1 - 17/04/2015 Department of Chemical Engineering Lecture 4 Kjemisk reaksjonsteknikk Chemical...

1 - 04/18/23

Departm

ent of Chem

ical E

ngineering

Lecture 4

Kjemisk reaksjonsteknikk

Chemical Reaction Engineering

Review of previous lectures Stoichiometry Stoichiometric Table Definitions of Concentration Calculate the Equilibrium Conversion, Xe

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Reactor Mole Balances in terms of conversion

Reactor Differential Algebraic Integral

A

0A

r

XFV

CSTR

A0A rdV

dXF

X

0 A0A r

dXFVPFR

Vrdt

dXN A0A

Vr

dXNt

X

0 A0A Batch

X

t

A0A rdW

dXF

X

0 A0A r

dXFWPBR

X

W2

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dDcCbBaA

d

r

c

r

b

r

a

r DCBA

Da

dC

a

cB

a

bA

Last LectureRelative Rates of Reaction

3

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BAA CkCr

βα

β

α

OrderRection Overall

Bin order

Ain order

Last LectureRate Laws - Power Law Model

4

C3BA2 A reactor follows an elementary rate law if the reaction orders just happens to agree with the stoichiometric coefficients for the reaction as written.e.g. If the above reaction follows an elementary rate law

2nd order in A, 1st order in B, overall third order

B2AAA CCkr

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Last LectureArrhenius Equation

k Ae E RT

k is the specific reaction rate (constant) and is given by the Arrhenius Equation.

Where:

T k A

T 0 k 0

A1013

k

T5

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These topics build upon one another

Mole Balance Rate Laws Stoichiometry

Reaction Engineering

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How to find

rA f X

rA g Ci Step 1: Rate Law

Ci h X Step 2: Stoichiometry

rA f X Step 3: Combine to get

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We shall set up Stoichiometry Tables using species A as our basis of calculation in the following reaction. We will use the stochiometric tables to express the concentration as a function of conversion. We will combine Ci = f(X) with the appropriate rate law to obtain -rA = f(X).

Da

dC

a

cB

a

bA

A is the limiting Reactant.

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NA NA 0 NA 0X

Xa

b

N

NNXN

a

bNN

A

BAABB

0

0000

For every mole of A that react, b/a moles of B react. Therefore moles of B remaining:

Let ΘB = NB0/NA0

Then:

NB NA 0 B b

aX

NC NC 0 c

aNA 0X NA 0 C

c

aX

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Species Symbol Initial Change Remaining

Batch System Stoichiometry Table

B B NB0=NA0ΘB -b/aNA0X NB=NA0(ΘB-b/aX)

A A NA0 -NA0X NA=NA0(1-X)

Inert I NI0=NA0ΘI ---------- NI=NA0ΘI

FT0 NT=NT0+δNA0X

Where: 0A

0i

0A

0i

00A

00i

0A

0ii y

y

C

C

C

C

N

N

1a

b

a

c

a

dan

d

C C NC0=NA0ΘC +c/aNA0X NC=NA0(ΘC+c/aX)

D D ND0=NA0ΘD +d/aNA0X ND=NA0(ΘD+d/aX)

10δ = change in total number of mol per mol A reacted

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Constant Volume Batch

Note: If the reaction occurs in the liquid phaseor

if a gas phase reaction occurs in a rigid (e.g. steel) batch reactor

V V0Then

CA NA

VNA 0 1 X

V0

CA 0 1 X

CB NB

VNA 0

V0

B b

aX

CA 0 B

b

aX

etc.11

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Suppose

rA kACA2CB

Batch: 0VV

Stoichiometry

12

rA kACA 02 1 X 2 B

b

aX

Equimolar feed:

B 1

Stoichiometric feed:

B b

a

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A A FA0 -FA0X FA=FA0(1-X)

Species Symbol Reactor Feed Change Reactor Effluent

B B FB0=FA0ΘB -b/aFA0X FB=FA0(ΘB-b/aX)

i Fi0

FA 0

Ci00

CA 00

Ci0

CA 0

y i0

yA 0

Where:

Flow System Stochiometric Table

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Species Symbol Reactor Feed Change Reactor Effluent

0A

0i

0A

0i

00A

00i

0A

0ii y

y

C

C

C

C

F

F

Where:

Flow System Stochiometric Table

Inert I FI0=A0ΘI ---------- FI=FA0ΘI

FT0 FT=FT0+δFA0X

C C FC0=FA0ΘC +c/aFA0X FC=FA0(ΘC+c/aX)

D D FD0=FA0ΘD +d/aFA0X FD=FA0(ΘD+d/aX)

1a

b

a

c

a

dand

A

A

FCConcentration – Flow System14

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Species Symbol Reactor Feed Change Reactor Effluent

A A FA0 -FA0X FA=FA0(1-X)

B B FB0=FA0ΘB -b/aFA0X FB=FA0(ΘB-b/aX)

C C FC0=FA0ΘC +c/aFA0X FC=FA0(ΘC+c/aX)

D D FD0=FA0ΘD +d/aFA0X FD=FA0(ΘD+d/aX)

Inert I FI0=FA0ΘI ---------- FI=FA0ΘI

FT0 FT=FT0+δFA0X

0A

0i

0A

0i

00A

00i

0A

0ii y

y

C

C

C

C

F

F

1a

b

a

c

a

dWhere: and

A

A

FCConcentration – Flow System

Flow System Stochiometric Table

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A

A

FCConcentration Flow System:

0Liquid Phase Flow System:

X1CX1FF

C 0A0

0AAA

X

a

bCX

a

b

V

N

V

NC B0AB

0

0ABB

Flow Liquid Phase

etc.16

Liquid Systems

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BAA CkCr If the rate of reaction were

X

a

bX1Cr B

20AAthen we would have

XfrA This gives us

FA 0

rA

X17

Liquid Systems

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For Gas Phase Flow Systems

We obtain:

0

0

0T

T0 T

T

P

P

F

F

Combining the compressibility factor equation of state with Z = Z0

CT P

ZRT

CT 0 P0

Z0R0T0

FT CTV

FT 0 CT 0V0

Stoichiometry:

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T

T

P

P

F

FF

T

T

P

P

FF

FFC

T

jT

T

T

jjj

0

00

00

0

00

CB CT 0

FB

FT

P

P0

T0

T

CT 0 FT 0 0

For Gas Phase Flow Systems

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XFFF 0A0TT The total molar flow rate is:

For Gas Phase Flow Systems

P

P

T

T

F

XFF 0

00T

0A0T0

P

P

T

TX

F

F1 0

00T

0A0

P

P

T

TXy1 0

00A0

P

P

T

TX1 0

00

Substituting FT gives:

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21

For Gas Phase Flow Systems

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Gas Phase Flow System:

Concentration Flow System:

0

00A

0

00

0AAA P

P

T

T

X1

X1C

PP

TT

X1

X1FFC

A

A

FC

P

P

T

TX1 0

00

0

0B0A

0

00

B0AB

B P

P

T

T

X1

Xab

C

P

P

TT

X1

Xab

FF

C

22

For Gas Phase Flow Systems

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For Gas Phase Flow Systems

0

00

0

00

0

11 P

P

T

T

X

Xab

C

PP

TT

X

Xab

FF

CjAjA

jj

Cj=f(Fj, T, P) =f(x, T,P)

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2

0

0

B2

0AAA T

T

P

P

X1

Xab

X1

X1Ckr

If –rA=kCACB

This gives us

FA0/-rA

X24

For Gas Phase Flow Systems

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Consider the following elementary reaction with KC=20 dm3/mol and CA0=0.2 mol/dm3. Calculate Equilibrium Conversion or both a batch reactor (Xeb) and a flow reactor (Xef).

Example: Calculating the equilibrium conversion for gas phase reaction in a flow reactor, Xef

C

B2AAA K

CCkr

BA2

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Consider the following elementary reaction with KC=20 m3/mol and CA0=0.2 mol/m3. Xe’ for both a batch reactor and a flow reactor.

Calculating the equilibrium conversion for gas phase reaction,Xe

C

B2AAA K

CCkr

BA2

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Batch Reactor Example

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Step 1:

dX

dt rAVNA 0

Lmol 2.0C 0A

KC 20 L mol

Step 2: rate law, BB2AAA CkCkr

Calculate Xe

B

AC k

kK

C

B2AAA K

CCkr

28

Batch Reactor Example

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Symbol Initial Change Remaining

B 0 ½ NA0X NA0 X/2

A NA0 -NA0X NA0(1-X)

Totals: NT0=NA0 NT=NA0 -NA0 X/2

@ equilibrium: -rA=0 C

Be2Ae K

CC0

Ke CBe

CAe2

CAe NAe

VCA 0 1 Xe

CBe CA 0

X e

229

Calculate Xe

Batch Reactor Example

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Departm

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A NA0 -NA0X NA=NA0(1-X)

B 0 +NA0X/2 NB=NA0X/2

NT0=NA0 NT=NA0-NA0X/2

Solution:

2Ae

BeC C

CK

C

Be2AeAA K

CCk0rAt equilibrium

0VV 2/BAStoichiometry

Constant volumeBatch

Calculating the equilibrium conversion for gas phase reaction

30

Batch Reactor Example

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2e0A

e2

e0A

e0A

eX1C2

X

X1C2

XC

K

82.0202

X1

XCK2 2

e

e0Ae

X eb 0.703

BR Example Xeb

31

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Gas Phase Example Xef

2A B

B2

1A

X ef ?

Rate law:

C

B2AAA K

CCkr

X eb 0.703

32

Solution:

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Species Fed Change Remaining

A FA0 -FA0X FA=FA0(1-X)

B 0 +FA0X/2 FB=FA0X/2

FT0=FA0 FT=FA0-FA0X/2

Gas Flow Example Xef

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A FA0 -FA0X FA=FA0(1-X)

B 0 FA0X/2 FB=FA0X/2

Stoichiometry: Gas isothermal T=T0, isobaric P=P0

V V0 1X

CA FA 0 1 X V0 1X

CA 0 1 X 1X

CB FA 0 X 2

V0 1X CA 0 1 X 2 1X

Gas Flow Example Xef

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Pure A yA0=1, CA0=yA0P0/RT0, CA0=P0/RT0

C

0A

2

0AAA KX12

XC

X1

X1Ckr

2e

ee0AC

X1

X1XCK2

yA 0 1 1

2 1

1

2

@ eq: -rA=0

Gas Flow Example Xef

35

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8dm

mol2.0

mol

dm202CK2

3

3

0AC

yA 0 11

2 1

1

2

8 Xe 0.5Xe

2

1 2Xe Xe2

8.5X e2 17X e 8 0

Gas Flow Example Xef

X ef 0.757Flow: Recall

X eb 0.70Batch:36