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343731210-Solution-Manual-LevenSpiel-by-13BCH-ITNU

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By Octave Levenspiel
PREFACE
This solution manual is result of hard work done by students of B. Tech.
Chemical Engineering (Batch of 2k13), Nirma University. This manual
provides solution to unsolved numerical from book “Chemical Reaction
Engineering” by Octave Levenspiel. We hope this will help you in solving
your queries regarding numericals. For any corrections or suggestions
mail us at [email protected]
Credit goes to….
Aashish
Abhijeet
Akshada
Aman
Ankit
Arpit
Rahul
Bhuvan
Brijesh
Samarpita
Harsh
Nirav
Harshit
Jay
Prashant
Yash
Shardul
Yagnesh
Krish
Deep
Manish
Chirag
Maulik
Mehul
Dhruv
Jigar
Parthvi
Darshan
Prasann
Shivang
Rohan
Vedant
Vishal
Huma
Mihir
Prerak
Jay
Rochesh
Sarthak
Fenil
Siddharth
Ravi
Sunil
Ruchir
Parth
Ravindra
Viren
Sunny
Yash
Yuga
Vikas
Hardik
Nirmal
Chapters Solved:

Chapter-2

Chapter-3

Chapter-4

Chapter-5

Chapter-7

Chapter-8

Chapter-9

Chapter-10

Chapter-11

Chapter-12

Chapter-13

Chapter-14

Chapter-18

Chapter-23

Chapter-24

Chapter-25

Chapter-26
Problem 2.1:
A reaction has the stoichiometric equation A + B = 2R. What is the order of reaction and rate
expression?
Solution 2.1:
We require experimental kinetics data to answer this question, because order of reaction need
not to match the stoichiometry, most oftenly it doesn’t.
If this reaction is elementary, the order of reaction is 2. And rate expression is,
− 𝑟𝐴 = 𝐾 𝐶𝐴 𝐶𝐵
Problem 2.2:
𝟏
Given the reaction 2NO2 + 𝟐 O2 = N2O5, what is the relation between the rates of formation and
disappearance of the three reaction components?
Solution 2.2:
The relation between the rates of formation and disappearance of the three reaction
components is,
−𝑟𝑁𝑂2
−𝑟𝑂2 −𝑟𝑁2 𝑂5
=
=
2
1/2
1
Problem 2.3:
𝟏
𝟏
A reaction with stoichiometric equation 𝟐 A + B = R + 𝟐 S has the following rate expression,
− 𝑟𝐴 = 2 𝐶𝐴0.5 𝐶𝐵
What is the rate expression for this reaction if the stoichiometric equation is written as A + 2B =
2R + S?
Solution 2.3:
It does not affect how we write the stoichiometry equation. The rate of reaction remains
unchanged,
− 𝑟𝐴 = 2 𝐶𝐴0.5 𝐶𝐵
Problem 2.4:
𝑒𝑛𝑧𝑦𝑚𝑒
For the enzyme-substance 𝐴 →
𝑅, the rate of disappearance of any substrate is given by
1760[A][𝐸0 ] 𝑚𝑜𝑙
, 3
6+𝐶𝐴
𝑚 𝑠
− 𝑟𝐴 =
What are the units of the two constants?
Solution 2.4:
Assume 𝐶𝐴 is in mol/m3
Constant 6 has to be the same unit as of 𝐶𝐴 , then and then only summation is possible.
 6 mol/m3
1760
𝑚𝑜𝑙 𝑚𝑜𝑙

𝑚3 𝑚3 = 𝑚𝑜𝑙
𝑚𝑜𝑙
𝑚3  𝑠
3
𝑚
 1760 𝑠 −1
Problem 2.5:
For the complex reaction with stoichiometry A + 3B  2R + S and with second order rate
expression
 rA = k1[A][B]
are the reaction rates related as follows: rA = rB = rR? If the rates are not so related, then how are
they related? Please account for the signs, + or .
Solution 2.5:
𝑟𝑅 = − 2𝑟𝐴 −
2
𝑟
3 𝐵
𝑜𝑟
1
1
− 𝑟𝐴 = − 𝑟𝐵 = 𝑟𝑅
3
2
Problem 2.6:
A certain reaction has a rate given by
𝑚𝑜𝑙
𝑟𝐴 = 0.005𝐶𝐴2 , 𝑐𝑚3 .𝑚𝑖𝑛
If the concentration is to be expressed in mol/liter and time in hours, what would be the
value and unit of the rate constant?
Solution 2.6:
Given rate expression:
𝑟𝐴 = 0.005𝐶𝐴2
𝑚𝑜𝑙
𝑐𝑚3 . 𝑚𝑖𝑛
Change the unit
−𝑟
𝐴=0.005𝐶𝐴2
𝑚𝑜𝑙
1
(10−3 )𝑙𝑖𝑡∗ ℎ𝑟
60
−𝑟 𝐴 = 300𝐶𝐴2
𝑚𝑜𝑙
𝑙𝑖𝑡.ℎ𝑟
Value of rate constant =300
Unit of rate constant=
𝑚𝑜𝑙
𝑙𝑖𝑡.ℎ𝑟
Problem 2.7:
For a gas reaction at 400 K the rate is reported as
−
𝑑𝑝𝐴
𝑎𝑡𝑚
= 3.66𝑝𝐴2 ,
𝑑𝑡
ℎ𝑟
(a) What are the unit of the rate constant?
(b)What is the value of the rate constant for this reaction if the rate equation is expressed as
−𝑟
𝐴=
−1 𝑑𝑁𝐴
=𝑘𝐶𝐴2 ,
𝑉 𝑑𝑡
𝑚𝑜𝑙
𝑚3 .𝑠
Solution 2.7:
(a)
Here the rate equation is expressed in homogeneous & order of the reaction is second ,so the
unit of the rate constant K can be expressed by
K=(𝑡𝑖𝑚𝑒)−1 (𝑐𝑜𝑛𝑠)1−2
K=3.66(𝑡𝑖𝑚𝑒)−1 (𝑐𝑜𝑛𝑠)−1
(b)
here
−𝑟
𝐴=
−1 𝑑𝑁𝐴
=𝑘𝐶𝐴2 ,
𝑉 𝑑𝑡
𝑚𝑜𝑙
𝑚3 .𝑠
-𝑟𝐴 =k𝐶𝐴2
So, now using ideal gas low
PV=nRT
𝑃𝐴 =
−
𝑑𝐶𝐴 𝑅𝑇
𝑑𝑡
𝑛𝐴
𝑉
𝑅𝑇 = 𝐶𝐴 𝑅𝑇 
−𝑑𝑃𝐴
= 3.66(𝐶𝐴 𝑅𝑇)2 
𝑑𝑡
𝑑𝐶𝐴
𝑑𝑡
= 3.66𝑃𝐴2
= 3.66𝐶𝐴2 𝑅𝑇
 −𝑟𝐴 = 3.66(0.0820) ∗ 400 ∗ 𝐶𝐴2 (
𝑚𝑜𝑙
)
𝑙𝑖𝑡. ℎ𝑟
𝑚𝑜𝑙
 K=120.13 (ℎ𝑟 −1) ( 𝑙𝑖𝑡 )-1
Problem 2.8:
The decomposition of nitrous oxide is found to be proceed as follows
𝑁2 O— ›𝑁2 +
1
2
𝑘 [𝑁2 𝑂]²
1
𝑂2 ; −𝑟𝑁2 𝑂 = 1+𝑘
What is the order of reaction WRT 𝑁2 O, and overall?
2 [𝑁2 𝑂]
Solution 2.8:
If, 𝑘2 concentration is low, then order of reaction is 2nd order.
If Its concentration is high, the order of this reaction is 1st order.
Overall order of reaction = 2-1 = 1.
Therefore it is of 1st order.
Problem 2.9:
The pyrolysis of ethane proceeds with an activation energy of about 300
𝐾𝐽
𝑚𝑜𝑙
How much
faster is the decomposition at 650°C than at 500°C?
Solution 2.9:
E=300 𝑘𝐽⁄𝑚𝑜𝑙.
Let. Kinetic express by= -𝑟𝐴 =k𝑐𝐴𝑛
where 𝑘0 𝑒 −𝑒⁄𝑅𝑇
-𝑟𝐴 ∞𝑘0 𝑒 −𝑒⁄𝑅𝑇 for same can be
−𝑟𝐴1 𝑒 −𝑒⁄𝑅𝑇
−𝑟𝐴2
=𝑒 −𝑒⁄𝑅𝑇 = 𝑒
−𝑟𝐴650
𝑟𝐴500
−𝑒⁄ 1
𝑅(
𝑝𝑖
1
−𝑒
𝑇 −𝑇
− 𝑇 2 ) = 𝑒 𝑅 [ 𝑇1 𝑇 2 ]
1 2
300 650−500
= exp [2.314(650×500)]
= 7.58 Time faster
Problem 2.10:
A 1100-K n-nonane thermally cracks (breaks down into smaller molecules) 20 times as rapidly
as at 1000K. Find the activation energy for this decomposition.
Solution 2.10:
𝑟1100
𝑟1100𝐾 =20𝑟1000𝐾
𝑟1000
=20
1 Arrhenius theory
K=𝑘0 𝑒 −𝑒⁄𝑅𝑇
In same concentration 𝑘1 =𝑘2
−𝐸
𝑟1100
𝑟1000
=
𝑘1 𝑒 𝑅𝑇1100
−𝐸
= 20
𝑘2 𝑒 𝑅𝑇1000
Log20 =
−𝐸
𝑅
1
1
(1100 − 1000)
𝑙𝑜𝑔20∗𝑅
E=9.09∗.00001
𝐽
E=1189.96 𝑚𝑜𝑙
Problem 2.11:
In the mid nineteenth century the entomologist henri fibre noted that French ants busily
bustled about their business on hot days but rather sluggish on cool days. Checking this
result with Orezon ants, I find
Running
150
160
230
295
370
16
22
24
28
speed (m/hr)
Temperature 13
C
What activation represents this change in bustliness?
Solution 2.11:
1
Speed
lnS
T
(𝑇*10−3 )
150
5.01
13
3.4965
160
5.07
16
3.4602
230
5.44
22
3.3898
295
5.69
24
3.3670
370
5.91
28
3.3220
𝐸
Slope of graph = − 𝑅=-5420
E=(5420)*(8.314)
=45062 J/mol
lnS vs. ((1/T)*(10^-3))
6
5.9
5.8
5.7
5.6
5.5
5.4
5.3
5.2
5.1
5
4.9
3.3
3.35
3.4
3.45
3.5
3.55
Problem 2.12:
The maximum allowable temperature for a reactor is 800K. At present our operating set
point is 780K, The 20-K margin of safety to account for fluctuating feed, sluggish controls,
etc. Now, with a more sophisticated control system we would be able to raise our set point
to 792 K with the same margin of safety that we now have. By how much can the reaction
rate, hence, production rate, be raised by this change if the reaction taking place in the
reactor has an activation energy of 175 KJ/mole?
Solution 2.12:
As we know according to Arrhenius Law
(-ra) = ko Toe-Ea/RT
Here
To = 1
ln(r1/r2) = ln(k1/k2) = (Ea/R )[1/T1-1/T2] Here
T1= 780K
T2= 800 K
Ea= 175000 J/mol
R =8.314
Put in formula we get
r2 = .51 r1 -------- equation 1
similarly
for temperature
T3= 792K
T2= 800 K
We get
r3 = .77 r1------------ equation 2
Hence from Eq. 1 and 2 we get
r3 = 1.509 times of r2
Problem 2.13:
Every may 22 I plant one watermelon seed. I water it, I fight slugs, I pray, I watch my beauty
grow, and finally the day comes when the melon ripens, I than harvest and feast. Of course,
some years are sad, like 1980, when blue jay flew off with the seed. Anyway, 6 summers
were a pure joy and for this I have tabulated, the number of growing days vs. the mean day
time temperature, during the growing season. Does the temperature affect the growth
rate? If so, represent this by an activation energy.
Year
1976
1977
1982
1984
1985
1988
Growing
87
85
74
78
90
84
22.0
23.4
26.3
24.3
21.1
22.7
days
Mean
temp °C
Solution 2.13:
Temperature
1
( ∗ 10−3 )
𝑇
Days
ln(rate)=ln(𝑑𝑎𝑦𝑠)
22
3.390
87
-4.466
23.4
3.374
85
-4.443
26.3
3.341
74
-4.304
24.3
3.364
78
-4.357
21.1
3.400
90
-4.500
22.7
3.382
84
-4.431
𝐸
Slope of graph = − 𝑅 = 3392
E = 3392 * 8.314 = 28201 J/mol
1
Chart Title
-4.25
3.33
3.34
3.35
3.36
3.37
3.38
3.39
3.4
3.41
-4.3
-4.35
-4.4
-4.45
-4.5
y = -3.3922x + 7.0323
R² = 0.9441
-4.55
Example 2.14:
On typical summer days, field crickets nibble, jump, and chirp now and then, But at a night,
when great numbers congregate, chirping seems to become a serious business and tends to
be in unison.in 1897, A.E. Dolbear reported that this social chirping rate was dependent on
the temperature as given by
(Numbers of chirps in 15 s) + 40 = (temperature °F)
Assuming that the chirping rate is directly measured of the metabolic rate, find the
activation energy in kJ/mol of these crickets in the temperature range 60-80°F.
Solution 2.14:
(Numbers of chirps in 15 s) + 40 = (temperature °F)
(Numbers of chirps in 15 s) = (temperature °F)-40
For 60 °F
(Numbers of chirps in 15 s) = 20
For 80 °F
(Numbers of chirps in 15 s) = 40
dN = k *(metabolism rate)
after integration we get
40 – 20 =k(80-60)
k= 1 Numbers of chirps/°F
(-ra = metabolism rate) = ko Toe-Ea/RT
Here To = 1
For 60 °F
(-rA = metabolism rate) = 60
For 80 °F
(-rA = metabolism rate) = 80
ln(r1/r2) = = (Ea/R )[1/T1-1/T2]
ln(60/80) = (Ea/8.314 )[1/60-1/80]
Ea = -69.04 KJ/mol
Problem 2.15:
On doubling the concentration of the reactant, the rate of reaction triples. Find the reaction
order.
For the stoichiometry A + B (products) find the reaction order with respect to A and B.
Solution 2.15:
−𝑟𝐴 = 𝑘𝐶𝐴𝑛
𝑟2
𝑟1
=
𝑘(𝐶𝐴 )𝑛
𝑘𝐶𝐴𝑛
= 2𝑛
 n=1.585
CHAPTER NO. : - 2
UNSOLVED EXAMPLE: - 2.16 to 2.23
Submitted by:11BCH044 (SUREJA VISHAL)
13BCH044 (NAGPARA KUMAR)
13BCH156 (PATEL KISHAN)
12BCH157 (SUVAGIYA SHARAD)
2.16
CA
CB
-rA
4
1
2
1
1
1
1
8
4
Solution
Guess that,
–rA = K*CAp*CBq
In this Equation, There are three unknown k,p and q
From question we can make three equations
2 = K (4)p (1)q --------------------------------(1)
1 = K (1)p (1)q --------------------------------(2)
1 = K (1)p (8)q --------------------------------(3)
Now Equation (3) divided by Equation (2),
𝟒 𝐊 (𝟏)𝐩 (𝟖)𝐪
=
𝟏 𝐊 (𝟏)𝐩 (𝟏)𝐪
4 = (8)q
(2)2 = (2)3q
So, 3q = 2
q = 2/3
Now Equation (1) divided by Equation (2),
𝟐 𝐊 (𝟒)𝐩 (𝟏)𝐪
=
𝟏 𝐊 (𝟏)𝐩 (𝟏)𝐪
2 = (4) p
2 = (2)2p
So, 2p = 1
p=½
Put the values of p, q in Equation (2),
1 = K (1)1/2 (1)2/3
1 = K (1)1/3
K=1
So, from value of p, q, and K we get rate equation,
-rA = CA1/2 CB2/3
2.17
CA 2
2 3
CB 125 64 64
-rA 50 32 48
Solution
Guess that,
–rA = K*CAp*CBq
In this Equation, There are three unknown k,p and q
From question we can make three equations
50 = K (2)p (125)q --------------------------------(1)
32 = K (2)p (64)q --------------------------------(2)
48 = K (3)p (64)q --------------------------------(3)
Now Equation (3) divided by Equation (2),
𝟒𝟖 𝐊 (𝟑)𝐩 (𝟔𝟒)𝐪
=
𝟑𝟐 𝐊 (𝟐)𝐩 (𝟔𝟒)𝐪
3/2 = (3/2)p
So, p = 1
P=1
Now Equation (1) divided by Equation (2),
𝟓𝟎 𝐊 (𝟐)𝐩 (𝟏𝟐𝟓)𝐪
=
𝟑𝟐
𝐊 (𝟐)𝐩 (𝟔𝟒)𝐪
(5/4)2 = (5/4) 3q
So, 3q = 2
q = 2/3
Put the values of p, q in Equation (1),
50 = K (2)1 (125)2/3
50 = K (2) (5)2
K=1
So, from value of p, q, and K we get rate equation,
-rA = CA CB2/3
EXAMPLE 2.18:The decomposition of reactant A at 4000 𝑐 for pressure between 1 and 10 atm follows a
first order rate law.
A) Show that a mechanism similar to azo-methane decomposition
𝐀 + 𝐀 ↔ 𝐀∗ + 𝐀
𝐀∗ → 𝐑 + 𝐒
Is consistent with observed kinetics
Different mechanisms can be proposed to explain first order kinetics. To claim that this
method is correct in the face of other alternatives requires additional evidence.
B) For this purpose, what further experiments would you suggest we run and what
results we expect to find?
Solution
𝐊𝟏
𝐀 + 𝐀 → 𝐀∗ + 𝐀
𝐊𝟐
𝐀∗ + 𝐀 → 𝐀 + 𝐀
𝐊𝟑
𝐀∗ → 𝐑 + 𝐒
Rate of disappearance of A,
−𝐫𝐀 = 𝐊 𝟏 [𝐀]𝟐 − 𝐊 𝟐 [𝐀][𝐀∗ ]
And rate of formation of A,
𝐫𝐀∗ = 𝐊 𝟏 [𝐀]𝟐 − 𝐊 𝟐 [𝐀][𝐀∗ ] + 𝐊 𝟑 [𝐀∗ ]
Also the rate of formation of intermediate is zero therefore,
𝐫𝐀∗ = 𝟎
𝐊 𝟏 [𝐀]𝟐 = [𝐀∗ ](𝐊 𝟐 [𝐀] + 𝐊 𝟑 )
K1 [A]2
[A =
K 2 [A] + K 3
∗]
So,
K1 [A]2
−rA = K1 [A] − K 2 [
]
K 2 [A] + K 3
2
−𝐫𝐀 =
𝐊 𝟑 𝐊 𝟏 [𝐀]
𝐊𝟐
Hence it is a first order reaction
Now at low enough concentration of CA the rate of decomposition of A will reduce to
−𝐫𝐀 = 𝐊 𝟏 [𝐀]𝟐
That is it would be a second order reaction.
EXAMPLE 2.19
Show that the following scheme
𝐊𝟏
𝐍𝟐 𝐎𝟓 → 𝐍𝐎𝟐 + 𝐍𝐎𝟑∗
𝐊𝟐
𝐍𝐎𝟐 + 𝐍𝐎𝟑∗ → 𝐍𝟐 𝐎𝟓
𝐊𝟑
𝐍𝐎𝟑∗ → 𝐍𝐎∗ + 𝐎𝟐
𝐊𝟒
𝐍𝐎∗ + 𝐍𝐎𝟑∗ → 𝟐𝐍𝐎𝟐
Proposed by R.Ogg J.Chem.Phys., 15,337(1947) is consistent with and can explain the
first order decomposition of[𝑁2 𝑂5 ].
Solution
Rate of formation of NO2,
𝐫𝐍𝐎𝟐 = 𝐊 𝟏 [𝐍𝟐 𝐎𝟓 ] − 𝐊 𝟐 [𝐍𝐎𝟐 ][𝐍𝐎∗𝟑 ] −
𝐊𝟒
[𝐍𝐎∗ ][𝐍𝐎∗𝟑 ]
𝟐
Rate of disappearance ofN2 O5 ,
−𝐫𝐍𝟐 𝐎𝟓 = −𝐊 𝟏 [𝐍𝟐 𝐎𝟓 ] + 𝐊 𝟐 [𝐍𝐎𝟐 ][𝐍𝐎∗𝟑 ] ---- (1)
Rate of intermediate,
∗
𝐫𝐍𝐎
= 𝐊 𝟏 [𝐍𝟐 𝐎𝟓 ] − 𝐊 𝟐 [𝐍𝐎𝟐 ][𝐍𝐎𝟑∗ ] − 𝐊 𝟑 [𝐍𝐎𝟑∗ ] − 𝐊 𝟒 [𝐍𝐎𝟑∗ ][𝐍𝐎∗ ]
𝟑
Also,
∗
𝐫𝐍𝐎
=𝟎
𝟑
Therefore;
K1 [N2 O5 ] = K 2 [NO2 ][NO∗3 ] + K 3 [NO∗3 ] + K 4 [NO∗3 ][NO∗ ]
K1 [N2 O5 ] = [NO∗3 ] (K 2 [NO2 ] + K 3 + K 4 [NO∗ ])
[𝐍𝐎𝟑∗ ] =
𝐊 𝟏 [𝐍𝟐 𝐎𝟓 ]
𝐊 𝟐 [𝐍𝐎𝟐 ] + 𝐊 𝟑 + 𝐊 𝟒 [𝐍𝐎∗ ]
Rate of intermediate,
Also
∗
𝐫𝐍𝐎
=𝟎
And,
∗
rNO
= K 3 [NO∗3 ] − K 4 [NO∗3 ][NO∗ ]
[NO∗ ] =
K 3 [NO∗3 ]
K 4 [NO∗3 ]
[𝐍𝐎∗ ] =
𝐊𝟑
𝐊𝟒
Therefore,
NO∗3 =
K1 [N2 O5 ]
K 2 [NO2 ] + K 3 + K 4 [
𝐍𝐎𝟑∗ =
K3
]
K4
𝐊 𝟏 [𝐍𝟐 𝐎𝟓 ]
𝐊 𝟐 [𝐍𝐎𝟐 ] + 𝟐𝐊 𝟑
So, now put value in equation (1),
−rN2 O5 = −K1 [N2 O5 ] + K 2 [NO2 ][NO∗3 ]
−rN2 O5 = −K1 [N2 O5 ] + K 2 [NO2 ](
−𝐫𝐍𝟐 𝐎𝟓 = [𝐍𝟐 𝐎𝟓 ](𝐊 𝟐 [𝐍𝐎𝟐 ] (
−rN2O5 = [N2 O5 ]((
K1 [N2 O5 ]
)
K 2 [NO2 ] + 2K 3
𝐊𝟏
) − 𝐊𝟏)
𝐊 𝟐 [𝐍𝐎𝟐 ] + 𝟐𝐊 𝟑
K1 K 2 [NO2 ]
) − K1 )
K 2 [NO2 ] + 2K 3
−rN2O5 = K1 [N2 O5 ]((
−𝐫𝐍𝟐𝐎𝟓 = 𝐊 𝟏 [𝐍𝟐 𝐎𝟓 ] (
K 2 [NO2 ]
) − 1)
K 2 [NO2 ] + 2K 3
𝐊 𝟐 [𝐍𝐎𝟐 ] − 𝐊 𝟐 [𝐍𝐎𝟐 ] − 𝟐𝐊 𝟑
)
𝐊 𝟐 [𝐍𝐎𝟐 ] + 𝟐𝐊 𝟑
Assume K3 >>> K2, K 2 ≅ 0
−𝐫𝐍𝟐𝐎𝟓 = −𝐊 𝟏 [𝐍𝟐 𝐎𝟓 ]
Therefore this is a first order reaction
EXAMPLE-2.20
Experimental analysis shows that the homogeneous decomposition of ozone proceeds
with a rate
-r𝐎𝟑 = k [O3]2 [O2]-1
[1] What is the overall order of reaction?
[2] Suggest a two-step mechanism to explain this rate.
Solution:The homogeneous decomposition of ozone proceeds as,
2 O3
2 O2
And follows the rate law
-r𝐎𝟑 = K [O3]2 [O2]-1
The two step mechanism, consistent with the rate suggested is,
Step 1:- Fast, at equilibrium
O3
K1
O2 + O
O3
K2
O2 + O
STEP 2:- Slow
K3
O + O3
2 O2
The step 2 is the slower, rate determining step and accordingly, the reaction rate is
-r𝐎𝟑 =
−𝐝 [𝐎𝟑 ]
𝐝𝐭
= K3 [O3] [O]
Thus step 1 is fast and reversible, for this equilibrium step, we have
𝐊=
𝐊𝟏 [𝐎𝟐 ] [𝐎]
=
𝐊𝟐
[𝐎𝟑 ]
[𝐎 ] =
𝐊 [𝐎𝟑 ]
[𝐎𝟐 ]
Putting value of [O] from equation (2) into equation (1), we get
-rO3=
Let,
𝐝 [𝐎𝟑 ]
[𝐎𝟑 ]𝟐
𝐝𝐭
[𝐎𝟐 ]
= K3 * K
K = K3 * K
-rO2 = k [O3]2 [O2]-1
Overall order of reaction is 2 – 1 = 1
EXAMPLE 2.21:Under the influence of oxidizing agents hypo phosphorus acid is transformed into
phosphorus acid
H3PO2
oxidizing agent
H3PO3
The kinetics of this transformation present the following features. At a low concentration
of oxidizing agent.
rH3PO3 = K [oxidizing agent] [H3PO2]
At a high concentration of oxidizing agent
rH3PO3 = k [H*] [H3PO2]
To explain the observed kinetics it has been postulated that, with hydrogen ions as
catalyst normal unreactive H3PO2 is transformed reversibly int an active from the nature
of which is unknown, this intermediate then react with the oxidizing agent to give H 3PO3
show that this scheme does explain the observed kinetics.
Solution :Hypothesize
H3PO4 + H
K1
X* + H
H3PO4 + H
K2
X* + H
X* + Ox
K3
H3PO4
X* + Ox
K4
H3PO4
[ where X* is an unstable intermediate]
rH3PO3 = k3 [X*] [Ox] – k4 [H3PO3]
………(1)
r[X*] = k1 [H3PO2] [H+] – k2 [X*] [H+] - k3 [X*] [Ox] + k4 [H3PO3] =0
(steady state assumption)
Thus,
[X*] =
𝐊 𝟏 [𝐇𝟑 𝐏𝐎𝟐 ][𝐇+ ] +𝐊 𝟒 [𝐇𝟑 𝐏𝐎𝟑 ]
𝐊 𝟐 [𝐇 + ]+𝐊 𝟑 [𝐎𝐱 ]
………(2)
Assuming k4 = 0, replacing (2) in (1) then gives,
rH3PO3 =
𝐊 𝟏 𝐊 𝟑 [𝐇𝟑 𝐏𝐎𝟐 ][𝐇+ ] [𝐎𝐱 ]
𝐊 𝟐 [𝐇 + ]+𝐊𝟑 [𝐎 𝐱 ]
……...(3)
now when k3 [ox] >> k2 [H+]…i.e. high oxidized concentration equations (3) gives
rH3PO4 = K1 [H3PO2] [H+]
………(4)
on the other hand wen k2 [H+] >> k3 [Ox]..i.e. low oxidized concentration equation (3)
gives,
rH3PO3 =
𝐊𝟏 𝐊𝟑
𝐊𝟐
[H3PO2] [Ox]
………(5)
Equation (4) & (5) is fit in the hypohesized model is accepted.
EXAMPLE 2.22:Come up with (guess and then verify) a mechanism that is consist with the experimentally
found rate equation for the following reaction
2A+B
A 2B
With
r [A2B] = K [A] [B]
Solution
It is the non-elementary reaction. So, assume reaction mechanism,
𝐀 + 𝐁 ↔ 𝐀𝐁 ∗
𝐀𝐁 ∗ + 𝐀 ↔ 𝐀𝟐 𝐁
𝟐𝐀 + 𝐁 ↔ 𝐀𝟐 𝐁
2A
2A
A2* + B
A2B
K1
A2*
……...(1)
K2
A2
………(2)
K3
A2 B
………(3)
K4
A2* + B ………(4)
From reaction (3) & (4) rate of formation of r [A2B],
r [A2B] = K3 [A2*] [B] – K4 [A2B]
In reaction (4) A2B is dispersed.
Rate of disappearance,
𝐫𝐀 = 𝐊 𝟏 [𝐀]𝟐
Rate of formation of A,
……….(5)
𝐊 𝟏 [𝐀]𝟐
𝐫𝐀 =
𝟐
(2 moles of A consume give 1 mole of A * so ½)
𝐫[𝐀∗𝟐 ]
𝐊 𝟏 [𝐀]𝟐
=
− 𝐊 𝟐 [𝐀∗𝟐 ] − 𝐊 𝟑 [𝐀∗𝟐 ][𝐁] + 𝐊 𝟒 [𝐀𝟐 𝐁]
𝟐
𝐫[𝐀∗𝟐 ] = 𝟎
K1 [A]2
0=
− K 2 [A∗2 ] − K 3 [A∗2 ][B] + K 4 [A2 B]
2
K 3 [A∗2 ][B] + K 2 [A∗2 ] = K 4 [A2 B] +
[A∗2 ]
K1 [A]2
2
K1 [A]2
[K 3 [B] + K 2 ] = K 4 [A2 B] +
2
[𝐀∗𝟐 ] =
𝐊 𝟏 [𝐀]𝟐
𝟐
[𝐊 𝟑 [𝐁] + 𝐊 𝟐 ]
𝐊 𝟒 [𝐀 𝟐 𝐁 ] +
Put value of [A2*] in equation (5),
𝐫𝐀𝟐𝐁
rA2B = K 3
𝐊 𝟏 [𝐀]𝟐
𝐊 𝟒 [𝐀 𝟐 𝐁] +
𝟐 ) [𝐁] − 𝐊 [𝐀 𝐁]
= 𝐊𝟑 (
𝟒
𝟐
[𝐊 𝟑 [𝐁] + 𝐊 𝟐 ]
K1 [A]2
K [ A B] K 3 [ B]
[ B] + K 3 K 4 [ A 2 B] [ B ] − 4 2
− K 4 K 2 [ A 2 B]
[K 3 [B] + K 2 ]
2
𝐫𝐀𝟐𝐁
If K4 is very small, K 4 ≅ 0
𝐊 𝟏 [𝐀]𝟐
[𝐁 ] + 𝐊 𝟐 𝐊 𝟒 [𝐀 𝟐 𝐁]
𝐊𝟑
𝟐
=
[𝐊 𝟑 [𝐁] + 𝐊 𝟐 ]
𝐊 𝟏 [𝐀]𝟐
[𝐁]
𝐊𝟑
𝟐
=
[𝐊 𝟑 [𝐁] + 𝐊 𝟐 ]
𝐫𝐀𝟐𝐁
If K2 is very small, K 2 ≅ 0
𝐊 𝟏 𝐊 𝟑 [𝐀]𝟐 [𝐁]
=
𝐊 𝟑 [𝐁]
𝐫𝐀𝟐𝐁
So, from above equation we can conclude that our first assumption is wrong so,
mechanism also wrong.
Now, we can assume for second order reaction
𝐀 + 𝐁 ↔ 𝐀𝐁 ∗
𝐀𝐁 ∗ + 𝐀 ↔ 𝐀𝟐 𝐁
𝟐𝐀 + 𝐁 ↔ 𝐀𝟐 𝐁
2A + B
K1
AB*
……...(1)
AB*
K2
A+B
………(2)
A2 B
………(3)
A2* + B
………(4)
A2* + B
A2B
K3
K4
From reaction (3) and (4) rate of formation of r[A2B],
𝐫𝐀𝐁∗ = 𝐊 𝟏 [𝐀][𝐁] − 𝐊 𝟐 [𝐀𝐁 ∗ ] − 𝐊 𝟑 [𝐀𝐁 ∗ ][𝐁] + 𝐊 𝟒 [𝐀𝟐 𝐁]
𝐫[𝐀𝐁 ∗ ] = 𝟎
0 = K1 [A][B] − K 2 [AB ∗ ] − K 3 [AB ∗ ][B] + K 4 [A2 B]
[𝐀𝐁 ∗ ] =
𝐊 𝟏 [𝐀][𝐁] + 𝐊 𝟒 [𝐀𝟐 𝐁]
[𝐊 𝟑 [𝐀] + 𝐊 𝟐 ]
Put [AB*] value in equation (5),
𝐫𝐀𝟐𝐁 = 𝐊 𝟑 (
𝐊 𝟏 [𝐀][𝐁] + 𝐊 𝟒 [𝐀𝟐 𝐁]
) [𝐀] − 𝐊 𝟒 [𝐀𝟐 𝐁]
[𝐊 𝟑 [𝐀] + 𝐊 𝟐 ]
…(5)
𝐫𝐀𝟐𝐁 = 𝐊 𝟑 𝐊 𝟏 [𝐀]𝟐 [𝐁] + 𝐊 𝟑 𝐊 𝟒 [𝐀𝟐 𝐁] [𝐀] −
𝐊 𝟒 𝐊 𝟑 [𝐀][𝐀𝟐 𝐁]
− 𝐊 𝟒 𝐊 𝟐 [ 𝐀 𝟐 𝐁]
[𝐊 𝟑 [𝐀] + 𝐊 𝟐 ]
If K4 is very small, K 4 ≅ 0
𝐫𝐀𝟐𝐁
𝐊 𝟑 𝐊 𝟏 [𝐀]𝟐 [𝐁]
=
[𝐊 𝟑 [𝐀] + 𝐊 𝟐 ]
𝐫𝐀𝟐𝐁
𝐊 𝟏 𝐊 𝟑 [𝐀]𝟐 [𝐁]
=
𝐊 𝟑 [𝐀]
If K2 is very small, K 2 ≅ 0
𝐫𝐀𝟐𝐁 = 𝐊 𝟏 [𝐀][𝐁]
EXAMPLE 2.23:Mechanism for enzyme catalyzed reactions. To explain the kinetics of enzyme-substratereactions, Michaelis and Menten came up with the following mechanism, which uses an
equilibrium assumption
𝐊𝟏
𝐀+𝐄 → 𝐗
𝐊𝟐
𝐗 → 𝐀 +𝐄
with 𝐊 =
𝐗
𝐀𝐄
=
𝐊𝟏
𝐊𝟐
𝑲𝟑
𝐗 → 𝐑+𝐄
𝑬𝟎 = 𝐄 + 𝑿
And where [E0] represents the total enzyme and [E] represents the free unattached
enzyme.
G.E. Briggs and J.B.S Haldane, Bochem.J. 19,338 (1925), on the other hand employed a
steady state assumption in place of the equilibrium assumption
𝑲𝟏
𝐀+𝐄 → 𝐗
𝑲𝟐
𝐗 → 𝐀 +𝑬
𝑬𝟎 = 𝐄 + 𝑿
What final rate from -ra in terms of [A], [E0], K1, K2 and K3 does
1) The Michalels and Menten mechanism gives?
2) The Briggs-Haldane mechanism give?
Solution
𝑲𝟏
𝐀 + 𝐄 → 𝐗 ------ (1)
𝑲𝟐
𝐗 → 𝐀 + 𝑬 ------ (2)
𝑲𝟑
𝐗 → 𝐑 + 𝐄 ------ (3)
𝐄𝟎 = 𝐄 + 𝐗 ------ (4)
M-M assume that the reverse reaction of (1) approach equilibrium quickly, or
𝐊=
B-H assume that quickly
𝐝𝐲
𝐝𝐭
𝐗
𝐊𝟏
=
𝐀𝐄 𝐊 𝟐
= 𝟎 ------ (5)
for Michacli’s – Menten
From equation (3)
𝐫𝐑 = 𝐊 𝟑 𝐗
From equation (5)
𝐗=
𝐊𝟏
𝐊𝟐
𝐀𝐄 ------ (6)
Eliminate E with (4)
𝐗=
𝐊𝟏
𝐊𝟐
𝐀(𝐄𝟎 − 𝐗)------ (7)
Or
𝐗=
Equation (8) in (7) gives
𝐊𝟏
𝐊𝟐
𝐀𝐄𝟎 / 𝟏 +
𝐊𝟏
𝐊𝟐
𝐀 ------ (8)
𝐫𝐑 =
𝐊 𝟑 𝐀𝐄𝟎
𝐊𝟐
+𝐀
𝐊𝟏
These equation gives essentially the same result.
(a) For the Briggs - Haldane Mechanisms
From equation (3),
𝐫𝐑 = 𝐊 𝟑 𝐗 ----- (9)
From equation (6)
𝐝𝐲
= 𝟎 = 𝐊 𝟏 𝐀𝐄 − (𝐊 𝟐 + 𝐊 𝟑 )𝐗
𝐝𝐭
Eliminate E with (A)
𝐊 𝟏 𝐀(𝐄𝟎 − 𝐗) − (𝐊 𝟐 + 𝐊 𝟑 )𝐗 = 𝟎
Or
𝐗=
𝐊 𝟏 𝐀𝐄𝟎
𝐊𝟏𝐀 + 𝐊𝟐 + 𝐊𝟑
Equation (10) in (9) gives
𝐫𝐑 =
𝐊 𝟑 𝐀𝐄𝟎
𝐊𝟐+ 𝐊𝟑
+𝐀
𝐊𝟏
---------- (10)
These equation give essentially the same result.
𝐊𝟐+ 𝐊𝟑
𝐊𝟏
(Michaclis const.)
CRE-1 Term Paper
Example 3.1-3.11
Submitted By:
Yagnesh Khambhadiya (13BCH025)
Chirag Mangrola (13BCH032)
Fenil Shah (13BCH052)
Ravindra Thummar (13BCH058)
Since the reaction order, hence concentration dependency is not known we
are not given enough information to find the rate of reaction at higher
concentration.
-ra = kCa
0.2 = k * 1
K= 0.2 sec-1
-ra = kCa =0.2*10 = 2 mol/l.s
For given first order reaction by equation,
(
) = Kt
For Xa = 0.5 then t = 5 min.
So, K = 0.1386
Now for 75% conversion Xa= 0.75 & K = 0.1386
(
) = 0.1386 * t
t = 10 min
So, for 75% conversion time required is 10min
For the second order disappearance of single reactant gives,
kt=
*(
)
Now, for 50% conversion
*(
T50 =
( )
( )
) = 5 min.
Now, for 75% conversion
*(
T75 =
( )
( )
) = 15 min.
So, extra time needed is 10 minutes.
-ra = k
Cao
=k
∫
=k
2[Case I : Xa= 0.75 when t = 10 min
2[k’ = 0.1
Case II : Xa= ? when t = 30 min
2[Xa=0.75
Since the fractional disappearance is independent of initial concentration we have
first order rate equation,
(
) = Kt
Where, Ca = Monomer concentration
We also can find the rate constant, thus replacing values.
Case I: Cao= 0.04 mol/l
= k*34
So, k= 0.00657
.
Case II:. Cao= 0.8 mol/l
= k*34
k= 0.00657
.
Hence the rate of disappearance of monomer is given by,
-ra = (0.00657
)*C
Cao = 1
t = 8min then conversion 80%
t = 18 mins then conversion 90%
Assume the reaction is second order
-ra = k*
=k
t
= k* *8
k=0.5 (mol/l)-1(min)-1
= k* *18
k=0.5 (mol/l)-1(min)-1
From the same value of k in both case, the reaction is second order
-ra = k*
-ra = 0.5*
Magoos batting habits and his losses can be described by,
( ) = kA
Where, A = money in hand
Now, t=0 then A0 = 180
t=2hrs then A= 135
K= *
( )= *
(
) = 0.144
After raise we have,
t= 0 then A0=?
t= 3hrs then A= 135
For unchanged habits,
(
) = 0.144*3
A0= 208
Hence his raise is 208-180 = 28
Equimolor quantities of H2 and NO are used…
Cao= Cbo
Co = Cao + Cbo
Co = 2Cao
Ca α Pa
Pao = Po /2
Cao=
Cao=
Po= 200 mmHg= 0.263 atm
=5.414 * 10-3
Cao =
ln Cao = -5.22
t1/2 = 265 sec
ln t1/2 = 5.58
P(mm Hg)
P(atm)
Cao* 10^-3
ln Cao
t1/2
ln t1/2
200
0.263
5.414
-5.22
265
5.58
240
0.316
6.5
-5.04
186
5.23
280
0.368
7.57
-4.88
115
4.74
320
0.421
8.67
-4.75
104
4.64
360
0.474
9.76
-4.63
67
4.2
6
5
y = 2.2756x - 6.2815
R² = 0.9786
ln t1/2
4
3
2
1
0
4.6
4.8
5
-ln Ca0
Slope = (n-1)=2.275
n = 3.275
5.2
5.4
For a first order reversible reaction,
A
R
Where,
= Reversible reaction
K1 = Forward reaction rate const.
K2= Backward reaction rate const.
Ca0 = 0.05
Cr0 = 0
M= 0/0.5=0
Xae = 0.67
The integral conversion in a batch reactor is given by,
-ln(1-
) = k1 * t *
Replacing values we then find,
-ln(1-
) = k1 * t *
…..(1)
k1= 0.0577
Now from thermodynamics we know that,
K=
=
=
So, K1= (0.667/0.332) K2…… (2)
From equation (1) & (2)
K1= 0.057750
K2= 0.028875
Thus, rate expression for the disappearance of A.
-ra = 0.057750 Ca - 0.028875 Cr
For the second order disappearance of single reactant gives,
k*t=
*(
) ….. (1)
Here it is given that Ca0 = 2.03
So, Xa =
& Ca = 1.97
by putting values of Ca & Cao ,Xa will be 0.029
So, putting value of Xa and Cao in equation (1)
K (1) =
*(
)
K = 0.0147
So, for second order rate expression is –ra = 0.0147(
)
From the table of data,
Reaction is second order.
0.006
y = 1E-05x + 0.001
R² = 1
0.005
Ca
0.004
1000
500
333
250
200
0.003
0.002
0.001
0
0
200
400
k = 1 * 10-5 (mol/l)-1(min)-1
t = 5*60 =300 min
Cao = 500 mol/m3
= 0.003 +
Ca = 200 mol/m3
Xa =
t
=
= 0.6
600
0
100
200
300
400
1/ca
0.001
0.002
0.003003
0.004
0.005
Ex - 3.12
Aqueous A at a concentration C,, = 1 mollliter is introduced into a batch reactor where it reacts
away to form product R according to stoichiometry A
R. The concentration of A in the reactor
is monitored at various
times, as shown below:
T (min)
Ca mol/m3
0
1000
100
500
200
333
Find the rate for the reaction
Ans: For batch reactor t = -cao∫
, Cao = 100 mol/m3
= ln
ln(cao/ca)
0
.693
1.099
1.386
1.609
t
0
100
200
300
400
slope =
0.0045 =
0.0045 =
K = 4.5 min-1
300
250
400
200
Ex – 3.13
Betahundert Bash by likes to play the gaming tables for relaxation. He does not expect to win,
and he doesn't, so he picks games in which losses area given small fraction of the money bet. He
plays steadily without a break, and the sizes of his bets are proportional to the money he has. If at
"galloping dominoes" it takes him 4 hours to lose half of his money and it takes him 2 hours to
lose half of his money at "chuk-a-luck," how long can he play both games simultaneously if he
starts with $1000 and quits when he has $10 left, which is just enough for a quick nip and carfare
home?
Ans:
Money → Game 1
Money → Game2
Game 1:
Game 2:
K1 =0 .693 / (t1/2 )1
K1 = 0.693 / (t1/2 )2
=0 .693 / 4
= 0.693 / 2
= 0.17325 hour-1
= 0.3465 hour-1
(- rate of reaction) = (K1 + K2).CA
= M e-(k1+k2)
M = M0 e-(k1+k2)t
10 = 1000 e-(0.3465 + 0.17325)t
0.01 = e-(0.5197t)
t = 8.82 hour
Ex - 3.14
For the elementary reactions in series
A
R
S
, K1 = K2 at t = 0
CA0 = CA , CRO = CSO = 0
find the maximum concentration of R and when it is reached.
Ans :
-ra = K1Ca
K2CR
= K1Ca
Ca = Ca0e-Kt
CR = ∫
+ra = K1Ca – K2CR
= K1Cao e-K1t - K2Cr
+K2Cr =K1Ca0 e-K1t
K1 CA0e-k1t e-k2t. dt + c
CReK2t= K1CA0 t + c
at t = 0,
CR = K1CA0t . e-kt
CR = 0,
C = 0,
k1cA0e-kt+ k1cA0e-kt .t. (-k) = 0
k1 – k12.t = 0
t= ⁄
+rS =
. e(-1/k1) . k1
(cr)max= k1cA0 .
⁄
Cr =
Ex – 3.15
At room temperature sucrose is hydrolyzed by the catalytic action of the enzyme sucrase as
follows:
Sucrose → product
Starting with a sucrose concentration CAo = 1.0 millimol/liter and an enzyme concentration
CEO = 0.01 millimollliter, the following kinetic data are obtained in a batch reactor
(concentrations calculated from optical
rotation measurements):
CA
t
.84
1
.68
2
.53
3
.38
4
.27
5
.16
6
.09
7
.04
8
.018
9
.006
10
.0025
11
Determine whether these data can be reasonably fitted by a kinetic equation of the MichaelisMenten type, or
-rA =
If the fit is reasonable, evaluate the constants K3 and Cm. Solve by the integral method.
Ans:
-rA =
=
∫
=∫
-(CA – CA0) – CM ln
T=
= K3 . t. CE0
.(
– CA) +
Compare with , Y = mx + c
m=
c=
– CA]
x = ln [
Y=time
-.665
-.294
0.104
0.587
1.039
1.672
2.317
3.178
3.999
5.109
1
2
3
4
5
6
7
8
9
10
Slop = 1.5511
Intercept = 2.885
C = 2.885 =
K3 =
K3 = 35.02
M = slope = 1.5511 =
Cm = 1.5511 * 35.02 * 0.01
Cm = 0.5431
Ex - 16.
At room temperature sucrose is hydrolyzed by the catalytic action of the enzyme sucrase as
follows:
Sucrose → product
Starting with a sucrose concentration Cao = 1.0 millimol/liter and an enzyme concentration CEO
= 0.01 millimollliter, the following kinetic data are obtained in a batch reactor (concentrations
calculated from optical rotation measurements):
CA
.84
.68
.53
.38
.27
.16
.09
.04
.018
.006
.0025
t
1
2
3
4
5
6
7
8
9
10
11
Determine whether these data can be reasonably fitted by a kinetic equation of the MichaelisMenten type, or
-rA =
If the fit is reasonable, evaluate the constants K3 and Cm. Solve by the diffrential method.
Ans:
-rA =
=
=
+
Plot CA v/s t & take slope at different CA
CA
0.38
0.225
0.1
0.075
0.139
0.097
0.052
0.042
2.63
4.44
10
12.82
Slope = 1.623
Intercept = 3.0035
=
7.2
10.3
19.23
23.8
=
K = 33.33
Slope = 1.623 =
M = 1.623 * 33.33*0.01
M = 0.54
Ex - 17
Enzyme E catalyzes the transformation of reactant A to product R as follows:
A
-rA =
If we introduce enzyme (CEO = 0.001 mollliter) and reactant (CAo = 10mollliter) into a batch
reactor and let the reaction proceed, find the time needed for the concentration of reactant to drop
to 0.025 mollliter. Note that the concentration of enzyme remains unchanged during the reaction.
Ans :
CEO = 0.001 M
CA0 = 10 M
For batch reactor, t = -CAO∫
= -CAO ∫
Ceo = .001M
= -CAO ∫
+
= - CAO [
* ln CA +
8 CA ]
= -10 [ -3.678 - 7.305 ]
t = 109.87 min.
Ex - 18
An ampoule of radioactive Kr-89 (half life = 76 minutes) is set aside for a day. What does this do
to the activity of the ampoule? Note that radioactive decay is a first-order process.
Ans :
t1/2 = 76 min.
k=
= 9.1184 * 10-3
t = 1440min-1
CA = CAO e
–(.00911*1440)
= CAO . 1.98*10-6
% concentration = ( CAO -
⁄
) * 100
= (
)* 100
= 99.99 %
Activity reduced to 99.99 %
Ex – 19
Find the conversion after 1 hour in a batch reactor for
A→R , -ra = √
, CA0 = 1M
Ans: For batch reactor,
t
= -CAO∫
CAO = 1M
1 = -1 ∫
-1 = 2 [√
-0.5 = √
]
-1
CA
= 0.25
XA
= 1=1 -
⁄
⁄
XA = 0.75
Ex - 20.
M. Hellin and J. C. Jungers, Bull. soc. chim. France, 386 (1957), present the data in Table P3.20
on the reaction of sulfuric acid with diethylsulfate in aqueous solution at 22.9 degree Celsius:
H2SO4 + (C2H5)2SO4 → 2C2H5SO4H
Initial concentrations of H2S04 and (C2H5)2SO4 are each 5.5 mollliter. Find a rate equation for
this reaction.
t, (min.)
0
41
48
55
75
96
127
146
162
C2H5SO4H,(M)
0
1.18
1.38
.63
2.2
2.75
3.31
3.76
3.81
t, (min.)
180
194
212
267
318
368
379
410
∞
C2H5SO4H,(M)
4.11
4.31
4.45
4.86
5.15
5.32
5.35
5.42
5.8
Ans:
H2SO4 + (C2H5)2SO4 → 2 C2H5SO4
A +
B → 2R
CAO = CBO = 5.5 M
-ra = K CA CB
= K CAO . CBO. (1-XA). (1-XB)
= K CAO. (1-XA).(1-MXA)
= K. (1-XA)2
= K. (1-XA)2
= (1 – Xa)2 .K
∫
=K∫
=
Xa
0.107
0.125
0.148
0.204
0.25
0.346
0.392
0.119
0.142
0.173
0.256
0.33
0.529
0.644
Slope = .0027
= .0027
K = 5.5 * 0.0027
( CAO.XA = CBO. XB )
K = 0.01485 min
-ra = 0.01485 CA. CB
Ex – 21
A small reaction bomb fitted with a sensitive pressure-measuring device is flushed out and then
filled with pure reactant A at 1-atm pressure. The operation is carried out at 250C, a temperature
low enough that the reaction does not proceed to any appreciable extent. The temperature is then
raised as rapidly as possible to 100°C by plunging the bomb into boiling water, and the readings
in Table are obtained. The stoichiometry of the reaction is 2A → B, and after leaving the bomb
in the bath over the weekend the contents are analyzed for A; none can be found. Find a rate
equation in units of moles, liters, and minutes which will satisfactorily fit the data.
T,(min)
1
2
3
∏ (atm)
1.14
1.04
0.982
T,(min)
7
8
9
∏ (atm)
0.85
0.832
0.815
4
5
6
0.94
0.905
0.87
10
15
20
0.8
0.754
0.728
Ans:
Constant volume reaction
V1 = V2
P2 = (
⁄
). P1
⁄
= (
). 1
= 1.25 atm
PA = PA0 – ( ⁄
)(P – P0)
= 1.25 –
). (1.14 – 1.25 )
⁄
= 2P -2P0 + P0
= 2P – P0
⁄
CA1 =
= 0.041
t
0
CA 0.041
1
0.034
2
0.027
3
0.023
4
0.021
5
0.018
6
0.016
7
0.015
Assume 2nd order reaction
K={[ ]–[
K1 = [
-
]*
]}*
⁄)
= 5.02 M-1 min.-1
8
0.013
9
0.012
10
0.0114
15
0.0089
20
0.0067
K2 = [
-
]*
= 6.32 M-1 min.-1
K3 = [
-
]*
= 6.36 M-1 min.-1
K4 = [
-
]*
= 5.81 M-1 min.-1
K5 = [
-
]*
= 6.23 M-1 min.-1
K6 = [
-
]*
= 6.35 M-1 min.-1
There is not much variation in the value of K.
Avg. K =
K =
6.067
. min-1
Rate equation - rA = 6.067 CA2
Ex. 4.1
Gaseous Feed,
A+B→R+S
CA0=100 , CB0=200, XA =0.8
For XA=0,VI=100A+200B+0R+0S=300,
For XA=1, VF=0A+100B+100R+100S=300,
So εa=0
CA= CA0(1- XA) = 100(1-0.8)=20 …ans(1)
XBCB0= bCA0XA
XB= CA0XA/ CB0=(100/200)(0.8)=0.4
…ans(2)
CB= CB0(1- XB)=200(1-0.4)=120 …ans(3)
Ex. 4.2
Dilute aqeous feed, So εa=0
A+2B→R+S
CA0=100 , CB0=100,CA=20
CA= CA0(1- XA) , So XA =1-(20/100)=0.8 …ans(1)
XBCB0= bCA0XA,
XB= CA0XA/ CB0=2(100/100)(0.8)=1.6(impossible) …ans(2 & 3 )
Ex. 4.3
Gaseous Feed,
A+B→R
CA0=200 , CB0=100, CA =50
For XA=0 ,VI=200A+100B+0R=300, So εa=( VF- VI)/ VI =(100-300)/300 = -0.667
For XA=1 , VF=0A-100B+200R=100,
XA=(CA0-CA)/(CA0+ εaXA) = (200-50)/(200-0.667*50) = 0.9
XBCB0= bCA0XA
XB= CA0XA/ CB0=(200/100)(0.9)=1.8 (impossible)…ans(2)
CB= CB0(1- XB)=200(1-1.8)= impossible …ans(3)
Ex. 4.4
Gaseous Feed,
A+2B→R
CA0=100 , CB0=100, CA =20
For XA=0, VI=100A+100B+0R=200, So εa=( VF- VI)/ VI =(0-200)/200 = -1
For XA=1, VF=0A-100B+100R=0,
XA=(CA0-CA)/(CA0+ εaXA) = (100-20)/(100-2*50) = 1
XBCB0= bCA0XA
XB= CA0XA/ CB0=(100/100)(1)=2 (impossible)…ans(2)
CB= CB0(1- XB)=100(1-2)= impossible …ans(3)
Ex. 4.5
Gaseous Feed,
A+B→2R
CA0=100 , CB0=200, T0=400 K , T= 300 K, π0=4 atm , π=3 atm , CA =20
For XA=0, VI=100A+200B+0R=300, So εa=( VF- VI)/ VI = 0
For XA=1, VF=0A+100B+200R=300,
XA=(CA0-CA)(Tπ0/T0π)/(CA0+ εaXA) (Tπ0/T0π)
= (100-20)(300*4/400/3)/(100+0*20)( 300*4/400/3) = 0.8
XBCB0= bCA0XA
XB= CA0XA/ CB0=(100/200) = (0.8) …ans(2)
CB= CB0(1- XB)=200(1-0.4) = 120 …ans(3)
Ex. 4.6
Gaseous Feed,
A+B→5R
CA0=100 , CB0=200, T0=1000 K , T= 400 K, π0=5 atm , π=4 atm , CA =20
For XA=0, VI=100A+200B+0R=300, So εa=( VF- VI)/ VI = 2
For XA=1, VF=0A+100B+500R=600,
XA=(CA0-CA)(Tπ0/T0π)/(CA0+ εaXA) (Tπ0/T0π)
= (100-20)(400*5/1000/4)/(100+2*20)( 400*5/1000/4) = 0.75 …ans(1)
XBCB0= bCA0XA
CB/CA0= ((CB0/ CA0)-b XA/a )/(Tπ0/T0π)=((200/100)-0.75)/( 400*5/1000/4) ,
So CB= 100 …ans(2)
XBCB0= bCA0XA
XB=0.75 …ans(3)
Ex. 4.7
1 lit/min →
→28 lit/min
A→31R
X=Popcorn produced
(1-X)=disappearance of A
So, outlet = 31X+(1-X)1=28 lit/min
X=27/30=0.9
Ans : 90% popcorn popped.
Ex. 5.1
2A→R+2S
XA=0.9
Space velocity(s) = 1/min
So, space time(τ)=1/s
=1 min
Here, εa=(3-2)/2 = 0.5
…ans(1)
VF = V0(1+ εa XA)
=V0(1+0.5*0.9)
=1.45 V0
So. Holding time, ŧ = V/Vf
= V/V0*1.45
=0.69 min
But in the plug flow reactor since the gas reacts progressively as it passes through the reactor,
expands correspondly , therefore holding time = 1 min.
…ans (2)
Ex 5.2
Liquid phase reaction , XA=0.7 , time = 13 min , first order reaction
For first order reaction,
-rA=k CA
-ln (1- XA) = kt , so, k=0.0926 / min
Case -1(for mixed flow reactor),
Here τm= CA0*XA/(-rA)
= CA0*XA/(k*CA0(1- XA))
=(0.7/(0.926*0.3))
τm= 25.2 min
… ans(1)
Case -2(for plug flow reactor),
Same as batch reactor, τp = τ = 13 min
…ans(2)
Ex. 5.3
Aqueous monomer A , V = 2 litre , ν = 4 lit/min , CA0 = 1 mol/lit , CA = 0.01 mol/lit
A +A → R + A → S + A → T + …
Here, τ = V/ν = 2/4 = 0.5 min
For reaction rate of A,
-rA = (CA0 – CA) / τ
=(1-0.01)/0.5 = 1.98 mol/(lit*min)
For W,
…ans(1)
CW = 0.0002 mol/lit
-rW = (-CW0 +CW) / τ
= (0.0002-0)/0.5 = 0.0004 mol/(lit*min)
…ans(2).
Ex 5.4
-rA=k CA1.5
XA=0.7
τ = V/ V0
For mixed flow reactor ,
= XA/ (-rA)
= XA*CA0/(k*CA01.5*(1- XA)1.5)
From above equation ,
…equ (1)
k = XA/(( CA00.5*(1- XA)1.5)*( V/ V0))
Putting XA=0.7
So, k = 4.260/( CA00.5(V/ V0))
Now if volume is doubles so V’ = 2V
So , τ = V’/ V0
= XA*CA0/(k*CA01.5*(1- XA)1.5)
8.52 = XA/(1- XA)1.5
… from equ (1)
Take square both side ,
72.6 = XA2/(1- XA)3
So, 72.6 – 72.6 XA3 + 216 XA2 - 217 XA =0
Solving by numerical method we get XA=0.7947
… ans
Ex. 5.5
A +B → R
-rA=200 CA*CB
CA0= 100 , CB0 = 200 , ν = 400 lit/min , XA=0.999
For second order reaction , (different reactants)
k τ *CA0*(M-1)= ln((M- XA)/(M* (1- XA))
here M=2, so substituting values,
V/ν =ln((2-0.999)/2(1-0.999)) / (200*0.10*1)
= 0.31
…ans.
So V=0.31 * 400 = 124 litre
Ex. 5.6
A↔R (reversible reaction)
-rA=0.04 CA*0.01 CB
so k1 = 0.04 , k2 = 0.01, M= CR0/ CA0 = 0
k1/ k2 = XAe/(1- XAe) from there XAe = 0.8
…ans (1)
For first order plug flow reactor (reversible reation),
XAf
V/(ν* CA0) = ∫0
𝑑𝑋𝐴/(-rA)
-rA=0.04 CA0(1- XA) - 0.01 CA0(M + XA)
So by integrating term we get,
-ln (1 - XA/ XAe )
= k1* V/ν*(M + 1) / (M + XAe )
Substituting all values,we get
-ln (1 - XA/ XAe ) = 0.64
XA/ XAe = 0.48
XA = 0.38
…ans
Ex 5.7
Half time for this reaction (τ1/2) = 5.2 days
Mean resident time (τ) = 30 days
So k = 0.693/ τ1/2 =0.693/5.2 = 0.1333 /day
For mixed flow reactor, CA/ CA0 = 1 / ( 1 + k τ) = 0.2
So fraction of activity = 1 - CA/ CA0 = 0.8
Ans = 80%
Ex. 5.8
A↔R (reversible reaction)
-rA=0.04 CA*0.01 CB
so k1 = 0.04 , k2 = 0.01, M= CR0/ CA0 = 0
k1/ k2 = XAe/(1- XAe) from there XAe = 0.8
For first order mixed flow reactor (reversible reation),
V/(ν* CA0) = XA /(-rA)
-rA=0.04 CA0(1- XA) - 0.01 CA0(M + XA)
V/ν = 0.10 XA / (0.04 CA0(1- XA) - 0.01 CA0(M + XA))
Substituting all values,we get
XA= 0.8-0.8 XA -0.01 XA
…ans
XA = 0.40
Ex. 5.9
A + E(enzyme) → R
-rA=0.1 CA/(1+0.5 CA)
Here, ν = 25 lit/min , CA0 = 2 mol/ litr , XA= 0.95
CAf
Now, V/(ν) = ∫CA0 𝑑𝐶𝐴/(-rA)
Substituting -rA=0.1 CA/(1+0.5 CA) and CA= CA0 (1 - XA)
So, V/(ν) = [-10(ln (0.10/2) - 5(0.10-2)] = 39.452
So, V = 986.25 lit
…ans.
…ans (1)
Ex 5.10
A → 2.5(all products)
-rA=10 CA , plug flow reactor
Here, ν = 100 lit/min , CA0 = 2 mol/ litr , V= 22 L
From the stoichiometry εa = (2.5-1)/1 = 1.5
If we write equation for plug flow reaction,
XAf
(V/ν)* CA0 = ∫0
𝑑𝑋𝐴/(-rA)
CA = CA0(1- XA)/(1+ εa XA)
So (V/ν)* CA0 * k = - (1+ εa) ln(1- XA) -1.5 XA
4.4 = -2.5 ln (1- XA) - 1.5 XA
XA = 0.733
…ans.
Group 12
13bch024, 13bch033, 13bch048, 13bch059
Solution 5.11:
Solution-
A→R
−rA = 0.1CA / (1 + 0.5CA )
According to the performance equation of mixed flow reactor,
τm
CA0
X
= −rA
A
- (1)
Now substituting the value of XA from the equation,
CA = CAO (1 − XA ) - (2)
We get,
τm =
CAO −CA
0.1CA
(1 + 0.5CA )
- (3)
Now, conversion is 95 % so, XA = 0.95
Therefore, substituting the values in the equation, we get
τm =
(2−0.1)(1+0.5(0.1))
0.1(0.1)
= 199.5 min
Therefore,
V = τm vo = 199.5(25) L = 4987.5 L = 5 m3
V = 5 𝐦𝟑
1
Problem 5.12:
Solution-
A+B→R
−rA = 200CA CB (mol)/(litre).(min)
Now,
CA = 0.1 mol/L
CB = 0.4 mol/L
vo = 400 L/min
Hence, from the equation: CA = CAO (1 − XA ) , we get XA = 0.99
τm =
XA .CA0
−rA
,
CA = CAO (1 − XA )
CB = CAO (MB − (b/a)XA )
Here, MB = 200/100 = 2
Now, b = a = 1
−rA = 200(0.1)2 (1 − XA )(2 − XA )
= 0.0202
τm =
XA .CA0
−rA
= τm =
(0.99)(0.1)
0.0202
= 49.9 min
Therefore,
V = τm vo = 49.9(400) L = 19960 L = 19.96 m3
V = 19.96 𝐦𝟑
2
Problem 5.13
Solution-
−rPH3 = 10CPH3 h−1 ; k = 10
4PH3 → P4 (g) + 6H2
XA = 0.75
P = 11.4 atm
FAO = 10 mol/hr
PH3 = (2/3) of feed
Inerts = (1/3) of feed
As it is a gaseous reaction we need to find ε𝐴 Hence,
ε𝐴 = (1 + 6 - 4)(2) / 4(3) = 0.5
The reaction is of a first order hence, the integral law can be written as
τp =
1
[(1 + ε𝐴 )(-ln(1-XA ) - ε𝐴 XA ]
10
τp = 0.17
CAO =
v0 =
𝑃𝐴𝑂
𝑅𝑇𝑜
𝐹𝐴0
𝐶𝐴0
=
11.4(2/3)
0.082(649+273)
= 0.1 mol/L
= 10/0.1 = 100 L/h
Therefore,
V = τp vo = 0.17(100) L = 17 L
V = 17 L
3
Problem 5.14
Solution-
3A → R
−rA = 54 mmol/ L.min = 0.054 mmol/ L.min
CAO = 0.6 mol/L
CA = 0.33 mol/L
FAO = 0.54 mol/min
Now,
ε𝐴 = 2/3
Hence,
CA =
𝐶𝐴0 (1− XA )
(1+ ε𝐴 XA )
(1−XA )
330 = 660
therefore, XA = 0.75
2
(1+(3)XA )
Therefore,
0.75 𝑑𝑋𝐴 𝐶𝐴0 XA
τp = ∫
= −𝑟 = 9.17 min
0
−𝑟𝐴
𝐴
F
V = τp vo = τp CAO = 9.17(540/660) = 7.5 L
AO
4
Problem 5.15
Solution-
−𝑟𝐴 = 0.05 CA 2 mol/L s
2A→R
Now,
τm =
XA .CA0
−rA
ε𝐴 = 0.5
From the equation,
CA =
𝐶𝐴0 (1− XA )
(1+ ε𝐴 XA )
(1−XA )
330 = 660
, therefore on solving we get X A = 0.66
(1− 0.5XA )
Now on expanding the rate expression we get,
(1−XA) 2
−𝑟𝐴 = 0.05𝐶𝐴0 2 ((1− 0.5X
)
A)
Hence from,
τm =
XA .CA0
−rA
= 54.442 min
Therefore,
Vo =
𝑉
τm
= 2/54.442 = 0.036 L/min
5
Problem 5.16
Solution-
−𝑟𝐴 = 0.6 CA 𝑚𝑖𝑛−1
A→3R
The feed contains of 50% A and 50% inerts
vo = 180 litre/min
CA0 = 0.3 mol/litre
V = 1000 L = 𝑚3
Now,
τm =
XA .CA0
−rA
=
V
v0
ε𝐴 = 2(0.5) = 1,
Hence the rate expression can be expanded as,
(1−X )
(1−X )
−𝑟𝐴 = 0.6 CA (1+ ε AX ) = 0.6 CA (1+ XA ),
A A
A
τm =
XA .CA0
−rA
=
V
v0
=
XA (1+ XA )
CA0 (1− XA )
we obtain,
3𝑋𝐴2 + 13𝑋𝐴 − 10 = 0
XA =
−13± √169+40(3)
2(3)
= 0.67
6
Problem 5.17
Solution-
2O3 → 3O2
−𝑟𝑜𝑧𝑜𝑛𝑒 = kCozone 2 , k = 0.05 litre/ (mol.s)
The rate expression shows that it is a second order reaction so, the integral expression is
defined as,
𝑘𝜏𝑝 𝐶𝐴0 = 2𝜀𝐴 (1 + 𝜀𝐴 ) ln(1 − 𝑋𝐴 ) + 𝜀𝐴2 𝑋𝐴 + (𝜀𝐴 + 1)2
𝑋𝐴
⁄1 − 𝑋
𝐴
Now,
𝐶𝐴0 =
𝑃𝐴0
𝑅𝑇0
= 1.5(0.2)/(0.082)(95 + 273) = 0.01 mol/L
𝜀𝐴 = (3-2)/2 x (0.2) = 0.1
Hence, putting in the equation we get
𝜏𝑝 =
𝑉
1
=
{2[0.1(1.1)𝑙𝑛0.5 + 0.12 (0.5) + 1.12 (1))}
𝑣0 0.05(0.01)
𝜏𝑝 = 2125.02 s
V = 2125.02 (1 L/s) = 2125 L = 2.125 m3
V = 2.125 𝐦𝟑
7
Problem 5.18
Solution-
−𝑟𝐴 = 0.05CA 2 , mol/litre.s
2A → R
FA0 = 1 mol/litre
V=2L
𝑣0 = 0.5 litre/min
Therefore,
𝑉
𝜏𝑝 = 𝑣 =
0
2𝐿
0.5 𝐿/𝑚𝑖𝑛
= 4(60) s = 240 s
The integral rate expression for the 2nd order reaction is given by,
𝑘𝜏𝑝 𝐶𝐴0 = 2𝜀𝐴 (1 + 𝜀𝐴 ) ln(1 − 𝑋𝐴 ) + 𝜀𝐴2 𝑋𝐴 + (𝜀𝐴 + 1)2
Now,
𝜀𝐴 = 0
Therefore, the equation will reduce to
𝑘𝜏𝑝 𝐶𝐴0 =
𝑋𝐴
1 − 𝑋𝐴
So, substituting the values we get,
𝑘𝜏𝑝 𝐶𝐴0 =
𝑋𝐴 =
𝑋𝐴
1 − 𝑋𝐴
𝑘𝜏𝑝 𝐶𝐴0
1+𝑘𝜏𝑝 𝐶𝐴0
=
0,05(240)(1)
1+(0.05)(240)(1)
= 0.92
8
𝑋𝐴
⁄1 − 𝑋
𝐴
Problem 5.19
Solution
A → 3R
P = 3 atm
T = 30°𝐶
𝐶𝐴𝑂 = 0.12 mol/litre
𝑉𝑂
𝐶𝐴
0.06
0.48
1.5
8.1
30
60
80
105
Now,
τm =
XA .CA0
−rA
=
𝑉
𝑣0
– (1)
Similarly,
𝜀𝐴 =
(3−1)1
1
=2
Now, from the equation
𝐶𝐴0 (1−𝑋𝐴 )
𝐶𝐴 =
(1+𝜀𝐴 𝑋𝐴 )
We get different values of 𝑋𝐴 and from that we calculate different values of –𝑟𝐴
Hence, we obtain the following data
XA
−rA
0.5
3.6
9
0.25
0.143
0.045
14.4
25.74
44.18
Hence, from the rate expression
−rA = 𝑘𝐶𝐴𝑛
Taking log on both sides
ln(−rA ) = lnk + lnCA
Now plotting ln(−rA ) on Y-axis and lnCA on X-axis
We get,
Chart Title
5
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
0
0.5
1
1.5
2
2.5
Hence, slope = n = (3.78-1.2809)/(4.653-3.4011) = 2
Similarly, k = 250
Hence, final rate expression will be
−rA = 250𝐶𝐴2
10
3
3.5
4
Problem 5.20
Solution-
A→R
CA0 = 0.1 mol/litre
V = 1 m3
𝑣
1
16
24
𝐶𝐴
4
20
50
For mixed flow reactor, performance equation is written as
𝜏𝑚 =
𝐶𝐴0 − 𝐶𝐴
−𝑟𝐴
Therefore,
−𝑟𝐴 =
𝐶𝐴0 − 𝐶𝐴
𝜏𝑚
−𝑟𝐴 =
(100 − 𝐶𝐴 )𝑣0
𝑉
So,
−rA
96
480
1200
CA
4
20
50
ln(−rA )
4.5643
6.173
7.09007
−rA = 𝑘𝐶𝐴𝑛
Now taking log on both sides,
ln(−rA ) = lnk + lnCA
11
ln(CA )
1.386
2.995
3.91
Chart Title
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
0
2
4
6
8
Therefore, slope = n= 1
𝐶
50
Similarly the value of k can be evaluated by k = −𝑟𝐴 = 1200 = 0.0417 𝑚𝑖𝑛−1
𝐴
So,
−𝑟𝐴 = 0.0417𝐶𝐴 mmol/L.min
12
Problem 5.21
SolutionA→R
CA0 = 1.3 mol/litre
𝐶𝐴 = 0.3 mol/litre
𝐶𝐴
−𝑟𝐴
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
1.0
1.3
2.0
0.1
0.3
0.5
0.6
0.5
0.25
0.1
0.06
0.05
0.045
0.042
It is a batch reactor so, the performance equation of the reactor is,
𝐶𝐴0 𝑑𝐶𝐴
1.3 𝑑𝐶𝐴
=∫
=
0.3 −𝑟𝐴
−𝑟𝐴
∫𝐶
𝐴
Therefore, we need to find the area under the curve of −𝑟𝐴 and CA from 0.3 to 1.3
Hence, in this plot the area under the curve from 0.3 to 1.3 we can calculate the residence
time
1.2
1
0.8
0.6
0.4
0.2
0
0
50
100
150
200
250
300
13
350
400
450
1.3 𝑑𝐶𝐴
t=∫
= 12.6 min
0.3 −𝑟𝐴
Problem 5.22
SolutionGiven,
𝑋𝐴 = 0.8
𝐹𝐴0 = 1000 mol/hr
𝐶𝐴0 = 1.5 mol/litre
From, 𝐶𝐴 = 𝐶𝐴0 (1 − 𝑋𝐴 ) we get CA = 0.3 mol/litre
Now, the performance equation of the plug flow reactor is
𝐶
𝜏𝑝 = ∫𝐶 𝐴0
𝐴
𝑑𝐶𝐴
−𝑟𝐴
Similarly from the example 5.21 we obtain
1.3 𝑑𝐶𝐴
∫0.3
−𝑟𝐴
= 12.8 min
Now, the time can be found by finding out the area between 1.3 to 1.5
1.2
1
0.8
0.6
0.4
0.2
0
0
50
100
150
200
250
300
350
So, the area between 1.3 to 1.5 we obtain the area = 4.5 min
Therefore,
The total area = 12.8 +4.5 min
= 17.3 min
14
400
450
Problem 5.23
Solution-
𝑋𝐴 = 0.75
𝐹𝐴0 = 1000 mol/hr
𝐶𝐴0 = 1.2 mol/litre
(a) The performance equation of mixed flow reactor is
𝛕𝐦 =
𝐗 𝐀 .𝐂𝐀𝟎
−𝐫𝐀
Now from the equation, 𝐶𝐴 = 𝐶𝐴0 (1 − 𝑋𝐴 ) so, substituting the values we get 𝐶𝐴 = 0.3 mol/L
and −𝑟𝐴 = 0.5 mol/L
Therefore,
𝑉
(𝐶𝐴0 − 𝐶𝐴 )
=
FA0
−𝑟𝐴
V = 1000(1.2-0.3)/1.2(0.5) (60) = 1500 L
(b)
𝐹𝐴0 = 2000 mol/hr
Hence, when the feed rate is doubled the volume also doubles.
Therefore, V = 50 ltr
(c)
𝐶𝐴0 = 2.4 mol/litre
𝐶𝐴 = 0.3 mol/litre
Therefore putting it in the performance equation we get,
𝑉
(𝐶𝐴0 − 𝐶𝐴 )
=
FA0
−𝑟𝐴
V = 1000(2.4-0.3) /(2.4)(0.5)(60) = 29.167 ltr
15
Problem 5.24
SolutionA→5R
V = 0.1 litre
CA0 = 100 mmol/litre = 0.1 mol/litre
𝐹𝐴0 mmol/ltr
300
1000
3000
5000
𝐶𝐴 mmol/ltr
16
30
50
60
εA = 4
τm =
−rA =
XA .CA0
=
−rA
v0 XA .CA0
V
𝑉
𝑣0
= 10FA0 X A
Therefore,
𝑋𝐴 =
𝐶
1− 𝐴
𝐶𝐴0
𝐶
1+𝜀𝐴 𝐴
𝐶𝐴0
, from this we generate the following data
FA0
CA
300
1000
3000
5000
XA
16
30
50
60
0.512
0.318
0.167
0.118
−rA = 𝑘𝐶𝐴𝑛
Now taking log on both sides,
ln(−rA ) = lnk + nlnCA
Chart Title
10
8
6
4
2
0
0
1
2
3
4
5
Slope = 1; k = 0.01
−𝐫𝐀 = 𝟎. 𝟎𝟏𝐂𝐀
16
−rA
1536.6
318.8
5000
5882.4
Problem 5.25
Solution-
XA = 0.75;
It is a batch reactor
CA0 = 0.8 mol/litre
CA in feed stream
CA in exit stream
2
2
2
1
1
0.48
0.48
0.48
0.65
0.92
1
0.56
0.37
0.42
0.28
0.20
Holding time
300
240
250
110
360
24
200
560
Hence, holding time = residence time = 𝜏
𝐶
𝜏 = − ∫𝐶 𝐴
𝐴0
𝑑𝐶𝐴
−𝑟𝐴
;
So, we generate the following data, by using the correlation of 𝜏 =
CA
𝐶𝐴0 −𝐶𝐴
1/(−rA )
0.65
0.92
1
0.56
0.37
0.42
0.28
0.2
222
222
250
250
572
400
1000
2000
On plotting the graph of 1/(−rA) vs CA
1.2
1
0.8
0.6
0.4
0.2
0
0
50
100
150
200
17
250
−𝑟𝐴
Therefore, area under the curve will give the residence time
𝝉 = 300 s
Problem 5.26
SolutionIf it is a mixed flow reactor
τm =
XA .CA0
−rA
=
𝑉
𝑣0
Following the same procedure and plotting the same graph of 1/(−rA) vs CA we find the area
of the complete rectangle because it is a MFR so,
1.2
1
0.8
0.6
0.4
0.2
0
0
50
100
150
200
𝜏 = 900 s
18
250
Problem 5.27
Solution(a) The conversion drops from 80 to 75 % so, it is quite possible that Imbibit fell into the vat
containing alcohol. The reason for the drop in conversion is due to decrease in the volume of
liquid available for the reaction.
(b) The revelation couldn’t be made because Watson asked for some dill (for smoking) but he
didn’t have any idea that smoking near a vat of ethanol isn’t a very good idea.
Problem 5.28
Solution2A→ R + S
The PFR is operating at 100℃ and 1 atm
FA0 = 100 mol/hr
XA = 0.95
The feed contains 20% inerts and 80% A
So,
−rA = kCAn
The reaction is of 1st order
So, the integral form of equation in terms of conversion is obtained which is
𝐶
𝑘𝑡 = −ln(1 − 𝑋𝐴 ), which can be also written as 𝑋𝐴 = 1 − 𝐶 𝐴 – (1)
𝐴0
Now,
CA =
𝑝𝐴
𝑅𝑇
So, replacing the term in the equation (1) we get,
𝑘𝑡 = 𝑙𝑛
𝑝𝐴0
𝑝𝐴
Hence, we now generate a data
19
ln(𝑝𝐴0 /𝑝𝐴 )
1
1.25
1.47
1.78
2.22
2.702
4
7.14
12.5
25
50
T
0
20
40
60
80
100
140
200
260
330
420
Chart Title
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
0
50
100
150
200
250
300
350
400
450
Therefore, slope = k = 0.01116 s −1
The reaction is 1st order reaction hence, the rate expression is given by
−rA = 0.01116CA
We can also write the equation like
V
kτp = k v = −ln(1 − XA )
Therefore, V = −
0
𝑣𝑜 =
𝐹𝐴0 𝐹𝐴0 𝑇0 𝑅 100(373)(0.082)
𝐿 1.06𝐿
=
=
= 3823.25 =
𝐶𝐴0
𝑝𝐴0
0.8
ℎ
𝑠
V = 1.06ln (1-0.95)/0.01116 = 284.54 L = 2.84 m3
20
𝑣0
𝑘
ln(1 − 𝑋𝐴 )
Problem 5.29
Solution-
Now, for the same problem if we take a MFR then, the performance equation is written as
Plotting Pa v τ
1.2
1
0.8
0.6
0.4
0.2
0
0
50
100
150
200
250
300
350
400
Slope = rate = 0.000375
τm =
𝑣𝑜 =
(pA0 −pA )
−rA
= (0.8-0.04)/ (0.000375) = 2026.67 s
𝐹𝐴0 𝐹𝐴0 𝑇0 𝑅 100(373)(0.082)
𝐿 1.06𝐿
=
=
= 3823.25 =
𝐶𝐴0
𝑝𝐴0
0.8
ℎ
𝑠
V = 2026.67(1.06) = 2.148 L = 2.15 m3
21
450
Problem 5.30
SolutionA→R
CA0 = 0.8 mol/litre
FA0 = CA0 (V) = 0.8 mol/s
0.8 mol/litre
XA = 0.75
V = 1 litre/s
(a) The reactor is a plug flow reactor
The performance equation is
XA
V
τ
dX A
=
=∫
FA0 CA0
−rA
0
It can be modified as
𝐶𝐴
𝑉
𝑑𝐶𝐴
=𝜏= − ∫
𝐹𝐴0
−𝑟𝐴
𝐶𝐴0
Hence, now from the equation, 𝜏
τ
50
16
60
22
4.8
72
40
1122
=
𝐶𝐴0 −𝐶𝐴
−𝑟𝐴
we can generate the following table
CA0
CA
2
1.2
2
1
0.48
1
0.48
0.48
1
0.8
0.65
0.56
0.42
0.37
0.28
0.2
Hence, on plotting the graph of 1/( −rA ) vs CA
22
−rA
0.02
0.025
0.0225
0.02
0.0125
0.00875
0.005
0.0025
450
400
350
300
250
200
150
100
50
0
0
0.2
0.4
0.6
0.8
So, for PFR we need to find the area under the curve.
Therefore, Area = τ = 70 s
So, putting in the performance equation we get,
V= (70)0.8/0.8 = 70 ltr
(2) The reactor is MFR
For MFR, we need to find the rectangular area which is
Area = τ = 320 s
V = 320(0.8)/ (0.8) = 320
Therefore, volume required = 320 ltr
23
1
1.2
Solution:-
Case 1:- n1=1, n2=2, n3=3
Use MFR with some particular concentration of A.
Case 2) n1=2, n2=3, n3=1
Use PFR with low XA
Case 3) n1=3, n2=1, n3=2
Use MFR with high XA
7.2
Solution:A+B→R
A→S
Low CA, high CB
7.3
Solution:A+B→R
2A → S
2B → T
Use low CA and low CB
Slowly add A & B
Don’t add excess of A or B.
7.4
A+B→R
A→S
First add B completely then add A slowly & steadily.
7.5
A+B→R
2A → S
Keep CA low , CB high.
Solution:-
FOR TANK-1
𝜓𝑅/𝐴 =
0.4 − 0
0.4
=
= 0.67
1 − 0.4
0.6
𝜏1 =
𝐶𝑅 − 𝐶𝑅0
𝑘1 ∗ 𝐶𝐴
2.5 =
0.4
𝑘1 ∗ 0.4
K1 = 0.4
𝜏1 =
𝐶𝑆 − 𝐶𝑆0
𝑘2 ∗ 𝐶𝐴
2.5 =
0.2
𝑘1 ∗ 0.4
K2 = 0.2
For tank – 2
𝜏2 =
5=
𝐶𝐴1 − 𝐶𝐴2
−𝑟𝐴
0.4 − 𝐶𝐴2
(𝐾1 + 𝐾2 )𝐶𝐴2
3 𝐶𝐴2 = 0.4 − 𝐶𝐴2
𝐶𝐴2 = 0.1
𝑚𝑜𝑙
ℎ𝑟
𝐶𝑅2 = 𝐶𝑅1 + 𝜓𝑅 (𝐶𝐴1 − 𝐶𝐴2 )
𝐴
= 0.4 +
𝐾1
𝐾1 +𝐾2
(0.4 − 0.1)
= 0.6 𝑚𝑜𝑙/ℎ𝑟
𝐶𝑆2 = 𝐶𝑆1 + 𝜑 𝑆 (𝐶𝐴1 − 𝐶𝐴2 )
𝐴
= 0.2 +
𝐾1
𝐾1 +𝐾2
= 0.3 mol/hr
(0.3)
Solution:FOR TANK-1
𝑁𝑅1 − 𝑁𝑅0
𝑁𝐴0 − 𝑁𝐴1
𝜓𝑅 =
𝐴
0.2 − 0
1 − 0.7
=
= 0.2857
𝜏=
=
𝐶𝐴0 ∗ 𝑋𝐴
−𝑟𝑅
𝐶𝑅 − 𝐶𝑅0
𝑘1 ∗ 𝐶𝐴 2
2.5 =
0.2−0
𝐾1 ∗0.4 2
K1= 0.5
𝜏=
𝐶𝑆 − 𝐶𝑆0
𝑘2 ∗ 𝐶𝐴
2.5 =
0.7 − 0.3
𝐾2 ∗ 0.4
𝐾2 = 0.4
FOR TANK -2
𝜏2 =
10 =
𝐶𝐴1 − 𝐶𝐴2
−𝑟𝐴
0.4 − 𝐶𝐴2
𝐾1 𝐶𝐴2 2 + 𝐾2 𝐶𝐴2
5𝐶𝐴2 2 + 4𝐶𝐴2 = 0.4 − 𝐶𝐴2
5𝐶𝐴2 2 + 5𝐶𝐴2 − 0.4 = 0
𝐶𝐴2 = 0.074
𝐶𝑅2 = 𝐶𝑅1 + 𝜓𝑅 (𝐶𝐴1 − 𝐶𝐴2 )
𝐴
= 0.2 +
𝐾1 𝐶𝐴2
(0.4 − 0.074)
𝐾1 𝐶𝐴2
= 0.2 + 0.5∗0.074(0.0326)
0.5∗0.074+0.4
= 0.227
𝐶𝑆2 = 𝐶𝑆1 + 𝜓𝑆⁄ (𝐶𝐴1 − 𝐶𝐴2 )
𝐴
= 0.7 +
𝐶𝑆2 = 0.998
𝐾2
𝐾1𝐶𝐴2 +𝐾2
(0.4 − 0.074)
Solution::--CR = CR0 + ψ R/A (CA0-CA)
=0+
0.4∗4
0.4∗4+2
(40 − 4)
= 1.6*36/3.6
= 16 mol/L
CS = 36-16 = 20 mol/L
τ=
τ=
𝐶𝐴0 −𝐶𝐴
−𝑟𝐴
40−4
0.4(4)2 +2(4)
τ = 36/14.4
τ = 2.5 sec
Solution::--𝑑𝐶𝑅
= 𝑟𝑅 = 𝑘1 𝐶𝐴 2 = 0.4𝐶𝐴 2
𝑑𝑡
𝑑𝐶𝑆
= 𝑟𝑆 = 𝑘2 𝐶𝐴 = 0.4𝐶𝐴
𝑑𝑡
𝜓𝑅⁄
=
𝐴
𝑟𝑅
−𝑟𝐴
=
=
𝐾1 𝐶𝐴 2
𝐾1 𝐶𝐴 2 + 𝐾2 𝐶𝐴
0.4𝐶𝐴
0.4𝐶𝐴 + 2
𝐶𝐴0
𝐶𝑅 = ∫ 𝜓 𝑑𝐶𝐴
𝐶𝐴
40
=∫ (
4
0.4𝐶𝐴
) 𝑑𝐶𝐴
0.4𝐶𝐴 + 2
40
= ∫ (1 −
4
= (40 − 4) − [
2
) 𝑑𝐶𝐴
0.4𝐶𝐴 + 2
2
ln(0.4𝐶𝐴 + 2)]4 40
0.4
= 36 − 5 𝑙𝑛
18
3.6
𝐶𝑅 = 27.953
𝐶𝑆 = 36 − 𝐶𝑅 = 8.047
40
𝜏=∫
4
𝑑𝐶𝐴
−𝑟𝐴
40
= ∫
4
𝑑𝐶𝐴
4𝐶𝐴 2 + 2𝐶𝐴
40
= 2.5 ∫
4
𝑑𝐶𝐴
5
5
𝐶𝐴 2 + 5𝐶𝐴 + ( )2 − ( )2
2
2
40
= 2.5 ∫
4
𝑑𝐶𝐴
𝜋 2
5 2
(𝐶𝐴 + ) − ( )
2
2
= 0.346 sec
Solution::--𝑟𝑅 = 0.4 ∗ 𝐶𝐴 2
𝑟𝑆 = 2 ∗ 𝐶𝐴
𝜓𝑆⁄ =
𝐴
𝑑𝐶𝑆
2𝐶𝐴
1
10
=
=
=
−𝑑𝐶𝐴
1 + 0.2𝐶𝐴
10 + 2𝐶𝐴
2𝐶𝐴 + 0.4𝐶𝐴 2
Ca
0
1
10
20
40
ψ
1
0.833333
0.333333
0.2
0.111111
ψ
1.2
1
ψ
0.8
0.6
0.4
0.2
0
-10
0
10
20
30
40
50
Ca
400 − 10𝑐𝐴𝑓
10
𝐴=(
) (40 − 𝑐𝐴𝑓 ) =
10 + 2𝑓𝐴
10 + 2𝑐𝐴𝑓
𝑑𝐴
= 10 + 2𝐶𝐴𝑓 (−10) − (400 − 10𝐶𝐴𝑓 )(2)
𝑑𝐶𝐴
0 = −100 − 20𝐶𝐴𝑓 − 800 + 20𝐶𝐴𝑓
−100 − 20𝐶𝐴𝑓 = 800 − 20𝐶𝐴𝑓
𝐶𝐴𝑓 = 0
XA = 100%
𝜏
1
=
40 2(0) + 0.4(0)
𝜏= ∞
SOLUTION::--𝜓𝑅⁄ =
𝐴
𝐾1 𝐶𝐴 2
2
𝐾1 𝐶𝐴 + 𝐾2 𝐶𝐴
𝑑𝜓𝑅⁄
𝐴
𝑑𝐶𝐴
=
0.4𝐶𝐴
𝐶𝐴
=
0.4𝐶𝐴 + 2 𝐶𝐴 + 5
= (𝐶𝐴 + 5) ∗ 1 − 𝐶𝐴 ∗ 1
Ca
ψ
0
0
1
0.166667
10
0.666667
20
0.8
40
0.888889
ψ
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
-5
0
5
10
15
20
25
30
35
40
40𝐶𝐴𝑓 − 𝐶𝐴𝑓 2
𝐴=
𝐶𝐴𝑓 + 5
40𝐶𝐴𝑓 − 𝐶𝐴𝑓 2
𝑑𝐴
= (𝐶𝐴𝑓 + 5)(40 − 2𝐶𝐴𝑓 ) − (
)
𝑑𝐶𝐴𝑓
(𝐶𝐴𝑓 + 5)2
40𝐶𝐴𝑓 −2𝐶𝐴𝑓 2 + 200−10𝐶𝐴𝑓 −40𝐶𝐴𝑓 + 𝐶𝐴𝑓 2 = 0
𝐶𝐴𝑓 2 + 10𝐶𝐴𝑓 − 200 = 0
𝐶𝐴𝑓 =
−10 + 30
= 10
2
𝑋𝐴 = 1 − (
10
)
40
𝜏
0.75
=
40 0.4𝐶𝐴 2 + 2𝐶𝐴
𝜏=
30
0.4 ∗ (10)2 + 2 ∗ (10)
30
τ = 40+20 = 0.5
45
CRE-I SPECIAL ASSIGNMENT
QUESTION: [7.12.]
Reactant A in a liquid either isomerizes or dimerizes as follows:
𝐴 → 𝑅𝑑𝑒𝑠𝑖𝑟𝑒𝑑
𝑟𝑅 = 𝑘1 𝐶𝐴
𝐴 + 𝐴 → 𝑆𝑢𝑛𝑤𝑎𝑛𝑡𝑒𝑑 𝑟𝑆 = 𝑘2 𝐶𝐴2
𝑅
𝑅
𝐴
(𝑅+𝑆)
(a) Write 𝜑 ( ) and 𝜑 [
].
With a feed stream of concentration CA0, find CR,max which can be
formed:
(b) In a plug flow reactor.
(c) In a mixed flow reactor.
A quantity of A of initial concentration CA0 = 1 mol/liter is dumped
into a batch reactor and is reacted to completion.
(d) If Cs = 0.18 mol/liter in the resultant mixture, what does this tell
of the kinetics of the reaction?
ANSWER:
(a) 𝜑 [
𝑅
(𝑅+𝑆)
]=
𝑟𝑅
𝑟𝑅 + 𝑟𝑆
𝑅
𝑟𝑅
𝐴
−𝑟𝐴
𝜑( ) =
=
=
𝑘1 𝐶𝐴
𝑘1 𝐶𝐴 + 𝑘2 𝐶𝐴2
𝑘1 𝐶𝐴
𝑘1 𝐶𝐴 + 2𝑘2 𝐶𝐴2
(b)
We get CR,max when we have CAF = 0.
𝐶
𝐶
So, CR,max = ∫0 𝐴0 𝜑𝑑𝐶𝐴 = ∫0 𝐴0
CR,max =
𝑘1
2𝑘2
{ln( 1 +
2𝑘2
𝑘1
1
2𝑘
1 + 𝑘 2 𝐶𝐴
1
𝑑𝐶𝐴 =
𝐶𝐴0 ) − ln 1} =
𝑘1
2𝑘2
𝑘1
2𝑘2
{ln( 1 +
{ln( 1 +
2𝑘2
𝑘1
2𝑘2
𝑘1
𝐶𝐴 )}
𝐶𝐴0 )}
𝐶𝐴0
0
(c)
CRm = 𝜑𝑓 (𝐶𝐴0 − 𝐶𝐴 )
CRm, max = 1(𝐶𝐴0 − 0) = 𝐶𝐴0
(d)
CS = 0.18 gives CR = CA0 – CA – CS = 1 – 0 – 0.18 = 0.82
Also,
CR can be calculated from the following equation:
CR,max =
𝑘1
2𝑘2
{ln( 1 +
2𝑘2
𝑘1
𝐶𝐴0 )}
Let’s say, K = k1/k2 and assume K = 4 and K = 5.
So, CR calculated from the above equation is 0.84 and 0.81
respectively.
K
4
5
CR calculated
0.84
0.81
Hence, for different values of k1 and k2, CR,max is different. It may be
more or less than the CR value.
QUESTION: [7.13.]
In a reactive environment, chemical A decomposes as follows:
𝐴 → 𝑅, 𝑟𝑅 = 𝐶𝐴
𝐴 → 𝑆, 𝑟𝑆 = 1
𝑚𝑜𝑙
𝑙𝑖𝑡𝑒𝑟𝑠
𝑚𝑜𝑙
𝑙𝑖𝑡𝑒𝑟
For a feed stream 𝐶𝐴0 = 4
𝑚𝑜𝑙
𝑙𝑖𝑡𝑒𝑟𝑠
what size ratio of two mixed flow
reactors will maximize the production rate of R? Also give the
composition of A and R leaving these two reactors.
ANSWER:
𝜑=
𝑟𝒓
𝐶𝐴
=
−𝑟𝐴
1 + 𝐶𝐴
𝐶𝐴 → 0,
𝜑 →0
𝐶𝑅 = 𝜑∆𝐶𝐴 = (
𝑎𝑛𝑑 𝐶𝐴 → ∞,
𝜑 →1
𝐶𝐴1
𝐶𝐴2
) (4 − 𝐶𝐴1 )+ = (
) (𝐶𝐴1 − 𝐶𝐴2 )
1 + 𝐶𝐴1
1 + 𝐶𝐴2
CA1 and CA2 are not known, but there is a fixed CA1 and CA2 value that
maximizes CR and is what makes dCR/dCA1 = 0
2 )
(1 + 𝐶𝐴1 )(4 − 2𝐶𝐴1 ) − (4𝐶𝐴1 − 𝐶𝐴1
𝑑𝐶𝑅
(1)
=0=
(1 + 𝐶𝐴1 )2
𝑑𝐶𝐴1
𝐶𝐴2
+
(1)
1 + 𝐶𝐴2
So,
2
4 − 2𝐶𝐴1 − 𝐶𝐴1
𝐶𝐴2
=
−
(1 + 𝐶𝐴1 )2
1 + 𝐶𝐴2
If CA2 = 0.5 mol,
2
4 − 2𝐶𝐴1 − 𝐶𝐴1
0.5
=
−
(1 + 𝐶𝐴1 )2
1 + 0.5
2
13 − 4𝐶𝐴1 − 2𝐶𝐴1
=0
2
𝐶𝐴1
4 ± √42 − 4(13)(−2)
𝑚𝑜𝑙
=
= 1.7386
2(−2)
𝑙𝑖𝑡𝑒𝑟
𝐶𝑅2 =
1.7386
0.5
(4 − 1.7386) +
(1.7386 − 0.5)
1 + 1.7386
1 + 0.5
= 1.8485
QUESTION:
Consider the parallel decomposition of A of different orders as
follows:
𝐴 → 𝑅,
𝑟𝑅 = 1
𝐴 → 𝑆,
𝑟𝑆 = 2𝐶𝐴
𝐴 → 𝑇,
𝑟𝑇 = 𝐶𝐴2
Determine the maximum concentration of desired product
obtainable in (a) plug flow, (b) mixed flow.
[7.14.]: R is the desired product and𝐶𝐴0 = 2.
ANS:
𝜑𝑅 =
1
1 + 2𝐶𝐴 + 𝐶𝐴2
(a)
2
𝐶𝑅,𝑚𝑎𝑥
2
0
= −
0
1
1
2
+ =
1+2 1
3
(b)
𝜑𝑅 =
1
1 + 2𝐶𝐴 + 𝐶𝐴2
When, 𝐶𝐴 = 0; 𝐶𝑅 = 𝐶𝑅,𝑚𝑎𝑥
So, 𝐶𝑅𝑚,
2
(𝐶𝐴 + 1)−1
𝑑𝐶𝐴
𝑑𝐶𝐴
= ∫
= ∫
=(
)
(𝐶𝐴 + 1)2
−1
1 + 2𝐶𝐴 + 𝐶𝐴2
0
𝑚𝑎𝑥
= 𝜑𝐶𝐴 = 0 (2 − 0) = 1(2) = 2 𝑚𝑜𝑙/𝑙𝑖𝑡𝑒𝑟
[7.15.]: S is the desired product and𝐶𝐴0 = 4.
ANS:
𝜑𝑅 =
2𝐶𝐴
=
1 + 2𝐶𝐴 + 𝐶𝐴2
(a)
When, 𝐶𝐴 = 0; 𝐶𝑆 = 𝐶𝑆𝑃,
4
𝐶𝑆𝑃,
𝑚𝑎𝑥
= ∫0
∫
𝑑𝐶𝐴
𝐶
1
+ 1+ 2𝐴
2𝐶𝐴
𝑚𝑎𝑥
4 𝐶𝐴 𝑑𝐶𝐴
(𝐶𝐴 + 1)2
= 2 ∫0
𝑥𝑑𝑥
1
𝑎
(
)
=
[𝑙𝑛
𝑎
+
𝑏𝑥
+
]
(𝑎 + 𝑏𝑥 )2
𝑏2
𝑎 + 𝑏𝑥
So,
𝐶𝑆𝑃,
𝑚𝑎𝑥
𝐶𝑆𝑃,
𝑚𝑎𝑥
4
1
1
= 2 { [𝑙𝑛(1 + 𝐶𝐴 ) +
]}
1
1 + 𝐶𝐴 0
1
1
𝑚𝑜𝑙
= 2 { [𝑙𝑛(1 + 4) + − 𝑙𝑛(1 + 0) − 1]} = 1.6188
1
5
𝑙𝑖𝑡𝑒𝑟
(b)
𝐶𝑆𝑃,
𝑚𝑎𝑥
= 𝜑𝐶𝐴𝑓 (𝐶𝐴0 − 𝐶𝐴𝑓 )
𝐶𝑆𝑚
2𝐶𝐴
𝐶𝐴 (4 − 𝐶𝐴 )
(
)
=
4 − 𝐶𝐴 = 2 {
}
2𝐶𝐴 + 1 + 𝐶𝐴2
2𝐶𝐴 + 1 + 𝐶𝐴2
𝑑𝐶𝑆𝑚
𝑑𝐶𝐴
(2𝐶𝐴 + 1 + 𝐶𝐴2 )[𝐶𝐴 (−1) + 4 − 𝐶𝐴 ] − (4𝐶𝐴 − 𝐶𝐴2 )(2 + 2𝐶𝐴 )
= 2{
}
(2𝐶𝐴 + 1 + 𝐶𝐴2 )2
= 0
So,
2{(2𝐶𝐴 + 1 + 𝐶𝐴2 )(2 − 𝐶𝐴 ) − (4𝐶𝐴 − 𝐶𝐴2 )(1 + 𝐶𝐴 ) } = 0
3𝐶𝐴2 + 𝐶𝐴 − 2 = 0
Therefore,
2
−1 ± √1 − 4(3)(−2)
2
𝐶𝐴 =
=
2(3)
3
𝐶𝑆𝑚,
𝑚𝑎𝑥
2
2( )
2
𝑚𝑜𝑙
3
=
(4
−
)
=
1.6
3
𝑙𝑖𝑡𝑒𝑟
2
2 2
2( ) + ( ) + 1
3
3
[7.16.]: T is the desired product and𝐶𝐴0 = 5.
ANS:
𝐶𝐴2
1
𝜑𝑇 =
=
2
1
1 + 2𝐶𝐴 + 𝐶𝐴2
1+
+ 2
𝐶𝐴 𝐶𝐴
(a)
CTP is maximum when CAf = 0.
5
𝐶𝑇𝑃,
𝑚𝑎𝑥
𝐶𝐴2 𝑑𝐶𝐴
= ∫
=∫
2
2
1
0 (𝐶𝐴 + 1)
+
1
+
0 𝐶
𝐶𝐴2
𝐴
𝑑𝐶𝐴
5
𝑥 2 𝑑𝑥
1
𝑎2
∫
= 3 [𝑎 + 𝑏𝑥 − 2𝑎 𝑙𝑛(𝑎 + 𝑏𝑥 ) −
]
(𝑎 + 𝑏𝑥 )2
𝑏
𝑎 + 𝑏𝑥
So,
𝐶𝑇𝑃,
𝑚𝑎𝑥
5
1
= {1 + 𝐶𝐴 − 2𝑙𝑛(1 + 𝐶𝐴 ) −
}
1 + 𝐶𝐴 0
1
𝑚𝑜𝑙
= 6 − 2𝑙𝑛6 − − [1 − 2𝑙𝑛1 − 1] = 2.2498
6
𝑙𝑖𝑡𝑒𝑟
(b)
𝐶𝑇𝑚
𝐶𝐴2
(5 − 𝐶𝐴 )
=
1 + 2𝐶𝐴 + 𝐶𝐴2
So,
𝑑𝐶𝑇𝑚
𝑑𝐶𝐴
(1 + 2𝐶𝐴 + 𝐶𝐴2 )[(5 − 𝐶𝐴 )(2𝐶𝐴 ) + 𝐶𝐴2 (−1)] − 𝐶𝐴2 (5 − 𝐶𝐴 )(2𝐶𝐴 + 2)
=
(1 + 2𝐶𝐴 + 𝐶𝐴2 )2
=0
So,
𝐶𝐴 {(𝐶𝐴 + 1)2 [−𝐶𝐴 + 2(5 − 𝐶𝐴 )] − 𝐶𝐴 (5 − 𝐶𝐴 )(2𝐶𝐴 + 2)} = 0
(𝐶𝐴 + 1){(𝐶𝐴 + 1)(− 𝐶𝐴 + 10 − 2𝐶𝐴 ) − (5𝐶𝐴 − 𝐶𝐴2 )(2)} = 0
(𝐶𝐴 + 1)(10 − 3𝐶𝐴 ) − 2(5𝐶𝐴 − 𝐶𝐴2 ) = 0
𝐶𝐴2 + 3𝐶𝐴 − 10 = 0
2
−3 ± √9 − 4(1)(−10)
𝐶𝐴 =
= 2 𝑚𝑜𝑙/𝑙𝑖𝑡𝑒𝑟
2
𝐶𝑇𝑚,
𝑚𝑎𝑥
22
8
= 2
=
= 0.89 𝑚𝑜𝑙/𝑙𝑖𝑡𝑒𝑟
2 + 2(2) + 1
9
QUESTION:
Under ultraviolet radiation, reactant A of CA0 = 10 kmol/m3 in a
process stream (𝜗 = 1 𝑚3 /min) decomposes as follows:
𝐴 → 𝑅,
𝐴 → 𝑆,
𝐴 → 𝑇,
𝑟𝑅 = 16𝐶𝐴0.5
𝑟𝑆 = 12𝐶𝐴
𝑟𝑇 = 𝐶𝐴2
We wish to design a reactor setup for a specific duty. Sketch the
scheme selected, and calculate the fraction of feed transformed into
desired product as well as the volume of reactor needed.
[7.17.]: Product R is the desired material.
ANS:
The reaction of the desired product is the lowest order, so it is better
to use a mixed reactor with high conversion.
CRm, max is obtained when CAf = 0.
𝐶𝑅𝑚,
𝑚𝑎𝑥
= 𝜑(𝐶𝐴0 − 𝐶𝐴𝑓 ) = 1(10 − 0) = 10 𝑚𝑜𝑙/𝑙𝑖𝑡𝑒𝑟
16𝐶𝐴0.5
𝜑𝑅 =
16𝐶𝐴0.5 + 12𝐶𝐴 + 𝐶𝐴2
𝐶𝑅𝑚 = 𝜑𝑅 (𝐶𝐴0 − 𝐶𝐴 )
𝜏𝑚 =
𝐶𝐴0 − 𝐶𝐴
16𝐶𝐴0.5 + 12𝐶𝐴 + 𝐶𝐴2
𝑉 = 𝜏𝑚 (𝜗0 )
Select a high conversion and make the calculations for each of them.
XA
CA
𝜏
V
𝜑
CR
0.98
0.2
1.0130
1.0130
0.7370
5.8960
0.99
0.995
0.1
0.05
1.5790
2.3803
1.5790
2.3803
0.8070
0.8558
7.9894
8.5159
As we can see above, going from XA = 0.99 to 0.995, ΔCR = 0.5265
mol / L and to achieve this, ΔV = 0.8013 m3 (almost 1 m3) is required,
then XA = 0.995 could be selected.
[7.18.]: Product S is the desired material.
ANS:
The desired reaction is the middle order, so there corresponds an
intermediate concentration, which makes maximum performance.
𝜑𝑆 =
16𝐶𝐴0.5
12𝐶𝐴
+ 12𝐶𝐴 + 𝐶𝐴2
(A)
If you cannot recirculate unreacted A, then use a mixed reactor until
the concentration giving 𝜑máx and thereafter a piston.
(B)
If the A can be recirculated unreacted economically, then use a mixed
reactor with concentration giving 𝜑máx.
(A)
𝑑𝜑𝑆
12(16𝐶𝐴0.5 + 12𝐶𝐴 + 𝐶𝐴2 ) − 12𝐶𝐴 (12 + 8𝐶𝐴−0.5 + 2𝐶𝐴 )
=
2
𝑑𝐶𝐴
(16𝐶𝐴0.5 + 12𝐶𝐴 + 𝐶𝐴2 )
=0
So,
18𝐶𝐴0.5 − 𝐶𝐴2 = 0
𝐶𝐴 = 4
𝑘𝑚𝑜𝑙
𝑚3
𝐶𝑆𝑚 = 0.5(10 − 4) = 3
𝜏𝑚 =
𝑘𝑚𝑜𝑙
𝑚3
10 − 4
= 0.0625 𝑚3 = 62.5 𝑙𝑖𝑡𝑒𝑟
0.5
2
16(4) + 12(4) + 4
𝐶𝐴
𝜑𝑆
4
3
2
1
0.6
0.4
0.11
0.02
0.5 0.4951 0.4740 0.4138 0.3608 0.2501 0.1988 0.0959
Suppose, XA = 0.998, implies CA = 0.02
𝑓−1
4
𝐶𝑆𝑃 = ∫ 𝜑𝑑𝐶𝐴 =
0.02
𝐶𝑆𝑃 =
∆𝐶𝐴
[𝜑0 + 𝜑𝑓 + 2 ∑ 𝜑𝑖 ]
2
𝑖=1
1
[0.4138 + 0.5 + 2(0.4740 + 0.4951)]
2
0.4
[0.4138 + 0.2501 + 2(0.3608)]
+
2
0.09
[0.2501 + 0.0959 + 2(0.1988)]
+
2
𝐶𝑆𝑃 = 1.7367 𝑚𝑜𝑙/𝑚3
𝐶𝑆,
𝑡𝑜𝑡𝑎𝑙
= 3 + 1.7367 = 4.7367 𝑚𝑜𝑙/𝑚3
𝐶𝐴𝑓
𝑓−1
𝑑𝐶𝐴
∆𝐶𝐴
−1
−1
𝜏𝑃 = ∫
≈
{(−𝑟𝐴 )−1
0 + (−𝑟𝐴 )𝑓 + 2 ∑ (−𝑟𝐴 )𝑖 }
−𝑟𝐴
2
𝑖=1
𝐶𝐴0
𝐶𝐴
−𝑟𝐴
So,
4
96
3
2
72.71 50.62
1
29
0.6
19.95
0.2
9.60
0.11
6.64
0.02
2.50
𝜏𝑃 =
1 1
1
1
1
[ +
+ 2(
+
)]
2 29 96
72.71 50.62
0.4 1
1
1
+
[ +
+ 2(
)]
2 29 9.60
19.95
0.09 1
1
1
+
[
+
+ 2(
)] = 0.1399 𝑚𝑖𝑛
2 9.6 2.5
6.64
(B)
At point D, balance the flow:
𝜗0 (𝑅 + 1)(4) = 0 + 𝜗0 𝑅 (10) which gives R = 2/3.
Now,
𝑉𝑚
𝜗0 (𝑅+1)
=
10−4
96
which gives 𝑉𝑚 = 0.104 𝑚3 = 104 𝑙𝑖𝑡𝑒𝑟𝑠
[7.19.]: Product T is the desired material.
ANS:
The reaction occurs where T is of the highest order. So we should use
a flow reactor.
𝐶𝐴2
𝜑𝑆 =
16𝐶𝐴0.5 + 12𝐶𝐴 + 𝐶𝐴2
𝐶𝐴𝑓
𝐶𝑇𝑃 = ∫ 𝜑𝑑𝐶𝐴 ≈
𝐶𝐴0
𝑓−1
∆𝐶𝐴
[𝜑0 + 𝜑𝑓 + 2 ∑ 𝜑𝑖 ]
2
𝑖=1
𝐶𝐴𝑓
𝑓−1
𝑑𝐶𝐴
∆𝐶𝐴
−1
−1
𝜏𝑃 = ∫
≈
{(−𝑟𝐴 )−1
0 + (−𝑟𝐴 )𝑓 + 2 ∑ (−𝑟𝐴 )𝑖 }
−𝑟𝐴
2
𝑖=1
𝐶𝐴0
Most T is formed when CAf = 0, but that requires 𝜏 = ∞, so let’s
choose XA = 0.998.
𝐶𝐴
0.02
0.11
0.2
0.6
1
2
3
4
5
6
7
8
9
10
So,
𝜑
0.0959
0.1988
0.2501
0.3601
0.0345
0.0790
0.1238
0.1667
0.2070
0.2446
0.2795
0.3118
0.3418
0.3696
−𝑟𝐴
2.5031
6.6387
9.5954
0.3608
29.0000
50.6274
72.7128
96.0000
120.7771
147.1918
175.3320
205.2548
237.0000
270.5964
1
{[0.0345 + 0.3696
2
+ 2(0.079 + 0.1238 + ⋯ + 0.3118 + 0.3418)]
+ (1 − 0.05)[0.0345 + 0.0598]}
𝐶𝑇𝑃 =
𝐶𝑇𝑃 = 1.9729
10
𝜏𝑃 = ∫
4
𝜏𝑃 =
𝑘𝑚𝑜𝑙
𝑚3
4𝑓
𝑑𝐶𝐴
𝑑𝐶𝐴
+ ∫
−𝑟𝐴
−𝑟𝐴
0.02
1
{[270.6−1 + 96−1
2
+ 2(120.8−1 + 147.2−1 + 175.3−1 + 205.3−1
+ 237−1 ) + 0.1399]} = 0.0369 + 0.1399
𝜏𝑃 = 0.1768 𝑚𝑖𝑛
This gives, V = 177 litres
SOLUTION 7.20
Since we know that fractional yield is defined by,
𝐶𝑅
𝐶𝑅
Φm=𝐶𝐴0−𝐶𝐴=100−𝐶𝐴
CA
90
80
70
60
50
40
30
20
10
0
CR
7
13
18
22
25
27
28
28
27
25
Given,
CA0=100
CAf=20
𝐶𝐴𝑂
CRp=∫𝐶𝐴𝑓 ΦmdCA
=
ΔCA
=
(100−20)
*(0.75+0.25)
2
2
(Φ0 + Φf)
=40 mol/ lit
Φm
0.7
0.65
0.6
0.55
0.5
0.45
0.4
0.35
0.3
0.25
SOLUTION 7.21
CRm = Φm (ΔCA) =0.35 (100 – 20) = 28 mol/lit
SOLUTION 7.22
y = mx + b
Φm = 0.25 + (0.4/80) CA
CR = Φm (100 – CA)
=(0.25 + 0.005 CA)(100 – CA)
CR= 25+0.25CA-0.005CA2
𝑑𝐶𝑅
=0
𝑑𝐶𝐴
(for maximum CR)
=0.25-0.005(2)CA
Therefore, CA=25 mol/lit
Now putting this value in equation, we get
CR= 28.125 mol/lit
SOLUTION 7.23
Given,
A+B→ R+T
rR = 50 CA
A+B→ S+U
rS = 100 CB
Given that CA0+CB0=60mol/m3
And they both are equimolar, therefore CA0=30 mol/m3 = CB0
Also given, FA0=FB0=300 mol/hr
and, XA=0.9
Now we know,
𝐹𝐴0
CA0= 𝑣0
or, vo=
𝐹𝐴0 300
= =10
𝐶𝐴0 30
m3/hr
Now, rate of disappearance of A= rate of formation of R and S
Therefore, -rA=rR+rS
=50CA+100CB
=150CA→ (1)
Now,
Performance equation curve for MFR is given by:
τ
𝑋𝐴
𝑉
=−𝑟𝐴=𝐹𝐴0
𝐶𝐴0
Therefore,
τ=
𝑋𝐴∗𝐶𝐴0
150𝐶𝐴
Now, τ=
𝑋𝐴∗𝐶𝐴0
=
=
0.9
150𝐶𝐴0(1−𝑋𝐴) 150∗0.1
=0.06hr
𝑉
𝑣0
or, V=Volume of reactor
= 0.06*10= 0.6 m3= 600 litres.
SOLUTION 7.24
Performance equation for a PFR or Plug Flow Reactor is given by:
τ
𝐶𝐴0
=
𝑉
∫ 𝑑𝑋𝐴
=
𝐹𝐴0
−𝑟𝐴
Here, we know
-rA=150CA
=150*CA0*(1-XA)
Therefore,
τ
𝐶𝐴0
=
∫ 𝑑𝑋𝐴
150 ∗ 𝐶𝐴0 ∗ (1 − 𝑋𝐴)
or, τ = −
=-
ln(1−XA)
150
ln 0.1
=0.0153hr
150
Therefore, V= τ ∗ (v0)
= 0.0153*10=0.15 m3= 150 litres.
SOLUTION 7.25
Given,
CBeverywhere=3 mol/m3
50𝐶𝐴
Φ(R/A)=50𝐶𝐴+100𝐶𝐵
Therefore,
𝐶𝐴0
CRF= ∫𝐶𝐴𝐹 Φ(R/A)dCA
30
=∫3
𝐶𝐴
dCA
𝐶𝐴+2𝐶𝐵
Given CB=3
CRF=18.68
𝐶𝐴0
S
A
Similarly, CSF=∫𝐶𝐴𝐹 Φ ( ) dCA
30
=∫3
2𝐶𝐵
dCA
𝐶𝐴+2𝐶𝐵
or, CSF=8.32
Finally,
𝐶𝐴𝐹 𝑑𝐶𝐴
V=(FA0/CA0)∫𝐶𝐴0
=
−𝑟𝐴
300
300
𝑑𝐶𝐴
∗ ∫30 50𝐶𝐴+100(3)
30
= 0.2773 m3= 277.3 litres.
SOLUTION 7.26
𝑑𝐶𝑅
φ=−𝑑𝐶𝐴
CA
90
80
70
60
50
40
30
20
10
0
CR
1
4
9
16
25
35
45
55
64
71
dCA
10
10
10
10
10
10
10
10
10
10
Φ
0
0.3
0.5
0.7
0.9
1
1
1
0.9
0.7
Φ
1.2
1
0.8
0.6
Φ
0.4
0.2
0
0
20
40
60
80
100
a) For MFR,
CR=φ10(CA0-CAf)
=0.9(100-10)
=0.9*90
=81 mol/lit
b) For PFR,
φNmixed=
𝜑1(𝐶𝐴0−𝐶𝐴1)+𝜑2(𝐶𝐴1−𝐶𝐴2)+⋯
(𝐶𝐴0−𝐶𝐴𝑁)
10
=100(0.3+0.7+0.9+0.5+3+0.7+0.9)
= 0.1*7= 0.7
Therefore,
CR= 𝜑(𝐶𝐴0 − 𝐶𝐴𝑓)
= 0.7*(100-10)= 0.7*90=63 mol/lit
c) Given,
CA0=70
Therefore,
For MFR,
CR= φ10(CA0-CAf)
=0.9(70-10)
=0.9*60
= 54 mol/lit
d) For PFR,
φNmixed=
𝜑1(𝐶𝐴0−𝐶𝐴1)+𝜑2(𝐶𝐴1−𝐶𝐴2)+⋯
(𝐶𝐴0−𝐶𝐴𝑁)
10
= 70(0.7+0.9+3+0.7+0.9+0.5)
= 0.142*6.7=0.9571
CR= φNmixed(CA0-CAf)
=0.9571*(70-10)=57.426 mol/lit
SOLUTION 7.28
Given,
A→ R
rR=1
A→ S
rS=2CA
A→ T
rT=CA2
Now, given flow rate of the reactor = 100 lit/sec
i.e. v0=100
Now –rA = rR+rS+rT
= 1+2CA+CA2
= (1+CA)2
Now, taking the first case i.e. MFR from 2-1
Therefore, here, CA0=2;
CAf=1
For MFR,
𝐶𝐴0−𝐶𝐴𝑓 2−1
= 4
−𝑟𝐴
τm=
= 0.25
therefore, V1= τ*(v0)
=0.25*100= 25 litres
Now taking the second case for PFR,
Here CA1=1; CA2=0
Therefore,
0 𝑑𝐶𝐴
τp=∫1
=-[
−𝑟𝐴
0
=∫1
1
]=
1+𝐶𝐴
𝑑𝐶𝐴
(1+𝐶𝐴)2
1
2
1- = 0.5
Therefore,
We get, V2= 0.5*(100)
= 50 litres
Hence, total volume required = V1+V2
= 25+50=75 litres
SOLUTION 7.27
Let B be the no of british ships and F be the no of French Ships and if French wins
dF/dt=-KF…..(1)
dB/dt=-KB….(2)
Div 1 by 2
dF/dB=F/B
Integrating we get,
𝐹
𝐵
∫ 𝐹𝑑𝐹 = ∫ 𝐵𝑑𝐵
𝐹0
𝐵0
F2-F02=B2-B02
Since british is losing B=0
F2=332-272=360
F=19 ships(approx)
8.1 (a) r1=k1CACB2
r2=k2CRCB
(b) r1=k1CACB
(c) r1=k1CACB
r2=k2CRCB2
r2=k2CR2CB
(d) r1=k1CA2CB
r2=k2CRCB
Sketch the best contacting patterns for both continuous and non-continuous
operations.
Solution 8.1
Continuous
A
PFR
PFR
B
(a)
A
PFR
B
(b)
A
PFR
B
(c)
A
PFR
B
(d)
Non-Continuous
Instantaneously
A
B
Batch
(a)
B
(Drop by drop)
Instantaneously
B
A
A
(b)
B
(Slowly)
A
(c)
(d)
Solution 8.2
a) PFR
b) MFR
CA0 = 1 mol/L CR0 = CS0 = 0
𝐶𝑅,𝑚𝑎𝑥
𝐶𝐴0
= e-1
1
𝑒
1
tmax = k-1 = 0.1 = 10 min
10 =
𝐶𝐴0
1
=
[(𝑘2 ⁄𝑘1)
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑜𝑟
𝑣𝑜𝑙𝑢𝑚𝑎𝑡𝑟𝑖𝑐 𝑓𝑙𝑜𝑤𝑟𝑎𝑡𝑒
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑜𝑟
1000⁄
60
1/2
2
+1]
1
CRmax =
CR,max = = 0.3679 mol/L
recidance time =
𝐶𝑅,𝑚𝑎𝑥
2
[(0.1⁄0.1)1/2 +1]
CRmax = 0.25 mol/L
1
tmax = √𝑘
1 𝑘2
tmax = 10 min
Volume of reactor = 166.67 L
Volume of reactor = 166.67 L
Here the volume to get maximum of R is 166.67 L but the maximum
concentration of R in PFR is 0.3679 mol/L and in MFR 0.25 mol/L. So
maximum concentration of R can be obtained from PFR.
Solution8.3
Assuming parallel reaction A
CA0 =100 mol/L CR0=CS0=0
Run
CA
CR
CS
1
2
75
25
15
45
10
30
τ=
𝐶𝑅 −𝐶𝑅0
τ=
𝐶𝑆 −𝐶𝑆0
𝑘1 𝐶𝐴
𝑘2 𝐶𝐴
---(1)
---(2)
Residence time will be equal from both equations
From eq. (1) and (2)
𝐶𝑅
𝑘1
=
𝐶𝑆
𝑘2
&
𝐶𝑅
𝐶𝑠
=
𝑘1
𝑘2
For both the runs the value of
𝑘1
𝑘2
is same which is 1.5
Therefore, the assumption of parallel reaction is correct.
𝑪𝑹
𝒌𝟏
=
𝑪𝒔
𝒌𝟐
1.5
1.5
Solution8.4
CA0 =100 mol/L
𝐶𝑅0 =CS0=0
Assuming parallel reaction taking place
Run
CA
CR
CS
1
2
75
25
15
45
10
30
τ=
𝐶𝑅 −𝐶𝑅0
τ=
𝐶𝑆 −𝐶𝑆0
𝑘1 𝐶𝐴
𝑘2 𝐶𝐴
𝑪𝑹
𝒌𝟏
=
𝑪𝒔
𝒌𝟐
1.976
0.666
---(1)
---(2)
Residence time will be equal from both equations
From eq. (1) and (2)
𝐶𝑅
𝐶𝑆
=
𝑘1
𝑘2
𝐶𝑅
𝑘1
=
𝐶𝑠
𝑘2
For both the runs the value of
𝑘1
𝑘2
is coming different
Therefore, the assumption of parallel reaction is not correct.
Now, assuming series reaction taking place
Run
CA
CR
CS
𝑪𝑹
𝒌𝟏
=
𝑪𝒔
𝒌𝟐
1
50
33.3
16.7
1.976
0.5
0.33
0.167
0.8
2
25
30
45
0.666
0.25
0.3
0.45
0.8
CA/CA0 CR/CA0 CS/CA0
𝒌𝟐
𝒌𝟏
Solution8.5
Assuming parallel reaction
Run
CA
CR
CS
1
2
50
20
40
40
10
40
τ=
𝐶𝑅 −𝐶𝑅0
τ=
𝐶𝑆 −𝐶𝑆0
𝑪𝑹
𝒌𝟏
=
𝑪𝑺
𝒌𝟐
4
1
---(1)
𝑘1 𝐶𝐴
---(2)
𝑘2 𝐶𝐴
Residence time will be equal from both equations
From eq. (1) and (2)
𝐶𝑅
𝐶𝑆
=
𝑘1
𝑘2
𝐶𝑅
𝑘1
=
𝐶𝑠
𝑘2
For both the runs the value of
𝑘1
𝑘2
is coming different
Therefore, the assumption of parallel reaction is not correct.
Now, assuming series reaction
Run
CA
CR
CS
𝑪𝑹
𝒌𝟏
=
𝑪𝑺
𝒌𝟐
CA/CA0
CR/CA0
CS/CA0
𝒌𝟐
𝒌𝟏
1
50
40
10
4
0.5
0.4
0.1
0.3
2
20
40
40
1
0.2
0.4
0.4
0.3
Solution 8.6
As grinding will progress as,
large particles → appropriate particles→ small particles
denoting this process as,
A→R→S
a) Basis: 100 particles (10 A, 32 R and 58 S)
Too small particles so you have to reduce the residence time,
increasing the feed flow to make an estimate, assume that a series of
first reaction order may represent milling process.
XA = 0.9 and CR/CA0 = 0.32 is that k2/k1≈0.7
If
𝑘2
𝑘1
≈ 0.7 ⇒CRmax / CA0 = 0.5 and XA= 0.75 and will get 25%
of very large particles, 50% of particles of appropriate size and 25%
of very small particles.
b) Multi-stage is better, as single stage grinder will act as single MFR
and multi-stage grinder will act as MFR in series which will act as
PFR and in PFR we get more uniform product. So using multistage
grinder is best option.
Solution 8.7
a) A0 = 1
B0=3
B consumed is 2.2 so ∆B=0.8
S=0.2
𝑘2
𝑘1
= 0.85
𝑘
For S=0.6 and 2 = 0.85 from graph R=0.3
𝑘1
A+B→ R
&R+B→ S
To produce 0.6 mol of S, B required is 0.6 and R required is 0.6
The final mol of R is 0.3 so total mol of R must be produced is 0.9
and to produce it 0.9 bole of A and B required
Therefore ∆B=0.6+0.9 = 1.5
The final concentration of A, B, R and S is 0.1, 1.5, 0.3 and 0.9
respectively.
b)
Only S is found so as the R is formed it is directly converted to S
and it indicates that k2>> k1 which cannot be determined form the graph
shown above.
c)
∆B
𝐴
= 1 and S=0.25 from graph
𝑘2
𝑘1
= 0.4
Solution 8.8
a) Here,
𝑘2
𝑘1
= 0.8
From graph concentration of diethyl-aniline is 0.2mol and
mono-ethyl-aniline is 0.4mol
Water will be formed in second reaction is 0.2mol
For producing 0.2 mol of di-ethyl-aniline 0.2 mol of monoethyl-aniline will be consumed so total mono-ethyl-aniline produced
in first reaction is 0.6 mol and 0.6 mol of water will be formed
Total mol of water formed is 0.8
0.6 mol of ethanol in first and 0.2 mol ethanol in second reaction will
be consumed so total ethanol consumed is 0.8 mol
0.6 mol of aniline will be consumed in first reaction
Final composition of mixture from batch reactor
Aniline
Ethanol
0.4
0.2
Mono-ethylaniline
0.4
Di-ethylaniline
0.2
Water
0.8
b)
For 70% conversion of ethanol value of ∆B = 1.4
∆B
𝐴
= 1.4
For this data final moles of mono-ethyl-aniline and di-ethyl-aniline are
0.2 mol and 0.6 mol respectively.
Ratio of mono to di-ethyl-aniline is 0.333
c)
From graph for equimolar feed in plug flow reactor for highest
concentration of mono-ethyl-aniline the conversion of A is 67%
From graph ∆B = 0.92
Therefore, overall conversion of B is 92% considering both reactions
Solution 8.9
Considering aniline (A), ethanol (B), mono-ethyl-aniline (R) and
di-ethyl-aniline (S)
A0 = 1 mol, B0 = 2 mol
∆B
𝐴
= 0.56
After reaction conversion of A is 40% so 60%will remain unconverted
A = 0.6
B reacted in first reaction is 0.4 mol
The remaining final mixture R is 3 parts and S is 2 parts to produce 2
parts of S, 2 parts of R must react in second reaction so total R formed in
first reaction is 5 parts which is 0.4 mol.
2
2 parts reacted so remaining moles of R is 0.4-( * 0.4) = 0.24
5
S formed is 0.16 mol
The value from graph
𝑘2
𝑘1
=1
The final composition in outlet of MFR
Aniline
Ethanol
0.6
0.44
Mono-ethylaniline
0.24
Di-ethylaniline
0.16
Water
0.56
Solution 8.10
pH
2
3
4
5
6
a) Here, the value of
k1
11
8
7
8
11
𝑘1
𝑘2
k2
3
4
5
6
7
k1/k2
3.666667
2
1.4
1.333333
1.571429
is highest at pH=2 so the pH which should be
maintained should be slight higher than 2 and should be maintained
throughout the reactor to get maximum of R as the k 1 will be higher
compare to k2 then R will be more formed. MFR should be used to
maintain pH throughout the reactor as 100% mixing and uniform pH
will be there in MFR.
b) As, the pH required to be constant no PFR is used and in MFR no
stages with different pH should be kept.
CRE-I Term paper
Sums 8.11-8.21
Group members: 13BCH012, 13BCH013, 13BCH037, 13BCH045, 13BCH055
Problem 8.11
As S is desired it requires both k1 as high as k2
pH
k1
k2
2
3
11
3
4
8
4
5
7
5
6
8
6
7
11
When pH=6, k1=7(maximum value) and k2=11(maximum value), so it is best to work with pH=6
and keep it constant.
Problem 8.12
(a)R is in series with A and S, so it is best to use the plug flow reactor.
(b) As in parallel reactions CRis the area under the curve Ф vs CA. So, let’s take Ф = f (CA)
𝑑𝐶
Ф = − 𝑑𝐶𝐴 = (k1* CA – k4*CR) / k12*CA
𝑅
, where k12= k1+k2
It is not possible to separate variables and integrate because Ф is also a function of CR; but it is a
linear differential equation of first order with integral factor.
𝑑𝐶𝑅
𝑑𝐶𝐴
𝑘
– (k4*CR)/ (k12*CA) =-𝑘 1
12
Comparing this equation with,
𝑑𝑦
𝑑𝑥
+ P(x) y = Q(x)
y = CR , x = CA , P(x) = - 𝑘
y = е−
𝑃 𝑥 𝑑𝑥
е−
𝑃 𝑥 𝑑𝑥
𝑃 𝑥 𝑑𝑥
е
𝑄 𝑥 е
𝑃 𝑥 𝑑𝑥
𝑄(𝑥)е
𝑃 𝑥 𝑑𝑥 = − (𝑘
𝑘4
𝑘4
12 ∗𝐶𝐴
,
12 ∗𝐶𝐴
𝑘
Q(x) = -𝑘 1
12
𝑑𝑥
)dCA = ln 𝐶𝐴 𝑘 4 /𝑘 12
= CA
= 𝐶𝐴 −𝑘 4 /𝑘 12
𝑃 𝑥 𝑑𝑥
𝑘1
𝑑𝑥 = -𝑘
12
𝐶𝐴
𝑘 4 /𝑘 12
−𝑘 1
dCA = ( 𝑘 ) ∗
𝐶𝐴
12
𝑘
(1− 4 )
𝑘 12
𝑘
(1− 4 )
𝑘 12
Put all these values in main equation,
CR= 𝐶𝐴
−𝑘 4 /𝑘 12
−𝑘 1
{( 𝑘 )*( 𝐶𝐴
𝑘
(1− 4 )
𝑘 12
12
𝑘
/ (1- 𝑘 4 )) + constant}
12
𝑘1
When CA = CA0 then CR = 0 and constant = (𝑘 ) ∗
12
CR= 𝐶𝐴
𝑘 4 /𝑘 12
(
−𝑘 1
𝑘 12
)∗
𝐶𝐴
𝑘
(1− 4 )
𝑘 12
𝑘
(1− 4 )
+ (
𝑘 12
𝑘1
𝑘 12
)∗
𝐶𝐴 0
𝐶𝐴 0
𝑘
(1− 4 )
𝑘 12
𝑘
(1− 4 )
𝑘 12
𝑘
(1− 4 )
𝑘 12
𝑘
(1− 4 )
𝑘 12
Dividing the whole equation for CA0 and removing the common factor constant ratio within the
key and performing the indicated multiplication,
𝐶𝑅
𝐶𝐴 0
=
𝑘1
𝑘 12
𝑘
1− 4
𝑘 12
𝑘4
𝑘4
−
𝐶𝐴 0 𝐶𝐴 0 𝑘 12 𝐶𝐴 𝑘 12
𝐶𝐴 0
−
𝐶𝐴
𝑘
𝑘
(1− 4 + 4 )
𝑘 12 𝑘 12
𝐶𝐴 0
𝐶𝑅
𝑘1
=
𝐶𝐴0
𝑘12 − 𝑘4
𝐶𝐴
𝐶𝐴0
𝑘4
𝑘 12
−
𝐶𝐴
𝐶𝐴0
Note that you could have obtained from the equation 8.48 (p. 195) by k34 = k3+ k4= k4 (because
k3= 0)
𝐶𝑅
𝐶𝐴 0
=
𝑘1
𝑘 12
𝑘4
𝑘 12 𝑘 4 −𝑘 12
𝑘4
3
4
=4
0.2
0.2−4
= 0.865
0.2
CR = 2(0.865) = 1.73mol
Problem no. 8.13
(a)
k1=40, k2=10, k3=0.1, k4= 0.2
CS = 0.2CA0
k12 = k1 + k2 = 50; k34 = k3 + k4 = 0.1 + 0.2 = 0.3
𝐶𝑆
𝐶𝐴 0
=𝑘
𝑘1 𝑘3
34 −𝑘 12
е1−𝑘 34 𝑡
(
𝑘 34
−
е−𝑘 12 𝑡 exp ⁡
(−k12t)
𝑘 12
k12
)+𝑘
𝑘1 𝑘3
34 −𝑘 12
We have, CA0 + CR0 + CT0 + CU0 = CA + CR + CT + CU
Or CA0 = CA + CR + CT + CU
Therefore,
0.2𝐶𝐴 0
𝐶𝐴 0
(40)(0.1) е−0.3𝑡
=
0.3−50
(
0.3
-
е−50 𝑡
50
40∗0.1
) + 50∗0.3 +0 +0
𝐶
𝑘
𝐶
+ 𝐶 𝑅0𝑘 3 (1 – е−𝑘 34 𝑡 ) + 𝐶 𝑆0
𝐴 0 34
𝐴0
→ 0.2 =
−0.0804
15
(50e-0.3t – 0.3e-50t) + 0.2667
Or 12.44 = 50e-0.3t – 0.3e-50t
→ ln12044 = ln50 – 0.3t – ln0.3 + 50t
→2.52 = 3.91 + 49.7t + 1.2
So, t = -0.0521 minutes.
Negative value of time is not possible. Thus there is some error in the data.
(b)
The above rate constants show that the first reaction is a slow reaction and the second reaction is
a fast reaction.
CA0 + CR0 + CT0 + CU0 = CA + CR + CT + CU
𝐶𝑆
𝐶𝐴 0
=𝑘
𝑘1 𝑘3
34 −𝑘 12
е1−𝑘 34 𝑡
(
𝑘 34
−
е−𝑘 12 𝑡
𝑘 12
)+𝑘
𝑘1 𝑘3
34 ∗𝑘 12
Let CA0 = 100 mol/lit
So, CS = 0.2*100 = 20 mol/lit
0.2𝐶𝐴 0
→
𝐶𝐴 0
(0.2)(10) е−30 𝑡
= 30−0.03 (
30
Or 0.02 = 6.67x10-3(
-
е−0.03 𝑡
0.03
0.02∗10
) + 0.03∗30
0.03е−30 𝑡 −30е−0.03𝑡
30∗0.03
) + 0.222
Therefore, 0.2 – 0.222 = 7.414*0.03*10-3e-30t – 70417*30*e-0.03t*10-3
0.0222 = -2.224x10-4e-30t + 0.2224xe-0.03t
-8.411 – 30t – 1.4961 + 0.03t = -3.867
-8.417 – 1.4961 + 3.8167 = 29.97t
Thus t =
−6.0904
29.97
= - 0.2032 minutes
Which is negative so the data given is incorrect.
Problem 8.14
(a) Arguably k3> k4 and k5> k6. we cannot conclude anything about k1 and k2 because although
the branch of R there are fewer moles than by the branch of S may be that k1> k2 and k3 and k4 to
be small and has accumulation of R may also be that k1<k2 and k1<k3 and k4 so that all R formed
pass T and U.
(b)If the reaction was now complete and is T, U, V and W only by the branch above 5 moles of T
and 1mole of U formed, so it was 6 moles of R transformed to U and T, whereas below the
branch 9 mol of V and 3 mol of W were formed, meaning that there were 12 mol of S. In this
case can be concluded that,
k1<k2
Formation rate of R =
Formation rate of S =
𝑘
𝑘
𝑑𝐶𝑅
𝑑𝑡
𝑑𝐶𝑆
𝑑𝑡
𝐶
= k1CACB
= k2CACB
6
dCR = 𝑘 1 dCs ═>𝑘 1 = 𝐶𝑅 = 12 ═> k2=2k1
2
2
𝑆
Formation rate of T =
𝑑𝐶𝑇
Formation rate of U =
𝑘
𝑘
𝑑𝑡
= k3CACB
𝑑𝐶𝑈
𝑑𝑡
= k4CACB
𝐶
5
dCT = 𝑘 3 dCU ═>𝑘 3 = 𝐶 𝑇 = 1 ═> k2=5k1
4
4
Formation rate of V =
Formation rate of W =
𝑘
𝑘
𝑈
𝑑𝐶𝑉
𝑑𝑡
= k5CACB
𝑑𝐶𝑊
𝑑𝑡
= k6CACB
𝐶
9
dCV = 𝑘 5 dCW ═>𝑘 5 = 𝐶 𝑉 = 3 ═> k2=3k1
6
6
valid for (a)
𝑊
valid for (a)
Problem 8.15
For series reaction, the overall rate constant is given by,
1
𝑘13
1
1
1
3
1
1
= 𝑘 + 𝑘 = 0.21 + 4.2 = 5
So, k13= 0.2
Overall rate constant is given by,K=k13+k2= 0.2 + 0.2 = 0.4
Since S is desired product, if we use mixed flow reactor, there will be more mixing and hence
intermediate formation will be less. So we can use plug flow reactor.
By using equation,
𝑘4
𝐶𝑆𝑚𝑎𝑥 𝑘123 𝑘4−𝑘123
=𝑘
𝐶𝐴0
4
0.004
0.4
=0.004 0.004 −0.4
= 0.9545
CSmax= 0.9545 CA0
𝑘
ln ⁡
( 2)
ԏ=𝑘
𝑘1
2 −𝑘1
0.004
ln ⁡
(
)
0.4
= 0.004−0.4
= 8.128
[therefore k1 = k123 = 0.4]
As we know equation,
CA = CA0е−𝑘1 𝑡
But CA0 = 100 е- (0.4*8.128)
=100*0.03872
=3.872 mol/L
XA =
𝐶𝐴0 −𝐶𝐴
𝐶𝐴0
=
100−3.872
100
= 96.12%
Problem no. 8.16
The initial gravel handling rate of shovel is 10 ton/min. It is also mentioned that the shovel’s
gravel handling rate is directly proportional to the size of the pile still to be moved. So it shows
that the removal of the sand using shovel can be compared to the first order reaction.
The conveyor on the other hand works at the uniform rate of 5 ton/min. This shows that the
working of the conveyor is independent of the amount of sand in the hopper. So it can be
considered as the zero order reaction.
So for shovel,
10 ton/min=k1 * 20000……… (1)
k1= 10/20000= 5 * 10-4
For Conveyor,
5 ton/min= k2
(a)
K= 𝐶
𝐶𝑏𝑖𝑛
=
𝐶𝐴0
𝑘2
𝐴 0 𝑘1
1-K(1-ln K)
5
= 20000 ∗5∗10 −4 = 0.5
Cbin= 20000 * (1-0.5(1-ln 0.5)) = 3068.52 tonnes
(b)
tRmax= ln(1/K)/k1= ln(1/0.5)/(5*10-4)= 1386.29 min
(c)
Input= Output
k1 *CA0= k2
5*10-4 *CA0= 5
CA0= 10000 tonne
(d)
When Cbin=0,
𝑘2 𝑡
𝐶𝐴 0
+ е−𝑘1𝑡 = 1
By using trial and error method,
t=0.0132min
Problem no. 8.17
A→R→S
Where, A = Garbage still to be collected
A0 = 1140 tonnes (initial garbage load)
R = Garbage in bin
S = Garbage in incinerator
Since the rate of collection is proportional; to amount of garbage still to be collected –
Therefore, A → R is a First order reaction.
Since the conveyor is operated at uniform rate to the incinerator –
R → S is a zero order reaction.
Therefore,
−𝑑𝐴
𝑑𝑡
= k1A
At t = 0,
−𝑑𝐴𝑜
𝑑𝑡
k1=
= 6 tons/min
−𝑑𝐴
𝑑𝑡
A
=
6
1
= 240 min-1
1140
A → R … (n1 = 1, k1 = (1/240) min-1)
R → S … (n2 = 0, k2 = 1 ton/min)
(a) For 95% of A collected –
𝑑𝑅
= k1A – k2
𝑑𝑡
From the given equation:
𝐶𝐴
𝐶𝐴 0
= е−𝑘1𝑡
A0 = 1140 tonnes
A (at 95%) = 0.05*1440 = 72 tonnes
1
k1 = 240
𝐴
𝐴𝑜
72
= 1140 = e-t/240
1
Therefore, 20 = e-t/240
1
−𝑡
Or, ln (20 ) = 240
So, t = 718 min = 12 hrs.
(b) For maximum collection in bin,
𝑅𝑚𝑎𝑥
𝐴𝑜
= 1-k (1 - lnk)
Thus, k =
(𝑘 2 /𝐴𝑜)
𝑘1
=
(1/1440 )
(1/240)
= 0.1667
Therefore, Rmax = 1140[1- (0.1667) (1 – ln (0.1667))] = 769.9 ton
1
1
1
(c) tmax =𝑘 ln (𝑘 ) = 240 ln (0.1667 ) = 430 min
1
(d) Bin empties when R = 0
Therefore, R = 0 = A0 (1 – e-k1*t) – k2t
Implies, 0 = 1440(1 – e-t/240) – t
𝑡
Or, e-t/240 + 1440 = 1
By using Goal Seek method,
t = 1436 min
Problem no. 8.18
Suppose Upper Slobbovians = U and Lower Slobbovians = L
Here it is given that,
−𝑑𝑈
𝑑𝑡
= L and
−𝑑𝐿
𝑑𝑡
=U
Last week, after the encounter of 10U and 3L, 8 U survived
It means that 2 U died in this encounter and all the 3 L died.
So, when 1U dies 1.5 L die
(a) Upper Slobbovians are better fighters. And one Upper Slobbovian is as good as 1.5
Lower Slobbovians.
(b) When 10 U and 10 L encountered
1.5L → 1U
10L →
1∗10
1.5
= 6.6 ≈ 7U
So here we can conclude that, after this encounter 3 U lived and 7 U died. Whereas all the 10 L
were killed.
Problem no. 8.19Since chemical X is added slowly and it dissolves quickly, this shows that it is a zero order
reaction.
Since Y decomposes slowly it shows that it is a 1st order reaction.
Thus it can be said that, the overall reaction is zero order followed by 1st order.
X→Y … (n=1)
Y→Z … (n=2)
CX0 = 100 mol/m3
Now,
−𝑑𝐶𝑋
𝑑𝑡
−𝑑𝐶𝑌
= k1 =
100
0.5
= 200 mol/ (m3.hr) … as n=0
= k2CY
𝑑𝑡
Overall rate constant can be given by ‘k’ where –
K=
𝑘 2 ∗𝐶0
𝑘1
=
1.5∗100
200
= 0.75
(a) At half hour Ymax
𝐶𝑌𝑚𝑎𝑥
𝐶𝑋 0
1
1
= 𝑘 (1– e-k) = 0.75(1 – e-0.75) = 0.7035
Or, CYmax = 100(0.7035) = 70.35 mol/m3
Thus, tmax =
𝐶𝑋 0
𝑘1
100
= 200 = 0.5 hours
(b) After 1 hour, CX0 = 0
𝐶𝑌
1
𝐶𝑋 0
1
= 𝑘 (е𝑘1−𝑘1𝑡 - е−𝑘 2 𝑡 ) = 0.75(e-0.75-1.5-e-1.5) = 0.3323
CY = 100(0.3323) = 33.23 mol/m3
CZ = CX0 – CX - CY = 100 – 0 – 33.23 = 66.7 mol/m3
Problem no. 8.20
(1)
𝑑𝐶𝑋
𝑑𝑡
≠ 0 and
𝑑𝐶𝑅
𝑑𝑡
≠0
CA = CA0 - kt
CA - CA0 = kt
From initial slope of CX Vs t and CR Vs t, which is not zero we can say that A gives X and R in a
Parallel reaction.
A→X
A→R
(2) But when A is completely exhausted, still CX decreases and CR and thus X decomposes to
give the final product R.
A→X→R
Therefore A→X, A→R, and X→R
We know, CA0 = CX + CA + CR
Therefore CX = 100 – CA – CR
t(min)
0
0.1
2.5
5
7.5
10
20
∞
CA(mol/m3)
100
95.8
35
12
4
1.5
-
0
CR(mol/m3)
0
1.4
26
41
52
60
80
100
CX(mol/m3)
0
2.8
39
47
44
38.5
20
0
CA = CA0e-kt
Or lnCA = ln CA0 – kt
Thus from the graph of lnCA vs t we get a straight line with intercept lnCA0. Thus it follows 1st
order kinetics.
𝐶
We know, ln 𝐶𝐴 0 = (k1 + k2)*t
𝐴
And k1 + k2 = k12
ln 95.8 −ln ⁡
(12)
Therefore, k12 =
0.1−5
Or k12 = 0.42 min-1
𝑑𝐶
1.4
( 𝑑𝑡𝑅 )t=0 = 0.1 = k2CA0
1.4
Thus k2 =
𝑑𝐶
0.1 ∗(100)
= 0.14 min-1
2.8
( 𝑑𝑡𝑋 )t=0 = 0.1 = k1CA0
So k1 =
2.8
0.1 ∗(100)
= 0.28 min-1
Therefore, X is the intermediate product and has a maximum at 5 minutes.
→ for CR0 = CX0 = 0,
𝑘 34
𝐶𝑋𝑚𝑎𝑥
𝑘 1 𝑘 12 𝑘 34 − 𝑘 12
= 𝑘 *𝑘
𝐶𝐴 0
2
34
Here K34 = k3
Thus
𝐶𝑋𝑚𝑎𝑥
𝐶𝐴 0
47
= 100 = 0.47
𝑘3
0.28
Or 0.47 = 0.42 ∗
0.42 𝑘 3 − 0.42
𝑘3
By using seek method we get, k3 = 0.067
To verify whether the method is correct,
Recalculating CR,
𝐶𝑅
𝐶𝐴 0
=𝑘
𝑘1 𝑘3
3 −𝑘 12
(
е−𝑘 3 𝑡
𝑘3
-
0.28(0.067) е−0.067 ∗5
= 0.067−0.42 (
0.067
е−𝑘 12 𝑡
𝑘 12
−
𝑘
𝑘
) + 𝑘 1 + 𝑘 2 (1 – е−𝑘 12 𝑡 )
2
е−0.42 ∗5
0.42
12
0.28
0.14
) + 0.42 + 0.42(1-е−0.42∗5 )
𝐶
→ 𝐶 𝑅 = 0.4087
𝐴0
Therefore CR = 0.4087x100 = 40.87 or 41 mol/m3
At CXmax, i.e. t = 5 minutes
CR = 41 mol/m3
Problem no. 8.21
k1 =6 hr-1
k2 = 3hr-1
k3 = 1hr-1
CA0 = 1 mol/litre
𝐶𝑅𝑚𝑎𝑥
𝐶𝐴 0
𝑘1
𝑘 34
𝑘 12 𝑘 34 − 𝑘 12
=𝑘 ∗𝑘
12
34
Since k1 = k12 for one case,
𝐶𝑅𝑚𝑎𝑥
𝐶𝐴 0
3
6 6
= 6(4)(4/6)
2
= 23
4
Therefore, CRmax = 9 = 0.44 mol/litre
tmax =
ln ⁡
(4/6)
4−6
= 0.202 hours.
Cre Term Paper
Submitted by – 13bch005, 13bch023,13bch041,13bch042
Problem 9.1
For the reacting system of example 9.4
A) What τ is required for the 60 % conversion of reagent using the optimal progression of
temperature in a flow reactor in piston?
B)Find the outlet temperature of the reactor. Use k10=33*10^6 , k20= 1.8*10^18
Use any information you need from the example 9.4
Solution
A)
The treaty system in the example 9.4 A↔ R with -rA = k1 CA - K2 CR, where k1 =exp
(17,34 - 48900/RT) and k2 = exp (42.04 - 124200/RT) and C0= 4 mol/L. The chart shown in
the example, to not be squared, makes the collection of data from it is very vague and that is
why we are going to develop the necessary data, without using the chart.
If you want to find the optimal profile is considered that in the same
With the previous equations and data taken from the example you can assess the temperature
of the optimal profile for each XA and then see how varies - r with XA along the optimal
profile
XTo
Tópt (K)
-rto
1/-RTO
0
368
15.54
0.06
0.1
386,47
12.29
0.08
0.2
373,54
9.04
0.11
0.3
365,43
5.89
0.17
0.4
359,02
3.79
0.26
0.5
353,35
2.43
0.41
0.6
347,84
1.49
0.67
0.7
342,04
0.84
1.19
0.8
335,22
0.40
2.5
With these values are estimated the volume of flow reactor in piston, using area under the
curve
τ= CA0
𝑋𝑎 𝑑𝑋𝐴
0.6 𝑑𝑋𝐴
=
4
0 −𝑟𝐴
0 −𝑟𝐴 = 0.558 min
B)
In the table above we see that if X = 0.6 The temperature in the optimal profile is 347,84 K =
74,84°C
Problem 9.2 (p. 238)
You want to convert the concentrated aqueous solution of previous example (C0 = 4 mol/L;
F0 = 1000 mol/min) up to 70 % with the smaller size of reactor for thorough mixing. Make
an outline of recommended system, indicating the temperature of the input and output current
and the required space weather
Solution
In the table that appears in the problem 9.1 we take the temperature and the speed of the
optimal profile X = 0.7
T = 342.04 - 273 = 69°C, -rA = 0.84 mol/l min
𝜏m= CA0XA/-rA= 4(0.7)/.84 = 3.33 min
XA= Cp’(T-T0)/-∆Hr , T0= T+( XA∆HR)/CP= 20◦C
Problem 9.3 (p. 238)
With regard to the flow reactor in piston which operates on the optimal profile
The example 9.4 (C0 = 4 mol/L; F0 = 1000 mol/min; X = 0.8; Tmin = 5°C; Tmax = 95°C)
and power and current product to 25°C, much heat or
Cooling will be required
(A)to the supply current
(B)In the reactor itself
(C)For the output current
Solution
In the table that appears in the problem 9.1 The temperature of the optimal profile for XA =
0.8 is 335,22 K = 62,22°C
25°C 95°C
62.2°C
Q1
25°C
Q3
Q2
Q1= cp’(Tsat—Texit)=250*4187*70◦C= 73272,5J/mol A= 73272,5 *1000= 73272.5kJ/min
Q3= cp’(Tsat—Texit )=250 *4187*(25-62.2)*1000=-38939.1kJ/min
Q2= cp’(Tsat—Texit )=250*4180*(62.2-95)*1000=34333.4kJ/min
There is that to provide the power 73272.5 kJ/min, while the output current there is that
extract 38939,1 J/min. There are also removed from the reactor 34333.4 kJ/min. In total there
are that remove 73272.5 kJ/min. I suggest the following
Problem 9.4 (p. 238)
It plans to carry out the reaction of the example 9.4 (C0 = 4 mol/L; F0 = 1000 mol/min) in a
flow reactor in piston which is maintained at 40°C to XTO = 90%. Find the required volume
Solution
Constant density system
k1τ /XA = ln(XA/(X*-XA))
From example 9.4
k1=exp (17.34-48900/RT)min-1
K= exp( 75300/RT-24.7)
R=8.314 J/mol K
T = 40 + 273 = 313
k313 = 0.2343 min-1 ,
K313 = 69.1405
X AE = K/ (K+1) = 0.98
τ p=
0.98
0.98
Ln
=10.48 min
0.2343 0.98 -0.9
Problem 9.5 (p. 238)
Rework the example 9.4 replacing C0 with 1 mol/L/h.
Example 9.4. Using the optimal progression of temperature in a flow reactor in piston to the
reaction of the previous examples. Tmax = 95 °C
(a)Calculate the time space and the volume required for the 80 % conversion of 1000 mol
to/min with C0 = 4 mol/L
B)Plot the temperature and conversion profile along the reactor
Solution
Constant density system because it is Liquid
-rA = k1 CA0 (1 - X) - k2 CA0 XA = CA0[k1 (1 – XA) - k2 XA]
(-rA)1 = [k1 (1 – XA) - k2 XA]
(-rA)4 = 4 [k1 (1 – XA) - k2 XA] = 4 (-rA)1
0.8
∫
dXA
0.8
=∫
(-rA )1
1
0
dXA
dXA
0.8
= 4∫
(-rA )4
0
(-r
A)4
0
4
0.8
∫
dXA
=0.405 taken from the example 9.4 ( p. 230)
(-rA )4
0
0.8
∫
dXA
(-rA )
1
=4(0.405)
0.8
τ p = CA0
∫
0.8
dX A
(-rA )1
=1 ∫
dXA
(-rA)
=(1)(4)(0.405)=1.63
min
0
v0=FA0/CA0
=1000L/min
V= τv0=1620L
As they are reactions of first order the change in the value of C0 did not affect the value of τ.
The volume if it is affected because F0 remained constant and therefore v0 increased 4
times,causing the volume is 4 times more.
Volume
(L)
Problem 9.6 (p. 238)
Rework the example 9.5 replacing CA0 with 1 mol/L
Example 9.5. The concentrated solution of the previous examples (C0 = 4 mol/L; F0 = 1000
mol/min is going to be 80% converted into a complete mix reactor
(A)What size reactor is required?
(B) What should be the transfer of heat if the power is at 25°C and the output current must be
at the same temperature?
Solution
A) in the table of the problem 9.1 appears which the speed in reported the optimal profile X =
0.8 is 0.4 mol/l min; but for CA0 = 4 mol/l and that is needed is the corresponding to C0 = 1
mol/L
-rA)1=-rA)4/4 =0.4/4= 0.1 mol/Lmin
τ=CA0XA/-rA=1(0.8)/0.1= 8 min, v0=1000L/min
V = τv0= 8(1000)= 8000 L
Case1
Q =cp(T-T0’)+ HXA
0=250*4187*(62,22-T0’)+(-18000)(4187)(0.8)
T0’= 4.62◦C
Q1= 250(4187)(4.62-25)=21332.765J/mol A
Q2=250(4187)(25-62.22)=-38960.035J/mol A
Case 2
Q1' = 250(4,187)(62,22 -25)+(18000)(4,187)(0.8)= -21332,765 J / molA
It can be seen that both forms of exchange of heat are equivalent
Problem 9.7 (p. 238)
Rework the example 9.6; but with C0 = 1 mol/l instead of C0 = 4 mol/L and recital FA0 =
1000 mol to/min
Example 9.6. Find the size of the flow reactor in piston required to convert up to 80 % the
1000 mol to/min with C0 = 4 mol/L, which is used in the example 9.5
Solution
In the example 9.6 appears that
0.8
dXA
Τ4mol / L = 4 ∫ (-rA )4mol/L = 8.66 min
0.8
dX A
Τ1mol / L = (1)∫ (-rA )1mol/L
0
(-rA )4mol / L =
4 (-rA )1mol / L
0.8
dX A
Τ1mol / L = (1)∫ (-rA )1mol/L
(-rA )1mol / L
= (-rA )4mol / L/4
(-r )
A
0
(-rA )4mol / L =
4 (-rA )1mol / L
0.8
(-rA )1mol / L
=
4mol / L
4
0.8
dX A
Τ1mol / L = (1)∫ (-rA )
dX A
/
4mol L
= 4 ∫ (-rA)
0
0
= 8.66 min
4mol / L
V1mol/L= τ1mol/L(v0)= 8.66(1000/1)=8660L=8.66m3
As is noted for a reaction of the first order, the τ is not dependent on CA0; but the volume yes
because FA0 remains constant, that is to say that v0 varied.
Problem 9.8 (p. 238)
Solution
(-rA)4mol/L = 4 (-rA)1 mol/L
1
(-rA )
1
1
= 4 (-rA
)
4mol / L
1mol / L
1
1
(-rA )
= 4 (-rA )
1mol / L
4mol / L
The scale of the fig. E9.7 (p. 234) is multiplied by 4 if it reduces the C0 from 4 to 1 mol/L
Area under the curve of 1/-Rto vs Xto the example 9.7 with 4 mol/L CA0 = 1.2 Area under
the curve of 1/-Rto vs X with 1 mol/L CA0 = 1.2 (4) = 4.8=τ/C0
Τ =4.8 (1) = 4.8 min (the same τ the example)
V = τ v0 = 4.8 (1000) = 4800 L (4 times larger than that of the example)
Space weather is not affected by the variation of the concentration; but if F0 remains constant
volume v0 Varies
Problem 9.9 (p. 238)
You want to carry out the reaction of the example 9.4 in a reactor complete mix up to 95 %
conversion of a power supply with CA0 = 10 mol/L and a volumetric flow of 100 L/min What
size reactor is required?
Solution
T=
𝐸2−𝐸1
𝑅
𝑘02𝐸2
𝑋𝐴
𝑙𝑛
+𝑙𝑛
𝑘01𝐸1
1−𝑋𝐴
124200 −48900
8.314
= 𝑒 42.04 (124200
)
𝑒 17.34 (48900 )
+𝑙𝑛
.95
1−0.95
= 316.94 K
-rA=10 𝑒𝑥𝑝 17.34 −
48900
𝑅𝑇
1 − 𝑋𝑎 − 𝑒𝑥𝑝 42.04 −
124200
𝑅𝑇
(𝑋𝑎)
-rA= 0.0897L/mol min
Τm = CA X A /-rA = 10(0.95)/.0897=105.91min
0
V =τmv0 =10590,85 L ≈10.6 m
3
Problem 9.10 (p. 239)
As E1 < E2, low temperature at the beginning of the reaction is good to produce more of R
As E3 > E4, increasing the temperature when the reaction has already advanced induces to
produce S which is the desired product. As E3 < E5 also E3 < E6,lowering the temperature in
the final stages of the reaction while deduce the disintegration of S hence we can maximize
concentration of S.
Problem 9.11 (p. 239)
If we analyze the E / R is concluded that in the first stage reaction (decomposition of A) the
temperature should be in the high and then in low end. Let the values of the kinetic constants
in the 2 extreme temperatures
T(° C)
K1
K2
K1/K2
K3
K4
K3/K4
10
0.62
7.27
0.085
1.51 10-6
3.84 10-7
4.00
90
66.31
163.82
.4
1.71 10-3
4.40 10-3
0.39
An analysis of the values of the constants is concluded what we already knew and more;
We know that the profile should be decreasing because the first reaction must occur in the
decomposition of A and R is favored with high temperatures (k1 / k2 = 0.4 at 90 ° C), after
lowering the temperature must
because the formation of S is favored by low temperatures (k3 / k4 = 4.00 at 10 ° C). Add to
this, we did not know that k1 and k2 >> K3 and K4, then A is exhausted practically without
R being reacted yet. It can therefore assumed that in the first tank only decomposition of A
occurs and in the second of R. Thus the first tank is kept at 90 ° C and second at 10 ° C
𝜑𝑟 =
𝑟𝑟
𝑘1
66.31
=
=
= 0.288
−𝑟𝑎 𝑘1 + 𝑘2 163.82 + 66.31
𝑟𝑠
𝑘3
1.51 ∗ 10−6
𝜑𝑠 =
=
=
= 0.8
−𝑟𝑟 𝑘3 + 𝑘4 1.51 ∗ 10−6 + 3.84 ∗ 10−7
𝜑
𝑆
= 𝜑𝑅 ∗ 𝜑𝑆 = 0.288 ∗ 0.8 = 0.2304
𝐴
Problem 9.12 (p. 239)
The reversible reaction in the gaseous phase to ↔
​​
R is going to be carried out in a reactor
of thorough mixing. If operated at 300 K the required volume of the reactor is 100 L for a
60 % conversion. What should be the volume of the reactor for the same power and the
same conversion; but operating at 400 K.
Data:
To Pure
K​
1​
= 10​
3​
exp (­2416/T)
K = 10 to 300 K
H​
r​
= ­8000 cal/mol to 300 K
C​
p​
'​
=0
Solution
With the data to 300 K you can calculate v​
0​
and v​
0​
the required volume to 400 K. It
should be noted that v​
0​
varies by varying the temperature and that H​
R​
is constant because
C​
p​
'​
=0
tm =
Vm
v0
=
CA0XA
−rA
V0 = Vm (
=
CA0XA
K1CA0(1− XA)− K2C40XA
=
XA
K1(1− XA)− K2X4
K1(1− XA)−K2XA
)
XA
K 1(300k) = 0.318 min−1
K 2(300k) =
V0 =
K1
K
=
0.318
10
= 0.0318 min−1
100[0.318(1−0.6)−0.0318(0.6)]
0.6
= 18.2 L/min
1 − 1 )] = 10exp [− 8000 ( 1 − 1 )] = 4.485
K 400k = K 300k exp [− HRr (400
300
8314 400 300
K 1(400k) = 2.382 min−1
K 2(400k) =
2.382
4.485
= 0.531 min−1
400 ) = 18.02 ( 400 ) = 24.03 L/min
V 0(400k) = V 0(300k)( 300
300
X
0.6
V = V 0( K (1− X A)−K X ) = 24.03 [ 2.382(1−0.6)−0.531(0.6)
] = 22.73L
1
A
2 A
Example 10.1
Given the two reaction
→
-r1 =k1 CACB
→
-r2 =k2 CRCB
Where R is the desired product and is to be maximized. Rate the four schemes shown in
Fig.P10.1-either “good” or “not so good,” Please, no complicated calculations, just reason it out.
Answer:
The distribution of products (R / S) is determined by the ratio of k1 / k2 because reactions are of
the same order with respect to A and B, so if I want more R, as in a simple reaction, which XA
require is greater. This is achieved by working with maximum speeds. Assuming isothermal
operation, -rA grows when the concentration is high, so considering all of the above
(D) The CA and CB better because high ( Very good)
(A) is eqully good because here CA is in excess quantity. so formation of R is really high
compare to formation of S.
(C) Not so good. beacuse C B is in excess quantity. So rate of formation is S is very high compare
to rate of formation of R.
(B) The worse because both concentrations are low.
Example 10.2
Repeat Problem 1 with just one change
-r2 = k2 CRCB2
Answer:
It is a typical reaction series and parallel performance tells us that precisely. We must analyze
separately the components in series and components in parallel.
A → R → S are in series so it should not be high.
The reagent parallel, B, has a lower order than the desired reaction unwanted, so B should be
maintained at low concentrations and also we take difference is high so A has as high as
possible.
(A) The best Design because high CA and low CB.
(B) and (D) Intermediate because in (B) C A and CB and lower (D) C A and CB high,
only meet one requirement both
(C) The worse Design because CA low and CB high means difference is less so Desired quantity
is R which is less form because Formation R it instantaneously react with B (It is in excess
quantity ) and form S. Which is undesired for us.
Example: 10.3
Repeat Problem 1 with just one change
-r2 = k2 CR2 CB
Answer:
It is a parallel reaction and performance series tells us just that. We must analyze separately the
components in series and components in parallel.
A → R → S are in series so C A should be high
The reagent parallel, B, has the same order that the desired reaction the unwanted, so B does not
influence the distribution of products.
(A) and (d) are the best because high
(B) and (C) worse (not so good) design because rate is mainly dependent on C R and also C A is in
denominator so A is low necessary but after it decreases concentration of R. and formation of S
is high in (C).
Example: 10.4
For the reaction
→
-r1 = k1 CACB
→
-r2 = k2 CRCB2
Where R is the desired product, which of the following ways of running a batch reactor is
favourable, which is not ? See Fig. P10.4
Answer:
(A) better because high CA and CB low
(B) Intermediate because CA = CB low. considering the third alternative where the contents of
the two beakers are rapidly mixed together, the reaction being slow enough so that it does not
proceed to any appreciable extent before the mixture becomes uniform. During the first few
reaction increments R finds itself competing with a large excess of A for B and hence it is at a
disadvantage. Carrying through this line of reasoning, we find the same type of distribution curve
as for the mixture in which B is added slowly to A.
(C) The CA worse because low CA and high CB, is not suitable. For the first alternative pour A a
little at a time into the beaker containing B, stirring thoroughly and making sure that all the A is
used up and that the reaction stops before the next bit is added. With each addition a bit of R is
produced in the beaker. But this R finds itself in an excess of B so it will react further to form S.
The result is that at no time during the slow addition will A and R be present in any appreciable
amount. The mixture becomes progressively richer in S and poorer in B. This continues until the
beaker contains only S.
Example: 10.5
The Oxydation of Xylene. The violent oxidation of xylene simply produces CO 2 and H2 O;
however, when oxidation is gentle and carefully controlled, it can also produce useful quantities
of valuable phthalic anhydride as shown in Fig.P10.5.Also,because of the danger of
explosion,the fraction of xylene in the reacting mixture must be kept below 1%.Naturally,the
problem in this process is to obtain a favourable product distribution.
(a) In a plug flow reactor what values of the three activation energies would require that we
operate at the maximum allowable temperature?
(b) Under what circumstances should the plug flow reactor have a falling temperature
progression?
Answer:
a) If E1> E2 and E1> E3
The most desired is the activation energy and will be favored by high temperatures, so it must
work to the maximum allowable temperature.
b) If E1> E3 and E1 <E2
At first you must work with high temperatures to encourage 1 against the reaction 3 and then the
temperature should drop to not favor the profile step 2. It is then to be used is decreasing.
Example 10.6
The Trambouze Reactions - Reactions in parallel. Given the set of elementary reactions with a feed
of CA0 =1moll/liter and V = 100 literslmin we wish to maximize the fractional yield, not the
production of S, in a reactor arrangement of your choice.
a) Do you judge that arrangement of Fig.P10.6 is the best set up?
b) If not,suggest a better scheme. Sketch your scheme and calculate the volume of the
reactors you plan to use.
Answer:
A→R
rR = k0
k0 = 0.025 mol/L min
A→S
rS = k1 CA
k1 =0.2 min-1
A→T
rS = k2 CA
k2 = 0.4 L/mol min
The order of the desired reaction determines how they should be the concentrations, whether
high or low. In this case
Order A→R
<
Order A → S
<
Order A → T
The order of reaction is the desired intermediate. C A favor high A → R and CA low to A → T. It
is obvious that there must be an intermediate concentration A → S. To find that CA makes
performance maximum must be raised that dφ (S / A) / dC A = 0
Φ( )=
=
(
)
=0
0.025 + 0.2 CA + 0.4 CA2 + 0.2 CA – 0.8 CA2 = 0
The only possible solution is C A = 0.25 mol / L and the concentration value making the most
appropriate performance
It is best to work with a mix Flow reactor that throughout the reactor instant performance equals
0.5 which is its maximum value.
b)
CS = Φ ( )(CA0 - CA) = 0.5(1-0.25) = 0.375
Example 10.7
For the set of elementary reaction of problem 10.6, with a feed of C A0 = 1 mol/liter and v = 100
liters/min we now wish to maximize the production rate of intermediate S (not the fractionl
yield) in a reactor arrangement of your choice. Sketch your chosen scheme and determine C S max
obtainable.
Answer:
Given the shape of the curve φ vs CA there to work with a reactor will complete mixture CA= 1 to
0.25 mol / L, which will have the maximum performance and with this a series of plug flow
reactor that CA = 0.25 go from to 0 mol / L, to take advantage as much as possible high yields.
The final concentration is 0 because if ΔCA ↑ Δ CR↑ and CR wants maximum
Cs max = CS m + CS P
Cs max = 0.5(1-0.25) = 0.375 mol/liter
Cs p = ∫
= ∫
Dividing numerator and denominator by 0.4
Cs p = 0.5∫
Cs p = 0.5 {
=
= 0.5∫
[ln(0.25 + CA) +
]0.25 0 }
Example 10.8
Automobile Antifreeze Ethylene glycol and diethylene glycol are usedas automobile antifreeze,
and are produced by the reactions of ethylene oxide with water, as follows:
A mole of either glycol in water is as effective as the other in reducing the freezing point of
water; however, on a molar basis the diethylene glycol is twice as expensive as the ethylene
glycol. So we want to maximize ethylene glycol, and minimize the diethylene glycol in the
mixture.
One of the country's largest suppliers produced millions of kilograms of antifreeze annually in
reactors shown in Fig. P10.8a. One of the company's engineers suggested that they replace their
reactors with one of the type shown in Fig. P10.8b. What do you think of this suggestion?
Answer:
H2 O+ ethylene → Ethylene Oxide
Ethylene + ethylene → Diethylene Oxide
If H2 O = A, Y = ethylene oxide, ethylene glycol =R and diethylene = S
The reaction can be expressed as
A + B → R (reaction 1)
R + B → S (reaction 2)
Analyzing the components in series A → R → S is the most convenient plug flow reactor and the
reactor of Figure (b) may be considered as diameter / length such by their relationship.
As for the addition of B, the component parallel can not conclude nothing because the reaction
order of this component is not known in the desired and undesired reaction. If the desired out the
higher order, then the addition of B, which raises the concentration of this component is
appropriate.
We select PFR because length/diameter ratio is high in PFR and also undesired quantity
(diaethyelene glycol ) is less.
Example 10.9
The Homoeneous catalytic Reaction. Consider the elementary reaction
A + B → 2B,
-rA = kCACB with k = 0.4 liter/mol min
For the following feed and reactor space time
Flow rate
Feed compositin
Space time
v = 100 liters/min
{
τ = 1 min
We want to maximize the concentration of B in the product stream. Our clever computer [see
problem 8, Chem. Eng. Sci., 45, 595-614 (1990)] gives the design of Fig. P10.9 as its best try.
Do you think that this is the best way to run this reaction? If not, suggest a better scheme. Do not
bother to calculate reactor size, recycle rate, etc. Just indicate a better scheme.
Answer:
As this is a simple reaction system criteria used is As the production efficiency, the more
efficient is the system having higher speeds. So to know what is more suitable reactor is need to
know how the conversion varies with –rA
-rA = kCACB (Elementary system)
CA = CA0 (1-XA)
CB =CB0 -CA0 XA+2CA0 XA (for stoichiometry)
CB = CA0 (M + XA) ,where M =
=
= 1.22
-rA = k CA0 2 (1-XA)(M + XA)
If, XA↑, (1-XA)↓ so you can increase or decrease, according to the relative weight of the factors
One way to know how varies –rA with XA is to find the derivative of the function.
= kCA0 2 [(1-XA)(1) + (M + XA)(-1)] = kCA0 2 (1-M-2XA) < 0
If we consider that M = 1.22 and analyze the derivative we see that it is negative over the entire
range of conversions, indicating that the function is decreasing for any value of XA, i.e, the speed
decreases to increase conversion.
Another way to know how to vary the speed conversion, less accurate, it is to evaluate the
function in the interval.
-rA = 0.4 (0.45)2 (1 – XA) (1.22 + XA)
-rA = 0.081 (1 – XA) (1.22 + XA)
XA
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
-rA*
9.9
9.6
9.2
8.6
7.8
7.0
5.9
4.7
3.3
1.7
10-2
The higher speed is XA = 0 (high concentrations) so that the reactor It is best piston without
recycling, because the recycle low profile concentrations occurring in the reactor and thus lower
the speed occurring in the reactor.
Compare both PFR and MFR
Here Plug flow is best design. but here recycling is not required if recycle then it is against the
assumption of PFR.
Example 10.10
To Color Cola Drinks. When viscous corn syrup is heated it caramelizes (turns a deep dark
brown). However, if it is heated a bit too long it transforms into carbon
The caramelized liquid is sent by railroad tank cars to the cola syrup formulators, who then test
the solution for quality. If it is too light in color-penalty; if it has too many carbon particles per
unit volume, then the whole tank car is rejected. There is thus a delicate balance between
underreacting and overreacting.
At present a batch of corn syrup is heated at 154°C in a vat for a precise time. Then it is rapidly
discharged and cooled, the vat is thoroughly cleaned (very labor intensive), and then is
recharged.
The company wants to reduce costs and replace this costly labor intensive batch operation with a
continuous flow system. Of course it will be a tubular reactor (rule 2). What do you think of their
idea? Comment. please, as you sit and sip your Coke or Pepsi.
Answer:
The substitution is theoretically substantiated that over piston take place the same story of
concentrations, and therefore speeds, which occur with time in the batch, and It was further
added advantage of continuous operation.
However, in this system is formed inevitably solid, which adhere to the reactor wall to some
extent (which is why the cleaning batch reactor was a daunting) task is that if the liquor is
viscous and heated through the walls will create a temperature gradient and coal to be formed
will adhere to the pipe walls preventing proper operation. or that think about stopping and clean
the pipe.
10.11 A-R-S
T
U
Cao = 6 CRo=0.6 CTo=0
The feed has CR/CT=∞
a) So to maximise CR/CT do not react at all, the design of figure is not
good.
b) To minimise CR/CT , run to completion. All the R will disappear.
Next reaction 1 is second order
Reaction 2 is first order
keep CA low
So, use a large MFR you will end up with more CT
10.12
Cao = 90 mol/m3, Cbo = 10 mol/m3
A+B  B+B -rA=kCaCb
If we want minimum volume, which will be efficient in terms of production and
with maximum speed, we have to know how rate varies with speed
-rA = k CA CB
CA = CA0 (1 – XA) = CA0 (1-XA)
CB = CB0 – CA0XA + 2CA0XA = CA0 (M + XA) where M = CA0/CB0
-rA = k CA02 (1- XA) (M + XA)
 d ( ra) 
2

  kCao (1  Xa )(M  Xa )
dXa


M = 1/9 = 0.11, so the derivative is positive for low values XA and
negative for high values and maximum -rA the derivative is 0, so
XA = (1-M)/2 = 0.44
Speed will be : (-rA/k) = 8100 (1 – XA) (1/9 + XA)
XA
0
0.1 0.2
0.3 0.4 0.5 0.6 0.7 0.8 0.9
-rA/k 900 1600 2100 2400 2500 2400 2100 1600 900 810
Y-Values
0.0014
0.0012
0.001
0.0008
Y-Values
0.0006
0.0004
0.0002
0
0
0.2
0.4
0.6
Use a MFR
10.13
A+B
B +B
-rA=KCACB CAf = 9 CBf= 91
CAo= 90 CBo=10
0.8
1
Y-Values
0.0014
0.0012
0.001
0.0008
Y-Values
0.0006
0.0004
0.0002
0
0
0.2
0.4
0.6
0.8
1
Use a MFR followed by a PFR
10.14
𝐴 + 𝐵 → 𝐵 + 𝐵, -rA = kCACB
CA0 = 90 mol/m3
CAf = 72 mol/m3
CB0 = 10 mol/m3
CBf = 28 mol/m3
XA
0
0.1 0.2
0.3 0.4 0.5 0.6 0.7 0.8
-rA/k 900 1600 2100 2400 2500 2400 2100 1600 900
Y-Values
0.0014
0.0012
0.001
0.0008
Y-Values
0.0006
0.0004
0.0002
0
0
0.2
0.4
0.6
0.8
1
As curve k/-rA vs XA downward slope in the range XA = 0 to 0.2, it is best to
work with a mixing reactor complete, which will have the higher speed range
and therefore will require the least τ. And for the 20% conversion it will give
the lower volume as compared the other reactors.
10.15
Since E1>E2 use a high temperature or T = 900C
At 900C, k1 = 30e-20000/8.314(363) = 0.0397 min-1
k2 = 1.9e-15000/8.314(363) = 0.0132 min-1
Therefore, k2/k1 = 0.3322
Therefore, Form the above fig, CRmax/CA0 = 0.58
topt = [ln(k2/k1)]/[k2-k1] = (ln 0.3322)/(0.0132-0.0397)
= 4.16 min-1
10.16
The following system of elementary reactions have
C6H6 + Cl2 → C6H5Cl + HCl
C6H5Cl + Cl2 → C6H4Cl2 + HCl
k1 = 0,412 L/kmol h
k2 = 0,055 L/kmol h
A = C6H6
B = Cl2
C = C6H5Cl
D = C6H4Cl2
A+B→R
R+B→S
It is best to use a plug flow reactor. You should not use the first recycle reactor.
It should work at low conversions of benzene. For example,
XA =0.4,
with k2 / k1 = 0.1335, CR / CA0 = 0.3855, CS / CA0 = 0.0145 and CR / CS = 26.68.
10. 17- Let A = C3H6 B=O2 R=C3H4O
Then A+B  R
A+4.5B S
K2/K1=0.1
K3/K1=0.25
AR S
From the Denbigh reaction
CRmax/CAo = K1/K2(K12/K3)^(K3/K3-K12)
=10/11*(11/25) ^(2.5/2.5-11)
=0.588
10.18
It is a series parallel system , so selection is based on analyzing their series and
parallel components as separate components . Parallel components are A → R
and A → T and if we see the orders of the reactions, lower order is desired ,
therefore taking into account parallel components concentration must be kept
low during the course of the reaction .
Series components are A → R → S and R is the desired as having into account
the components in series the concentration of A must remain as high as possible
during the course of the reaction .
There is a contradiction and the final decision depends on the relative weight of
reactions .
Considering the kinetic constants
T (K)
360
396
k1
0.1306
0.9700
k2
0.0560
0.1530
k3
0.00006
0.00120
k1/k3
2176
808.3
k1/k2
2.33
6.3
Analysis of this table we see that between 360 and 396 K ,k3 <<k1 and k1 and
k2 are almost similar.
The reaction A → T ½ can be neglected because it occurs slowly compared to
the other . if so then the plug flow reactor is the most convenient .
b ) Working with a piston 396 K
k2
Cr  k1  k 2 k1
 
Cao  k 2 
Cr=0.707 mol/L
k2 

 ln

k1   2.252s
 
 k 2  k1 




Xa=1-exp(k1τ)=0.888
QUES: 10.19 Phthalic Anhydride from Naphthalene.
The accepted mechanism for the highly exothermic solid catalysed oxidation of
naphthalene to produce
Phthalic anhydride is
R
A
S
T
Where
k1 = k2 = 2 * 1013 exp (-159000/RT) [hr-1]
k3 = 8.15 * 1017 exp (-209000/RT) [hr-1]
k4 = 2.1 * 105 exp (-83600/RT) [hr-1]
And where
A = naphthalene (reactant)
R = naphtha Quinone (postulated intermediate)
S = phthalic anhydride (desired product)
T = CO2 + H2O (waste products)
and the Arrhenius activation energy is given in units of J/mol. This reaction is to
be run somewhere between 900 K and 1200 K.
A local optimum reactor setup discovered by the computer [see example 1,
Chem. Eng. Sci., 49, 1037-1051 (1994)] is shown in Fig. P10.19.
(a) Do you like this design? Could you do better? If so, how?
(b) If you could keep the whole of your reactors at whatever temperature and τ
value desired, and if recycle is allowed, how much phthalic anhydride could be
produced per mole of naphthalene reacted?
Suggestion: Why not determine the values of k1, k2, k3, and k4 for both
extremes of temperature, look at the values, and then proceed with the solution?
SOLUTION:
Given,
k1 = k2 = 2 * 1013 exp (-159000/RT) [hr-1]
k3 = 8.15 * 1017 exp (-209000/RT) [hr-1]
k4 = 2.1 * 105 exp (-83600/RT) [hr-1]
R
A
S
T
Since k4 value is much smaller than the other values of k we use the highest
allowable temperature of 1200K.
k1 = 2.4 * 106 hr-1
k2 = 2.4 * 106 hr-1
k3 = 650 * 106 hr-1
k4 = 48.2 hr-1
Now,
Since k3 value is much bigger than k2 value we can approximate the reaction as
follows.
4.8 * 106
48.2
A
R
T
Now we find CRmax
𝐾6
CRmax
CA0
=
𝐾5 𝐾6−𝐾5
(𝐾6)
ln(
tplug =
𝑘2
)
𝑘1
𝑘2−𝑘1
= (
4.8∗106
48.2
48.2
)48.2−(4.8∗106)
= 0.99988
= 2.4 * 10-6 hr = 0.0086 sec = time of residence
With such a small time and high conversion practically we can use any type of
reactor but we are using a straight plug flow reactor with no recycle which is the
best.
Now we find CR for this residence time
𝐶𝑅
𝐶𝐴𝑜
=
𝑘5
𝑘5−𝑘6
[𝑒 −𝑘1𝑡 − 𝑒 −𝑘2𝑡 ] = (-1) [0 – 0.9762] = 0.9762 = CR
10.20. Professor Turton dislikes using reactors in parallel, and he cringed when he saw my
recommended "best" design for Example 10.1. He much prefers
using reactors in series, and so for that example he suggests using the design of Figure El0.la,
but without any recycle of fluid. Determine the fractional yield of S, Φ (S/A), obtainable with
Turton's design, and see if it matches that found in Example 10.1
SOLUTION:
Here,
τ1 =
τ2 =
τ3 =
𝑉
25
𝑉
50
𝑉
75
=
=
=
𝐶𝐴0−𝐶𝐴1
(−𝑟𝐴)1
𝐶𝐴1′−𝐶𝐴2
(−𝑟𝐴)2
𝐶𝐴2′−𝐶𝐴3
(−𝑟𝐴)3
where CA1’ =
1
where CA2’ =
2
2
3
𝐶𝐴1 +
𝐶𝐴2 +
1
2
1
3
τ4 =
𝑉
100
=
𝐶𝐴3′−0.25
(−𝑟𝐴)25
where CA3’ =
3
4
𝐶𝐴3 +
1
4
It is a set of 4 equations and 4 unknowns , so it is determined ; but must be solved by trial and
error and once you know the CR concentrations determine the output of each reactor. It is
clear that the procedure is quite cumbersome and graphical method for solving the material
balance of the reactor mixture can help to the solution. According to the method the operating
point reactor located where the curve ( -rA ) vs CA is cut with balance materials in the plane (
-rA ) -CA is a straight line passing through the reactor inlet concentration and whose slope -1 /
τi .
(-rA) = 0.025 + 0.2 CA + 0.4 (CA)2
CA 0.6 0.55
0.5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
(- 0.625 0.256 0.225 0.196 0.169 0.144 0.121 0.100 0.081 0.064 0.049 0.036
rA)
The steps followed in solving the problem are:
1. With these values the curve ( -rA ) is obtained vs CA
2. It is assumed CA1 and τ1 is calculated by equation 1
3. τ1 is calculated V
4. V is calculated τ2 , and τ1 τ3
5. CA ! '= ½ CA1 + 1/2, and τ2 operating point is calculated
reactor 2
6. CA2 '= 2/3 +1/3 CA2 and τ3 the operating point is calculated
reactor 3
7. CA3 '=¾ CA3 +1/4 and τ4 operating point is calculated
reactor 4
8. If CA4 = 0.25, the assumed value of CA1 is correct, if not return to
Step 2
Suppose CA1 = 0.3
(1 −0.3)(25)
V = 0.025 + 0.2(0.3)+ 0.4(0.32 )=144.6281 L
144.6281
τ2 =
1
⇒
50 = 2.8925 min
τ3 = 144.6281=1.9283 min
τ2 = 0.3457
1= 0.5186 min −1
⇒
τ3
75
τ2 =
144.6281
1 = 0.6914
min −1
⇒
=1.4462 min
τ2
100
C′A1 = 02,3 + 12 = 0.65 mol / L
C′A2 = 2(03,3) + 13 = 0.5333 mol / L
C′A3 = 3(04,3) + 14 = 0.475 mol / L
Suppose CA1 = 0.25
(1 −0.25)(25)
V=
τ2 =
(
0.025 + 0.2(0.25)+ 0.4 0.252
187.5
= 3.75 min
⇒
τ3 = 187.5 = 2.5 min
1 = 0.2667
min −1
⇒
1 = 0.4 min −1
τ3
75
187.5
100
=187.5 L
τ2
50
τ2 =
)
=1.875 min
⇒
1
τ2
= 0.5333
C′A1 =
0.25
1
+
2
2
C′A2 =
C′A3 = 3(0.25)
4
= 0.625 mol / L
2(0.25) + 1 =
0.5 mol / L
3 3
+ 1= 0.4375 mol / L
4
CRE 1 Unsolved examples chapter 11
11.1. A pulse input to a vessel gives the results shown in Fig. P1l.l.
(a) Check the material balance with the tracer curve to see whether the results are consistent.
(b) If the result is consistent, determine t, V and sketch the E curve.
Ans. Area Under the curve=Cpulse×t= 0.05×5= 0.25 mol min. /lit.
a. M= 1 mol at t=0 (given); ν= 4 lit. / min.
𝑀
Therefore, ν = ¼ = 0.25 mol min. / lit.
Since Area Under the curve equals
b. Taverage =
(Ʃ ti Ci)
(5×0.05)
Ʃ Ci
0.05
V
=
Therefore, Taverage=
𝑀
ν
the results are consistent.
= 5 min.
ν
V = Taverage × ν = 5 × 4 = 20 liters
E = (Cpulse × ν)/ M = (0.05 × 4)/ 1 = 0.2
11.2. Repeat Problem P1l.l with one change: The tracer curve is now as shown in Fig. P11.2.
𝑀
Ans. From the previous sum ν = ¼ = 0.25 mol min. / lit.
Area Under the curve=Cpulse×t= 0.05×6= 0.3 mol min. /lit.
Therefore,
𝑀
ν
≠ Cpulse×t
The tracer comes out late hence the experiment is not done correctly.
11.3. A pulse input to a vessel gives the results shown in Fig. P11.3.
(a) Are the results consistent? (Check the material balance with the experimental tracer curve.)
(b) If the results are consistent, determine the amount of tracer introduced
M,
Ans. ν= 4 cm3/s V= 60 cm3
V
In Pulse Input taverage= ν = 60/4 = 15 s
From the graph (assuming the value of tip to be C)
1
𝐶
Area = 2 [(3 × 𝐶) + (6 × 𝐶)] = 9 × 2
taverage=
Ʃ𝑡𝐶∆𝑡
𝐴𝑟𝑒𝑎
=
[(16×0)+(19×𝐶)+(25×0)]
9𝐶
2
=
38
9
Since taverage is not the same the result if not consistent
11.4. A step experiment is made on a reactor. The results are shown in Fig. P11.4.
(a) Is the material balance consistent with the tracer curve?
(b) If so, determine the vessel volume V, 7
Ans. Since Value of Cmax is not given this problem cannot be solved
11.6. A pipeline (10 cm I.D., 19.1 m long) simultaneously transports gas and liquid from here to
there. The volumetric flow rate of gas and liquid are 60 000 cm3/s and 300 cm3/s, respectively.
Pulse tracer tests on the fluids flowing through the pipe give results as shown in Fig. P11.6. What
fraction of the pipe is occupied by gas and what fraction by liquid?
Ans. V= 𝜋/4*D2H
= 𝜋/4*102*10-4*19.1 [Given D=10cm and H= 19.1m]
∴ VT= 0.15
VG=60000 cm3/s
tRG = 2s
VG = 60000 * 10-6 * 2 m3
=
0.12 m3
VL = 300 cm3/s
tRL = 100s
∴ VL = 300 * 10-6 * 100 m3
=
0.03 m3
∴ % Volume occupied by gas =0.12/0.15 * 100 =80%
% Volume occupied by liquid =0.03/0.15 * 100=20%
A liquid macro fluid reacts according to A + R as it flows through a vessel. Find the conversion
of A for the flow patterns of Figs. P11.7 to P1l.ll and kinetics as shown.
11.7)
∞
Ans. ∫0 𝐸. 𝑑𝑡 =1
∴ 2*E=1
∴E= 0.5
∞
C
= C/C0 = ∫0 (C0) . Edt
(C/C0) for a batch material is given by Eqn
Here n = 0.5 and k= 2 mol0.5/litre0.5 * min
(n-1) kCA0nt = (CA/CA0)1-n -1
∴ -1/2 * 2 * 1 * t = (CA/CA0)1/2 -1
∴ (CA/CA0) = (1-t)2
2
∴ C/C0 = ∫0 (1 − 𝑡)2 . (½). dt
= -1/2* [(1-t)3/3]20
= -1/2*[-1/3-1/3]
= -1/2 * -2/3
= 1/3
= XA= 1-1/3 = 2/3
11.8)
∞
Ans. C/C0 =∫0 (C/C0). E.dt
∞
∫0 𝐸. 𝑑𝑡 = 1
∴ 0.5= 0.5 * 0.5 * E
∴E=2
Here n = 2, k=2 litre/mol * min and CA0= 2 mol/litre
(n-1) kCA0nt = (CA/CA0)1-n -1
1 * 2 * 22 * t = (CA/CA0)-1 -1
1 + 8t = (CA/CA0)-1
=
(CA/CA0) = 1/1+8t
0.5
=2∫0 1/1 + 8𝑡 * dt
=1/4 [ln (1+8t)]0.5
0
=1/4 [ln (5)]
= XA = 0.6
11.9
CA0 = 6 mol/Lit
-rA= k
k = 3 mol/lit min
∞ 𝐶𝐴
𝐶𝐴
̅̅̅̅
Ans. The Performance Equation is 𝐶𝐴0 = ∫0 𝐶𝐴0 𝐸𝑑𝑡
For t< CA0 / k = 6/3 = 2min; we have CA / CA0 = 1 – ( kt / CA0 )
For t> 2 min; we have CA / CA0= 0
Replacing it in the above equation,
2
̅̅̅̅̅̅
𝐶𝐴
𝑘𝑡
= ∫(1 −
)𝐸𝑑𝑡
𝐶𝐴0
𝐶𝐴0
0
Putting E =1/3,
2
̅̅̅̅̅̅
𝐶𝐴
1
𝑡
= ∫(1 − 3 ∗ )𝑑𝑡
𝐶𝐴0 3
6
0
2
̅̅̅̅̅̅
𝐶𝐴
1
= ∫( 2 − 𝑡 )𝑑𝑡
𝐶𝐴0 6
0
̅̅̅̅̅̅
𝐶𝐴
1
=
𝐶𝐴0 3
So the conversion is XA = 1 – CA / CA0 = 2/3 = 0.666 = 66.6 %
11.10
CA0 = 4 mol /lit
-rA= k
k = 1mol/lit min
Ans. Here Dirac Delta function is used. It is Zero everywhere except one point and that point is
at Infinite value. The Equation to be used here is:
∞
̅̅̅̅̅
CA
= ∫ f(t) δ (t − t0) dt
CA0
0
∞
We know that, ∫0 𝑓(𝑡) 𝛿 (𝑡 − 𝑡0) 𝑑𝑡 = 𝑓(𝑡0)
For t< CA0 / k = 4/1 = 4 min; CA / CA0 = 1 – ( kt / CA0 )
For t> 4min ;CA / CA0 = 0
Here f (t) = 1 – (kt / CA0)
So putting it in the above equation and by Dirac Delta function,
4
̅̅̅̅̅̅
𝐶𝐴
𝑘𝑡
= ∫(1 −
)𝛿 (𝑡 − 3) 𝑑𝑡
𝐶𝐴0
𝐶𝐴0
0
̅̅̅̅̅̅
𝐶𝐴
𝑘𝑡
= 1−
| 𝑎𝑡 𝑡 = 3
𝐶𝐴0
𝐶𝐴0
̅̅̅̅̅̅
𝐶𝐴
= 0.25
𝐶𝐴0
So XA = 0.75 = 75%
11.11
CA0 = 0.1 mol /lit
-rA= k
k = 0.03mol/lit min
Ans. The Performance Equation is :- ̅̅̅̅
= ∫0
𝐶𝐴0
𝐶𝐴
∞ 𝐶𝐴
𝐶𝐴0
𝐸𝑑𝑡
For t< CA0 / k = 0.1/0.03 = 3.33 min; CA / CA0 = 1 – ( kt / CA0 )
For t> 3.33 min ;CA / CA0 = 0
𝐶𝐴
̅̅̅̅
Here Nothing is leaving the Reactor after t = 3.33 min. so everything has been reacted .so𝐶𝐴0= 0
and XA = 1.
11.12-11.14. Hydrogen sulfide is removed from coal gas by contact with a moving bed of iron
oxide particles which convert to the sulfide as follows:
Fe203 -+ FeS
In our reactor the fraction of oxide converted in any particle is determined by its residence time t
and the time needed for complete conversion of
11.12
The reaction occurring here is Fe2O3 → FeS.
1-X = (1-t/𝜏 )3
X=1
where t<1 and 𝜏 = 1 hr
where t ≥ 1 hr
3
1
t
̅̅̅̅̅̅̅
1
− 𝑋 = ∫0 ( 1 − 𝜏) Edt
1
̅̅̅̅̅̅̅
1
− 𝑋 = ∫0 ( 1 − t )3 Edt
Here E=1 [ From figure Area = E*t so we have 1=E*1 which results to E=1 ]
1
̅̅̅̅̅̅̅
1
− 𝑋 = ∫0 ( 1 − 𝑡 3 − 3𝑡 + 3𝑡 2 ) dt
̅̅̅̅̅̅̅
1
− 𝑋 = 0.25
̅ = 0.75 = 75 %
𝑿
Find the conversion of iron oxide to sulfide if the RTD of solids in the contactor is approximated by the
curve
Ans.
For solid particles,
∞
̅̅̅̅̅
CA
1−X=
= ∫ f(t) δ (t − t0) dt
CA0
0
3
1
t
̅̅̅̅̅̅̅
1
− 𝑋 = ∫0 ( 1 − 𝜏) E dt ,
∞
We know that, ∫0 𝑓(𝑡) 𝛿 (𝑡 − 𝑡0) 𝑑𝑡 = 𝑓(𝑡0)
1 − 𝑋 = ( 1 − 𝑡 )3 |𝑎𝑡 𝑡 = 0.5
1 − 𝑋 = 0.125
Ans: X= 0.875
where E= 𝛿 (𝑡 − 𝑡0)
Ans.
For solid particles,
∞
̅̅̅̅̅
CA
1−X=
= ∫ f(t) δ (t − t0) dt
CA0
0
3
1.5
t
̅̅̅̅̅̅̅
1
− 𝑋 = ∫0.5 ( 1 − 𝜏) E dt
1.5
̅̅̅̅̅̅̅
1
− 𝑋 = ∫0.5 ( 1 − t )3 E dt
Here E=1 [ From figure Area = E*t so we have 1=E*1 which results to E=1 ]
1.5
̅̅̅̅̅̅̅
1
− 𝑋 = ∫0.5 ( 1 − 𝑡 3 − 3𝑡 + 3𝑡 2 ) dt
̅̅̅̅̅̅̅
1
−𝑋 =0
̅ = 1 = 100 %
Ans: 𝑿
Q 11.15
Cold solids flow continuously into a fluidized bed where they disperse rapidly enough so that
they can be taken as well mixed. They then heat up, they devolatilize slowly, and they leave.
Devolatilization releases gaseous A which then decomposes by first-order kinetics as it passes
through the bed. When the gas leaves the bed decomposition of gaseous A stops. From the
following information determine the fraction of gaseous A which has decomposed. Data: Since
this is a large-particle fluidized bed containing cloudless bubbles, assume plug flow of gas
through the unit. Also assume that the volume of gases released by the solids is small compared
to the volume of carrier gas passing through the bed.
Mean residence time in the bed: 6 = 15 min, 6 = 2 s for carrier gas For the reaction:
A →products, -rA = kcA, k = 1 s-l
Ans.
Here A is released uniformly.
1
2
From the curve area under curve must be 1, therefore E=0.5
∞ 𝑪𝑨
̅̅̅̅
𝑪𝑨
The Performance Equation is 𝑪𝑨𝟎 = ∫𝟎 𝑪𝑨𝟎 𝑬𝒅𝒕
For first order reaction,
CA
= e−kt
CA0
CA
̅̅̅̅
CA0
∞
= ∫0 e−kt Edt
CA
= [e−t ]20
CA0
CA
= 0.4323
CA0
𝑪𝑨
Ans: X=1 - 𝑪𝑨𝟎= 0.5677
Q 11.16
Reactant A (CAo = 64 mol/m3) flows through a plug flow reactor (r = 50 s), and reacts away as
follows:
Determine the conversion of A if the stream is:
(a) a microfluid,
(b) a macrofluid.
A 11.16
(a) For Microfluid:
𝐶𝐴
∫ (𝑑𝐶𝐴 )
𝜏
̅̅̅̅̅̅
= 𝐶𝐴𝑜1.5
𝐶𝐴0 𝑘𝐶𝐴 𝐶𝐴𝑜
𝜏=
[𝐶𝐴−1.5+1 ]𝐶𝐴
𝐶𝐴0
−𝑘[−1.5+1]
1
1
𝐴
𝐴𝑜
50*0.005*0.5 = 𝐶 0.5 - 𝐶 0.5
CA = 16,
X=
𝐶𝐴𝑂− 𝐶𝐴
𝐶𝐴𝑂
Since CAO= 64
= 0.75
Ans: X = 0.75
CRE-2
Group No-11
Chapter –13
(13BCH054, 055, 056, 057)
13.4 is not done as per the talk with you sir
13.5
Solving it with the help of dispersion model –
Using equation 8;
σ2 = 2(DL/u3)
Or; σ2 is proportional to L
Therefore;
σc2 - σb2 = σb2 - σa2
Or σc2-1024 = 1024-256
Thus;
σc2=1792
Hence Width = kσc = 42.3 meters.
13.7
For finding σ2 for this flow –
Re = (dup/ µ) = (0.225)(1.1)/(0.9x10-6) = 3.2x105
Now since (D/udt) = 0.22
From Eqation 8 – (D/uL)=(D/udt)(dt/L) = 0.22(0.255/1000000) = 5.61x10-8
(σ2/t2) = (σ0)2 = 2(D/uL) = 2(561x10-8) = 11.22x10-8
Therefore; σ = (11.22x10-8)0.5x(100000/1.1) = 11022x10-8
So the width at 1σ = (304.5)(1.1) = 335m
From 5/95 to 95/5 Figure 12 shows that this includes –
(1.65σx2) of width
Therefore; the 5/95 to 95/5 % width is – (335m)(1.655x2) = 1105m.
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Chapter 14: Tanks in Series Model
14.1 From experiment
T
0-2
2-4
4-6
6-8
8-10
10-12
t mean
1
3
5
7
9
11
C
2
10
8
4
2
0
The last observation was interpolated linearly.
C vs T
12
10
8
6
C vs T
4
2
0
1
t average =
σ2 =
1
𝑁𝑁
Ʃ𝑡𝑡 2 𝐶𝐶
Ʃ𝐶𝐶
Ʃ𝑡𝑡𝑡𝑡
Ʃ𝐶𝐶
3
1(2)+ 3(10)+ 5(8)+ 7(4)+ 9(2)+ 11(0)
=
2
− � t(avg)�
= σ2 θ =
5
𝜎𝜎 2
�𝑡𝑡(𝑎𝑎𝑎𝑎𝑎𝑎)�
N= 4.68 tanks
2
=
2+10+8+4+2
=
118
26
= 4.538
12 (2)+ 32 (10)+ 52 (8)+ 72 (4)+ 92 (2)+ 112 (0)
4.4038
4.5382
= 0.2138
14.2
For N= 10
C MAX = 100 mmol ; σ2= 1 min.
a. E θMAX ∝ C MAX
=
2+10+8+4+2
- (4.538)2 = 4.4038
E θMAX =
𝑁𝑁 (𝑁𝑁−1)(𝑁𝑁−1)
(𝑁𝑁−1)!
e-(N-1)
E θMAX (for 10 tanks) = 1.317
E θMAX (for 20 tanks) = 1.822
𝐸𝐸𝜃𝜃𝜃𝜃𝜃𝜃𝜃𝜃 (𝑓𝑓𝑓𝑓𝑓𝑓 10 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡)
𝐸𝐸𝜃𝜃𝜃𝜃𝜃𝜃𝜃𝜃 (𝑓𝑓𝑓𝑓𝑓𝑓 20 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡)
1.317
1.822
=
=
100
𝐶𝐶𝑀𝑀𝑀𝑀𝑀𝑀 (𝑓𝑓𝑓𝑓𝑓𝑓 10 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡)
𝐶𝐶𝑀𝑀𝑀𝑀𝑀𝑀 (𝑓𝑓𝑓𝑓𝑓𝑓 20 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡)
𝑥𝑥
X = 138.3 mmol
The maximum concentration of leaving tracer is 138.3 mmol
b. σ2 ∝ N
𝜎𝜎 2 (𝑓𝑓𝑓𝑓𝑓𝑓 10 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡)
𝜎𝜎 2 (𝑓𝑓𝑓𝑓𝑓𝑓 20 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡)
1
𝑥𝑥
=
10
=
10
20
20
Therefore x= 2
2
The spread is √2 = 1.414 min.
c. The relative spread increases with increase in number of tanks. (ans)
14.3
The above problem is similar to the mixed flow reactor
V= 1.25 × 109 bills; ν = 109 bills/year
T average = 1.25 × 109/ 109 = 1.25 years
E(t) =
𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 𝑜𝑜𝑜𝑜 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑜𝑜𝑜𝑜 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏
𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌
=
𝑏𝑏
𝑏𝑏0
Where b is number of number of bills and b 0 is total number of bills
a. F(t) = 1- et/t(average) = 1- et/1.25
b. The number of bills in circulation which are older than 21 years is by the following formula:
∞
b = ∫21 𝑏𝑏𝑜𝑜 𝐸𝐸(𝑡𝑡)𝑑𝑑𝑑𝑑 where b 0 = 1.25 × 109
E(t) = ∫ 𝐹𝐹(𝑡𝑡)𝑑𝑑𝑑𝑑 = ∫(1 − exp �
Therefore E(t) =
1
𝑡𝑡𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
𝑡𝑡
𝑡𝑡(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)
�) 𝑑𝑑𝑑𝑑
𝑒𝑒 𝑡𝑡/𝑡𝑡(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) = 0.8 e-0.8t
∞
Therefore b = ∫21 0.8 exp(−0.8𝑡𝑡)𝑑𝑑𝑑𝑑 = 63.2 bills ≈ 63 bills
14.4
V=1.25*10^15 bills
v=10^9 bills
T=(V/v)= 1.25*10^6
(a)
t
F(t)=𝑒𝑒 −(1.25∗10^6)
t=0 ; F(t)=0
t=∞ ; F(t)=1
(b)
∞
b=∫10 b0 E(t)
Where b0 = 1.25 ∗ 10^15
t
E(t)=∫ F(t)dt = ∫ 1 − exp � � dt
T
∞
b=∫10 (1.25 ∗ 1015 )(
=1.009*10^9 bills
1
1.25∗10^6
exp�
−t
1.25∗10^6
�
=109 bills
14.5
kT=ln
cA0
cA
= ln
1000
1
= 6.9078
For the small deviation from plug flow , by the tanks in series model first calculate ỽ2 from the tracer
curve.
α=8-12=4
ỽ2 =
T=10
α
2
=
24 3
2
ỽ2 � 3 �
1
2
= ỽ∅ = 2 = 2 = 0.67 ∗ 10−2
T
10
N
N=150 tanks
kT=6.9078/150=0.0461
For tank in series
cA0
1
=
cA (1+kT)N
=1/863
cA0
= 0.00116
cA
cA = 1.16
1
= (1+0.0461)150
14.6
fully suspended
solid
particle leaves
v=1 m^3/min
1m^3 slurry
Conversion complete after 2 minutes in the reactor.
So there is no any dead zone is present because 1 minutes for each reactor.
First approximate each pulse by plug flow pattern.
𝐴𝐴2 1
= =
𝑅𝑅
𝐴𝐴1 3 𝑅𝑅+1
R=0.5
𝑉𝑉𝑉𝑉1
(𝑅𝑅+1)𝑣𝑣
= 1;
Putting value R;
𝑉𝑉𝑉𝑉1
(0.5+1)1
= 1;
3
𝑉𝑉𝑉𝑉1 = ;
𝑉𝑉𝑉𝑉1
(𝑅𝑅+1)𝑣𝑣
2
+
𝑉𝑉𝑉𝑉2
𝑅𝑅𝑅𝑅
= 3;
Putting values and find Vp2;
3/2
(0.5+1)1
Vp2=1;
𝑉𝑉𝑉𝑉2
+ (0.5)1 = 3;
Vactual =(3/2)+1=2.5;
Vdead= 5-2.5= 2.5;
Now considering that the pulse output has width,
∆𝜃𝜃
𝜃𝜃𝜃𝜃𝜃𝜃𝜃𝜃
2/3
=
1
2
=
;
√(𝑁𝑁1−1)
2
√(𝑁𝑁1−1)
;
N1=10 tanks;
N2=(2/3)*N1;
N2=6.67 tanks;
∑ 𝑡𝑡𝑡𝑡
Tavg= ∑ ;
𝑐𝑐
Tavg=2.49=2.5;
𝑉𝑉1+𝑉𝑉2
= 1.5 + 1.0 = 2.5;
𝑣𝑣
14.8
Find out number of tanks.
Tavg =
∑ 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡
∑ 𝑐𝑐𝑐𝑐𝑐𝑐
320
=
40
;
;
=8 min
𝜎𝜎 2 =
2
∑ 𝑡𝑡𝑡𝑡
∑ 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡
� ;
−�
∑ 𝑐𝑐
∑ 𝑐𝑐𝑐𝑐𝑐𝑐
=�
250+490+810+1210
=5;
𝜎𝜎 2
1
=
;
𝑁𝑁 (𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡)2
N=
64
5
;
N=12.8
40
�−�
320 2
40
� ;
N=13.
14.9 For N tanks in series.
Finding out number of tanks
N=1+4(
𝜃𝜃𝜃𝜃𝜃𝜃𝜃𝜃
∆𝜃𝜃
)^2;
N1peak=104 tanks;
N2peak=101 tanks;
N3peak=101 tanks;
N4peak=97 tanks;
So that average is 100 tanks.
Model
14.10
(a) Area under curve c vs t curve
∞
� cidti = � ci∆ti
0
=3000
∑ citi∆ti
Mean resident time = ∑
(b) E curve
C
35
38
40
40
39
38
35
ci∆ti
= 34.66
Time
0
10
20
30
40
50
60
E=C/q
0
0.00333
0.00666
0.0133
0.0166
0.022
0.0233
CA/CAo
0
0.0202
0.0245
0.0223
0.0183
0.0136
0.0099
(c) Variance=
∑ citi2 ∆ti
∑ ci∆ti
− Ti2
=22.45
(d) How many tanks in series in this vessel equipment to?
ỽ2 (22.45)^2
1
= ỽ2∅ = 2 =
T
N
34.662
N=2.38 tanks
Example:-14.11
A reactor has flow characteristics given by the nonnormalized C curve in Table P14.11, and by the
shape of this curve we feel that the dispersion or tanks-in-series models should satisfactorily
represent flow in the reactor.
(a) Find the conversion expected in this reactor, assuming that the dispersion model holds.
(b) Find the number of tanks in series which will represent the reactor and the conversion expected,
assuming that the tanks-in-series model holds.
Time
1
2
3
4
5
6
7
8
10
15
20
30
41
52
70
Concentration
0
9
57
81
90
90
86
77
67
47
32
15
7
3
0
(c) Find the conversion by direct use of the tracer curve.
(d) Comment on the difference in these results, and state which one you think is the most reliable.
Solution:To find out non ideal characteristics of reactor determine proper D/uL
To use for dispersion model, or the proper N value to use for tank in series model.
This is done by two way :- matching tracer concentration curve with dispersion model or with tank in
series model , or by calculating ỽ^2 fromthat D/uL or N.
Lets us use the latter procedure. So first calculate T and ỽ^2 from the table of data with the help of
given equ. 𝑇𝑇 =
∞
∫0 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡
∞
∫0 𝐶𝐶𝐶𝐶𝐶𝐶
=
∑ 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡∆𝑡𝑡𝑡𝑡
∑ 𝐶𝐶𝐶𝐶∆𝑡𝑡𝑡𝑡
2
ỽ =
∞
∫0 (𝑡𝑡 − 𝑇𝑇)2𝐶𝐶𝐶𝐶𝐶𝐶
∞
∫0 𝐶𝐶𝐶𝐶𝐶𝐶
ƩC=213
T=2149/213=10.09min
Ʃtc=2149
ỽ2 = �
37685
213
∞
=
∫0 𝑡𝑡 2 𝐶𝐶𝐶𝐶𝐶𝐶
∞
∫0 𝐶𝐶𝐶𝐶𝐶𝐶
� − (10.09)2 = 75.11
− 𝑇𝑇 2
ỽ2∅ =
Ʃt^2c=37685
ỽ2
𝑇𝑇 2
=
75.116
(10.09)^2
Next determine the behavior of ideal PFR
k=o.456min−1
= 0.7378
kT=4.6
so for PFR x𝑎𝑎 = 1 − 𝑒𝑒 kT = 1 − e4.6 = 0.99
(a)
Use the dispersion model
Here related ỽ2∅ with D/uL so,
uL
D
D
ỽ2∅ = 0.7378 = 2 � � − 2( )2 [1 − 𝑒𝑒 − D ]
uL
uL
Solve by trial and error.this gives D/uL=1
v/vp
constant kt=4.6 and D/uL=1
xa=0.89
From fig x(disp)=0.89
(b)
Use tank in series model
1
N= 2 =
ỽ∅
1
0.7385
= 1.35 Tanks
From fig 6.5we can find the X(tank)=83%
(c)
Use the tracer data directly
From
c
c0
c
= ∑( )E∆t
c0
To find out E curve make the area under C curve unity,
Time Concentration
1
2
3
4
5
6
7
8
10
15
20
30
41
52
70
E=
c
area
=
0
9
57
81
90
90
86
77
67
47
32
15
7
3
0
citi
𝑒𝑒 −0.465(213)
0
0.0506
0.0456
0.0158
0.0050
0.0015
0.0005
0.0002
0.0001
16*10^-6
5*10^-6
1*10^-6
0
0
0
c
∑ ci∆t
Now we know that
c
c0
c
= ∑ � � E∆t
c0
∑ c =213
∑ e =0.1911
X=0.88(from curve)
(d)
Which answer is most reliable
Naturally the direct use of tracer gives most reliable answer. In this given model RTD
come from dispersion model D/uL=1. Thus wind accepted that ans to part (a) and (c)
should agree.
Solid Catalysed Reactions
Solution 18.1
We cannot tell just by looking at the reactor and the catalyst bed height whether the reactor is integral or
differential. It isn’t the depth or the packing or the conversion level that counts but the assumptions used when
the data from the reactor are analysed. If we assume it as a constant rate of reaction throughout the reactor, then
it is a differential reactor but if we assume a changing rate of reaction with changing position, then it becomes a
integral reactor. It is just the matter of how we are looking at the rate.
Solution 18.2
Since the volume is made 3 times the original volume but the temperature, feed flow rate, the amount of the
catalyst remains the same, according to the performance equation of the mixed flow reactor,
𝑊
𝑋𝐴
=
𝐹𝐴0
−𝑟𝐴
Thus there will be no change in the conversion.
Solution 18.3
a) We have been given data for the weight of the catalyst, molar flow rate and the conversion. Since it is a
packed bed reactor, the performance equation is as follows.
W
FAO
XA
= ∫
0
dX A
−rA
Writing it in the differential form,
𝑑𝑊
𝑑𝑋𝐴
=
𝐹𝐴0
−𝑟𝐴
Thus making the subject as –rA,
−𝑟𝐴 =
𝑑𝑋𝐴
𝑊
𝑑 (𝐹 )
𝐴
Thus we see that the slop of the cure at the xA= 0.4 gives us the rate of the equation.
Thus taking the slope of the curve at the point Xa= 0.4
−𝑟𝐴 =
0.482 − 0.23
𝑘𝑔𝑚𝑜𝑙𝑒 𝑐𝑜𝑛𝑣𝑒𝑟𝑡𝑒𝑑
= 0.36
0.8 − 0.1
𝑘𝑔 𝑐𝑎𝑡 . ℎ𝑟
0.5
0.45
0.4
0.35
XA
0.3
0.25
0.2
0.15
0.1
0.05
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
(W/FAO)
b) For finding the amount of catalyst required if the molar flow is 400 kmol/hr and the conversion required is
Xa=0.4,
From the curve we can see that for Xa= 0.4, the ration of W/FAO is 0.575.
Therefore, W= 0.575 X 400 = 230 kg Cat.
c) For completely mixed flow reactor,
W
FAO
W
FAO
=
=
XA
−rA at 0.4
XA
−rA at 0.4
=
0.4
0.36
W = 400 X (0.4/0.36) = 444 kg catalyst
Solution 18.4
AR
n=1/2 th order reaction with infinite (very large) recycle hence R = ∞
W = 3 kg of catalyst
CA0 =2 mol/m3 , CA = CAout = 0.5 mol/m3
V = 1 m3/hr
A = 1-1/1 = 0
XA = (CA0 − CA)/(CA0 + A ∗ CA)
2−0.5
XA = 2+00.5
XA=0.75
FA0 = molar flow rate of gas A
FA0 = 2 mol/m3  1 m3/hr
= 2 mol/hr
W
 FA0 =
XAout
(for MFR) ______________ (1)
−rAout
Here, -rAout = k × CAout1/2 _________________(2)
 k =
=
FA0∗XAout
W∗CAout
_________________________(3)
20.75
30.5
= 1 m3/kg of catalyst hr
Put the value of k in equation (2) to get value of -rAout
-rAout = 1 m3/kg of catalyst hr  0.51/2 mol/m3
-rAout = 0.707 mol / kg of catalyst hr
Hence, -rA (mol / kg of catalyst hr ) = (1 m3/kg of catalyst hr) (CA mol / m3)1/2
SOLUTION 18.5
Very large recycle, A→3R, n=1and feed contains 50% A and 50% inerts.
Here Ɛ=1, FA0=2 mol/hr, W=3 kg, CAout = 0.5 mol/m3, CAo=2 mol/m3
With large recycle,use the performance equation of a mixed flow reactor,
𝑊
𝐹𝐴0
∴
=
𝑊
𝐹𝐴0
𝑋𝐴𝑜𝑢𝑡
−𝑟𝐴𝑜𝑢𝑡
, -rAout=k’CAout … first order decomposition of A
𝐶𝐴𝑜𝑢𝑡
=
1− 𝐶𝐴𝑜
( 𝐶𝐴𝑜𝑢𝑡
)
1+ 𝐶𝐴𝑜
×
1
𝑘′𝐶𝐴𝑜𝑢𝑡
∴ k’= 0.8 m3/(h∙ kg ∙cat)
Therefore, -rA’
=-
1 𝑑𝑁𝐴
𝑊 𝑑𝑡
= 0.8 m3/(h∙ kg ∙cat) ∙ CA mol/m3
SOLUTION 18.6
No recycle, A→3R, n=2 and feed contains 25% A and 75% inerts, 4 volumes
ƐA=(6-4)/4 =0.5
1−𝑋𝐴
CA = CAo(1+0.5XA)
With No recycle, it becomes a plug flow reactor.
So,
𝑋𝐴 𝑑𝑋𝐴
𝑊
=∫0
𝐹𝐴0
−𝑟𝐴′
-rA’= k’CA2
𝑋𝐴
𝑊
𝐹𝐴0
=∫0
𝑑𝑋𝐴
2
1−𝑋𝐴
(1+0.5XA)
𝑋𝐴 (1+0.5𝑋𝐴)2 𝑑𝑋𝐴
=∫
0
(1−𝑋𝐴)2
𝑋𝐴 (1+𝑋𝐴+0.25𝑋𝐴2 )𝑑𝑋𝐴
(1−𝑋𝐴)2
=∫0
𝑋𝐴 𝑑𝑋𝐴
=∫0 (1−𝑋𝐴)2
+
𝑋𝐴 (𝑋𝐴)𝑑𝑋𝐴
∫0 (1−𝑋𝐴)2
𝑋𝐴 𝑑𝑋𝐴
I=∫
0 (1−𝑋𝐴)2
=
𝑋𝐴 (𝑋𝐴)𝑑𝑋𝐴
II = ∫
0 (1−𝑋𝐴)2
=
𝑋𝐴
(1−𝑋𝐴)
𝑋𝐴
(1−𝑋𝐴)
𝑋𝐴 𝑋𝐴2 𝑑𝑋𝐴
III = 0.25 ∫
0 (1−𝑋𝐴)2
I+II+III = 2.25
𝑊
=
𝐹𝐴0
1
𝑋𝐴
(1−𝑋𝐴)
[2.25
𝑘 ′ 𝐶𝐴𝑜
+
𝑋𝐴 𝑋𝐴2 𝑑𝑋𝐴
0.25 ∫
0 (1−𝑋𝐴)2
+ ln(1-XA)
= 0.25
𝑋𝐴
(1−𝑋𝐴)
+ 0.50 ln (1-XA) + 0.25XA
+ 1.50 ln (1-XA) + 0.25XA
𝑋𝐴
(1−𝑋𝐴)
+
1.50 ln (1 − XA) + 0.25XA]
XA = 0.67
𝑋𝐴
k’= [2.25
(1−𝑋𝐴)
+
1.50 ln (1 − XA) + 0.25XA]
×
𝐹𝐴0
𝐶𝐴02 ×𝑊
k’= 0.512 m6/(mol.h.kg cat)
-rA’ = -
1 𝑑𝑁𝐴
𝑊 𝑑𝑡
= 0.512 m6/(mol.h∙ kg ∙cat) ∙ (CA mol/m3)^2
Solution 18.7
Mixed Flow Behaviour
A→R
∈a=
1−1
CAO =10
1
mol
m3
=0
; W=4 gm
Performance Equation
W
XA
−rA
=
FAO
For No. 1:
vo = 3
m3
h
, XA = 0.2
FAO = Vo * CAO
= 3 * 10
= 30
mol
h
-rA = XA *
FAO
W
= 0.2 * 30/4
mol
= 1.5 h g cat
CA = CAO * (1-XA) since ∈ a = 0
= 10 (0.8)
CA = 8
mol
m3
For No. 2:
Vo = 2
m3
XA = 0.3
h
FAO = Vo * CAO
= 2 * 10
= 20
mol
-rA = XA *
h
FAO
W
= 0.3 * 20/4
mol
= 1.5 h g cat
CA = CAO * (1-XA) since ∈ a = 0
= 10 (0.7)
CA = 7
vo
FAO
XA
CA
-rA
mol
m3
3
30
0.2
8
1.5
2
20
0.3
7
1.5
1.2
12
0.5
5
1.5
So the reaction is independent of CA (0 th Order)
-rA = K CA n
mol
-rA = 1.5
h g cat
Solution 18.8
Plug Flow
mol
CAO = 60
m3
lt
vO = 3 min = 3 * 10-3
m3
min
FAO = vO * CAO
= 60 * 3 * 10-3
= 0.18
W
FAO
mol
min
XA
dXA
−rA
= ∫
0
A→R , ∈a=
1−1
1
=0
CA = CAO * (1-XA)
C
XA = 1- C A
AO
30
XA1 = 1- 60 = 0.5
20
XA2 = 1- 60 = 0.67
10
XA3 = 1- 60 = 0.833
Guessing the rate as
First order
-rA = k * CA
= k * CAO * (1-XA)
W
FAO
XA
= ∫
0
XA
= ∫
0
=
dXA
−rA
dXA
k ∗ CAO ∗ (1 − XA )
− ln(1− XA )
k∗ CAO
W1
FAO
W2
FAO
W3
FAO
0.5
= 0.18 = 2.78
1
= 0.18 = 5.56
2.5
= 0.18 = 13.89
W
1
=
FAO
1
* ln ( 1−𝑋 )
𝑘∗𝐶𝐴𝑂
𝐴
y = mx
W
FAO
1
ln ( 1−𝑋 )
1
2.78
2
5.56
3
13.89
0.693
1.098
1.79
𝐴
ln ( 1/(1-XA ) )
First order
2
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
2
4
6
8
10
12
14
16
W/(FAO)
W
1
vs ln ( 1−𝑋 ) is not coming straight line. So our assumption of First order is wrong.
FAO
𝐴
Guessing Second order:
-rA = k * CA2
= k * CAO2 * (1-XA)2
W
FAO
XA
= ∫
0
dXA
−rA
XA
= ∫
k ∗
0
W
FAO
=
1
2
𝑘∗ 𝐶𝐴𝑂
2
𝐶𝐴𝑂
𝑋
* 1− 𝐴𝑋
W
FAO
𝑋𝐴
1 − 𝑋𝐴
dXA
∗ (1 − 𝑋𝐴 )2
𝐴
1
2.78
2
5.56
3
13.89
1
2.03
4.998
Second order
6
XA/(1-XA )
5
4
3
2
1
0
0
2
4
6
8
10
12
14
16
W/(FAO)
W
FAO
𝑋
vs 1−𝑋𝐴 is coming straight line. So our assumption of Second order is correct.
𝐴
1
Slope =
2
𝑘∗ 𝐶𝐴𝑂
1
𝑘∗ 602
= 2.773
= 2.773
m6
k = 10-4 mol min g cat
-rA = 10-4 * CA2
Solution 18.9
We have reaction A→R, CAO= 10mol/m3
Ca
8
6
4
2
1
XA
0.2
0.4
0.6
0.8
0.9
FA0
0.14
0.42
1.67
2.5
1.25
1/-rA
3.571429
0.595238
0.0998
0.05
0.088889
4
3.5
3
1/-rA
2.5
2
1.5
1
0.5
0
0
1
2
3
4
5
6
7
8
9
Ca
For packed bed reactor with CAO= 8 mol/m3 and XA=0.75, CA= 2 mol/m3.
Also the area under the curve for 1/-rA vs CA from CA0= 8 mol/m3 to CA= 2 mol/m3 is 5.08.
According to the performance equation of the Packed bed reactor,
𝑊
𝑑 𝑋𝐴
= ∫
𝐹𝐴𝑂
−𝑟𝐴
We get, W= 1000 X 5.08/8 = 615 kg of catalyst is required.
Solution 18.10
For Mixed Flow reactor, the concentration inside the reactor will be same as that of the leaving stream. Given
that the initial reactant concentration is 10 mol/m3 and the conversion is 90 %, the outlet concentration of the
stream will be 1 mol/m3. Thus from the above tabulated data, the rate constant can be found out.
𝑘=
𝑇ℎ𝑢𝑠 𝑘 =
11.25
1
−𝑟𝐴
𝐶𝐴
= 11.25 m3/mol. Min
Thus the amount of the catalyst required is given by
𝑊=
𝐹𝐴𝑂 𝑋𝐴
𝑋
𝑘
𝐶𝐴
𝑊 = 1000 𝑋 0.9/(11.25 𝑋 1)
𝑊 = 80 𝑘𝑔
Term Paper
Of Cre-II,
Chapter No:18
Solid Catalysed Reactions
Sum No: 18.11-18.20
Submitted By:
Samarpita Chakraborty
(13BCH013)
Harsh Dodia
(13BCH014)
Nirav Donga
(13BCH015)
Harshit Goyal
(13BCH018)
18.11)
V=1000 m3/hr
Cao=100 mol/m3
XB=0.8
50𝐶𝐴
-rA=1+.02𝐶𝐴 Mol/kg.hr
For Plug Flow
𝑊
𝑋𝐴𝑓
T’= = ∫0
𝑣
100 1+.02𝐶𝑎
𝑑𝑋𝐴/−𝑟′𝐴=1/k’ ∫20
50𝐶𝑎
𝑑𝐶𝑎
K’T’= [ln (Cao/Ca) + k(Cao-Ca)]

1000
100
50
20
[ln
+ .02 (100-20)]
 𝑊 = 64.19 𝐾𝑔𝑐𝑎𝑙
18.12)
V=1000m3/hr
CS.=100mol/m3
XB=0.8
-rA= 8 CA2 Mol/kg.hr
For Plug Flow
𝑊
𝑋𝐴𝑓
T’= 𝑣 = ∫0
100
100
𝑑𝑋𝐴/−𝑟′𝑎=1/k’ ∫20 d𝐶𝑎 /−𝑟′𝑎
∫20 d𝐶𝑎
T’= 1/8
8𝐶𝑎2
 -1/8 [1/Ca0-1/Ca]
 -1/8 [1/100 -1/20]
 T’= W/V= 0.005
 𝑊 = 0.005 𝑥 1000 = 5 𝑘𝑔 𝑐𝑎𝑡𝑎𝑙𝑦𝑠𝑡
18.13)
v=10 m3/hr
Cao=0.1 mol/m3 and Cbo= 10 mol/m3
W= 4 kg
Xa= ?
-ra’=.6 CACB
For CAo=0.01 CBo
We can assume that CB= Constant
 -rA’ = CACB = .6 CA. 10 = 6 CA
 T’=
𝑤
𝑣
4
= 10= .4 kg.hr/m3
 For First order

CA
𝐶𝐴𝑜
= 𝑒 −𝑘′𝑇′ = 𝑒 −6 .4= .0907
 XA=90.9%
 XB= .909%
18.14)
Mol.wt of feed= 0.255 kg/kmol
Mol wt. of product= 0.070 kg/kmol
Feed enters at 630 ⁰C and 1 atm (vaporized)
Density of liquid feed =869 kg/m3
Denisty of catalyst particles = 950 kg/m3
Half the feed is cracked at rate 60 m3 liq/ m3 reactor. Hr
For Plug flow
𝑑𝑋𝑎
𝜏/Cao = V/W∫ −𝑟𝑎′
Where V= volume of catalyst and W= weight of catalyst
We know, Cao = Pao/RT
= 1 atm/0.082 x (630 + 273)
= 0.01350 mol/litre
= 13.50 mol / m3
For Plug flow and 1st order kinetics,
k’’’ 𝜏’’’= ln[Cao/Ca]
k’’’ (1/60)= ln (13.5/7.75)
k’’’ = 41.5888 hr-1
We know,
k’’’ 𝜏’’’= k’ 𝜏’ [𝜏’ is 𝜏′′′x (Density of the catalyst particles-bulk Density of packed bed) - as in
solid catalyzed reactions volume of catalysts can be taken as the void volume of reactor.]
(41.588)(1/60) =k’ ((950-700)/60)
k’= 0.166 m3 liquid/kg catalyst. Hr
18.15)
8atm
A
ε=2
3R
700⁰C
𝑊
𝑋𝑎
=
𝐹𝑎𝑜 −𝑟𝑎′
𝐶𝑎
=
1−𝑋𝑎
𝐶𝑎𝑜 1+2𝑋𝑎
8
Ca= PA/RT= .08206∗973=0.1mol/lit
Ca
Ca/Cao
Xa=(1-
Fao=vCao -ra
Ca/Cao)/(1+2Ca/Cao)
=Xa
Fao/W
0.08
0.8
0.076923077
10
0.769
0.05
0.5
0.25
2.2
0.55
0.02
0.2
0.571428571
0.4
0.228
0.01
0.1
0.75
0.1
0.075
0.863636364
0.05
0.0482
0.005 0.005
-ra vs Ca
0.9
y = 9.9132x
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.02
0.04
0.06
The Trendline of the curve gives us the relation:
-rA’= 9.9132CA
0.08
0.1
18.16)
For 1st order solid catalyzed reaction, without any complex piping conversion is XA = 0.8 and
thus
 k’ 𝜏’ =ln [CAO/CA]
 k’ 𝜏’ = -ln [1-Xa]
 k’ 𝜏’ = -ln[1-0.8]
 k’ 𝜏’= 1.6094 (Damkohler number for 1st order reactions)
at points 2, 3, 4 the conversion is same at steady state.
𝑑𝑋𝑎
𝜏’/CA0 =∫ −𝑟𝑎
=∫
𝑑𝑋𝑎
𝑘′ 𝐶𝑎
𝑋𝑎1
 k’ 𝜏′/CA0 = ∫𝑋𝑎2
𝑑𝑋𝑎
𝐶𝑎𝑜(1−𝑋𝑎)
 k’ 𝜏′= - ln [(1-Xa1)/(1-Xa2)]
 1.6094 = - ln [(1-Xa1)/(1-Xa2)]……………………………………………… 1
Taking mole balance,
CA0v0 + CA1v1 = CA3v3
And CA2= CA3= CA4
 CA0v0 + CA1v1 = CA3v3
 (100)(100) + CA1(100) = CA3(200)………………………………………………2
Solving equation 1 and 2,
1.6094 = - ln [(1-Xa1)/(1-Xa2)]
 𝑒 1.6094 =[(1-Xa2)/(1-Xa1)]
 5(1 − 𝑋𝐴1)= 1 − XA2
 5CA1/CA0 = CA2/CA0
 CA2/CA1 = CA3/CA1 = CA4/CA1 = 5
Therefore solving equation 2 we get,
CA4 = 11.11 x 5 = 55.55
Thus, XA4 =1-(55.55/100) = 0.444
18.17) (a)
V=1lit/hr
W=3gm
Cao=2mol/lit
Caout=0.5mol/lit
V0=1lit/hr
A
R
R=∞
Second order
- ra´=k´Ca2
T’=
𝑊𝐶𝑎𝑜 𝑊
𝐹𝑎𝑜
𝑉0
k´= 𝑊 .
1
=𝑣0=
𝐶𝑎0−𝐶𝑎
− ra´
𝐶𝑎0−𝐶𝑎
𝐶𝑎∗𝐶𝑎
2−0.5
=3 . 0.5∗0.5 = 2lit2/mol.gm.hr
(b) For packed bed (assume plug flow)
V0=1000lit/hr
Ca0=1mol/lit
Xa= 0.8
W= ?
For plug flow
𝐶𝑎𝑜 𝑑𝐶𝑎
𝑊
T’=𝑣0 = ∫𝐶𝑎
1000
W=
2
.
1
= .
𝑘𝐶𝑎.𝐶𝑎 𝑘
1
[𝐶𝑎1 − 𝐶𝑎𝑜
]
[0.21 − 11]=2000gm=2kg
(c) Add inert solid
This does not change the rate equation or performance equation
Therefore, adding inert make the reactor bigger but does not change the weight of the
needed for the process
18.18)
A  R in a packed bed reactor.
Cao
mol/m3
Ca
Vo
mol/m3 (lit/hr)
Fao
Xa
10
1
5
0.05
W kg cat
2
20
0.2
1
3
65
0.65
6
133
1.33
9
540
5.4
0.9
0.8
0.7
0.4
0.1
(-)ra'
0.045
0.16
0.455
0.532
0.54
1/(-)ra'
22.22222
6.25
2.197802
1.879699
1.851852
25
1/-ra'
20
15
10
5
0
0
0.2
0.4
0.6
Xa
0.8
1
catalyst
25
20
1/-ra'
15
10
5
0
0
2
4
6
8
10
Ca
( Area under the curve for PFR i.e with no recycle of exit fluid for
𝑑𝐶𝑎
∫ −𝑟𝑎 =14
Xa= 0.75)
(Ca0-Ca)/-ra = 36
(Area under the curve for CSTR i.e with very high recycle and for
Xa= 0.75)
Hence,
Part 1: For PFR,
𝑑𝐶𝑎
W = FAO/CAO ∫ −𝑟𝑎 = (1000 𝑥 14)/8 = 1750 𝑘𝑔 𝑐𝑎𝑡𝑎𝑙𝑦𝑠𝑡
Part 2 : For CSTR,
W= FAO/CAO x (CA0-CA)/-ra
= (1000/8) x 36
= 4500 kg catalyst
18.19)
1atm
2A
R
V=100 cm3
This batch circulation system is a constant volume system, so έa=0
101325
Ca= PA0/RT= 8.314∗609=20mol/m3
𝑎
PA= PA0 + ∆𝑛 (𝜋𝑜 − 𝜋)
2
=1+−1 (1 − 𝜋)
=2 𝜋 − 1……………………(1)
t
0
4
8
6
36
𝜋
1
0.75
o.67
0.60
0.55
PA
1
0.5
0.33
0,2
0.1
Ca
20
10
6.67
4
2
lnCa
3
2.3
1.9
1.38
0.693
1/Ca
0.05
0.1
0.135
0.25
0.5
609K
Slope of 1/Ca vs t is 0.0125.
Therefore rate equation is ra=0.0125Ca
2
18.20)
Mw= µT2∈= L2(-rA’’’/CA)obs/De
Mw=
.15
-rA’’’/CA= .88
De= 2*10-6
.15=
L2*.88/2*10-6
L=
5.83* 10-4 m
=
58.3mm
L=
R/3
L=
D/6
D=
350 mm
(Sphere)
18.21. A reaction A→R is to take place on a porous catalyst pellet (d = 6mm, de = 10m3/m cat.
s). How much is the rate slowed by pore diffusional resistance if the concentration of reactant
bathing the particle is100 mol/m3 and the diffusion-free kinetics are given by
-r a = 0.1c a 2
ans.
For nth order reaction M T =
=
𝑑𝑑𝑝𝑝
6
�
𝑚𝑚𝑚𝑚𝑚𝑚
𝑚𝑚3 .𝑐𝑐𝑐𝑐𝑐𝑐.𝑠𝑠
(𝑛𝑛+1)∗𝐾𝐾′′ 𝐶𝐶𝑎𝑎𝑛𝑛−1
6∗10−3
6
= 3.873
2∗𝐷𝐷𝑒𝑒
�
(2+1)∗0.1∗1002−1
2∗10−6
From the Graph value of E = 0.23
18.22 In the absence of pore diffusion resistance a particular first-order gas- phase reaction
proceeds as reported below.
What size of spherical catalyst pellets ( g e 10^-3= cm3/cm cat as) would ensure that pore
resistance effects do not intrude to slow the rate of re- action?
Given P= 1atm
C Ao =P/RT
C Ao = 1/(0.08206*673)
C Ao = 1.8*10^-5 mol/cm^3
X Ae = C Ao - C A / C Ao
X Ae = (1.8-1)10^-5/1.8*10^-5
X Ae = 0.444
M t = L(K’’’/D e *X Ae )^(1/2)
By Solving above equation
We get
For Spherical particle
M T = 0.4
L=0.4cm
R =3L
R= 0.0795 cm
R = 0.8mm
18.23. The first-order decomposition of A is run in an experimental mixed flow reactor. Find the
role played by pore diffusion in these runs in effect determine whether the runs were made under
diffusion-free, strong resistance or intermediate conditions.
dp
w
C ao
v
xa
3
1
100
9
0.4
12
4
300
8
0.6
A → 𝑅𝑅
ans.
No
dp
w
C AO
v
xa
CA
K’
1
3
1
100
9
0.4
60
6
2
12
4
300
8
0.6
120
3
For Mixed Flow
(1)
(2)
𝐶𝐶𝐴𝐴 = 𝐶𝐶𝐴𝐴0 (1 - 𝑋𝑋𝐴𝐴 )
𝐶𝐶𝐴𝐴 = 𝐶𝐶𝐴𝐴0 (1 - 𝑋𝑋𝐴𝐴 )
= 100(1 – 0.4)
= 300(1 – 0.6)
= 60
= 120
For No diffusion
𝐾𝐾2
𝐾𝐾1
=1
For Strong resistance
𝐾𝐾2
𝐾𝐾1
=
𝑑𝑑𝑝𝑝2
𝑑𝑑𝑝𝑝1
=
1
4
From the given data :
𝜏𝜏 ′ =
𝑤𝑤
𝑣𝑣
𝐾𝐾1 ′ =
=
=
𝐶𝐶𝐴𝐴0 − 𝐶𝐶𝐴𝐴
𝑊𝑊 ′ 𝐶𝐶𝐴𝐴
𝑉𝑉∗(𝐶𝐶𝐴𝐴0 − 𝐶𝐶𝐴𝐴 )
𝐾𝐾2 ′ =
𝑊𝑊 ′ ∗𝐶𝐶𝐴𝐴
(100−60)∗9
=6
=
1∗60
=3
𝑉𝑉∗(𝐶𝐶𝐴𝐴0 − 𝐶𝐶𝐴𝐴 )
𝑊𝑊 ′ ∗𝐶𝐶𝐴𝐴
(300−120)∗8
4∗120
𝐾𝐾2 3 1
= =
𝐾𝐾1 6 2
So that, diffusion take place at intermediate condition
18.24 The first-order decomposition of A is run in an experimental mixed flow reactor. Find the
role played by pore diffusion in these runs in effect determine whether the runs were made under
diffusion-free, strong resistance or intermediate conditions.
dp
w
C ao
v
xa
4
1
300
60
0.8
8
3
100
160
0.6
A → 𝑅𝑅
ans
No
dp
w
C A0
V
xa
CA
K’
1
4
1
300
60
0.8
60
240
2
8
3
100
160
0.6
40
80
For Mixed Flow
(1)
(2)
𝐶𝐶𝐴𝐴 = 𝐶𝐶𝐴𝐴0 (1 - 𝑋𝑋𝐴𝐴 )
𝐶𝐶𝐴𝐴 = 𝐶𝐶𝐴𝐴0 (1 - 𝑋𝑋𝐴𝐴 )
= 300(1 – 0.8)
= 100(1 – 0.6)
= 60
= 40
For No diffusion
𝐾𝐾2
𝐾𝐾1
=1
For Strong diffusion
𝐾𝐾2
𝐾𝐾1
=
𝑑𝑑𝑝𝑝2
𝑑𝑑𝑝𝑝1
=
1
2
From the given data :
𝜏𝜏 ′ =
𝑤𝑤
𝑣𝑣
𝐾𝐾1 ′ =
=
=
𝐶𝐶𝐴𝐴0 − 𝐶𝐶𝐴𝐴
𝑊𝑊 ′ 𝐶𝐶𝐴𝐴
𝑉𝑉∗(𝐶𝐶𝐴𝐴0 − 𝐶𝐶𝐴𝐴 )
𝑊𝑊 ′ ∗𝐶𝐶𝐴𝐴
(300−60)∗60
= 240
1∗60
𝐾𝐾2 ′ =
=
𝑉𝑉∗(𝐶𝐶𝐴𝐴0 − 𝐶𝐶𝐴𝐴 )
𝑊𝑊 ′ ∗𝐶𝐶𝐴𝐴
(100−40)∗160
= 80
3∗40
𝐾𝐾2
80
1
=
=
𝐾𝐾1 240 3
So that, diffusion take place at strong resistance condition
18.25 The first-order decomposition of A is run in an experimental mixed flow reactor. Find the
role played by pore diffusion in these runs in effect determine whether the runs were made under
diffusion-free, strong resistance or intermediate conditions.
dp
w
C ao
v
xa
2
4
75
10
0.2
1
6
100
5
0.6
A → 𝑅𝑅
ans
No
dp
w
C A0
V
xa
CA
1
2
4
75
10
0.2
60
2
1
6
100
5
0.6
40
For Mixed Flow
(1)
(2)
𝐶𝐶𝐴𝐴 = 𝐶𝐶𝐴𝐴0 (1 - 𝑋𝑋𝐴𝐴 )
𝐶𝐶𝐴𝐴 = 𝐶𝐶𝐴𝐴0 (1 - 𝑋𝑋𝐴𝐴 )
= 75(1 – 0.2)
= 60
= 40
For No diffusion
𝐾𝐾2
𝐾𝐾1
=1
For Strong diffusion
𝐾𝐾2
𝐾𝐾1
=
𝑑𝑑𝑝𝑝2
𝑑𝑑𝑝𝑝1
=
1
2
From the given data :
𝜏𝜏 ′ =
𝑤𝑤
𝑣𝑣
= 100(1 – 0.6)
=
𝐶𝐶𝐴𝐴0 − 𝐶𝐶𝐴𝐴
𝑊𝑊 ′ 𝐶𝐶𝐴𝐴
K’
𝐾𝐾1 ′ =
=
𝑉𝑉∗(𝐶𝐶𝐴𝐴0 − 𝐶𝐶𝐴𝐴 )
𝐾𝐾2 ′ =
𝑊𝑊 ′ ∗𝐶𝐶𝐴𝐴
(75−60)∗10
=
4∗60
= 0.625
𝑉𝑉∗(𝐶𝐶𝐴𝐴0 − 𝐶𝐶𝐴𝐴 )
𝑊𝑊 ′ ∗𝐶𝐶𝐴𝐴
(100−40)∗5
6∗40
= 1.25
𝐾𝐾2
1.25
=
=2
𝐾𝐾1 0.625
So that, diffusion take place at strong pore diffusion
Solution 18.26
dp
1
2
2
CA
20
40
40
-r A ’
1
1
3
T,K
480
480
500
A→ R; C A0 = 50
Now, d p2 / d p1 = k 1 /k 2 = k 0 e-E/RT 1 / k 0 e-E/RT 2 = 2
Similarly,
d p3 / d p2 = k 2 /k 3 = k 0 e-E/RT 2 / k 0 e-E/RT 3 = 1
d p3 / d p1 = k 1 /k 3 = e-E/RT 1 / e-E/RT 3 = 2
On solving we get,
E = -69139.224 J = -69.13 kJ
Solution 18.27
Quantity of
Catalyst
1
4
1
4
Pellet Diameter
1
1
1
2
Flow rate of
given feed
1
4
1
4
Recycle Rate
High
Higher still
Higher still
High
Measured
reaction rate
4
4
3
3
All these runs were made at the same W/F Ao . We are also told that the recycle rate is high
enough to have mixed flow throughout. Thus changing the recycle rate from run 1 to run 2, or
from run 3 to run 4 with no change in rate shows that film resistance doesn’t influence the rate.
Next, if particles were non-porous then the smaller size (runs 1 & 2) should have double the rate
of the larger size (runs 3 & 4). This is not what is measured hence the results are not consistent
with the guess of non-porous particles. (1) If in the regime of no pore diffusion resistance the
rates should be same for large and small pellets. (2) If in the regime of strong pore diffusion the
rate should halve from going 4 to 2 in going to the larger particle.
Our results are half-way between these two extremes, hence we must be operating under
conditions where pore diffusion is just beginning to influence the rate. Hence we can conclude
that, porous particles, no film resistance, pore resistance just beginning to intrude and effect the
rate.
Solution 18.28
dp
4
8
A→ R → S
W/F A0
1
2
C R,max /C A0
0.5
0.5
Under the best possible condition we may accept C R, max /C A0 to be 0.5 and to obtain that we may
use the plug flow pattern and catalyst size to be small.
Solution 18.29
For mixed flow the reactor performance equation is: W/F A0 = X A /-r A ’ and since the data show
that the rate is inversely proportional to pellet size shows that pore diffusion controls in both runs
shows:
k 2 /k 1 = 0.5 and by eliminating pore diffusion in backmix flow we can get – C R /C A0 = 0.44
and by eliminating pore diffusion and using plug flow we can get - C R /C A0 = 0.63
So, use smaller pellets in a packed and we should be able to achieve as high as C R /C A0 = 0.63
Solution 18.30
C A0 = 10 mol/m3
A packed bed catalytic reactor is used.
(a) It should be operated in the strong diffusion regime.
(b) We should use a plug flow.
Solution: First order catalytic reaction, A → R (Strong pore diffusion resistance)
-ln
𝐶𝐶𝐴𝐴
𝐶𝐶𝐴𝐴𝐴𝐴
= 𝑘𝑘𝑘𝑘
-ln(1-𝑋𝑋𝑎𝑎 ) = kt
1 − 𝑋𝑋𝑎𝑎 = 𝑒𝑒 −𝑘𝑘𝑘𝑘
In the strong pore diffusion resistance, Double the pellet size rate
constant is same
K1 = K2 = K
𝑋𝑋𝑎𝑎1 = 0.632 , 𝑘𝑘1 = ?
1 − 𝑋𝑋𝑎𝑎1 = 𝑒𝑒 −𝑘𝑘1𝑡𝑡1
1 – 0.632 = 𝑒𝑒 −𝑘𝑘𝑡𝑡1
Kt1 = 1
For 18 mm pellets, t2= 0.5t1 so Kt2= 0.5.
1 − 𝑋𝑋𝑎𝑎2 = 𝑒𝑒 −𝑘𝑘2𝑡𝑡2
1 – Xa2 = 𝑒𝑒 −0.5
𝑋𝑋𝑎𝑎2 = 0.393
Solution:
P= 1atm, T=336 ºC, Xa= 0.8, De= 2*10-6
L=
1.2∗10−2
Cao=
2∗3
𝑃𝑃𝑎𝑎𝑎𝑎
𝑅𝑅𝑅𝑅
=
= 0.002𝑚𝑚
101325
8.314∗(336+273)
𝜏𝜏
𝑑𝑑𝑑𝑑𝑎𝑎
= �
𝐶𝐶𝑎𝑎𝑎𝑎
−𝑟𝑟𝑎𝑎
= 20
𝑚𝑚3
𝑚𝑚 𝑐𝑐𝑐𝑐𝑐𝑐.𝑠𝑠
𝑚𝑚𝑚𝑚𝑚𝑚
𝑚𝑚3
2000 𝑋𝑋𝑎𝑎 ∗ (1 +∈𝑎𝑎 𝑋𝑋𝑎𝑎 )2
𝜏𝜏
� 2
=
0.01
𝐶𝐶𝑎𝑎𝑎𝑎
𝐶𝐶𝑎𝑎𝑎𝑎 ∗ (1 − 𝑋𝑋𝑎𝑎 )2
𝜏𝜏 =
τ=
1
1
1
1
1
𝑑𝑑𝑑𝑑𝑎𝑎
�
=
�
−
1�
=
�
− 1� = 0.1
40 1 − 𝑋𝑋𝑎𝑎
40 1 − 0.8
20 ∗ 𝐶𝐶𝑎𝑎𝑎𝑎 (1 − 𝑋𝑋𝑎𝑎 )2
𝑊𝑊
𝐹𝐹𝐴𝐴𝐴𝐴
→ 𝑊𝑊 = 0.1 ∗ 1 = 0.1𝑚𝑚3 = 200 𝑘𝑘𝑘𝑘
Solution:
P= 1atm, T= 336 ºC, v= 4cm3/s, Xa= 0.8 when 10 gm of 1.2 mm catalyst
Design a fluidized bed of 1-mm particles (assume mixed flow of gas) and
packed bed
of 1.5 cm particles which need minimum catalyst?
Cao=
𝑃𝑃𝑎𝑎𝑎𝑎
𝑅𝑅𝑅𝑅
101325
=
𝑋𝑋𝑎𝑎
𝜏𝜏
=
𝐶𝐶𝑎𝑎𝑎𝑎
−𝑟𝑟𝑎𝑎
τ=
𝑊𝑊
𝐹𝐹𝐴𝐴𝐴𝐴
-ra =
Mw=
=
8.314∗(336+273)
0.01
4∗10−6
20∗0.8
2500
= 2500 𝑠𝑠𝑠𝑠𝑠𝑠
= 6.4 ∗ 10−3
𝐿𝐿2 ∗(−𝑟𝑟𝑎𝑎)
𝐶𝐶𝑎𝑎 ∗𝐷𝐷𝑒𝑒
=
= 20
𝑚𝑚𝑚𝑚𝑚𝑚
𝑘𝑘𝑘𝑘.𝑠𝑠
12.8∗(0.0002)2
20∗0.2∗10−6
For Packed bed:
𝑚𝑚𝑚𝑚𝑚𝑚
k=
𝑚𝑚3
= 12.8
−𝑟𝑟𝑎𝑎
𝐶𝐶𝑎𝑎
=
12.8
20∗(1−0.8)
= 3.2 𝑠𝑠𝑠𝑠𝑠𝑠 −1
𝑚𝑚𝑚𝑚𝑚𝑚
𝑚𝑚3 .𝑠𝑠
= 0.128
L= 2.5*10-3 m
𝐾𝐾
MT = L� = 2.5*10-3�
𝐷𝐷
3.2
10−6
= 4.472 → ↋ =
𝑡𝑡𝑡𝑡𝑡𝑡ℎ𝑀𝑀𝑇𝑇
𝑀𝑀𝑇𝑇
=
tanh(4.472)
4.472
= 0.2236
𝑊𝑊
1
𝐶𝐶𝑎𝑎𝑎𝑎
𝑘𝑘𝑘𝑘
=
𝑙𝑙𝑙𝑙
= 4500 3
𝑉𝑉
𝑘𝑘 ∗∈ 𝐶𝐶𝑎𝑎
𝑚𝑚 . 𝑠𝑠
For Fluidized Bed:
L= 1.67*10-4 m
𝐾𝐾
MT = L� = 1.67*10-4�
𝐷𝐷
3.2
10−6
= 0.2987 → ↋ =
1
𝑘𝑘𝑘𝑘
𝐶𝐶𝑎𝑎𝑎𝑎 𝑋𝑋𝑎𝑎
𝑊𝑊
=
= 2500 3
𝑘𝑘 ∗∈ 𝐶𝐶𝑎𝑎 (1 − 𝑋𝑋𝑎𝑎)
𝑚𝑚 . 𝑠𝑠
𝑉𝑉
𝑡𝑡𝑡𝑡𝑡𝑡ℎ𝑀𝑀𝑇𝑇
𝑀𝑀𝑇𝑇
=
tanh(0.298)
0.298
= 0.98
Fluidized bed is better because it’s required less amount of catalyst. It needs
only 55.55% of the packed bed catalyst.
Problem 18.34 & 18.35
In aqueous solution, and in contact with the right catalyst, reactant A is
converted to product R by the elementary reaction AR. Find the mass of
catalyst needed in a packed bed reactor for 90% conversion of 104 mol A/hr of
feed having CAo= 103 mol/m3. For this reaction
34. k"' = 8 x 10-4 m3/m3 bed. s
35. k"' = 2 m3/m3 bed. s
Additional data:
Diameter of porous catalyst pellets= 6 mm
Effective diffusion coefficient of A in the pellet = 4 X 104 m3/m cat. s
Void age of packed bed = 0.5
Bulk density of packed bed = 2000 kg/m3 of bed
Solution 35:
k’’’= 2 (
=4 (
𝐦𝐦𝟑𝟑
𝐦𝐦𝟑𝟑 bed sec
m3
k′′′
ε=
D
6 ∗ 10^ −3
= 10
)
m3 cat sec
MT = L √
=
6
√
1
MT
𝐶𝐶𝐴𝐴0∗𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉
=∫
𝐹𝐹𝐴𝐴0
𝑑𝑑𝑑𝑑𝐴𝐴
𝑘𝑘 ′′′ ∗ 𝐶𝐶𝐴𝐴 ∗
4
4 ∗10^−8
= .1
τ’’’=
)* 2(
ε
𝐦𝐦𝟑𝟑 bed
)
1 m3 catalyst
1
=
𝐶𝐶𝐴𝐴0
ln
𝑘𝑘 ′′′ ∗ ε
𝐶𝐶𝐴𝐴
𝐹𝐹𝐹𝐹0
vcat =
𝐶𝐶𝐴𝐴0
ln
𝐶𝐶𝐶𝐶0∗ 𝑘𝑘 ′′′ ∗ 𝜀𝜀
𝐶𝐶𝐴𝐴
10^4
=
ln
3600∗1000∗4∗.1
10
= 0.016
1
W = Vcat * Pcat
= Vcat *(
= 0.016 *
Pbulk
)
voidage
2000
= 64 kg
0.5
Solution 34:
m3
k’’’= 8 * 10^-4 (
m3 bed sec
= 16 * 10^-4 (
k′′′
MT = L √
D
6 ∗ 10^ −3
=
= 0.2
√
6
m3
4 ∗10^−8
ε =1
τ’’’=
𝐶𝐶𝐴𝐴0∗𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉
=∫
=
𝐹𝐹𝐴𝐴0
𝑑𝑑𝑑𝑑𝐴𝐴
𝑘𝑘 ′′′ ∗ 𝐶𝐶𝐴𝐴 ∗
1
𝑘𝑘 ′′′ ∗
Vcat =
ε
ln
𝐹𝐹𝐹𝐹0
ε
𝐶𝐶𝐴𝐴0
𝐶𝐶𝐴𝐴
𝐶𝐶𝐶𝐶0∗ 𝑘𝑘 ′′′ ∗ 𝜀𝜀
)
m3 cat sec
16∗10^−4
ln
𝐶𝐶𝐴𝐴0
𝐶𝐶𝐴𝐴
) * 2(
m3 bed
)
1 m3 catalyst
=
10^4
3600∗1000 ∗ 16∗10^−4∗1
=4
W
ln
10
1
= Vcat * Pcat
= Vcat *(
=4*
2000
Pbulk
)
voidage
0.5
= 16000 kg
Problem 18.36
A first-order catalytic reaction A(l) → R(l) is run in a long, narrow vertical
reactor with up flow of liquid through a fluidized bed of catalyst particles.
Conversion is 95% at the start of operations when the catalyst particles are 5
mm in diameter. The catalyst is friable and slowly wears away, particles
shrink and the fine powder produced washes out of the reactor. After a few
months each of the 5-mm spheres has shrunk to 3-mm spheres. What should
be the conversion at this time? Assume plug flow of liquid.
(a) Particles are porous and allow easy access for reactants (no resistance to
pore diffusion).
(b) Particles are porous and at all sizes provide a strong resistance to pore
diffusion.
Solution 18.36:
a)
–𝑟𝑟𝑟𝑟0
𝑟𝑟𝐴𝐴′
=
𝑘𝑘𝑘𝑘𝑘𝑘
𝑘𝑘𝑘𝑘𝑘𝑘′
=1 =
𝐶𝐶𝐶𝐶0(1−𝑋𝑋𝑋𝑋)
𝐶𝐶𝐶𝐶0(1−𝑋𝑋𝐴𝐴′ )
1 – XA = 1 – XA’
1 – XA’ = 1 – 0.05
XA’ = 0.95
b) rA= k CA
𝑟𝑟𝑟𝑟1
𝑟𝑟𝑟𝑟2
=
𝑘𝑘 𝐶𝐶𝐶𝐶1
𝑘𝑘𝑘𝑘𝑘𝑘2
𝐶𝐶𝐶𝐶0(1−𝑋𝑋𝑋𝑋)
𝐶𝐶𝐶𝐶0(1−𝑋𝑋𝐴𝐴′ )
1−.95
1−𝑋𝑋𝑋𝑋′
=
𝑅𝑅2
=
3
= 0.6
𝑅𝑅1
5
1 – XA’ = 0.0833
XA’ = 0.916
Soln: Denoting catalyst palate without grains as 1 and with grains as 2. Therefore, in
strong pore diffusion regime (as grains are having strong pore diffusion);
𝑀𝑀𝑇𝑇2 𝜀𝜀1 𝑅𝑅2
𝑟𝑟 ′𝐴𝐴1 𝑊𝑊2 𝑉𝑉2 ∗ .75
= =
= 𝟎𝟎. 𝟎𝟎𝟎𝟎 = ′ =
=
𝑀𝑀𝑇𝑇1 𝜀𝜀2 𝑅𝑅1
𝑟𝑟 𝐴𝐴2 𝑊𝑊1
𝑉𝑉1
• Reason for last 3 terms;
But,
𝑟𝑟 ′𝐴𝐴 =
𝑋𝑋𝐴𝐴 ∗ 𝐹𝐹𝐴𝐴
𝑊𝑊
𝑁𝑁𝐴𝐴 ∗ 𝑣𝑣𝐴𝐴
𝑉𝑉
• As between grains there is 25% of void age and there exist free pore diffusion
hence in only 75% of total volume there will be strong pore resistance.
𝐹𝐹𝐴𝐴 = 𝐶𝐶𝐴𝐴 ∗ 𝑣𝑣𝐴𝐴 =
𝑊𝑊2
= 0.01
𝑊𝑊1
• Hence 1% of earlier catalyst weight is required to obtain same
results.
𝑉𝑉2 ∗ .75
= 0.01
𝑉𝑉1
𝑉𝑉2
= 0.0133
𝑉𝑉1
• Therefore 1.33% of earlier volume is required.
Soln: As catalyst is still in strong pore diffusion regime even after
impregnating only outer surface hence formulas used in previous
numerical are applicable. Denoting throughout impregnated
catalyst as 1 and catalyst impregnated only at outer surface as 2.
𝑀𝑀𝑇𝑇2 6.3
=
= 1.05
6
𝑀𝑀𝑇𝑇1
1
𝑊𝑊2
𝜀𝜀2
=
= 0.9524 =
𝜀𝜀1 1.05
𝑊𝑊1
𝑊𝑊2 = 0.9524 ∗ 𝑊𝑊1
• Assuming W kg of Pt was used earlier then amount of Pt saved by
impregnating only outer surface will be;
= 𝑊𝑊1 − 𝑊𝑊2 = 0.0476 ∗ 𝑊𝑊
• Means 4.76% kg of earlier amount of Pt used can be saved.
Soln:
𝐶𝐶𝐴𝐴 =
𝑝𝑝𝐴𝐴
𝑅𝑅 ∗ 𝑇𝑇
𝑋𝑋𝐴𝐴 = 0.5 = 1 −
𝑝𝑝𝐴𝐴0
𝑝𝑝𝐴𝐴
• Therefore, pA at 50% conversion is 361.9≅362 mm-Hg.
PA v/s Distance
800
700
y = -0.0006x3 + 0.1517x2 - 15.055x + 759.7
PA (in mm-Hg)
600
500
400
300
200
100
0
0
10
20
30
40
50
60
70
80
90
Distance (in m)
• From given data we can obtain above curve and 50% conversion
(362 mm-Hg) is achieved at distance 39.97 m.
• Now each 1m length of tube consists 1cm3 of bulk catalyst. Hence
till distance 39.97 cm3 of bulk catalyst will be there.
Now,
• From ideal gas law,
𝑉𝑉𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝑉𝑉𝑟𝑟
=
𝐶𝐶𝐴𝐴0 ∗ 𝑣𝑣0
𝐹𝐹𝐴𝐴0
PV=nRT
𝐶𝐶𝐴𝐴0 =
𝑃𝑃𝐴𝐴0
𝑛𝑛
1
𝑚𝑚𝑚𝑚𝑚𝑚
=
=
𝑉𝑉 𝑅𝑅 ∗ 𝑇𝑇 22400 𝑐𝑐𝑐𝑐3
Therefore,
39.97 𝑐𝑐𝑐𝑐3
𝑉𝑉𝑟𝑟
=
100 ∗ 103 𝑚𝑚𝑚𝑚𝑚𝑚
1
𝑚𝑚𝑚𝑚𝑚𝑚
𝑐𝑐𝑐𝑐3
∗ 10
22400 𝑐𝑐𝑐𝑐3
3600
𝑠𝑠
𝑠𝑠
Vr = 2.48*106 cm3
Vr = 2.48 m3
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Group no: 7: 13bch036, 13bch037, 13bch038, 13bch040
Cre II- term paper II
Chapter 24: Fluid-Fluid reactors design
24.1
24.2
A = agitated tank
MH =
√𝐷𝑎𝑏∗𝑘∗𝐶𝑏
𝐾𝑎𝑙
𝐸𝑖 = 1 +
HA=0, K=0
=
√3.6∗10−6 ∗0∗𝐶𝑏
0.114
=0
𝐶𝑏 ∗ 𝐻𝑎
=1
𝑃𝐴1
𝐸𝑖 ≥ 1 , 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝐸𝑖 = 𝐸 = 1
𝐺𝑎𝑠 𝑓𝑖𝑙𝑚 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 =
𝑘𝑖𝑛𝑒𝑡𝑖𝑐 𝑟𝑒𝑖𝑠𝑖𝑡𝑎𝑛𝑐𝑒 =
1
1
=
= 2.778
𝐾𝑎𝑔 ∗ 𝑎 0.36
𝐻𝑎
=0
𝑘 ∗ 𝐶𝑏 ∗ 𝑓𝑙
𝐿𝑖𝑞𝑢𝑖𝑑 𝑓𝑖𝑙𝑚 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 =
−𝑟𝐴 =
𝐻𝑎
=0
𝑘𝑎𝑙 ∗ 𝑎 ∗ 𝐸
𝑃𝐴
1000
=
= 360
1
2.778
𝑘𝑎𝑔 ∗ 𝑎
Fg (Yin - Yout) = -ra*Vr = Fg*(PAin - PAout)/Π
(i) Volume:
90000 (1000 − 100)
= 22.5 𝑚3
360
105
𝑉𝑟 =
(ii) 100% gas film resistance
24.3
Straight mass transfer, counter current, HA= 18, PA2 = 100
By material balance between points 1 & 2, we can find CA1
90000
900000
(1000
−
100)
=
(𝐶𝐴1 − 0)
105
55556
Or CA1 = 50 mol/m3
At equilibrium, PA*= HA*CA or PA1* = 18*50= 900 Pa
Now from the PA vs CA diagram we can see that the operating and equilibrium lines are
parallel. Thus the rate of transfer is the same everywhere. Hence,
−𝑟𝐴′′′′ =
1
1
𝐻𝐴
+
𝑘𝑎𝑔 ∗ 𝑎 𝑘𝑎𝑙 ∗ 𝑎
(𝑃𝐴 − 𝑃𝐴 ∗) =
1
1
18
0.36 + 72
−𝑟𝐴′′′′ = 33 𝑚𝑜𝑙/(𝑚3 ∗ ℎ𝑟)
(i) Volume
𝑉𝑟 =
𝐹𝑔 𝑑𝑝𝑎
90000 (1000 − 100)
ʃ
=
= 24.5 𝑚3
𝛱 −𝑟𝑎′′′′
105
33
(ii) 100% gas film resistance
24.4
Tower
(KAg*a*PA) top = 0.36*100 = 36
(KAg*a*PA) bottom = 0.36*1000 = 360
(1000 − 900)
(KAl*a*CB) top = 5.556*72 = 400
(KAl*a*CB) bottom = 72*555556 = 4000
CBin = 55.55 mol/m3
CBout = 10% of Cbin = 5.55 mol/m3
(KAl*a*CB) bottom > (KAg*a*PA) bottom
(KAl*a*CB) top > (KAg*a*PA) top
Therefore gas film resistance is controlling one.
−𝑟𝐴′′′′ =
1
𝑃𝐴
1
𝑘𝑎𝑔 ∗ 𝑎
(i) Volume
103
𝑉𝑟 =
𝐹𝑔 𝑑𝑋𝑎
𝐹𝑔
𝑑𝑝𝑎
90000 10
ʃ
=
∫
=
𝑙𝑛
= 5.75 𝑚3
𝛱 −𝑟𝑎
𝑘𝑎𝑔 ∗ 𝑎 ∗ 𝛱
𝑝𝑎
0.36
1
102
(ii) 100% gas film resistance
24.5
By doing material balance,
Fg (YA1- YA2) = Fl (XA1- XA2)
0.01
90000 (0.99 −
0.001
0.999
) = 900000(0.001 − 𝑋B1)
CB1 = 5 mol/m3
We need to calculate Ei and MH to find out the rate
𝐸𝑖 = 1 +
𝐷𝑏 ∗ 𝐶𝑏
𝐶𝑏
𝐶𝑏 ∗ 𝐻𝑎
105 ∗ 𝐶𝑏
=1+
= 1+
=1+
𝑏 ∗ 𝐷𝑎 ∗ 𝐶𝑎
𝐶𝑎
𝑃𝑎𝑖
𝑃𝑎𝑖
MH =
√𝐷𝑎𝑏∗𝑘∗𝐶𝑏
𝐾𝑎𝑙 2
=∞
Since 𝐸𝑖 ≪ 𝑀H , we have E=Ei
At top point 2,
Assuming that all the resistance lies in the gas film and liquid film resistance is negligible so
taking PA2*= PA2
55.56 ∗ 105
𝐸𝑖 = 1 +
= 5.556 ∗ 104
100
Hence E = 5.556*104. Now rate equation,
−𝑟𝐴′′′′ =
=
1
𝑃𝐴
1
𝐻𝐴
𝐻𝐴
+
+
𝑘𝑎𝑔 ∗ 𝑎 𝑘𝑎𝑙 ∗ 𝑎 ∗ 𝑒 𝑘 ∗ 𝐶𝑏 ∗ 𝑓𝑙
1
𝑃𝐴 = 0.36 ∗ 𝑃𝐴
1
105
105
0.36 + 72(5.556 ∗ 104 ) + ∞
We can see that gas film resistance is not negligible.
At bottom (point 1),
𝐸𝑖 = 1 +
MH =
5 ∗ 105
= 501
1000
√𝐷𝑎𝑏∗𝑘∗𝐶𝑏
𝐾𝑎𝑙 2
=∞
Since 𝐸𝑖 ≪ 𝑀h, we have E=Ei so rate equation is,
−𝑟𝐴′′′′ =
1
1
𝑃𝐴 = 0.18 ∗ 𝑃𝐴 =
𝑃𝐴 = 0.18 ∗ 𝑃𝐴
5
5
1
1
1
10
10
0.36 + 0.36 + 0
0.36 + 72(501) + ∞
This shows that gas and liquid film resistance is 50-50%
Trying 2nd time,
Guessing PAi = 500, and repeating the procedure, we get Ei = 1001 and from the rate
expression 1/3rd of the resistance is in the gas film. Thus our guess was wrong.
Trying 3rd time, guessing PAi = 500 or 10% resistance in the gas film. Then Ei = 5001
−𝑟𝐴′′′′ =
1
1
1
0.36 + 3.6 + 0
𝑃𝐴 = 0.32 ∗ 𝑃𝐴
So −𝑟𝐴′′′′ mean = (0.36+0.33)/2 = 0.345 PA
Hence,
(i) Volume
103
𝐹𝑔 𝑑𝑝𝑎
90000
𝑑𝑝𝑎
90000 10
𝑉𝑟 =
ʃ
=
∫
=
𝑙𝑛
= 6 𝑚3
𝛱 −𝑟𝑎′′′′
105
0.345 ∗ 𝑝𝑎
34500
1
102
(ii) Gas film resistance is 91% of total, and 9% is from liquid film.
24.6
Agitated Tank, HA = 105 , k = 2.6*107
Fg (YOUT – YIN) = Fl (XIN – XOUT)
1000 100
5556
𝐶𝑏𝑜𝑢𝑡
90000 ( 5 −
) = 900 (
−
)
5
10
10
55556 55556
CBout = 555.96 mol/m3
MH =
√𝐷𝑎𝑏∗𝑘∗𝐶𝑏
𝐾𝑎𝑙 2
=
√3.68∗10−6 ∗2.6∗107 ∗555.96
0.722
= 300.46
PAi = 100 KPa
𝐸𝑖 = 1 +
𝐶𝑏 ∗ 𝐻𝑎
500 ∗ 105
𝐸𝑖 = 1 +
= 5 ∗ 105
𝑃𝑎𝑖
100
𝐸𝑖 ≥ 𝑀H, MH = E = 300.46
−𝑟𝐴′′′′ =
−𝑟𝐴′′′′ =
1
𝑃𝐴
1
𝐻𝐴
𝐻𝐴
+
+
𝑘𝑎𝑔 ∗ 𝑎 𝑘𝑎𝑙 ∗ 𝑎 𝑘 ∗ 𝐶𝑏 ∗ 𝑓𝑙
1
105
1
105
+
+
0.72 144 ∗ 300 2.6 ∗ 105 ∗ 555.56 ∗ 0.9
𝑃𝐴 = 0.710 ∗ 𝑃𝐴
(i) Volume
𝑉𝑟 =
𝐹𝑔 𝑑𝑝𝑎
90000 1000 − 100
ʃ
=
(
) = 11.26 𝑚3
𝛱 −𝑟𝑎′′′′
105 100 ∗ 0.719
1
0.72
(𝑖𝑖)𝑇ℎ𝑢𝑠 𝑔𝑎𝑠 𝑓𝑖𝑙𝑚 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 =
= 99%
1
(0.719)
24.7
Fg (YOUT – YIN) = Fl (XIN – XOUT)
0.01 0.001
900000
90000 (
−
)=
(55.56 − 𝐶𝑏𝑜𝑢𝑡)
0.99 0.999
55556
CB = 5 mol/m3
At exit conditions, MH =
𝐸𝑖 = 1 +
√3.68∗10−6 ∗2.6∗105 ∗500
0.722
= 30
105 ∗ 500
> 𝑀𝐻
1000
Therefore, Ei = MH =30
−𝑟𝐴′′′′ =
𝑃𝐴
105
1
105
+
+
5
0.72 144 ∗ 40 2.6 ∗ 10 ∗ 500 ∗ 0.9
−𝑟𝐴′′′′ = 0.041*PA
Therefore,
𝑉𝑟 =
𝐹𝑔
∆𝑃
90000(1000 − 100)
∗
=
= 197.6 𝑚3
𝛱 −𝑟𝐴′′′′
105 (0.041) ∗ 100
(i) Volume = 197.6 m3
(ii) Liquid film dominates (94.3%)
24.8
HA= 1000, k=2.6*103, Tower
𝐹𝑔
𝐹𝑙
(𝑃𝐴1 − 𝑃𝐴2) =
(𝐶𝑏2 − 𝐶𝑏1)
𝛱
𝑏 ∗ 𝐶𝑡
90000
900000
(1000
−
900)
=
(𝐶𝑏2 − 𝐶𝑏1)
105
55556
CBout = 5.556 mol/m3
At top, MH =
√3.68∗10−6 ∗2.6∗103 ∗55.56
0.72
=1
55.56 ∗ 105
𝐸𝑖 = 1 +
= 55561 > 𝑀𝐻
100
−𝑟𝐴′′′′ =
−𝑟𝐴′′′′ =
1
𝑃𝐴
1
𝐻𝐴
𝐻𝐴
+
+
𝑘𝑎𝑔 ∗ 𝑎 𝑘𝑎𝑙 ∗ 𝑎 𝑘 ∗ 𝐶𝑏 ∗ 𝑓𝑙
100
105
1
105
+
+
0.36 0.72 ∗ 100 2.6 ∗ 1000 ∗ 55.56 ∗ 0.9
−𝑟𝐴′′′′ = 0.0718
1/−𝑟𝐴′′′′ = 13.92
At bottom, MH =
√3.68∗10−6 ∗2.6∗103 ∗55.56
0.722
= 0.3167
But MH can’t be less than 1. Therefore it is equal to 1.
105 ∗ 5.556
𝐸𝑖 = 1 +
= 55561 > 𝑀𝐻
1000
−𝑟𝐴′′′′ =
100
105
1
105
+
+
0.36 0.72 2.6 ∗ 1000 ∗ 5.556 ∗ 0.9
−𝑟𝐴′′′′ = 0.714
1/−𝑟𝐴′′′′ = 1.4
At the middle, at PA = 3000 Pa, CA = 44.44 mol/m3
MH =
√3.68∗10−6 ∗2.6∗103 ∗44.44
𝐸𝑖 = 1 +
0.722
= 0.895
105 ∗ 44.44
> 𝑀𝐻
1000
So E = MH = 1
−𝑟𝐴′′′′ =
100
1
105
105
+
+
0.36 0.72 2.6 ∗ 1000 ∗ 44.44 ∗ 0.9
−𝑟𝐴′′′′ = 0.359
1/−𝑟𝐴′′′′ = 2.58
103
𝐹𝑔
𝑑𝑝𝑎
𝑉𝑟 =
∫
𝛱
−𝑟𝐴′′
102
𝑉𝑟 = 2685.6 m3
24.9
Fg (Yin-Yout) = Fl ( XBOUT – XBIN)
0.01 0.001
900000
90000 (
−
)=
(55.56 − 𝐶𝑏𝑜𝑢𝑡)
0.99 0.999
55556
CB = 5 mol/m3
At top, MH =
𝐸𝑖 = 1 +
√3.68∗10−6 ∗2.6∗107 ∗55.56
0.722
= 100
105 ∗ 55.56
> 𝑀𝐻
1000
Therefore, Ei = MH =100
−𝑟𝐴′′′′ =
100
105
1
105
+
+
0.36 0.72 ∗ 100 2.6 ∗ 107 ∗ 55.56 ∗ 0.9
−𝑟𝐴′′′′ = 6
At bottom, MH =
𝐸𝑖 = 1 +
√3.68∗10−6 ∗2.6∗107 ∗5
0.722
= 30
105 ∗ 5
> 𝑀𝐻
1000
Therefore, Ei = MH =30
−𝑟𝐴′′′′ =
100
105
1
105
+
+
0.36 0.72 ∗ 30 2.6 ∗ 1000 ∗ 5 ∗ 0.9
−𝑟𝐴′′′′ = 20.4
In the middle, at PA = 547 Pa, CA = 30.56 mol/m3
MH =
√3.68∗10−6 ∗2.6∗107 ∗30.56
0.722
= 74
105 ∗ 30.56
𝐸𝑖 = 1 +
> 𝑀𝐻
1000
So E = MH = 74
100
−𝑟𝐴′′′′ =
105
1
105
+
+
7
0.36 0.72 ∗ 74 2.6 ∗ 10 ∗ 30.56 ∗ 0.9
−𝑟𝐴′′′′ = 25.4
1/−𝑟𝐴′′′′ = 0.0393
103
𝐹𝑔
𝑑𝑝𝑎
𝑉𝑟 =
∫
𝛱
−𝑟𝐴′′
102
(i) Volume
𝑉𝑟 = (9000/105)*66= 59.4 m3
(ii) liquid film dominates (84%)
24.10
HA= 1000, k=2.6*105 , Tower
𝐹𝑔
𝐹𝑙
(𝑃𝐴1 − 𝑃𝐴2) =
(𝐶𝑏2 − 𝐶𝑏1)
𝛱
𝑏 ∗ 𝐶𝑡
90000
900000
(1000 − 900) =
(𝐶𝑏2 − 𝐶𝑏1)
5
10
55556
CBout = 5.556 mol/m3
At top, MH =
𝐸𝑖 = 1 +
√3.63∗10−6 ∗2.6∗105 ∗55.56
0.72
= 10
55.56 ∗ 103
= 556.6 > 𝑀𝐻
100
Therefore, Ei = MH =10
−𝑟𝐴′′′′ =
100
103
1
103
+
+
5
0.36 0.72 ∗ 10 2.6 ∗ 10 ∗ 55.56 ∗ 0.9
−𝑟𝐴′′′′ = 23.95 mol/m3*hr
1/−𝑟𝐴′′′′ = 0.0417
At bottom, MH =
√3.68∗10−6 ∗2.6∗105 ∗5.56
0.722
= 3.16
But MH can’t be less than 1. Therefore it is equal to 1.
103 ∗ 5.556
𝐸𝑖 = 1 +
= 6.55 > 𝑀𝐻
1000
Therefore, Ei = MH =3.16
−𝑟𝐴′′′′ =
100
1
103
105
+
+
5
0.36 0.72 ∗ 3.16 2.6 ∗ 10 ∗ 5.556 ∗ 0.9
−𝑟𝐴′′′′ = 137.93
1/−𝑟𝐴′′′′ = 0.0073
At the middle, at PA = 300 Pa, CA = 44.44 mol/m3
MH =
√3.68∗10−6 ∗2.6∗107 ∗44.44
0.722
= 8.95
105 ∗ 44.44
𝐸𝑖 = 1 +
> 𝑀𝐻
1000
So E = MH = 8.95
−𝑟𝐴′′′′ =
300
1
105
105
+
+
5
0.36 0.72 ∗ 8.95 2.6 ∗ 10 ∗ 44.44 ∗ 0.9
−𝑟𝐴′′′′ = 69.136
1/−𝑟𝐴′′′′ = 0.0144
103
𝐹𝑔
𝑑𝑝𝑎
𝑉𝑟 =
∫
𝛱
−𝑟𝐴′′
102
𝑉𝑟 = 0.9*10.8 = 9.72 m3
24.11
𝐶𝐴 =
𝑃𝐴𝑖𝑛
101325
=
= 28.95 𝑚𝑜𝑙/𝑚3
𝐻𝐴
3500
CBin = 300 mol/m3, since CB is greater than CA we have excess of B, so CB can be taken as 300
mol/m3
MH =
√1.4∗10−9 ∗0.433∗300∗120
0.25
𝐸𝑖 = 1 +
= 2.05
300 ∗ 3500
= 11.36 > 𝑀𝐻
101325
So E = MH = 2.05
101325
−𝑟𝐴′′′′ =
3500
3500
+
0.025 ∗ 2.05 0.433 ∗ 300 ∗ 0.08
0+
−𝑟𝐴′′′′ = 1.476
1/−𝑟𝐴′′′′ = 0.677
𝑋𝐴
𝑉𝑟 = 𝐹𝐴0 ∫
0
𝑑𝑋𝐴
−𝑟𝐴′′
FA0 = V0*CA0 = 0.0363*18.95 = 1.05 m3/hr
𝑋𝐴 =
𝑉𝑟
0.6042
(−𝑟𝐴′′′′ ) =
∗ 1.48 = 85% 𝑐𝑜𝑛𝑣𝑒𝑟𝑠𝑖𝑜𝑛
𝐹𝐴0
1.05
24.12
Cl2 + 2NaOH  P
K= 2.6*109 km3/mol.hr, HA = 105 Pa m3/mol.hr
Acs = 55 m3
Fg = 100, Fl = 250 mol/s*m2
D = 1.5*109 m2/s
CB (IN) = 10 % * 250 = 25 mol/m3
𝐹𝑔
𝐹𝑙
(𝑃𝐴1 − 𝑃𝐴2) =
(𝐶𝑏2 − 𝐶𝑏1)
𝛱
𝑏 ∗ 𝐶𝑡
100
250
(2.36 ∗ 0.99) =
(25 − 𝐶𝐵2)
5
10
2 ∗ 2736
CB2 = 24.9 m3 / mol
MH =
√1,5∗10−9 ∗2.6∗109 ∗24.9
0.45
𝐸𝑖 = 1 +
= 0.218
24.5 ∗ 105
= ∞ > 𝑀𝐻
0
But MH can’t be less than 1. Therefore E is equal to 1.
When PA = 1000,
−𝑟𝐴′′′′ =
𝑃𝐴
105
1
105
+
+
9
133 133 ∗ 1 2.6 ∗ 10 ∗ 24.9 ∗ 0.9
= 1.329 ∗ 10−3 𝑃𝐴
103
𝐹𝑔
𝑑𝑃𝑎
100
𝑙𝑛10
𝐻=
∫
=
∗
= 3.14 𝑚
𝐴𝑐𝑠 ∗ 𝜋
0.1329 ∗ 𝑃𝑎
55 ∗ 1 0.1329
102
24.13
For agitated tank,
K= 2.6*105, HA = 105
At the beginning, CB = 555.6 mol/m3
MH =
√3.6∗10−6 ∗2.6∗105 ∗555.6
𝐸𝑖 = 1 +
1.44
= 15.84
555.6 ∗ 105
= 55561 > 𝑀𝐻
1000
So E = MH = 15.84
−𝑟𝐴′′′′ =
−𝑟𝐴′′′′ =
𝑃𝐴𝑖𝑛
𝛱 ∗ 𝑉𝑟
1
𝐻𝑎
𝐻𝑎
𝐹𝑔 + 𝐾𝑎𝑔 ∗ 𝑎 + 𝐾𝑎𝑙 ∗ 𝑎 ∗ 𝐸 + 𝐾 ∗ 𝐶𝑏 ∗ 𝑓𝑙
1000
∗ 1.62
1
105
105
+
+
+
9000 ∗ 0.9 0.72 144 ∗ 15.84 2.6 ∗ 105 ∗ 555.6 ∗ 0.9
105
−𝑟𝐴′′′′ = 15.33 mol/m3*hr
1/−𝑟𝐴′′′′ = 0.0652
At the end of the run, MH =
√3.68∗10−6 ∗2.6∗105 ∗55.6
1.44
= 5.01
105 ∗ 55.56
𝐸𝑖 = 1 +
= 5561 > 𝑀𝐻
1000
Therefore, Ei = MH =5.01
1000
−𝑟𝐴′′′′ =
20 + 1.3889 +
105
105
+0
∗ 144 ∗ 5.01
−𝑟𝐴′′′′ = 6.25 mol/m3.hr
1/−𝑟𝐴′′′′ = 0.16
At some intermediate run, CA = 138.9 mol/m3
MH = 15.84/2 = 7.92
105 ∗ 138.9
𝐸𝑖 = 1 +
= 13890 > 𝑀𝐻
1000
So E = MH = 7.92
−𝑟𝐴′′′′ =
1000
105
20 + 1.3889 + 144 ∗ 7.92 + 0
−𝑟𝐴′′′′ = 9.168 mol/m3.hr
1/−𝑟𝐴′′′′ = 0.109
The length of the time that we have to bubble gas through the liquid is
103
𝑡=
𝐹𝑙
𝑑𝐶𝑏
∫
= 𝐹𝑙 ∗ 𝑎𝑟𝑒𝑎 = 0.9 ∗ 51.38 = 46.2 ℎ𝑟
𝑏
−𝑟𝐴′′
102
The minimum time needed if all the A reacts with B,
𝑡 min =
𝑉𝑙(𝐶𝑏0 − 𝐶𝑏𝑓)
1.62(555.6 − 55.56)
=
= 8.91 ℎ𝑟
𝐹𝑔(𝑃𝐴/(𝛱 − 𝑃𝑎))
9000(1000/99000)
Therefore, the efficiency of the use of A = 8.91/46.24 = 19.26%
24.14
For agitated tank,
K= 2.6*109, HA = 105
At the beginning, CB = 555.6 mol/m3
MH =
√3.6∗10−6 ∗2.6∗109 ∗555.6
𝐸𝑖 = 1 +
1.44
= 1583.6
555.6 ∗ 105
= 55561 > 𝑀𝐻
1000
So E = MH = 1583.6
−𝑟𝐴′′′′ =
−𝑟𝐴′′′′ =
𝑃𝐴𝑖𝑛
𝛱 ∗ 𝑉𝑟
1
𝐻𝑎
𝐻𝑎
𝐹𝑔 + 𝐾𝑎𝑔 ∗ 𝑎 + 𝐾𝑎𝑙 ∗ 𝑎 ∗ 𝐸 + 𝐾 ∗ 𝐶𝑏 ∗ 𝑓𝑙
1000
105 ∗ 1.62
1
105
105
+
+
+
9
9000 ∗ 0.9 0.72 144 ∗ 1583.6 2.6 ∗ 10 ∗ 555.6 ∗ 0.9
−𝑟𝐴′′′′ = 45.81 mol/m3*hr
1/−𝑟𝐴′′′′ = 0.0218
At the end of the run, MH =
√3.68∗10−6 ∗2.6∗109 ∗55.6
1.44
105 ∗ 55.56
𝐸𝑖 = 1 +
= 5561 > 𝑀𝐻
1000
Therefore, Ei = MH =506.5
1000
−𝑟𝐴′′′′ =
20 + 1.3889 +
105
+0
144 ∗ 506.5
−𝑟𝐴′′′′ = 43.93 mol/m3.hr
1/−𝑟𝐴′′′′ = 0.0227
At some intermediate run, CA = 138.9 mol/m3
MH = 1583.6/2 = 791.8
= 506.5
105 ∗ 138.9
𝐸𝑖 = 1 +
= 13890 > 𝑀𝐻
1000
So E = MH = 395.9
−𝑟𝐴′′′′ =
1000
105
20 + 1.3889 + 144 ∗ 791.8 + 0
−𝑟𝐴′′′′ = 44.9 mol/m3.hr
1/−𝑟𝐴′′′′ = 0.022
The length of the time that we have to bubble gas through the liquid is
𝑡=
𝐹𝑙
𝑑𝐶𝑏
∫
= 𝐹𝑙 ∗ 𝑎𝑟𝑒𝑎 = 0.9 ∗ 10.987 = 9.88 ℎ𝑟
𝑏
−𝑟𝐴′′
The minimum time needed if all the A reacts with B,
𝑡 min =
𝑉𝑙(𝐶𝑏0 − 𝐶𝑏𝑓)
1.62(555.6 − 55.56)
=
= 8.91 ℎ𝑟
𝐹𝑔(𝑃𝐴/(𝛱 − 𝑃𝑎))
9000(1000/99000)
Therefore, the efficiency of the use of A = 8.91/9.88 = 90.1%
24.15
For agitated tank,
K= 2.6*103, HA = 105
At the beginning, CB = 555.6 mol/m3
MH =
√3.6∗10−6 ∗2.6∗103 ∗555.6
𝐸𝑖 = 1 +
1.44
= 1.584
555.6 ∗ 105
= 55561 > 𝑀𝐻
1000
So E = MH = 1.58
−𝑟𝐴′′′′ =
−𝑟𝐴′′′′ =
𝑃𝐴𝑖𝑛
𝛱 ∗ 𝑉𝑟
1
𝐻𝑎
𝐻𝑎
𝐹𝑔 + 𝐾𝑎𝑔 ∗ 𝑎 + 𝐾𝑎𝑙 ∗ 𝑎 ∗ 𝐸 + 𝐾 ∗ 𝐶𝑏 ∗ 𝑓𝑙
1000
∗ 1.62
1
105
105
+
+
+
9000 ∗ 0.9 0.72 144 ∗ 1.584 2.6 ∗ 105 ∗ 555.6 ∗ 0.9
105
−𝑟𝐴′′′′ = 2.175 mol/m3*hr
1/−𝑟𝐴′′′′ = 0.459
At the end of the run, MH =
√3.68∗10−6 ∗2.6∗103 ∗55.6
1.44
= 0.5
105 ∗ 55.56
𝐸𝑖 = 1 +
= 5561 > 𝑀𝐻
1000
Therefore, Ei = MH =0.5
1000
−𝑟𝐴′′′′ =
20 + 1.3889 +
105
105
+0
∗ 144 ∗ 0.5
−𝑟𝐴′′′′ = 0.7096 mol/m3.hr
1/−𝑟𝐴′′′′ = 1.41
At some intermediate run, CA = 555.6/4= 138.9 mol/m3
MH = 15.84/4 = 0.792
𝐸𝑖 = 1 +
105 ∗ 138.9
= 13890 > 𝑀𝐻
1000
So E = MH = 1
−𝑟𝐴′′′′ =
1000
105
20 + 1.3889 + 144 ∗ 1 + 0
−𝑟𝐴′′′′ = 1.66 mol/m3.hr
1/−𝑟𝐴′′′′ = 0.6
The length of the time that we have to bubble gas through the liquid is
103
𝐹𝑙
𝑑𝐶𝑏
𝑡=
∫
= 𝐹𝑙 ∗ 𝑎𝑟𝑒𝑎 = 0.9 ∗ 350 = 315 ℎ𝑟
𝑏
−𝑟𝐴′′
102
The minimum time needed if all the A reacts with B,
𝑡 min =
𝑉𝑙(𝐶𝑏0 − 𝐶𝑏𝑓)
1.62(555.6 − 55.56)
=
= 8.91 ℎ𝑟
𝐹𝑔(𝑃𝐴/(𝛱 − 𝑃𝑎))
9000(1000/99000)
Therefore, the efficiency of the use of A = 8.91/315 = 2.82%
24.16
K= 2.6*1011 m3/mol.hr, HA = 103 Pa/m3*mol
MH =
√1.5∗10−9 ∗2.6∗1011 ∗24.9
45
= 2.18
Ei = ∞, therefore, Ei =MH = 2.18
−𝑟𝐴′′′′ =
=
1
𝑃𝐴
1
𝐻𝐴
𝐻𝐴
+
+
𝑘𝑎𝑔 ∗ 𝑎 𝑘𝑎𝑙 ∗ 𝑎 ∗ 𝐸 𝑘 ∗ 𝐶𝑏 ∗ 𝑓𝑙
1
103
1
103
+
+
133 133 2.6 ∗ 1011 ∗ 24.09 ∗ 0.9
103
𝐻=
𝑃𝐴 = 0.1328 ∗ 𝑃𝐴
𝐹𝑔
𝑑𝑃𝑎
100
𝑙𝑛10
∫
=
∗
= 31.524 𝑚
𝐴𝑐𝑠 ∗ 𝜋
0.1328 ∗ 𝑃𝑎
55 ∗ 1 0.1328
102
Chapter 25 : Fluid-Particle Reactions:Kinetics
Roll Number : Rohan Patel (13BCH041)
Shivang Patel (13BCH042)
Vedant Patel (13BCH043)
Vishal Patel (13BCH045)
Q 25.1 A batch of solids of uniform size is treated by gas in a uniform environment. Solid is
converted to give a nonflaking product according to the shrinking-core model. Conversion is
about 7/8 for a reaction time of 1 h, conversion is complete in two hours. What mechanism is
rate controlling?
A 25.1
Assume spherical particles
τ=2 hr t= 1 hr, Xb=0.875
Film Diffusion
Xb
τ
Film Diffusion not dominating.
Ash Diffusion
τ
Ash Diffusion is dominating.
Reaction
0.5=τ
Reaction control is also dominating
Hence, Ash Diffusion and Reaction controls are dominating.
Q 25.2 In a shady spot at the end of Brown Street in Lewisburg, Pennsylvania, stands a Civil
War memorial-a brass general, a brass cannon which persistent undergraduate legend insists may
still fire some day, and a stack of iron cannonballs. At the time this memorial was set up, 1868,
the cannonballs were 30 inches in circumference. Today due to weathering, rusting, and the
once-a-decade steel wire scrubbing by the DCW, the cannonballs are only 29.75 in. in
circumference. Approximately, when will they disappear completely?
A 25.2
R=4.78 inch
rc=4.74 inch
t=148 years
As it is a reaction controlling mechanism.
Now, τ
Hence the time in another 17,538 years.
Q 25.3 Calculate the time needed to burn to completion particles of graphite (R, = 5 mm, p, = 2.2
gm/cm3, k" = 20 cm/sec) in an 8% oxygen stream. For the high gas velocity used assume that
film diffusion does not offer any resistance to transfer and reaction. Reaction temperature =
900°C.
A 25.3
As it is a shrinking core particle ash film diffusion will not participate in the reaction. Also it is
given that film diffusion is negligible. Hence only reaction control is applicable.
R=5 mm
ρb=2.2 gm/cm3=0.1833mol/cm3
k’’=20 cm/sec
b=1
ρ
Hence 1.53 hours are needed for complete combustion of graphite particles.
Q 25.4 Spherical particles of zinc blende of size R = 1 mm are roasted in an 8% oxygen stream
at 900°C and 1 atm. The stoichiometry of the reaction is Assuming that reaction proceeds by the
shrinking-core model calculate the time needed for complete conversion of a particle and the
relative resistance of ash layer diffusion during this operation.
Data
Density of solid, p, = 4.13 gm/cm3 = 0.0425 mol/cm3
Reaction rate constant, k" = 2 cm/sec
For gases in the ZnO layer, De = 0.08 cm2/sec
Note that film resistance can safely be neglected as long as a growing ash layer is present.
A 25.4
Fro the reaction, b= 2/3
τtotal = τash + τreaction
For ash layer control,
CAG = Pa / RT = 0.08*101325/(8.314*1173) = 0.83118*10-2 mol/L
τash =
= 1598.02 sec
For Reaction control,
τreaction =
= 3834.90sec
τtotal = 1598.02 + 3834.90 = 5432.94sec
= 90.54min
Relative Resistance of ash layer =
= 29.41 %
Q On doubling the particle size from R to 2R the time for complete conversion triples. What is
the contribution of ash diffusion to the overall resistance for particles of size,
Q 25.5 R?
A 25.5
Let 1 refer to particle size of R
2 refer to particle size of 2R
Then, τ2 = 3τ1……. (1)
τ1 = τ1ash + τ1reaction…….(2)
τ2 = τ2ash + τ2reaction……(3)
Put τ2ash = 4τ1ash…..(4)
τ2reaction = 2τ1reaction……(5)
So, from equation (1),(2),(3),(4) and (5),
3τ1 = 4τ1ash + 2τ1reaction….(6)
So from equation (2) and (6),
τ1ash = τ1reaction
therefore % contribution of ash diffusuion for R = 50%
Q 25.6 R?
A 25.6
Let 1 refer to particle size of R
2 refer to particle size of 2R
Then, τ2 = 3τ1……. (1)
τ1 = τ1ash + τ1reaction…….(2)
τ2 = τ2ash + τ2reaction……(3)
Put τ2ash = 4τ1ash…..(4)
τ2reaction = 2τ1reaction……(5)
So, from equation (1),(2),(3),(4) and (5),
3τ1 = 4τ1ash + 2τ1reaction….(6)
So from equation (2) and (6),
τ1ash = τ1reaction
put values from equation (4) and (5),
τ2ash /4 = τ2reaction/2
so, τ2reaction = τ2ash /2
therefore % contribution of ash diffusion for R = 66.66%
Q Spherical solid particles containing B are roasted isothermally in an oven with gas of constant
composition. Solids are converted to a firm nonflaking product according to the SCM as follows:
From the following conversion data (by chemical analysis) or core size data (by slicing and
measuring) determine the rate controlling mechanism for the transformation of solid.
Q 25.7
Dp (mm)
1
1.5
Xb
1
1
t (min)
4
6
A 25.7
For finding out the corresponding rate control mechanism, considering the constant size
spherical particles if it is:
Film diffusion controlling : The value of R/τ will remain constant for both cases.
Reaction controlling: The value of R/τ will remain constant for both cases.
Ash diffusion control : The value of R2/τ will remain constant for both cases.
Substituting the value of τ from both cases:
Dp
(mm)
Xb
t
(min)
Film Diffusion
control
Ash film diffusion control
Reaction Control
1
1.5
1
1
4
6
0.25
0.1667
0.0625
0.0938
0.125
0.125
From the above table it implies that the rate controlling mechanism for the transformation of
solid is Reaction Control.
Q 25.8
Dp (mm)
1
1
Xb
0.3
0.75
t (sec)
2
5
A 25.8
For finding out the corresponding rate control mechanism, considering the constant size
spherical particles if it is:
Film diffusion controlling : The value of R/τ will remain constant for both cases.
Reaction controlling: The value of R/τ will remain constant for both cases.
Ash diffusion control : The value of R2/τ will remain constant for both cases.
Substituting the value of τ from both cases:
Dp
(mm)
Xb
t (sec)
Film Diffusion
control
Ash film diffusion control
Reaction Control
1
0.3
2
0.15
0.00436
0.0280
1
0.75
5
.015
0.01547
0.037
From the above table it implies that the rate controlling mechanism for the transformation of
solid is Film diffusion control.
Q 25.9
Dp (mm)
1
1.5
Xb
1
1
t (sec)
200
450
A 25.9
For finding out the corresponding rate control mechanism, considering the constant size
spherical particles if it is:
Film diffusion controlling : The value of R/τ will remain constant for both cases.
Reaction controlling: The value of R/τ will remain constant for both cases.
Ash diffusion control : The value of R2/τ will remain constant for both cases.
Substituting the value of τ from both cases:
Dp
Xb
(mm)
1
1
1
1
t
Film Diffusion Ash film diffusion control
(sec) control
200
450
0.005
0.0022
Reaction Control
0.00125
0.00125
0.0025
0.00167
From the above table it implies that the rate controlling mechanism for the transformation of
solid is Ash film diffusion control.
Q 25.10
Dp (mm)
2
1
Xb
0.875
1
t (sec)
1
1
A 25.10
For finding out the corresponding rate control mechanism, considering the constant size
spherical particles if it is:
Film diffusion controlling : The value of R/τ will remain constant for both cases.
Reaction controlling: The value of R/τ will remain constant for both cases.
Ash diffusion control : The value of R2/τ will remain constant for both cases.
Substituting the value of τ from both cases:
Dp
(mm)
Xb
t (sec)
Film Diffusion
control
Ash film diffusion control
Reaction Control
2
0.875
1
0.875
0.5
0.5
1
1
1
0.5
0.25
0.5
From the above table it implies that the rate controlling mechanism for the transformation of
solid is Reaction control.
Q 25.11 Uniform-sized spherical particles UO, are reduced to UO, in a uniform
environment with the following results:
If reaction follows the SCM, find the controlling mechanism and a rate
equation to represent this reduction.
A 25.11
Method : 1
For
Gas Film Controlling : τ =
Reaction Controlling : τ =
Ash Film Controlling : =
τ
Now, Find the value of R.H.S of all equation...
t, hr
0.18
0.347
0.453
0.567
0.733
0.45
0.68
0.8
0.95
0.98
0.1807
0.316
0.4125
0.6316
0.7286
0.0861
0.2365
0.374
0.6928
0.819
Then, Plot R.H.S. vs time(hr) and whichever is nearer to diagonal line it will be controlling
mechanism.
1
R.H.S. of equations
0.8
0.6
Gas Film Controlling
0.4
Reaction Controlling
Ash Film Controlling
0.2
0
0
0.2
0.4
0.6
0.8
1
time (hr)
The assumption of reaction control gives a better straight line fit than does the assumption for
gas film and ash film.
Method 2 :
Find the value of τ for any 2 or 3 time for all three mechanism and in whichever mechanism all τ
are same then it will be controlling mechanism.
Time,hr
0.18
0.453
0.733
Xb
0.45
0.8
0.98
Gas Film Controlling
0.4
0.56625
0.7479
Reaction Controlling
0.996
1.09
1.004
Ash Film Controlling
2.09
1.211
0.894
From the above table it implies that the rate controlling mechanism for the transformation of
solid is Reaction control.
Q 25.12 A large stockpile of coal is burning. Every part of its surface is in flames. In a 24-hr
period the linear size of the pile, as measured by its silhouette against the horizon, seems to
decrease by about 5%.
(a) How should the burning mass decrease in size?
(b) When should the fire burn itself out?
(c) State the assumptions on which your estimation is based.
A 25.12
(a). As it is written that surface is in flames, the burning mass should decrease in size by
Shrinking Core Model (SCM)
(b). The coal fire will burn out when entire coal will be covered by the ash layer and it will no
longer be able to diffuse oxygen through the ash layer and the fire should burn itself out.
(c). The outer surface of the solid particle and the deeper layers do not take part in the reaction
until all the outer layer has transformed into solid or gaseous product. The reaction zone then
moves inward(into solid), constantly reducing the size of core of unreacted solid and leaving
behind completely converted solid(solid product) and inert material (inert constituent of the solid
reactant). And we refer to converted solid and inert material as ash. At any time the solid
comprises of a core surrounded by a envelope. The envelope consist of a solid product and inert
material. Ash layer may contribute to resistance.
CHAPTER 26
FLUID-PARTICLE REACTORS: DESIGN
SOLUTION OF UNSOLVED EXAMPLES
PREPARED BY
CHIRAG MANGAROLA (13BCH032)
MAULIK SHAH (13BCH033)
MEHUL MENTA (13BCH034)
DHRUV MEWADA (13BCH035)
26.1
Solution
Given: Shrinking core model with ash diffusion as controlling mechanism
Conversion = 80%
To find conversion when the size of reactor is doubled
Feed rate, flow pattern, gas environment are considered constant
For plug flow, SCM, Ash diffusion
𝑡𝑝
= 1 − 3 1 − 𝑋𝐵
𝜏
= 1 − 3 1 − 0.8
2
2
3
3
+ 2 1 − 𝑋𝐵
+ 2 1 − 0.8
= 0.374
On doubling the size of reactor 𝑡𝑝 ′ = 2𝑡𝑝
2𝑡𝑝
= 1 − 3 1 − 𝑋𝐵 ′
𝜏
2
3
+ 2 1 − 𝑋𝐵 ′
So, 𝑋𝐵 ′ = 96.5%
26.2
Solution
For MFR same gas environment and same feed rate so no change in conversion.
26.3
Solution
Reactor type: tubular
Hard solid product (SCM/Reaction control)
𝜏 = 4hr for 4mm particles
To find: residence time for 100% conversion of solids
For plug flow SCM/reaction control
𝜏=
𝜌𝐵 𝑅
𝑘 ′′ 𝑏 𝐶𝐴𝑔
This shows that 𝜏 α R
𝜏2𝑚𝑚 = 2𝑕𝑜𝑢𝑟𝑠
𝜏1𝑚𝑚 = 2𝑕𝑜𝑢𝑟𝑠
For 100% conversion residence time should be more than 4hours.
26.4
Solution
We need to find conversion of 1 mm, 2mm & 4mm particle sizes in 15 min of time.

For 4mm particle
15
= 1 − 1 − 𝑥𝑏
4 ∗ 60
1
3
1 − 𝑥𝑏 = 0.824
𝑥𝑏 = 0.176

For 2 mm particle
15
= 1 − 1 − 𝑥𝑏
2 ∗ 60
1
3
1 − 𝑥𝑏 = 0.669
𝑥𝑏 = 0.331

For 1mm particle
15
= 1 − 1 − 𝑥𝑏
1 ∗ 60
1
3
1 − 𝑥𝑏 = 0.423
𝑥𝑏 = 0.577

Over all conversion
1 − 𝑥𝑏 = 0.5 ∗ 0.176 + 0.3 ∗ 0.331 + 0.2 ∗ 0.577
𝑥𝑏 = 0.6973
Solution
Given: SCM with reaction control
Conversion 60%
To find: conversion when reactor is made twice as large keeping amount of solids and gas
environment constant
W1=W2 (Amount of solids)
Concentration CA01=CA02
𝑊
𝑡
=
𝐹𝐴0
𝐶𝐴0
Here same gas environment so FA01=FA02
Therefore, 𝑡1 = 𝑡2
Conversion is same
Size of reactor does not play role but the weight of solids passed play a role in changing
conversion in this case.
𝑡=
𝑤
= 100𝑠𝑒𝑐
𝐹0
1 − 𝑥𝑏 =
1 𝜏 1 𝜏
∗ −
4 𝑡 20 𝑡
2
+
1
𝜏
∗
120 𝑡
3
𝜏
= 1.246
𝑡
𝜏 = 124.65
1 − 𝑥𝑏 = 0.02
1 𝜏 1 𝜏
0.02 = ∗ −
4 𝑡 20 𝑡
𝑡=
2
1
𝜏
+
∗
120 𝑡
124.6
= 1538.27
0.081
𝑡› 𝜏 because of presence of non idealities.
𝑤 = 𝑡 ∗ 𝐹0
𝑤 = 15382.7 𝑔
3
Information from Example 26.3
 30% of 50μm radius particles (τ=5min)
Modification
 Ash film control
 40% of 100μm radius particles (τ=10min)
 τ(R=100μm) =10min
 30% of 200μm radius particles (τ=20min)
 Reactor type: Fluidized bed reactor (MFR)
 Unchanging particle (SCM) with reaction control
 Feed rate = 1kg solids/min
 Solids in bed = 10kg
 Mean residence time = 10min
Solution
For ash controlling mechanism τ α R2
So, τ(R=50μm) =
τ R=100μm ∗(50 2 )
100 2
And, τ(R=200μm) =
=
10∗(50 2 )
100 2
τ R=100μm ∗(200 2 )
100 2
=
= 2.5min
10∗(200 2 )
100 2
= 40min
For ash diffusion
1-𝑋𝐵 =
1 τ(R i )
5
= 0.3
𝑡
1
5
−
∗
τ(R i ) 2
19
420
2.5
10
−
𝑡
𝐹(𝑅𝑖 )
𝐹
19
420
∗
2.5 2
10
0.3
= 0.0989
Conversion = 0.901 = 90.1%
+ 0.4
1
5
∗
40
10
1
5
∗
−
10
10
−
19
420
∗
19
420
∗
40 2
10
10 2
10
+
Information from Example 26.4
 A(gas) + B(solid)  R(gas) + S(solid)
 τ = 1hr (with environment of CA0 gas concentration)
 Gas and solid both in mixed flow
 CA0 = CB0
 𝑋𝐵 = 0.9
 FB0 = 1 ton/hr
 Stoichiometric feed rate of A and B
Modification
 Twice gas to solid
Stoichiometric ratio
Solution
Material balance
2(CA0 – CA) FA0 = (CB0 – CB) FB0
2(CA0 – CA) = 0.9 CA0
Therefore, CA = 0.55 CA0
The solid will be in the environment of the gas of concentration 0.55CA0
τα
1
from equation τ =
𝐶𝐴 0
So τ =
1
0.55
𝜌𝐵 𝑅
𝑏 𝑘 𝐶𝐴 0
= 1.818h
For reaction control;
1 − 𝑋𝐵 =
1τ
4𝑡
−
On solving this equation
𝑡=
τ
0.435
=
1.818
0.435
τ
𝑡
1
τ 2
20
𝑡
= 0.1
= 0.435
= 4.18 𝑕𝑟
W = 𝑡 FB0 = 4.18 tons
Weight of bed required is 4.18tons.
Information from Example 26.4
 A(gas) + B(solid)  R(gas) + S(solid)
 τ = 1hr (with environment of CA0 gas concentration)
 Gas and solid both in mixed flow
 CA0 = CB0
 𝑋𝐵 = 0.9
 FB0 = 1 ton/hr
 Stoichiometric feed rate of A and B
Modification
 Gas is in plug flow
Solution
CAf = 0.1CA0
Taking logarithmic mean for the gas
𝐶𝐴 0 −𝐶𝐴𝑓
𝐶𝐴 =
𝐶
𝑙𝑛 𝐴0
𝐶 𝐴𝑓
𝐶𝐴 0 −0.1𝐶𝐴 0
=
𝑙𝑛
𝐶𝐴 0
0.1𝐶 𝐴0
= 0.39𝐶𝐴0
The solid will be in the environment of the gas of concentration 0.55CA0
τα
1
𝐶𝐴 0
So τ =
from equation τ =
1
0.39
𝜌𝐵 𝑅
𝑏 𝑘 𝐶𝐴 0
= 2.564h
For reaction control;
1 − 𝑋𝐵 =
1τ
4𝑡
−
On solving this equation
𝑡=
τ
0.435
=
2.564
0.435
τ
𝑡
1
τ 2
20
𝑡
= 0.1
= 0.435
= 5.89 𝑕𝑟
W = 𝑡 FB0 = 5.89 tons
Weight of bed required is 5.89tons.
Reaction step is rate controlling (SCM)
Assuming that Fibers are cylinder in shape with same particle size and the useful product is
solid (unchanging particle size)
Solution
1-XB = 1 −
1-𝑋𝐵 =
𝜏
0
t 2
τ
1 − 𝑋𝐵 𝐸 𝑑𝑡
=
𝜏
0
1−
=
𝜏
0
1−
=
1+
t 2
τ
2𝑡
𝜏
−𝑡
+
𝑡 2
𝜏
𝑒
−
𝑒 −𝑇
𝑇
+
𝜏
2𝑡𝑡
𝜏2
𝑒
−
2𝑒 −𝑇
2(1−𝑒 −𝑇 )
𝑇2
𝑑𝑡
𝑡 2
= −𝑒 −𝑇 + 𝑒 −𝑇 −
1-𝑋𝐵 = −
𝑡
𝑇
−𝑡
2𝑡 2
𝜏2
−
𝑡
𝑑𝑡
+
2𝑡
𝜏
2𝑒 −𝑇
𝑇2
+
𝑡
𝑒
𝜏
−𝑡
𝜏
𝑡
0
+ 2𝑒 −𝑇 +
+ 2𝑒 −𝑇 + 1 −
1
𝑇
𝑒 −𝑇
𝑇
1
2
𝑇
𝑇2
+1− −
𝜏
[where T = 𝑡 ]
Solution
26.11
23.12
For SCM/Reaction Control,
t/τ=1-(1-XB)1/3
so from above equation τ=1 hr(60min)
For SCM/Reaction Control,
t/τ=1-(1-XB)1/3
so from above equation τ=1 hr(60min)
𝜏
𝜏
1-𝑋B= 0 (1 − 𝑋B)Edt
1-𝑋B= 0 (1 − 𝑋B)Edt
1-𝑋B= 0 (1 − 𝑡/𝜏)3*(60/90)dt
1-𝑋B=
𝜏
1
=(2/3)* 0 (1 − 𝑡)3*dt
=(2/3)*(1/4)
=0.167
𝑋B=83.3%
0.75
(1 − 𝑡/𝜏)3*(60/30)dt
0.25
0.75
=2* 0.25 (1 − 𝑡)3*dt
=2*0.078
=0.156
𝑋B=0.8437
26.13
26.14
For SCM/Reaction Control,
t/τ=1-(1-XB)1/3
so from above equation τ=1 hr(60min)
For SCM/Reaction Control,
t/τ=1-(1-XB)1/3
so from above equation τ=1 hr(60min)
𝜏
1-𝑋B= 0 (1 − 𝑋B)Edt
1-𝑋B=
=
1
(1 − 𝑡/𝜏)3*(60/60)*dt
0.5
1
(1 − 𝑡)3*dt
0.5
=0.0156
𝑋B=0.9843
𝜏
1-𝑋B = 0 (1 − 𝑋 B)Edt
𝜏
1-𝑋B = 0 (1 − 𝑡/𝜏)*1*dt
=1∗
0.75
(1
0
=1*0.4687
=0.4687
𝑋B=0.5313
− 𝑡)*dt
Compiled by
Manish Makwana
[email protected]
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