chemical reaction engineering 1 chapter 4 isothermal reactor design 반응공학 1

Chemical Reaction Engineering 1Chemical Reaction Engineering 1

Chapter 4Chapter 4

Isothermal Reactor DesignIsothermal Reactor Design

반응공학 반응공학 11

• Describe the algorithm that allows the reader to solve chemical reaction engineering problems through logic rather than memorization.

• Sizing batch reactors, semi-batch reactors, CSTRs, PFRs, and PBRs for isothermal operation given the rate law and feed conditions.

• Studying a liquid-phase batch reactor to determine the specific reaction rate constant needed for the design of a CSTR.

• Design of a tubular reactor for a gas-phase reaction.

• Account for the effects of pressure drop on conversion in packed bed tubular reactors and in packed bed spherical reactors.

• The principles of unsteady operation and semi-batch reactor.

Objectives

Fig. 4-1 Isothermal Reaction Design Algorithm for Conversion

Algorithm for isothermal reactor designAlgorithm for isothermal reactor design1. Mole balance and design equation2. Rate law3. Stoichiometry4. Combine5. Evaluate

4.1 Design structure for isothermal reactors

We can solve the equations in the We can solve the equations in the combine step eithercombine step either

A. Graphically (Chapter 2)

B. Numerical (Appendix A4)

C. Analytical (Appendix A1)

D. Software packages (polymath)

Algorithm for Isothermal Reactors

French Menu Analogy

Scale-up of Liquid-Phase Batch Reactor to the Design of a CSTR

To make this jump successfully requires a through understanding of the chemical kinetics and transport limitations.

Pilot plant Pilot plant OperationOperation

Full-scale Full-scale ProductionProduction

Laboratory Laboratory ExperimentExperiment

PAST

Full-scale Full-scale ProductionProduction

MicroplantMicroplant(Lab-bench-scale unit) (Lab-bench-scale unit)

High cost of a pilot-plant leads to jump pilot plant operation

FUTURE

Algorithm for isothermal reactor designAlgorithm for isothermal reactor design

1. Mole balance and design equation

2. Rate law

3. Stoichiometry

4. Combination

5. Analytical Evaluation

Batch Operation

tCCk

dtC

dC

k

dtkC

dC

kCdt

dC

AinorderndleirreversibkCrBA

rdt

dC

dt

VNd

dt

dN

V

rdt

dN

V

AA

tC

CA

A

A

A

AA

AA

AAAA

AA

A

A

0

02

2

2

2

0

0

111

1

2,

/1

1

0

CA=CAo(1-X)

CAo dtdX

=kCAo2(1-X)2

kCAo

1 0

X dX

(1-X)2 = dt 0

t

t= kCAo

1 ( ) 1-X X

Batch Reaction Times

BA

Mole balance

)1(00

0

XCV

NC

VN

r

dt

dX

AA

A

A

A

Rate law

Xkt

Xkdt

dX

kCr AA

1

1ln

1

)1(

order-First

)1(

)1(

order Second

0

20

2

XkC

Xt

XkCdt

dX

kCr

A

A

AA

Stoichiometry (V=V0)

Combine

Integration

Batch Reaction Times

hr

s

k

k

Xkt

s

R

4.6

sec000,23

10

3.2

3.2

9.01

1ln

1

1

1ln

1

)10k 0.9,(Xorder -1st

14

1-4

hr

s

kC

kC

XkC

Xt

s

A

A

AR

5.2

sec000,9

10

9

9

)9.01(

9.0

)1(

)10kC 0.9,(Xorder 2nd

13

0

0

0

13A0

Batch Reaction Times (Table 4-2)

1st-orderk (s-1)

2nd-orderkCA0 (s-1)

Reaction timetR

10-4 10-3 Hours

10-2 10-1 Minutes

1 10 Seconds

1,000 10,000 Milliseconds

The order of magnitude of time to achieve 90% conversion

For first- and second-order irreversible batch reactions

Reaction Time in Batch Operation (Table 4-3)

0

111

AA CCkt

This time is the time t needed to reduce the reactant concentration in a batch reactor from an initial value CA0 to some specified value CA.

1. Heat to reaction temperature, te

2. Charge feed to the reactor and agitate, tf

3. Carry out reaction, tR

4. Empty and clean reactor, tc

1.5-3.01.0-2.0

Varies (5-60)0.5-1.0

Total cycle time excluding reaction 3.0-6.0

Activity Time (h)

2nd orderIsothermal

Liquid-phaseBatch reaction

Typical cycle times for a batch polymerization process

tt = tf + te + tR + tc

Decreasing the reaction time with a 60-h reaction is a critical problem. As the reaction time is reduced, it becomes important to use large lines and pumps to achieve rapid transfer and to utilize efficient sequencing to minimize the cycle time

Example 4-1: Design a Reactor Producing Ethylene Glycol

Design a CSTR to Produce 200 million pounds of ethylene glycol per year by hydrolyzing ethylene oxide. However, before the design can be carried out , it is necessary to perform and analyze a batch reactor experiment to determine the specific reaction rate constant (kA). Since the reaction will be carried out isothermally, kA will need to be determined only at the reaction temperature of the CSTR. At high temperature there is a significant by-product formation, while at temperature below 40oC the reaction does not proceed at a significant rate; consequently, a temperature of 55oC has been chosen. Since the water is usually present in excess, its concentration may be considered constant during the course of the reaction. The reaction is first-order in ethylene oxide.

O CH2-OH

CH2-CH2 + H2O CH2-OHH2SO4

A + B C Catalyst

Example 4-1: Determining k from Batch Data

Time Concentration of EG (min) (kmol/m3) 0.0 0.000 0.5 0.145 1.0 0.270 1.5 0.376 2.0 0.467 3.0 0.610 4.0 0.715 6.0 0.84810.0 0.957

In the lab experiment, 500mL of a 2 M solution (2 kmol/m3) of EO in water was mixed with 500mL of water containing 0.9 wt % sulfuric acid catalyst. At T=55oC, the CEG was recorded with time.

Problem Solving AlgorithmExample 4-1 Determining k from Batch Data

A. Problem statement. Determine the kA

B. SketchC. Identify

C1. Relevant theoriesRate law:

Mole balance:C2. Variables

Dependent: concentrationsIndependent: time

C3. Knowns and unknownsKnowns: CEG = f(t)Unknowns: 1. CEO = f(t) 2. kA

3. Reactor volumeC4. Inputs and outputs: reactant fed

all at once a batch reactorC5. Missing information: None

AAA Ckr

Vrdt

dNA

A

D. Assumptions and approximations:Assumptions1. Well mixed2. All reactants enter at the same time3. No side reaction4. Negligible filling and emptying time5. Isothermal operationApproximations1. Water in excess (CH2O~constant)

E. Specification. The problem is neither overspecified nor underspecified.

F. Related material. This problem uses the mole balances developed in Chap. 1 for a batch reactor and the stoichiometry and rate laws developed in Chap. 3.

G. Use an Algorithm.

dtdCA = -kCA

CA0

CA dCA-CA

0

t= k dt ln

CA

CAo = kt CA = CAoe-kt

A + B C

NC=NAoX=NAo-NA

CC=NC/V=NC/Vo= CAo-CA=CAo- CAoe-kt =CAo(1-e-kt)

Rearranging and taking the logarithm of both side yields

We see that a plot ln[(CA0-CC)/CA0] as a function of t will be a straight line with a slope –k.

ktC

CC

A

CA

0

0lnAA Cr

ttk

)min311.0(

min311.055.195.8

3.210ln

1

1

12

Design of CSTR

Design Equation for a CSTR

A

AA

A

AA

exitA

A

r

CC

v

V

r

CCvV

r

XFV

0

0

00

0

)(

For a 1st-order irreversible reaction, the rate law is

A

AA

AA

kC

CC

kCr

0

Mole balance

Rate law

Combine

Solving for the effluent concentration of A, we obtain

k

kX

XCk

CC A

AA

1

)1(1 0

0

the space time

Relationship between space time and

conversion for a 1st-order liquid-phase rxn

Reaction Damköhler number

rate" convection a"

rate"reaction a"

A of Rate Flow Entering

Entranceat Reaction of Rate

0

0

A

A

F

VrDa

The Damkohler is a dimensionless number that can give us a quick estimate of the degree of conversion that can be achieved in continuous-flow reactor.

For 1st-order irreversible reaction

For 2nd-order irreversible reaction

Da 0.1 will usually give less than 10% conversion.Da 10.0 will usually give greater than 90% conversion.

For first order reaction, X = =

kCv

VkC

F

VrDa

A

A

A

A

00

0

0

0 0

000

20

0

0A

A

A

A

A kCCv

VkC

F

VrDa

k1 + k

Da1 + Da

CSTRs in SeriesCA1, X1

CA2, X2

CA0

v0

-rA1, V1 -rA2, V2

For first-order irreversible reaction with no volume change (v=v0) is carried out in two CSTRs placed in series. The effluent concentration of A from reactor 1 is

11

01 1 k

CC A

A

From a mole balance on reactor 2,

22

210

2

212

A

AA

A

AA

Ck

CCv

r

FFV

CSTRs in SeriesSolving for CA2, the concentration exiting the second reactor, we get

1122

0

22

12 111 kk

C

k

CC AA

A

If instead of two CSTRs in series we had n equal-sized CSTRs connected in series (1 = 2 = … = n = i = (Vi/v0)) operating at the same temperature (k1 = k2 = … = kn = k), the concentration leaving the last reactor would be

nA

nA

AnC

k

CC

Da1100

The conversion and the rate of disappearance of A for these n tank reactors in series would be

nkX

1

11 n

AAnAn

k

kCkCr

10

X= 1-CA/CAo

1(1+Da)n

= 1 -

Conversion as a Function of Reactors in Series for different Damkohler numbers for a first-order recation

Da 1, 90% conversion is achieved in two or three reactors; thus the cost of adding subsequent reactors might not be justifiedDa ~0.1, the conversion continues to increase significantly with each reactor added

Da=k=1

Da=k=0.5

Da=k=0.1

CSTRs in Parallel

-rA1, V1, X1

-rAi, Vi

-rAn, Vn

FA0

FA0 i

1

i

n

A balance on any reactor i, gives the individualreactor volume

n

FF

n

VV

rrrr

XXXX

r

XFV

AiAi

AAnAA

n

Ai

iiAi

00

21

21

0

The volume of each individual reactor, Vi, is related to the total volume, V, of all the reactors, and similar relationship exists for the total molar flow rate

CSTRs in Parallel

-rA1, V1

-rAi, Vi

-rAn, Vn

FA0

FA0i

1

i

n

Substituting these values into Eq (4-12) yields

A

A

Ai

iA

Ai

iA

Ai

iiAi

r

XF

r

XFV

r

X

n

F

n

V

r

XFV

00

0

0

The conversion achieved in any one of the reactors in parallel is identical to what would be achieved if the reactant were fed in one stream to one large reactor of volume V

n

FF A

iA0

0

n

VVi

FA01

-rA1, V1, X1

-rAi, Vi, Xi

-rAn, Vn, Xn

FA0n

A Second-Order Reaction in a CSTR

20

20

0

200

)1( XkC

X

kC

CC

v

V

kC

XF

r

XFV

A

A

AA

A

A

A

A

For a 2nd-order liquid-phase reaction being carried out in a CSTR, the combination of the rate law and the design equation yields

For const density v=v0, FA0X=v0(CA0-CA)

We solve the above eq. for X:

Using our definition of conversion, we have

Da2

Da41Da21

2

4121

2

22121

0

00

0

20

200

A

AA

A

AAA

kC

kCkC

kC

kCkCkCX

The minus sign must be chosen in the quadratic equation because X cannot be greater than 1.

(4-16)

(4-15)

(4-14)

A Second-Order Reaction in a CSTR

606

0.670.88

Example 4-2: Producing 200 Million Pound/Year in a CSTR

A 1 lb mol/ft3 solution of ethylene oxide (EO) in water is fed to the reactor together with an equal volumetric solution of water containing 0.9 wt% of the catalyst H2SO4. The specific reaction rate constant is 0.311 min-1.

(a) If 80% conversion is to be achieved, determine the necessary CSTR volume.

(b) If two 800-gal reactors were arranged in parallel, what is the corresponding conversion?

(c) If two 800-gal reactors were arranged in series,what is the corresponding conversion?

Tubular Reactors

- 2nd-order gas-phase reaction- Turbulent- No dispersion- No radial gradients in T, u, or C

PLUG-FLOW REACTOR

PFR mole balance

AA rdV

dXF 0

must be used when there is a DP or heat exchange between PFR & the surrounds. In the absence of DP or heat exchange, the integral form of the PFR design equation is used.

X

AA r

dXFV

00

X

AA

kC

dXFV

0 20

Rate law

For Liquid-Phase Reaction

X

AA

kC

dXFV

0 20 0

X= FAo

dX

kCAo2(1-X)2

= kCAo

vo ( ) 1-X X

V/vo= = kCAo

1 ( ) 1-X X

kCAo

1+kCAo

Da2

1 + Da2

X= =

(Da2=Damköhler number for a 2nd-order RXN)

For n-th order RXN, Dan=kCAon-1

For Gas-Phase Reaction

X

AA

kC

dXFV

0 20

Conversion as a Function of Distance Down the Reactor

0)5.01( vXv

The volumetric flow rate decreases with increasing conversion, and the reactant spends more time in the reactor than reactants that produce no net change in the total number of moles.

the reactant spends more time

0)21( vXv the reactant spends less time

öV(m3)

v=vo(1+X)

Change in Gas-Phase Volumetric Flow Rate Down the Reactor

v=vo(1+X)

Example 4-3: Determination of a PFR Volume

Determine the PFR volume necessary to produce 300 million pounds of ethylene a yearfrom cracking a feed stream of pure ethane. The reaction is irreversible and follows an elementary rate law. We want to achieve 80% conversion of ethane, operating the reactor isothermally at 1100K at a pressure of 6 atm.

C2H6 (A) C2H4 (B) + H2 (C)

FB=300x106 lb/year=0.340 lb-mol/sec

FB=FAoX

FAo=FB/X=0.340/0.8=0.425 lb-mol/sec

Pressure Drop in Reactors

In liquid-phase reaction

- the concentration of reactants is insignificantly affected by even

relatively large change in the total pressure

- ignore the effect of pressure drop on the rate of reaction

when sizing liquid-phase chemical reactors

- that is, pressure drop is ignored for liquid-phase kinetics calculations

In gas-phase reaction

- the concentration of the reacting species is proportional to

the total pressure

- the effects of pressure drop on the reaction system are a key factor

in the success or failure of the reactor operation

- that is, pressure drop may be very important for gas-phase reactions

Pressure Drop and Rate Law

• For an ideal gas,

- determine the ratio P/P0 as a function of V or W

- combine the concentration, rate law, and design equation

- the differential form of the mole balance (design equation) must be used

)/)(/)(1( 000

0

TTPPXv

XvF

v

FC iiAi

i

00 1 P

P

X

XvCC ii

Ai

기상반응에서는 반응 성분의 농도가 반응압력에 비례

하므로 압력강하에 대한 고려가 필수적이다 .

To

T(4-18)


• For example,

- the second order isomerization reaction in a packed-bed reactor

2A B + C

-the differential form of the mole balance

- rate law

-from stoichiometry for gas-phase reactions

mincatalyst g

gmoles0 AA r

dW

dXF

2AA kCr

T

T

P

P

X

XCC AA

0

00 1

1

Pressure drop and the rate law

• Then, the rate law

- the larger the pressure drop from frictional losses, the smaller the reaction rate

• Combining with the mole balance and assuming isothermal operation (T=To)

• Dividing by FA0

20

00 1

1

T

T

P

P

X

XCkr AA

2

0

20

0 1

)1(

P

P

X

XCk

dW

dXF A

A

2

0

2

0

0

1

1

P

P

X

X

v

kC

dW

dX A

(4-20)


• For isothermal operation (T =T0)

-a function of only conversion and pressure

-Another equation is needed to determine the conversion as a function of

catalyst weight

- that is, we need to relate the pressure drop to the catalyst weight

),( PXfdW

dX

)(WfP

(4-21)

Flow Through a Packed Bed

• The majority of gas-phase reactions are catalyzed by passing the reactant through a packed of catalyst particles

• Ergun equation: to calculate pressure drop in a packed porous bed

• The gas density is the only parameter that varies with pressure on the right-hand side. So, calculate the pressure drop through the bed laminar turbulent

G

DDg

G

dz

dP

ppc

75.1)1(1501

3

laminar turbulent

(4-22)

G=u=superficial mass velocity [g/cm2s]; u=superficial velocity [cm/s]; Dp=diameter of particle in the bed [cm]; =porosity=volume of void/total bed volume; 1- =volume of solid/total bed volume

Flow through a Packed Bed

• Equation of continuity

- steady state the mass flow rate at any point is equal to the entering

mass flow rate

• Gas-phase volumetric flow rate

• Then,

vv

mm

00

0

00

00

T

T

F

F

T

T

P

Pvv

T

T

F

F

T

T

P

P

v

v 00

00

00

mo = m

(3-41)

(4-23)

Pressure Drop in a Packed Bed Reactor

• then, Ergun equation

• Simplifying

• The catalyst weight,

00

00

T

T

F

F

T

T

P

P

dz

dP

cc zAW )1(

00

03

0

75.1)1(150)1(

T

T

ppc F

F

T

T

P

PG

DDg

G

dz

dP

GDDg

G

ppc

75.1)1(150)1(

30

0

Volume ofsolid

Density of solid catalyst

dzAdW cc )1(

(4-24)

dW

dPWe need

(4-26)

(4-25)


• then, Ergun equation

• Simplifying

00

00

)1( T

T

cc F

F

T

T

P

P

AdW

dP

0

0

)1(

2

PA cc

00

0

0 /2 T

T

F

F

PP

P

T

T

dW

dP(4-27)

X

F

FFXFFF

T

ATTTT

0

0000 1 X

F

F

T

T 10

0

00

T

AA F

Fy

)1(/2 0

0

0

XPP

P

T

T

dW

dP

(4-28)

(4-29)

Let y=P/Po 2ydy FT

FTo

Pressure drop in a packed bed reactor

ε < 0, the pressure drop (P) will be less than ε = 0

ε > 0, the pressure drop (P) will be greater than ε = 0

(4-30)

• For isothermal operation

),( PXfdW

dP ),( PXf

dW

dXand

• The two expressions are coupled ordinary differential equations. We can solve them simultaneously using an ODE solver such as Polymath.

• For the special case of isothermal operation and ε = 0, we can obtain an analytical solution.

• Polymath will combine the mole balance, rate law and stoichiometry

(4-31)

)1(/2 0

0

0

XPP

P

T

T

dW

dP

2y(1+X)

dy


Analytical Solution

If ε = 0 or ε X ≪ 1, we can obtain an analytical solution to Eq. (4-30) for isothermal operation (i.e., T=T0). For isothermal operation with ε = 0, Eq. (4-30) becomes

)/(2 0

0

PP

P

dW

dP Isothermal (T=To)

with ε = 0

(4-30))1(/2 0

0

0

XPP

P

T

T

dW

dP

2y(1+X)

dy

dy

2y

2ydydW = -

dy2

dW = -

At W=0, y=1 (P/Po=1) y2= 1- w


If ε = 0 or ε X ≪ 1, we can obtain an analytical solution to Eq. (4-30) for isothermal operation (i.e., T=T0). For isothermal operation with ε = 0, Eq. (4-30) becomes

0

0

)1(

2

PA cc

Pressure ratio only for ε = 0

WP

P 1

0

0

0

0

21

P

z

P

P

cc zAW )1(

(4-33)

(4-34)

(4-26)

Pressure as a function of reactor length, z

y=

y= =f (z)

Pressure Drop in Pipes

D

fG

dL

dG

dL

dP

22

Integrating with P=P0 at L=0, and assuming that f = constant

Pressure drop along the length of the pipe

02 2

2

00

D

fG

PdL

dPG

dL

dP

P

P

P

P

D

Lf

PG

PP 0

0

0222

0 ln22

0

Rearranging, we get

DAP

fGV

P

P

cpp

00

20 4

1

Example 4-4: 1½” schedule 40 x1000-ft L (p=0.018), P<10%

Analytical Solution for Reaction with Pressure DropConversion as a function of catalyst weight

AA rdW

dXF 0

A B

2nd-order isothermal reaction

Mole balance:

2AA kCr

Rate law:

00 )1(

P

PXCC AA

Stoichiometry: Gas-phase isothermal with =0

WP

P 1

0

2/10 )1)(1( WXCC AA

22/12

0

20 1)1( WX

F

kC

dW

dX

A

A

Combining

21

1

0,0

1)1(

0

0

000

220

0

WW

X

X

kC

v

vCFandWXat

dWWX

dX

kC

F

A

AA

A

A

Separating variable and Integrating

211

21

0

0

0

0

W

v

WkC

W

v

WkC

XA

A

(4-38)y=

Reaction with Pressure DropConversion as a function of catalyst weight

211

21

0

0

0

0

W

v

WkC

W

v

WkC

XA

A

(4-38)

2/100 )1/(/)2(11 XXkCv

W A (4-39)

Catalyst weight for 2nd-order isothermal reaction in PFR with P

Reaction with Pressure DropConversion as a function of catalyst weight

For gas phase reactions, as the pressure drop increases, the concentration decreases, resulting in a decreased rate of reaction, hence a lower conversion when compared to a reactor without a pressure drop.

Example 4-5 and Example 4-6

지금까지 배운 지식을 활용하여 적들이 방심하고 있는 사이에

이 예제들은 집에서 풀어 봐야지 ...

Spherical Packed-Bed Reactors

Spherical Ultraformer Reactor (Amoco) for dehydrogenation reaction such as

Paraffin Aromatic + 3 H2

Spherical reactor

- minimize pressure drop

- inexpensive

- the most economical shape for high pressure

Coordinate system and variables used with a spherical reactor

Synthesizing a Chemical Plant

Always challenge the assumptions, constraints, and boundaries of the problem

The profit from a chemical plant will be the difference between income from sales and the cost to produce the chemical

Profit = (value of products) – (cost of reactants)

– (operating costs) – (separation costs)

The operating cost: energy, labor, overhead, and depreciation of equipment

Production of Ethylene Glycol

C2H4O(aq)

7 8 H2O, 0.9wt% H2SO4

200 millionlb EG/yr

9

V=197 ft3, X=0.8

CH2OHC2H4O + H2O CH2OH

Cat.

5

O2, C2H4, N2, C2H4O

6

separator

2

H2, C2H4

C2H6

separator

H2O C2H4O

absorber

C 2H 6 C 2H 4 + H 21

V=81 ft3, X=0.8

402 million lbC2H6 /yr

3C2H4

4 Air

W=45,440 lb, X=0.6

C2H4+ ½ O2 C2H4OAg

Synthesizing a Chemical Plant

Ethylene glycol = $0.38/lb (2x108 lb/yr)

Ethane = $0.04/lb (4x106 lb/yr)

Sulfuric acid = $0.043/lb (2.26x108 lb/yr)

Operating cost = $8x106/yr

Profit = ($0.38/lb x 2x108 lb/yr) – ($0.04/lb x 4x108 lb/yr)

-($0.043/lb x 2.26x106 lb/yr) – ($8x106/yr)

= $52 million

How the profit will be affected by conversion, separation, recycle stream, and operating costs?

Using CA (liquid) and FA (gas)

in the mole balance and rate laws

More convenient to work in terms of the number of moles or molar flow rate rather than conversion.

Must write a mole balance on each species when molar flow rates (Fi) and concentrations (Ci) are used as variables

Membrane reactor, multiple reaction, and unsteady state

BAAA

DCBA

CCkr

d

r

c

r

b

r

a

r

Da

dC

a

cB

a

bA

Liquid Phase

Mole balance for liquid-phase reactions

AA

AA

A

AA

AA

rdW

dCv

rdV

dCv

r

CCvV

rdt

dC

0

0

00 )(

AB

AB

A

BB

AB

ra

b

dW

dCv

ra

b

dV

dCv

rab

CCvV

ra

b

dt

dC

0

0

00

)/(

)(

Batch

CSTR

PFR

PBR

DCBA dcba

DCBAa

d

a

c

a

b

For liquid-phase reaction with no volume change Concentration is the preferred variable

Gas Phase For gas phase reactions

need to be expressed in terms of the molar flow rates

T

T

P

P

F

FCC

T

jTj

0

00

Total molar flow rate

n

jjT FF

1

jj r

dV

dF

y

(4-28, T=To)dydW

=2y-

FTo

FT

(1-11)

(3-42)

= FA + FB + FC + FD + FI

Algorithm for Gas Phase Reaction

dDcCbBaA

Mole balances:

Vrdt

dN

Vrdt

dN

Vrdt

dN

Vrdt

dN

DD

CC

BB

AA

D

DD

C

CC

B

BB

A

AA

r

FFV

r

FFV

r

FFV

r

FFV

0

0

0

0

DD

CC

BB

AA

rdV

dF

rdV

dF

rdV

dF

rdV

dF

Batch CSTR PFR


Rate law: BAAA CCkr

Stoichiometry:

DCBAT

T

DTD

T

CTC

T

BTB

T

ATA

ADACAB

DCBA

FFFFF

T

T

P

P

F

FCC

T

T

P

P

F

FCC

T

T

P

P

F

FCC

T

T

P

P

F

FCC

ra

drr

a

crr

a

br

d

r

c

r

b

r

a

r

0

00

0

00

0

00

0

00

Relative rate of reaction:

Concentration:

Total molar flow rate:

y


Combine:

T

B

T

ATA

B

T

B

T

ATA

C

T

B

T

ATA

B

T

B

T

ATA

A

F

F

F

FCk

a

d

dV

dF

F

F

F

FCk

a

c

dV

dF

F

F

F

FCk

a

b

dV

dF

F

F

F

FCk

dV

dF

00

00

- Specify the parameter values: kA, CT0, T0, a, b, c, d

- Specify the entering number: FA0, FB0, FC0, FD0, and final value: Vfinal

Use an ODE solver

Microreactors

Microreactors are used for the production of special chemicals, combinatorial chemical screening, lab-on-a-chip, and chemical sensors.

Example 4-7: Gas-Phase Reaction in a Microreactor

The gas phase reaction, 2NOCl 2NO + Cl2, is carried out at 425oC and 1641 kPa (16.2 atm). Pure NoCl is to be fed, and the reaction follows an elementary rate law. It is desired to produce 20 tons of NO per year in a micro reactor system using a bank of ten microreactors in parallel.Each microreactor has 100 channelswith each channel 0.2 mm square and250 mm in length. Plot the molar flowrates as a function of volume down thelength of the reactor. The volume of each channel is 10-5 dm3.

Rate constant and activation energy (given): k=0.29 dm3/mol-sec at 500K with E=24 kcal/molTo produce 20 tons/year of NO at 85% conversion would require a feed rate of 0.0226 mol/s.

Solution

For one channel,

FAo=22.6 mol/s

FB=19.2 mol/s, X=0.85

2NOCl 2NO + Cl2

2A 2B + CA B +1/2C

1. Mole balance: jj r

dV

dF

dFA

dV= rA

dFB

dV= rB

dFC

dV= rC

2. Rate law:

-rA=kCA2, k=0.29 dm3/mol-sec

3. Stoichiometry: Gas phase with T=To and P=Po, then v=vo(FT/FTo)

3-1. Relative rate

rA

-1=

rB

1=

rC

1/2

rB = -rA, rC= -0.5rA

3-2. Concentration

By applying Equation (3-42)

CA=CTo(FA/FT), CB=CTo(FB/FT), CC=CTo(FC/FT) with FT=FA+FB+FC

4. Combine

-rA=kCA2=kCTo

2(FA/FT)2

dFA

dVdFB

dVdFC

dV

= -kCTo2(FA/FT)2

= kCTo2(FA/FT)2

= 0.5kCTo2(FA/FT)2

5. Evaluate

CTo=Po/RT=(1641)/(8.314)(698K)=0.286 mol/dm3=0.286 mmol/cm3

Use Polymath to solve the ODE

FAFB

FC

Membrane Reactors

Membrane reactors can be used to achieve conversions greater than the original equilibrium value. These higher conversions are the result of Le Chatelier's Principle; you can remove one of the reaction products and drive the reaction to the right. To accomplish this, a membrane that is permeable to that reaction product, but is impermeable to all other species, is placed around the reacting mixture

By having one of the products pass throughout the membrane, we drive the reaction toward completion

What kinds of membrane reactors are available?

Membrane reactors are most commonly used when a reaction involves some form of catalyst, and there are two main types of these membrane reactors: the inert membrance reactor and the catalytic membrane reactor.

The inert membrane reactor allows catalyst pellets to flow with the reactants on the feed side (usually the inside of the membrane). It is known as an IMRCF, which stands for Inert Membrane Reactor with Catalyst on the Feed side. In this kind of membrane reactor, the membrane does not participate in the reaction directly; it simply acts as a barrier to the reactants and some products.

A catalytic membrane reactor (CMR) has a membrane that has either been coated with or is made of a material that contains catalyst, which means that the membrane itself participates in the reaction. Some of the reaction products (those that are small enough) pass through the membrane and exit the reactor on the permeate side.

Membrane Reactors

Inert membrane reactor with catalyst pellets on the feed side (IMRCF)

C6H12 C6H6 + 3H2

H2 molecule is small enough to diffuse through the small pore of the membrane while C6H12 and C6H6 cannot..

Membrane Reactors

Catalytic membrane reactor (CMR)

C6H12 C6H6 + 3H2

Membrane reactors are commonly used in dehydrogenation reactions (e.g., dehydrogenation of ethane), where only one of the products (molecular hydrogen) is small enough to pass through the membrane. This raises the conversion for the reaction, making the process more economical.

Dehydrogenation Reaction

A B + C

CH2CH3 CH=CH

+ H2 (B)

C4H10 C4H8 + H2 (B)

C3H8 C3H6 + H2 (B)

According to The DOE, an energy saving of 10 trillion Btu per year could result from the use of catalytic membrane reactors as replacements for conventional reactors for dehydrogenation reactions such as the dehydrogenation of ethylbenzene to styrene.

RB=kcCB

A B + CFA

FB

V

FA

FB

FCmembrane

V+V

KC=0.05 mol @227oCP=8.2 atmT=227oCFA0=10mol/min

RB

1. Mole balance:

AA

AVVAVA

rdV

dF

VrFF

0

BBB

BBVVBVB

RrdV

dF

VrVRFF

0

for a differential mole balance on A in the catalytic bed at steady state

for a differential mole balance on C in the catalytic bed at steady state C

C rdV

dF

There are two “OUT”

terms

for a differential mole balance on B in the catalytic bed at steady state

Basic Algorithm for Membrane Reactor (Example 4-8)

2. Rate law:

ACABC

CBAA rrrr

K

CCCkr

;;

kc is a transport coefficient. kc=f(membrane & fluid properties, tube diameter…) constant

3. Transport out the sides of the reactor:

BCB CkR

4. Stoichiometry:

CBA

CBAT

T

CTC

T

BTB

T

ATA

rrr

FFFF

F

FCC

F

FCC

F

FCC

000 ;;

Basic Algorithm for Solving Reaction in the Membrane Reactor

5. Combining and Summarizing:

6. Parameter evaluation: CT0=0.2 mol/L, k=0.7 min-1, KC=0.05 mol/L, kc=0.2 min-1

FA0=10 mol/min, FB0=FC0=0

CBAT

T

C

T

B

C

T

T

ATcA

AC

T

BTcA

BA

A

FFFF

F

F

F

F

K

C

F

FCkr

rdV

dF

F

FCkr

dV

dFr

dV

dF

00

0 ;;

7. Numerical solution:Solve with POLYMATH or MATLAB

Basic Algorithm for Solving Reaction in the Membrane Reactor

Reactor volume, V [L]

FA

FB

FC

10

5

00 100 200 300 400 500

FA

FB

FCkc=0.20 min-1

Effects of Side Stream, RB=kcCBin a membrane reactor


FA

FB

FC

10

5

00 100 200 300 400 500

FAFB

FCkc=20 min-1

Large side stream


FA

FB

FC

10

5

00 100 200 300 400 500

FA

FB FC

kc=0.0022 min-1

Little side stream

ConversionX=(10-4)/10=0.6

Unsteady-State Operation of Stirred Reactors

Determine the time to reach steady-state operation

Predict the concentration and conversion as a function of time

Semibatch reactor

(b) ammonolysis, chlorination, hydrolysis

(c) acetylation reaction, esterification reaction

A

B

A, B

C

Startup of a CSTR

Time to Reach Steady State for a First-Order Reaction in a CSTR

To determine the time to reach steady-state operation of a CSTR, we begin with the general mole balance equation applied to a well-mixed CSTR.

dt

dNVrFF A

AAA 0

Conversion does not have any meaning in startup because one cannot separate moles reacted from moles accumulated. Consequently, we must use concentration as the variable in our balance equation. For liquid-phase reactions V =V0 and for a constant overflow, v =v0. After dividing by v0 and replacing V/v0 by the space time , we find that

dt

dCrCC A

AAA 0

(4-45)

(4-46)

Utilizing the definitions of FA and NA, we have

dt

VCdVrvCvC A

AAA)(

00

Startup of a CSTR

AA kCr

01 AA

A CC

k

dt

dC

tk

k

CC A

A )1(exp11

0

For a first-order reaction:

Solution (4-47)

Letting ts be the time necessary to reach 99% of the steady state concentration, CAS:

k

CC A

AS

10

Rearranging (4-46) for CA=0.99CAS yields

ktS

16.4 (4-48)

Startup of a CSTR

For slow reactions :

For rapid reactions:

ktS

16.4 (4-48)

kt

t

S

S

6.4

6.4

(4-49)

(4-50)

Time to reach steady state in an isothermal CSTRFor most first-order system, steady state is achieved in three to four space times

Semibatch ReactorsCBA

B

A

Mole balance on specials A:

[ in ] - [ out ] + [ gen. ] = [ acc. ]

dt

dVC

dt

VdC

dt

VCdVr

dt

dNtVr

AAAA

AA

)(00

Mass balance of all specials:

[ in ] - [ out ] + [ gen. ] = [ acc. ]

00

00

for

)(00

vdt

dV

dt

Vdv

tVV 00

AAA

AAA

CV

vr

dt

dC

dt

VdCVrCv

0

0

(4-55)

(4-56)

(4-54)

(4-53)

(4-52)

(4-51)

B

A

0 0Semibatch

reactorvolume as a

function of time

Mole balance on specials B:

[ in ] - [ out ] + [ gen. ] = [ acc. ]

000

0

0

)(

)(0

BBBB

BB

BB

BBB

BBB

CvVrFVr

dt

dCV

dt

dVC

dt

VCd

dt

dN

FVrdt

dN

dt

dNtVrF

V

CCvr

dt

dC BBB

B )( 00

(4-57)

(4-58)

Example 4-9: Isothermal Semibatch Reactor with 2nd–order Reaction

CNBr + CH3NH2 CH3Br + NCNH2

A + B C + D

Isothermal elementary reaction in a semibatch reactor

t=0, CA=0.05 gmol/, CB=0.025 gmol/ℓ, v0=0.05ℓ/s, k=2.2ℓ/s·mol, V0=5ℓ

tvVV

CCV

vCkC

dt

dC

CV

vCkC

dt

dC

CkCr

BBBAB

ABAA

BAA

00

00

0

)(

CC

CCCC

ACC

Cvdt

dCV

dt

dVC

dt

dCV

dt

VCd

dt

dN

VrVrdt

dN

0

)(

Mole balance of A, B, C, and D

A

B

C CBAC C

V

vCkC

dt

dC 0

DBAD C

V

vCkC

dt

dC 0D

Conversion, X

00

00

0

0

VC

VCVC

N

NNX

A

AA

A

AA

Concentration-time Trajectoriesin Semibatch Reactor

CA

CC

CB

0.05

0.04

0.03

0.02

0.01

0.000 100 200 300 400 500


(A) (B) (C) (D)


Time

Co

nc

entr

ati

on

Reaction Rate-time Trajectoriesin Semibatch Reactor

0.0025

0.0020

0.0015

0.00010

0.00005

0.000 50 100 150 200 250


(A) (B) (C) (D)


Re

act

ion

rat

e [

mo

le/s

•L)

Reactor Equations in terms of Conversionin semibatch reactor

DCBA The number of moles of A remaining at any time , t

XNNN AAA 00

B

A

B

A

0 0

t timeto

up reactedA of

moles ofnumber

initially

vatin theA of

moles ofnumber

tat time

vatin theA of

moles ofnumber

where X is the mole of A reacted per mole of A initially in the vat.

The number of moles of B remaining at any time , t

XNdtFNN A

t

BBiB 00

0

t timeto

up reacted B of

moles ofnumber

vat the

toadded B of

moles ofnumber

initially

vatin the B of

moles ofnumber

tat time

vatin the B of

moles ofnumber

For a constant molar feed rate

XNtFNN ABBiB 00 (4-61)

(4-60)

(4-59)

Reactor Equations in terms of Conversionin semibatch reactor

DCBA

A mole balance on specials A:

dt

dNVr A

A

B

A

B

A

0 0

The number of moles of C and D cab be taken directly from the stoichiometric table

The rate law (reversible 2nd-order reaction)

XNNN

XNNN

ADiD

ACiC

0

0

(4-62)

C

DCBAA K

CCCCkr (4-65)

The concentration of A and B

tvV

XN

V

NC

tvV

XN

V

NC

tvV

XNtFN

V

NC

tvV

XN

V

NC

ADA

ACC

ABBiBB

AAA

00

0

00

0

00

00

00

0 )1(

tvV

KXNXNtFNXk

dt

dX CAABBi

00

2000 /1

Combine (4-66)

(4-66) can be solved numerically

(4-63)

Equilibrium conversionin semibatch reactor

B

A

B

A

0 0

For reversible reactions carried out in a semibatch reactor, the maximum attainable conversion (i.e., the equilibrium conversion) will change as the reaction proceeds because more reactant is continually to the right.

The rate law (reversible 2nd-order reaction)

C

DCBAA K

CCCCkr

DCBA

)/)(/(

)/)(/(

VNVN

VNVN

CC

CCK

BeAe

DeCe

BeAe

DeCeC

))(1(

))(1(

))((

00

20

000

00

eABe

eA

eABeA

eAeA

BeAe

DeCeC

XNtFX

XN

XNtFXN

XNXN

NN

NNK

(4-68)

e

eeC

BC

A

X

XXK

FK

Nt

1

2

0

0 (4-69)

)1(2

)1(4110

0

2

0

0

0

0

C

A

BCC

A

BC

A

BC

e K

N

tFKK

N

tFK

N

tFK

X(4-70)

Equilibrium conversion in a semiba

tch reactor

Reactive Distillation

The distillation of chemically reacting mixtures has become increasingly common in chemical industries.

Carrying out these two operations, reaction and distillation, simultaneously in a single unit results in significantly lower capital cost and operating costs.

Reactive distillation is particularly attractive when one of the reaction products has a lower boiling point, resulting in its volatilization from the reacting liquid mixture.


By continually removing the volatile reaction product, methyl acetate, from the reacting liquid-phase reaction, the reverse reaction is negligible and the reaction continues to proceed towards completion in the forward direction.

An example of reactive distillation is the production of methyl acetate:

Reactive distillation is used with reversible, liquid phase reactions. Suppose a reversible reaction had the following chemical equation :


However, if one or more of the products are removed more of the product will be formed because of Le Chatlier's Principle :

For many reversible reactions the equilibrium point lies far to the left and little product is formed :

Removing one or more of the products is one of the principles behind reactive distillation. The reaction mixture is heated and the product(s) are boiled off. However, caution must be taken that the reactants won't boil off before the products.

Homework

P4-11B

P4-12B

P4-13B

P4-14C

Due Date: Next Week

chemical reaction engineering 1 chapter 4 isothermal reactor design 반응공학 1

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