gives :

Rate of accumulation of substrate

VrvCCvdt

dCV sss

s 00

Rate of substrate entering

Rate of substrate leaving

Rate of net generation of substrates

Mass balance on cellsFor a batch system, a mass balance on the microorganism gives :

Rate of accumulation of cells

Vrrdt

dCV dg

c )(

Rate of net generation of live cells

Mass balance on substratesFor a batch system, a mass balance on the substrate gives :

Rate of accumulation of substrate

VmCVrYVrdt

dCV cgcss

s )(/

Rate of substrate used for cell growth

Rate of substrate used for maintenance

In the growth phase

VmCVrYdt

dCV cpps

s )(/In the stationary phase

VrYVrdt

dCV sspp

p )(/ Rate of product formation

Example: bacteria growth in a batch reactor

A fermentation process is carried out in a batch reactor. Plot the concentrations of cells,substrate, and product and growth rates as functions of time. The initial concentration is1.0 g/dm3 and the substrate concentration is 250 g/dm3.

Values of the parameters:

)/()(03.0

/7.1

33.0

52.0

/93

3

1max

3*

hcellsgsubstrategm

dmgK

h

n

dmgC

s

p

1

/

/

/

01.0

/6.5

/45.0

/08.0

hk

ggY

ggY

ggY

d

cp

sp

sc

Mass balances

Cells: Vrrdt

dCV dg

c )(

Substrate: VrVrYdt

dCV smgcs

s )(/

Product VrYdt

dCV gcp

p )(/

Rate laws

ss

sc

p

pg CK

CC

C

Cr

52.0

*max 1

cdd Ckr

csm mCr

Stoichiometry

gcpp rYr /

cdss

sc

p

pc CkCK

CC

C

C

dt

dC

52.0

*max 1

css

sc

p

pcs

s mCCK

CC

C

CY

dt

dC

52.0

*max/ 1

gcpp rY

dt

dC/

Chemostats

• Chemostats are essentially CSTRs that contain microorganisms.

• One of the most important features of the chemostat is that is allows the operator to control the cell growth rate.– By adjusting the volumetric feed rate

Skip this part. Refer to Profs. Chen & Liu