Chemical Reactor Design Chemical Reactor Design

Chemical Reactor Design Chemical Reactor Design

Y W L

Youn-Woo Lee

School of Chemical and Biological Engineering Seoul National Universityy

155-741, 599 Gwanangro, Gwanak-gu, Seoul, Korea ywlee@snu.ac.kr http://sfpl.snu.ac.kr

Chapter 4 Chapter 4

Isothermal Reactor Design Isothermal Reactor Design

Chemical Reactor Design Chemical Reactor Design Chemical Reactor Design Chemical 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

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 c reaction 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 l

bed tubular reactors and in packed bed spherical reactors.

Fig. 4-1 Isothermal Reaction Design Algorithm Algorithm

for Conversion

Algorithm for isothermal reactor design 1. Mole balance and design equation

1. Mole balance and design equation 2. Rate law

3 Stoichiometry 3. Stoichiometry 4. Combine

5 E al ate 5. Evaluate

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

combine step either combine step either combine 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)

French Menu Analogy

Algorithm for isothermal reactors

French Menu Analogy

4.2.1 Batch Operation p

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

dN 



1 ^{1 dN 1 dN d}  ^{N / V}  ^dC

, in terms of concentration

A

A

r

dt dN

V  



 





1  

A A

A A

A

r

dt dC dt

V N

d dt

dN V

dt dN

V  

 

0

1 / 1

Generally, when analyzing laboratory experiments, it is best to process the data in terms of the measured variable. Because concentration is the measured variable for most liquid-phase concentration is the measured variable for most liquid phase reactions, the general mole balance equation applied to reactions in which there is no volume change becomes

A

A

r

dt

dC  



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

Reaction time in Batch Operation

Algorithm for isothermal reactor design

Algorithm for isothermal reactor design

A   B





A

r V

dt

N

dX

1. Mole balance

& Design equation





X C

C

A in order nd

le irreversib kC

r

) 1

(

2

2. Rate law

,







A

X dX kC

X C

C

2 0

) 1

(

) 1

(

4 C bi ti 3. Stoichiometry





dt dX kC

X dt kC

( 1 )

2nd order Isothermal Liquid-phase Batch reaction

4. Combination





 



X t

A

dX dt kC

X

1 ) 1

(



 X 1

Batch reaction

5. Analytical Evaluation





^{dt  } _kC

_ ^dX _X

0 0

( 1 )

1 



 





 

X X t kC

A

1

1

0

Reaction time in Batch Operation

Typical cycle times for a batch polymerization process

t = t _f + t + t + t

Activity Time (h)

t _t t _f + t _e + t _R + t _c

1. Charge feed to the reactor and agitate, t

2. Heat to reaction temperature, t

1.5-3.0 0.2-2.0 3. Carry out reaction, t

4. Empty and clean reactor, t

Varies

0.5-1.0 Total cycle time excluding reaction 3 0 6 0 Total cycle time excluding reaction 3.0-6.0

Batch polymerization reaction times may vary between 5 and 60 hours Batch polymerization reaction times may vary between 5 and 60 hours.

Decreasing the reaction time with a 60-h reaction is a critical problem. As the

reaction time is reduced (e.g. 2.5 h for a 2nd-order reaction with X=0.9 and

kC 10

s

) it becomes important to se large lines and p mps to achie e

kC

=10

s

), it becomes important to use large lines and pumps to achieve

rapid transfer and to utilize efficient sequencing to minimize the cycle time.

Batch Reaction Times

B A  

Mole balance

N V r dt

dX  

N

dt

kC r_A  _A



order -

First Second order

kC2

r_A  _A



) 1

0

(

0

X V C

C

 N





r_{A A} r_A kC_A

Stoichiometry (V=V₀)

X dt k

dX  (1 ) kC ₀(1 X)²

dt dX

A 

Combine 

X t k

 

1 ln 1 1

) 1

0( X kC

t X

A 

Integration 

Batch Reaction Times

t

s ln 1

1

) 10

k 0.9, (X

order -

1st







t X

s

R

) 10

kC 0.9, (X

order

2nd







X t

k

ln 1 1

ln 1



 

X t kC

A R

9 . 0

) 1

0

(



 

k 3 . 2

9 . 0 ln 1



 _kC

9

) 9 . 0 1

0

(



 

k 3 . 2

1



kC

9

0





s

sec 000 , 23

10







s sec 000 , 9

10





hr 4 .

 6 _{ 2 . 5 hr}

Batch Reaction Times

The order of magnitude of time

hi 90% i

Table 4-3

to achieve 90% conversion

For first- and second-order irreversible batch reactions

1

-order

k (s

) 2

-order

kC

(s

) Reaction time t

10

10

Hours

10

10

Mi t

10

10

Minutes

1 10 Seconds

1 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 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 (k

). Since the reaction will be carried out isothermally k will need to be determined only at the reaction carried out isothermally, k

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 40

C the reaction does not proceed at a significant rate; consequently, a temperature of 55

C 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 be considered constant during the course of the reaction. The reaction is first-order in ethylene oxide.

O CH OH

O CH

-OH CH

-CH

+ H

O CH

-OH

A + B C

Example 4-1 Determining k from Batch Data

In the lab experiment, 500mL of a 2 M solution (2 kmol/m

) of EO in water was mixed with 500mL of water containing 0 9 wt % sulfuric acid water was mixed with 500mL of water containing 0.9 wt % sulfuric acid catalyst. At T=55

C, the C

was recorded with time. Determine the specific reaction rate at 55

C

Time Concentration of EG

specific reaction rate at 55 C.

EO + H₂O → EG

(min) (kmol/m

)

0.0 0.000

0.5 0.145

A + B → C

1.0 0.270

1.5 0.376

2.0 0.467

2.0 0.467

3.0 0.610

4.0 0.715

6 0 0 848

6.0 0.848

10.0 0.957

Problem Solving Algorithm

Example 4-1 Determining k from Batch Data

A. Problem statement. Determine the k_A B Sketch

D. Assumptions and approximations:

Assumptions B. Sketch

C. Identify

C1. Relevant theories

Rate law: r_A  k_AC

Assumptions 1. Well mixed

2. All reactants enter at the same time 3. No side reactions

batch A, B, C

Rate law:

Mole balance:

C2. Variables

A A A k C r 

V dt r

dN

A A 

3. No side reactions

4. Negligible filling and emptying time 5. Isothermal operation

Approximations Dependent: concentrations, C_i

Independent: time, t C3. Knowns and unknowns

pp

1. Water in excess (C_H2O~constant) C_B~ C_BO

E. Specification. The problem is neither Knowns: C_EG = f(time)

Unknowns:

1. C_EO = f(time)

overspecified nor underspecified.

F. Related material. This problem uses the 2. k_A

3. Reactor volume

C4. Inputs and outputs: reactant fed

ll b h

mole balances developed in Chap. 1 for a batch reactor and the stoichiometry and rate laws developed

i Ch 3

all at once a batch reactor C5. Missing information: None

in Chap. 3.

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

Problem Solving Algorithm

Example 4-1 Determining k from Batch Data

A

A

r

dt dN

V 1  

1. MOLE BALANCE Batch reactor that is well-mixed

dt V

2 RATE LAW

r kC

Since water is present in such excess, the concentration of water at any time t is virtually the same as the initial concentration and the rate

2. RATE LAW

 r

 kC

(C_B~C_B0)

3. STOICHIOMETRY

Species symbol Initial Change Remaining Concentration CH₂CH₂O A N_A0 - N_A0X N_A=N_A0(1-X) C_A=C_A0(1-X) H₂O B Θ_BN_A0 - N_A0X N_B=N_A0(Θ_B-X) C_B=C_A0(Θ_B-X) H₂O B Θ_BN_A0 N_A0X N_B N_A0(Θ_B X) C_B C_A0(Θ_B X)

C_B~ C_A0 Θ_B= C_B0 (CH₂OH)2 C 0 N_A0X N_C =N_A0X C_C=C_A0X

N_T0 N_T =N_T0 - N_A0X

Problem Solving Algorithm

Example 4-1 Determining k from Batch Data

A

A

r

dt dN

V 1  

A

r

dC   r  kC

4 COMBINING

dC

kC

A



dt   r  r

 kC

4. COMBINING

Mole balance Rate law

kC

dt 

, 

5. EVALUATE

For isothermal operation k is constant:

For isothermal operation, k is constant:

t t A

C A

C kt dt

k dC kdt

A

  

    ^ln

kt C A A

e C

C

C kt dt

k C kdt

A



 



ln

0

A

A

C e

C 

Problem Solving Algorithm

Example 4-1 Determining k from Batch Data

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

A + B C

N N

X N

N

) 1

0

(

0 0 0

kt A

A A

C C

C

A A

A C

e C

C N C

C N

N N

X N

N



















)

0

(

0 0

A A

A

C

V V

) 1

0 (

kt A

C C e

C   ^

C k C _{A 0}  _C

l kt

C _A

C

A  

0

ln 0

Example 4-1 Determining k from Batch Data

Rearranging and taking the logarithm of both side yields

C C

A

A C

r  (0.311min^1)



We see that a plot ln[(C_A0-C_C)/C_A0] as C kt

C C

A C

A   

0

ln 0 The rate law can now be used in the design

of an industrial CSTR. Note that this rate law was obtained from the lab-scale batch We see that a plot ln[(C_A0 C_C)/C_A0] as

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

1

3 2 10

ln

reactor (1000 mL).

Time CA0 CC

CA0

1 0.6

1 2

min 311 . 55 0

. 1 95 . 8

3 . 2 10

ln _

 

 

 t t k

(min)

0.0 1.000

0.5 0.855

1 0 0 730

0

CA

0.1

(CA0-CC)/CA

0.06

1.0 0.730

1.5 0.624

2.0 0.533

3.0 0.390

(

4.0 0.285

6.0 0.152

0.01

0 2 4 6 8 10 12

t ( min)

1.55 8.95

4.3 Design of CSTR

Design Equation for a CSTR

A X V  F ⁰ Mole balance

A A

exit A

X V C

X C V v

V r









 

0 0

0

) Mole balance (

the space time

0 0

0 A

A v C

F 

A

A v r

r 

 ₀

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

the space time

A A A

X X C

kC r

1

 

 







Rate law

C bi

A A

kC X

k

0

1  

 

  



Combine

Rearrangingg g

k X k





  1

CSTR relationship between space time and conversion for a 1^st-order

 k

 1

liquid-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

C k X

C

C



( 1  ) 

  1    

concentration of A, C

:

X  k

k C k

k C k

C k X

C C

A A

A A

A



 

 

 















 

 

   





1 1

1

1 1 )

1 (

0 0

0

0 1k

C C

k k

A A





 

 1   1

0

C

k



 1

k C

C



  1

exit concentration of A

 k



1

4.3.2 CSTRs in Series

C_A1, X₁ C_A0

v₀

C_A2, X₂ v₀

-r_A1, V₁ -r_A2, V₂

For 1^st-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

1 1

0

1

1 k

C

C



 

From a mole balance on reactor 2,





2 0

1 A A A

A

F v C C

F    

2 2

2 1

0 2

2 1

2

A A A

A A A

C k

C C

v r

F

V F 

 

CSTRs in Series

Solving for C_A2, the concentration exiting the second reactor, we get







2 2

1

2

1 1 k 1 k

C k

C

C







 



 

1 1

0

1 1 k

C_A C^A



 

If instead of two CSTRs in series we had n equal-sized CSTRs connected in series (₁ = ₂ = … = _n = _i = (V_i/v₀)) operating at the same temperature (k₁ =

k k k) th t ti l i th l t t ld b

k₂ = … = k_n = k), the concentration leaving the last reactor would be

A

A

C

C  C



 

 

An

k

C

Da 1

1  



 

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

X 1

1 kC

kC

 ^k 

X     1 1 1





An

An

k

kC

r    



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

F V Da r



, ;

thus the cost of adding subsequent reactors might not be justified

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

Reaction Damköhler number

rate"

reaction a

"

Entrance at

Reaction of

0

 Rate 

  r

V

Da 

 Entering Flow Rate of A  " a convection rate"

F

Da

The Damköhler 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 1^st-order irreversible reaction

k

C v

V kC F

V Da r

A A A

A

  

 

0 0

0 0

0

For 2^nd-order irreversible reaction ₀

0 0

2 0 0

0

A A

A A

A

kC

C v

V kC F

V

Da  r   



RULE OF THUMB

0 0

0 A

A

v C

F

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, the combination of the rate law and the design equation yields

We solve the above eq. for X:



 

 



X

2 0 0

kC X F r

X

V  F



     

 

2

4 1 2

1

2

0 0

0

0 0

0









 

 

A A

A

A A

A

kC

kC kC

X kC (4-14)

kC

r



For const density v=v₀, F_A0X=v₀(C_A0-C_A)

 

Da 2

Da 4 1 Da

2 1

2 ₀





 

kC_A

(4-16)

2 0

0

kC

C C

v V

A A A









quadratic equation because X cannot

 X



Using our definition of conversion, we have

q q

be greater than 1.

(4 15)





2 0

( 1 X ) kC







 

Da 2

Da 4 1 Da

2

1 

 X

A Second-Order Reaction in a CSTR

V

0 0 A A

F V Da r



0.67

0.88

At high conversion, a 10-fold increase in Da will increase the increase in Da will increase the conversion only to 88% due to lowest value of reactant concentration in CSTR

60 6

concentration in CSTR.

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

~91 ton/yr

It is desired to produce 200 x 10

pounds per year

C_A0=1 v_A0=v_B0

p p p y

of EG. The reactor is to operated isothermally.

A

solution of ethylene oxide (EO) in water is fed to the reactor together with

water is fed to the reactor together with

of water containing 0.9 wt%

of the catalyst H

SO

. The specific reaction rate constant is 0 311 min

as determined in Ex 4-1 constant is 0.311 min as determined in Ex 4-1.

(a) If 80% conversion is to be achieved, determine

th CSTR l

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?

M.W. of EG=62

M.W. of EO=58

v₀ = v_A0 + v_B0

=1

3m 10’

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=0.81

800gal 800gal

X₁=0.68 X₂=0.90 1480gal

X=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 than The two equal sized CSTRs in series will give a higher conversion than

two CSTRs in parallel of the same size when the reaction order is greater than zero.

4.4 Tubular Reactors

-2^nd-order liquid-phase rxn

(A Products) Rate law: r_A  kC_A²

(A Products) -Turbulent,

- No dispersion

- No radial gradients in T, u, or C

^ 

A

A

kC

F dX V

0 2

0

St i hi t f li h PLUG-FLOW REACTOR

PFR mole balance

Stoichiometry for liq. phase rxn T & P = constant

) 1

( X

C

PFR mole balance

C

A

A

r

dV

F

dX  

) 1

0

( X

C

C





Combination





The differential form of PFR design equation must be used when there is a

P or heat exchange between PFR &

th d I th b f P

 

 





 

 _kC ^F   _X ^dX _kC ^{v X} _X V

A X

A A

1 )

1 (

1

0 0

0 2

2 0 0

the surrounds. In the absence of P or heat exchange, the integral form of the PFR design equation is used.

2 2 0

0

1

1 Da

Da kC

X kC

A A

 

 





 _



r F dX

V

2

0

1

1   kC

 Da

Da₂ is the Damkohler number

4.4 Tubular Reactors

-2^nd-order gas-phase rxn

(A Products) Rate law: ²

A

A

kC

r 

(A Products)



-Turbulent,

- No dispersion

- No radial gradients in T, u, or C

^{V  F}



_kC ^dX

PLUG-FLOW REACTOR PFR mole balance



kC

Stoichiometry for gas phase rxn PFR mole balance

A

A

r

dV

F

dX  

Stoichiometry for gas phase rxn T & P = constant

) 1

) ( 1

0

( X

X C F

C F



 



The differential form of PFR design



equation must be used when there is a

P or heat exchange between PFR &

th d I th b f P

) 1

( )

1

(

0

C X

X v

C

v



 



 



Combination the surrounds. In the absence of P or

heat exchange, the integral form of the PFR design equation is used.

Combination

X dX F

V ^



^{( 1}

^{  )}

 _



r F dX

V



kC

_ X

0

0

( 1 )

4.3 Tubular Reactors

X dX F

V



_kC ^{( 1   )} _X

^dX

F V

A

A

 _



0 2 2

0

0

( 1 )

) (

C_A0 is not function of X; k=constant (isothermal)

dX X

X kC

V F

A X

 ^{ }



0 2

2 2

0

) 1

(

) 1

(

A0 ; ( )

0 0

0 C v

F_A  _A

X kC



_

0

( 1 )





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

 





 









 













 X

X X kC X

V v

A

1

) 1 ) (

1 ln(

) 1 ( 2

2 2

0 0

 

   





 X

X v X

L ( 1 )

) 1

l ( ) 1 ( 2

2 0 2

Reactor length will be

L A V  _c

 

 

         

 X X X

A L kC

c

A

1

) ) (

1 ln(

) 1 (

2

0

0 V _c

Conversion as a function of distance down the reactor

1 1.2

A  0.5B (=-0.5) A B ( 0)

0.8

on ( X )

A  2B (=1) A  3B (=2)

0.4 0.6

o nv e rs i

0.2

C o

3 0

0

2 dm . 0 kC

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 length will have the highest conversion for a fixed reactor length.

)

5 . 0 1

( X v

v  

the reactant spends more time the reactant spends more time

)

2 1

( X v

v  

the reactant spends less time the reactant spends less time

The volumetric flow rate decreases with increasing conversion 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.

Change in Gas-Phase Volumetric Flow Rate Down the Reactor

v=v

(1+X)



 







 







 X

X X kC X

V v

A 1

) 1 ) (

1 ln(

) 1 ( 2

2 2

0

0

   

=1 : (A→2B) v = v

(1+X) ( )

 = 0 : (A→B) ( ) v = v

 =0.5 : (2A→B) v = v

(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 conversion

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.

Example 4-3: Determination of a PFR Volume

Determine the PFR volume necessary to produce 300 million

d f h l f ki f d f

pounds of ethylene a year from 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 law. We want to achieve 80% conversion of ethane, operating the reactor isothermally at 1100K at a pressure of 6 atm.

C

H

 C

H

+ H

A  B + C

F 300 x ¹⁰

^{lb/ 0 340 lb l/}

A  B + C

F

= 300 x ¹⁰

lb/year = 0.340 lb-mol/sec F

= F

X

F

= F

/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)

C

H

 C

H

+ H

4.5 Pressure Drop in Reactors





- 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

th 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





- the concentration of the reacting species is proportional to total pressure - the effects of pressure drop on the reaction system are a key factor p p y y

in 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 law p

• for an ideal gas

의 농도가 반응압력에 비례

for an ideal gas,

 

) / )(

/ )(

1

(

0

0

T T P P X v

X v F

v

C

F







 



의 농도가 반응압력에 비례 하므로 압력강하에 대한

고려가 필수적이다.

• For isothermal operation

0

1 P

P X

X C v

C





 











 

i

 F

   y_A0

1 X  P

  

0 A

i F

  y_A0 a

B b

A  1,   /



- determine the ratio P/P

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

Pressure drop and the rate law p

• For example

• For example,

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

2A B C

2A  B + C -the mole balance (differential form)

 

 





 



 g catalyst min gmoles

0 A

A

r

dW F dX

The differential form of PFR design equation

must be used when there is a P

- rate law

2 A A

kC r  



there is a P

- stoichiometry for gas-phase reactions

T P C X

C 1

 

  

T P C X

C

0

0

1 

 

  



Pressure drop and the rate law p

• Then, the rate law

2 0 0 0

1

1 



 



 



 









 

 

T T P

P X

C X k

r

- 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)

2 2

0

( 1 )     

 C  X P

dX

• Dividing by F_A0(=v₀C_A0)

0 0

0

1

) 1

( 



 



 

 





 

P P X

X k C

dW

F

dX

2

0 2

0 0

1

1  

 



 



 









 

P P X

X v

kC dW

dX

0 0

 1   X   P  v

dW

Pressure drop and the rate law p

2

 

0 0

1

1 



 



 



 









 

P P X

X v

kC dW

dX

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

) , ( X P

dW dX  f

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

l i h h i d l h d h l

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

) (W f

P  dP

We need

) (W f P 

dW

We need

Flow through a packed bed g p