l17-1 slides courtesy of prof m l kraft, chemical & biomolecular engr dept, university of...

L17-1

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

Review: Unsteady State Nonisothermal Reactor Design

An open system (for example, CSTR)

Q

W

Fin

Hin

Fout

Hout

n nsysi i in i i out

i 1 i 1

dE Q W FE FE

dt

rate of heat flow from

surroundings to system

Rate of accumulation of energy in

system

=

Rate of work done by

system on surroundings

- +

Rate of energy added to

system by mass flow in

-

Rate of energy leaving system by mass flow

out

sysdE 0 steady state

dt

sysdE 0 unsteady state

dt

Goal: develop EB for unsteady state reactor

L17-2


Review: Simplified EB for Well-Mixed Reactors

n

S i0 i i0 RX Ai 1

n

i pii 1

Q W F H H H T r VdTdt

NC

Energy balance for unsteady state reactor

with phase change:

n

S i0 pi i0 RX Ai 1

n

i pii 1

Q W F C T T H T r VdTdt

NC

Energy balance for

unsteady state reactor without phase change:

n n n n

i iS i i i i i i

i 1 i 1 i 1 i 1in out

dH dN dQ W FH FH N H PV

dt dt dt

Total Energy Balance for unsteady state

ipi

dH dTC

dt dt i

i0 i i AdN

F F r Vdt

dPV 0

dt

Constant PV variationMade following substitutions & solved for dT/dt:

L17-3


Review: Unsteady State EB, Liquid-Phase Reactions

For liquid-phase reactions, often Cp = S iCpi is so small it can be neglectedWhen Cp can be neglected, then:

If the feed is well-mixed, it is convenient to use:

Plug these equations & Ti0 = T0 into the EB:

This equation for the EB is simultaneously solved with the mass balance (design eq) for unsteady state, nonisothermal reactor design

S i0 RX A

n

i pii 1

n

i0 pii 1

Q W T T H T r V

NC

FT

dt

Cd

S

n

i pi A0 ps ps i pii 1

where is the heat capacity of the solNC N C C C ution

S i0 pi A0 psF C F C

S i0 RA

A0

s X

p

0 Ap

s

Q W T T H T

N C

VF C r dTdt

L17-4


Review: Nonisothermal Batch Reactor Design

n

S i0 pi i0 RX Ai 1

n

i pii 1

Q W F C T T H T r VdTdt

NC

No flow, so:

0

p

S RX

1i i

An

i

Q W H T

CN

r V

p

S RX An

i 1A0 i Api

Q W H T r V dT

CNt

Cd

X

Put the energy balance in terms of conversion:

Solve with the batch reactor design equation using an ODE solver (Polymath):

AA0 A

dXN r V

dt

i A0 i i Ai0

iA

i0

i PpNN

where & CNN

X C

L17-5


Review: Adiabatic Nonisothermal Batch Reactor Design

S RX An

A0 i pi p Ai 1

Q W H T r V dTdt

N C C X

For negligible stirring work:

i pi psC CSubstitute:Rearrange

00

T

T RX R p R0

XA A

pS p AX 0A0

dT

H T C T T

dX

C C X

p 0 p 0S SA A

RXRX R p R

C T T C T TX X

H TH T C T T

Solve for T: RX 0 A RX 0 A

0 0 np A ps i p A pi

i 1

H T X H T XT T T T

C X CC X C

Solve with the batch reactor design equation using an ODE solver (Polymath)

XAA

A0A0

dXt N

r V

Integrate & solve for XA:

L17-6


L17 Basic Catalysis & Reaction Mechanisms

•Though we have discussed the use of catalyst in a PBR, we have not discussed the process of catalysis itself

•An understanding of catalysis, the mechanisms, and catalytic reactor design are the subject of the next few lectures

•Catalyst properties•Steps involved in a catalytic reaction•Development of a rate law using steps in catalytic reaction•Different types of catalyst mechanisms•Design of catalytic reactors

L17-7


Catalysts & Catalysis• ~1/3 of the GNP of materials produced in the US involve a catalytic process• A Catalyst is a substance that speeds up the rate of reaction but is not

changed by the reaction• A catalyst lowers the energy barrier by promoting a different molecular

pathway (mechanism) for the reaction

Many catalyst are porous (high surface area)

• Homogeneous catalysis: catalyst is in solution with at least 1 reactant• Heterogeneous catalysis: more than 1 phase, usually solid and fluid or

solid and gas is present. Reaction occurs at solid/liquid or gas interface.

catalyst

L17-8


1. Mass transfer of A to surface

2. Diffusion of A from pore mouth to

internal catalytic surface

3. Adsorption of A onto catalytic surface

4. Reaction on surface

5. Desorption of product B from surface

6. Diffusion of B from pellet interior to pore mouth

7. Diffusion of B from external surface to the bulk fluid (external diffusion)

Steps in a Heterogeneous Catalytic Reaction

Ch 10 assumes steps 1,2,6 & 7 are fast, so only steps 3, 4, and 5 need to be considered

L17-9


Adsorption Step

A(g) + S ⇌ A·S

S: open (vacant) surface site A·S: A bound to a surface site

The adsorption of A (gas phase) on an active site S is represented by:AI

-S-S-S-

Rate of adsorption = rate of attachment – rate of detachment

AD A A v A A Sr k P C k C partial pressure of A

• Rate is proportional to # of collisions with surface, which is a function of PA

• Rate is proportional to # of vacant (active) sites, Cv, on the surface• Active site: site on surface that can form a strong bond with adsorbed species

Molar conc of vacant sites on surface

A

-S-S-S-

AD A A v A A Sr k P C k C

In terms of the adsorption equilibrium constant KA where AA

A

kK

k

AAD A A v A S

A

kr k P C C

k

A SAD A A v

A

Cr k P C

K

Equation I

Conc of sites occupied by A

L17-10


Site BalanceCt: Total number of active sites per unit mass of catalyst divided by

Avogadro’s # (mol/g cat)Cv: Number of vacant sites per unit mass of catalyst divided by Avogadro’s #

Ct = Cv + CA·S + CB·S

Surface

Vacant active site

A

Active site occupied by A

B

Active site occupied by B

Site balance:

In the absence of catalyst deactivation, assume the total number of active sites remains constant:

We will use the site balance equation to put Cv in terms of measurable species

Cv is not measurable, but the total number of sites Ct can be measured

L17-11


Langmuir Isotherm AdsorptionAdsorption of carbon monoxide onto a surface: CO + S ⇌ CO·S

AD A CO v A CO Sr k P C k C CO S

AD A CO vA

Cr k P C

K

AA

A

kK

k

Put Cv in terms of Ct using the site balance; only CO can absorb on the surface:

t v CO SC C C

At equilibrium, rAD = 0:CO S

AD A CO vA

Cr 0 k P C

K

Determine the concentration of CO adsorbed on the surface at equilibrium

Rearrange & solve for CCO·S

CO SCO v

A

CP C

K

t CO S vC C C Insert into eq for CCO·S from rxn rate

CO S A CO vC K P C CO S A CO t CO SC K P C C Solve for CCO·S

CO S A CO t A CO CO SC K P C K P C CO S A CO CO S A CO tC K P C K P C

A CO tCO S

A CO

K P CC

1 K P

Concentration of CO adsorbed on surface vs PCO→ Langmuir Isotherm

CO S A CO vC K P C

L17-12


Surface Reaction StepAfter the molecule is adsorbed onto the surface, it can react by a few different mechanisms1. Singe site mechanism: Only the site to which the reactant is absorbed is

involved in the reactionAI

-S-⇌

BI

-S-

A·S ⇌ B·SB S

S S A SS

Cr k C

K

SS

S

kwhere K

k

2. Dual site mechanism: Adsorbed reactant interacts with another vacant site to form the product

AI

-S-S-S⇌

B I

-S-S-S-

A·S + S ⇌ S + B·S

B S vS S A S v

S

C Cr k C C

K

Equation IIa

Equation IIb

3. Eley-Rideal mechanism: reaction between adsorbed reactant and a molecule in the gas phase

⇌

C I

-S-S-S-

A·S + B(g) ⇌ C·S

C SS S A S B

S

Cr k C P

K

Equation IIcAI

-S-S-S

B

L17-13


Desorption StepProducts are desorbed into the gas phase

C I

-S-S-S-⇌

C

-S-S-S-

C·S ⇌ C + S

C vD,C D C S

D,C

P Cr k C

K

DD,C

D

kwhere K

k

Equation III

Note that the desorption of C is the reverse of the adsorption of C

D,C AD,Cr r

Also the desorption equilibrium constant KD,C is the reciprocal of the adsorption equilibrium constant KC

D,CC

1K

K

D,C D C S C C vr k C K P C Substituting 1/KC for KD,C in the rate equation for product desorption gives:

L17-14


Derive a Rate Law for Catalytic Rxn• Postulate catalytic mechanism, and then derive the rate law for that

mechanism• Assume pseudo-steady state hypothesis (rate of adsorption = rate of

surface reaction = rate of desorption) • No accumulation of species on the surface or near interface• Each species adsorbed on the surface is a reactive intermediate• Net rate of formation of species i adsorbed on the surface is 0, r i·S=0

• One step is usually rate-limiting• If the rate-limiting step could be sped up, the entire rxn would be faster• Although reactions involve all 7 steps, (for chapter 10) only adsorption,

surface reaction, or desorption will be rate limiting• The surface reaction step is rate limiting ~70% of the time!

• Steps to derive the rate law• Select among types of adsorption, surface reaction, and desorption• Write rate laws for each individual step, assuming all are reversible• Postulate which step is rate limiting• Use non-rate-limiting steps to eliminate the surface concentration

terms that cannot be measured

L17-15


Consider A B and assume the following mechanism is correct:⇌

A SAD A A v

A

Cr k P C

K

1. Adsorption:

2. Surface reaction:

A(g) + S ⇌ A·S

A·S + S ⇌ S + B·SB S v

S S A S vS

C Cr k C C

K

3. Desorption: B·S ⇌ B + SB v

D D B SD

P Cr k C

K

We need to select one of these 3 reactions as the rate limiting step, then derive the corresponding rate equation, and see if this rate eq matches experimental data. Which step is the most logical to start with? a) Adsorptionb) Surface reactionc) Desorptiond) None of the abovee) Any of these would be “logical” - they all have equal probability of being

the rate limiting step

The surface reaction step as is rate limiting ~70% of the time

L17-16



A SAD A A v

A

Cr k P C

K

1. Adsorption:

2. Surface reaction:

A(g) + S ⇌ A·S

A·S + S ⇌ S + B·SB S v

S S A S vS

C Cr k C C

K

3. Desorption: B·S ⇌ B + SB v

D D B SD

P Cr k C

K

Derive the rate equation for when the surface reaction is rate limiting (true ~70% of the time)

• For surface reaction-limited mechanisms, kS is small and kA and kD are relatively large

• Therefore rAD/kA and rD/kD are very small, and can be approximated as equal to zero

B S vA S S A S v

S

C Cr ' r k C C

K

1. CA·S, Cv, and CB·S need to be expressed in terms of measurable quantities

2. Use this relationship to eliminate CA·S and CB·S from their respective rate equations and the site balance to eliminate CV

L17-17



A SAD A A v

A

Cr k P C

K

1. Adsorption 2. Surface reaction

B S vS S A S v

S

C Cr k C C

K

3. Desorption

B vD D B S

D

P Cr k C

K

Derive the rate equation for when the surface reaction is rate limiting

B S vA S S A S v

S

C Cr ' r k C C

K

Use rAD/kA =0 and rD/kD =0 to eliminate CA·S and CB·S from their respective rate equations and the site balance to eliminate CV

A SAD A A v

D

Cr k P C

K

A SADA v

A A

Cr0 P C

k K

A SA v

A

CP C

K A S A A vC K P C

B vD D B S

D

P Cr k C

K

D B vB S

D D

r P C0 C

k K B vB S

D

P CC

K

Use rAD/kA =0 & rAD equation to solve for CA·S:

Use rD/kD =0 & rD equation to solve for CB·S:

L17-18



A SAD A A v

A

Cr k P C

K


B S vS S A S v

S

C Cr k C C

K

3. Desorption

B vD D B S

D

P Cr k C

K


B S vA S S A S v

S

C Cr ' r k C C

K

A S A A vC K P C rAD/kA =0 & rD/kD =0B v

B SD

P CC

K

Use site balance to solve for CV: t v A S B SC C C C

v t A S B SC C C C Make substitutions for CA·S & CB·S, solve for Cv

B vv t A A v

D

P CC C K P C

K B v

v A A v tD

P CC K P C C

K

Bv A A t

D

PC 1 K P C

K

t

vB

A AD

CC

P1 K P

K

L17-19



A SAD A A v

A

Cr k P C

K


B S vS S A S v

S

C Cr k C C

K

3. Desorption

B vD D B S

D

P Cr k C

K


B S vA S S A S v

S

C Cr ' r k C C

K

A S A A vC K P C B vB S

D

P CC

K tv

A A B D

CC

1 K P P K

Substitute in CA·S, CB·S, &Cv

2 2t tB

A S S A AA A B D S D A A B D

C CPr ' r k K P

1 K P P K K K 1 K P P K

2

t BA S S A A

A A B D S D

C Pr ' r k K P

1 K P P K K K

This is the rate equation in terms of measurable species and rate constants for the mechanism given in the problem statement

L17-20


Evaluating a Catalytic Reaction Mechanism

• Collect experimental data from test reactor• See if rate law is consistent with data• If not, then try other surface mechanism (i.e., dual-site

adsorption or Eley-Rideal) or choose a different rate-limiting step (adsorption or desorption)

L17-21



A SAD A A v

A

Cr k P C

K


B S vS S A S v

S

C Cr k C C

K

3. Desorption

B vD D B S

D

P Cr k C

K

Now derive the rate equation for when adsorption is rate limiting:

v

AA AD A A

A

Sr ' r k PK

CC

Conc of vacant and occupied sites must be eliminated from the rate equation

A S

B

SS

S vv

S CC

r CC

k K0

B S v

vS

A SKC

C CC

S

AB S

SKC

C

BB

D

v

D

DS

r P0

C

k KC B v

B SD

P CC

K

If adsorption is rate limiting, kS>>kAD, so rS/kS can be approximated as 0. Then:

B vS S v

S

SA S

Cr k C

KC

C

If adsorption is rate limiting, kD>>kAD, so rD/kD can be approximated as 0. Then:

BB S

vD D

D

P Cr Ck

K

Need to put CB·S in measureable terms

L17-22



A SAD A A v

A

Cr k P C

K


B S vS S A S v

S

C Cr k C C

K

3. Desorption

B vD D B S

D

P Cr k C

K


Conc of vacant and occupied sites must be eliminated from the rate eq

SA

B SSK

CC B

B SD

vPC

K

C

Make substitutions for CA·S & CB·S

B S Bv

Dt

v

SC

C P C

K KC

Bv

B

Dt

v v

DSK K

CP P CC

KC

B B

tS D D

vP P

C 1K K K

C

tv

B B

S D D

CC

P P1

K K K

Solve for Cv using the site balance equation

A B Sv St C CCC

Substitute Cv into the expression for CB·S:

t

B B

S D

B

D

B S

D

C

P P1

K K K

PC

K

Substitute CB·S into CA·S:

B t

B BD

S

A S

DS

D

P C

P P1

KK K

K K

C

v

AA AD A A

A

Sr ' r k PK

CC

L17-23



A SAD A A v

A

Cr k P C

K


B S vS S A S v

S

C Cr k C C

K

3. Desorption

B vD D B S

D

P Cr k C

K


B tA S

B BS D

S D D

P CC

P PK K 1

K K K

B tB S

B BD

S D D

P CC

P PK 1

K K K

tv

B B

S D D

CC

P P1

K K K

AAD A

A

B t

B BS D

S D D

t

B B

S D D

P C

P PK K

CP P

1K K

1K

KKK K

Pr k

Use these eqs to replace CA·S

& Cv in rAD:

A BAD

B B B BA S D

A t

S D D S D D

P Pr

P P P P1 K K K 1K K K K

C

K

k

K

k

v

AA AD A A

A

Sr ' r k PK

CC

Factor out Ct:

L17-24


The gas phase hydromethylation of toluene: C6H5CH3 + H2 → C6H6 + CH4 is to be carried out in a PBR. Plot the conversion and the partial pressures of toluene, hydrogen and benzene as a function of catalyst weight.

FA0 = 50 mol toluene/min, P0 = 40 atm, T= 913K, a= 9.8 x 10-5 kg-1, Feed is 30% toluene (species tol), 45% hydrogen (species H) and 25% inerts (I)

Rate law: ' H tolT

tol tol B B

kP Pr

1 K P K P

k= 0.00087 mol/atm2·kg cat·minKB = 1.39 atm-1 Ktol= 1.038 atm-1

Mole balance:'

tol tol

A0

dX r

dW F

Rate law: ' H tol

Ttol tol B B

kP Pr

1 K P K P

Stoichiometry: tol 0tol tol,0

tol 0

1 X TPC C

1 X P T

tolT T0

tol 0

1 X PC C

1 X P

1Need concentrations in terms of pressure

tol tol tolP V n RT tol tolP C RT toltol

PC

RT tol,0

tol,0P

CRT

tol,0tol tol

tol 0

PP 1 X PRT RT 1 X P

toltol tol,0

tol 0

1 X PP P y where y=

1 X P

tol,0y 1 1 1 1 0 tol,0

Tol,0T0

F 0.3y 0.3

F 1 0.3 0 0

Total P

Total P at inlet

L17-25




Rate law: ' H tolT

tol tol B B

kP Pr

1 K P K P


Mole balance:'

tol tol

A0

dX r

dW F

Rate law: ' H tol

Ttol tol B B

kP Pr

1 K P K P

Stoichiometry: tol tol,0 tol0

PP P 1 X y where y=

P

0 tol,0y 0.3

0 tol,0 tol,0 0P y P tol,0P 0.3 40atm 12atm

Use the Ergun equation to evaluate y:

Isothermal rxn & =0, so: 1 2y 1 Wa 1 2tol tol,0 tolP P 1 X 1 Wa

Need an equation for PB & PH

B tol,0 tolP P X y H tol,0 H tol2 2P P X y H2

H2tol

F 0.451.5

F 0.3

L17-26




Rate law: ' H tolT

tol tol B B

kP Pr

1 K P K P


Mole balance:'

tol tol

A0

dX r

dW F

Rate law: ' H tol

Ttol tol B B

kP Pr

1 K P K P

Stoichiometry:

0 tol,0y 0.3

0

tol,0P 12atm 1 2tol tol,0 tolP P 1 X 1 Wa

B tol,0 tolP P X y H tol,0 tol2P P 1.5 X y

Finally, calculate the kg cat where P0 = 1atm: 1 2

0

P1 W

Pa 1 25 11atm

1 9.8 10 kg W40atm

1 2y 1 Wa

5 10.000625 1 9.8 10 kg W finalW 10197.7kg

Plug equation in boxes into Polymath to solve

L17-27


L17-28


L17-29


l17-1 slides courtesy of prof m l kraft, chemical & biomolecular engr dept, university of...

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