5a- Isothermal Designs

Chemical Reaction Engineering (CRE) is thefield that studies the rates and mechanisms of

chemical reactions and the design of the reactors inwhich they take place.

Lecture 5

Lecture 5

2

Block 1: Mole Balances Block 2: Rate Laws Block 3: Stoichiometry Block 4: Combine

Review of Blocks 1, 2 and 3Examples

Reactor Differential Algebraic Integral

V = FA0XrACSTR

=

X

AA

r

dXFV0

0PFR

VrdtdXN AA =0Batch

X

t

FA 0 dXdW = rAPBR

X

W3

Building Block 1: Mole Balancesin terms of conversion, X

Review Lecture 2

Building Block 2: Rate Laws

4

Power Law Model:BAA CkCr =

CBA 32 +A reactor follows an elementary rate law if the reactionorders just happens to agree with the stoichiometriccoefficients for the reaction as written.e.g. If the above reaction follows an elementary rate law

Review Lecture 4

5Review Lecture 4

Building Block 3:Stoichiometry

Part 1: Mole Balances in terms ofConversion

6

Algorithm for Isothermal Reactor Design1. Mole Balances and Design Equation2. Rate Laws3. Stoichiometry4. Combine5. Evaluate

7Building Block:Combine

8Building Block 5:Design Algorithm

Scale-up of Liquid-Phase BatchReactor Data to the Design of a CSTR

9

One of the jobs in which chemical engineers are involved isthe scale-up of laboratory experiments to pilot-plantoperation or to full-scale production.

To make this jump successfully requires a thouroughunderstanding of the chemical kinetics and transportlimitations.

Batch Operation

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For most liquid-phase reactions, the density change withreaction is usually small and can be neglected (i.e., )

In addition, for gas phases in which the batch reactorvolume remains constant, we also have V = Vo.

Lets calculate the time necessary to achieve a givenconversion X for the irreversible second-order reaction:

Batch Operation

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Example 4-1

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It is desired to design a CSTR to produce 200 million pounds of ethylene glycol peryear by hydrolyzing ethylene oxide. However, before the design can be carried out, it isnecessary to perform and analyze a batch reactor experiment to determine the specificreaction rate constant.The reaction is first-order in ethylene oxide.

In the laboratory experiment, 500 mL of a 2 Msolution (2 kmol/m3) of ethylene oxide inwater was mixed with 500 mL of watercontaining 0.9 wt % sulfuric acid, which is acatalyst. The temperature was maintained at55C. The concentration of ethylene glycol wasrecorded as a function of time. From these data,determine the specific reaction rate at 55C.

Example 4-1: Solution

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Design of Continuous Stirred TankReactors (CSTRs)

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General Single CSTR

CSTRs in Series

Damkohler Number

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For a first-order reaction the product k is often referred to as thereaction Damkohler number. The Damkohler is a dimensionlessnumber that can give us a quick estimate of the degree ofconversion that can be achieved in continuous-flow reactors.

The Damkohler number is the ratio of the rate of reaction of A tothe rate of convective transport of A at the entrance to thereactor.

For first- and second-order irreversible reactions the Damkijhlernumbers are

Damkohler Number

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It is important to know what values of the Damkohlernumber, Da, give high and low conversion in continuous-flow reactors.A value of Da = 0.1 or less will usually giveless than 10% conversion and a value of Da = 10.0 orgreater will usually give greater than 90% conversion.

The rule of thumb is:

CSTRs in Series

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CSTRs in Parallel

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A second-order reaction in a CSTR

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Example 4-2: Producing 200 MillionPounds per Year in a CSTR

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It is desired to produce 200 million pounds per year ofethylene glycol.The reactor is to be operated isothermally.A 1lb mol/ft3 solution of ethylene oxide in water is fed to thereactor together with an equal volumetric solution of watercontaining 0.9 wt % of the catalyst H2SO4.The specificreaction rate constant is 0.311 min-1, as determined inExample 4- 1.a) If 80% conversion is to be achieved, determine the

necessary CSTR volume?


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Tubular Reactors

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Gas-phase reactions are carried out primarily in tubular reactorswhere the flow is generally turbulent.

By assuming that there is no dispersion and there are no radialgradients in either temperature, velocity, or concentration, we canmodel the flow in the reactor as plug-flow.

The differential form of the design equation

must be used when there is a pressure drop in the reactor or heatexchange between the PFR and the surroundings. In the absence of pressure drop or heat exchange the integral formof the plug flow design equation is used

Conversion as a function of distancedown the tubular reactor

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The effect of the change in the numberof moles in the gas phase reactions

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For constant temperature and pressure:

When there is no change in the number of moles with reaction(e.g., A B) = 0 and = 0, then the fluid moves through thereactor at a constant volumetric flow rate ( = 0) as theconversion increases.

When there is a decrease in the number of moles ( < 0, < 0) inthe gas phase (e.g., 2A B), the volumetric gas flow ratedecreases as the conversion increases; for example:

Consequently, the gas molecules will spend longer in the reactorthan they would if the flow rate were constant, = 0. As a result,this longer residence time would result in a higher conversion than ifthe flow were constant at 0.

The effect of the change in the numberof moles in the gas phase reactions

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On the other hand, if there is an increase in the totalnumber of moles ( > 0, > 0) in the gas phase (e.g., A 2B), then the volumetric flow rate will increase as theconversion increases; for example:

and the molecules will spend less time in the reactor thanthey would if the volumetric flow rate were constant. As aresult of this smaller residence time in the reactor, theconversion will be less than what would result if thevolumetric flow rate were constant at 0.

Example 4-3: Producing 300 Million Pounds per Year ofEthylene in a Plug-Flow Reactor: Design of a Full-scaleTubular Reactor

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Determine the plug-flow reactor volume necessary to produce300 million pounds of ethylene a year from cracking a feedstream of pure ethane. The reaction is irreversible and followsan elementary rate law. We want to achieve 80% conversion ofethane, operating the reactor isothermally at 1100 K at apressure of 6 atm.


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Todays example Example for Liquid Phase Undergraduate Laboratory

Experiment

EnteringVolumetric flow rate v0 = 0.0033 dm3/sAcetic Anhydride 7.8% (1M)Water 92.2% (51.2M)Elementary with k 1.95x10-4 dm3/(mol.s)

Case I CSTRCase II PFR

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CSTR Laboratory ExperimentExample: (CH2CO)2O + H20 2CH3OOH

1) Mole Balance: CSTR:A

A

r

XFV

=0

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A + B 2C

2) Rate Law: BAAA CCkr =

3) Stoichiometry:

B FA0B -FA0X FB=FA0(B-X)

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CSTR Laboratory Experiment

( ) ( )XCXFC BABAB == 00

0

( ) ( ) 000 2.512.51 BAAB CCXCC =32


( ) ( ) ( )XkXV

XkXV

XCXkCV

A

A

==

=

=

111 00000

kkX

+

=

1

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1) Mole Balance:0A

A

Fr

dVdX

=

2) Rate Law:BAA CkCr =

34

A + B 2C

s

dm300324.0

PFR Laboratory Experiment

4) Combine: ( ) ( )XkCXCCkr AABA == 11' 000

kX

=

11ln

keX = 1

sec0.96sec00324.0

311.03

3

0

===

dmdmV

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101.0 = sk

PFR Laboratory Experiment

Todays example Example for Gas Phase : PFR and Batch Calculation

Pure NOCl fed with CNOCl,0 = 0.2 mol/dm3 follows anelementary rate law with k = 0.29 dm3/(mol.s)Case I PFR with v0 = 10 dm3/s

Find space time, with X = 0.9Find reactor volume, V for X = 0.9

Case II Batch constant volume

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2 NOCl 2 NO + Cl2

1) Mole Balance:0A

A

Fr

dVdX

=

2) Rate Law: 2AA kCr =

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2A 2B + C

Gas Flow PFR Example

3) Stoichiometry:(Gas Flow)

( )X += 10

4) Combine: ( )( )222

0

11

XXkC

r AA +

=

A B + C

38`1


( ) ( ) ( )X

XXXkCA

++++=

111ln12

22

0

02.170 =AkC

sec29402.170

==

AkC

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2) Rate Law: 2AA kCr =

( )220 1 XkCr AA =

3) Stoichiometry:(Gas Flow)

0VV =

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Gas Phase 2A 2B + C t=?

Constant Volume Batch Example

( ) ( )200

220 11 XkCC

XkCdtdX

AA

A=

=4) Combine:

( ) dtkCXdX

A021=

tkCX A01

1=

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Constant Volume Batch Example

Mole Balance

Rate Laws

Stoichiometry

Isothermal Design

Heat Effects

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End of Lecture 5a

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5a- Isothermal Designs

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