Fundamentals of reaction engineering worked exam

FUNDAMENTALS OF
REACTION ENGINEERING
– WORKED EXAMPLES
F R EE EE

SST U D Y

BBO
OO
OK S

RAFAEL KANDIYOTI

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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
ô 3 1 ằ
ơô m s ẳằ

> 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:
ẵ
ê 1
RT
ư
(Eq. 1.37)
VR

nT 0 đ( 1  y A0 )ln ô
ằ  y A0 x Ae ắ
Pk


ơ 1  x Ae ẳ
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

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 ẵ
PT đk1nA0  k1nA0 xA 
ắ
nA0 RT (1  x A ) ¿
¯

Simplifying,
§ nA0 RT ·
x A (1  x A )
.
ă
á
2
â PT ạ ưk  k x  k2 PT xA ẵ
đ 1 1 A
ắ
RT 1  x A ¿

¯
kP
To further simplify, we define a  2 T .
RT
VR

VR

§ nA0 RT ·
x A (1  x A ) 2
ă
á
2
2
â PT ạ 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

ôơ 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 ;

ư 4 u 1017 ẵ
. Solving, T
ln đ
12 ắ

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