Fundamentals of Reaction Engineering - Worked Examples

RAFAEL KANDIYOTI

FUNDAMENTALS OF REACTION ENGINEERING ‐ EXAMPLES

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Rafael Kandiyoti

Fundamentals of Reaction Engineering Worked Examples

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Fundamentals of Reaction Engineering – Worked Examples © 2009 Rafael Kandiyoti & Ventus Publishing ApS ISBN 978-87-7681-512-7

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Fundamentals of Reaction Engineering – Worked Examples

Contents

Contents Chapter I:

Homogeneous reactions – Isothermal reactors

5

Chapter II:

Homogeneous reactions – Non-isothermal reactors

13

Chapter III:

Catalytic reactions – Isothermal reactors

22

Chapter IV:

Catalytic reactions – Non-isothermal reactors

31

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Fundamentals of Reaction Engineering – Worked Examples

I. Homogenepus reactions - Isothermal reactors

Worked Examples - Chapter I Homogeneous reactions - Isothermal reactors Problem 1.1 Problem 1.1a: For a reaction AoB (rate expression: rA = kCA ) , taking place in an isothermal tubular reactor, starting with the mass balance equation and assuming plug flow, derive an expression for calculating the reactor volume in terms of the molar flow rate of reactant ‘A’. Solution to Problem 1.1a: Assuming plug flow, the mass balance over a differential volume element of a tubular reactor is written as:

(MA nA )V Taking the limit as 'VR  0

- (MA nA)V+'V

dn  A  rA dVR

MA rA 'VR

-

0 ; rA

dn  A ; VR dVR

= 0



n Ae

dn A rA

³

n A0

Problem 1.1b: If the reaction is carried out in a tubular reactor, where the total volumetric flow rate 0.4 m3 s-1, calculate the reactor volume and the residence time required to achieve a fractional conversion of 0.95. The rate constant k=0.6 s-1. Solution to Problem 1.1b: The molar flow rate of reactant “A” is given as,

n A0 ( 1  x A ) . kn A . Substituting, kC A vT

nA Also nA

C AvT . The rate is then given as rA

v  T k

VR

VR



n Ae

³

n A0

dn A nA

vT k

x Ae

³

0

n A0 dx A n A0 ( 1  x A )

vT k

x Ae

³

0

dx A ( 1  xA )

vT ln( 1  x Ae ) # 2 m3 k

In plug flow the residence time is defined as

W{

VR vT

ª m3 º « 3 1 » ¬« m s ¼»

> s@ ;

W

2 / 0.4

5 s

Problem 1.2 The gas phase reaction Ao2S is to be carried out in an isothermal tubular reactor, according to the rate expression rA = k CA. Pure A is fed to the reactor at a temperature of 500 K and a pressure of 2 bar, at a rate of 1000 moles s-1. The rate constant k= 10 s-1. Assuming effects due to pressure drop through the reactor to be negligible, calculate the volume of reactor required for a fractional conversion of 0.85.

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Fundamentals of Reaction Engineering – Worked Examples

Data: R, the gas constant P the total pressure nT0 total molar flow rate at the inlet k, reaction rate constant

I. Homogenepus reactions - Isothermal reactors

: 8.314 J mol-1 K-1 = 0.08314 bar m3 kmol-1 K-1 : 2 bar : 1000 moles s-1= 1 kmol s-1 : 10 s-1

Solution to Problem 1.2

In Chapter I we derived the equation for calculating the volume of an isothermal tubular reactor, where the first order reaction, A o 2S, causes volume change upon reaction: ½° ª 1 º RT °­ (Eq. 1.37) VR nT 0 ®( 1  y A0 )ln « »  y A0 x Ae ¾ Pk °¯ °¿ ¬ 1  x Ae ¼ In this equation, for pure feed “A”, yA0 { nA0/nT0 = 1. ( 0.08314 )( 500 ) VR ( 1 )^2 ln( 6.67 )  0.85` ( 2 )( 10 ) VR

6.1 m3

Let us see by how much the total volumetric flow rate changes between the inlet and exit of this reactor. First let us review how the total molar flow rate changes with conversion: = nI0 nI nA0 – nA0 xA nA = nS0 + 2 nA0 xA nS = ----------------------------------= nT0 + nA0xA nT

for the inert component for the reactant for the product nT is the total molar flow rate

nT 0 RT ( 1 )( 0.08314 )( 500 ) / 2 20.8 m3 s-1 P vT ,exit nT ,exit RT / P; but nT,exit nT0  n A0 x A,exit vT 0

and nT 0 vT ,exit

n A0 ; therefore

( 1  0.85 )( 0.08314 )( 500 ) / 2 38.5 m3 s 1

vT 0

20.8 m3 s 1 ; vT ,exit

38.5 m3 s 1

The difference is far from negligible! In dealing with gas phase reactions, rates are often expressed in terms of partial pressures: rA= kpA, as we will see in the next example.

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Fundamentals of Reaction Engineering – Worked Examples

I. Homogenepus reactions - Isothermal reactors

Problem 1.3 The gas phase thermal decomposition of component A proceeds according to the chemical reaction k1 ZZZ X A YZZ Z BC k2

with the reaction rate rA defined as the rate of disappearance of A: rA k1C A  k2 CB CC .The reaction will be carried out in an isothermal continuous stirred tank reactor (CSTR) at a total pressure of 1.5 bara and a temperature of 700 K. The required conversion is 70 %. Calculate the volume of the reactor necessary for a pure reactant feed rate of 4,000 kmol hr-1. Ideal gas behavior may be assumed. Data k1 108 e10,000 / T s-1 (T in K)

k2

108 e8,000 / T

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Gas constant, R = 0.08314

m3 kmol-1 s-1 (T in K) bar m3 kmol-1 K-1

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Fundamentals of Reaction Engineering – Worked Examples

I. Homogenepus reactions - Isothermal reactors

Solution to Problem 1.3 Using the reaction rate expression given above, the “design equation” (i.e. isothermal mass balance) for the CSTR may be written as:

VR

n A0  n A k1C A  k2CB CC

n A0 x AvT , where k2 k1n A  ( )nB nC vT

ni vT

Ci

(Eq. 1.1.A)

We next write the mole balance for the reaction mixture:

nA

nA0  nA0 x A

nB

 nA 0 x A

nC

 nA0 x A

nT

nA0  nA0 x A

nA0 (1  x A ) .

This result is used in the “ideal gas” equation of state:

vT

nT RT PT

n A0 RT ( 1  xA ) . PT

Substituting into Eq. 1.1.A derived above, we get:

VR

nA0 x A (nA0 RT )(1  x A ) . ­ k2 PT nA2 0 x A2 ½ PT ®k1nA0  k1nA0 xA  ¾ nA0 RT (1  x A ) ¿ ¯

Simplifying, § nA0 RT · x A (1  x A ) . ¨ ¸ 2 © PT ¹ ­k  k x  k2 PT xA ½ ® 1 1 A ¾ RT 1  x A ¿ ¯ kP To further simplify, we define a  2 T . RT VR

VR

§ nA0 RT · x A (1  x A ) 2 ¨ ¸ 2 2 © PT ¹ k1  k1 x A  k1 x A  ax A  k1 x A

(Eq. 1.1.B)

Next we calculate the values of the reaction rate constants k1 62.49 s 1 ; xA,exit = 0.70 ; T = 700 K k2 = 1088 m3s-1kmol-1; PT = 1.5 bar

Substituting the data into Eq. 1.1.B 2 4000  0.08314  700 ª 0.7 1  0.7  VR « 2 2  3600 1.5 «¬ 62.49  k1 x A  a  0.7 VR

º »;a »¼

 1088  1.5  0.08314 700

28

 43.11  0.11 # 4.74 m3

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Fundamentals of Reaction Engineering – Worked Examples

I. Homogenepus reactions - Isothermal reactors

Problem 1.4 The liquid phase reaction for the formation of compound C k1 A  B o C

with reaction rate rA1

k1C AC B , is accompanied by the undesirable side reaction k2 A  B o D

with reaction rate rA2

k2 C ACB . The combined feed of 1000 kmol per hour is equimolar in A and B. The two

components are preheated separately to the reactor temperature, before being fed in. The pressure drop across the reactor may be neglected. The volumetric flow rate may be assumed remain constant at 22 m3 hr-1. The desired conversion of A is 85 %. However, no more than 20 % of the amount of “A” reacted may be lost through the undesirable side reaction. Find the volume of the smallest isothermally operated tubular reactor, which can satisfy these conditions. Data k1 = 2.09u1012 e-13,500/T k2 = 1017 e-18,000/T

m3 s-1 kmol-1 m3 s-1 kmol-1

; ;

(T in K) (T in K)

Solution to Problem 1.4 For equal reaction orders: (n1 / n2 ) (k1 / k2 ) , where n1 is moles reacted through Reaction 1, and n2 is moles reacted through Reaction 2. Not allowing more than 20 % loss through formation of by-product “D” implies: (n1 / n2 ) (k1 / k2 ) t 4 . Since 'E2 > 'E1, the rate of Reaction 2 increases faster with temperature than the rate of Reaction 1. So the maximum temperature for the condition ^ n1 / n2 t 4` will be at (n1 / n2 ) (k1 / k2 ) 4 . This criterion provides the equation for the maximum temperature:

2.09 u 10 12 4500 / T e 10 17

4 ;

­ 4 u 1017 ½ . Solving, T ln ® 12 ¾ ¯ 2.09 u 10 ¿

4500 T

4500 12.16

370 K .

The temperature provides the last piece of information to write the integral for calculating the volume. 0.85

n dx

A ; ³0  k1  A0 k2 C ACB

VR

VR

0.85

³0

vT2 n A0 dx A kn 2A0  1  x A

2

define k  k1  k2

vT2

0.85

dx

A ³ 0 kn A0  1  x A 2

­° ½°  22 1  1¾ VR ® k  3600 (1000)  0.5 ¯° 1  0.85 ¿° 2.69 u 104  5.67 ; (k + k ) = 2.98u10-4 + 0.74u10-4 = 3.72u10-4 2

VR

k

1

2

VR = 4.1 m3

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I. Homogenepus reactions - Isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

Problem 1.5 A first order, liquid phase, irreversible chemical reaction is carried out in an isothermal continuous stirred tank reactor A o products. The reaction takes place according to the rate expression rA = k CA. In this expression, rA is defined as the rate of reaction of A (kmol m-3 s-1), k as the reaction rate constant (with units of s-1) and CA as the concentration of A (kmol m-3). Initially, the reactor may be considered at steady state. At t = 0, a step increase takes place in the inlet molar flow rate, from nA0,ss to nA0. Assuming VR (the reactor volume) and vT (total volumetric flow rate) to remain constant with time, a. Show that the unsteady state mass balance equation for this CSTR may be expressed as:

n A0 dn A ª 1 º  «  k » nA , dt ¬W W ¼ Also write the appropriate initial condition for this problem. The definitions of the symbols are given below. b. Solve the differential equation above [Part a], to derive an expression showing how the outlet molar flow rate of the reactant changes with time. c. How long would it take for the exit flow rate of the reactant A to achieve 80 % of the change between the original and new steady state values (i.e. from nA,ss to nA) ? DEFINITIONS and DATA: nA0,ss steady state inlet molar flow rate of reactant A before the change : 0.1 kmol s-1 new value of the inlet molar flow rate of reactant A : 0.15 kmol s-1 nA0 reactor volume : 2 m3 VR k reaction rate constant : 0.2 s-1 total volumetric flow rate : 0.1 m3 s-1 vT nA,ss nA t

W

rA CA

steady state molar flow rate of reactant A exiting from the reactor, before the change, kmol s-1 the molar flow rate of reactant A exiting from the reactor at time t, kmol s-1 time, s average residence time, VR/vT , s rate of reaction of A , kmol m-3 s-1 concentration of A, kmol m-3

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I. Homogenepus reactions - Isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

Solution to Problem 1.5 a. CSTR mass balance with accumulation term: material balance on component A, over the volume VR for period ' t:

^n

A0

`

(t )  n A (t )  r A (t )VR 't

N A t ' t  N A t

Bars indicate averages over the time interval 't, and NA(t) is the total number of moles of component A in the reactor at time t. Also C A n A / vT N A / VR , and the average residence time, W VR / vT . Taking the limit as 't o dt in the above equation, we get:

dNA dt

nA0  nA  rAVR Combining this with N A { VR C A

VR (nA / vT ) leads to the equation: VR dnA nA0  nA  kC AVR vT dt

Rearranging,

nA0  nA 

knAVR vT

dnA , and, dt

W

dnA ª 1 º  «  k » nA dt ¬W ¼

nA 0

W

with initial condition: at t=0,

ª1

nA=nA,ss

º

b. We define E { «  k » and write the complementary and particular solutions for the first order ordinary ¬W ¼ differential equation derived above. nA,complementary = C1e -E t ; nA,particular = C2 (const) where the solution is the sum of the two solutions: nA (t) = nA,c (t)+ nA,p (t). We substitute the particular solution into differential equation to evaluate the constants:

ª1 º «¬W  k »¼ C2

nA0

W

The most general solution can then be written as n A

; C2

C1e E

t



nA0

EW n A0

EW

. . To derive C1, we substitute the

initial condition: nA=nA,ss at t=0, into the general solution:

nA, ss

C1 

nA0

EW

;

Ÿ

C1

nA, ss 

nA0

EW

The solution of the differential equation can then be written as:

nA

c. lim n A t of

n A0

EW

­ nA0 ½  E t nA0  ®nA, ss  ¾e EW ¿ EW ¯ where EW =1+kW.

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I. Homogenepus reactions - Isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

The relationship linking t, nA,ss and nA is derived from the solution of the differential equation ­ § ­ § nA 0 ·½ nA 0 · ½  nA, ss ¸ ° ° ¨ nA, ss  ° ¨ ¸° EW ¹° W ° © W ° © EW ¹° t ®ln ¾ ®ln ¾ 1  kW ° § 1  kW ° § nA0 · ° nA 0 · °  nA ¸ n  °¯ ¨© A E W ¸¹ °¿ °¯ ¨© E W ¹ °¿

W

2 0.1

W 1 kW

nA, final

20 s

;

nA, ss

20 1  (0.2)(20) nA 0

EW

nA0 1  kW

nA0, ss 1 kW

0.1 1  (20)(0.2)

0.02 kmol s 1

4s

0.15 1  (20)(0.2)

0.03 kmol s 1

For 80 % of the change to be accomplished: nA = nA,ss + 0.8 (nA,final - nA,ss) nA= 0.8 nA,final + 0.2 nA,ss = 0.8(0.03) + 0.2 (0.02) = 0.028 kmol s-1 º ªn º °½ W °­ ª nA0  nA, ss »  ln « A0  nA » ¾ t ®ln « 1  kW ¯° ¬ E W ¼ ¬EW ¼ ¿° t

4 ^ln > 0.03  0.02@  ln > 0.03  0.028@`

6.44 s

6.44 s

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II. Homogeneous reactions - Non-isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

Worked Examples - Chapter II Homogeneous reactions – Non-isothermal reactors Problem 2.1 Problem 2.1a: Identify the terms in the energy balance equation for a continuous stirred tank reactor: T

¦ ni0 ³ C pi dT  VR ¦ H fi ( T )ri  Q 0 T0

1st term: heat transfer by flow 2nd term: heat generation/absorption by the reaction(s) 3rd term: sensible heat exchange Steady state: no accumulation

Solution to Problem 2.1a:

Problem 2.1b: What form does the equation take for the single reaction: A o B + C Solution to Problem 2.1b:

6 Hfi ri= HfA rA + HfB rB + HfC rC

and rA = - rB = -rC.

= rA {HfA – HfB – HfC} = - rA'Hr

Then the energy balance equation takes the form: T

¦ ni0 ³ C pi dT  VR ' H r rA  Q 0 T0

Problem 2.1c: For a simple reversible-exothermic reaction A œ B, qualitatively sketch the locus of maximum reaction rates on an xA vs. T diagram (together with the xA,eq line). Solution to Problem 2.1c:

xA xA,eq

Conversion

xA,OPT

Temperature

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II. Homogeneous reactions - Non-isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

Problem 2.1d: The chemical reaction A o B is carried out in a continuous stirred tank reactor. Calculate the adiabatic temperature rise for 100 % conversion. The mole fraction of A in the input stream yA0 = (nA0/nT0) = 1. The total heat capacity, C p = 80 J mol-1 K-1 The heat of reaction, ' H r  72 kJ mol-1 Solution to Problem 2.1d: From notes, the simplified energy balance for an adiabatic CSTR

xA Solving for T  T0

­° C p ½° ® ¾ T  T0 . °¯ y A0 ' H r °¿

'H r

( T  T0 )

72,000 80

Cp

900 K

Problem 2.2 Problem 2.2a: Sketch the operating line on an xA vs. T diagram (i) for an endothermic reaction carried out in an adiabatic CSTR; (ii) for an exothermic reaction carried out in an adiabatic CSTR. Solution to Problem 2.2a: xA 'Hr > 0

'Hr < 0

To

T

Problem 2.2b: Sketch the operating line on an xA vs. T diagram for an exothermic reaction carried out in a CSTR, with cooling (Q < 0 ): Solution to Problem 2.2b: xA

Q<0

(heat removal)

B

A

Q = 0 (adiabatic)

T

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II. Homogeneous reactions - Non-isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

Problem 2.3 The gas phase reaction k1

 oB A m k2

with reaction rate expression rA k1 p A  k2 pB will be carried out in a continuous stirred tank reactor. The feed is pure A. The required conversion of xA is 0.62. Calculate a) the minimum reactor size in which the required conversion can be achieved, and, b) the amount of heat to be supplied or removed for steady state operation.

Ideal gas behaviour may be assumed.

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DATA: Volumetric flow rate of feed vT0 : 10 m3 sec-1 Average heat capacities of A and B : 42 kJ kmol-1 K-1 : 573 K Feed temperature, T0 Pressure of feed line and reactor : 10 bar Gas constant, R : 0.08314(bar m3K-1kmol-1) = 8.314 kJ K-1 kmol-1 Heat of reaction, 'Hr : - 41570 kJ kmol-1 3 kmol m-3 s-1 bar-1 k1 = 10 exp{-41570 / RT} 6 k2 = 10 exp{-83140 / RT} kmol m-3 s-1 bar-1 Activation energies are given in kJ kmol-1

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II. Homogeneous reactions - Non-isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

Solution to Problem 2.3 The minimum reactor size for achieving a given conversion implies operating at the temperature where the reaction rate is a maximum. The reaction rate expression is written as:

PT  1  x A and pB

k1 p A  k2 pB , where p A

rA

k1 PT  1  x A  k2 PT x A

rA

PT x A .

PT ^k1   k1  k2 x A `

§ dr · The temperature for the maximum reaction rate can be found from ¨ A ¸ © dT ¹ x A

§ drA · ¨ ¸ © dT ¹ x A

PT

xA dk2 dT

where and

k20

' E2 RT

2

dk1 ­ dk1 dk2 ½ ®  ¾ PT x A dT ¯ dT dT ¿

0

0

1  dk2 / dT 1  dk1 / dT e  ' E2

RT

;

dk1 dT

k10

' E1 RT

2

e ' E1

RT

­  ' E2  ' E1 ½ k20 ' E2 exp ® ¾ k10 ' E1 RT ¯ ¿ ­  83140  41570 ½ 106 § 83200 · § 5000 · exp ® ¾ 2000 exp ¨ ¸ 3 ¨ 41600 ¸ 8.314 T ¹ © T ¹ 10 © ¯ ¿ 1 xA . 1  2000 e 5000 / T dk2 dT dk1 dT

dk2 dT dk1 dT

Solving for the temperature:

 5000

T

­§ 1 · 1 ½ ln ®¨¨  1 ¸¸ ¾ ¹ 2000 ¿ ¯© x A

618 K

The reactor volume is obtained from the mass balance equation

VR n A0

PT vT 0 RT

n A0 x A PT ^k1   k1  k2 x A `

 10  10  0.08314  573

2.099kmol / sec

At 618 K, k1 = 0.306 ; k2 = 0.094. Combining, we find:

VR = 2.24 m3 b) The heat balance:

Q

' H r n A0 x A  nT 0 C p T  T0

Q

54098  3967

50131 # 50000 kJ / sec

The amount of heat to be removed is

Q # 50000 kJ / sec

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II. Homogeneous reactions - Non-isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

Problem 2.4 A continuous stirred tank reactor is used for carrying out the liquid phase reaction: A + B o products. The reaction rate expression is rA = k(T) CA CB kmol h-1 m-3. The total molar flow rate at the inlet is 3 kmol h1 . The inlet flow is equimolar in A and B. The total volumetric flow rate (vT) is: 1 m3 h-1. Calculate the operating temperature required for a fractional conversion of 0.7. Also calculate how much heat must be added to the reactor to maintain the system at steady state, if the feed is introduced at the reactor operating temperature. The vessel is designed to withstand the vapour pressure of the reaction mixture. No reaction takes place in the vapour phase. Density changes due to (i) the chemical reaction and (ii) changes in temperature may be ignored. Data: Reactor volume, VR Heat of reaction, 'Hr Reaction rate constant, k(T),

: 3 m3 : 35,000 kJ (kmol A reacted)-1 : 1015 e-13000/T m3 kmol-1 h-1 (T in K)

Solution to Problem 2.4

The material balance equation for a CSTR may be written as:

VR

n A0  n A rA

n A0 x AvT2

2 k( T )n A0 ( 1  x A )2

where ni ni0 vT xA nA

= molar flow rate of component i = initial molar flow rate of component i = total volumetric flow rate = fractional conversion of A = CAvT

Solving for k(T), we get:

k( T )

x AvT2

( 0.7 )( 1 )2 2

VR n A0 ( 1  x A )

1015 exp{ 13000 / T }

2

( 3 )( 1.5 )( 0.3 )

13000 T 382.3 K

1.73 ; 

T

1.73 m3 kmol 1 h 1

­ 1.73 ½ ln ® 15 ¾ 34.0 ¯ 10 ¿

Heat balance

nT 0 C p ( T0  T )  ' H R x An A0  Q 0 for 'T Q

0, heat must be added

' H R x An A0 35000( 0.7 )( 1.5 ) 36750 kJ h 1 Q

36750 kJ h 1

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II. Homogeneous reactions - Non-isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

Problem 2.5 The gas phase reaction A o B + C is carried out in an adiabatic continuous stirred tank reactor. The reaction rate expression is given by the equation: rA = k(T) pA, where k(T) is the reaction rate constant and pA the partial pressure of the reactant A. The feed stream is pre-heated to 600 K; it contains an equimolar amount of reactant A and inert component D and. The total pressure is 1 bar; it may be assumed constant throughout the system. Calculate the steady state reactor operating temperature and the conversion of the reactant. (Note: There are two equally valid solutions to this problem. One of the two solutions will suffice as an adequate answer.) Ideal gas behaviour may be assumed. You may also assume total heat capacities of the feed and product streams to be equal. Data:

: 10 m3 : 10 kmol s-1 : 1.0 x 1020 e-30,000/T kmol bar-1 m-3 s-1 :  48 000 kJ (kmol A reacted)-1 : 200 kJ kmol-1 K-1

Reactor volume, VR Inlet total molar flow rate, nT0 Reaction rate constant, k(T) Heat of reaction, 'Hr Heat capacity of the feed stream, C p

(T in K)

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II. Homogeneous reactions - Non-isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

Solution to Problem 2.5 The mass and energy balances must be solved simultaneously. The energy balance equation:

nT 0 C p ( T0  T )  nA0 x A ' H R where xA : vT : VR: ni : nT0 :

0 ; nT 0 2nA0

fractional conversion total volumetric flow rate Reactor volume molar flow rate of component i total inlet molar flow rate

From the energy balance equation

nT 0 C P T ( 2 )( 200 ) ( T0  T ) ( T  600 ) ; xA 5 n A0 ' H r 48000 120 n A0 x A The mass balance equation: VR . Next we need to calculate nT as a function of xA. k pA xA

= nA0 – nA0 xA nA nI = nI0 = nA0xA nB nC = nA0xA ___________________ = nT0 + nA0xA nT

pA

m inert component

nA ptotal nT

n A0 ( 1  x A ) ptotal ; then nT 0  n A0 x A

­ ½ ­ nT 0  nA0 x A ½ xA 1 ® ¾® ¾ ¯VR k (T ) ptotal ¿ ¯ 1  x A ¿

Solve 1

­ ½ ­10  5 x A ½ xA ® 21 30000 / T ¾ ® ¾ with ) ¿ ¯ 1  xA ¿ ¯10 (e

o

VR

nA0 x A nT 0  nA0 x A ; k (T ) nA0 (1  x A ) ptotal

­ ½ ­10  5 x A ½ xA 1 ® ¾ 20 30000 / T ¾ ® ) ¿ ¯ 1  xA ¿ ¯10(1)(1)(10 )(e

xA

T  5 with 120

640 0.333 3.884 4.390u10-21 1.326

645 0.375 4.453 6.313u10-21 1.129

Two possible solutions T xA numerator exp(-30000/T) RHS

635 0.292 3.346 3.035u10-21 1.557

650 0.417 5.039 9.029u10-21 0.957

655 0.458 5.629 1.284u10-20 0.809

ANS: T = ~ 647 K ; xA = ~0.39

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II. Homogeneous reactions - Non-isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

Alternative solution: T 601 xA 0.00833 numerator 0.0836 exp(-30000/T) 2.096u10-22 RHS 0.402

602 0.0167 0.168 2.277u10-22 0.750

603 0.025 0.281 2.437u10-22 1.183

604 0.0333 0.339 2.686u10-22 1.306

605 0.0417 0.426 2.916u10-22 1.524

ANS: T ~ 602.5 ; xA= ~ 0.021

Problem 2.6 The rate of the first order irreversible gas-phase reaction, A o B is given by rA = k pA. The reaction will be carried out in a tubular adiabatic reactor. The reaction is exothermic. Calculate the reactor volume, VR (m3), required for a fractional conversion of “A”: xA = 0.9. Plug flow and ideal gas behaviour may be assumed. Definitions & data: Inlet mole fractions: yA0 = 0.20; yI0 (inerts) = 0.8 T0 nT0 PT k 'Hr

Feed temperature inlet total molar flow rate total pressure reaction rate constant Heat of Reaction

:400 K :1 kmol s-1 :1 bar :1.8 x 104 exp{- 5000/T} kmol s-1 m-3 bar-1 (T in K) :- 50,000 kJ (kmol A reacted)-1

Cp

Heat capacity of the reaction mixture

: 40 kJ kmol-1 K-1

Q rA pA nA0

Heat exchange with surroundings Reaction rate of component A, kmol m-3 s-1 partial pressure of A, bar molar flow rate of A, kmol s-1

Note 1. The operating line for a tubular reactor assumed to operate in plug flow is given by:

dT dx A

n A0 nT 0

­° Q ( ' H r  ® Cp ¯° rA C p

) ½° ¾ ¿°

Note 2. The Trapezoidal Rule for integration is given by x

³ f ( x )dx |

0

f ½ ­f ( ' x ) ® 0  f1  f 2  ...  f n1  n ¾ 2¿ ¯2

where, the interval of integration is divided into n equal segments and 'x= x/n . For the purposes of this exercise, use n = 3. Solution to Problem 2.6 The mass balance equation for a plug flow reactor is given by: n A ,exit

VR



³

n A0

dnA rA

Download free books at BookBooN.com 20

II. Homogeneous reactions - Non-isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

where rA kp A k( 1  x A ) p A0 , VR is the reactor volume (m3), nA the molar flow rate of “A” (kmol m-3 s-1) and pA0 the inlet partial pressure of “A” (bar). x A ,exit

VR

³

0

VR

x A ,exit

n A0 dx A

³

k( 1  x A ) p A0

n A0 p A0 k0

x A,exit

³

0

0

n A0 dx A [ k0 exp( 5000 / T )]( 1  x A ) p A0

dx A

;

( 1  x A ) exp( 5000 / T ) 0.9

VR

( 5.55 u 10 5 ) ³ 0

x A,exit

0.9

dx A ( 1  x A ) exp( 5000 / T )

(Eq. 2.6.A)

To express T in terms of xA, we use the operating line for an adiabatic tubular reactor, assumed to operate in plug flow [i.e Q=0]:

dT dx A

nA0 (  ' H r ) nT 0 C p

( 0.2 )( 50 000 ) ( 1 )( 40 )

250 K .

Integrating: T = T0 + 250 xA . This equation can then be substituted for T in Equation Eq. 2.6.A.

VR

( 5.55 u 10

5

0.9

)³ 0

dx A ( 1  x A ) exp[ 5000 /( T0  250x A )]

We use the Trapezoidal Rule for the integration, with

f ( xA )

1

;

( 1  x A ) exp[ 5000 /( 400  250x A )] [f(xA)]-1 3.73 x 10-6 1.88 x 10-5 4.51 x 10-5 3.35 x 10-5

xA 0 0.3 0.6 0.9

'xA= xA/3 = (0.9/3) = 0.3 f(xA)/2 13.4 x 104 1.49 x104

f3 ½ ­ f0 ®  f1  f 2  ¾ 2¿ ¯2

f(xA) 5.33 x 104 2.22 x 104

2.24 x 105

0.9

VR

f ½ ­f ( 5.55 u 10 5 ) ³ f ( x )dx | ( 5.55 u 10 5 ) ( ' x ) ® 0  f1  f 2  3 ¾ 2¿ ¯2 0 VR

(5.55 u105 ) (0.3) (2.24 u105 ) # 3.73 m3

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III. Catalytic reactions - Isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

Worked Examples - Chapter III Catalytic reactions – Isothermal reactors Problem 3.1 The first order irreversible gas phase reaction

AoB + C

is to be carried out in an isothermal tubular fixed bed catalytic reactor operating at atmospheric pressure. The intrinsic reaction rate expression is rA k C A . Pure “A” will be fed to the reactor at a rate of 0.13 kmol s-1. a. Compare the relative magnitudes of the rates of external mass transport and chemical reaction. b. Calculate the mass of catalyst required for a conversion of 80 %. For purposes of calculating km, an average Reynolds number of 500 may be assumed over the length of the reactor. Plug flow and ideal gas behaviour may be assumed. The pressure drop along the length of the reactor may be neglected. Data: Reaction rate constant, k: 0.042 Average gas density, Ug: Average gas viscosity, μg : Average diffusivity of “A”, DA :

m3 (kg-cat s)-1 1.0 kg m-3 -5 kg m-1 s-1 5.0 x 10 5.0 x 10-5 m2 s-1

External surface area of catalyst pellets, am:

0.667

Bed void fraction, HB: Mass flow rate, G : Effectiveness factor, K : Reactor temperature, T :

0.5 8.3 0.12 423

jD

0.46

HB

Re

0.4

where Re

m2 kg-1 kg m-2 s-1 K

d p G / P B and jD

km U B ­ P B ½ ® ¾ G ¯ Ub DAB ¿

2/ 3

where km is defined by the equation: rA,p = km am (Cb - Cs). R (gas constant) = 8.314 kJ kmol-1 K-1 = 0.08314 bar m3 K-1 kmol-1 Solution to Problem 3.1 From the course notes rp Solving for Cs , Cs

km am ( Cb  Cs ) K kCs .

km am Cb and rp K k  km am

­ 1 1½  ® ¾ ¯ km am K k ¿

1

Cb .

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III. Catalytic reactions - Isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

1 1 and , given Re constant at a value of 500. km am Kk ­ 0.458 ½ 0.4 0.077 ® ¾ u ( 500 ) ¯ 0.5 ¿

(a) Compare

jD Sc

P

5 u 10 5

U DA

1 u 5 u 10 5

1.

( 0.077 ) u ( 8.3 )( 1 ) 0.64 m s 1 1 U 1 1 1 1 2.34 ; 198.4 K k ( 0.12 ) u ( 0.042 ) km am ( 0.64 ) u ( 0.667 )

km

jD G

( Sc )2 / 3

Comparing magnitudes, chemical reaction turns out to be the controlling (slower) step.

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III. Catalytic reactions - Isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

­ 1 1 ½  (b) Let us define the overall rate constant as . { ® ¾ ¯ km am K k ¿ rp . Cb , and since n A C AvT , we can write rp

1

. Then,

. ( n A / vT ) .

(Eq. A)

Also ( 2.34  198.4 )1

.

4.98 u 10 1

(Eq. B)

Since the reaction “A o B + C” in the gas phase leads to a molar change upon reaction, we next need to express vT in terms of the conversion. We start with the change in molar flow rate.

nA

n A0  n A0 x A

nB

 n A0 x A

nC

 n A0 x A

 nT n A0 ( 1  x A ) Here, nT is the total volumetric flow rate expressed as a function of the fractional conversion of “A”. Using the ideal gas law: nT RT n A0 RT vT ( 1  xA ) . PT PT Substituting in Eq. (A), we get

. n A0 ( 1  x A )

rp

n A0 RT ( 1  xA ) PT

. PT ( 1  x A )

RT ( 1  x A )

(Eq. C)

To calculate the mass of catalyst required: n A,exit



W

³

n A0 x A,exit

³

0

( 1  xA ) dx A ( 1  xA )

W



W



x A,exit

³

0

dn A rA

n A0 RT . PT

x A,exit

³

0

( 1  xA ) dx A ( 1  xA )

­ 1 ½ 1 1 ® ¾ dx A ( 1  xA ) ¿ ¯( 1  xA )

 2 ln( 1  x A,exit )  x A,exit

n A0 RT ^2 ln( 1  x A,exit )  x A,exit ` . PT

( 0.13 )( 0.08314 )( 423 )^2 ln( 0.2 )  0.8` [ 4.98 u 10 3 ]( 1 )

# 2220 kg

Weight of catalyst: 2220 kg

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III. Catalytic reactions - Isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

Problem 3.2 A tubular reactor packed with spherical pellets will be used for carrying out the catalytic reaction: 2A oB + C . Reactant “A” is available as 20 (mole) % A in an inert gas and will be fed to the reactor at the rate of 0.02 kmol s-1. The intrinsic reaction rate given by

rA,V

kV C A2 kmol m3 s 1 ,

where CA denotes the concentration of A. The intended conversion is 75 %. (a) The reactor is to be operated isothermally at 600 K and 1 bar pressure. Estimate whether intraparticle diffusion resistances affect the overall reaction rate. For the purposes of this part of the calculation CA,s # CA,b may be assumed. (b) Estimate whether bulk to catalyst surface mass transport resistances affect the overall reaction rate. Due to dilution by the inert gas, bulk properties of the gas mixture may be considered as approximately constant. The pressure drop over the length of the reactor and deviations from the ideal gas law may be neglected. Data Catalyst pellet density, U p :

2200

kg m-3

Effective diffusivity of A within the pellet, DA,eff:

5 x 10-3 1 x 10-6

m m2 s-1

Molecular diffusivity of A in the gas mixture, DA:

1 x 10-4

Reaction rate constant, kv:

3.46 x 104

m2 s-1 m3 kmol-1 s-1

Mass flow rate, G: Density of the gas mixture, U g :

10 1

kg m-2 s-1 kg m-3

Viscosity of the gas mixture, μ: Void fraction of the bed, H b:

1.5 x 10-5 0.42

kg m-1 s-1

Catalyst pellet diameter, dp

For integral reaction rate orders the effectiveness factor may be expressed as 1/ 2

K

Sx 2 V p k [ C A,s ] n

­C A,s ½ ° ° n D k [ C ] dC ® ³ A,eff ¾ °¯ 0 ¿°

.

The dimensionless mass flux jD is given by the correlation

jD

( 0.46 / H ) (Re)0.4 , where Re

d pG / P

where jD is related to km, the bulk stream to particle surface mass transfer coefficient, by

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III. Catalytic reactions - Isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

­° P ½° ( km U g / G ) ® ¾ ¯° U g DA ¿°

jD

2/ 3

.

R (gas constant) = 8.314 kJ kmol-1 K-1 = 0.08314 bar m3 K-1 kmol-1 Solution to Problem 3.2 If intraparticle diffusion resistances are significant, K # 1 / ) where ) is usually >5. Now let us calculate ) by assuming that K # 1 / ) . From the main text (Chapter 6), we have: C ½ V p k [ C A,s ] n ­° A,s ° n ® ³ DA,eff k [ C ] dC ¾ Sx 2 °¯ 0 °¿

)

)

V p kV [ C A,s ] n ­° k [ C A,s ] n1 ½° D ® A,eff ¾ Sx n1 2 °¯ °¿

1 / 2

1 / 2

V p ­° 2Deff 1 ½° ® n1 ¾ S x °¯ n  1 kV C A,s °¿

(Eq. 6.74) 1 / 2

For n = 2, we get 1/ 2

½° d p ­° 3 k C ® V A,s ¾ 6 °¯ 2Deff °¿

)

/2 1.9 u 10 2 C 1A,s

(Eq. A)

Neglecting external transport resistances, at the reactor inlet:

CA

p A,partial

nA vT

RT

0.2( 1 ) ( 0.08314 ) ( 600 )

4 u 10 3 kmol m 3

At the reactor exit:

CA

( 0.05 )( 1 ) 1 u 10 3 kmol m 3 ( 0.08314 )( 600 )

Substituting these values into Eq. A above, we get

) | 12 at the reactor inlet and ) | 6 at the reactor exit. So we must conclude that diffusion resistances are not negligible within this reactor. Part (b): At steady state, the rate of reaction within a particle must be equal to the net rate of diffusion of the reactant.

net flux

From jD

5 u 10 3 ( 10 )

3333 1.5 u 10 5 ( 0.46 / H ) (Re)0.4 we can calculate jD 0.043 . Meanwhile jD G 1 km 2/ 3 Ug ­ ° P ½° ® ¾ ¯° U g DA ¿°

Let us calculate km am .

Re

d pG

km am ( Cb  Cs )

P

°­ P °½ ® ¾ °¯ U g DA °¿

2/ 3

°­ 1.5 u 10 5 °½ ® 4 ¾ °¯ 1 u 1 u 10 °¿

2/ 3

0.28 Download free books at BookBooN.com

26

III. Catalytic reactions - Isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

( 0.043 )( 10 ) 1 1.54 and 1 0.28 4S rp2 3 6 am 3  4 3 S rp U p rp U p 5 u 10 u 2200 3 km am 0.84 m3 kg 1 s 1

km

0.545 m 2 kg 1

The catalyst bed density is calculated from U B ( 1  H ) U p ( 0.58 ) u ( 2200 ) 1276 kg m 3 Then we use the concept that, at steady state, net diffusion of reactant to the catalyst pellet equals the amount of reactant that has been converted. 2 U B km am ( C Ab  C As ) K k C As

Define a constant D . At the inlet D

kmol /( m2 s ) U B km am ( 1276 ) u ( 0.84 ) 0.372 kmol m 3 Kk § 1 · 4 ¨ ¸ ( 3.46 u 10 ) © 12 ¹

The equation to solve now takes the form: 2 D ( C Ab  C As ) C As

(Eq. B)

and at the inlet C Ab 4 u 10 3 kmol m 3 . Eq. B can be solved as a quadratic equation or by trial and error.

C As # 3.95 u 10 3 kmol m 3 Comparing C Ab and C As , we can see the difference is small. Therefore, in this case, the bulk-stream to pellet surface mass transport resistance appears negligible at the reactor inlet.

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III. Catalytic reactions - Isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

Problem 3.3 Differential reactors are used to simulate an infinitesimally thin (“slice”) section of, say, a tubular reactor. In the laboratory, this is done to measure reaction rates and other parameters of catalytic reactions, in the context of a small amount of conversion.

The first order irreversible reaction A o B  C will be carried out over 0.002 kg packed catalyst in a differential test reactor. The intrinsic rate expression for the reaction is

kmol ( kg  cat u s )1

rA 7 u 1015 exp ^18,000 / T ` C A

where the temperature T is given in K and CA, the concentration of the reactant “A”, is expressed in kmol m-3. Assuming isothermal operation at 500 K and given the data listed below, (a) Estimate whether intraparticle diffusion resistances would be expected to affect the overall rate. (b) Estimate whether bulk to surface diffusion resistances would be expected to affect the overall rate. (c) Calculate the conversion through the reactor. Data: Catalyst pellet diameter, dp:

0.001

m

Catalyst pellet density Up : Effective diffusivity of A within the pellet, DA,eff :

1500

kg m-3

1 x 10-6

m2 s-1

Molecular diffusivity of A in the bulk stream, DA :

1 x 10-4

Mass flow rate, G: Viscosity of the gas mixture, μ: Density of the gas mixture, Ug: Void fraction of the bed, H: Total volumetric flow rate, vT :

0.1 1 x 10-6 1 kg m-3 0.42 0.005

m2 s-1 kg m-2 s-1 kg m-1 s-1

m3 s-1

The dimensionless mass flux jD is given by the correlation

jD

( 0.46 / H ) (Re)0.4 , where Re

d pG / P

where jD is related to km, the bulk stream to particle surface mass transfer coefficient, by

jD

­° P ½° ( km U g / G ) ® ¾ ¯° U g DA ¿°

2/ 3

.

The Thiele modulus for spherical catalyst pellets is given by:

· °­ k U p ¸® ¹ °¯ DA,eff

1/ 2

°½ )s ¾ ¿° where Rs is the pellet radius, k the reaction rate constant; Up and Deff have been defined above. § Rs ¨ © 3

(Eq. A)

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III. Catalytic reactions - Isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

Solution to Problem 3.2 Part (a)

Rs

0.0005 m .

At 500 K, the reaction rate constant k

0.0005 3

)s

( 1.62 )( 1500 ) 1 u 10 6

1.62. Substituting into Eq. A, we can calculate ) s .

8.225

) s is large. We conclude that intraparticle diffusion resistances are significant and would be expected to affect the overall rate. Part (b):

4S rp2

am

3 rp U p

4 3 S rp U p 3

3 5 u 10

4

u 1500

4.0 m 2 kg 1

To calculate km , we first need the Reynolds number. Re

jD

( 0.001 )( 0.1 ) 1 u 10 6

100 . Using this value we can get

0.174. Then

km

jD G ­° P ½° ® ¾ U g ¯° U g DA ¿°

2 / 3

0.375

m s 1

The basic equation to use for comparing external and intraparticle diffusion effects is:

rp

1 1 1  K k km am

km am

C Ab . We now compare

1 1 and . Kk km am

1 0.66 km am 1 1.62 / 8.225 0.197; 5.1 Kk

( 4 )( 0.375 ) 1.5 m3 s 1 ;

Kk # k / )s

External diffusion resistance less significant! Part (c): For a “differential reactor” the mass balance equation dW

'W # 

'n A rp

, where 'n A



 n A0 ' x A . We also have rp

dn A can be written as: rp

1 C Ab and 1 1  K k km am

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III. Catalytic reactions - Isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

CA

n A / vT . Thus we can write C Ab

n A0 ( 1  x A ) and rp vT

In the case of the differential reactor, x A # ' x A . Thus

1 1 1  K k km am

u

n A0 ( 1  x A ) . vT

­1 1 ½ ®  ¾ ¯K k km am ¿ 0.005 ^5.1  0.667` 14.42 0.002

1  x A vT # 'W xA

'W

­1 1  xA n A0 ' x A 1 ½ u®  ¾ vT ; then xA n A0 ( 1  x A ) ¯K k km am ¿

xA

1 0.066 15.41

xA

0.066

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IV. Catalytic reactions - No isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

Worked Examples - Chapter IV Catalytic reactions – Non-isothermal reactors Problem 4.1 The exothermic catalytic gas phase cracking reaction A o light products is carried out with a feed of 10 % “A” in an inert diluent, “B”. The intrinsic rate for the reaction is given by:

rA

kmol s 1 ( kg  cat )1

125 exp ^5000 / T ` p A

where pA denotes the partial pressure (bar) of “A” and T denotes the local temperature in K. The catalyst pellets (dia. 0.004 m) may be considered isothermal and the value of the effectiveness factor is estimated as 0.08 throughout the reactor. We select a point within the reactor where, G, the superficial mass velocity of the gas stream is 0.8 kg m-2 s-1, the bulk temperature, Tb, is 500 K, and the partial pressure of “A” in the bulk, pA,b, is 0.05 bar. Calculate: (i) (ii) (iii)

the global rate of reaction, the external concentration gradient (expressed in terms of the partial pressures), and, the temperature gradient between the bulk gas stream and external catalyst pellet surfaces,

at the selected point in the reactor. Bulk gas properties may be taken as those of the diluent. The pressure drop along the reactor axis may be neglected. Data: Effectiveness factor, K : Total reactor pressure, pT :

0.08 1 bar

Void fraction of the packed bed, HB : Surface area of catalyst pellets, am: Heat of reaction, 'Hr : Molecular diffusivity of A in B, DAB :

(see statement of problem)

0.5 0.4 m2 kg-1  20 000 kJ (kmol A reacted)-1 2 u 10-4 m2s-1

Density of “B”, U B :

0.1 kg m-3

Viscosity of “B”, P B : Heat capacity of “B”, Cp,B :

1.0 u 10-5 kg m-1 s-1 1.0 kJ kg-1 K-1

Thermal conductivity of “B”, OB :

jD

jD

1.5 u 10-5 kJ m-1 s-1 K-1

( 0.7 ) jH

0.46

HB

km U B ­ P B ½ ® ¾ G ¯ Ub DAB ¿

Re 0.4 ; Re

2/ 3

;

jH

d p G / PB

­ C p PB ½ ® ¾ C p G ¯ OB ¿ h

2/ 3

.

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IV. Catalytic reactions - No isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

In these equations, km, the bulk to surface mass transfer coefficient is defined in terms of the equation: rA,p = (kmam/RTb) ( pb - ps ), and, h is the surface to bulk heat transfer coefficient, defined by h am (Ts - T b) = (  'Hr) rA,p . Finally, R (gas constant) = 8.314 kJ kmol-1K-1= 0.08314 bar m3 K-1 kmol-1 Solution to Problem 4.1

( 0.004 ) u ( 0.8 )

Re

d p G / PB

jD

9.16 u 10 2

320 ; substituting this value into the correlations, we get

1 u 10 5 and jH 0.131 .

Next, we calculate the Prandtl and Schmidt numbers

Pr

C pB P B

1 u 1 u 10 5

OB

1.5 u 10 5

PB

0.67 and Sc

U B DAB

1 u 10 5

0.5 .

0.1 u 2 u 10 4

On the other hand

ham ( Ts  Tb ) rA,p u ' H r , leads to

h

jH C p G (Pr)2 / 3

( Ts  Tb )

rA,p (  ' H r )(Pr)2 / 3

( rA,p ) ( 20,000 ) ( 0.67 )2 / 3

jH C p G am

( 0.131 )( 1 )( 0.8 )( 0.4 )

recalling that Tb

3.65 u 10 5 rA,p

500 K , Ts

500  3.65 u 10 5 rA,p .

(Eq. A)

Similarly

km am ( pb  ps )A RTb

rA,p and km

jD G

1

U B Sc 2 / 3

Then

( pb  ps )A

leads to ( pb  ps )A

rA,p U B ( Sc 2 / 3 ) R Tb

rA,p ( 0.1 ) ( 0.5 )2 / 3 ( 0.08314 )( 500 )

am jD G

( 0.4 )( 9.16 u 10 2 )( 0.8 )

89.1 u rA,p , where p A,b

0.05 leads to p A,s

The intrinsic reaction rate was given as rA

0.05  89.1 u rA,p

(Eq. B)

125 exp ^5000 / T ` p A

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IV. Catalytic reactions - No isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

Since rA,p

K rA,s K rA ( at surface conditions ) , and the intrinsic reaction rate expression was

given in the problem statement as rA can be written as

rA,p K rA,s

( 125 ) exp ^5000 / T ` p A , the global reaction rate expression

( 0.08 ) u ( 125 ) exp ^5000 / Ts ` p A,s .

Substituting expressions for Ts from Eq. (A) and p A,s from Eq. (B), we get rA,p

K rA,s

^

`

( 0.08 ) u ( 125 ) u ª¬0.05  89.1 u rA,p º¼ u exp 5000 /( 500  3.65 u 10 5 rA,p )

p

p

p A,s

Ts

This equation can be solved by a simple trial and error procedure. rA,p 2.6 u 10 5 kmol ( kg  cat u s )1 Then Ts

500  3.65 u 10 5 [ 2.6 u 10 5 ] # 509.5

'T ps

0.05  89.1 rA,p

9.5 K

Not negligible!

4.77 u 10 2

' p 0.05  0.0477 0.0023 bar o ~ 5 % of total pressure.

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IV. Catalytic reactions - No isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

Problem 4.2 The exothermic gas phase reaction

A+BoC is carried out over a porous catalyst, in a non-isothermal/non-adiabatic fixed bed catalytic reactor. The feed stream is equimolar in A and B. (a) Using a one-dimensional, pseudo-homogeneous model framework, derive steady-state mass and energy balance equations and state boundary conditions to describe the behaviour of the reactor. Define your terms carefully. (b) Calculate the maximum temperature that the exit gas stream would attain if the reactor were to be operated adiabatically. Judging by this result, what change would you recommend in reactor operating conditions. (c) Discuss briefly for the present case, advantages and disadvantages of using a one-dimensional reactor model over a two-dimensional model.

Data: Heat of reaction, 'Hr : Average heat capacities

CpA # CpB : CpC :

 300 000

kJ (kmol A reacted)-1

30

kJ kmol-1 K-1 kJ kmol-1 K-1

60

Solution to Problem 4.1 Part A: Mass balance: The mass balance over an infinitesimally thin (“slice”) element of the catalyst packed reactor (dW) gives: dn A  rA dW , where n A is defined by the equation n A n A0 ( 1  x A ) and dn A n A0 dx A . In these equations W denotes the mass of catalyst and nA the molar flow rate of “A”. nT 0 C p is the total heat capacity of the inlet stream. It is assumed not to change with conversion; so we shall use it as the total heat capacity of the reaction mixture. From dn A

 rA dW we can then write the full mass balance equation as:

Energy balance:

ª ¦ ni C pi º dT ¬ ¼

where Q is the external heat transfer term and

dx A dW

rA . n A0

(Eq. A)

dW ( Q  rA ' Hr )

¦ ni C pi #

nT 0 C p as explained in Chapter 3 of the main

text. The energy balance equation may then be written as:

dT dW

Q  rA ' H r nT 0 C p

(Eq. B)

The initial condition is given as: At W = 0, T = T0 = Twall .

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IV. Catalytic reactions - No isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

There are many ways of formulating Q, the heat loss term. One common way would be: Q A h ( T  Twall ) , where A denotes the contact surface area and h the heat transfer coefficient. Part B: Combining Eqs. A and B, we can write

n A0 rA

dT dx A For adiabatic operation, Q = 0 .

dT dx A



­° Q  r ' H ½° A r . ® ¾ ¯° nT 0 C p ¿°

'H r [ nT 0 C p ]

n A0

Integrating

'T



' H r n A0 C p nT 0

( 0.5 )( 3 u 10 5 ) 30

5000 K

Unacceptably high! Improve cooling and/or reduce reactant concentrations and/or dilute the catalyst. Part C: The main difficulty arises from ignoring the radial temperature distribution. In dealing with highly exothermic reactions, when the maximum temperature along the central axis turns out to be much larger than the radially averaged temperature, runaway may occur under conditions where the 1-dimensional model would predict “safe” operation. The disadvantage of 2-dimensional operation is the requirement for accurate additional data for radial heat (and mass) transport: Orad ,effective , Drad ,effective .

Problem 4.3 The exothermic reaction

A+BoC

was carried out using two different catalyst particle sizes: 8 mm diameter catalyst pellets and 50 μ (average) diameter catalyst particles. Both sets of experiments were carried out under conditions where external temperature and concentration gradients were negligible. Global rates of reaction, measured as a function of temperature, using approximately similar external surface concentrations of A and B for all experiments, are given in the table below. Assuming the 50 μ particles to be isothermal under all operating conditions, (a) Calculate from this data, the energy of activation most likely to approximate the true value for the chemical reaction step at the catalytic sites. (b) For the 8 mm catalyst pellets, calculate the effectiveness factor at each temperature. State your assumptions clearly. c. Briefly explain observed changes in the values of effectiveness factors as a function of the experimental temperature. What do values above unity imply?

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IV. Catalytic reactions - No isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

Table A. Experimental global reaction rates as a function of catalyst particle size and temperature.

Catalyst Particles Rate, kmol s-1

Temp (K) 425.5 458.7 495.1 549.5 637.0

9.2 x 10-8 2.5 x 10-7 6.95 x 10-7 1.85 x 10-6 4.12 x 10-6

8 mm pellets rA,p (kg cat)-1 5.12 x 10-7 5.95 x 10-7 7.18 x 10-7 8.58 x 10-7 9.39 x 10-7

Solution to Problem 4.3 Part (a): “Series 2” Arrhenius plot for 50 Pm particles (Figure A) shows the line bending above 495 K. The slope of the straight component of the line gives the correct ('Ea/R) value for chemical control, i.e. for K=1. Using

values of the rate between 425 and 495 K, we can calculate the slope of the straight part of the line. ln r1  ln r2  1.42  1.62 °­ °½ Slope  ® 3 ¾ 1 1 °¯ ( 2.02  2.35 ) u 10 °¿  T2 T2

' Ea R

8.314 kJ /( kmol u K )

5.04 kJ kmol 1

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' Ea

6060 K ; the gas cons tan t R

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IV. Catalytic reactions - No isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

ln r vs. 1000 x (1/T) 0 0

0.5

1

1.5

2

2.5

ln r

-5 Series1

-10

Series2

-15 -20 1000 x (1/T) [T in K]

Figure A. Arrhenius plot based on data for 50 Pm particles in Table A

Part (b): The first three rates in rows 1-3 of column-2 in Table A and Table B are the same. The data corresponds to the part of the Series 2 curve in Figure A before it begins to bend. Since the line is straight up to and including 495 K, we can safely conclude we have chemical control and K = 1. Above that temperature, the line bends. If there had been no diffusive limitations the line would have continued as a straight line. That is why the extrapolated line still corresponds to K = 1. We can read off (in an enlarged diagram) that the extrapolated rate for 549.5 K is 2.26u10-6 and for 637 K, the rate corresponding to K = 1 is 1.07u10-5. We then use Eq. 6.76 (Chapter 6) from the main text.

(robs )1 (Eq. 6.76) Ș1 /Ș2 (robs )2 where (robs )1 is the second column in Table B and (robs )2 third column in Table B. With K1 taken as unity, we can calculate the effectiveness factors for the 8 mm pellets at different temperatures. Table B. Experimental global reaction rates as a function of catalyst particle size and temperature.

Rate, kmol s-1 Temp (K) 425.5 458.7 495.1 549.5 637.0

rA,p (kg cat)-1 8 mm pellets (data from Table A)

50 Pm catalyst particles (K1=1) 9.2 x 10-8 2.5 x 10-7 6.95 x 10-7 2.26 x 10-6 1.07 x 10-6

5.12 x 10-7 5.95 x 10-7 7.18 x 10-7 8.58 x 10-7 9.39 x 10-7

K Effectiveness factor for 8 mm pellets

5.56 2.38 1.03 0.38 0.228

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IV. Catalytic reactions - No isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

Part (c): For exothermic reactions K=1 means that the temperature at the centre of the pellet is greater than the surface temperature Ts. That in turn implies that the rate at the centre of the pellet is greater than the rate at the

surface. In this example, the effect was more visible at relatively low surface temperatures. This explains the change in the nature of the value of effectiveness factor with increasing temperature.

Problem 4.4 Differential reactors are used to simulate an infinitesimally thin (“slice”) section of, say, a tubular reactor. In the laboratory, this is done to measure reaction rates and other parameters of catalytic reactions, in the context of a small amount of conversion.

The intrinsic rate for the endothermic catalytic gas phase cracking reaction “A o products”, is given by the first order reaction rate expression

rA

4,000 ^exp(  10,000 / T )` p A

kmol /( kg  cat u s )1 ,

where pA is the partial pressure (bar) of A, and T denotes the local temperature in K. The reaction is carried out in a differential reactor, where reactant “A” is supplied, mixed 1:1 with an inert diluent. The superficial mass velocity of the gas stream flowing over the catalyst particles, G, is 1 kg m-2 s-1, and the temperature of the bulk stream is 750 K. The overall conversion in the reactor has been determined experimentally as 3.6 %. The catalyst pellets have a diameter of 0.004 m and may be considered isothermal. Calculate the global rate of reaction, the magnitude of the effectiveness factor, and the external concentration and external temperature gradients. The bulk gas properties can be taken as those of the diluent, B. The pressure drop through the reactor may be neglected. Data: Molecular diffusivity of A in B, DAB: Density of diluent B, UB: Viscosity of B, μB: Heat capacity of B, Cp,B: Thermal conductivity of B, OB: Reactor pressure, pT: Void fraction of the packed bed, H: Surface area of the catalyst pellets, am: Heat of reaction, 'HR: Pellet effective diffusivity, Deff: Pellet density, Uc:

jD

jD

( 0.7 ) jH

3 x 10-4 0.1 1.5 x 10-5 1.0 2.5 x 10-5 1 0.48 0.4 25,000 1 x 10-6 2,000

0.46

HB

km U B ­ P B ½ ® ¾ G ¯ Ub DAB ¿

m2 s-1 kg m-3 kg m-1 s-1 kJ kg-1 K-1 kJ m-1 s-1 K-1 bar m2 kg-1 kJ (kmol A)-1 m2 s-1 kg m-3

Re 0.4 ; Re

2/ 3

;

jH

d p G / PB

­ C p PB ½ ® ¾ C p G ¯ OB ¿ h

2/ 3

.

In these equations, km, the bulk to surface mass transfer coefficient is defined in terms of the equation: rA,p = (kmam/RTb) ( pA,b – pA,s ), Download free books at BookBooN.com 38

IV. Catalytic reactions - No isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

and, h is the surface to bulk heat transfer coefficient, defined by h am (Ts - T b) = (  'Hr) rA,p . Finally, R (gas constant) = 8.314 kJ kmol-1K-1= 0.08314 bar m3 K-1 kmol-1 Solution to Problem 4.4

h am ( Ts  Tb ) rA,p ( ' H r ) , where h ( Ts  Tb ) Pr

Re

C pB P B

OB

1 u 1.5 u 10 5

d pG

2.5 u 10 ( 0.004 )( 1 )

PB

1.5 u 10 5

5

0.6;

( Ts  Tb )

(Pr)2 / 3

(Eq. A)

jH C pB G am 0.71

( rA,p ) u ( 25000 ) u ( 0.71 ) ( 0.147 )( 1 )( 1 )( 0.4 ) 750  3.02 u 10 5 rA,p

0.103 and jH

0.147 .

 3.02 u 10 5 rA,p

(Eq. B)

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Ts

(Pr)2 / 3

rA,p ( ' H r )(Pr)2 / 3

267 ; from equations in the data section jD

Substituting in Eq. A, we get

jH C p G

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IV. Catalytic reactions - No isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

km am ( p A,b  p A,s ) ; we can write RTb

Using the mass transfer equation: rA,p

(Eq. C)

2 / 3

km km

jD G ­ P B ½ jD G ( Sc )2 / 3 ® ¾ U B ¯ Ub DAB ¿ UB º ( 0.103 )( 1 ) ª 1.5 u 10 5 1.59 « 4 » 0.1 ¬« ( 0.1 ) u ( 3 u 10 ) ¼»

Substituting this result in Eq. C, we get:

p A,b  p A,s

rA,p RTb

( 0.08314 )(750 )rA,p

km am

( 1.59 )( 0.4 )

98.04 rA,p

Reactor pressure: 1 bar. Conversion is 3.6%. Exit stream composition assumed equal to reactor composition. This basically means that we treat the “differential reactor” as a mini-CSTR. The feed is 50 % “A”. Therefore:

pb

( 0.5 )( 1  0.036 ) 0.482 bar . Thus: ps 0.482  98.04 rA,p rA,p K rA,s and K

1 ­ 1 1 ½  ® ¾ ; rearranging ) s ¯ tanh 3) s 3) s ¿ 1/ 2

­° k RTs Uc ½° )s ® ¾ °¯ Deff °¿ ) s in the above equation is form of the Thiele modulus appropriate for the given reaction rate Rs 3

expression.

)s

0.002 3

1/ 2

ª 10000 º ( 0.08314 )( 2000 ) °½ °­ u u T ®4,000 exp « ¾ » s 1 u 10 6 ¬ Ts ¼ ¯° ¿°

)s

1/ 2

544 ^Ts exp( 10000 / Ts )`

(Eq. D)

At Ts 750 K , ) s # 14 ; therefore we can safely consider K # 1 / ) s . Going back to the global reaction rate expression: rA,p 4000 K u exp ^10,000 / Ts ` u p A,s (Eq. E)

rA,p

^

`

4000 K u exp 10,000 /(750  3.02 u 10 5 rA,p u ( 0.482  98.04 rA,p )

(Eq. F)

We need to substitute for K in Eq. E. As already mentioned, K # 1 / ) s ; ) s is given by Eq. C, and Ts by

Eq. B. Then: 1/ 2

­ ª º °½ 10,000 ° 5 »¾ ) s 544 ®(750  3.02 u 10 rA,p ) u exp « «¬ (750  3.02 u 10 5 rA,p ) »¼ ° °¯ ¿ The easier way to solve for rA,p is by trial-and-error between K # 1 / ) s , Eq. G and Eq. F.

(Eq. G)

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IV. Catalytic reactions - No isothermal reactors

Fundamentals of Reaction Engineering – Worked Examples

Close enough answers:

rA,p # 1.15 u 10 4

K 0.075 Ts

'T p A,s

715.3 K 34.7 K 0.47 bar

p A,b  p A,s

0.011 bar

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