Chemical Reactor Designocw.snu.ac.kr/sites/default/files/NOTE/7230.pdf · 2018-01-30 · Objectives • Describ hibe the alihlgorithm thll h d l hilhat allows the reader to solve

Chemical Reactor DesignChemical Reactor Design

Y W LYoun-Woo LeeSchool of Chemical and Biological Engineering

Seoul National Universityy155-741, 599 Gwanangro, Gwanak-gu, Seoul, Korea [email protected] http://sfpl.snu.ac.kr

Chapter 4Chapter 4

Isothermal Reactor DesignIsothermal Reactor Design

Chemical Reactor DesignChemical Reactor DesignChemical Reactor DesignChemical Reactor Design

化學反應裝置設計化學反應裝置設計化學反應裝置設計化學反應裝置設計

Seoul National University

Objectives

ib h l i h h ll h d l h i l

j

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

• Size batch reactors, semibatch 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 Study g a qu d p ase batc eacto to dete e t e spec creaction rate constant needed for the design of a CSTR.

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

• Account for the effects of pressure drop on conversion in packed b d b l d i k d b d h i lbed tubular reactors and in packed bed spherical reactors.


Fig. 4-1 Isothermal Reaction Design AlgorithmAlgorithm for Conversion


Algorithm for isothermal reactor design

1. Mole balance and design equation1. Mole balance and design equation2. Rate law3 Stoichiometry3. Stoichiometry4. Combine5 E al ate5. Evaluate

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

A. Graphically (Chapter 2)

B N i l (A di A4)B. Numerical (Appendix A4)

C. Analytical (Appendix A1)

D. Software packages (polymath)Seoul National University

French Menu Analogy


Algorithm for isothermal reactorsFrench Menu

Analogy


4.2.1 Batch Operationp

For constant volume batch reactor, the mole balance can be written

dN 1 dCVNddNdN /11

,in terms of concentration

AA r

dtdN

V

1

AAAAA r

dtdC

dtVNd

dtdN

VdtdN

V 0

0

/11

Generally, when analyzing laboratory experiments, it is best toprocess the data in terms of the measured variable. Becauseconcentration is the measured variable for most liquid-phaseconcentration is the measured variable for most liquid phasereactions, the general mole balance equation applied to reactionsin which there is no volume change becomes

AA r

dtdC

This is the form we will use in analyzing reaction rate data in Chap 5.Seoul National University

Reaction time in Batch Operation

Algorithm for isothermal reactor designAlgorithm for isothermal reactor design BA

AA VrdtdXN 00

1. Mole balance

& Design equation

AA

XCC

AinorderndleirreversibkCr 2

)1(

2,2. Rate law

AA

XkCdX

XCC

2

0

)1(

)1(

4 C bi ti

3. Stoichiometry

A

dtkCdX

XkCdt 0 )1(

2nd orderIsothermal

Liquid-phaseBatch reaction

4. Combination

Xt

A

dX

dtkCX 02

1

)1(

X1

Batch reaction

5. Analytical Evaluation

X

A

t

XdX

kCdt

0 200 )1(

1

X

XkC

tA 1

1

0Seoul National University

Reaction time in Batch Operation

Typical cycle times for a batch polymerization processt = tf + t + t + tActivity Time (h)

tt tf + te + tR + tc

1. Charge feed to the reactor and agitate, tf2. Heat to reaction temperature, te

1.5-3.00.2-2.0V i3. Carry out reaction, tR

4. Empty and clean reactor, tc

Varies0.5-1.0

Total cycle time excluding reaction 3 0 6 0Total cycle time excluding reaction 3.0-6.0

Batch polymerization reaction times may vary between 5 and 60 hoursBatch polymerization reaction times may vary between 5 and 60 hours.Decreasing the reaction time with a 60-h reaction is a critical problem. As thereaction time is reduced (e.g. 2.5 h for a 2nd-order reaction with X=0.9 andkC 10-3 s-1) it becomes important to se large lines and p mps to achie ekCA0=10-3 s-1), it becomes important to use large lines and pumps to achieverapid transfer and to utilize efficient sequencing to minimize the cycle time.


Batch Reaction Times

BA

Mole balance VN

rdtdX A

0Ndt A

Rate lawkCr AA

order-First order Second2kCr AA

)1(00

XCVNC A

AA

kCr AA kCr AA

Stoichiometry (V=V0)

XkdtdX

)1( )1( 20 XkC

dtdX

A Combine

Xkt

11ln1

)1(0 XkCXt

A Integration



t

s

1ln1

)10k 0.9,(Xorder -1st 1-4

Xt

s

R

)10kC 0.9,(Xorder 2nd 13A0

XktR

1ln11

ln

XkCt

AR

9.0

)1(0

k3.2

9.01ln

kCA

9

)9.01(0

k3.2

14

kCA

90

ssec000,23

10 14

ssec000,9

10 13

hr4.6 hr5.2Seoul National University


The order of magnitude of time hi 90% i

Table 4-3

to achieve 90% conversionFor first- and second-order irreversible batch reactions

1st-orderk (s-1)

2nd-orderkCA0 (s-1)

Reaction timetR

10-4 10-3 Hours

10 2 10 1 Mi t10-2 10-1 Minutes

1 10 Seconds1 10 Seconds

1,000 10,000 Milliseconds


Design a Reactor to Produce of ethylene glycol

Design a CSTR to produce 200 million pounds of ethylene glycol per yearby hydrolyzing ethylene oxide. However, before the design can be carriedout, it is necessary to perform and analyze a batch reactor experiment todetermine the specific reaction rate constant (kA). Since the reaction will becarried out isothermally k will need to be determined only at the reactioncarried out isothermally, kA will need to be determined only at the reactiontemperature of the CSTR. At high temperature there is a significant by-product formation, while at temperature below 40oC the reaction does notproceed at a significant rate; consequently, a temperature of 55oC has beenchosen. Since the water is usually present in excess, its concentration maybe considered constant during the course of the reaction The reaction isbe considered constant during the course of the reaction. The reaction isfirst-order in ethylene oxide.

O CH OHO CH2-OH

CH2-CH2 + H2O CH2-OHH2SO4

A + B C Catalyst


Example 4-1 Determining k from Batch Data

In the lab experiment, 500mL of a 2 M solution (2 kmol/m3) of EO inwater was mixed with 500mL of water containing 0 9 wt % sulfuric acidwater was mixed with 500mL of water containing 0.9 wt % sulfuric acidcatalyst. At T=55oC, the CEG was recorded with time. Determine thespecific reaction rate at 55oC

Time Concentration of EG

specific reaction rate at 55 C.

EO + H2O → EG(min) (kmol/m3)0.0 0.0000.5 0.145

A + B → C

1.0 0.2701.5 0.3762.0 0.4672.0 0.4673.0 0.6104.0 0.7156 0 0 8486.0 0.848

10.0 0.957Seoul National University

Problem Solving AlgorithmExample 4-1 Determining k from Batch Data

A. Problem statement. Determine the kAB Sketch

D. Assumptions and approximations:AssumptionsB. Sketch

C. IdentifyC1. Relevant theories

Rate law: AA Ckr

Assumptions1. Well mixed2. All reactants enter at the same time3. No side reactions

batchA, B, C

Rate law:

Mole balance:C2. Variables

AAA Ckr

Vrdt

dNA

A

3. No side reactions4. Negligible filling and emptying time5. Isothermal operationApproximations

Dependent: concentrations, CiIndependent: time, t

C3. Knowns and unknowns

pp1. Water in excess (CH2O~constant)

CB~ CBOE. Specification. The problem is neither

Knowns: CEG = f(time)Unknowns:

1. CEO = f(time)

overspecified nor underspecified.

F. Related material. This problem uses the2. kA3. Reactor volume

C4. Inputs and outputs: reactant fedll b h

mole balances developed in Chap. 1for a batch reactor and thestoichiometry and rate laws developedi Ch 3all at once a batch reactor

C5. Missing information: Nonein Chap. 3.

G. Use an Algorithm.(figs 4-1 & 4-2)Seoul National University


AA r

dtdN

V

11. MOLE BALANCE Batch reactor that is well-mixeddtV

2 RATE LAW kCrSince water is present in such excess, theconcentration of water at any time t is virtuallythe same as the initial concentration and the rate2. RATE LAW AA kCr the same as the initial concentration and the ratelaw is independent of the concentration of H2O.(CB~CB0)

3. STOICHIOMETRY

Species symbol Initial Change Remaining Concentration

CH2CH2O A NA0 - NA0X NA=NA0(1-X) CA=CA0(1-X)H2O B ΘBNA0 - NA0X NB=NA0(ΘB-X) CB=CA0(ΘB-X)H2O B ΘBNA0 NA0X NB NA0(ΘB X) CB CA0(ΘB X)

CB~ CA0 ΘB = CB0

(CH2OH)2 C 0 NA0X NC =NA0X CC=CA0X

NT0 NT =NT0 - NA0X



AA r

dtdN

V

1

A rdC kCr 4 COMBINING

A kCdCAr

dt AA kCr 4. COMBINING

Mole balance Rate law

AkCdt

,

5. EVALUATE

For isothermal operation k is constant:For isothermal operation, k is constant:

At tC

A ktC

dtkkdtdCA 0ln

kt

AC A

eCC

ktC

dtkkdtCA

0 0ln

0

AA eCC 0



The concentration of EG at any time t can be obtained from the reaction stoichiometry

A + B C

NNXNN

)1(00

00

ktAAA

CCC

AAAC

eCCCNN

C

NNXNN

)(000

AAAC VV

)1(0kt

AC eCC

kCC CA 0l kt

CA

CA 0

0lnSeoul National University

Example 4-1 Determining k from Batch Data

Rearranging and taking the logarithmof both side yields

CCAA Cr )min311.0( 1

We see that a plot ln[(CA0-CC)/CA0] as

ktC

CC

A

CA

0

0ln The rate law can now be used in the designof an industrial CSTR. Note that this ratelaw was obtained from the lab-scale batchWe see that a plot ln[(CA0 CC)/CA0] as

a function of t will be a straight line witha slope –k.

1

3210ln

reactor (1000 mL).

Time 0 CA CC

CA0

0.61

12

min311.055.195.8

3.210ln

tt

k

(min)0.0 1.0000.5 0.8551 0 0 730

0AC0.1

(CA0-C

C)/

CA

0.061.0 0.7301.5 0.6242.0 0.5333.0 0.390

(

4.0 0.2856.0 0.152

10.0 0.043

0.01

0 2 4 6 8 10 12

t (min)1.55 8.95Seoul National University

4.3 Design of CSTRDesign Equation for a CSTR

A XFV 0Mole balance

AA

exitA

XCVXCvV

rV

000

)(Mole balance

the space time000 AA CvF

AA rvr 0

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

the space time

A

AA

XCX

kCr

01

Rate law

C biA

A

kCXk0

1

Combine

Rearrangingg g

kkX

1

CSTR relationship between space time and

conversion for a 1st-order k1liquid-phase rxn


4.3.1 A single CSTR

We could also combine Eq (3-29) and (4-8) to find the exit t ti f A C

kCXCC AAA

1)1( 00

concentration of A, CA:

kX

kC

kkkC

kCXCC

AA

AAA

111

11)1(

00

00 k1

CC

kk

AA

11

0

kCA 1

kC

C AA

10exit concentration of Ak1


4.3.2 CSTRs in Series

CA1, X1CA0

v0

CA2, X2

v0

-rA1, V1 -rA2, V2

For 1st-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,

21021 AAAA CCvFF 22

210

2

212

A

AA

A

AA

CkCCv

rFFV


CSTRs in Series

Solving for CA2, the concentration exiting the second reactor, we get

1122

0

22

12 111 kk

Ck

CC AAA

11

01 1 k

CC A

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 = k k k) th t ti l i th l t t ld bk2 = … = kn = k), the concentration leaving the last reactor would be

AA CCC 00

nnAnk

CDa11

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

X 11 AkCkC 0

nkX

111

nA

AnAnk

kCr

1

0


Conversion as a function of reactors in series for different Damköhler numbers for a first-order reaction

k=1

k=0.1

k=0.5

Da 1, 90% conversion is achieved in two or three reactors; 0

0

A

A

FVr

Da

, ;thus the cost of adding subsequent reactors might not be justified

Da ~0.1, the conversion continues to increase significantly with each reactor added Seoul National University

Reaction Damköhler number

rate"reaction a"Entranceat Reaction of Rate0

A VrDa

rate" convection a"A of Rate Flow Entering0

AFDa

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

For 1st-order irreversible reaction kCv

VkCF

VrDa

A

A

A

A

00

0

0

0

For 2nd-order irreversible reaction 000

20

0

0A

A

A

A

A kCCv

VkCF

VrDa

RULE OF THUMB

000 AA CvF

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


4.3.4 A Second-Order Reaction in a CSTR

For a 2nd-order liquid-phase reaction We solve the above eq. for X:being carried out in a CSTR, thecombination of the rate law and thedesign equation yields

We solve the above eq. for X:

22121 20

200 AAA kCkCkC

X

200

kCXF

rXF

V AA

2

41212

00

0

000

AA

A

AAA

kCkCkC

kCX

(4-14)2kCr AA

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

Da2Da41Da21

2 0

AkC

(4-16)

20

0 kCCC

vV

A

AA The minus sign must be chosen in the

quadratic equation because X cannot

X

Using our definition of conversion, we have

q qbe greater than 1.

(4 15) Da41Da21 2

0 )1( XkCA (4-15)

Da2Da41Da21

X


A Second-Order Reaction in a CSTR

V

0

0

A

A

FVr

Da

0.670.88

At high conversion, a 10-foldincrease in Da will increase theincrease in Da will increase theconversion only to 88% due tolowest value of reactantconcentration in CSTR

606concentration in CSTR.


Example 4-2: Producing 200,000,000 lb/yr in a CSTR~91 ton/yr

It is desired to produce 200 x 106 pounds per year

CA0=1 vA0=vB0

p p p yof EG. The reactor is to operated isothermally.A 1 lb mol/ft3 solution of ethylene oxide (EO) inwater is fed to the reactor together with an equalwater is fed to the reactor together with an equalvolumetric solution of water containing 0.9 wt%of the catalyst H2SO4. The specific reaction rateconstant is 0 311 min-1 as determined in Ex 4-1constant is 0.311 min as determined in Ex 4-1.

(a) If 80% conversion is to be achieved, determineth CSTR lthe 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?


M.W. of EG=62

M.W. of EO=58


v0 = vA0 + vB0

=1

10’3m

5’1.5m

~2 gps

1 gal ~3.78 L

5 gal~19.8L

ga 3 8







Producing 200,000,000 lb/yr of EG in a CSTR

two equal-sized CSTRs in series two equal-sized CSTRs in parallel one CSTR

800gal

X=0.81

800gal

800gal 800gal

800gal

X=0.81800gal 800gal

X1=0.68 X2=0.901480galX=0.8

Conversion in the series arrangement is greater than in parallel for CSTRs.The two equal-sized CSTRs in series will give a higher conversion thanThe two equal sized CSTRs in series will give a higher conversion thantwo CSTRs in parallel of the same size when the reaction order is greater than zero.


4.4 Tubular Reactors

-2nd-order liquid-phase rxn(AProducts)

Rate law: 2AA kCr

(AProducts)-Turbulent,- No dispersion- No radial gradients in T, u, or C

X

AA kC

dXFV0 20

St i hi t f li hPLUG-FLOW REACTOR

PFR mole balance

Stoichiometry for liq. phase rxnT & P = constant

)1( XCCPFR mole balance

AA rdVdXF 0

)1(0 XCC AA

Combination

The differential form of PFR designequation must be used when there is aP or heat exchange between PFR &th d I th b f P

XX

kCvdX

XkCFV

A

X

A

A

1)1(1

0

00 22

0

0

the surrounds. In the absence of P orheat exchange, the integral form of thePFR design equation is used.

2

2

0

0

11 DaDa

kCkCX

A

A

X

AA r

dXFV0

0

20 11 DakCA Da2 is the Damkohler numberfor a second-order reactionSeoul National University


-2nd-order gas-phase rxn(AProducts)

Rate law: 2AA kCr

(AProducts)-Turbulent,- No dispersion- No radial gradients in T, u, or C

XA kC

dXFV0 20

PLUG-FLOW REACTOR

PFR mole balance

AkC0 2

Stoichiometry for gas phase rxnPFR mole balance

AA rdVdXF 0

Stoichiometry for gas phase rxnT & P = constant

)1()1(0 XCXFFC AA

The differential form of PFR designequation must be used when there is aP or heat exchange between PFR &th d I th b f P

)1()1( 00 X

CXvv

C AA

Combinationthe surrounds. In the absence of P orheat exchange, the integral form of thePFR design equation is used.

Combination

dXXFVX

A

22

2

0)1(

X

AA r

dXFV0

0XkCA

A 0 220

0 )1(Seoul National University


dXXFVX

2)1( dX

XkCFV

AA

0 22

00 )1(

)(

CA0 is not function of X; k=constant (isothermal)

dXXX

kCF

VX

A

0 2

2

20

)1()1(

A0 ; ( )

000 vCF AA XkCA

00 )1(

2

Integration yields (see Appendix A.1 Eq. (A-7) @ page1009)

X

XXXkC

vV

A 1)1()1ln()1(2

22

0

0

XXX

vL )1()1l ()1(2

220

Reactor length will be

LAV c

X

XXAkC

LcA 1

)()1ln()1(2 2

0

0

Cross sectional area

V c


Conversion as a function of distance down the reactor

1

1.2A 0.5B (=-0.5)

A B ( 0)

0.8

on (

X) A B (=0)

A 2B (=1)

A 3B (=2)

0.4

0.6

onve

rsi

0.2

Co

3

0

0 0.2 dmkC

v

A

0

0 2 4 6 8 10 12 14

L (m)L (m)


The reaction that has a decrease in the total number of moles will have the highest conversion for a fixed reactor lengthwill have the highest conversion for a fixed reactor length.

0)5.01( vXv the reactant spends more timethe reactant spends more time

0)21( vXv the reactant spends less timethe reactant spends less time

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


Change in Gas-Phase Volumetric Flow Rate Down the Reactor

v=vo(1+X)

X

XXXkC

vVA 1

)1()1ln()1(22

2

0

0

=1 : (A→2B)v = vo(1+X)o( )

= 0 : (A→B)( )v = vo

=0.5 : (2A→B)v = vo(1-0.5X)Complete

conversion

When there is a decrease in the number of moles in the gase phase,the volumetric flow rate decreases with increasing conversionthe volumetric flow rate decreases with increasing conversion,and the reactant spends more time in the reactorthan reactants that produce no net change in the total number of moles.


Example 4-3: Determination of a PFR Volume

Determine the PFR volume necessary to produce 300 milliond f h l f ki f d fpounds of ethylene a year from cracking a feed stream of pure

ethane. The reaction is irreversible and follows an elementary ratelaw We want to achieve 80% conversion of ethane operating thelaw. We want to achieve 80% conversion of ethane, operating thereactor isothermally at 1100K at a pressure of 6 atm.

C2H6 C2H4 + H2

A B + C

F 300x106 lb/ 0 340 lb l/

A B + C

FB = 300x106 lb/year = 0.340 lb-mol/secFB = FAoXFAo = FB/X = 0.340/0.8 = 0.425 lb-mol/sec


.2)

Ind. Eng. Chem. Process Des. Dev., 14, 218 (1975)Ind. Eng. Chem., 59(5), 70 (1967)






C2H6 C2H4 + H2


4.5 Pressure Drop in Reactors

In liquid-phase reaction In liquid phase reaction- the concentration of reactants is insignificantly affected by evenrelatively large change in the total pressure

- ignore the effect of pressure drop on the rate of reactionwhen sizing liquid-phase chemical reactorsth t i d i i d f li id h ki ti l l ti- that is, pressure drop is ignored for liquid-phase kinetics calculations

In gas phase reaction In gas-phase reaction- the concentration of the reacting species is proportional to total pressure- the effects of pressure drop on the reaction system are a key factorp p y yin the success or failure of the reactor operation

- that is, pressure drop may be very important for gas-phase reactions(Micro-reactors packed with solid catalyst)


Pressure drop and the rate lawp

• for an ideal gas 기상반응에서는 반응 성분의 농도가 반응압력에 비례 for an ideal gas,

)/)(/)(1( 000

0

TTPPXvXvF

vF

C iiAii

의 농도가 반응압력에 비례하므로 압력강하에 대한

고려가 필수적이다.

• For isothermal operation

0 1 PP

XXv

CC iiAi

0ii

F 0Ay

01 PX 0A

i F 0Ay

abBA /,1

- 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



• For example• For example,

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

2A B C2A B + C

-the mole balance (differential form)

mincatalyst g

gmoles0 AA r

dWdXF

The differential form of PFR design equationmust be used when

there is a P- rate law

2AA kCr

there is a P

- stoichiometry for gas-phase reactions

TPXCC 01

TPXCC AA

0

00 1



• Then, the rate law2

0

00 1

1

TT

PP

XXCkr AA (4-20)

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

C bi i ith th l b l ( i i th l ti T T )• Combining with the mole balance (assuming isothermal operation: T=T0)

220 )1(

PXCdX A

• Dividing by FA0(=v0CA0)0

00 1

)1(

PP

XXC

kdWdXF A

A

2

0

2

0

0

11

PP

XX

vkC

dWdX A

00 1 PXvdW



22 2

0

2

0

0

11

PP

XX

vkC

dWdX A

-The right-hand side is a function of only conversion and pressure

),( PXfdWdX

(4-21)

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

l i h h i d l h d h lcatalyst weight: that is, we need to relate the pressure drop to the catalyst

weight)(WfP dP

We need)(WfP dW

We need


Flow through a packed bedg p


Flow through a packed bed dWdP

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

dW

through a packed of catalyst particles

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

d )(

G

DDgG

dzdP

ppc75.1)1(1501

3 (4-22)

laminar turbulent

G=u=superficial mass velocity [kg/m2s]; u=superficial velocity [m/s]; Dp=diameter of particlein the bed [m]; =porosity=volume of void/total bed volume; 1 =volume of solid/total bed

• The gas density () is the only parameter that varies with pressure on the right

in the bed [m]; =porosity=volume of void/total bed volume; 1- =volume of solid/total bedvolume, gc=1.0 [m/s2]; =viscosity [kg/ms]

• The gas density () is the only parameter that varies with pressure on the right-hand side. We are now going to calculate the pressure drop through the bed.


Flow through a packed bedg p


Flow through a packed bed dWdP

We need g p

• Equation of continuity

dW

vv

mm

00

0

- The reactor is operated at steady state, the mass flow rate at any point is equal to the entering mass flow rate

• Gas-phase volumetric flow rate

FTP

v

• Then,00

00

T

T

FF

TT

PP

vv

vv0

0

T

FF

TT

PP

vv 00

00

0

TFTPv 0


Pressure drop in a packed bed reactor

• then, Ergun equation

00

03

075.1)1(150)1(

T

T

ppc FF

TT

PP

GDDg

GdzdP

• Simplifying

00

00

T

T

FF

TT

PP

dzdP

(4-24)

00 T

GD

G 75.1)1(150)1(30dP (4-25)

DDg ppc3

00

dWdP

We need

( )

Ac

dWdz

dzdP

dWdP

• The catalyst weight, cc zAW )1(

Volume of Density of

z

solid solid catalyst

dzAdW cc )1( (4-26)Seoul National University

Pressure drop in a packed bed reactor

00 TFTPdP

dzdPdP cc AdW

dz)1(

1

00

00

)1( T

T

cc FTPAdW

dWdz

dzdP

dWdP

cc )(

00

00

T

T

FF

TT

PP

dzdP

• Simplifying

00

0

0 /2 T

T

FF

PPP

TT

dWdP

(4-28)

0

)1(2

PA

(4-29)0)1( PA cc

XFFXFFF A

TATT 0000 1 XFT 1 0

0A

AF

y

X

FFXFFF

TTATT

0000 1 X

FT1

0

00

TA F

y

0PTdP (4 30))1(/2 0

0

0X

PPP

TT

dWdP

(4-30)


Pressure drop in a packed bed reactorp p

)1(0 XPTdP

(4-30)

< 0 th d (P) ill b l th 0

)1(/2 00

XPPTdW

(4 30)

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

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

• For isothermal operation

),( PXfdWdP

),( PXfdWdX

and (4-31))(fdW

),(fdW

• The two expressions are coupled ordinary differential equations.

We can solve them simultaneously using an ODE solver such as Polymath.

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

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



PTdP

01For the isothermal operation and ε = 0,

Analytical Solution

)1(/2 0

0

0X

PPP

TT

dWdP

(4-30)

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

PPP

dWdP

Isothermal withε = 0

Rearranging gives us

dW

PPdPP )/(

2 0

0

dWPPd 2

0 )/(

Taking P/P0 inside the derivative(4-32)


Pressure Drop in a Packed Bed Reactorp

PPd 2)/(

dWPPd 2

0 )/(

Integrating with P=P @ W=0

WPP 1)/( 20

Integrating with P=P0 @ W=0

WP 10 WP

10

(4-33)ε = 0T = T0



If ε = 0 or ε X ≪ 1, we can obtain an analytical solution to Eq. (4-30) for isothermalIf 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

Pressure ratio WP 1 (4-33)

only for ε = 0 W

P 1

0(4-33)

0

0

)1(2

PA cc

02 zW

cc zAW )1( 0PW

021

zP Pressure as a

GDDg

G

ppc75.1)1(150)1(

30

0

0

0

01

PP

(4-34)function of reactor length, z


4.5.3 Pressure Drop in Pipes

Pressure drop for gases along the length of the pipe w/o packing

DfG

dLduG

dLdP

22 02 2

2

00

DfG

PdLdPG

dLdP

PP/Gu

0 PP

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

0

0P

PP

DLf

PG

PP 0

0

0222

0 ln22

0

0

Rearranging, we get

fGVP 2

0 41DAP

fGVP c

pp00

0 41

Example 4 4: 1½” schedule 40 x1000 ft L ( =0 018) P<10%Example 4-4: 1½ schedule 40 x1000-ft L (p=0.018), P<10%However, for high volumetric flow rates through microreactors, the P may be significant. Example 4-4 Seoul National University

4.5.4 Analytical Solution for Reaction with Pressure Drop

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 toresulting in a decreased rate of reaction, hence a lower conversion when compared toa reactor without a pressure drop.


Reaction with Pressure DropConversion as a function of catalyst weighty g

A B

2nd-order isothermal reaction

Combining

dX

A B

Mole balance: 22/122

0 1)1( WXF

kCdWdX A

Combining

AA rdWdXF 0

Rate law:

0FdW A

Separating variable and Integrating

2AA kCr

S

1)1( 22

0

0 dWWX

dXkCF

A

A

0 )1(PPXCC AA

Stoichiometry: Gas-phase isothermal with =0

1

0,0

0

000

WWXv

vCFandWXat AA

0P

WPP

10

2

110

0 WXkCA

0

2/10 )1)(1( WXCC AA (4-37)


Reaction with Pressure DropConversion as a function of catalyst weightConversion as a function of catalyst weight

Conversion for 2nd-order isothermal reaction in PFR with P

210 WWkCA

211

2

0

0

0

Wv

WkCv

XA

(4-38)

20v

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

2/1

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

W A (4-39)


The Optimum Catalyst Particle Diameter

Internal diffusion inside catalyst

dominant

Pressure dropdominant

Conversion k ~ 1/DpdominantX

D t

Particle Diameter, Dp

Dp, opt

Problems with large diameter tubes:Why not pack the catalyst g(1) Bypassing or Channeling(2) Little efficient of heat transfer rate

y p yinto a large diameter

to reduce P?Seoul National University

Example 4-6 Production of ethylene glycol

H2, C2H4402 million lbC2H6 /yr

O C H N C H O

2

C2H6separator

C2H6 C2H4 + H21

V=81 ft3, X=0.8

lbC2H6 /yr

O2, C2H4, N2, C2H4O

3C2H4

C2H4+ ½ O2 C2H4OAg

3/23/1BAA PkPr C

hcatlbatmlbmolk [email protected]

56

separator

H O C H O

3

4 Air

W=45,440 lb, X=0.6

260oC, 10bar

C H O( )

7 8 H2O, 0.9wt% H2SO4

H2O C2H4O

C2H4O(aq)200 millionlb EG/yr9

V=197 ft3 X=0 8V 197 ft , X 0.8

CH2OHC2H4O + H2O

CH2OHCat.

absorber


Example 4-6 Calculating X in a Reactor with Pressure Drop

yXX

Fk

dWdX

A 11

0

Xdy

A

)1(

0

ydW 2










Negligible Pressure Drop in Pipes

Pressure drop along the length of the pipe

DfG

dLdG

dLdP

22 02 22

00

DfG

PdLdPG

dLdP

PP

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

0

PP

DLf

PG

PP 0

0

0222

0 ln22

0

0

Rearranging, we get2

DAPfGV

PP

cpp

00

20 41

Example 4-4: 1½” sch. 40 x1000-ft L (p=0.018), P<10%Seoul National University

4.6 Synthesizing a Chemical Planty g

Always challenge the assumptions, constraints, andboundaries of the problem

The profit from a chemical plant will be the differencebetween income from sales and the cost to produce thechemicalchemical

Profit = value of products cost of reactantsProfit = value of products – cost of reactants

– operating costs – separation costs

The operating cost: energy, labor, overhead, anddepreciation of equipmentdepreciation of equipment


Production of ethylene glycol

H2, C2H4402 million lbC2H6 /yr

O C H N C H O

2

C2H6separator

C2H6 C2H4 + H21

V=81 ft3, X=0.8

lbC2H6 /yr

O2, C2H4, N2, C2H4O

3C2H4

C2H4+ ½ O2 C2H4OAg

56

separator

H O C H O

3

4 Air

W=45,440 lb, X=0.6

C H O( )

7 8 H2O, 0.9wt% H2SO4

H2O C2H4O

C2H4O(aq)200 millionlb EG/yr9

V=197 ft3 X=0 8V 197 ft , X 0.8

CH2OHC2H4O + H2O

CH2OHCat.

absorber


Production of ethylene glycol

bp(oC) 1 2 3 4 5 6 7 8 9lb mol M.W. $/lbbp( C) 3 5 6 8 9s0.425C2H6

C2H4

H2

$/ b

0.040

O2

N2

EO0.102EG 0.380

-H2SO4

H2O0.425 0.102Total

0.043


Synthesizing a Chemical Planty g

Ethylene glycol = $0 38/lb (2x108 lb/yr) Ethylene glycol = $0.38/lb (2x10 lb/yr)Ethane = $0.04/lb (4x106 lb/yr)S lf i id $0 043/lb (2 26 108 lb/ )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 How the profit will be affected by conversion, separation, recyclestream, and operating costs?


Fig. 4-1 Isothermal Reaction Design AlgorithmAlgorithm for Conversion


4.11 The Practical Side

A practical guidelines for the operation of chemical reactors have beenA practical guidelines for the operation of chemical reactors have beenpresented over the years, and tables and some of these descriptions aresummarized and presented on the CD-ROM and web. The articles arelisted in Table 4-7.s ed ab e

For example, Mukesh gives relationships between the CSTR tank diameter,T, impeller size diameter, D, tank height, H, and the liquid level, l. To scalep g qup a pilot plant (1) to a full scale plant (2), the following guidelines are given

RHTD 2222

And the rotational speed, N2, is

RHTD

1111

Wh l f f diff t i iti d F d b

nRNN 12

Where values of n for different pumping capacities and Froude numbers aregiven in Mukesh’s article.


Closure

After completing this chapter, you should be able to apply the algorithm building blocks

Evaluation

Stoichiometry

Combine

Mole Balance

Rate Law

Mole Balance

To batch reactor, CSTR, PFR, PBR., , ,


Documents

Chemical Reactor Designocw.snu.ac.kr/sites/default/files/NOTE/7230.pdf · 2018-01-30 · Objectives • Describ hibe the alihlgorithm thll h d l hilhat allows the reader to solve