lecture 3 - reactor sizing

7/24/2019 Lecture 3 - Reactor Sizing

1/41

Conversionand

Reactor Sizing

Suchithra Thangalazhy Gopakumar

H83RED Reactor

Design

Autumn Semester 2015


2/41

Outline

Characterisation of reactor feed

Identification of Limiting reactant

Definition of the conversion

Conversion as a measurable design parameter

Design equations in terms of conversion

Levenspiel Plot

Levenspiel Plot as a reactor sizing tool

Sizing reactors in different configurations to

achieve a given conversion


3/41

Limiting reactant is the one that get consumed faster in areaction

Excess reactant is the surplus amount of a reactant overthe stoichiometric amount required for the reaction tocomplete.

Characterisation of the Reactor Feed

+ +

=


4/41

Characterisation of the Reactor Feed: Example

The gas phase reaction + 2 takes place inside a constant volume batchreactorat isothermal conditions. Initially, the reactor contains 1.5 kmol of CO, 1 kmol of O2and 1 kmol ofCO2 and 0.5 kmol ofN2at a pressure 5 atm.

(a) Identify the limiting reactant

(b) Determine the excess amount of the other reactant.

Solution

For reactants calculate theratios

= 1.52

= 0.75 = 1.01 = 1.0

Values of the ratiosshow

Therefore CO is the limitingreactant

One mole of CO2reacts with 0.5 moles of O2

Stoichiometric amount ofO2required

= 0.5 1.5 = 0.75

= .

. =.


5/41

Characterisation of the Reactor Feed

When we consider the reactions taking place inside a reactor,the stoichiometric equation should be considered.

Consider following reaction

The stoichiometric equation is

Consider a chemical reaction as shown below

-a, -b, cand dare called stoichiometric coefficients. Negative and

positive signs indicate the consumption and generation ofmaterials respectively.

This can be rewritten in the form

Stoichiometric coefficients


6/41

Conversion of species A is defined as the number of moles A that have

reacted per mole of A fed into the system:

fedAofmoles

reactedAofmolesAX (1) XA(max)

= 1 for irreversible reaction

=XAeqfor reversible reaction

In general, for reaction of any complexity we are able to follow conversion of any

substance via stoichiometry. For example, in case of the following reaction:

aA + bB cC + dD

every quantity could be put on a per mole A basis

A + B C + Da

b

a

c

a

d

Conversion of reactants correlates with product generation via stoichiometry

Conversion


7/41

Conversion in a Batch Reactor

Batch reactors are operated for a given period of time. Consider ageneral form of the reaction within a batch reactor as

At timet= 0 the amount of A (the limiting reactant) charged into the batchreactor inNA0. After timet= tthe amount of A remaining in the reactor is NA.

Amount of A reacted during the period t

The number of moles of A that remain in the reactor after a time t:

[2]

[3]

[A]

XNA

0

reacted

Aofmoles


8/41


The design equation for the batch reactor is (in which reaction [A] takes

place)

=

[4]

Substituting equation [3] in [4] =

=

The design equation indifferential formin term ofXA is

=

We call eq [5] the design equationfor a batch reactor, because we

have written the mole balance in terms of conversionXA .

[5]


9/41


=

The design equation in differential form in term of XA is

= (

)

Integration with the limits at reaction start: t = 0,X = 0

= (

)

The design equation inintegral formin term of XA is

[6]

[7]


10/41

Levenspiel* Plot

* Named after Octave Levenspiel, Oregon State University, who suggested this graphical method first

1

When we inspect the equations derived above, it is clear that wehave to carry out an integration of the form

It is clear from the rate lawthat

The above integral can be estimated graphically if we know the values of

or for various values of conversion X.

ConversionX

1.00

The rate of reaction for variousconversions of a limiting reactantcan be measure in a series ofcontrolled experiments.

Curve that shows

against

conversion x is called the LEVENSPIELPLOT


11/41

Levenspiel Plot as a Reactor Sizing Tool

Batch Systems Graphical analysis

=

Consider a batch reactor

Therefore

As a result lowest at

Nearer to the end of the

reaction

have higher value that at()

Area=

1.00

Levenspiel Plot

ConversionX

At the start of the reaction (t=0)


12/41


=

When we consider aconstant volume batchreactor

=

=

= 1

=

= 1

= 1

and

= 1

()

Area=

Levenspiel Plot

Concentration

[8]

[9]


13/41

Conversion: Example

A liquid phase reaction A + B C + D takes place in a constant volumeBatchReactor. The initial concentration of A, the limiting reactant, is 7 mol dm-3. Thereaction is non-elementary so that rA=k CAwith the specific rate constant k = 0.06

min-1. what is the conversion of A if the reactor has been operational for 30 minutes?

Solution

The reaction is

+ + With =

For a Batch reactor =

=

Reason out how the concentration is

included

Integrating gives

=

Reason out the limitsuses

=

= 1

=

= =

=

= 1

= 1

= 1

= .Therefore


14/41

Conversion: Example

A liquid phase reaction A C + D takes place in a constant volumeBatch Reactor. The initialamount of A, the limiting reactant, fed to the reactor is 20 mol. The reaction is elementary so thatrA=k CAwith the specific rate constant k = 0.06 min

-1.

How long would it take to achieve 90% conversion?

How long would it take to achieve 99 % conversion?

Solution The reaction is + With = = 0.06

= 1 11

For a Batch reactor =

Fill the missing steps= ()

Substituting forNA = 1 ()

=

Therefore when XA=0.9

=1

0.06

1

1 0.9 min =

2.3

0.06 min

Therefore when XA=0.99

= 10.06

1

1 0.99 m in =

2.3

0.06 min


15/41

Conversion in Flow Reactors

In batch reactors:X = f(t). In flow systems, reaction time increaseswith increasing reactor volume, therefore X = f(V)

Conversion is defined as

=

gives the molar rate at whichspecies A is reacting within the entiresystem at steady state

=

Molar flowrate

of A leaving

the system

=Molar

flowrate

of A fed

[10]

[11]

Rearranging equation [10]


16/41

Flow Systems

The entering molar flow rate of species A, FAO(mol/s) is the product of the entering

concentration, CAO(mol/dm3) and the entering volumetric flow rate,v0 (dm3/s):

000 vCF AA

Liquid phase

CAOis given in term of molarity. Example: CAO =2 mol/dm3

Gas phase

CAOand FAOcan be calculated by the entering temperature and pressure using the

ideal gas law:

0

00

0

0

0 RT

Py

RT

PC AA

A

0

000000

RT

PyvCvF AAA

where NA0= entering number of mole of A

yA0 = entering mole fraction of AP0 = entering total pressure, e.g., kPa

PA0= yA0 P0 = entering partial pressure of A, e.g., kPa

T0 = entering temperature, K

R = ideal gas constant (e.g., R = 8.314 )mol.K

kPa.dm3

0000 RTNVP AA



17/41


Since the exit composition fromthe reactor is identical to thecomposition inside the reactor(perfect mixing), the rate ofreaction is evaluated at the exitconditions.

Continuous Stirred Tank Reactor (CSTR)

Following equation was derived from molebalance

[12]

[11]

For flow reactor systems, from equation[11]

Substituting eq [11] in eq [12]

[13]


18/41


Plug Flow Reactor (PFR)

The mole balance applied toPFR provided

[11]

[14]

Substituting eq [11] in eq [14]

Packed Bed Reactor (PBR)

The mole balance applied toPFR provided

[11]

[16]

Substituting eq [13] in eq[18]

[15] [17]


19/41

Conversion: Example

A liquid phase reaction A C + D takes place in a Packed Bed Reactor. Thereaction is elementary so thatrA=k CAwith the specific rate constant k = 0.005 dm

3

min-1 [kg(cat)]-1 . The expected conversion is 80%.

a. Find the mass of the catalyst as a function of the inlet flowrate.

Solution The reaction is + With = = 0.005 ( )

= 11

For a PBR =

Fill the missing steps=

()

Substituting forNA = 1

()

=

Therefore when XA=0.8 =

1

0.005

1

1 0.8 = 200 1.61

=


20/41


CSTR Graphical analysis

Rearranging the design equation

1.00

()

Area=

Levenspiel Plot

ConversionX

The area of the rectangleshown in the figure gives

the ratio.


21/41


Tubular Flow Reactors

Plug Flow Reactor Packed Bed Reactor

=

=

1.00

()

Area=

Levenspiel Plot

ConversionX

1.00

()

Area=

Levenspiel Plot

ConversionX

= 1

= 1

[16] [17]


22/41

Space Time and Space Velocity

Thespace time(theholding timeormean residence time) is the time

necessary to process one reactor volume of feed fluid based on entranceconditions.

[18] Vthe reactor volume

0the volumetric flow rate entering the reactor

20 m20 m

Reactor

Consider the tubular reactor, which is 20 m long and 0.2 m 3 in volume. The dashed

line represents 0.2 m3 of fluid directly upstream of the reactor The time it takes for thisvolume to enter the reactor completely is the space time. For instance, if the space

time is 5 min, it means that every 5 minutes one reactor volume of feed at specified

conditions is being treated by the reactor.

Thespace velocity(SV) is defined as:

1so0 SV

VSV [19]

SVand

are not just the reciprocal parameters since they often refer to different

conditions:

to the entrance conditions; SV might be referred to the conditions at

specific location in the reactor

0

V



23/41

CSTR PFR

Design equations exit

0

)( A

A

r

XF

V

[20]

X

AA r

dX

FV0

0 [21]

Divide both sides of theequations by the entering

volumetric flow rate

exit0

0

0 )( A

A

r

XFV

X

A

A

r

dXFV

00

0

0

Put the left-hand sidesin terms of space times exit

0

)r(

XC

A

A

X

AA r

dX

C 00

[22] [23]

Area=/CA0=

XAX

ArX

1Ar

1

exitAr1

XAX

Area =/CA0=A

r

1

X

Ar

dX

0



24/41

Mole Balance for Different Reactors

Reactor Type Differential Algebraic Integral

Batch

CSTR

PFR

PBR

Vrdt

dXN AA 0

)(

0

0

tX

A

A

A

Vr

dXNt

exit

0

)( A

A

r

XFV

AA rdV

dXF0

out

in

X

X A

Ar

dXFV 0

AA rdW

dXF0

out

in

X

X A

Ar

dXFW 0

exit)r(

XC

A

A

0

out

in

X

X A

Ar

dXC 0

Conversion: Summary


25/41

Constant Density Systems

Since the reaction volume does not change in the reactor (q= q0), molar flux

of species A into the reactor:

Lets express conversion from the equation (13):

In CSTR operated at steady-state

conditions, conversion does not change

through the reactor volume:

000 AA CF

)1(0 XFF AA 0

0

A

AA

F

FFX

0

0

00

000

0

0

A

AA

A

AA

A

AA

C

CC

C

CC

F

FFX

In PFR systems conversion changes

depending on location. The conversion at

specific point can be expressed viadifferentiation:

0

0

A

AA

C

CCX

0A

A

C

dCdX



26/41

CSTR PFR

Design equations

Replace conversionusing its relation withconcentrations

Design equations viaconcentration term

exit

0

)r(

XC

A

A

X

AA r

dX

C 00[26] [27]

Ar

1

A

r

1

0

00

A

A

A

A

C

CC

r

C

A

A

C

C A

A

A

AC

dC

rC

0 0

0

1

A

AA

r

CC

0

0A

A

C

C A

A

r

dC

[28] [29]

CACA(t)

Area ==

CA0

Constant densityonly

CACA(t)

Area ==

CA0

Constant densityonly

r

CC AA

0

0A

A

C

C A

A

r

dC


Finding Space Time


27/41

Methodology

If the rate of reaction is available as a function of conversion, -rA=f(X) (or in

some specific cases as a function of concentration, rA=f(CA)), or if it can begenerated by some intermediate calculation, one can size the reactor usingthe set of design equations by determining reactor volume or space time.

Task: The laboratory measurements of reaction rate for the gas phasereaction

A B + Cgave the following result - rates are in mol/(dm3 s)

0.0010.001250.00180.00250.00330.00400.00450.00500.00520.0053-rA

0.90.80.70.60.50.40.30.20.10.0X

T = 422.2 K, total pressure = 1013 kPa, initial charge 50 % inerts, 50 % A.Compare the volumes of CSTR and PFR required for the same conversion

of 60%, if the volumetric flow rate is 34.65 dm3

/s.

Reactor Sizing: Graphical Method


28/41

X 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

-rA 0.0053 0.0052 0.0050 0.0045 0.0040 0.0033 0.0025 0.0018 0.00125 0.001

Solution

mol

sdm24060400

1 3

0

.X

rF

V

AA

The CSTR volume necessary to

achieve 60% conversion is:

33

dm1200mol

sdm240

s

mol5

V

is the area of rectangle withvertices0A

F

V

(0, 0), (0, 400), (0.6,400), and(0.6, 0)

Reactor Sizing: Graphical Method -Example

0.80.60.40.2

200

400

600

0

ConversionX

Ar

1

CSTR

exit

0

)( A

A

r

XFV

exit)r(

X

F

V

AA

0

-1/rA

189 192 200 222 250 303 400 556 800 1000


29/41

A BPFR

Solution (continuation)

X 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

-rA 0.0053 0.0052 0.0050 0.0045 0.0040 0.0033 0.0025 0.0018 0.00125 0.001

-1/rA

189 192 200 222 250 303 400 556 800 1000

X

AA r

dX

F

V

00

).(r).(r)(r

X

r

dX

AAA

.

A 60

1

30

4

0

1

3

60

0

400]2224[1893

3.0

mol

sdm148

3

numerical evaluation of integralbased on Simpson's One-ThirdRule

The PFR volume necessary toachieve 60% conversion is:

33

dm740mol

sdm148

s

mol5

V

0.80.60.40.2

200

400

600

0

ConversionX

Ar

1



30/41

A B

Graphical Analysis

Solution (continuation)

X 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

-rA 0.0053 0.0052 0.0050 0.0045 0.0040 0.0033 0.0025 0.0018 0.00125 0.001

-1/rA

189 192 200 222 250 303 400 556 800 1000

0.80.60.40.2

200

400

600

0

ConversionX

Differencebetween

CSTR and PFRPFRAr

1

For isothermal reactions ofgreater than zero order, the PFRwill always require a smallervolume that the CSTR toachieve the same conversion



31/41

Reactor in Series

The same conversion may be reached in two stages, for example, one reactor will

provide conversion to 30% while second one will increase the conversion from 30%to 60%:

Two CSTR

X2=0.6

FA2

FA0

X1=0.3

X1=0.3

X2=0.6

FA0

Two PFR

FA1

V1

V2

V1

V2

FA1

FA2

Xi = Total moles of A reacted up to point i

Moles of A fed to the first reactor

i=1 i= 1

i= 2

i= 2

-rA1

-rA2

-rA1

-rA2

Conversion is defined in terms of location at a point downstream rather than with

respect to any single reactor

Reactor Sizing: Different Configurations


32/41

Two CSTR

X2=0.6

FA2

FA0

X1=0.3

FA1

V1

V2

i=1

i= 2-rA1

-rA2

V = V1+ V2

?V

Xr

FV

2

1

A1

A01

A mole balance on the second reactor:

In Out + Generation = 0

Reactor 2:

A2

A2A12

2A2A2A1

r-FF

rFF

V

0V

A2

12AO2

A2

2AO1AO2

A2

2AOAO1AOAO2

r-

)X( XFV

r-XFXF-V

r-

)XF-( F)XF-( FV

2AOAOA2

1AOAOA1

XF-FF

XF-FF



33/41

Two CSTR in series Two PFR in series

122

01

1

021 XX

r

FX

r

FVVV

A

A

A

A

0.80.60.40.2

200

400

600

0

ConversionX

CSTR 1

CSTR 2Ar

1

0.80.60.40.2

200

400

600

0

ConversionX

PFR 1

PFR 2Ar

1

Volume of two CSTR in series smaller

than the volume of one CSTR

Volume of two PFR in series is

identical to that of one PFR

The difference between overall volume of two CSTR in series and two (or one)

PFR decreases

2

1

1

0

021

X

X A

X

A

Ar

dX

r

dXFVVV



34/41

1) CSTR-PFR 2) PFR-CSTR

Ar

1

0.80.60.40.2

200

400

600

0

ConversionX

CSTR

PFR Ar

1

0.80.60.40.2

200

400

600

0

Conversion X

CSTR

PFR

Reactor size can be optimised by the selection of number of stages, type of

reactors, and their combinations

2

1

01

1

021

X

X A

A

A

A

rdXFX

rFVVV 12

2

0

0

021

1

XXr

F

r

dXFVVV

A

AX

A

A



35/41

3) PFR - CSTR - PFR

X1

X=0

V1

V2

V3

X2

X3

FA0

Ar

1

200

400

600

Conversion X

CSTR

PFR

V1

/ FA0

X1

X2

X3

V2

/ FA0

V3

/ FA0

122

02 XX

r

FV

A

A

1

0

01

X

A

Ar

dXFV

3

2

03

X

X A

Ar

dXFV



36/41

Sequencing of reactors

Question:Which arrangement is the best?

Answer:It depends on the:

a) Shape of the Levenspiel plots

b) Relative reactor sizes

For autocatalytic reactions:

CSTR is more efficient at low conversions,

PFR is more efficient at high conversions



37/41

Reactor Staging

Lets analyse multistage reactor system, say 5 stages of similar reactor:

A PFR can modelled as a

number of CSTR in series

1 2 3 4 5

31 2 4 5

X4X3X2X10 X5

V1V2

V3

V4

V5A

A

r

F

0 As we make the volume of each

CSTR smaller and increase the

number of CSTRs,

Total number of the CSTRs in series

Volume of the PFR



38/41

Consider the reaction A B . It is carried out adiabatically in the liquidphase and following data were obtained.

X 0.0 0.2 0.4 0.6 0.65

-rA (kmol/m3

h)39 53 59 38 25

The reactor scheme is shown below

X2=0.6

FAex

FA0

X1=0.2

X2=0.65

FA1FA0

V1V2

V3

Calculate the volume of each of the reactors for an entering molar flowrate ofA of 50 kmol/h

Reactor Sizing: Different Configurations -Example


39/41

Solution

For CSTR 1

X2

=0.6

FAex

FA0

X1

=0.2

X3=0.65

FA1FA0

V1V2

V3

CSTR 1

CSTR 2

PFR 1

10

1 Xr

FV

A

A

For PFR 1

dXr

FV

X

X A

A

2

1

02

For CSTR 2

230

3 XXr

FV

A

A

X 0.0 0.2 0.4 0.6 0.65

-rA (kmol/m3

h)39 53 59 38 25

[FA0/-rA] (m3) 1.28 0.94 0.85 1.32 2.0

hmolFA /50AofflowrateMolar 0



40/41

1.28

0.940.85

1.32

2

0

0.5

1

1.5

2

2.5

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

310

1 188020940 m...Xr

FV

A

A

dXr

FV.

. A

A

60

20

02

60

0

40

0

20

02 4

3.xA

A

.xA

A

.xA

A

r

F

r

F

r

FXV

3

2 380 m.V

3230

3 106065002 m....XXr

FV

A

A

X 0.0 0.2 0.4 0.6 0.65

-rA(kmol/m3 h)

39 53 59 38 25

[FA0/-rA](m3)

1.28 0.94 0.85 1.32 2.0 CSTR 1

PFR

CSTR 2


X2=0.6

FAex

FA0

X1=0.2

X3=0.65

FA1FA0

V1V2

V3

CSTR1

CSTR

2

PFR 1


41/41

Summary

lecture 3 - reactor sizing

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