chemical reaction engineering 1 제 2 장 conversion and reactor sizing 반응공학 1

Chemical Reaction Engineering 1Chemical Reaction Engineering 1

제제 22 장장ConversionConversion

and Reactor Sizingand Reactor Sizing

반응공학 반응공학 11

Objectives

a. To define conversion (X) and space time ()

b. To rewrite the mole balances in terms of conversion for a batch reactor, CSTR, PFR, and PBR.

c. In expressing -rA as a function of conversion (X), a number of reactors and reaction system can be sized or a conversion be calculated from a given reactor size.

- To relate the relative rates of reaction of reactants and products.

Definition of ConversionConsider the general equation

Choose A as our basis of calculation(The basis of calculation is most always the limiting reactant )

Question - How can we quantify how far a reaction has progressed ? - How many moles of C are formed for every mole A consumed ?

The convenient way to answer these question is to define conversion.

DCBA dcba

DCBAa

d

a

c

a

b

fedAofmole

reactedAofmoleXA

Design Equations

The longer a reactant is in the reactor, the more reactant is converted to product until either equilibrium is reached or the reactant is exhausted. consequently, the conversion X is a function of reaction time

XNconsumed

Aofmole

fedAofmole

reactedAofmoles

fed

Aofmole

consumed

Aofmole

A

0

XNNN

reactionchemicalby

consumedbeenhave

thatAofmoles

tatreactor

tofedinitially

Aofmoles

ttimeat

reactorin

Aofmoles

AAA 00

0

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

Batch system

The number of moles of A in the reactor after a conversion X

The mole balance on species A for a batch system

In term of conversion by differentiating equation

The design equation for a batch reactor in differential form is

)1(000 XNXNNN AAAA

Vrdt

dNA

A

dt

dXN

dt

dNA

A00

Vrdt

dXN AA 0

The differential form for a batch reactor

For constant-volume batch reactor

For a variable volume batch reactor

When the volume is varied by

For the most common batch reactors where volume is not predetermined,

the time necessary to achieve a conversion X is

A

AAA rdt

dC

dt

VNd

dt

dN

V

/1

Vr

dXNdtor

r

dXNVdt

AA

AA

00

X

AA

t

r

dXNdtV

000

tX

AA Vr

dXNt

00The integral form for a batch reactor

If FA0 is the molar flow rate of species A fed to a system at steady state,

the molar rate at which species A is reacting within the entire system will be FA0X.

The molar flow rate

Rearranging gives

XFF AA 10

Flow systems

time

reactedAofmolesXF

fedAofmoles

reactedAofmoles

time

fedAofmolesXF

A

A

0

0

AAA FXFF

systemtheleaves

Awhichat

rateflowmolar

systemthewithin

consumed

whichatratemolar

systemthetofed

isAwhichat

rateflowmolar

00

The entering molar flow rate, FA0 (mol/s)

• For liquid systems : CA0 is commonly given in term of molarity

• For gas systems : CA0 can be calculated from the entering T and P

using the ideal gas law or some other gas law

• For an ideal gas (see Appendix B) :

000 vCF AA

CA0 : the entering concentration

v0 : the entering volumetric flow rate

PTfCA ,0

0

00

0

00 RT

Py

RT

PC AA

A

- The design equation for a CSTR

- conversion of flow system

- Combining (2-12) with (2-11)

CSTR or Back-mixing Reactor

XFFF AAA 00

A

AA

r

FFV

0

exitA

A

r

XFV

0

(2-11)

(2-12)

Equation to determine the CSTR volume necessary to achieve a specified conversion X. Since the exit composition from the reactor is identical to the composition inside the reactor, the rate of reaction is evaluated at the exit condition.

FA0

FA

(2-13)design equation

for a CSTR

X

-rA

1

Area

- General mole balance equation


- The differential form of the design equation

- Volume to achieve a specified conversion X

Tubular Flow Reactor (PFR)

AA r

dV

dF

XFFF AAA 00

AA rdV

dXF 0

X

AA r

dXFV

00

FA0 FA

(2-14)

(2-15)

(2-16) X

-rA

1

Area

- General mole balance equation


- The differential form of the design equation

Packed-Bed Reactor (PBR)

'0 AA r

dW

dXF

X

AA r

dXFW

0 '0

XFFF AAA 00

'A

A rdW

dF

FA0 FA

(2-17)

(2-18)

-The catalyst weight W to achieve a specified conversion X

tX

AA Vr

dXNt

00Design equation

for a batch reactor

Summary of Design Equation

exitA

A

r

XFV

0

FA0

FA

Design equation for a CSTR

X

AA r

dXFV

00FA0 FA

Design equation for a PFR

X

AA r

dXFW

0 '0FA0 FA

Design equation for a PBR

tX

AA Vr

dXNt

00

Summary of Design Equation

반응시간은

- NA0 에 비례

- X 에 비례

- 반응속도 (rA) 에 반비례

- 반응기 부피에 반비례

exitA

A

r

XFV

0

FA0

FA

X

AA r

dXFV

00FA0 FA

X

AA r

dXFW

0 '0FA0 FA

반응기 부피 ( 촉매의 무게 ) 는

- FA0 에 비례

- X 에 비례

- 반응속도 (rA) 에 반비례

Applications of the design equation for continuous-flow reactor

XkCkCr AAA 10For a first-order reaction :

The rate of disappear of A, -rA, is almost always a function of the concentrations of the various species present. When a single reaction is occurring, each of the concentrations can be expressed as a function of the conversion x; consequently, -rA, can be expressed as a function of X.

X

AA r

dXFV

00FA0 FA

V = FA0

kCA00

X dX

1-X =

FA0

kCA0

ln (1-X)-

Consider the isothermal gas-phase isomerization

A B

How to use the raw data of chemical reaction rate?

The laboratory measurements give the chemical reaction rate as a function of conversion.

(at T=500K, 830kPa(=8.2atm), Reactant=Pure A)

X -rA (mol/m3-sec) 1/-rA (m3-sec/mol) FAo/-rA (m3)

00.10.20.40.60.70.8

0.450.370.30

0.1950.1130.0790.050

2.222.703.335.138.8512.720.0

0.891.081.332.053.545.068.0

Levenspiel Plot

Greatest rate

Small rate

- For irreversible reactions,

the maximum value of X is that for complete conversion, i.e. X=1.0.

- For reversible reactions,

the maximum value of X is the equilibrium conversion, i.e. X=Xe.

11

Xasr

CBA

A

eA

XXasr

CBA

1

How to use the raw data of chemical reaction rate?

For vs. X, the volume of a CSTR and the volume of a PFR

can be represented as the shaded areas in the Levenspiel plots.A

A0

r

F

• Given –rA as a function of conversion.

• Constructing a Levenspiel plot.

• Here we plot either or as a function of X.Ar

1

A

A0

r

F

Reactor Size

Example 2-2. Sizing a CSTR

The reaction, AB, is carried out in a CSTR. Molar flow rate of A is 0.4 mol/sec.

(1) Using data in the previous Table, calculate the reactor volume necessary to achieve 80% conversion in a CSTR(2) Shade a area in Figure 2-2 that would give the CSTR volume necessary to achieve 80% conversion

(1)

exitA

A

r

XFV

0

rA

1( )X=0.8

= 20 m3-sec/mol

V=(0.4 mol/sec)(0.8)(20 m3-sec/mol) =6.4 m3=6400 liter

(2)

Example 2-3. Sizing a PFR

The reaction, AB, is carried out in a PFR. Molar flow rate of A is 0.4 mol/sec.

(1) Using data in the previous Table, calculate the reactor volume necessary to achieve 80% conversion in a PFR(2) Shade a area in Figure 2-2 that would give the PFR volume necessary to achieve 80% conversion(3) Make qualitative sketches of conversion (X) and rate of reaction (-rA) with respect to reactor volume

(1)

X

AA r

dXFV

00

V = 0

0.8dX

-rA = 0

0.8dX

-rA FA0

FA0

By applying Appendix A-23 (Five Point Quadrature Formula): X=0.8/4=0.2

30.2( )V= [0.89+4(1.33)+2(2.05)+4(3.54)+8] =2.165m3=2165 liter


(b)


(c) By applying Simpson’s rule in Appendix A.4, we can calculate V for X=0.2, 0.4, 0.6, 0.8(See the text, page 52). The results are as follows.

X -rA (mol/m3-sec)V (dm3)

00.450

0.20.30218

0.40.195551

0.60.1131093

0.80.052165

전환율을 조금 더 높이기 위해서는

반응기 부피가 많이 늘어나야

한다 .

Example 2-4. Comparing CSTR and PFR Sizes

For isothermal reaction of greater than zero order, the PFR will always require a smaller volume than the CSTR to achieve. What if zero order reaction?

Reactors in series

Define conversion

The conversion X defined as the “total number of moles” of A that

have reacted up to that point per mole of A fed to the “first” reactor.

(assumption : no side stream withdrawn and the feed stream enters

only the first reactor in the series)

reactor first to fed A of moles

i point to up reacted A of moles totalX i

PFR-CSTR-PFR in series

The relationships between conversion and molar flow rate

V 1

X=0FA0

X1

FA1

V 3

X2

FA2 V2

X3

FA3

FA1 = FA0 - FA0 X1

FA2 = FA0 - FA0 X2

FA3 = FA0 - FA0 X3

reactor first to fed A of moles

2 point to up reacted A of moles totalX 2 where similar definitions

exist for X1 and X3

1

001

X

AA r

dXFV

0

0.

2221

VrFF

genoutin

AAA

2

1202

)(

A

A

r

XXFV

3

203

X

XA

A r

dXFV

V 1

X=0FA0

X1

FA1

V 3

X2

FA2 V2

X3

FA3

Reactor 1:

Reactor 2 :

Reactor 3 :

FA1 = FA0 - FA0 X1

FA2 = FA0 - FA0 X2

FA3 = FA0 - FA0 X3

-rA2 is evaluatedat X2 for the CSTRIn this seriesarrangement

FAe

X2=0.8

Different Schemes of Reactors in Series

Two CSTRs in series

Two PFRs in series

a PFR and a CSTR in series

FA0

X1=0.4

FAe

X2=0.8

FAe

X2=0.8

FA0X1=0.4

FA0

FAe

X2=0.8

X1=0.5

X1=0.5

FA0

a CSTR and a PFR in series

11

01

1X

rFV

AA

2

1202

)(

A

A

r

XXFV

Two CSTRs in Series

FA0

X1=0.4

FAe

X2=0.8

Reactor 1

Reactor 2

=FAo (X1-Xo)

-rA1 0

Example 2-5: Two CSTRs in Series

FA0

X1=0.4FAe

X2=0.8

What is the volume of each of Two reactors?

XA

[FAo/-rA] (m3)0.0 0.1 0.2 0.4 0.6 0.7 0.80.89 1.09 1.33 2.05 3.54 5.06 8.0

Reactor 1

[FAo/-rA]x=0.4=2.05 m3

V1=([FAo/-rA]x=0.4)(X1-X0)=(2.05)(0.4-0)=0.82 m3

Reactor 2

[FAo/-rA]x=0.8=8.0 m3

V1=([FAo/-rA]x=0.8)(X2-X1)=(8.0)(0.8-0.4)=3.2 m3


Therefore, V1+V2=0.82+3.2=4.02 m3

What is the reactor volume to achieve 80%Conversion in a single CSTR?

[FAo/-rA]x=0.8=8.0 m3

V1=([FAo/-rA]x=0.8)(X1-X0) =(8.0)(0.8-0)=6.4 m3

The sum of the two CSTR reactor volumes (4.02 m3) in series is less than the

volume of one CSTR (6.4 m3) to achieve the same conversion (X=0.8)

[FAo/-rA](m3)

FA0

X1=0.4

FAe

X2=0.8

FA0

FA

X=0.8

Vtotal = 4.02 m3

Vtotal = 6.4 m3


A PFR by a Large Number of CSTRs in Series

Approximating a PFR with a number of small, equal-volume CSTRs of Vi in series

54321

1 2 3 4 5

A PFR by a Large Number of CSTRs in Series

80

60

40

20

A

A

r

F

0

.35 .53 .65 .74 .8X

1 2 3 4 554321

As we make the volume of each CSTR smaller and increase the number of CSTRs, the total volume of the CSTRs and the PFR will become identical. The performance of a PFR is equal to that of a number of (N) CSTRs in Series.

Can you verify this mathematically?

1

001

X

AA r

dXFV

2

102

X

XA

A r

dXFV

Two PFRs in Series

Reactor 1

Reactor 2

FAe

X2=0.8

FA0X1=0.4

Two PFRs in Series

0

X1

dX +-rA FA0

VTotal= V1 + V2= X2

-rA FA0 dX =

X1

X2

-rA FA0

0

Sizing PFR in Series

FAe

X2=0.8

FA0X1=0.4

What is the volume of each of Two reactors?Molar flow rate of A is 0.4 mol/sec.

XA

[FAo/-rA] (m3)0.0 0.1 0.2 0.4 0.6 0.7 0.80.89 1.09 1.33 2.05 3.54 5.06 8.0

Reactor 1

By applying Simpson’s rule in Appendix A.4 (Text page 60),

30.2( )V1= [0.89+4(1.33)+2.05] =0.551 m3=551 liter

Reactor 2

By applying Simpson’s rule in Appendix A.4 (Text page 60),

0.2( )V2= [2.05+4(3.54)+8.0] =1.614 m3=1614 liter3Therefore, V1 + V2=0.551 m3 + 1.614 m3=2.165 m3 < 4.02 m3 (Two CSTR in Series)

Combination of CSTR and PFR in Series

V3 = X2

dX-rA FA0

=FAo (X1-Xo)

-rA1 0

Reactor 1

V1

=FAo (X2-X1)

-rA2

Reactor 2

V2

Reactor 3

X3

Example 2-7: Liquid-Phase Isomerization

n-C4H10 i-C4H10

X 0.0 0.2 0.4 0.6 0.65

-rA (kmol/m3-h) 39 53 59 38 25

Calculate the volume of each of the three reactors for an entering molar flow rate n-butene of 50 kmol/h.


FAo = 50 kmol/h

X 0.0 0.2 0.4 0.6 0.65

-rA (kmol/m3-h) 39 53 59 38 25

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

(a) Reactor 1 (X1=0.2)

=FAo (X1-Xo)

-rA1 0V1

= (0.94)(0.2)=0.188 m3

(b) Reactor 2 (X2=0.6)

V2 = 0.2dX

-rA FA0

0.6


By applying Simpson’s three point formula in Appendix A.4 (Text page 64),

0.2( )V2= [0.94+4(0.85)+1.32] =0.38 m3

3

(c) Reactor 3 (X3=0.65)

=FAo (X3-X2)

-rA3

V3= (2.0)(0.65-0.6)=0.1 m3

2

10

X

XA

A r

dXFV

exitA

A

r

XFV

0

Reactor Sequence

FA0

FAe

X2=0.8

X1=0.5

Reactor 1

Reactor 2

1

0

PFR

CSTR

FA0

FAe

X2=0.8

X1=0.5

Total volume= Vtotal=V1+V2= 305 dm3

Scheme A

Reactor Sequence

CSTR

PFR

Total volume= Vtotal=V1+V2= 262.3 dm3

FAe

X2=0.8X1=0.5

FA0

Reactor Sequence

Scheme B

FAe

X2=0.8X1=0.5

FA0

FA0

FAe

X2=0.8

X1=0.5Scheme A

Scheme B

Vtotal=V1+V2= 262.3 dm3

Vtotal=V1+V2= 305 dm3

Scheme B will give the smaller total volume for an intermediate

conversion of 50%.

However, the relative sizes of the reactors depend on the intermediate conversion.

Reactor Sequence

What if zero order reaction?

Choosing Reactor Sequence

X

[FAo/-rA]

0

MFR PFR

time

conditionspecifiedat

measuredfeedofvolumereactor

oneprocesstorequiredtime

v

V

0

X

AA

X

A

A

r

dXC

r

dX

v

F

v

V000

0

0

0

Space TimeSpace-time :

The time necessary to process one reactor volume of fluid based on

entrance conditions. Also called the holding time or mean residence time.

A space-time of 2 min means that every 2 min one reactor volume of

feed at specified condition is being treated by the reactor.

Space TimeSpace-time :

The time necessary to process one reactor volume of fluid based on

entrance conditions. Also called the holding time or mean residence time.

Consider the tubular reactor, which is 20m long and 0.2 m3 in

volume. The dashed line represents 0.2 m3 of fluid directly upstream

of the reactor. The time it takes for this fluid to enter the reactor

completely is the space time.

20m 20m

• A space-velocity of 5 hr-1 means that five reactor volumes of feed at specified condition are being fed into the reactor per hour.

• Difference in the definitions of SV and

- space time : the entering volumetric flow rate is measured at the entrance condition

- space velocity : other conditions are often used

10 1

time

volumeunitintreatedbecan

whichconditionspecifiedatfeed

ofvolumesreactorofnumber

V

vSV

Space Velocity

Definition of Space-velocity

• LHSV ( liquid hour space velocity)

- v0 is frequently measured as that of a liquid at 60 or 75 0F, even though the feed to the reactor may be a vapor at some higher temperature.

• GHSV ( gas hour space velocity)

- v0 is normally measured at standard temperature and pressure

LHSV and GHSV

• It is usually convenient to report –rA as a function of concentration rather than conversion.

- PFR

design equation :

molar flow rate :

flow system conversion :

• For the special case when v = v0

X

AA r

dXFV

00

000 AA CvF

0

0

A

AA

F

FFX

0

0

00

00

0

0

A

AA

A

AA

A

AA

C

CC

vC

vCvC

F

FFX

AA

AA

CCXXwhen

CCXwhen

,

,0 0

For reaction rate depending only on the concentration

• Differentiating yields• A typical curve for determining the space time,

0

0

A

A

C

CA

A

r

dCvV

0A

A

C

CA

A

r

dC

0A

A

C

dCdX

For reaction rate depending only on the concentration

Homework

P2-7B

P2-8B

P2-9B

Due Date: Next Week

chemical reaction engineering 1 제 2 장 conversion and reactor sizing 반응공학 1

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