Chapter 1 (Notes) (2)

7/21/2019 Chapter 1 (Notes) (2)

http://slidepdf.com/reader/full/chapter-1-notes-2 1/76

!

Lecturer and Unit Coordinator (Clayton):

Prof. Sankar Bhattacharya (Department of Chemical Engineering)

Email: [email protected]

Lecturer and Unit Coordinator (Malaysia):

Dr. Chai Siang Piao (School of Engineering)

Email: [email protected]

Tutor: Ms. Tan Lling Lling ([email protected]) Ms. Yeo Zee Ying ([email protected])

Lab demonstrator: Mr. Oh Kai Siang ([email protected])

ACADEMIC STAFF

1

7/21/2019 Chapter 1 (Notes) (2)


#

Tutorial will start in week 2.Laboratory will start in week 6.

SCHEDULE OF LECTURES, LABORATORY SESSIONS AND PROBLEMS CLASSES

2

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<=>

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01123 -7890128**)(

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01123 -780<8+$$%

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$%&'!()*+,*-#*.,/ 0>?4=3>6$':@(

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$%&'!()*+,*-#*.,/ <=4.3C1D6$)

01123 -78<=4.3C1D8+%$&

$+6$":()6$&; $"6'$:((6'$

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01123 -78<=4.3C1D8+%$&

$"6$":(+6$&; '$6'$:()6'$

7/21/2019 Chapter 1 (Notes) (2)


'UNIT OUTLINE

PRESCRIBED TEXTBOOK: H. Scott Fogler, Elementsof Chemical Reaction Engineering , 4th edition, Prentice

Hall, Upper Saddle River, New Jersey, USA, 2005.

OTHER REFERENCE: Octave Levenspiel, Chemical

Reaction Engineering , 3rd edition, Wiley, New York,USA, 1999.

3

7/21/2019 Chapter 1 (Notes) (2)


)

4

Chapter Topic Sections in text*

1

2

34

5

6

78

Mole balancesConversion and Reactor sizing

Rate laws and stoichiometryIsothermal reactor design

Collection and analysis of rate dataMultiple reactions

Reaction mechanisms, Bioreactions and

BioreactorsNon-isothermal reactors and Energy Balance

Catalysis and catalytic reactors

Heterogeneous reactionsDiffusion and reactions

1.1-1.52.1-2.6

3.1-3.64.1-4.4, 4.6-4.10

5.1-5.76.1-6.3

7.1-7.3,

8.2-8.6, 9.1-9.34.5,

10.1-10.4,10.7

11.1-11.512.1-12.7

UNIT OUTLINE

7/21/2019 Chapter 1 (Notes) (2)


UNIT SCHEDULE

Week Lecture (chapters in Fogler)/Test Tutorialclasses

Laboratory

1 Introduction - -

2 Mole balances: 1.1-1.5

Conversion and reactor sizing: 2.1-2.6

Homeworkfrom week 1

-

3 Rate laws and stoichiometry: 3.1-3.6 Homeworkfrom week 2

-

4 Isothermal reactor: 4.1-4.4, 4.6-4.10

Collection/analysis of raw data: 5.1-5.7

Homeworkfrom week 3

-

5 Collection/analysis of raw data: 5.1-5.7 – continued

Multiple Reactions: 6.1-6.6

Test 1

Homeworkfrom week 4

-

6 Multiple Reactions: 6.1-6.6 - continued Homeworkfrom week 5

Yes

5

7/21/2019 Chapter 1 (Notes) (2)


(

7 Reaction mechanisms: 7.1-7.3

Non-isothermal reactor: 8.2-8.6, 9.1-9.3

Homeworkfrom week 6

Yes

8 Non-isothermal reactor: 8.2-8.6, 9.1-9.3 - continued Homeworkfrom week 7

Yes

9 Non-isothermal reactor: 8.2-8.6, 9.1-9.3 - continued

Catalysis and catalytic reactor: 4.5. 10.2-10.4, 10.7, catalyst

characterization Test 2

Homeworkfrom week 8

Yes

10 Catalysis and catalytic reactor: 4.5. 10.2-10.4, 10.7, catalystcharacterization – continued

Catalysis and catalytic reactor - Heterogeneous reactions:

11.1-11.5

Homeworkfrom week 9

Yes

11 Catalysis and catalytic reactor - Heterogeneous reactions:

11.1-11.5 – continued

Homework

from week 10

-

12 Catalysis and catalytic reactor - Diffusions and Reactions:12.1 – 12.7

Take home assignment

Revision, Q & A

Homeworkfrom week 11

-

6

UNIT SCHEDULE

7/21/2019 Chapter 1 (Notes) (2)


5ASSESSMENT

To pass this unit, you must achieve at least a minimum of 45% on the class-work (totalcontinuous assessment), and a minimum of 45% on the final exam. Otherwise you will

receive a maximum mark of “45” as the final grade. The overall pass mark is still 50%.

7

Week Activity Value

5 Test 1 – on materials taught up to week 4

Closed book

7.5%

9 Test 2 – on materials taught up to week 8

Closed book

7.5%

12 Take home assignment – on materials taught from week 1–11 5%

One Laboratory report covering both experiments 10%

Total continuous assessment 30%

Final Examination (3 hours closed book) 70%

Total assessment 100%

7/21/2019 Chapter 1 (Notes) (2)


CHAPTER 1

INTRODUCTION

6

8

7/21/2019 Chapter 1 (Notes) (2)


What is Reaction?

7

9

7/21/2019 Chapter 1 (Notes) (2)


10

7/21/2019 Chapter 1 (Notes) (2)


Differentiate chemical

engineers from otherengineers

11

7/21/2019 Chapter 1 (Notes) (2)


Physicaltreatment

processesReaction

Physicaltreatment

processes

To purify raw materials:

Distillation, UF, RO, separatorTo separate products from reactants:

Distillation, separator

Recycle

Typical chemical plant

12

7/21/2019 Chapter 1 (Notes) (2)


13

! Definition - chemical reaction

"

A chemical species is said to have reacted when it haslost its chemical identity.

" Three main ways a chemical species can lose its

chemical identity:- decomposition (dehydrogenation)

- combination

- isomerization

22233 CH C H H CH CH =+!

NOO N 222 !+

232252 )(CH C CH CH CH H C =!=

7/21/2019 Chapter 1 (Notes) (2)


! Types of chemical reaction:

1) Combustion: CH4 + O2 CO2 + 2H2O

2) Synthesis: A + B # AB: CH4 + H2O CO + 3H2

3) Decomposition: AB # A + B: CH4 C + H2

4) Single displacement: A + BC # AC + B: Mg + 2 H2O Mg(OH)2 + H2

5) Double displacement: AB + CD # AD + CB: Pb(NO3)2 + 2KI PbI2 +

2KNO3

6) Acid-base: HA + BOH # H2O + BA: HBr + NaOH NaBr + H2O

14

7/21/2019 Chapter 1 (Notes) (2)


! Classification of reactions:

- According to the number of phases involved in the reaction

a) Homogeneous – chemical reactions in which the reactants

(or catalyst and reactants) are in the same phase

b) Heterogeneous – chemical reactions in which the reactants

(or catalyst and reactants) are in two or more phases

Catalytic reactions: Ammonia synthesis, hydrocracking of crude oil,

Gas-To-Liquid (GTL)Non-catalytic reactions: Burning of coal, attack of solids by acids, etc.

15

7/21/2019 Chapter 1 (Notes) (2)


! Definition - Catalyst

- A substance that initiates and accelerates a chemical

reaction without itself being consumed.

- Homogeneous catalysts, heterogeneous catalysts and

biocatalyst

16

7/21/2019 Chapter 1 (Notes) (2)


- Equipment used to carry out reaction - Reactor

a) Batch reactor

b) Flow reactor – CSTR, tubular reactor

c) Catalytic reactor – Fixed bed, moving bed, fluidised bed

17

Batch CSTRTubular

7/21/2019 Chapter 1 (Notes) (2)


18

7/21/2019 Chapter 1 (Notes) (2)


" Industrial reactors

19

Type of Reactor Characteristics

Simple Batch Reactor is charged via two holes in the top of the tank; while reaction is

carried out, nothing else is put in or taken out until the reaction is done;

tank easily heated or cooled by jacket

Kinds of Phases

Present

Usage Advantages Disadvantages

1. Gas phase

2. Liquid phase

3. Liquid Solid

1. Small scale

production

2. Intermediate or one

shot production

3. Pharmaceutical

4. Fermentation

1. High conversion per unit

volume for one pass

2. Flexibility of operation-

same reactor can produce

one product one time and a

different product the next

3. Easy to clean

1. High operating cost

2. Product quality

7GG=HIJ./G

'K L. C/M.N .3 .=4 M.N F7$

O F7 O $

(K A>DD HCP>B

7/21/2019 Chapter 1 (Notes) (2)


20


Continuously Stirred TankReactor (CSTR)

Run at steady state with continuous flow of reactants and products; thefeed assumes a uniform composition throughout the reactor, exit stream

has the same composition as in the tank

Kinds of Phases

Present


1. Liquid phase

2. Gas-liquid rxns

3. Solid-liquid rxns

1. When agitation is

required

2. Series configurations

for different

concentration streams

1. Continuous operation

2. Good temperature control

3. Easily adapts to two phase

runs

5. Simplicity of construction

6 Low operating (labor) cost

7. Easy to clean

1. Lowest

conversion perunit volume

2. By-passing and

channeling

possible with poor

agitation

Assumptions1) Steady state therefore

2) Well mixed therefore r A is the same throughout the reactor

7/21/2019 Chapter 1 (Notes) (2)


21


Plug flow reactor (PFR) Arranged as one long reactor or many short reactors in a tube bank ; noradial variation in reaction rate (concentration); concentration changes

with length down the reactor

Kinds of Phases

Present


1. Primarily Gas

Phase

1. Large Scale

2. Fast Reactions

3. Homogeneous

Reactions

4. Heterogeneous

Reactions

5. ContinuousProduction

6. High Temperature

1. High Conversion per

Unit Volume

2. Low operating (labor)

cost)

3. Continuous

Operation

4. Good heat transfer

1. Undesired thermal

gradients may exist

2. Poor temperature

control

3. Shutdown and

cleaning may be

expensive

7/21/2019 Chapter 1 (Notes) (2)


22


Tubular fixed bed reactor Tubular reactor that is packed with solid catalyst particles

Kinds of Phases

Present


1. Gas phase/ solid

catalyzed

2. Gas-solid rxns

1. Used primarily in

heterogeneous has

phase reactions with

a catalyst

1. High conversion

per unit mass of

catalyst

2. Low operating cost

3. Continuous

operation

1. Undesired thermal

gradients may exist

2. Poor temperature

control

3. Channeling may

occur

4. Unit may be

difficult to serviceand clean

7/21/2019 Chapter 1 (Notes) (2)


23


Fluidized Bed Reactor Heterogeneous reactions; like a CSTR in that the reactants are well

mixed

Kinds of Phases

Present


1. Gas-solid

2. Gas-solid catalyzed

1. Heterogeneous has

phase reactions with a

catalyst

1. Good mixing

2. Good uniformity oftemperature

3. Catalyst can be

continuously

regenerated with the

use of an auxiliary loop

1. Bed-fluid mechanics

not well known

2. Severe agitation can

result in catalyst

destruction and dust

formation

3. Uncertain scale-up

7/21/2019 Chapter 1 (Notes) (2)


24

7/21/2019 Chapter 1 (Notes) (2)


!

Elementary & non-elementary reactions

i) Elementary reaction

ii) Non-elementary reaction

25

2

2

B A A

B A A

C kC r

R B A

C kC r

R B A

=!

"+

=!

"+

The rate equation corresponds to a stoichiometric

equation

[ ][ ][ ] [ ]

22

21

221

22

/

2

Br HBr k

Br H k r

HBr Br H

HBr

+

=

!+

No direct correspondence between stoichiometryand rate

7/21/2019 Chapter 1 (Notes) (2)


1.1 Rate of Reaction

The rate of reaction can be expressed as the rate of

disappearance of a reactant or as the rate of appearance

(formation) of a product .

The rate of reaction is ( the number of moles of reactant

disappearing ) or ( the number of moles of product formation )

per ( unit time ) per ( unit volume ).

A single-phase reaction aA + bB# cC + dD

26

!"

#$%

&='='

sm

mol

dt

dN

V r

A

A.

,ime)(volume)(t

ng)disappeariAof (amount13

Rate of disappearance of A

The minus sign means disappearance

7/21/2019 Chapter 1 (Notes) (2)


! A single-phase reaction aA + bB# cC + dD

The minus sign means disappearance

!"

#$%

&='='

sm

mol

dt

dN

V r

B

B.

,ime)(volume)(t

ng)disappeariBof (amount13

!"

#$%

&==

sm

mol

dt

dN

V r

C

C

.,

ime)(volume)(t

forming)Cof (amount13

!"

#$%

&==

sm

mol

dt

dN

V r

C

D

.,

ime)(volume)(t

forming)Dof (amount13

d

r

c

r

b

r

a

r DC B A

==

!

=

!

Rate of formation of D

#5

Relative rates

27

7/21/2019 Chapter 1 (Notes) (2)


A # B

r A = the rate of formation of species A per unit volume-r A = the rate of a disappearance of species A per unit volumer B = the rate of formation of species B per unit volume

Example: A B

If B is being created at a rate of 0.2 moles/dm3/s. What are the rate offormation of B, rates of formation and disappearance of A ?

Rate of formation of B: r B = 0.2 mole/dm3/s

Rate of disappearance of A: -r A = 0.2 mole/dm3/s

Rate of formation of A: r A = -0.2 mole/dm3/s

28

7/21/2019 Chapter 1 (Notes) (2)


Example 1.1A rocket engine burns a stoichiometric mixture of fuel (liquid hydrogen) in oxidant (liquid oxygen).The combustion chamber is cylindrical, 75 cm long and 60 cm in diameter, and the combustion

process produces 108 kg/s of exhaust gases (H2O). If combustion is complete, find the rate of

reaction of hydrogen and oxygen. Complete combustion: H2 + ! O2 H2O

Solution

Rate of reaction: and

Reactor volume: V = !/4 (0.6)2(0.75) = 0.2121 m3

H2O produced = 108 kg/s (1kmol/18 kg) = 6 kmol/s

According to the stoichiometry equation, to produce 6 kmol/s of H2O requires 6 kmol/s of H2 and

3 kmol/s of O2

Thus, H2 consumed= 6 kmol/s and O2 consumed = 3 kmol/s

Rate of equation:

29

dt

dN

V r

H

H

2

2

1!=!

dt

dN

V r

O

O

2

2

1!=!

.sm

mol10x2.829

s

kmol6)-(0

m0.2121

13

4

32=!=!

H r

.sm

mol10x1.415

s

kmol3)-(0

m0.2121

13

4

32=!=!

Or

2/11

22 O H r r !

=

!

7/21/2019 Chapter 1 (Notes) (2)


1.2 Definition of Conversion

"

7 GC/QD>:IE1G> 3>1?J./ 17 R 2S# ?T R BU

" FD.N G5G4>HG

F7$ O >/4>3C/Q H.D13 M.N 314> VH.D6GK

T7$ O H.D13C45 VC/CJ1D ?./?>/431J./; H.D6BH%K VW.3 DCX=CBK

Y$ O >/4>3C/Q Z.D=H>43C? M.N 314> VBH%6GK

30

A0

AA0A

N

N N

fed)Aof (moles

reacted)Aof (molesX

!

==

Initial amount of A

Amount of A at time t

X N N N N

N NX

A0A0A

A0

AA0

A

!=

!

=

Moles of A that havebeen consumed

X)( F F

F

F F X

A A

A

A A

!=

!

=

10

0

0

000 vC F A A

=

X)1( N N A0A !=

7/21/2019 Chapter 1 (Notes) (2)


!

For gas phase

C A0 = entering concentration (mol/dm3)

y A0 = entering mole fraction of A

P0 = entering total pressure, kPa

T0 = entering temperature, K

P A0 = entering partial pressure, kPa

R = ideal gas constant

31

0

00

0

0

0

0

0

000

RT

P yC

RT

P C

V

n

nRT V P

A A

A A

A

A A

=

==

=Idea gas law

7/21/2019 Chapter 1 (Notes) (2)


Example 1.2A gas mixture consists of 50% A and 50% inerts at 10 atm (1013 kPa) and entersthe reactor with a flow rate of 6 dm3 /s at 422.2 K. Calculate the entering

concentration of A, CA0, and the entering molar flow rate, FA0. The ideal gas

constant is R = 0.082 dm3.atm/mol.K.

Solution

Given, To calculate C A0 and F A0

P0 = 10 atm

y A0 = 0.5

P A0 = 0.5 x 10 = 5 atm

T0 = 422.2 K

V0 = 6 dm3/s

R = 0.082 dm3.atm/mol.K

32

mol/L0.144C

mol/dm14442.0K)(422.2.atm/mol.K dm0.082

atm)0.5(10C

RT

PyC

A0

3

3A0

0

0A0A0

=

==

=

mol/s867.0F

)/dm)(6.0mol/dm(0.14442vCF

A0

33

0A0A0

=

== s

7/21/2019 Chapter 1 (Notes) (2)


1.3 Space Time

!

.89:;<1:# [I1?> JH> V41=K CG 4E> JH> />?>GG135 4. I3.?>GG ./>

Z.D=H> .W 3>1?4.3 M=CB 14 4E> >/431/?> ?./BCJ./\

AE>3>; Y O 3>1?4.3 Z.D=H> VBH%K

Z$ O >/4>3C/Q Z.D=H>43C? M.N 314> VBH%6HC/K

[I1?> Z>D.?C45; VJH>:'K

33

0v

V =!

!

10==

V

vSV

(time)

7/21/2019 Chapter 1 (Notes) (2)


Example 1.3

The space time necessary to achieve 80% conversion in a CSTR is 5 h.Determine the reactor volume required to process 2ft3/min entering flow. What are

the reactor volume and space velocity for this system ?

Solution

t = 5 hrs Reactor volume, V = 5 hrs (2 ft3/min)(60 min/1 hr)

X = 0.8 V = 600 ft3

v0 = 2 ft3/min

34

!

1VvSVvelocity,Space 0

==

1-hr 2.0

5

1SV ==

7/21/2019 Chapter 1 (Notes) (2)


1.4 Basic Batch Reactor Design Equations

dt

dN dV r F F

dV r V r G

dt

dN

F G F

j

V

j j jo

V

j j j

j

j j jo

=+!

="=

=!+

#

#

35

REACTORFj,0 Fj

[Rate of flow

of j into the +

system]

[Rate of flow of j out

the system] =

[Rate of generation

of j by chemical -

reaction within the

system]

[Rate of

accumulation of j

within the system]

!

-.D> S1D1/?> 9X=1J./

7/21/2019 Chapter 1 (Notes) (2)


! Batch Reactor

In – out + generation = accumulation

F A0 and F A = 0; Assumption: the reactant is well-mixed

36

dt

dN G F F A

A A A =+!

0

V r dV r G A A A

== !

dt

dN G F F A

A A A =+!

0

0 0

dt

dN V r

A

A =

dt

dN

V r

A

A

1=

!

! !

"=

"=

0

0

1

1

0

A

A

A

A

N

N A

A

N

N A

At

r

dN

V t

r

dN

V dt

Design equation

7/21/2019 Chapter 1 (Notes) (2)


Constant-volume reactor

Constant-pressure reactor (volume varies)

37

dt

dC r

dt

V N d

dt

dN

V r

A

A

A A

A

=

==

)/(1

dt

V d C

dt

dC r

dt

dV C

V dt

VdC

V r

dt

V C d

V dt

dN

V r

A A

A

A A

A

A A

A

ln

11

)(11

+=

+=

==

NA = CAV

NA= mole of A (mol)

CA = concentration of A (mol/m3)

V = volume of reactor (m3)

Design equation

Design equation

7/21/2019 Chapter 1 (Notes) (2)


Example 1.4

The gas phase reaction A # B + C is carried out isothermally in a 20 dm3 constant-volume batch reactor. Twenty moles of pure A is initially placed in thereactor. The reactor is well mixed. Calculate the time necessary to reduce thenumber of moles of A in the reactor to 0.2 mol.

-r A = kC A with k = 0.865 min-1

Solution

Design equation for constant-volume batch reactor

V = 20 dm3

N A0 = 20 mol

N A

= 0.2 mol

38

dt

dC r

A

A =

=

=

==

t

t

V N

V N

k C

C

k

t

A

A

A

A

2.0

20ln

865.0

1

/

/ln

1ln

100

dt

dC r

A

A =

A

C

C A

A

C

C A

t

C d kC

t

C d

r

dt

A

A

A

A

! ! !

"=

=

0

0

1

1

0

5.32 min-1

7/21/2019 Chapter 1 (Notes) (2)


!

Batch reactor

39

X)(1 N N

X N N N

N

N NX

A0A

A0A0A

A0

AA0A

!=

!=

!

=

dt

dX N

dt

dN

dt

dX N0

dt

dN

A0

A

A0

A

!=

!=

dt

dN

V

1A

= Ar Design equation for batch reactor

dt

dX NVr

A0A !=

!

! !

=

=

X

0

A

A0

X

0A

A0

t

0

Vr -

dX Nt

Vr -

dX Ndt

Time required for the reaction toachieve conversion X

7/21/2019 Chapter 1 (Notes) (2)


1.5 Basic Continuous Flow Reactor Design Equations

!

Continuous Stirred Tank Reactor (CSTR)

In – out + generation = accumulation

At steady state,

40

dt

dN dV r F F AV

A A A =+! " 00

0=dt

dN A

A

A A

A A A

A A

V

A

r

F F V

F F V r

F F dV r

!

!=

!=!

!=!

"

0

0

0

F A = C A x v

F A = molar flow rate (mol/s)

C A = concentration (mol/m3)

v = volumetric flow rate (m3

/s)

X F F F

X F F F

F

F F X

A A A

A A A

A

A A

A

00

00

0

0

=!

!=

!

=

A

A

r

X F V

!

=0 Design equation

7/21/2019 Chapter 1 (Notes) (2)


! Plug Flow Reactor (PFR)

At steady state, ; Thus,

Differentiate with respect to V

41

dt

dN dV r F F

AV

A A A =+! " 00

0=dt

dN A ! =+"

V

A A A dV r F F 0

0

A

Ar

dV

dF !=!0

A

Ar

dV

dF =

! =

1

0

A

A

F

F A

A

r

dF V

! ! =

1

00

A

A

F

F A

AV

r

dF dV

dX F dF

X F F F

A A

A A A

0

00

!=

!=

!

! !

"=

=

"

=

"

X

A

A

X V

A

A

A

A

r

dX F V

dV dX r

F

r dV

dX F

00

0 0

0

0

Design equation asa function of FA

Design equation asa function of X

7/21/2019 Chapter 1 (Notes) (2)


! For packed-bed reactor (PBR)

,

W = mass of catalyst

42

! =

1

0'

A

A

F

F A

A

r

dF W

!

! !

"=

=

"

=

"

X

A

A

X W

A

A

A

A

r

dX F W

dW dX r

F

r dW

dX F

0 '0

0 0'

0

'0

dX F dF

X F F F

A A

A A A

0

00

!=

!=

'

A

Ar

dW

dF =

Design equation

7/21/2019 Chapter 1 (Notes) (2)


1.6 Reactors in Series and Parallel

43

)( 10 X

r

F V

A

A

CSTR

!

=

! "=

1

00

X

A A PFR r

dX F V

7/21/2019 Chapter 1 (Notes) (2)


Example 1.5

<E> >P.4E>3HC? 3>1?J./ V7 R S # TK N1G ?133C>B .=4 1BC121J?1DD5 1/B 4E> W.DD.NC/Q B1413>?.3B>B#

<E> >/4>3C/Q H.D13 M.N 314> .W 7 N1G %$$ H.D6HC/\ AE14 13> 4E> T[<] 1/B @F] Z.D=H>

/>?>GG135 4. 1?EC>Z> )$^ ?./Z>3GC./ _

!"#$%"&

U>GCQ/ >X=1J./#

Y@F] O %$$ P V13>1 =/B>3 4E> ?=3Z>K

`GC/Q [CHIG./aG ./>:4EC3B 3=D> 4. ?1D?=D14> 4E> 13>1

44

X 0 0.2 0.4 0.5 0.6 0.8 0.9

-rA (mol/dm3.min) 10 16.67 50 50 50 12.5 9.09

A

A

CSTR

r

X F V

!

=10

3

3

dm2.4 .minmol/dm50

.4)mol/min)(0(300

=

=CSTR

V

! "=

1

00

X

A

A PFR

r

dX F V

3

0

dm7.2

50

1

67.16

4

10

1)2.0)(

3

1(300

)4.0(

1

)2.0(

4

)0(

1)

3(

=

!"

#$%

&++=

!"

#$%

&

'+

'+

'=

PFR

A A A

A PFR

V

r r r

h F V

In this case, VCSTR < VPFR. Thus,using CSTR is suitable

7/21/2019 Chapter 1 (Notes) (2)


! Two CSTR in series

Mole balance on R1 Mole balance on R2

45

7/21/2019 Chapter 1 (Notes) (2)


! Two plug reactors in series

46

V1 V2

F A0 F A1F A2

X1 X2

! "=

1

001

X

A

Ar

dX

F V ! "=

2

102

X

X A

Ar

dX

F V

! ! ! "

+

"

=

"

=

2

1

12

00

00

0

X

X

A

A

X

A

A

X

A

ATotal

r

dX F

r

dX F

r

dX F V

7/21/2019 Chapter 1 (Notes) (2)


! PFR and CSTR in series

47

1

10

1

A

A

r

X F V

!

= ! "=

2

12

0

2

X

X A

AdX

r

F V

3

2303

)(

A

A

r

X X F V

!

!

=

7/21/2019 Chapter 1 (Notes) (2)


Example 1.6

The adiabatic exothermic irreversible gas phase reaction is to be carried out in a flow

reactor for a stoichiometric feed of A and B.

a) What PFR volume is necessary to achieve 50% conversion? (150 dm3)

b) What CSTR volume is necessary to achieve 50% conversion? (250 dm3)

c) What CSTR volume must be added to raise the conversion in Part (b) to 80%? (30 dm3)

d) What PFR volume must be added to raise the conversion in Part (b) to 80%? (90 dm3)

48

7/21/2019 Chapter 1 (Notes) (2)


)7

Read:

Examples 1.1, 2.1 – 2.8

49

7/21/2019 Chapter 1 (Notes) (2)


1.7 Rate Laws

!

Rate laws: algebraic equation that relates rate of reaction (-r A) to thespecies concentrations.

!

The rate of a reaction is a function of temperature (through the rate

constant) and concentration.

!

Relative rates of reaction

50

dDcC bBaA +!+

d

r

c

r

b

r

a

r DC B A

==!

=!

For example, in the reaction

22 22 NOO NO !+

smmol r NO !

=

3

/42

smmol r NO

!"=

3/4

smmol r O

!"=3

/22

! "

B A A C kC r =#

7/21/2019 Chapter 1 (Notes) (2)


Example 1.7

The Reaction: 2A + 3B -> 5C is carried out in a reactor. If at aparticular point, the rate of disappearance of A is 10 mol/dm3/s, what

are the rates of B and C ?

Solution

51

The rate of disappearance of A, -r a, is

The rate of formation of species A is

Species BThe rate of formation of species B is

The rate of disappearance of B, -r b,is

Species C

The rate of formation of C, r c, is

7/21/2019 Chapter 1 (Notes) (2)


" Power law

Power law model

k = rate constant

The reaction is " order with respect to reactant A, and

# order with respect to reactant B. Overall reaction order = " + #

Example:

Gas phase synthesis of phosgene

The reaction is first order with respect to carbon monoxide and

three-halves order with respect to chlorine.

52

! "

B A A C kC r =#

2/3

2Cl COCO C kC r !=!

7/21/2019 Chapter 1 (Notes) (2)


53

r 0 = k 4

n

CHP

!"#$

%&

+++++

=' +

'

424242422

4

23

51

23

521

21

23

5321

25

54321

25

221

5

25

543216

CH H CH H CH H CH H H

CH T

S C

P P K K P P K K K P P K K K K P K K K K K P P K

P C K K K K K k r

7/21/2019 Chapter 1 (Notes) (2)


54

C A -r A Reaction

OrderRate Law k

(mol/dm3) (mol/dm3*s) zero -r A = k (mol/dm3*s)

(mol/dm3) (mol/dm3*s) 1st -r A = kC A s-1

(mol/dm3) (mol/dm3*s) 2nd -r A = kC A2 (dm3/mol*s)

b.= ?1/ 4>DD 4E> .Z>31DD 3>1?J./ .3B>3 25 4E> =/C4G .W c

O H NaCl HCl NaOH 2

+!+

ik

O H NaCl HCl NaOH K k k k k

2

====

F.3 3>1?J./G C/ NEC?E 4E> GJ.?EC.H>43C? ?.>d?C>/4 CG ' W.3 1DD GI>?C>G; W.3

>P1HID>;

4E>/; 1DD 13> >X=1D; .3

7/21/2019 Chapter 1 (Notes) (2)


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

55

RT E

Aek !

=

k = rate constant

E = activation energy

R = gas constant

T = absolute temperature A = frequency factor

!"

#$%

&'=

'=

211

2 11

ln

lnln

T T R

E

k

k

RT

E Ak

Activation energy:The minimum energy that must be

possessed by reacting molecules

before the reaction will occur

7/21/2019 Chapter 1 (Notes) (2)


Example 1.8

Calculate the activation energy for the following reaction

A# B + C

Solution

56

e VG:'K $\$$$)% $\$$'"$

< VeK %'%\$ %(%\$

!"

#$%

&'=

211

2 11

lnT T R

E

k

k

!"

#$%

&'=!"

#$%

&

323

1

313

1

00043.0

00180.0ln

R

E mol kJ E

E

/120

)00310.000319.0(

)314.8(432.1

=

!

=

R = universal gas constant = 8.314 J/mol K

7/21/2019 Chapter 1 (Notes) (2)


Example 1.9 (Ex. 3.1 Fogler)

Calculate the activation energy for the decomposition of benzene

diazonium chloride to give chlorobenzene and nitrogen:

57

k(s-1) 0.00043 0.00103 0.00180 0.00355 0.00717

T (K) 313.0 319.0 323.0 328.0 333.0

Solution

RT E

Aek !

= RT

E Ak != lnln

k 0.00043 0.00103 0.0018 0.00355 0.00717

T 313 319 323 328 333

ln k -7.75173 -6.8782 -6.31997 -5.64081 -4.93785

1/T 0.003195 0.003135 0.003096 0.003049 0.003003

7/21/2019 Chapter 1 (Notes) (2)


58

y = -14017x + 37.12R$ = 0.998

-9

-8.5

-8

-7.5

-7

-6.5

-6

-5.5

-5

-4.5

-4

0.00295 0.003 0.00305 0.0031 0.00315 0.0032 0.00325

ln k

1/T

-E/R = -14,017K

E = 14,017 x 8.314

E = 116.5 kJ/mol

ln A = 37.12

A = 1.32 x 1016 s-1

RT

E Ak != lnln

7/21/2019 Chapter 1 (Notes) (2)


1.8 Reversible Reaction

The net rate of formation of any species is equal to its rate of formation on the

forward reaction plus its rate of formation in the reverse reaction

Rate = rate forward + rate reverse

At equilibrium, net rate = 0 and the rate law must reduce to an equation that isthermodynamically consistent with the equilibrium constant for the reaction.

For the reaction: aA + bB # cC + dD

Equilibrium constant,

Unit of KC= (mol/dm3)d+c-a-b

59

7 R S ⇌ T

7/21/2019 Chapter 1 (Notes) (2)


How to write rate laws for reversible reaction?

The rate of disappearance of A

The rate of formation of A

60

(T+f+ ⇌ T'(f'$ R f(

(7 ⇌ S R T

:37; W.3N13B

O c7T

7

(

37; 3>Z>3G> O c:7TSTT

k A

k-A

S R T# 7

(7# S R T

k A

k-A

37; />4 O 37; W.3N13B R 37; 3>Z>3G>

O :c7T7( R c:7TSTT

: 37; />4 O c7T7( : c:7TSTT

Equilibrium constant

Thermodynamically consistent at

equilibrium

7/21/2019 Chapter 1 (Notes) (2)


Example 1.10Suggest a rate law for the following reversible liquid-phase reaction and calculate

the equilibrium conversion and concentration

(A + B C) with CA0 = CB0 = 2 mol/dm3 and KC = 10 dm3 /mol

Solution

61

37; />4 O 37; W.3N13B R 37; 3>Z>3G>

O :c7T7TS R c:7TT

: 37; />4 O c7VT7TS : TT6eTK NE>3> eT O c76c:7

74 >X=CDC23C=H; 37 O $

<E=G TTOeTT7TS

T7 O T7$V':g>K

TS O T7$V':g>K

TT O T7$g>

T7$

g>

O eT

T7$

(V':g>

K(

(g> O '$V)KV':g>K(

[.DZ> W.3 g>

g> O $\"

CA = 2(1-0.8) = 0.4 mol/dm3

CB = 2(1-0.8) = 0.4 mol/dm3

CC = 2(0.8) = 1.6 mol/dm3

7/21/2019 Chapter 1 (Notes) (2)


1.9 Stoichiometry!

So far, we have shown how the rate law can be expressed as a function ofconcentrations. Then, we need to express the concentration as a function of

conversion in order to size reactors.

!

Stoichiometry plays a major role if the rate law depends on more than one

species, we must relate the concentration of different species by

stoichiometric balance.

! Consider the general reaction

Mole of A reacted =

Mole of B reacted =

=

62

Da

d C

a

c B

a

b A

dDcC bBaA

+!+

+!+

The mole of B remaining in the system,

7/21/2019 Chapter 1 (Notes) (2)


! Stoichiometric Table for Batch System

63

Species Symbol Initial Change Remaining

A A

B B

C C

D D

Inert I

<.41D

X N N N

a

b

a

c

a

d

AT T 00

1

!

!

+=

""+=

0

0

0

0

0

0

A

i

A

i

A

ii

y

y

C

C

N

N ===!

X N a

b

a

c

a

d N N

AT T 00 )1( !!++=

7/21/2019 Chapter 1 (Notes) (2)


! Constant volume batch reactor

64

7/21/2019 Chapter 1 (Notes) (2)


Example 1.11

Write the rate law for the liquid phase reaction (follow the elementary rate law)solely in terms of conversion. The feed to the batch reactor is equal molar A and B

with CA0 = 2 mol/dm3 and kA= 0.01 (dm3 /mol)4 /s.

Solution

Rate Law: -r A=kC3

AC2

B

Liquid phase, v = vo (no volume change)

65

Species A is the limitingreactant because the feedis equal molar in A and B,

and two moles of Bconsumes 3 moles of A.

7/21/2019 Chapter 1 (Notes) (2)


!

Flow system

66

0

0

0

0

00

00

0

0

A

i

A

i

A

i

A

ii

y

y

C

C

vC

vC

F

F ====!

7/21/2019 Chapter 1 (Notes) (2)


! Flow system

67

Species Symbol Reactor Feed Change Reactor

Effluent

A A

B B

C C

D D

Inert I

Liquid Phase Flow System:

If the rate of reaction were -rA = kCACB then we would have

7/21/2019 Chapter 1 (Notes) (2)


! Volume change with reaction

68

!"

#$%

&!!"

#$$%

&

+

+'=

!

"

#$

%

&!!

"

#$$

%

&!!

"

#$$

%

&

+

+'!!

"

#$$

%

&=

!"

#$%

&!!"

#$$%

&!!"

#$$%

&

+

+'=

!"

#$%

&!!"

#$$%

&!!"

#$$%

&=

!"

#$%

&!!"

#$$%

&!!"

#$$%

&!!"

#$$%

&=!!

"

#$$%

&=

!!"

#$$%

&!"

#$%

&!!"

#$$%

&!!"

#$$%

&=!!

"

#$$%

&=

T

T

P

P

X

X vC C

T

T

P

P

X F F

X v

F

F C C

T

T

P

P

X F F

X v F C C

T

T

P

P

F

F C C

T

T

P

P

F

F

v

F

v

F C

N

N

T

T

P

P

v

F

v

F C

j j A

j

T A

j j

T

AT j

AT

j j A

T j

T

j

T j

T

jT j

j

T

T j j

j

0

0

0

0

0000

00

0

000

0

0

0

0

0

0

00

0

00

00

1

)(

)/(1

)(

(

)

)

For gas phase reaction,

X y X N

N

N

N

X N N N

N

N

Z

Z

T

T

P

P vv

RT N z V P

RT zN PV

A

T

A

T

T

AT T

T

T

T

T

0

0

0

0

00

000

00

00000

11

0)(at t

factor ilitycompressibzt),(at time

! !

!

+=""#

$%%&

'+=""

#

$%%&

'

+=

"

"

#

$

%

%

&

'

"

"

#

$

%

%

&

'

"

"

#

$

%

%

&

'"

#

$%

&

'=

==

==

)1(00

00 X

Z

Z

T

T

P

P vv ! +""

#

$%%&

'""#

$%%&

'"#

$%&

'=

!!"

#$$%

&!"

#$%

&+=

0

00 )1(

T

T

P

P X vv '

! " 0 A y=

!!"

#

$$%

&=

!!"

#

$$%

&

T

T

T

T

F

F

N

N 00

000 T A A C yC =

! " 0 A y=

7/21/2019 Chapter 1 (Notes) (2)


69

Gas phase (volume change)

Incompressible liquid, gas phase at

constant volume in batch system

7/21/2019 Chapter 1 (Notes) (2)


For the gas phase reaction 2A + B # C

The feed is equal molar in A and B. Calculate %

Solution

A is the limiting reactant

A + &B # &C

' = & - & -1 = -1

% = y A0' = & (-1) = -0.5

70

Example 1.12

7/21/2019 Chapter 1 (Notes) (2)


Example 3.5 (Book)

71

Gas phase reaction that does not have anequal number of product and reactant moles

7/21/2019 Chapter 1 (Notes) (2)


Example 1.13

7 HCP4=3> .W ("^ [h( 1/B ,(^ 1C3 CG ?E13Q>B 4. 1 M.N 3>1?4.3 C/ NEC?E [h( CG .PCBCi>B\

([h( R h( # ([h%

<E> 4.41D I3>GG=3> CG ')"* c@1 1/B 4>HI>314=3> CG ?./G41/4 14 ((,.T\ T1D?=D14> 4E> Q1G

?./?>/431J./G V[h(; h(; [h% 1/B L(K 14 4E> 3>1?4.3 .=4D>4 CW 4E> ?./Z>3GC./ CG $\*\

!"#$%"&

<1cC/Q [h( 1G 21GCG .W ?1D?=D1J./# [h( R '6(h( # [h%

F[h(;$ O V$\("KVF<$K

Fh(;$ O V$\,(KV$\('KVF<$K

FL(;$ O V$\,(KV$\,&KVF<$K

jh( O Fh(;$6F[h(;$ O $\*)

jL( O FL(;$6F[h(;$ O (\$%

j[h% O $

72

mol/dm31.0

K(500K)dm3.kPa.8.314

kPa148528.0

RT

PyCyC

0

0SO2,0T0SO2,0SO2,0

=

!"

#$%

&=

!"

#$%

&==

7/21/2019 Chapter 1 (Notes) (2)


! " 0 A y=

14.0)2/111(28.0 !=!!="

At X = 0.5

! O2

73

7/21/2019 Chapter 1 (Notes) (2)


74

T7

TS

TT

TU

Tk

Summary variable volume gas flow system

7/21/2019 Chapter 1 (Notes) (2)


75

7/21/2019 Chapter 1 (Notes) (2)


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3.1-3.6

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Chapter 1 (Notes) (2)