L11-1 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois,...

L11-1

Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.

Review: Rate Equation for Enzymatic Reaction

maxP

m

SVv rSK

experimentally determined reaction rate

P2P ES

dCrate of product formation : v r Ck

dt

E E0 ES E0 E,t 0conservation of enzyme C C C where C = C

k k1 2k 1

E S ES E P

E: enzyme S: substrateES: enzyme-substrate complex

ES1 1 2S E0 ES ES

dC0 C C C C k k k

dt

S E0ES

1 2S

1

C CC

k k Ck

2 maxE0 S SPP P

1 2 m SS

1

C C CdC k Vr rdt K Ck k C

k

max 2 E0V k C

1 2m

1

k kKk

Where:

L11-2


Review: Lineweaver-Burk Equation

2

Lineweaver & Burk: inverted the MM equation

max SP

m S

CVrK C

m S

maxP S

K C1r CV

m

max maxp S

K1 1 1r CV V

y m x b

By plotting 1/ V vs 1/CS, a linear plot is obtained:

Slope = Km/Vmax

y-intercept = 1/Vmax

x-intercept= -1/Km

L11-3


Review: Competitive Inhibitionmaxmax S S

p pm SI

m SI

V C CVr vs rK ' CC

K ' 1 CK

m,app m

I

IK ' K ' 1

K

-1 0 1 20

0.10.20.30.40.50.60.70.80.9

1

1/CS (mmol)-1

1/rP

Inhibited reaction

Uninhibited reaction

m

max maxP S

K1 1 1r CV V

Slope = Km/Vmax y-int = 1/Vmax

x-int= -1/Km

Can be overcome by high substrate concentration

Substrate and inhibitorcompete for same site

Km, app >Km

Vmax, app =Vmax

L11-4


max I I SP

S m

V 1 C K C v= r

C K

Review: Noncompetitive Inhibition

m

P m,app S m,app

K1 1 1 r V C V

maxmax,app

I

I

VV

C1

K

Vmax, app < Vmax

Km, app = Km

substrate and inhibitor bind different sites

higher CI

No I

CI

Vmax

Vmax,app

Vmax,app

CSKm

rP

max,app

1 y int

V

S

1C

m

1K

p

1r

Increasing CI

m

max,app

K 'm

V

L11-5


substrate & inhibitor bind different sites but I only binds after S is bound

max I I Sp

S m I I

V 1 C K Cv = r =

C K 1 C K

m,app

max,app

K slope

V

app max,Vint y

1

m, app

1 x-int

K 1/CS

1/v

Vmax, app < Vmax

Km, app <Km

maxmax,app

I

I

VV =

C1

K

mm,app

I

I

KK =

C1

K

m,app

P max,app S max,app

K1 1 1 r V C V

No rxn

Review: Uncompetitive Inhibition

L11-6


• Region 1: Lag phase – microbes are adjusting

to the new substrate• Region 2: Exponential

growth phase – microbes have

acclimated to the conditions

• Region 3: Stationary phase – limiting substrate or

oxygen limits the growth rate

• Region 4: Death phase – substrate supply is

exhaustedTime

log [X]32 41

Review: Kinetics of Microbial Growth (Batch or Semi-Batch)

CC,max

Log CC

CC0

L11-7


Review: Quantifying Growth Kinetics• Relationship of the specific growth rate to substrate concentration

exhibits the form of saturation kinetics

• Assume a single chemical species, S, is growth-rate limiting

• Apply Michaelis-Menten kinetics to cells→ called the Monod equation:

max Sg C

s S

Cr C

K C

• max is the maximum specific growth rate when S>>Ks

•CS is the substrate concentration

•CC is the cell concentration

•Ks is the saturation constant or half-velocity constant. Equals the rate-limiting substrate concentration, S, when the specific growth rate is ½ the maximum

•Semi-empirical, experimental data fits to equation

•Assumes that a single enzymatic reaction, and therefore substrate conversion by that enzyme, limits the growth-rate

L11-8


S

m

KS

mm

Exponential phase

deceleratingphase

m Sg C

s S

Cr C

K C

Review: Monod Model

m SS S g C

s

CC K r C

K

First-order kinetics:

S S g mC K r Zero-order kinetics:

L11-9


L11: Thermochemistry for Nonisothermal Reactor Design

• The major difference between the design of isothermal and non-isothermal reactors is the evaluation of the design equation – What do we do when the temperature varies along the length of

a PFR or when heat is removed from a CSTR?• Today we will start nonisothermal reactor design by reviewing

energy balances• Monday we will use the energy balance to design nonisothermal

steady-state reactors

Nonisothermal Energy balance

L11-10


Why do we need to balance energy?

kA B A A

A0

dX r

dV F

A Ar kC

FAXA = 0.7

Mole balance:

Rate law:

Stoichiometry: A A 0

0

A A0 A

F C

C C (1 X )

A A

0

dX k(1 X )

dV

ERTk Ae

Arrhenius Equation

E 1 1R T TA A1

10

dX (1 X )k exp

dV

Need relationships: X T V

Consider an exothermic, liquid-phase reaction operated adiabatically in a PFR (adiabatic operation- temperature increases down length of PFR):

FA0

We can get them from the energy balance

L11-11


Clicker Question

The concentration of a reactant in the feed stream (inlet) will be greatly influenced by temperature when the reactant is

a) a gasb) a liquidc) a solidd) either a gas or a liquid e) extremely viscous

Gas phase:

Liquid& solid phase:

A0 j j Aj

A 0

0C X P

C1 X P

T

T

j A0 j j A C C X

Hints:

L11-12


Thermodynamics in a Closed System

• First law of Thermodynamics– Closed system: no mass crosses the system’s boundaries

ˆdE Q W

dÊ: change in total energy of the systemdQ: heat flow to systemdW: work done by system on the surroundings

Q

W

L11-13


Ẇ

Fin

Hin

Fout

Hout

n nsysi i i iin out

i 1 i 1

ˆdE Q W FE FE

dt

Thermodynamics in an Open System

• Open system: continuous flow system, mass crosses the system’s boundaries

• Mass flow can add or remove energy

Q

Energy balance on system:

Rate of accum of energy in

system

work done by system

energy added to sys. by

mass flow in

energy leaving sys. by mass

flow out

Heat in

= - + -

Let’s look at these terms individually

L11-14


n nsysi i i iin out

i 1 i 1

ˆdE Q W FE FE

dt

The Work Term, Ẇ

• Work term is separated into “flow work” and “other work”.• Flow work: work required to get the mass into and out of system• Other work includes shaft work (e.g., stirrer or turbine)

other work (shaft work)

P : pressure

Ẇ: Rate of work done by the system on the surroundings

n n

i i i i sin outi 1 i 1

W FPV FPV W

Flow work

3mmol of species i

iV specific volume

Plug in:

n nsys

s i i i ii iin outi 1 i 1

ˆdE Q W F E PV F E PV

dt

Accum of energy in system

Other work

Energy & work added by flow in

Energy & work removed by flow out

Heat in= - + -

L11-15


The Energy Term, Ei

n nsys

s i i i ii iin outi 1 i 1

ˆdE Q W F E PV F E PV

dt

Accum of energy in system

Other work



Heat in= - + -

2i

i i iu

E U gz other2

Internal energy Kinetic energy

Potential energy

Electric, magnetic, light, etc.

Usually: 2i

i iu

U gz other2

i iE U

Plug in Ui for Ei:

n nsys

s i i i in i i i outi 1 i 1

ˆdEQ - W F(U PV ) - F(U PV )

dt

Internal energy is major contributor to energy term

L11-16


n nsyss i i i in i i i out

i 1 i 1

ˆdEQ - W F(U PV ) - F(U PV )

dt

Recall eq for enthalpy, a function of Ti i iH U PV

unit : (cal / mole)

n nsyss i in i out

ii i

1 i 1

ˆdEQ FHW FH

dt

n n

s i0 i0 i ii 1 i 1

0 Q W F H FH

Steady state:

Accumulation = 0 = in - out + flow in – flow out

Total Energy Balance

L11-17


In Terms of Conversion:

i i0 i A0 A i A0 i i AF F F X F F X i0i

A0

Fwhere

F

If XA0=0, then:

n n

s i0 i0 ii

i1 i 1

0 Q W F H HF Steady state:

n nssyyss

s i0 i A0 ii 1 i 1

A0 i i AFˆdE

Q W H F Hdt

X

i

n nssyysss i

iA0 0 i i i A0 A

i 1 i 1

ˆdEQ W H H H FF X

dt

n nssyyss

s i0 i i A0 i Ai 1 i 1

A0 A0i i

ˆdEQ W H H HF X

dtF F

n

i i RXi 1

H H T heat of rxn at temp T

Total energy balance (TEB)

Relates temperature to XA

Multiply out:

nssyyss

s A0 Ai i0 i 0R A1

Xi

H TH HˆdE

Q W F Xdt

F

Must use this

equation if a phase change occurs

L11-18


What is (Hi0 – Hi)? i RX

nssyysss A0 A

i 1i0 i0A

ˆdEQ W F XH

tF H TH

d

i Ri Qiheat of reaction H TH H

When NO phase change occurs & heat capacity is constant:

T2Qi pi pi RT1

H C dT C T T

Enthalpy of formation of i at reference temp (TR) of 25 °C

What is the heat of reaction for species i (Hi)?

Change in enthalpy due to heating from TR to rxn temp T

i R pi Ri H T TH C T

i0 i i R pi i0 R i R pi RH H H T C T T H T C T T

i0 i pi i0 R pi RH H C T T C T T

i0 i pi i0 pi R pi pi RH H C T C T C T C T

i0 i pi i0 piH H C T C T i0 i pi i0H H C T T

L11-19


What is ΔHRX(T)? i RX

nssyysss i0 i A0 A

i 1A0

ˆdEHQ W H H

dtF FT X

How do we calculate ΔHRX(T), which is the heat of reaction at temperature T?b c d

A B C Da a a

For the generic reaction:

RX D C B A

d c bH T H H H H

a a a ii piR Rwhere T TH CH T

D R C R B R PD PC PX PR R BA RA

d c bH T H T H TH

d c bC C C C

a a aH T

a aT

aT T

RX R D R C R B R A R

d c bH T H T H T H T H T

a a a

P PD PC PB PA

d c bC C C C C

a a a

RX RRX RPH T CT TH T

L11-20


Example: Calculation of ΔHRX(T) For the reaction N2 (g) + 3H2 (g) → 2NH3 (g), calculate the heat of reaction at 150 °C in kcal/mol of N2 reacted.

Extra info: 2 2 3N R H R NH RH T 0 H T 0 H T 11,020cal mol

H N NH2 2 3P P P

cal cal calC 6.992 C 6.984 C 8.92

mol K mol K mol K

RX R

2 cal 3H T 11,020 0 0

1 mol 1

RX R D R C R B R A R

d c bH T H T H T H T H T

a a a

2 2 31N 3H 2NH a 1 b 3 c 2 d 0

RXRX RPRH T CH TT T

RX R2

calH T 22,040

mol N reacted

P PD PC PB PA

d c bC C C C C

a a a

P2

2 3 calC 8.92 6.992 6.984

1 1 mol N reacted K

P2

calC 10.12

mol N reacted K

L11-21


Example: Calculation of ΔHRX(T) For the reaction N2 (g) + 3H2 (g) → 2NH3 (g), calculate the heat of reaction at 150 °C in kcal/mol of N2 reacted.


H N NH2 2 3P P P

cal cal calC 6.992 C 6.984 C 8.92

mol K mol K mol K

RXRX RPRH T CH TT T

RX R2

calH T 22,040

mol N reacted

P2

calC 10.12

mol N reacted K

T 150 C 150 273 K 423K

2 2

RX

cal10.12

mol

calH 22,040

mol N reacted KT 423K 298

N reac dK

te

RT 25 C 25 273 K 298K

Convert T and TR to Kelvins

RX2 2

cal kcalH T 23,310 23.31

mol N reacted mol N reacted

L11-22


Example: Calculation of ΔHRX(T) For the reaction N2 (g) + 3H2 (g) → 2NH3 (g), calculate the heat of reaction at 150 °C in kJ/mol of H2 reacted.


H N NH2 2 3P P P

cal cal calC 6.992 C 6.984 C 8.92

mol K mol K mol K

RXRX RPRH T CH TT T

RX R2

calH T 22,040

mol N reacted

P2

calC 10.12

mol N reacted K

T 423K

RT 298K

RX2

kcalH T 23.31

mol N reacted

RX2

23.31 kcal 4.184 kJH T

mol N reacted kcal

Convert kcal to kJ

RX2

kJH T 97.53

mol N reacted

Put in terms of moles H2 reacted

2RX

2 2

1 mol NkJH T 97.53

mol N reacted 3 mol H

2 2 31N 3H 2NH

RX2

kJH T 32.5

mol H reacted

L11-23


Q and Hi in Terms of T• Ignore enthalpy of mixing (usually an acceptable assumption)

•Look up enthalpy of formation, Hi◦(TR) in a thermo table, where the

reference temperature TR is usually 25◦C

•Compute Hi(T) using heat capacity and heats of vaporization/meltingT

i i R piTRno phase change: H H (T ) C dT

n n nT

RX i i i i R i piTRi 1 i 1 i 1H T H H (T ) C dT

Phase change at Tm

(solid to liquid):

T Tmi i R psi m,i pli

T TR m

H T H T C dT H C dT

Solid at TR

For Tm < T < Tb←boiling

If constant of average heat capacities are used, then:

i i R psi m R m,i pli MH T H T C T T H C T T

psi pliC : heat capacity of solid C : heat capacity of liquid

m,iH : enthalpy of melting

For Tm < T < Tb

melting

L11-24


Insert ΔHRX(T) & (Hi0 – Hi) into EB i RX

nssyysss A0 A0 Ai

ii0

1

ˆdEQ W F F X

tH H TH

d b c d

A B C Da a a

RXRX RPRH T CH TT T i0 i pi i0H H C T T

nssyysss A0 i pi i0 RX R P R A0 A

i 1

ˆdEQ W F C T T H T C T T F X

dt

Example calculations of ∆H°RX(TR) & ΔCp are shown on the previous slides

pA B pB C pC D

n

i p1

pDii

C CC C C

i0i

A0

Fwhere

F

If the feed does not contain the products C or D, then:

C0 D0C D

A0 A0

F F0 & 0

F F

n

i pii 1

pA B pBC C C

pi i

nssyysss A0 RP A0 A

iR

1X R0i

ˆdEQ W F T TH T C

dC T T F X

t

(Ti0 – T) = - (T – Ti0)

L11-25


Clicker Question

If the reactor is at a steady state, which term in this equation would be zero?

a) dEsys/dt

b)

c) Ẇ

d) FA0

e) ∆CP

Q

Accum of energy in

system

Other work



Heat in= - + -

RX

nssyysss A0 R A0 APpi i R

i 1i 0C T T H T

ˆdEQ W F T T F

tC X

d

n

s A0 R A0 Ai 1

RX Ri pi i P0 H0 Q W F T T FT T XCC T

At the steady state:

L11-26


How do we Handle Q in a CSTR? CSTR with a heat exchanger, perfectly mixed inside and outside of reactor

T, X

FA0

T, X

Ta

Ta

The heat flow to the reactor is in terms of:• Overall heat-transfer coefficient, U• Heat-exchange area, A

•Difference between the ambient temperature in the heat jacket, Ta, and

rxn temperature, T

aQ (UA T T)

L11-27


Integrate the heat flux equation along the length of the reactor to obtain the total heat added to the reactor :

A Va aQ U(T T)dA Ua(T T)dV

adQ

Ua(T - T)dV

Heat transfer to a perfectly mixed PFR in a jacket

a: heat-exchange area per unit volume of reactor

For a tubular reactor of diameter D, a = 4 / D

For a jacketed PBR (perfectly mixed in jacket):

ab b

1 dQ dQ Ua(T T)

dV dW

Heat transfer to a PBR

Tubular Reactors (PFR/PBR):

Aa

V

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L11-1 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois,...

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